Handbook of Environmental Engineering Calculations

Handbook of Environmental Engineering Calculations, 2nd edition by C. C. Lee, Shun Dar Lin

• ISBN: 0071475834 • Pub. Date: May 2007 • Publisher: McGraw-Hill Companies

Table of Contents Pt. 1

Calculations of Water Quality Assessment and Control

Basic Science and Fundamentals

Streams and Rivers

Lakes and Reservoirs

Groundwater

Fundamental and Treatment Plant Hydraulics

Public Water Supply

Wastewater Engineering

Pt. 2

Solid Waste Calculations

Thermodynamics used in Environmental Engineering

Basic Combustion and Incineration

Practical Design of Waste Incineration

Calculations for Permitting and Compliance

Calculational Procedures for Ash Stabilization and Solidification

Incineration Technologies and Facility Requirements

Pt. 3

Air Pollution Control Calculations

Air Emission Control

Particulate Emission Control

Wet and Dry Scrubbers for Emission Control

Air Toxic Risk Assessment

Fundamentals Of Fuel Cell Technologies

Source: HANDBOOK OF ENVIRONMENTAL ENGINEERING CALCULATIONS

●

CALCULATIONS OF WATER QUALITY ASSESSMENT AND CONTROL

Part 1 of this book is written for use by the following readers: students taking coursework relating to public water supply, waste-water engineering or stream sanitation, practicing environmental (sanitary) engineers; regulatory officers responsible for the review and approval of engineering project proposals; operators, engineers, and managers of water and/or wastewater treatment plants; and any other professionals, such as chemists and biologists, who have gained some knowledge of water/wastewater issues. This work will benefit all operators and managers of public water supply and of wastewater treatment plants, environmental design engineers, military environmental engineers, undergraduate and graduate students, regulatory officers, local public works engineers, lake managers, and environmentalists. The chapters in Part 1 present the basic principles and concepts relating to water/wastewater engineering and provide illustrative examples of the subject. To the extent possible, examples rely on practical field data. Each of the calculations provided herein are solved step-by-step in a streamlined manner that is intended to facilitate understanding. Calculations (step-by-step solutions) range from calculations commonly used by operators to more complicated calculations required for research or design. Advances and improvements in many fields are driven by competition or the need for increased profits. It may be fair to say, however, that advances and improvements in environmental engineering are driven instead by regulation. The US Environmental Protection Agency (EPA) sets up maximum contaminant levels, which research and project designs must reach as a goal. The step-by-step solution examples provided in this book are informed by the integration of rules and regulations on every aspect of waters and wastewaters. The author has performed an extensive survey of literature on surface and groundwaters encountered in environmental engineering and compiled them in the following chapters. Rules and regulations are described as simply as possible, and practical examples are given. The following chapters include calculations for basic science, surface waters ground water, drinking water treatment, and wastewater engineering. Chapter 1.1 covers conversion factors between the two measurement systems, the United States (US) customary system and the System International (SI), basic mathematics for water and wastewater plant operators, fundamental chemistry and physics, and basic statistics for environmental engineers.

CALCULATIONS OF WATER QUALITY ASSESSMENT AND CONTROL 1.2

PART 1

Chapter 1.2 comprises calculations for river and stream waters. Stream sanitation had been studied for nearly 100 years. By the mid-twentieth century, theoretical and empirical models for assessing waste assimilating capacity of streams were well developed. Dissolved oxygen and biochemical oxygen demand in streams and rivers have been comprehensively illustrated in this chapter. Apportionment of stream users and pragmatic approaches for stream dissolved oxygen models are also covered. From the 1950s through the 1980s, researchers focused extensively on wastewater treatment. In 1970s, rotating biological contactors also became a hot subject. Design criteria and examples for all of these are included. Some treatment and management technologies are no longer suitable in the United States. However, they are still of some use in developing countries. Chapter 1.3 is a compilation of adopted methods and documented research. In the early 1980s, the USEPA published Guidelines for Diagnostic and Feasibility Study of Public Owned Lakes (Clean Lakes Program, or CLP). This was intended to be used as a guideline for lake management. CLP and its calculation (evaluation) methods are present in this chapter. Hydrological, nutrient, and sediment budgets are presented for reservoir and lake waters. Techniques for classification of lake water quality, assessment of the lake trophic state index, and of lake use support are presented. Calculations for groundwater are given in Chapter 1.4. They include groundwater hydrology, flow in aquifers, pumping and its influence zone, setback zone, and soil remediation. Well setback zone is regulated by the state EPA. Determinations of setback zones are also included in the book. Well function for confined aquifers is presented in Appendix B. Hydraulics for environmental engineering is included in Chapter 1.5. This chapter covers fluid (water) properties and definitions; hydrostatics; fundamental concepts of water flow in pipes, weirs, orifices, and in open channel; and of flow measurements. Pipe networks for water supply distribution systems and hydraulics for water and wastewater treatment plants are included. Chapters 1.6 and 1.7 cover each unit process for drinking water and wastewater treatments, respectively. The USEPA developed design criteria and guidelines for almost all unit processes. These two chapters depict the integration of regulations (or standards) into water and wastewater design procedure. Water fluoridation and the CT values are incorporated in Chapter 1.6. Biosolids are discussed in detail in Chapter 1.7. These two chapters are the heart Part 1, providing the theoretical considerations of unit processes, traditional (or empirical) design concepts, and integrated regulatory requirements. Most calculations provided herein use U.S. Customary units. Readers who use the International System (SI) may apply the conversion factors listed in Chapter 1.1. Answers are also generally given in SI for most of problems solved using U.S. units. The current edition corrects certain computational, typographical, and grammatical errors found in the previous edition. Drinking water quality standards, wastewater effluent standards, and several new examples have also been added. The author also wishes to acknowledge Meiling Lin, Heather Lin, Robert Greenlee, Luke Lin, Kevin Lin, Jau-hwan Tzeng, and Lucy Lin for their assistance. Any reader suggestions and comments will be greatly appreciated. Shun Dar Lin

Source: HANDBOOK OF ENVIRONMENTAL ENGINEERING CALCULATIONS

CHAPTER 1.1

BASIC SCIENCE AND FUNDAMENTALS Shun Dar Lin

1 CONVERSION FACTORS 1.3 2 PREFIXES FOR SI UNITS 1.8 3 MATHEMATICS 1.8 3.1 Logarithms 1.9 3.2 Basic Math 1.10 3.3 Threshold Odor Measurement 1.13 3.4 Simple Ratio 1.14 3.5 Percentage 1.15 3.6 Significant Figures 1.19 3.7 Transformation of Units 1.21 3.8 Geometrical Formulas 1.26 4 BASIC CHEMISTRY AND PHYSICS 1.31 4.1 Density and Specific Gravity 1.31 4.2 Chemical Solutions 1.33 4.3 pH 1.37 4.4 Mixing Solutions 1.40 4.5 Chemical Reactions and Dosages 1.42

4.6

Pumpage and Flow Rate 1.44

5 STATISTICS 1.50 5.1 Measure of Central Value 1.50 5.2 The Arithmetic Mean 1.50 5.3 The Medium 1.50 5.4 The Mode 1.50 5.5 Moving Average 1.51 5.6 The Geometric Mean 1.51 5.7 The Variance 1.52 5.8 The Standard Deviation 1.53 5.9 The Geometric Standard Deviation 1.54 5.10 The Student’s t Test 1.54 5.11 Multiple Range Tests 1.56 5.12 Regression Analysis 1.59 5.13 Calculation of Data Quality Indicators 1.65 REFERENCES 1.69

1 CONVERSION FACTORS The units most commonly used by water and wastewater professionals in the United States are based on the complicated U.S. Customary System of Units. However, laboratory work is usually based on the metric system due to the convenient relationship between milliliters (mL), cubic centimeters (cm3), and grams (g). The International System of Units (SI) is used in all other countries. Factors for converting U.S. units to the SI are given below (Table 1.1) to four significant figures. EXAMPLE 1:

Find degrees in Celsius of water at 68F.

Solution: C (F 32)

5 5 (68 32) 20 9 9

1.3

BASIC SCIENCE AND FUNDAMENTALS 1.4

CHAPTER 1.1

TABLE 1.1 Factors for Conversions U.S. Customary units Length inches (in) feet (ft) yard (yd) miles

Area square inch (sq in, in2) square feet (sq ft, ft2) acre (a)

square miles (mi2)

Volume cubic feet (ft3)

cubic yard (yd3) gallon (gal)

million gallons (Mgal) quart (qt) acre feet (ac ft)

Weight pound (lb, #)

grain ton (short) ton (long) gallons of water (US) Imperial gallon

Multiply by

SI or U.S. Customary units

2.540 0.0254 0.3048 12 0.9144 3 1.609 1760 5280

centimeters (cm) meters (m) m in m ft kilometers (km) yd ft

6.452 0.0929 144 4047 0.4047 43,560 0.001562 2.590 640

square centimeters (cm2) m2 in2 square meters (m2) hectare (ha) ft2 square miles km2 acres

28.32 0.02832 7.48 6.23 1728 0.7646 3.785 0.003785 4 8 128 0.1337 3785 32 946 0.946 1.233 103 1233

liters (L) m3 US gallons (gal) Imperial gallons cubic inches (in3) m3 L m3 quarts (qt) pints (pt) fluid ounces (fl oz) ft3 m3 fl oz milliliters (mL) L cubic hectometers (hm3) m3

453.6 0.4536 7000 16 0.0648 2000 0.9072 2240 8.34 10

grams (gm or g) kilograms (kg) grains (gr) ounces (oz) g lb tonnes (metric tons) lb lb lb

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

TABLE 1.1 Factors for Conversions (contd.) U.S. Customary units Unit weight ft3 of water pound per cubic foot (lb/ft3)

Concentration parts per million (ppm) grain per gallon (gr/gal) Time day

hour minute Slope feet per mile Velocity feet per second (ft/sec)

inches per minute miles per hour (mi/h)

knot Flowrate cubic feet per second (ft3/s, cfs)

million gallons daily (MGD)

gallons per minute (gpm)

gallons per day MGD per acre ft acre feet per day Application (loading) rate pounds per square foot (lb/ft2)

Multiply by

SI or U.S. Customary units

62.4 7.48 157.09 16.02 0.016

lb gallon newton per cubic meter (N/m3) kg force per square meter (kgf/m2) grams per cubic centimeter (g/cm3)

1 8.34 17.4 142.9

mg/L lb/Mgal mg/L lb/Mgal

24 1440 86,400 60 60

hours (h) minutes (min) seconds (s) min s

0.1894

meter per kilometer

720 0.3048 30.48 0.6818 0.043 0.4470 26.82 1.609 0.5144 1.852

inches per minute meter per second (m/s) cm/s miles per hour (mph) cm/s m/s m/min km/h m/s km/h

0.646 448.8 28.32 0.02832 3.785 0.04381 157.7 694 1.547 3.785 0.06308 0.0000631 0.227 8.021 0.002228 3.785 0.4302 0.01427

million gallons daily (MGD) gallons per minutes (gpm) liter per second (L/s) m3/s m3/d (CMD) m3/s m3/h gallons per minute cubic feet per second (ft3/s) liters per minute (L/min) liters per second (L/s) m3/s m3/h cubic feet per hour (ft3/h) cubic feet per second (cfs, ft3/s) liters (or kilograms) per day gpm per cubic yard m3/s

4.8827

kilograms per square meter (kg/m2)

1.5

BASIC SCIENCE AND FUNDAMENTALS 1.6

CHAPTER 1.1

TABLE 1.1 Factors for Conversions (contd.) U.S. Customary units pounds per 1000 square foot per day (lb/1000 ft2 d) pounds per cubic foot (lb/ft3) pounds per 1000 cubic foot per day (lb/1000 ft3 d) pounds per foot per hour (lb/ft h)

Multiply by 0.00488 16.017 0.016 1.4882

pounds per horse power per hour (lb/hp h) pounds per acre per day (lb/acre d)

0.608

gallons per acre (gal/acre) million gallons per acre (Mgal/acre) million gallons per acre ft (Mgal/acre ft) gallons per square foot per day (gal/ ft2 d)

0.00935 0.93526 0.43

gallons per minute per square foot (gpm/ ft2) square root of gpm per square foot (gal/min)0.5/ft2 gallons per day per foot (gal/d ft) square foot per cubic foot (ft2/ ft3) cubic foot per gallon (ft3/gal) cubic foot per pound (ft3/lb) cubic foot per 1000 cubic foot per minute (ft3/1000 ft3 min) Force pounds

Pressure pounds per square inch (lb/in2, psi)

pounds per square foot (lb/ft2) pounds per cubic inch tons per square inch millibars (mb) inches of mercury

1.121

0.04074

SI or U.S. Customary units kilograms per square meter per day (kg/m2 d) kilograms per cubic meter (kg/m3) kilograms per cubic meter per day (kg/m3 d) kilograms per meter per hour (kg/m h) kilograms per kilowatts per hour (kg/kW h) kilograms per hectare per day (kg/ha d) m3/ha m3/m2 gpm/yd3

0.04356 58.674

cubic meter per square meter per day (m3/m2 d) Mgal/acre d m3/m2 d

2.7

(L/s)0.5/ m2

0.01242 3.28 7.48 0.06243 62.43 1

m3/d m m2/m3 m3/m3 m3/kg L/kg L/ m3 min

0.4536 453.6 4.448

kilograms force (kgf) grams(g) newtons (N)

2.309 2.036 51.71 6895

feet head of water inches head of mercury mmHg newtons per square meter (N/m2) pascal (Pa) kgf/cm2 kgf/m2 bars kgf/m2 N/m2 (Pa) gmf/cm3 gmf/L kg/mm2 N/m2 kg/m2 kg/cm2 bar psi (lb/in2)

0.0703 703.1 0.0690 4.882 47.88 0.01602 16.017 1.5479 100 345.34 0.0345 0.0334 0.491

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.7

TABLE 1.1 Factors for Conversions (contd.) U.S. Customary units inches of water atmosphere

pascal (SI)

Mass and density slug pound slug per foot3 density () of water specific wt () of water

Viscosity pound-second per foot3 or slug per foot second square feet per second (ft2/s) Work British thermal units (Btu)

hp-h kW-h Power horsepower (hp)

kilowatts (kW) Btu per hour

Temperature degree Fahrenheit (F) (C)

Multiply by

SI or U.S. Customary units

248.84 101,325 1013 14.696 29.92 33.90 1.0 1.0 105 1.0200 105 9.8692 106 1.40504 104 4.0148 103 7.5001 104

pascals (Pa) Pa millibars (1 mb 100 Pa) psi (lb/in2) inches of mercury feet of water N/m2 bar kg/m2 atmospheres (atm) psi (lb/in2) in, head of water cm head of mercury

14.594 32.174 0.4536 515.4 62.4 980.2 1.94 1000 1 1

kg lb (mass) kg kg/m3 lb/ft3 at 50F N/m3 at 10C slugs/ft3 kg/m3 kg/L gram per milliliter (g/mL)

47.88

newton second per square meter (Ns/m2) m2/s

0.0929 1.0551 778 0.293 1 2545 0.746 3413 1.34

kilo joules (kj) ft lb watt-h heat required to change 1 lb of water by 1F Btu kW-h Btu hp-h

550 746 2545 3413 0.293 12.96 0.00039

ft lb per sec watt Btu per h Btu per h watt ft lb per min hp

(F 32) (5/9) (C) (9/5) 32 C 273.15

degree Celsius (C) (F) Kelvin (K)

BASIC SCIENCE AND FUNDAMENTALS 1.8

CHAPTER 1.1

EXAMPLE 2:

At a temperature of 4C, water is at its greatest density. What is the degree of

Fahrenheit? Solution: F (C)

9 32 5

9 32 5 7.2 32 39.2 4

2 PREFIXES FOR SI UNITS The prefixes commonly used in the SI system are based on the power 10. For example, a kilogram means 1000 grams, and a centimeter means one-hundredth of 1 meter. The most used prefixes are listed in Table 1.2, together with their abbreviations, meanings, and examples.

3 MATHEMATICS Most calculations required by the water and wastewater plants operators and managers are depended on ordinary addition, subtraction, multiplication, and division. Calculations are by hand, by calculator, or by a computer. Engineers should master the formation of problems: daily operations require calculations of simple ratio, percentage, significant figures, transformation of units, flow rate, area and volume computations, density and specific gravity, chemical solution, and mixing of solutions. Miscellaneous Constants and Identities 3.14 (pi) e 2.7183 (Napierian) xo 1(x 0) 0x 0 (x 0) TABLE 1.2 Prefixes for SI Units Prefix

Abbreviation

tera giga mega myria kilo hecto deka

T G M my k h da

deci centi millimicro nano pico femto atto

d c m n p f a

Multiplication factor 1 000 000 000 000 1012 1000000000 109 1 000 000 106 10000 104 1000 103 100 102 10 101 1 100 0.1 101 0.01 102 0.001 103 0.000 001 106 0.000 000 001 109 0.000 000 000 001 1012 0.000 000 000 000 001 1015 0.000 000 000 000 000 001 1018

Example

km, kg

meter (m), gram (g) cm mm, mg m, g

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.9

3.1 Logarithms Every positive number x can be expressed as a base b raised to a power y: x by y is called the logarithm of x to the base b and is symbolized as y logb x The logarithm of a number is the power to which the base must be raised to equal that number. Logarithms and exponential functions have numerous uses in engineering. Logarithms simplify arithmetic and algebraic computations according to the following rules: logb(xy) logb x logb y x logb y logb x logb y logb xy ylogb x The two most common bases are 10 and e. The exponential e is an irrational number resulting from an infinite series and is approximately equal to 2.71828. Definitions: log base 10 log10 log log base e loge ln log base q logq where q is any positive number except 0 or 1. Antilogq, or log–1 , is the number whose log value being stated for logq x m. q EXAMPLE 1:

log x a S x 10a since 1000 103,

i.e. log 1000 3

0.001 10 , 3

i.e. log 0.001 3

ln y b S y eb logq z c S z qc Given m, find x logq–1 m EXAMPLE 2:

Solution:

Find log 3.46 and log 346.

The answer can be found in log10 tables or using a calculator or computer. We can obtain log 3.46 0.539,

i.e. 3.46 100.539

log 346 2.539,

i.e. 346 102.539

If a positive number N is expressed as a power of e, we can write it as N ep. In this case, p is the logarithm of N to the base e (2.7183) or the natural logarithm of N and can be written as p ln N or p lne N. More examples are as follows: ln 2 0.6931 ln 10 2.3026

BASIC SCIENCE AND FUNDAMENTALS 1.10

CHAPTER 1.1

ln y 2.3026 log y log x 0.4343 ln x log ab log a log b a log a log b b log an n log a log

n 1 log 2a n log a

3.2 Basic Math This section involves the basics of plus, minus, multiplication, and division, which may be useful for basic-level certification examinations of water or wastewater operators. Addition If a trickling filter has a recirculation ratio of 0.25 to 1, how many times the flow of raw wastewater passes through the filter? EXAMPLE 1:

Solution: 1 0.25 1.25 the flow of raw wastewater The field crew drives the company’s car 18.8 miles on Monday, 45 miles on Tuesday, 0 miles on Wednesday, 22.2 miles on Thursday, and 36.5 miles on Friday. What is the total mileage driven for the week and what is the average per day?

EXAMPLE 2:

Solution: Total (18.8 45 0 22.2 36.5) miles 122.5 miles Average total/5 day 122.5 miles/5 day 24.5 miles/day EXAMPLE 3: If an activated sludge unit has a recirculation of 0.4 to 1, how many times the flow of raw wastewater passes over the unit?

Solution: Answer 1 0.4 1.4 (times) Subtraction A flow meter reads 00023532 gal on June 1 and 06040872 gal on July 1. What is the average daily flow?

EXAMPLE 1:

Solution: (6040872 23532) gal/30 days 200,578 gal/day (gpd) If 5.2 mg/L of chlorine (Cl2) is added to a water sample, after 30 min of contact time the residual chlorine concentration is 0.9 mg/L. What is the chlorine demand of the water?

EXAMPLE 2:

Solution: Cl2 demand 5.2 mg/L 0.9 mg/L 4.3 mg/L

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.11

Multiplication The weight of a sand bag averages 46.5 lb. What is the total weight of 1200 bags of sand to fill a hole in the bank?

EXAMPLE 1:

Solution: Weight 46.5 lb/bag 1200 bags 55,800 lb If a water treatment plant uses 124 chlorine cylinders (cy) during a 30-day period, the average supply from each cylinder is 135 lb. What is the total weight of chlorine used in 30 days?

EXAMPLE 2:

Solution: Total weight 135 lb/cy 124 cy 16,740 lb EXAMPLE 3:

How many pounds in 5 gal of water? A gallon (gal) of water weighs 8.34 lb.

Solution: Weight 8.34 lb/gal 5 gal 41.7 lb EXAMPLE 4: Since 1 cubic foot (ft3) of a liquid is equal to 7.48 gal, how many gallons are equivalent to 25 ft3 of the liquid?

Solution: 25 ft3 7.48

gal 187 gal ft3

EXAMPLE 5: The minimum velocity (v) at average flow in a sanitary sewer is 2.0 ft/s. What is it in terms of m/s? (1 ft 0.3 m)

Solution: v 2.0 ft/s 0.3 m/ft 0.6 m/s or 60 cm/s EXAMPLE 6: Calculate the molecular weight of alum, Al2(SO4)3.14H2O. The approximate atomic weights are aluminum 27, sulfur 32, oxygen 16, and hydrogen 1.

Solution: Atomic weight 2(27) 3(32 4 16) 14(1 2 16) 54 3(96) 14(18) 594 A chemical storage room is to be painted. The room size is 16 ft wide, 20 ft long, and 10 ft high. The walls and the ceiling will use the same paint. The floor paint will be a different paint. Assume 1 gal of paint covers an area of 400 ft2. How many gallons of wall and floor paints are required for the painting job? EXAMPLE 7:

Solution: Step 1: Determine floor paint needed: Floor area 16 ft 20 ft 320 ft2 400 ft2 One gallon of floor paint is enough.

BASIC SCIENCE AND FUNDAMENTALS 1.12

CHAPTER 1.1

Step 2: For wall and ceiling: Wall area 10 ft (16 16 20 20) ft, including doors 720 ft2 Ceiling area 320 ft2, same as the floor Total area 720 ft2 320 ft2 1040 ft2 Gallon cans 1040 ft2/(400 ft2/gal) 2.6 gal Therefore, purchase three 1-gal cans. EXAMPLE 8:

If the water rates are

$1.846 per 100 cubic foot (cf) for the first 3000 cf $1.100 per 100 cf from 3001 cf to 60,000 cf $0.755 per 100 cf from 60,001 cf to 1,300,000 cf, and $0.681 per 100 cf for 1,300,001 cf and over In addition to the above monthly charge, there is a customer charge depending on the meter size provided and public fire protection service charge. Their rates are as follows: Size of meter, inches: 5/8 3/4 1 1.5 2 4 8 Customer charge, $/M: 9.50 12.5 20.5 45.5 70.5 225.5 710.50 Fire protection, $/M: 1.96 2.94 4.90 9.80 9.80 9.80 9.80 What is the monthly charge (excluding taxes, if any) for (a) 575 cf with 5/8-in meter (b) 86,000 cf with 2-in meter (c) 1,500,000 cf with 8-in meter Solution: Step 1: Solve for (a) (1) Water used $1.846 (2) Customer charge (3) Fire protection Total monthly charge

575 $10.61 100 $9.50 $1.96 $22.07

Step 2: Solve for (b) (1) Water charge: First 3000 cf $(1.846 3000)/100 $55.38 Next 57,000 cf $1.100(600 30) $627.00 Next 26,000 cf $0.755(860 600) $196.30 Total water charge $878.68 (sum of above) (2) Customer charge (2-in meter) $70.50 (3) Fire protection $9.80 Total monthly charge

$958.98

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.13

Step 3: Solve for (c) (1) Water charge: First 3000 cf $55.38 Next 57,000 cf $627.00 Next 1,240,000 cf $0.755(13000 600) $9,362.00 Next 200,000 cf $0.681 2000 $1,362.00 Total for 1,500,000 cf $11,406.38 (2) Customer charge (8-in meter) $710.50 (3) Fire protection $9.80 Total charge for $1,500,000 cf

$12,125.68

Division The storage tank of your plant has a 1.20-Mgal capacity and the demand is 0.40 million gallons per day (MGD). If the pump was broken, how many more days could your plant supply the water?

EXAMPLE 1:

Solution: 1.2 Mgal/0.4 MGD 3.0 days The intermittent sand filter needs to be drained and cleaned. If the treated wastewater volume in the filter is 18,000 gal, what is the time to empty the filter if the withdrawal rate is 500 gal/min?

EXAMPLE 2:

Solution: Time t

18,000 gal 500 gal/min

36 min EXAMPLE 3: If a membrane filter count is 69 fecal coliform (FC) colonies and 5.00 mL of a river water sample is filtered, what is the reported FC density?

Solution: The bacterial density (D) is commonly reported as number of bacteria per 100 mL of water: D (69 FC/5 mL) 13.8 FC/mL 13.8 100 FC/100 mL 1380 FC/100 mL Report: D 1400 FC/100 mL Note: Use two significant figures for reporting bacterial density. 3.3 Threshold Odor Measurement Taste and odor in drinking water supplies are the most common customer complaint. Test procedures are listed in Standard methods (APHA, AWWA, and WEF, 1995). The ‘threshold odor number’ (TON) is the greatest dilution of sample with odor-free water yielding a definitely perceptible odor. The TON is the dilution ratio at which taste or odor is just detectable. The sample is diluted with odor-free water to a total volume of 200 mL in each flask. The TON can be computed as TON

AB A

BASIC SCIENCE AND FUNDAMENTALS 1.14

CHAPTER 1.1

where A sample size, mL, and B volume of odor-free water, mL. If the total volume is 200 mL, then TON 200/A EXAMPLE: The first detectable odor is observed when a 25 mL sample is diluted to 200 mL with odor-free water. What is the TON of the water sample?

Solution: 200 mL 200 A 25 mL 8

TON

3.4 Simple Ratio A ratio is one number divided by another number. It is a pure number, such as 2, 5, or 4.6. A pure number multiplied by a physical unit (mg/L, or miles, etc.) is called a concrete number, such as 2.5 mg/L or 20 ft. The division of two concrete numbers may be either a pure number (if the same physical unit) or a rate (with different physical units). The ratios are shown in the following examples. 3 is the ratio of 3 to 4, a pure number (or 0.75) 4 12 m 4; the ratio is a pure number 3m 120 mg 1 mg ≤, a rate 24 mg/L 24 times¢ 1L 5L 1 mg 1 mg of a constituent in 1L of liquid, a physical unit 1L 24 oz 6 oz/gal, a rate 4 gal 4 ft 0.25 ft/s, a rate 16 s 3.2 lb/d 0.4, a pure number 8.0 lb/d EXAMPLE: If 1.5 L of activated sludge with volatile suspended solids (VSS) of 1880 mg/L is mixed with 7.5 L of raw domestic wastewater with BOD5 of 252 mg/L, what is the F/M (food/microbes) ratio?

Solution:

amount of BOD5 F M amount of VSS

252 mg/L 7.5 L 1890 2820 1880 mg/L 1.5 L

0.67 1

or 0.67

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.15

3.5 Percentage A percentage is a ratio with the denominator of 100; while the numerator is called the percent (%). It is also called parts per hundred. A proportion is the equality of two ratios. This equality with one denominator of 100 is used in calculating percentage (ratio times 100%). EXAMPLE 1:

Solution: 16 4 100 25

What percentage is 4/25? therefore 4 16% of 25

EXAMPLE 2:

What percentage is 18/40?

Solution 1: 18 0.45 40 then

100% 0.45 45%

Solution 2: 18 x 40 100 Cross-multiplying: 40x 18 100 x(%) EXAMPLE 3:

Solution:

1800 45 40

What percentage is 121/94?

y 121 94 100 y(%)

121 100 128.7 94

One liter of water weighs 1 kg (1000 g 1,000,000 mg). Hence milligrams per liter is parts per million (ppm); i.e. pounds per million pounds, grams per million grams, and milligrams per million milligrams are parts per million. EXAMPLE 4:

Solution:

What is the percentage of 1 ppm?

Since 1 part per million (ppm) 1 mg/L 1 mg 100% 1 L 1,000,000 mg/L 1 % 10,000

1 mg/L

0.0001% EXAMPLE 5:

How many mg/L is a 1.2% solution?

Solution: 1.2 , since the weight of 1 L water is 106 mg 100 1.2 1,000,000 mg/L 100 12,000 mg/L

1.2%

BASIC SCIENCE AND FUNDAMENTALS 1.16

CHAPTER 1.1

EXAMPLE 6:

Solution:

Compute pounds per million gallons (Mgal) for 1 ppm (1 mg/L) of water.

Since 1 gal of water 8.34 lb 1 ppm

1gal 106 gal 1 gal 8.34 lb/gal Mgal

8.34 lb/Mgal For a laboratory-scale test, 2.4 lb of activated carbon (AC) and 32.0 lb of sand are mixed. Compute the percentage of activated carbon in the mixture.

EXAMPLE 7:

Solution: therefore

Total weight (2.4 32.0) lb 34.4 lb. Let percent of AC x, x 2.4 lb 100% 34.4 lb x 100% 2.4/34.4 7.0%

EXAMPLE 8:

How many kilograms of AC are required to make 38 kg of mixture that is to be 28%

AC? Solution:

Let y be the weight of AC y kg 28 28% 0.28 100 38 kg y 38 kg 0.28 10.64 kg

EXAMPLE 9:

How many pounds of AC are needed with 38 lb of sand so that the mixture is 28%

AC? Solution: y 0.28 38 y y 0.28(38 y) y 10.64 0.28y (1 0.28)y 10.64 y

10.64 14.78 (lb) 0.72

EXAMPLE 10: A softening unit removes total hardness from 258 to 62 mg/L as CaCO3. What is the efficiency of the process (percent removed)?

Solution: amount removed 100% original concentration 258 62 100% 258 76% removal

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.17

EXAMPLE 11: A water treatment plant treats 996,000 gal/d (gpd). On average it uses 28,000 gpd for backwashing the filters. What is the net production of the plant and what is the percentage of backwash water?

Solution: Step 1: Net production of plant 996,000 gpd 28,000 gpd 968,000 gpd Step 2: 28,000 gpd 100% 996,000 gpd 2.9%

Percent of backwash water

EXAMPLE 12:

A pipe is laid at a rise of 120 mm in 20 m. What is the grade?

Solution: 120 mm 100(%) 20 m 120 mm 100% 20 1000 mm 0.6%

Grade

EXAMPLE 13: In the reverse osmosis process, water recovery is the percent of product water flow Qp (in gpd or m3/d) divided by feedwater flow Qf. If Qf 100,000 gpd and Qp 77,000 gpd, what is the water recovery?

Solution: Recovery

Qp Qf

100%

77,000 100% 100,000

77% The laboratory data show that the total suspended solids (TSS) for the influent and effluent of a sedimentation basin are 188 and 130 (mg/L), respectively. What is the efficiency of TSS removal?

EXAMPLE 14:

Solution: TSS removal 100% TSS influent TSSinf TSSeff 100% TSSinf 188 130 100% 188 30.9%

Percent removal

A motor is rated as 30 horsepower (hp). However, the output horsepower of the motor is only 24.6 hp. What is the efficiency of the motor?

EXAMPLE 15:

BASIC SCIENCE AND FUNDAMENTALS 1.18

CHAPTER 1.1

Solution:

hp output 100% hp input 24.6 hp 100% 30 hp

Efficiency

82% If 1000 m3 of residual (sludge) containing 0.8% (P1) of solids have to be thickened to 4% (P2) what is the final volume?

EXAMPLE 16:

Solution: Since the amount of solids is the same before and after thickening, V1P1 V2P2 1000 m 0.8% V2 4% 3

1000 0.8 3 m 4 200 m3

EXAMPLE 17: Sludge analysis shows that weights of water and dry solids are 88.46 and 1.22 g, respectively. What is the moisture content of the sludge?

Solution: weight of water 100 weight of wet sludge 88.46 100 88.46 1.22 98.6

Moisture content (%)

The Surface Water Treatment Rule requires that water treatment must achieve at least 3-log removal or inactivation of Giardia lamblia cysts and 4-log removal of viruses. What are the percents removal? EXAMPLE 18:

Solution: Let Ci initial concentration of cysts or virus plaques Cf final concentration of cysts or virus plaques after treatment (a) For G. lamblia cysts removal/inactivation: log Ci log Cf 3 Ci log 3 Cf taking anti-log

Ci log1 3 1000 Cf

Cf 1 Ci 1000 Removal

Ci Cf 999 1000 1 Ci 1000 1000

0.999 100% 99.9%

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.19

(b) For viruses removal: log Ci log Cf 4 log

Ci 4 Cf Ci 10,000 Cf Cf 1 Ci 10,000

Removal

Ci Cf 10,000 1 Ci 10,000

0.9999 100% 99.99% 3.6 Significant Figures In calculation, numbers may be either absolutes or measurements: absolutes indicate exact values, while measurements are readings of meters, balances, gages, scales, and manometers. Each measurement has a sensitivity limit below which a change of measurement is not registered; thus, every measurement is not exact or incomplete. All digits in a reported value are expected to be known definitely except for the last digit, which may be doubtful. Each correct digit of the approximation, except a zero which serves only to fix the decimal point, is called a significant figure. For example 321, 83.8, 7.00, 0.925, and 0.0375 each contain three significant figures. During measurements, a rounded-off value is usually recorded. Rounding off is carried out either by the operator or using an instrument with minimum error. As a general rule, the magnitude of the error is less than or equal to one-half unit of the place of the nth digit in the rounded number. For example 8.05349 8.0535, 8.053, 8.05, 8.1, and 8; but 8.0535 8.054 and 8.05 8.0. A general rule is to assume that a reading is correct within one-half unit in the last place recorded. For example, a reading of 38.8 means 38.8 0.05 or any number between 38.75 and 38.85: rounding off, by dropping digits which are not significant. If the digit 0, 1, 2, 3, or 4 is dropped, do not change the number to the left. If the number 6, 7, 8, or 9 is dropped, increase by one to the preceding digit. If the digit 5 is dropped, round off preceding digit to the nearest even number. EXAMPLE 1:

Round off to one decimal 38.73 S 38.7 38.77 S 38.8 38.75 S 38.8 38.45 S 38.4 38.35 S 38.4

The digit 0 may be recorded as a measured value of zero or may serve merely as a space to locate the decimal point. For example, in 420, 32.08, and 6.00, all the zeros are significant. The roundedoff values may be 9.00, 9.0, or 9. The zeros are also significant. For 38,500, the zeros may or may not be significant: if written as 38.5 103, the two zeros are not significant. When a calculated value (x) with a standard deviation (s) is known, it is recommended to report the value as x s.

BASIC SCIENCE AND FUNDAMENTALS 1.20

CHAPTER 1.1

If numbers are added or subtracted, the number which has the fewer decimal places, not necessarily the fewer significant figures, makes the limit on the number of places for the sum or difference. EXAMPLE 2:

Add the following numbers

Solution: 3.1472 32.05 1234 8.9426 0.0032

9.00 1287.1430

The sum should be rounded off to 1287, no decimals. EXAMPLE 3:

78.3 3.14 0.388

Subtraction

Solution: 78.3 3.14

78.3 3.1

0.388

0.4

74.772

74.8

The answer is 74.8 In case several numbers are multiplied or divided, the number of significant figures of the calculated value should use the fewest significant figures presented in the factors; i.e. the weakest link in the chain. For the residual chlorine titration with the iodometric method, use a 250-mL water sample, 2.3 mL of 0.01N Na2S2O3 added to the end point. While 0.1 mL of 0.01N Na2S2O3 is titrated for the blank. What is the residual chlorine in the water sample?

EXAMPLE 4:

Solution:

The result of residual chlorine analysis is computed by (2.3 0.1) 0.01 35.450 250 3.1196

MgCl as Cl2/L

The answer is 3.1 mg/L of residual chlorine: only take two significant figures. After considering significant figures or rounding off, the maximum relative error can be estimated by how far it can be trusted. EXAMPLE 5: The flow chart of the water plant reads 3.8 million gallons per day (MGD). What is the maximum relative error for the reading?

Solution: 3.8 MGD 3.8 0.05 MGD The maximum relative error

0.05 3.8

0.01316 0.013 100% 1.3%

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.21

EXAMPLE 6: When we multiply 3.84 0.36, the arithmetic product is 1.3824. Evaluate the maximum relative errors.

Solution: Step 1: The product 1.3824 is rounded off to 1.4 Step 2: Calculate errors. The product 1.4 means 1.4 0.05 has

0.05 100% 3.6% error 1.4

While 3.84 has

0.005 100% 1.3% error 3.84

and 0.36 has

0.005 100% 1.4% error 0.36 Total 2.7% error

3.7 Transformation of Units Since all measurements are products of pure numbers and physical units, through transformation by multiplication and/or division, the computed results may be a ratio (pure number) or a concrete number with a physical unit. Physical units, like numbers, are also included in calculation. EXAMPLE 1:

Length

2 ft 32 64 ft

Area

12 ft 14 ft 168 ft2

Volume

6 ft 8 ft 3.5 ft 168 ft3

Ratio

25 25 ft 5 5 ft 5 88 ft3 22 ft ft ft 22 1 1 ft ft ft 4 ft3 99 ft3/h ft3 1 99 33 ft/h 2 h 3 ft 3 ft2

Division

72 ft2 24 ft2 3 36 ft/s 1 36 ft/s

Multiplication

(any quantity divided by its equivalent equals one.)

28 ft3 5 140 ft3 ft3 68 s 6 s 408 ft3

EXAMPLE 2: An operator recorded that the water treatment plant treated 2.8 Mgal of water during his shift. What is the plant’s total volume treated per day, assuming the same treatment rate?

BASIC SCIENCE AND FUNDAMENTALS 1.22

CHAPTER 1.1

Solution: 2.8 Mgal 24 h 8h day 2.8 24 MGD 8 8.4 MGD 1 MGD equals how many cubic feet per second (cfs)? gallons per minute (gpm)? cubic meters per day (or metric tons per day)? cubic meters per second (m3/s)

EXAMPLE 3:

(a) (b) (c) (d)

Solution: (a)

1 MGD

106 gal 1 day

106 gal 0.1337 ft3/gal 1 day 86,400 s/day

133,700 ft3 86,400 s

1.547 cfs (b)

1 MGD

106 gal 1 day 24 h/day 60 min/h 1,000,000 gal 24 60 min

694.4 gpm (c)

1 MGD

106 gal 1 day

106 gal 3.785 L/gal 1 kg/L 1 day

3.785 106 kg day

3.785 106 kg 103 metric ton/kg day

3785 metric tons per day (m3/day) (d)

1 MGD 1 Mgal/day

1 Mgal 3785 m3/Mgal 1 day 86400 s/day

0.0438 m3/s

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.23

A blower has a capacity of 200 cfs. Compute the capacity in (a) cubic meters per second? (b) cubic meters per minute?

EXAMPLE 4:

Solution: (a)

200 ft3 (0.3048)3 m3/ft3 ft3 200 s s 200(0.02832) m3/s 5.66 m3/s

(b)

EXAMPLE 5:

ft3 200 s 5.66 m3/s 60 s/min 340 m3/min How many cubic inches are in 1 cubic foot?

Solution: 1 ft3 1 ft 1 ft 1 ft Q 1 ft

12 in 3 1 ft R

123 in3 1728 in3 A wastewater treatment plant receives an average flow of 626 gpm. What is the daily total flow to plant?

EXAMPLE 6:

Solution: Flow 626 gpm (gallons per minute) 626

gal min h 60 24 min h day

901,440

gal day

0.90 106 gallons per day (gpd) 0.90 million gallons per day (MGD) EXAMPLE 7:

What is the weight (in pounds, in tons, and in kilograms) of 200 gal of water?

Solution: (a)

Weight (lb) 200 gal 8.34

lb gal

1668 lb (b)

Weight (tons) 1668 lb

tons 2000 lb

0.834 tons (c)

Weight (kg) 1668 lb 0.454 757 kg

kg lb

BASIC SCIENCE AND FUNDAMENTALS 1.24

CHAPTER 1.1

EXAMPLE 8: How many gallons per minute/cubic yard (gpm/yd3) for 1 MGD/acre ft (million gallons per minute/acre foot)?

Solution: 1 MGD 1,000,000 gal/day (1 day/1440 min) 694.4 gpm 1 acre ft 43,560 ft2 1 ft 43,560 ft3 1 yd3/27 ft3 1613.3 yd3 therefore 1 MGD/acre ft 694.4 gpm/1613 yd3 0.430 gpm/yd3 EXAMPLE 9:

How many grains per gallon for 1 mg/L?

Solution: 1 mg/L 1

mg grain L 0.0154 mg 3.785 L gal

0.0583 grain/gal 1 mg/L 17.1 mg/L or 1 grain/gal 0.0583 EXAMPLE 10:

Solution:

How many lb/Mgal for 1 grain/gal?

Since 1 lb 7000 grains 1 grain/gal 1/7000 lb/gal 106/7000 lb/Mgal 142.9 lb/Mgal

EXAMPLE 11:

How many lb/Mgal for 1 mg/L?

Solution: 1 mg/L 1 gal/Mgal 1 gal 8.34 lb/gal/1 Mgal 8.34 lb/Mgal EXAMPLE 12:

Solution:

How many gallons is 50 lb of water?

Since 1 gal of water weighs 8.34 lb V 50 lb

1 gal 8.34 lb

6.0 gal EXAMPLE 13:

A 5-gal empty tank weighs 2.3 lb. What is the total weight of the tank filled with

5 gal of water? Solution: Weight of water 5 gal 8.34 lb/gal 41.7 lb

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.25

Total weight 41.7 lb 2.3 lb 44.0 lb EXAMPLE 14:

The discharge rate is 1.2 ft3/s. How many gallons of water is discharged in 5 min?

Solution: Total discharge 1.2 ft3/s 60 s/min 5 min 360 ft3 360 ft3 7.48 gal/ft3 2693 gal EXAMPLE 15: In a small activated sludge plant, the maximum BOD5 loading is in the range 25–35 lb BOD/1000 ft3/d. What is the BOD/m3 d range in grams?

Solution: Step 1: Convert 25 lb BOD/1000 (ft3 d) to g BOD/(m3 d) 25 lb BOD 453.6 g/lb 25 lb BOD 3 # 1000 (ft d) 1000 (ft3 # d) 0.02832 m3/ft3 25 16,017 g BOD/(m3 # d) 1000 400 g BOD/(m3 # d) Step 2: Convert 35 lb BOD/1000 ft3/d 35 35 lb BOD 16,017 g BOD/(m3 # d) 1000 1000 (ft3 # d) 560 g BOD/(m3 # d) Answer: 400–560 g BOD/(m3 d) In an activated sludge process, 2 ft3 of compressed air is needed for treating each gallon of domestic wastewater. What is this in terms of liter of air per liter of wastewater (L/L)?

EXAMPLE 16:

Solution: 2 ft3/gal

2 ft3 28.32 L/ft3 1 gal 3.785 L/gal

2 7.48 L air/L wastewater 14.96 L/L >15 L/L Note: 1 ft3/gal 7.48 L/L. The depth of sludge usually applied to the sludge drying bed each time is 8–12 in. What is the depth in centimeters?

EXAMPLE 17:

Solution: 8 in 8 2.54 cm (as 2.54 cm 1 in) 20.32 cm 12 in 12 2.54 30.48 cm For simplicity, the beds are 20–30 cm in depth.

BASIC SCIENCE AND FUNDAMENTALS 1.26

CHAPTER 1.1

3.8 Geometrical Formulas 1. Oblique triangle

1 Area 2 b h

12 b c sin A 2[l(l a)(l b)(l c)] where b base h altitude l (a b c)/2 2. Right-angled triangle

Area 12 a b where

∠ C 90 c2 a2 b2

3. Rectangle

Area a b

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

4. Trapezoid

Area 12 (a b)h

5. Circle and sphere

Area of circle pr2 Area of sphere 4pr2 4 Volume of sphere 3pr3

Circumference 2pr where r radius

6. Cylinder

Lateral area 2rh Volume r2h

1.27

BASIC SCIENCE AND FUNDAMENTALS 1.28

CHAPTER 1.1

7. Cone

Volume of cone

1 2 3 pr h

8. Frustrum

1 Volume of frustrum 3 (A1 A2 2A1 A2 )h

where A1 r 12 A2 r 22 The dimensions of a primary settling basin are 22 ft wide, 33 ft deep, and 66 ft long. Compute the volume of the basin.

EXAMPLE 1:

Solution: Volume 22 ft 33 ft 66 ft 47,916 ft3 A circular sedimentation tank has a radius of 18 ft. Its wastewater depth is 20 ft. What is the volume of the tank?

EXAMPLE 2:

Solution: Volume r2h 3.14 (18 ft)2 20 ft 20,347 ft3 EXAMPLE 3: The cone-shape lime applicator has a diameter of 18 ft and a height of 7 ft. What is the capacity of the tank?

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.29

Solution: r 12 d 1/2 18 ft 9 ft Volume of cone 13 pr2h 1/3 3.14 (9 ft)2 7 ft 593.5 ft3 EXAMPLE 4: A sedimentation tank has a diameter of 66 ft. Its effluent weir is located along the rim of the tank. What is the length of the weir?

Solution: L 2r or d 3.14 66 ft 207 ft EXAMPLE 5: Copper sulfate is commonly used for controlling algae in lakes and reservoirs. If an area of 2200 ft (670 m) long and 790 ft (240 m) wide is to be treated with an application rate of 1.5 lb per surface acre (1.83 kg/ha), how much copper sulfate is needed?

Solution: Surface area A 2200 ft 790 ft 1,738,000 ft2 1,738,000 ft2

1 acre 43,560 ft2

39.9 acre 16.1 ha Weight of CuSO4 needed 1.5 lb/acre 39.9 acre 59.85 lb 27.1 kg say, 60 lb are needed. EXAMPLE 6: A cylinder tank has a diameter of 20 ft. Liquid alum is filled to a depth of 7.8 ft. What is the volume (gal) of alum?

Solution:

Step 1. Compute the volume (v) in cubic feet: v area height r2h 3.14 (20/2)2 7.8 3.14 100 7.8 2449.2 (ft3) Step 2. Convert to gallons: v 2449.2 ft3

7.48 gal 1 ft3

18,320 gal EXAMPLE 7: A standpipe has an inside diameter of 96 in and is 18 ft high. How many gallons does it contain if the water is 15 ft deep?

BASIC SCIENCE AND FUNDAMENTALS 1.30

CHAPTER 1.1

Solution: Radius

96 1 ft in 4 ft 2 12 in 2 v pr h 3.14 4 ft 4 ft 15 ft

753.6 ft3 753.6 ft3 7.48 gal/ft3 5637 gal The diameter (d) of a circular sedimentation tank is 88 ft. What is the circumference (c) and area of the tank?

EXAMPLE 8:

Solution: Step 1.

c d 3.14 88 ft 376.3 ft

Step 2. Radius r d 2 88 ft/2 44 ft Area r2 3.14(44)2 6079 ft2 EXAMPLE 9:

The circumference of a storage tank is 113 ft. What is radius (r) of the tank?

Solution: c d 2r 113 2 3.14 r r 113/6.28 18 (ft) EXAMPLE 10: Your treatment plant occupies a rectangular area with a length of 488 ft (l) and a width of 95 ft (w). What is the length of fence needed for the lot (including gates)? What is the area of the lot (in square feet and acres)?

Solution: Step 1. Fence length 2(l w) 2(488 195) 2 683 1366 (ft) Step 2. Area l w 488 195 95,160 (ft2) 1 acre 95,160 ft2 43,560 ft2 2.18 acres EXAMPLE 11:

A rectangle has a perimeter of 1234 ft and a length of 390 ft. What is its width?

Solution: Perimeter 2(l w) 1234 2(390 w) 617 390 w w 617 390 227(ft)

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.31

A rectangular sedimentation basin is 20 m wide, 40 m long, and 4 m water depth. What is its volume in m3, ft3, and in gallons?

EXAMPLE 12:

Solution: Volume 20 m 40 m 4 m 3200 m3 or 3200 m3

1 ft3 0.02832 m3

112,990 ft3 or 112,990 ft3 7.48 gal/ft3 845,200 gal EXAMPLE 13:

excavated?

A trench is 2 ft wide, 5 ft deep, and 21 mile long. How many cubic yards of soil are

Solution: 0.5 mile 0.5 43,560 ft 21,780 ft Volume width depth length 2 ft 5 ft 21,780 ft 217,800 ft3 217,800 ft3 24,200 yd3

1 yd3 27 ft3

18,500 m3

4 BASIC CHEMISTRY AND PHYSICS 4.1 Density and Specific Gravity Density () is a weight for a unit volume at a particular temperature. It generally varies with temperature. The weight may be expressed in terms of pounds, ounces, grams, kilograms, etc. The volume may be liters, milliliters, gallons, or cubic feet, etc. The densities for fresh water under different temperature conditions are shown in Table 1.3. For salt water, density is 64 lb/ft3 at 0C. The densities of plain and reinforced concretes are 144 and 150 lb/ft3, respectively. Specific gravity (sp gr) is defined as the ratio of the density of a substance to the density of water. The temperature of both the substance and water be stated. The specific gravity of fresh water at 0C is 1.00, while that of salt water is 1.02. The specific gravities of vegetable oil, fuel oil, and lubricant oil are 0.91–0.94, 1.0, and 0.9, respectively. EXAMPLE 1:

The specific gravity of an oil is 0.98 at 62F. What is the weight of 1 gal of oil?

Solution: Weight specific gravity weight of water 0.98 8.34 lb/gal 8.17 lb

BASIC SCIENCE AND FUNDAMENTALS 1.32

CHAPTER 1.1

TABLE 1.3 Temperature and Density of Fresh Water Temperature C 0 4 16.7 21 100

Density (unit weight) F

lb/ft3

32 39.2 62 69.8 212

lb/gal

N/m3

8.34

9805 9806 9795 9787 9378

62.417 62.424 62.355 52.303 59.7

EXAMPLE 2: An empty 250-mL graduated cylinder weighs 260.8 g. The cylinder filled with 250 mL of water weighs 510.8 g. The cylinder filled with a given solution (250 mL) weighs 530.4 g. Calculate the specific gravity of the solution.

Solution: Weight of water 510.8 260.8 250.0 g Weight of solution 530.4 260.8 269.6 g weight of solution Specific gravity weight of water

269.6 g 250.0 g

1.078 The density of water varies slightly with temperature. Simon (1976) reported the relationship of water temperature and density (Table 1.4). EXAMPLE 3:

A liquid has a specific gravity of 1.13. How many pounds is 52 gal of this liquid?

Solution: Weight 52 gal 8.34 lb/gal 1.13 490 lb TABLE 1.4 Effect of Water Temperature on Density, Viscosity, Surface Tension, and Vapor Pressure Temperature, C 0 4 5 10 15 20 25 30 40 50 60 70 80 90 100

Density, g/mL

Viscosity, poise

Surface tension, dyne/cm

0.99987 1.00000

0.01792 0.01568 0.01519 0.01308 0.01140 0.01005 0.00894 0.00801 0.00656 0.00549 0.00469 0.00406 0.00357 0.00317 0.00284

75.6

0.006

74.22

0.012 0.0168 0.0231 0.0313 0.0419 0.0728 0.1217 0.1965 0.4675

0.99973 0.99913 0.99823 0.99707 0.99567

0.95838

72.75 71.18 69.56 67.91 66.18 64.4 62.6 58.9

Vapor pressure, atmosphere

1.000

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

EXAMPLE 4:

1.33

If a solid in water has a specific gravity of 1.25, how many percent is it heavier than

water? Solution: sp gr of solid sp gr of water 100 sp gr of water 1.25 1.0 100 1.0 25

Percent heavier

4.2 Chemical Solutions Solids, liquids, and gases may dissolve in water to form solutions. The amount of solute present may vary below certain limits, so-called solubility. The strength of a solution can be expressed in two ways: (1) weight (lb) of active solute per 100 pounds (i.e. %) and (2) weight of active solute per unit volume (gallons or liters) of water. Either expression can be computed to the other if the density or specific gravity is known. If the solution is dilute (less than 1%), the specific gravity can be assumed to be 1.0; i.e. 1 L of solution is equal to 1 kg and 1 gal of solution equals 8.34 lb. In water chemistry, molarity is defined as the number of gram-molecular weights or moles of substance present in a liter of the solution. If solutions have equal molarity, it means that they have an equal number of molecules of dissolved substance per unit volume. The weight of substance in the solution can be determined as follows: The molarity (M) of a solution can be expressed as: M

(mol/L)

moles of solute (mole) 1.0 L of solution

For uniform purpose, the normality (N) is used for preparation of laboratory solutions. The normality can be written as: N

(eq/L or meq/L)

equivalent of solute (eq or meq) 1.0 L of solution

where Equivalent mass (g/eq or mg/meq)

molecular (or atom) mass (g or mg) v (eq or meq)

Here v is the equivalence or the number of replacement hydrogen atoms; for oxidation–reduction reactions, v change in valence. It should be noted that 1.0 equivalent (eq) is 1000 milliequivalents (meq) and that 1.0 eq/m3 1.0 meq/L. In environmental engineering practices, parts per million (ppm) is a term frequently used. It may be expressed by ppm

mass of solute (g) 106 g of solution

or ppm

concentration, mg/L specific gravity of liquid

The relationship of molarity and weight of solute and concentrations can be expressed as follows: Weight in grams molarity volume of solution in liters gram-molecular weight

BASIC SCIENCE AND FUNDAMENTALS 1.34

CHAPTER 1.1

where unit of molarity mole per liter (mol/L) unit of the gram-molecular weight grams per mole (g/mol) It may be converted to milligrams per liter (mg/L). The relationship between mg/L and molarity is mg/L molarity (mol/L) L gram-molecular weight (g/mol mg/g) molarity gram-molecular weight 103 It also may be converted to concentration in parts per million. Since mg/L (ppm) density of solution (ppm) density of solution molarity gram-molecular weight 103 ppm

molarity gram-molecular weight 103 density of solution

EXAMPLE 1: How much of reagent grade calcium oxide (CaO) and aluminum sulfate [Al2(SO4)3 14H2O] is needed to prepare 0.10 mole solutions?

Solution: (a) The molecular weight of CaO 40.1 16.0 56.1 0.10 mole 0.10 56.1 5.61 (g/L) (b) The molecular weight of Al2(SO4)3 14H2O 27 2 3(32 16 4) 14(1 2 16) 594 0.10 mole 0.10 594 59.4 (g/L) Note: The two solutions have equal numbers of solute molecules. Laboratory-use sulfuric acid has 96.2% strength or purity. Its specific gravity is 1.84. Compute (a) the total weight in grams per liter, (b) the weight of acid in grams per liter, (c) pounds of sulfuric acid per cubic foot, (d) pounds of acid per gallon, (e) the molecular weight of acid (H2SO4), and (f ) the volume (mL) of concentrated acid required to prepare 1.0 L of 0.1 N solution.

EXAMPLE 2:

Solution: (a)

Total weight 1000 g/L sp gr 1000 1.84 g/L 1840 g/L

(b)

Weight of acid Total weight

% 100

96.2 100 1770 g/L 1840 (c) Since 1 ft3 of water weighs 62.4 lb, % 100 62.4 1.84 0.962 lb/ft3

Weight of acid 62.4 lb/ft3 sp gr 110 lb/ft3

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.35

(d) Since 1 gallon of water weighs 8.34 lb % 100 8.34 1.84 0.962 lb/gal

Weight of acid 8.34 lb/gal sp gr 14.8 lb/gal

(e) The standard atomic weights of elements are available from Standard methods (APHA, AWWA, and WEF, 1995). The molecular weight of H2SO4 (1 2) 32 (16 4) 98 (f ) For H2SO4, valence 2 98 49 g/eq 2 0.1 eq 1.0 L of solution 0.1 eq 49 g/eq 1.0 L 4.9 g 1L 4.9 g 1 L 1.84 g/mL

Equivalent mass 0.1 N (eq/L) Divided by sp gr

2.66 mL/L With correction for impurity, the volume needed is Volume EXAMPLE 3:

2.66 mL/L 2.77 mL/L 0.962

After adding 0.72 lb of chemical to 2 gal of water, what is the percent strength of the

solution? Solution: Total weight of the solution 2 gal 8.34 lb/gal 0.72 lb 17.40 lb Percent of solution

0.72 100% 17.40

4.14% EXAMPLE 4: Sodium hydroxide (NaOH) is used as the base in acid–base titrations. It has a molecular weight of 40 (23 for Na 16 for O 1 for H). How many grams of NaOH are needed to prepare 500 mL of a 0.25 N (normal) solution?

Solution: 1 N NaOH 40 g/L 0.25 N NaOH 0.25 40 g/L 10 g/1000 mL

BASIC SCIENCE AND FUNDAMENTALS 1.36

CHAPTER 1.1

10/2 g 1000/2 mL 5 g/500 mL

500 mL 0.25 N

Answer: 5 g NaOH will be required to prepare 500 mL 0.25 N solution. Caustic soda, NaOH, is an important chemical for pH adjustment and acid titration. It is often manufactured by the reaction of slaked lime, Ca(OH)2, and soda ash, Na2CO3. (a) What weight in kilograms of NaOH will be generated if 26.5 kg of soda ash is used? (b) How many kilograms of lime, CaO, is needed for the reaction? The atomic weights are Na 23, C 12, O 16, Ca 40.1, and H 1.

EXAMPLE 5:

Solution for (a): Step 1. Balance the formula and compute molecular weights (MWs):

Ca(OH)2 Na2CO3 (23 2) 12 (16 3) 40.1 2(16 1) 106 74.1

S 2NaOH 2(23 16 1) 80

CaCO3

Step 2. Solve NaOH generated, x, by proportion: 80 106 x 26.5 x

80 26.5 20 kg 106

Solution for (b): Step 3. Compute the weight of CaO, y: CaO H2O S Ca(OH)2 40.1 16 56.1 74.1

Also, by proportion, weight of CaO needed is y y 56.1 x 80 therefore 56.1 20 56.1x 80 80 14.03 (kg)

EXAMPLE 6: If calcium (Ca) is 56.4 mg/L and Mg is 8.8 mg/L, what is the total hardness of the water in mg/L as CaCO3? The atomic weights are Ca 40.05, Mg 24.3, C 12.01, and O 16.

Solution: Step 1. Compute molecular weight (MW) of CaCO3: MW 40.05 12.01 (16 3) 100.06

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.37

Step 2. Compute the factor (f1) of Ca equivalent to CaCo3: MW of CaCO3 100.06 f1 MW of Ca 40.05 2.498 Step 3. Compute the factor (f2) of Mg equivalent to CaCO3: f2

MW of CaCO3 100.06 24.3 MW of Mg

4.118 Step 4. Compute total alkalinity (T.alk.) T.alk. Ca (mg/L) f1 Mg (mg/L) f2 56.4 2.498 8.8 4.118 177 (mg/L as CaCO3) Note: According to Standard methods (APHA, AWWA, and WEF, 1995): T.alk. (mg/L as CaCO3) 2.497 Ca (mg/L) 4.118 Mg (mg/L) 4.3 pH The acid–base property of water is a very important factor in water and wastewater. Water ionizes to a slight degree to produce both hydrogen ion (H) and hydroxyl ion (OH) as below: H2O 4 H OH Water may be considered both as an acid and a base. The concentration of H(aq) in a solution is often expressed as pH. The pH is defined as the negative log to the base 10 of the hydrogen ion concentration (molecular weight in g/L). It can be written as pH log [H] log(1/[H]) It is noted that a change in [H] by a factor of 10 results in 1 unit change in pH as shown in Table 1.5. TABLE 1.5 Hydrogen Ion Concentration and pH Value [H+], g/L

1.0 0.1 0.01 0.001 0.0001 0.00001 0.000001 0.0000001 (107) 108 109 1010 1011 1012 1013 1014

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

increasing acidity c neutral T increasing basicity

BASIC SCIENCE AND FUNDAMENTALS 1.38

CHAPTER 1.1

When [H] [OH] 1 107, the pH of the solution is neutral at 7.0 pH log [H] log (1 107) 0 (7) 7 Because pH is simply another means of expressing [H], acidic and basic solutions have the relationships: pH 7 in acidic solutions pH 7 in neutral solutions pH 7 in basic solutions

The number of [H ] ions multiplied by the number of [OH] ions gives the same value (constant), i.e. K [H] [OH] 1.0 1014 Therefore, if the number of [H] ions is increased tenfold, then the number of [OH] ions will be automatically reduced to one-tenth. Calculate the pH values for two solutions: (a) in which [OH] is 0.001 mole; (b) in a solution [OH] is 2.5 1010 mole.

EXAMPLE 1:

Solution: Step 1. Solve for question (a) [H] [OH] 1.0 1014 1.0 1014 1.0 1014 [OH ] 0.001 1.0 1011 mole

[H]

then

pH log [H] log (1.0 1011) (11.00) 11.00

basic

Step 2. Solve for question (b) [H]

1.0 1014 1.0 1014 [OH] [2.5 1010]

4.0 105 mol pH log(4.0 105) (log 4 log 105) (0.602 5.00) 4.40 EXAMPLE 2:

acidic

Alum coagulated water has a pH of 4.56. Calculate [H].

Solution: pH log [H] 4.56 thus

log [H] 4.56

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.39

Find antilog [H] log1(4.56) 2.75 105 mole [H] log1(5 0.44) 1 105 2.75 2.75 105 mole

How many moles of water is contained in 1 L of water?

EXAMPLE 3:

Solution:

1 L of water is 1000 g 1 mole of water is 18 g

Thus 1000 g/L 55.55 mol/L 18 g/mol EXAMPLE 4:

Demonstration of ionization of water—pH

Solution: Step 1. H2O 4 H OH In 10,000,000 (or 107) liters of water, 1 gram-ion each of hydronium (H) and hydroxyl (OH) will be formed; i.e. 1/(1 107) 107 mol H and 107 mol OH in 1 liter of water. Step 2. Compute equilibrium constant K In 1 liter of water 1000 55.56 mol/L of water W 18 (107)(107) [H] [OH] K 1.80 1016 [H2O] 55.56 [H2O]

K W 1.80 1016 55.56 1 1014

or Step 3. Compute pH

In pure water, [H] [OH] 1 107 By definition, pH log

1 1 log 7 log 107 7.0 (pure water) [H] 10

EXAMPLE 5: What are the hydrogen ion concentration (in gram-ions per liter) and pH value of a 0.2 mol acetic acid (CH3COOH) solution at 25C? The ionization constant of acetic acid is 1.8 105 at 25C.

Solution: Step 1. Determine H CH3COOH 4 H CH2COOH K

[H] [CH2COOH] [CH3COOH]

BASIC SCIENCE AND FUNDAMENTALS 1.40

CHAPTER 1.1

Let [H] y mole then

[CH2COOH] y mole

and

[CH3COOH] (0.2 y) mole > 0.2 mole, since y is very small compared with the concentration of nonionized acid.

Simply (y) # (y) 0.2 y2 0.2 1.8 105 3.6 106

K 1.8 105

y 1.9 103 [H ] 1.9 103 gram-ion/L

Step 2. Determine pH pH log

1 1 log log 526 2.72 [H] 1.9 103

4.4 Mixing Solutions Solutions may be prepared by diluting a strong stock solution with distilled water or deionized (DI) water or by diluting a strong solution with a weak solution. The following relationships can be expressed for the mixture of solutions: Q Q 1 Q2 Q C Q1 C 1 Q2 C 2 where Q quantity (weight or volume) of mixture Q1 quantity (weight or volume) of solution 1 Q2 quantity (weight or volume) of solution 2 C concentration of mixture C1 concentration of solution 1 C2 concentration of solution 2 EXAMPLE 1: In order to prepare a solution of 3.0 mg/L from 0.10% stock solution, calculate how many milliliters of stock solution are needed to prepare 500 mL of solution from deionized water.

Solution: Initial concentration Ci 0.1%

1 103 106

Final concentration Cf 3.0 mg/L Let Vi and Vf be the initial and final volumes, respectively. Vf 500 mL Since Vi Ci Vf Cf 1 103 3 500 6 106 10 1500 1.50 mL Vi 1000 Vi

3 106

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.41

EXAMPLE 2: 50 gallons of a 4.0% solution is made by mixing of 6.8% and 2.8% solutions. (a) In what ratio should the 6.8% and 2.8% solutions be mixed? (b) How much of each is needed?

Solution: Step 1:

Q Q1 Q2 50 Q1 Q2 Q1 50 Q2 Q1 50 Q2 Q2 Q2

Step 2:

Q C Q1 C1 Q2 C2 50 4 Q1 6.8 Q2 2.8 200 (50 Q2)6.8 2.8Q2 200 340 6.8Q2 2.8Q2 4Q2 140 Q2 35 gal

Step 3: (a)

Q1 50 Q2 50 35 15 3 7 Q2 Q2 35 35

Step 4: (b)

Q2 35 gal Q1 50 Q2 50 35 15 (gal)

In the laboratory, 500 mL of hydrochloric acid (HCl) is being prepared by reaction of sulfuric acid (H2SO4) and sodium chloride. The density of HCl will be 1.20 with 40% purity by weight. The atomic weights are H 1, O 16, Na 23, S 32, and Cl 35.5. What are the weights of H2SO4 and NaCl needed? EXAMPLE 3:

Solution: Step 1. Compute HCl generated HCl 0.5 L 1.20 1000 g/L 0.40 240 g Step 2. Write the chemical equation and calculate the molecular weight (MW) MW

H2SO4 (1 2) 32 (16 4) 98

NaCl 23 35.5 58.5

2HCl Na2SO4 2(1 35.5) 73

Required X Step 3. Determine H2SO4 needed

X 240 g

98 322.2 g 73

Step 4. Determine NaCl needed Y 240 g

58.5 192.3 g 73

Note: 0.5 mole of H2SO4 and 1 mole of NaCl will produce 1 mole of HCl.

240

BASIC SCIENCE AND FUNDAMENTALS 1.42

CHAPTER 1.1

4.5 Chemical Reactions and Dosages Equations of chemical reactions commonly encountered in water and wastewater treatment plants are Cl2 H2O4 HCl HOCl NH3 HOCl 4 NH2Cl H2O NH2Cl HOCl 4 NHCl2 H2O NHCl2 HOCl4 NCl3 H2O Ca(OCl)2 Na2CO3 4 2NaOCl CaCO3 Al2(SO4)3 3CaCO3 3H2O 4 Al2(OH)6 3CaSO4 3CO2 CO2 H2O 4 H2CO3 H2CO3 CaCO3 4 Ca(HCO3)2 Ca(HCO3)2 Na2CO3 4 CaCO3 2NaHCO3 CaCO3 4 Ca2 CO32 CaCO3 H2SO4 4 CaSO4 H2CO3 Ca(HCO3 )2 H2SO4 4 CaSO4 2H2CO3 2S2 2O2 S SO2 S T 4 H2S Cl2 S 2HCl S T H2S 4Cl2 4H2O S H2SO4 8HCl SO2 H2O S H2SO3 HOCl H2SO3 S H2SO4 HCl NH2Cl H2SO3 H2O S NH4HSO4 HCl Na2SO3 Cl2 H2O S Na2SO3 2HCl Chlorine dosage at a treatment plant averages 114.7 lb/d. Its average flow is 12.5 MGD. What is the chlorine dosage in mg/L?

EXAMPLE 1:

Solution: Dosage

114.7 lb/d 114.7 lb 12.5 MGD 12.5 106 gal

114.7 lb 12.5 106 gal 8.34 lb/gal

1.1 106 1.1 ppm

1.1 mg/L EXAMPLE 2: Twenty-seven pounds (27 lb) of chlorine gas is used for treating 750,000 gallons of water. The chlorine demand of the water is measured to be 2.6 mg/L. What is the residual chlorine concentration in the treated water?

Solution: Total dosage 27 lb/0.75 Mgal 36 lb/Mgal 36 lb/Mgal 4.3 mg/L

1 mg/L 8.34 lb/Mgal

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.43

Residual Cl2 4.3 mg/L 2.6 mg/L 1.7 mg/L What is the daily amount of chlorine needed to treat an 8 MGD of water to satisfy 2.8 mg/L chlorine demand and provide 0.5 mg/L residual chlorine.

EXAMPLE 3:

Solution: Total Cl2 needed 2.8 0.5 3.3 mg/L Daily weight 8 106 gal/d 8.34 lb/gal 3.3 mg/L 1 L/106 mg 220 lb/d EXAMPLE 4: At a 11-MGD water plant, a hydrofluosilicic acid (H2SiF6) with 23% by wt solution is fed by a pump at the rate of 0.03 gpm. The specific gravity of the H2SiF6 solution is 1.191. What is the fluoride (F) dosage?

Solution: Pump rate 0.03 gal/min 1440 min/day 43.2 gal/day F appl. rate 43.2 gal/day 0.23 9.94 gal/day Wt of F 9.94 gal/day 8.34 lb/gal 1.191 Wt of water 11 106 gal/day 8.34 lb/gal 1.0 Dosage wt of F/wt of water

9.94 8.34 1.191 11 106 8.34 1.0

1.08 106

1.08 mg/L The average specific gravity of 25% hydrofluosilicic acid (H2SiF6) is 1.208 (Reeves, 1986). How much of 25% acid is required to dose 1 (mg/L) of fluoride (F) to 1 Mgal of water?

EXAMPLE 5:

Solution: Step 1. Determine the density of the acid 8.345 lb/gal sp gr 8.345 lb/gal 1.208 10.08 lb/gal Step 2. Determine F required Since 1 mg/L 1 gal/Mgal This means that 1 gal of full strength F is needed; however, the acid is only 25% strength. Therefore, the amount needed q is q or

1 gal 4 gal 0.25

4 gal 10.08 lb/gal 40.32 lb

12 mg/L of liquid alum with 60% strength is continuously fed to a raw water flow that averages 8.8 MGD. How much liquid alum will be used in a month (assume 30 days)?

EXAMPLE 6:

BASIC SCIENCE AND FUNDAMENTALS 1.44

CHAPTER 1.1

Solution: Since 1 mg/L 1 gal/Mgal Mgal gal 12 8.8 Required/day day 0.60 Mgal 176 gal/day day gal 30 Required/month 176 day month 5,280 gal/month

4.6 Pumpage and Flow Rate Operators of water and wastewater treatment plants encounter pumpage and flow rate during daily operations. In this chapter, some fundamental knowledge of pumpings and flow rates are illustrated for operators. Hydraulics for environmental engineering will be presented in Chapter 1.5. EXAMPLE 1:

Solution:

What is the pressure gage reading at the base of 1 ft (0.3048 m) water column?

Since 1 cubic foot of water is 62.4 lb 62.4 lb 62.4 lb 1 cu ft 1 ft 1 ft 1 ft

62.4 lb 12 in 12 in 1 ft

0.433 lb sq in 1 ft

0.433 psi/ft

pounds per square inch (lb/in2) per foot

Note: At the base of 1 ft depth of water is 0.433 psig (g means gage). EXAMPLE 2:

How many pounds of pressure (psi) will be produced by 50 ft of head?

Solution: Pressure 0.433 psi/ft 50 ft 21.65 psi EXAMPLE 3: A fire hydrant needs a nozzle pressure of 100 psi. What is the head of water to provide this pressure?

Solution: 100 psi H ft 0.433 psi/ft H

100 psi 0.433 psi/ft

231 ft EXAMPLE 4: A high-service pump is pumping the finish water to a storage tank of 120 ft elevation difference. The pressure gauge reading at the discharge line of the pump is 87.6 psi. Determine the head loss due to friction.

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.45

Solution: Gauge reading elevation pressure friction loss Friction head loss 87.6 psi 0.433 psi/ft 120 ft 87.6 psi 52.0 psi 35.6 psi EXAMPLE 5: What is the height (h) of a water column that produces one pound per square inch (1 psi) pressure at the bottom?

Solution:

Since the pressure at the base of a 1-ft water column is 0.433 psi, 0.433 psi 1 psi h 1 ft 1 or h 0.433 2.308 (ft)

144 62.4

When a water tank is filled with water to a height of 46.2 ft and the tank is 22 ft in diameter, what is the pressure (psi) at the bottom of the tank?

EXAMPLE 6:

Solution: height of water column 2.308 ft/psi 46.2 ft 2.308 ft/psi 20.0 psi

Pressure

Note: Pressure is affected by the height of the water column only, not the diameter. A positive displacement pump delivers 48 gpm at 25 strokes per minute. What does the pump deliver at 30 strokes per minute which is within the accurate regulated operating speed?

EXAMPLE 7:

Solution:

By proportioning x gpm 30 strokes 48 gpm 25 strokes x

30 48 25

57.6 (gpm) EXAMPLE 8: The static groundwater level, before pumping a well, is 18 ft below the ground surface. During pumping, the dynamic water level is 52 ft below the ground surface. What is the drawdown?

Solution: The drawdown dynamic level static level 52 ft 18 ft 34 ft A pump discharges at a rate of 125 gpm (Q). How long will it take (t) to fill up a 5000-gal (V ) tank?

EXAMPLE 9:

BASIC SCIENCE AND FUNDAMENTALS 1.46

CHAPTER 1.1

Solution: t

5000 gal V Q 125 gal/min

40 min EXAMPLE 10: A rectangular basin is 50 ft wide, 120 ft long, and 18 ft deep. When 187,000 gal of water are pumped into the basin, what is the height of water raised?

Solution: Step 1. Compute volume V in ft3 V 187,000 gal

1 ft3 7.48 gal

25,000 ft3 Step 2. Compute surface area A in ft2 A 50 ft 120 ft 6000 ft2 Step 3. Determine water raised h in ft (or inches) h V/A 25,000 ft3/6000 ft2 4.17 ft 4.17 ft 12 in/ft 50 in A wastewater plant has eight activated sludge units. It is designed so that each pipe from the primary settling basin to the secondary units has a 1.25 MGD rate capacity. The activated sludge units are capable of operating to a flow of 10% in excess of rate capacity. However, the velocity of sewage in the pipe is limited to the regulatory requirement of 2 fps. What standard pipe size from the basin should be selected? (1 MGD 1.547 cfs)

EXAMPLE 11:

Solution: Step 1. Determine the maximum allowable flow Q in cfs Q 1.25 MGD 1.547 cfs/MGD 8 (1 0.1) 17.02 cfs Step 2. Compute required area A A Q/V 17.02 cfs/2 fps 8.51 ft2 8.51 ft2 144 in2/ft2 1225 in2 Step 3. Select standard pipe size A r2 or r2 A/ r2 1225/3.14 390 r 19.7 in or diameter d 39.4 in However, the standard pipe sizes are 24, 30, 36, 42, and 48 inches. Thus, a 42-inch pipe will be selected.

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.47

A 5-hp motor runs 24 hours per day for aeration purpose. It has 88% efficiency. The electricity cost is 0.018 $/kW-h. What is the monthly (M) power cost?

EXAMPLE 12:

Solution: Work

5 hp kW 0.746 0.88 hp

4.24 kW M use 4.24 kW 24 h/day 30 days 3052.8 kW h M cost 0.018 $/kW h 3052.8 kW h $54.95 A pump delivers a flow of 600 gallons per minute against a head of 30.5 ft (10 m). What is the horsepower of the pump delivered?

EXAMPLE 13:

Solution: Step 1. Convert gpm into lb/s Q 600

gal lb 1 min 8.34 min gal 60 s

83.4 lb/s Step 2. Compute work (W ) done by the pump W Q h 83.34 lb/s 30.5 ft 2543.7 ft lb/s Step 3. Convert to horsepower hp W 2543.7 ft lb/s 4.62 hp

1 hp 550 ft # lb/s

A sodium hypochlorite (NaOCl) solution containing 20% chlorine is used for temporary disinfection when the gas chlorinator is being repaired. Five mg/L of chlorine are required for 2.4 million gallons pumped in 12 hours. How many gallons per hour of hypochlorite solution (assume its specific gravity is 1.0) should be used?

EXAMPLE 14:

Solution: 5 mg/L 5 gal of Cl2/Mgal of water 2.4 Mgal gal Pumpage 5 Mgal 12 h 1 gal/h of Cl2

1 gal/h of NaOCl 0.2

5 gal/h of NaOCl A filter has a filter bed area of 22 ft 22 ft. It is filled to the water level of 15 inches above the backwash-water trough with a water surface area of 22 ft 24 ft. After closing the influent valve, the water dropped 10 in in 4 min. What is the filter rate in gallons per minute per square foot?

EXAMPLE 15:

BASIC SCIENCE AND FUNDAMENTALS 1.48

CHAPTER 1.1

Solution: Step 1. Compute volume filtered per minute, Q Q 22 ft 24 ft (10/12) ft/4 min 110 ft3/min 110 ft3/min 7.48 gal/ft3 822.8 gpm Step 2. Compute filter area A A 22 ft 22 ft 484 ft2 Step 3. Determine filter rate R R Q/A 822.8 gpm/484 ft2 1.7 gpm/ft2 EXAMPLE 16: A rectangular sedimentation basin has a bottom of 60 ft 100 ft and wastewater height of 18 ft. The flow rate is 9.7 MGD. What is the detention time in the basin?

Solution: Step 1. Compute volume (V ) of the basin: V Ah 60 ft 100 ft 18 ft 108,000 ft3 108,000 ft3 7.48 gal/ft3 808,000 gal Step 2. Compute detention time t t V/Q 0.808 106 gal/9.7 106 gal/day 0.0833 days 0.0833 days 24 h/day 2.0 h EXAMPLE 17: The dimensions of a filter are 25 ft 20 ft. What is the backwash rise rate in inches per minute, when the backwash rate is 8230 gpm (gallons per minute)?

Solution: Step 1. Calculate the area of the filter A A 25 ft 20 ft 500 ft2 Step 2. Convert gpm to cubic feet per minute Wash rate 8230 gal/min 8230 gal/min

1 ft3 7.48 gal

1100 ft3/min Step 3. Determine the rise rate 1100 ft3/min wash rate Rise rate A 500 ft2 2.2 ft/min ft in 2.2 12 min ft 26.4 in/min

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.49

EXAMPLE 18: A filter has a surface area of 18 ft (6 m) 27 ft (9 m). The filtration rate (Q) is 800 gpm (0.0505 m3/s). What is the surface loading rate?

Solution: Step 1. Determine surface area A A 18 ft 27 ft 486 ft2 45.1 m2 Step 2. Calculate loading rate (LR) LR

800 gpm Q A 486 ft2

1.646 gpm/ft2 0.000935 m3/(s m2)

After a new main of 4500 ft of 24-in diameter pipeline is installed, it has to be chlorinated, then flushed at a rate of 96 gpm. How long will it take to flush the pipeline completely?

EXAMPLE 19:

Solution: Step 1. Determine the volume (V ) of the water in the pipeline. Diameter 24 in 24 in/12 in/ft 2 ft radius, r 1 ft Volume, V r2l 3.14(1)2 4500 ft3 14,130 ft3 7.48 gal/ft3 105,692 gal Step 2. Compute time of flushing (t) t V/Q 105,692 gal/96 gal/min 1101 min 1101 min/60 min/h 18.35 h 18 h 21 min

EXAMPLE 20: It is generally accepted that the average daily rate of water demand (consumption) is 100–200 gallons per capita per day (gpcpd). Assume that the maximum daily rate is 1.5 times the average daily rate, and the maximum hourly rate is approximately 2.0 times the maximum daily rate, or 300% of the average daily rate. If a city has a population of 140,000 and its water consumption is 130 gpcpd, what are the average and maximum daily rates and maximum hourly rate of water consumption?

Solution: Step 1. Compute average daily rate Qa Qa 130 gpcpd 140,000 18.2 106 gallons per day, gpd or 18.2 million gallons per day, MGD Step 2. Compute maximum daily rate Qm Qm 1.5 Qa 1.5 18.2 MGD 27.3 MGD

BASIC SCIENCE AND FUNDAMENTALS 1.50

CHAPTER 1.1

Step 3. Compute maximum hourly rate Hm Hm 2 Qm 2 2.73 MGD 54.6 MGD

5 STATISTICS 5.1 Measure of Central Value Central value refers to the location of the center of the distribution. There are several measures of central tendency, such as the median, the mid, and the geometric mean; most are used by environmental professionals. 5.2 The Arithmetic Mean The arithmetic mean or the mean (or average) of a set of N numbers X1, X2, X3, . . . Xn is denoted by X . The mean is the sum of the values of observations divided by the number of observations and is given by n

a Xi X1 X2 X3 c Xn i1 X N N EXAMPLE:

The mean of the values 9, 7, 21, 16, 11, 9, and 18 is X

9 7 21 16 11 9 18 91 13 7 7

5.3 The Medium The medium (Md) of a set of numbers is the middle value or the mean of the two middle values when they have been arranged in order of magnitude. EXAMPLE 1:

The set of numbers 7, 9, 9, 11, 16, 18, and 21 (previous example) has a median 11.

EXAMPLE 2:

The medium of 33, 41, 43, 44, 48, 51, 53, and 54 is 44 48 46 2

5.4 The Mode The mode Mo of a set of numbers is the value which occurs most frequently and is hence the most common value. It is the value corresponding to the maximum of the frequency curve of best fit. The mode may not exist, and even if it does exist it may not be unique. An important example of the use of the mode is in the determination of the most probable number of coliform bacteria in a water sample. EXAMPLE 1:

The mode of numbers 9, 11, 11, 17, 18, 18, 18, and 21 is 18.

EXAMPLE 2:

The set of numbers 31, 38, 41, 44, 46, 50, and 51 has no mode.

The set of numbers 3, 3, 8, 8, 8, 9, 11, 14, 14, 14, and 17 has two modes. The numbers 8 and 14 are called bimodal.

EXAMPLE 3:

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.51

5.5 Moving Average Moving averages are averages (simple or weighted) of a convenient number of successive terms. Given a set of numbers Y1, Y2, Y3, . . . Yn, the moving average of order k is calculated by the sequence of arithmetic means, i.e. Y2 Y3 c Yk Yk1 , k

Y1 Y 2 c Y k , k

Y3 Y4 c Yk2 ,... k

The sums in the numerators above are called moving totals of order k. Given a set of numbers 4, 6, 9, 5, 7, 8, 2, 3, 5, 7, and 9, find the moving averages of order 4 by the sequence.

EXAMPLE 1:

Solution:

Moving averages are 4695 , 4 7823 , 4

6957 , 4 8235 , 4

9578 , 4 2357 , 4

5782 4 3579 4

6.00, 6.75, 7.25, 5.50, 5.00, 4.50, 4.25, 6.00

EXAMPLE 2: If the weights 2, 1, 3, and 1 are used for the number set in Example 1, a weighted moving average of order 4 is calculated by the sequence

Solution: 2(4) 1(6) 3(9) 213 2(5) 1(7) 3(8) 7 2(2) 1(3) 3(5) 7 or

1(5) , 1 1(2) ,

2(9) 1(5) 3(7) 1(8) 2(6) 1(9) 3(5) 1(7) , , 7 7 2(7) 1(8) 3(2) 1(3) 2(8) 1(2) 3(3) 1(5) , , 7 7 1(7) 2(3) 1(5) 3(7) 1(9) , 7 6.57, 6.14, 7.43, 6.14, 4.43, 4.57, 4.14, 5.86

Moving averages tend to reduce the amount of variable in a set of data: they eliminate unwanted fluctuations and smooth time series. Using moving averages of appropriate orders, the trend movement can be found, which eliminates seasonal, cyclical, and irregular patterns. One disadvantage of the moving average method is that data at the beginning and the end portion of a series are lost. Another disadvantage is that the method may generate cycles or other movements and is strongly affected by extreme values. In the water and wastewater industries and water resources, data extending over a long period are usually available. Using the moving average method with a proper order may provide valuable trends and results for management purposes. If data are given daily, monthly, or yearly, a moving average of order k is a k-day moving average, a k-month moving average, or a k-year moving average, respectively. 5.6 The Geometric Mean The geometric mean Mg of a set of numbers is the Nth root of the product of the number: N Mg 2(X1 # X2 # X3 c Xn)

BASIC SCIENCE AND FUNDAMENTALS 1.52

CHAPTER 1.1

or common logarithm transformation to (Fair et al., 1963) iN

log Mg EXAMPLE 1:

1 ¢ a log Xi ≤ N i1

Find the Mg of the numbers, 5, 17, 58, 88, 150, and 220.

Solution: 6 Mg 2 (5 17 58 88 150 220) 6 2 14,316,720,000

49.28 This can be calculated with a computer or calculator; however, it also can be done using common logarithms: 1 log 14,316,720,000 6 1 (10.15584) 6 1.69264

log Mg

Mg 49.28

then Another solution: log Mg

1 (log 5 log 17 log 58 log 88 log 150 log 220) 6

1 (0.69897 1.23045 1.76343 1.94448 2.17609 2.34242) 6

1 (10.15584) 6

1.69264 Mg 49.28 EXAMPLE 2:

The arithmetic mean of the same set of numbers is 1 (5 17 58 88 150 220) 6 1 (538) 6

89.67 As shown in Examples 1 and 2, the arithmetic mean of a set of numbers is greater than the geometric mean. However, the two means can be equal to each other for the same cases. Analysis of geometrically normal distribution can also be plotted as the straight-line summation of a frequency distribution on logarithmic probability paper to determine the geometric mean and geometric standard deviation. 5.7 The Variance Because an average in itself does not give a clear picture of a distribution of a set of numbers, many different distributions may all have the same arithmetic mean. A measure of dispersion, spread, or variability is the variance. The variance is defined as the sum of squares of deviations of the observation

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.53

results from their mean (X ) divided by one less than the total number of observations (N). The variance denoted as s2 is given by N

2 a (Xi X)

For simpler calculation, an equivalent formula for s2 is used:

Q g Ni1Xi R

2 a Xi

s 2

Many books (Spiegel, 1961; REA, 1996; Clarke and Cooke, 1992) use N instead of N 1 in the denominator for s2 and later adjust this by multiplying by N/(N 1). For large values of N (generally N 30), there is practically no difference between the two definitions. 5.8 The Standard Deviation The standard deviation of a set of numbers X1, X2, X3, . . . XN is denoted by s and is defined as the positive square root of the variance. Thus the standard deviation is given by N

s ï

≥

2 a (Xi X)

For normal distribution, 68.27% of observations (samples) are in the range between X s and X s, i.e. one standard deviation on either side of the mean. Two and three standard deviations on either side of the mean include 95.45% and 99.73% of the observations, respectively. The standard deviation for the data of observations defined with (N 1) in the denominator is better than defined with N to estimate the standard deviation of a population. EXAMPLE: Find the standard deviation of monthly chemical usages (5, 4, 6, 8, 10, 12, 13, 11, 9, 7, 5, and 6 tons) in a year.

Solution:

X N 5 4 6 8 10 12 13 11 9 7 5 6 12

Mean X

8.0 (tons)

(X X)2 (5 8)2 (4 8)2 (6 8)2 (8 8)2 (10 8)2 (12 8)2 (13 8)2 (11 8)2 (9 8)2 (7 8)2 (5 8)2 (6 8)2 98 s

(X X)2 R N1 Å B

98 Ä 12 1

2.985 (tons)

BASIC SCIENCE AND FUNDAMENTALS 1.54

CHAPTER 1.1

5.9 The Geometric Standard Deviation The geometric standard deviation of a set of numbers X1, X2, X3, . . . XN is denoted by g and is defined by N

log sg ï

≥

2 a (log Xi log Mg)

1 ≤ ¢ a log2 Xi N log2 Mg ≤ B i1 Å N1 ¢

The value of g can be determined by a graphic method. On logarithmic probability paper, the straight line of best fit passes through the intersection of the geometric mean with the 50% percentile point and through the intersection of Mg g with the 84.1 percentile frequency, and Mg/g with the 15.9 percentile, respectively. EXAMPLE:

Find the geometric mean and standard deviation for the example in Sec. 5.8.

Solution:

log X N (log 5 log 4 log 6 log 8 log 10 log 12 log 13 log 11 log 9 log 7 log 5 log 6)/12 0.87444 Mg 7.489 (tons)

log Mg

log2 X 9.481

log2X NlogMg log sg a b N Å

9.481 12(0.87444)2 ≤ 12 1 Å 2 0.02775 0.16659 sg 1.468 (tons)

5.10 The Student’s t Test There are some useful statistical tools to compare means. The Student’s t test can be used for statistical comparison of two means to determine whether or not they are the same. Two sets of data may be such as one from a new experiment and one for control; two results of split samples from two laboratories; or a new analytical and an existing analytical method. To test difference of two means, the t value is calculated by X1 X2 t s 21/N1 1/N2 where N1, N2 the numbers of sample sizes in the first and second sets X1, X2 the means, respectively s1, s2 the standard deviations, respectively universal standard deviation N1s21 N2s22 ≤ ¢ Å N1 N2 2 v N1 N2 2 degrees of freedom

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.55

In a water treatment plant, 12-day observations are made on two parallel flocculation– sedimentation units using a new type of coagulant and an old coagulant. The results of turbidity removal (%) are given in Table 1.6. Find any improvement of treatment with new chemical.

EXAMPLE:

X1 1088 90.7 N 12

X2 1051 X2 87.6 N 12 X1

The s21 and s22 are calculated by

X21 ( X1)2/N N1 98,980 (1088)2/12 12 1

s21

30.42

X22 ( X2)2/N 92,367 (1051)2/12 s22 N1 12 1 28.81 The universal standard deviation is s

(N1 1)s21 (N2 1)s22 R N1 N2 2 Å B

(12 1)30.42 (12 1)28.81 R 12 12 2 Å B

5.442

The Student’s t value for the data is calculated by t

X1 X2 s 2(1/N1 1/N2)

90.7 87.6 5.442 2(1/12 1/12)

1.395

TABLE 1.6 Results of Turbidity Removal (%) New chemical used Day 1 2 3 4 5 6 7 8 9 10 11 12

Old chemical used X 12

X22

95 90 88 96 80 93 99 90 86 97 88 86

9025 8100 7744 9216 6400 8649 9801 8100 7396 9409 7744 7396

90 92 80 90 81 90 95 85 84 94 90 80

8100 8464 6400 8100 6561 8100 9025 7225 7056 8836 8100 6400

1088

98980

1051

92367

BASIC SCIENCE AND FUNDAMENTALS 1.56

CHAPTER 1.1

The critical value of t for 22 (N1 N2 2 12 12 2) degrees of freedom is 1.717 at a 95% confidence level, t0.05(22). The critical value can be found from the t distribution table of any statistical textbook. The calculated t value of 1.395 is less than the critical value of 1.717. Therefore, it can be concluded that the results from the two types of coagulants used are not significantly different at the 95% confidence level. 5.11 Multiple Range Tests Testing significance difference for a group of ranked means can employ multiple range tests. The multiple range tests may include the lsd (least significant difference) or the multiple t test, Student— Newman—Keuls test, Tukey’s test based on allowances, and Duncan’s multiple range test (Duncan, 1955; Federer, 1955). An example of Duncan’s multiple test is given below. EXAMPLE: A lake area has implemented the best management practices (soil conservation) in the watershed in 1984 and 1985. Water quality monitoring programs were carried out in 1981 (pre-implementation); in 1994–95 during implementation; in 1986 (after implementation); and in 1992–94 (long-term post-implementation). The observed total phosphorus (TP) concentrations and yearly mean values for a sampling station are listed in Table 1.7 (Lin and Raman, 1997).

Question: Is there any water quality improvement (decrease concentration) after implementation? Solution: This can be solved by using Duncan’s multiple range test to determine any significant differences among the means of each year. It is necessary to determine the standard error of the difference between two treatment means (Sd). The Sd is calculated as the square root of two times the error mean square divided by the number of replication (Carmer and Walker, 1985) as follows: Sd

2s2 ¢ r ≤ Å

where s2 is the estimated error variance and r is the number of replications. If the ith and jth groups (treatments) have different number of observations (this example) or unequal variances, the standard error of the difference between the ith and jth treatment means (each pair) is calculated by Sd(ij)

s2 s2 ¢r r ≤ i j Å

TABLE 1.7 Total Phosphorus Concentrations at Station 1

May Jun Jul Aug Sep Oct Mean X Total T n T 2/n

1981

1984

1985

1986

1992

1993

1994

0.74 0.28 0.33 0.6 0.77 1.2 1.11 0.96 1.44 1.43 0.26

0.088 0.074 0.268 0.801 0.857 0.91 0.873 1.671 0.733 0.143 0.209

0.125 0.17 0.194 0.151 0.453 0.389 0.473 0.555 0.557

0.065

0.135

0.319

0.874

0.048 0.027 0.131

0.626

0.167

0.971

0.119

1.066

0.162

0.312

0.051 0.131 0.128 0.124 0.115 0.214 0.289 0.33 0.204 0.757 0.265

0.096

0.083

0.146

0.829 9.12 11 7.5613

0.602 6.627 11 3.9925

0.338 3.379 10 1.1418

0.237 2.608 11 0.6183

0.524 3.143 6 1.6464

0.257 1.54 6 0.3953

0.202 1.008 5 0.2032

0.656

BASIC SCIENCE AND FUNDAMENTALS 1.57

BASIC SCIENCE AND FUNDAMENTALS

TABLE 1.8 Analysis of Variance Table for One Variable Sum of squares 2 T

n N i

Category means

T2i

Within group

X ij2 n i

Total

X 2ij

T2 N

Degree of freedom (df)

Mean square

s 2m

s 2p

F ratio F

Sm2 S p2

k number of categories or treatments n number of measurements (observations, or replication r) in each category i the ith category j the jth measurement in each category T1 X11 X12 X13 … Xln, etc. T T1 T2 … Tk the grand total N n1 n2 n3 … nk

where

Step 1. We shall make two-way analyses of variance by using the formulas in Table 1.8 (Dixon and Massey, 1957) to estimate the population 2 (the variance of the universe) in two ways, then compare these two estimates. The mean square for (or between) categories, sm2, is the sum of square for means (or between categories) divided by the degree of freedom for s2m. The computing formula for s2m if there are k categories is

s2m

2 k T2 T i a ni N i1

sp2 is

sometimes called the within-groups mean square, or within-groups variThe pooled variance ance. The sp2 is the within-groups sum of squares divided by the degrees of freedom (k categories) in s2p and is calculated by s2p

X2ij (T2i/ni)

ni k

These two values, s 2m and s 2p, may be tested for significant difference by using the F ratio (F sm2 /sp2). Step 2. We shall estimate 2. The computations are simple to perform with a calculator or a computer. The hypothesis of no difference in categories (i.e. equal means) will be tested at 5% level of significance (or 95% of confidence). Computation: Using data from Table 1.7 2 Ti a ni (7.5613 3.9925 1.1418 0.6183 1.6464 0.3953 0.2032) 15.5588 T 9.12 6.627 3.379 2.608 3.143 1.14 1.008 27.425 grand total N 11 11 10 11 6 6 5 60

X 2 (0.74)2 (0.28)2 (0.33)2 … (0.146)2 ij

22.2167

BASIC SCIENCE AND FUNDAMENTALS 1.58

CHAPTER 1.1

Sum of squares category means 2 2 Ti T a n N i (15.5588) (27.425)2/60

3.0233 Degree of freedom (df) for category k 1 7 1 6 s2m 3.0233/6 0.5039 Sum of squares within group Tit2

X 2ij n i

22.2167 15.5588 6.6579 df for within N k 60 7 53 sp2 6.6579/53 0.1256 Calculated F

s2m s2p

0.5039 4.012 0.1256

Critical F.95(6.53) 2.28 The results of computations are summarized in Table 1.9. The calculated F ratio is greater than the critical F ratio. We will reject the hypothesis of no difference in means. In other words, the groups are from populations having unequal means. The procedure for Steps 1 and 2 is called twoway analysis of variance. Step 3. We determine the difference in the means among the group. First, the means are rearranged with the order of magnitude. Each possible difference in means is determined. Table 1.10 presents the ranked means and all possible differences among the seven means. Step 4. We determine s2p/ri for each year. These are 0.0114, 0.0114, 0.0209, 0.0126, 0.0209, 0.0114, and 0.0251 for 1981, 1984, 1992, 1985, 1993, 1986, and 1994, respectively. Since s2 s2p, we then determine the standard error of the difference between each two means by Sd(ij)

2 s2p 1 sp B ¢r r ≤R j Å 2 i

The results are presented in Table 1.11, which is similar to Table 1.10. Step 5. For the present example, choose the significance level of 5%. From Duncan’s tables or by interpolation from the table, significant ranges (SR) for 5% significant level with df 53 are 3.25, 3.21, 3.15, 3.09, 2.99, and 2.88, respectively, for n 7, 6, 5, 4, 3, and 2. The critical ranges are computed by SR times Sd(ij) as shown in Table 1.12. Values of Sd(ij) are from the corresponding matrix in Table 1.11. Step 6. Comparing respective values in Tables 1.10 and 1.12, we may conclude the difference is significant with 95% confidence if the difference of mean is greater than the critical value of the multiple range test. It is concluded that the mean value in 1981 was significantly higher than that in 1994, 1986, 1993, and 1995. Also, the mean in 1984 was greater than that in 1986. There is no signficant difference in mean among 1985, 1986, and 1992–94. Note: This example not only illustrates Duncan’s multiple range test but also analysis of variance.

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.59

TABLE 1.9 Results of Analysis of Variance Sum of squares

Mean square

Category means

3.0233

0.5039

Within-group

6.6579

0.1256

Total

9.6811

TABLE 1.10

F ratio 0.5039 F 4.012 0.1256 F0.95(6.53) 2.28

Differences of Total Phosphorus Means

Year

Year Mean

1981 0.829

1984 0.602

1992 0.524

1985 0.338

1993 0.257

1996 0.237

1994 1986 1993 1985 1992 1984

0.202 0.237 0.257 0.338 0.524 0.602

0.627 0.592 0.572 0.491 0.305 0.227

0.400 0.365 0.345 0.264 0.078

0.322 0.287 0.267 0.186

0.136 0.101 0.081

0.055 0.020

0.035

TABLE 1.11 Standard Error of Means Year

Year s2p/ri

1981 0.0114

1984 0.0114

1992 0.0209

1985 0.0126

1993 0.0209

1986 0.0114

1994 1986 1993 1985 1992 1984

0.0251 0.0114 0.0209 0.0126 0.0209 0.0114

0.1351 0.1068 0.1272 0.1096 0.1272 0.1068

0.1351 0.1068 0.1272 0.1096 0.1272

0.1517 0.1272 0.1446 0.1295

0.1372 0.1096 0.1293

0.1517 0.1272

0.1351

TABLE 1.12 Critical Ranges for Multiple Range Test Year

1981

1984

1992

1985

1993

1986

1994 1986 1993 1985 1992 1984

3.25 3.21 3.15 3.09 2.99 2.88

0.439 0.343 0.401 0.339 0.381 0.343

0.439 0.343 0.401 0.339 0.381

0.479 0.393 0.433 0.368

0.424 0.328 0.368

0.454 0.362

0.380

5.12 Regression Analysis Data collected from studies of water and wastewater engineering may be evaluated from the relationships between two or more parameters measured: there are many mathematics functions to fit the data. Those most commonly encountered by environmental engineers for curve fitting are linear, semi-log, and log-log relations between parameters (or variables) X and Y. Curve fitting can be estimated by the graphical (plot) method and can be calculated by least-square regression analysis.

BASIC SCIENCE AND FUNDAMENTALS 1.60

CHAPTER 1.1

Linear equation (the straight line).

A straight line is described by the equation Y a bX

where a and b are constants and are computed from normal equations. Given any two points (X1, Y1) and (X2, Y2) on a straight line, the constants a and b can be computed. The following relation exists: Y Y1 Y2 Y1 X X1 X2 X1 then

Y Y1 ¢

Y2 Y1 ≤(X X1) X2 X1

Y Y1 m(X X1) where m is called the slope of the line. The constant a, which is the value of Y when X 0, is called the Y intercept. The least-square regression. Given the set of points (X1, Y1), (X2, Y2), (X3, Y3) . . . (Xn, Yn) is described by the equation Y a bX The normal equations for the least-square line are

Y aN b X

XY a X b X2 The constants a and b can be determined by solving simultaneously the above equations a

n( Y )( X2) ( X)( XY ) n X2 ( X )2

n XY ( X )( Y ) n X 2 ( X )2

The coefficient of correlation is given by r

(Y est. Y )2

(Y Y )2

where Yest. is the estimated Y value from the regression line and Y is the mean of Y values. The r lies between zero and one. It is more frequently used as r2, because it is always non-negative (US Department of Interior, 1968). EXAMPLE: Determine the regression line of Y (new chemical used) on X (old chemical) and coefficient of regression for the data of the example given in Sec. 5.10 (Student’s t test section)

Solution: Let Y a bX Step 1. Compute the following values (Table 1.13)

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.61

TABLE 1.13 Determination of Constants Y

Yest.

95 90 88 96 80 93 99 90 86 97 88 86

90 92 80 90 81 90 95 85 84 94 90 80

8100 8464 6400 8100 6561 8100 9025 9225 7056 8836 8100 6400

8550 8280 7040 8640 6480 6370 9405 7650 7224 9118 7920 6880

92.7 94.3 84.3 91.7 85.1 92.7 96.9 88.5 87.6 96.0 92.7 84.3

1051

92367

95557

1088

Y 1088/12 90.7

Step 2. Compute a and b a

( Y )( X2) (X ) (XY ) N X 2 ( X )2 (1088)(92,367) (1051)(95,557) (12)(92,367) (1051)2

17.06 b

N XY ( X )( Y ) N X 2 ( X )2 12(95,557) (1051)(1088) (12)(92,367) (1051)2

0.84 The regression equation is Y 17.06 0.84X Step 3. Calculate Yest. by the above equation and shown in Table 1.13. Step 4. Compute coefficient of correlation with Table 1.14. Then, coefficient of correlation is given by r and

(Y est. Y )2

Å (Y Y )2

223.22 0.817 Å 334.68

r2 0.667 Nonlinear equations. For two parameters without a linear relationship, three mathematical models are usually used in an effort to formulate their relationship. They include

BASIC SCIENCE AND FUNDAMENTALS 1.62

CHAPTER 1.1

TABLE 1.14 Determination of Coefficient of Correlation Yest. 92.7 94.3 84.3 92.7 85.1 92.7 96.9 88.5 87.6 96.0 92.7 84.3

(Yest. Y )2

(Y Y )2

4 12.96 40.96 4 31.36 4 38.44 4.84 9.61 28.09 4 40.96

95 90 88 96 80 93 99 90 86 97 88 86

18.49 8.49 7.29 28.09 114.49 5.29 68.89 0.49 22.09 39.69 7.29 22.09

223.22

quadratic

Y a bX cX2

logarithmic

Y aXb

and geometric

Y a log X b

334.68

where Y is a dependent variable, X is an independent variable, and a, b, c are constants determined by regression analysis. EXAMPLE: A river water is used as a supply source for a city. The jar test results show the algal removal with various alum dosage as below:

Alum dosage, mg/L 0 10 20 25 30 40

Algal reduction, % 951 5183 69100 7397 87100 93100

Determine the relationship of algal removal (X ) versus alum dosage (Y ). Solution:

(see Lin et al., 1971)

Applying the three models stated above for 25 test runs using regression analysis, it was found that model 1 (quadratic equation) expressed best the relationship of algal (and turbidity) removal with coagulant dosage. The relationships are for algal removal Y 36.54 3.2325X 0.04256X2 for turbidity removal Y 37.08 2.9567X 0.04355X2 Semi-log regression. Nonlinear relationships can sometimes be reduced to linear relationships by appropriate transformation of variables. If the data of two variables show a semi-log relationship, the formula is given by Y aebx

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.63

Taking logarithm transformation, it can be rewritten as log Y log a bx log e log Y log a (0.4343b)X This is similar to a linear equation. The constants a and b and the coefficient of correlation can be determined with the steps mentioned earlier (linear regression) after transformation of Y values to log Y values. On the other hand, a and b can also be roughly estimated by plotting data points (X1, Y1), (X2, Y2) . . . (Xn, Yn) on semi-log paper and drawing a straight line. Log-log regression. If the data of two variables are plotted on log-log paper and show a straightline relationship, the data can be expressed as Y aXb log Y log a b(log X)

Similarly, constants a and b as well as the coefficient of correlation r can be computed using the same manner of linear regression. The least-square parabola. A set of points (X1, Y1), (X2, Y2), (X3, Y3) . . . (Xn, Yn) may show a parabolic relationship. The equation can be written Y a bX cX2 where the constants a, b, and c are calculated by solving normal equations simultaneously for the least-square parabola as follows:

Y aN b X c X2

XY a X b X2 c X3

X2Y a X2 b X3 c X4 These three equations are obtained by multiplying equation Y a bX cX2 by 1, X, and X2, respectively, and summing both sides of the result. This technique can be extended to obtain normal equations for least-square cubic curves, leastsquare quartic curves, and so on. The relationship of some chemical parameters in water, population increase with time periods, and wastewater production against time may fit least-square parabola. EXAMPLE: The following table presents data on temperature and dissolved solids in lake water samples. Find the equation of a least-square parabola fitting the data.

Temperature, C DS, mg/L

6 38

9 48

12 56

15 63

18 69

21 76

24 83

27 88

30 92

Solution: Step 1. Let the variables X and Y denote the water temperature and the dissolved solids concentration, respectively. The equation of a least-square parabola fitting the data can be written as Y a bX cX2 where constants a, b, and c can be determined from the normal equations.

BASIC SCIENCE AND FUNDAMENTALS 1.64

CHAPTER 1.1

TABLE 1.15 Calculation of Data

X 2Y

38 48 56 63 69 76 83 88 92

6 9 12 15 18 21 24 27 30

36 81 144 225 324 441 576 729 900

216 729 1728 3375 5832 9261 13824 19683 27000

1296 6561 20736 50625 104976 194481 331776 531441 810000

228 432 672 945 1242 1596 1992 2376 2760

1368 3888 8064 14175 22356 33516 47808 64152 82800

613

162

3456

81648

2051892

12243

278127

Step 2. Perform the calculations shown in Table 1.15 to solve power of 2 equations.

Y aN b X c X2 a(9) b(162) c(3456) 613

XY a X b X2 c X3 a(162) b(3456) c(81,648) 12,243

X Y a X2 b X3 c X4 2

a(3456) b(81,648) c(2,051,892) 278,127 The equations are simplified as a 18b 384c 68.11 a 21.33b 504c 75.57 a 23.63b 593.7c 80.48 Eliminate a 3.33b 120c 7.46 56.3b 209.7c 12.37 or

b 36c 2.24 b 37.2c 2.20

Eliminate b 1.2c 0.04 c 0.033 Determine b b 36(0.033) 2.24 b 3.44 Determine a a 18(3.44) 384(0.033) 68.11 a 18.86

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.65

Step 4. The equation for the data set is Y 18.86 3.44X 0.033X 2 where the origin X 0 is at temperature of 0C. Another method used to solve the least-square parabola will be presented in an example of population projection in the section on water treatment (Chapter 1.6). Multiple linear regression analysis. In environmental engineering, a water quality parameter may be influenced by several factors. To evaluate the observed results, one can use multiple linear regression analysis. This analysis can be easily performed with a readily available computer program. The standard multiple linear relationship can be expressed as Z C0 C1U1 C2U2 … CnUn where

Z a parameter, such as algal density C0 . . . Cn regression constants U1 . . . Un independent water quality

For example, Lin et al. (1971) evaluated the algal density after the coagulation process and the factors that were possibly influenced, such as initial turbidity, total solids, nonfilterable residue, pH of untreated water, water temperature, time of floc formation, and initial algal concentration. A step-wise multiple linear regression analysis was used to evaluate data for all observed 25 runs. The result gave Z 1132 104.85U1 0.23U2 1.80U3 33.25U4 where Z algal concentration in the treated water, cells/mL U1 alum dosage, mg/L U2 initial algal concentration, cells/mL U3 nonfilterable residue, mg/L U4 water temperasture, C The coefficient of correlation is 0.857 and the standard error of estimate is 896 cells/mL. 5.13 Calculation of Data Quality Indicators Quality assurance. Quality assurance (QA) is a set of operating principles that should be followed during sample collection and analysis. It includes quality control and quality assessment. Any data collection activities performed for the US Environmental Protection Agency (EPA) are required to have a QA project plan to produce data with a high level of confidence. Standard methods (APHA, AWWA, and WEF, 1995), Simes (1991), and EPA (1991) have recommended guidelines for QA programs. The QA plan specifies the correction factors to be applied, as well as the procedures to be followed, to validate the result. Quality control. Quality control (QC) data production may be either or both internal and external. A good-quality control program should include at least certification of operator competence, recovery of known additions, analysis of externally supplied standards, analyses of reagent blank, calibration with standards, analyses of duplicates, and control charts. The main indicators of data quality are bias and precision. Bias is consistent deviation of the measured value from the true value caused by systematic errors which are due to the analytical method and to the laboratory’s use of the method. Precision is a measurement of closeness, in which the results of multiple analyses of a given sample agree with each other; it is usually expressed as the standard deviation.

BASIC SCIENCE AND FUNDAMENTALS 1.66

CHAPTER 1.1

The requirements of data quality indicators of EPA projects should at least include calculations for precision, accuracy, completeness, and method detection limit. In addition, equations must be given for other project-specific calculations, such as mass balance, emission rates, confidence ranges, etc. Precision. Precision includes the random error in sampling and the error in sample preparation and analysis. It is specified by the standard deviation of the results. Precision of the laboratory analysis should be assessed by comparing the analytical results for a sample and its duplicate. If calculated from duplicate measurements, the relative percent difference (RPD) is the normal measurement of precision (USEPA, 1996, 1997): RPD

(C1 C2) 100% (C1 C2)/2

where RPD relative percent difference C1 larger of the observed values C2 smaller of the observed values Find the RPD of the total alkalinity measurements (152 and 148 mg/L as CaCO3) of a duplicate water sample.

EXAMPLE 1:

Solution: RPD

(C1 C2)(100%) (C1 C2)/2 (152 148)(100%) (152 148)/2

2.67% If calculated from three or more replicates, relative standard deviation (RSD) is used rather than RPD. The RSD is given by RSD (s/Y ) 100% where RSD relative standard deviation s standard deviation Y mean of replicate analysis Standard deviation, defined previously, is n

B aB n1

(Yi Y)2 R n1

where s standard deviation Yi measured value of ith replicate Y mean of replicate measurements n number of replicates EXAMPLE 2:

Find the RSD for 6 replicates (56, 49, 54, 48, 47, and 51 mg/L) of a sample.

Solution: Step 1. Determine the mean Y Yi /n 305/6 51

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

1.67

Step 2. Determine s s

Å 3.55

g(Yi Y )2 R 2(52 22 32 32 42 0)/5 n1

Step 3. Calculate RSD RSD (s/Y ) 100% (3.55/51) 100% 7.0% EXAMPLE 3: A known concentration (1.10 mg/L of nitrate-nitrogen) is added to each of 25 water samples. For spiked samples, the precision calculation can be computed (Table 1.16).

Solution: Step 1. Calculate the concentration of recovery which is spiked (1) minus sample concentraction (2): shown in column (3). Step 2. The differences between calculated recovery (3) and known addition (4) are shown in column (5). TABLE 1.16 (1) Concentration for spiked sample

(2) Concentration for sample only

(3) Calculated recovery (1) (2)

(4) Known concentration added

(5) Deviation from expected (3) (4)

3.58 3.5 2.7 2.55 4.81 2.52 4.8 5.4 2.53 2.77 4.56 5.05 2.45 3.6 4.78 3.9 4.66 3.8 3.25 5.55 6.17 5.66 5.44 4.28 5.78 4 3.07

2.33 2.29 1.58 1.49 3.56 1.44 3.57 4.46 1.6 1.51 3.45 3.99 1.3 2.54 3.48 2.79 4.48 2.79 2.11 4.55 5.01 4.39 4.39 3.27 4.66 2.79 1.99

1.25 1.21 1.12 1.06 1.25 1.08 1.23 0.94 0.93 1.26 1.11 1.06 1.15 1.06 1.3 1.11 0.18 1.01 1.14 1 1.16 1.27 1.05 1.01 1.12 1.21 1.08

1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1 1.1

0.15 0.11 0.02 0.04 0.15 0.02 0.13 0.16 0.17 0.16 0.01 0.04 0.05 0.04 0.2 0.01 0.92 0.09 0.04 0.1 0.06 0.17 0.05 0.09 0.02 0.11 0.02

(deviation)2 1.1245

BASIC SCIENCE AND FUNDAMENTALS 1.68

CHAPTER 1.1

Step 3. Calculate s by

( deviation2) 1.1245 0.216 Å 25 1 Å n1

Accuracy. Accuracy is the combination of bias and precision of an analytical procedure and reflects the closeness of a measured value to a true value. In order to assure the accuracy of the analytical procedures, matrix spikes are analyzed. The spike sample is made by adding a known amount of analyte to a water sample or to deionized water. The percent recovery (%R) for the spiked sample is calculated by the following formula: %R

SU 100% C

where %R percent recovery S measured concentration in spiked aliquot U measured concentration in unspiked aliquot C actual concentration of known addition EXAMPLE:

In the previous (precision) example, the percent recovery for the first sample is %R

3.58 2.33 100% 13.6% 1.10

When a standard reference material (SRM) is used, the percent can be computed by %R

Cm 100% C srm

where %R percent recovery Cm measured concentration of SRM Csrm actual concentration of SRM Completeness. Completeness for all measures is defined as follows: %C

V 100% T

where %C percent completeness V number of measurements judged valid T total number of measurements Method detection limit. follows:

The method detection limit (MDL) for all measurements is defined as MDL t(n1,10.99) s

MDL method detection limit s standard deviation of replicate analyses t(n1,10.99) Student’s t value for a one-side 99% confidence level and a standard deviation estimate with n 1 degree of freedom. They are as follows:

where

df t(n1,10.99)

2 6.965

3 4.541

4 3.747

5 6 3.365 3.143

7 2.998

8 2.896

10 2.764

20 2.528

BASIC SCIENCE AND FUNDAMENTALS BASIC SCIENCE AND FUNDAMENTALS

TABLE 1.17

1.69

Determination of s

Concentration measured

Standard concentration

0.022 0.019 0.025 0.026 0.016 0.017 0.020 0.022

0.020 0.020 0.020 0.020 0.020 0.020 0.020 0.020

Deviation

(Deviation)2

0.002 0.001 0.005 0.006 0.004 0.003 0 0.002

0.000004 0.000001 0.000025 0.000036 0.000016 0.000009 0.000000 0.000004 0.000095

To determine the MDL, it is required that a standard solution must be measured for at least seven replicates. The concentration of standard solution should be as close as the MDL; generally 2 to 5 times of the MDL. EXAMPLE: A 0.02 mg/L of nitrate nitrogen was measured for eight replicates. The measured results are 0.022, 0.019, 0.025, 0.026, 0.016, 0.017, 0.020, and 0.022 mg/L. Find the MDL of nitrate nitrogen for your laboratory.

Solution: Step 1. Determine s (Table 1.17) From Table 1.17

(Deviation)2 0.00095 s

(Deviation)2 0.000095 0.00368 Å Å 81 n1

Step 2. Compute MDL MDL t(n1,10.99) s 2.998 0.00368 0.011 It will be reported that the MDL of nitrate nitrogen is 0.011 mg/L for the laboratory.

REFERENCES American Public Health Association (APHA), American Water Works Association (AWWA), and Water Environment Federation (WEF). 1995. Standard methods for the examination of water and wastewater. Washington, DC: American Public Health Association. Carmer, S. G. and Walker, W. M. 1985. Pair wise multiple comparisons of treatment means in agronomic research. J. Agronomic Educ. 14(1): 19–26. Clark, G. M. and Cooke, D. 1992. A basic course in statistics, 3rd edn. London: Edward Arnold. Dixon, J. and Masey, Jr., J. 1957. Introduction to statistical analyses, 2nd edn. New York: McGraw-Hill. Duncan, B. 1955. Multiple range and multiple F-tests. Biometrics, 11(1): 1–42.

BASIC SCIENCE AND FUNDAMENTALS 1.70

CHAPTER 1.1

Fair, G. M., Geyer, J. C. and Morris, J. C. 1963. Water supply and waste-water disposal. New York: John Wiley. Federer, W. T. 1955. Experimental design, theory, and application. New York: W. W. Norton. Friedman, D., Pisani, R., Purves, R. and Adhikor, A. 1991. Statistics, 2nd edn. New York: W. W. Norton. Lin, S. D. and Raman, R. K. 1997. Phase III, post-restoration monitoring of Lake Le-Aqua-Na. Contract Report 610, Champaign: Illinois State Water Survey. Lin, S. D., Evans, R. L. and Beuscher, D. B. 1971. Algal removal by alum coagulation. Report of Investigation 68. Urbana: Illinois State Water Survey. New York State Department of Health. 1961. Manual of instruction for water treatment plant operators. Albany, New York: State of New York. Reeves, T. G. 1986. Water fluoridation—A manual for engineers and technicians. Atlanta, Georgia: US Department of Health and Human Service. Research and Education Association. 1996. The statistics problem solvers. Piscataway, New Jersey: REA. Simon, A. L. 1976. Practical hydraulics. New York: John Wiley. Simes, G. F. 1991. Preparation aids for the development of category III quality assurance project plans. EPA/600/8-91/005. Washington DC: USEPA. Spiegel, M. R. 1961. Schaum’s outline of theory and problems of statistics. New York: McGraw-Hill. US Department of the Interior. 1968. Data evaluation and analysis—Training manual. Washington, DC: US Department of the Interior. US Environmental Protection Agency. 1991. Preparation aids for the development of category I quality assurance project plans. EPA/600/8–91/1003. Washington, DC: USEPA. US Environmental Protection Agency. 1996. EPA guidance for quality assurance project plans. EPA QA/G–5. External workshop draft, US Environmental Protection Agency, Washington, DC: USEPA. US Environmental Protection Agency. 1997. EPA request for quality assurance project plans for environmental data operations. EPA QA/R–5. Washington, DC: USEPA.

Source: HANDBOOK OF ENVIRONMENTAL ENGINEERING CALCULATIONS

CHAPTER 1.2

STREAMS AND RIVERS Shun Dar Lin

GENERAL 1.72

POINT SOURCE DILUTION 1.72

DISCHARGE MEASUREMENT 1.72

TIME OF TRAVEL 1.73

DISSOLVED OXYGEN AND WATER TEMPERATURE 1.74 5.1 Dissolved Oxygen Saturation 1.75 5.2 Dissolved Oxygen Availability 1.77

BIOCHEMICAL OXYGEN DEMAND ANALYSIS 1.78

STREETER–PHELPS OXYGEN SAG FORMULA 1.80

BOD MODELS AND K1 COMPUTATION 1.81 8.1 First-order Reaction 1.82 8.2 Determination of Deoxygenation Rate and Ultimate BOD 1.84 8.3 Temperature Effect on K1 1.95 8.4 Second-order Reaction 1.98

DETERMINATION OF REAERATION RATE CONSTANT K2 1.102 9.1 Basic Conservation 1.102 9.2 From BOD and Oxygen Sag Constants 1.103 9.3 Empirical Formulas 1.104 9.4 Stationary Field Monitoring Procedure 1.106 SEDIMENT OXYGEN DEMAND 1.108 10.1 Relationship of Sediment Characteristics and SOD 1.109 10.2 SOD Versus DO 1.109

12 PHOTOSYNTHESIS AND RESPIRATION 1.111 13 NATURAL SELF-PURIFICATION IN STREAMS 1.112 13.1 Oxygen Sag Curve 1.112 13.2 Determination of kr 1.118 13.3 Critical Point on Oxygen Sag Curve 1.119 13.4 Simplified Oxygen Sag Computations 1.130 14 SOD OF DO USAGE 1.131 15 APPORTIONMENT OF STREAM USERS 1.131 15.1 Method 1 1.132 15.2 Method 2 1.134 15.3 Method 3 1.136 16 VELZ REAERATION CURVE (A PRAGMATIC APPROACH) 1.137 16.1 Dissolved Oxygen Used 1.137 16.2 Reaeration 1.138 17 STREAM DO MODEL (A PRAGMATIC APPROACH) 1.141 17.1 Influence of a Dam 1.141 17.2 Influence of Tributaries 1.143 17.3 DO Used 1.145 17.4 Procedures of Pragmatic Approach 1.146 18 BIOLOGICAL FACTORS 1.153 18.1 Algae 1.153 18.2 Indicator Bacteria 1.154 18.3 Macroinvertebrate Biotic Index 1.162 REFERENCES 1.162

ORGANIC SLUDGE DEPOSITS 1.110

1.71

STREAMS AND RIVERS 1.72

CHAPTER 1.2

1 GENERAL This chapter presents calculations on stream sanitation. The main portion covers the evaluation of water assimilative capacities of rivers or streams. The procedures include classical conceptual approaches and pragmatic approaches: the conceptual approaches use simulation models, whereas Butts and his coworkers of the Illinois State Water Survey use a pragmatic approach. Observed dissolved oxygen (DO) and biochemical oxygen deman (BOD) levels are measured at several sampling points along a stream reach. Both approaches are useful for developing or approving the design of wastewater treatment facilities that discharge into a stream. In addition, biological factors such as algae, indicator bacteria, diversity index, macroinvertebrate biotic index, and fish habitat analysis are also presented.

2 POINT SOURCE DILUTION Point source pollutants are commonly regulated by a deterministic model for an assumed design condition having a specific probability of occurrence. A simplistic dilution and/or balance equation can be written as Cd

QuCu QeCe Qu Qe

(2.1)

where Cd completely mixed constituent concentration downstream of the effluent, mg/L Qu stream flow upstream of the effluent, cubic feet per second, cfs or m3/s Cu constituent concentration of upstream flow, mg/L Qe flow of the effluent, cfs or m3/s Ce constituent concentration of the effluent, mg/L Under the worst case, a 7-day, 10-year low flow is generally used for stream flow condition, for design purposes. EXAMPLE: A power plant pumps 27 cfs from a stream, with a flow of 186 cfs. The discharge of the plant’s ash-pond is 26 cfs. The boron concentrations for upstream water and the effluent are 0.051 and 8.9 mg/L, respectively. Compute the boron concentration in the stream after completely mixing.

Solution:

By Eq. (2.1) Cd

Qu Cu Qe Ce Qu Qe (186 27)(0.051) 26 8.9 (186 27) 26

1.29 (mg/L)

3 DISCHARGE MEASUREMENT Discharge (flow rate) measurement is very important to provide the basic data required for river or stream water quality. The total discharge for a stream can be estimated by float method with wind and other surface effects, by die study, or by actual subsection flow measurement, depending on cost, time, man-power, local conditions and so on. The discharge in a stream cross-section can be measured from a sub-section by the following formula:

STREAMS AND RIVERS STREAMS AND RIVERS

1.73

Q Sum (mean depth width mean velocity) n

1 1 Q a (hn hn1)(wn wn1) (yn yn1) 2 n1 2

(2.2)

If equal width w n

w Q a (hn hn1) (yn yn1) n1 4

(2.2a)

where Q discharge, cfs wn nth distance from initial point 0, ft hn nth water depth, ft n nth velocity, ft/s Velocity y is measured by a velocity meter, of which there are several types. The flow rate is 1559 cfs. EXAMPLE: Data obtained from the velocity measurement are listed in the first three columns of Table 2.1. Determine the flow rate at this cross-section.

Solution:

Summarized field data and complete computations are shown in Table 2.1.

4 TIME OF TRAVEL The time of travel can be determined by dye study or by computation. The river time of travel and stream geometry characteristics can be computed using a volume displacement model. The time of travel is determined at any specific reach as the channel volume of the reach divided by the flow as follows: t

V 1 Q 86,400

(2.3)

where t time of travel at a stream reach, days V stream reach volume, ft3 or m3 Q average stream flow in the reach, ft3/sec (cfs) or m3/s 86,400 a factor, s/d EXAMPLE: The cross-section areas at river miles 62.5, 63.0, 63.5, 64.0, 64.5, and 64.8 are, respectively, 271, 265, 263, 259, 258, and 260 ft2 at a surface water elevation. The average flow is 34.8 cfs. Find the time of travel for a reach between river miles 62.5 and 64.8.

STREAMS AND RIVERS 1.74

CHAPTER 1.2

TABLE 2.1 Velocity and Discharge Measurements 1 Distance from 0, ft

2 Depth, ft

3 Velocity, ft/s

0 1.1 1.9 2.7 3.6 4.5 5.5 6.6 6.9 6.5 6.2 5.5 4.3 3.2 2.2 1.2 0

0 0.52 0.84 1.46 2.64 4.28 6.16 8.30 8.88 8.15 7.08 5.96 4.20 2.22 1.54 0.75 0

0 2 4 7 10 14 18 23 29 35 40 44 48 50 52 54 55

4 Width, ft 2 2 3 3 4 4 5 6 6 5 4 4 2 2 2 1

5 Mean depth, ft

6 Mean velocity, ft/s

0.55 1.50 2.30 3.15 4.05 5.00 6.05 6.75 6.70 6.35 5.85 4.90 3.75 2.70 1.45 0.35

0.26 0.68 1.15 2.05 3.46 5.22 7.23 8.59 7.52 6.62 6.52 5.08 3.21 1.88 1.15 0.38

7 456 Discharge, cfs 0.3 2.0 7.9 19.4 56.1 104.4 349.9 302.3 302.3 210.2 152.2 99.6 24.1 10.2 3.3 0.1 1559.0*

*The discharge is 1559 cfs.

Solution: Step 1. Find average area in the reach 1 (271 265 263 259 258 260) ft2 6 262.7 ft2

Average area Step 2. Find volume

Distance of the reach (64.8 62.5) miles ft 2.3 miles 5280 mile 12,144 ft V 262.7 ft2 12,144 ft 3,190,000 ft3 Step 3. Find t V 1 Q 86,400 3,190,000 ft3 34.8 ft3/s 86,400 s/d 1.06 days

5 DISSOLVED OXYGEN AND WATER TEMPERATURE Dissolved oxygen (DO) and water temperature are most commonly in situ monitored parameters for surface waters (rivers, streams, lakes, reservoirs, wetlands, ocean, etc.). Dissolved oxygen concentration in

STREAMS AND RIVERS STREAMS AND RIVERS

1.75

milligrams per liter (mg/L) is a measurement of the amount of oxygen dissolved in water. It can be determined with a DO meter or by a chemical titration method. The DO in water has an important impact on aquatic animals and plants. Most aquatic animals, such as fish, require oxygen in the water to survive. The two major sources of oxygen in water are from diffusion from the atmosphere across the water surface and the photosynthetic oxygen production from aquatic plants such as algae and macrophytes. Important factors that affect DO in water (Fig. 2.1) may include water temperature, aquatic plant photosynthetic activity, wind and wave mixing, organic contents of the water, and sediment oxygen demand. Excessive growth of algae (bloom) or other aquatic plants may provide very high concentration of DO, so called supersaturation. On the other hand, oxygen deficiencies can occur when plant respiration depletes oxygen beyond the atmospheric diffusion rate. This can occur especially during the winter ice cover period and when intense decomposition of organic matter in the lake bottom sediment occurs during the summer. These oxygen deficiencies will result in fish being killed. 5.1 Dissolved Oxygen Saturation DO saturation (DOsat) values for various water temperatures can be computed using the American Society of Civil Engineers’ formula (American Society of Civil Engineering Committee on Sanitary Engineering Research, 1960): DOsat 14.652 0.41022T 0.0079910T 2 0.000077774T 3 where DOsat dissolved oxygen saturation concentration, mg/L T water temperature, °C

FIGURE 2.1

Factors affecting dissolved oxygen concentration in water.

(2.4)

STREAMS AND RIVERS 1.76

CHAPTER 1.2

This formula represents saturation values for distilled water (b 1.0) at sea level pressure. Water impurities can increase the saturation level ( b 1.0) or decrease the saturation level ( b 1.0), depending on the surfactant characteristics of the contaminant. For most cases, b is assumed to be unity. The DOsat values calculated from the above formula are listed in Table 2.2 (example: DOsat 8.79 mg/L, when T 21.3 C) for water temperatures ranging from zero to 30°C (American Society of Engineering Committee on Sanitary Engineering Research, 1960). Calculate DO saturation concentration for a water temperature at 0, 10, 20 and 30 C, assuming b 1.0.

EXAMPLE 1:

Solution: (a) at T 0°C DOsat 14.652 0 0 0 14.652 mg/L

TABLE 2.2 Dissolved Oxygen Saturation Values in mg/L Temp., C 0.0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

14.65 14.25 13.86 13.49 13.13 12.79 12.46 12.14 11.84 11.55 11.27 11.00 10.75 10.50 10.26 10.03 9.82 9.61 9.40 9.21 9.02 8.84 8.67 8.50 8.33 8.18 8.02 7.87 7.72 7.58 7.44

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

14.61 14.21 13.82 13.46 13.10 12.76 12.43 12.11 11.81 11.52 11.24 10.98 10.72 10.48 10.24 10.01 9.79 9.58 9.38 9.19 9.00 8.82 8.65 8.48 8.32 8.16 8.01 7.86 7.71 7.56 7.42

14.57 14.17 13.79 13.42 13.06 12.72 12.40 12.08 11.78 11.49 11.22 10.95 10.70 10.45 10.22 9.99 9.77 9.56 9.36 9.17 8.98 8.81 8.63 8.46 8.30 8.14 7.99 7.84 7.69 7.55 7.41

14.53 14.13 13.75 13.38 13.03 12.69 12.36 12.05 11.75 11.47 11.19 10.93 10.67 10.43 10.19 9.97 9.75 9.54 9.34 9.15 8.97 8.79 8.62 8.45 8.29 8.13 7.98 7.83 7.68 7.54 7.40

14.49 14.09 13.71 13.35 13.00 12.66 12.33 12.02 11.72 11.44 11.16 10.90 10.65 10.40 10.17 9.95 9.73 9.52 9.32 9.13 8.95 8.77 8.60 8.43 8.27 8.11 7.96 7.81 7.66 7.52 7.38

14.45 14.05 13.68 13.31 12.96 12.62 12.30 11.99 11.70 11.41 11.14 10.87 10.62 10.38 10.15 9.92 9.71 9.50 9.30 9.12 8.93 8.75 8.58 8.42 8.25 8.10 7.95 7.80 7.65 7.51 7.37

14.41 14.02 13.64 13.28 12.93 12.59 12.27 11.96 11.67 11.38 11.11 10.85 10.60 10.36 10.12 9.90 9.69 9.48 9.29 9.10 8.91 8.74 8.56 8.40 8.24 8.08 7.93 7.78 7.64 7.49 7.35

14.37 13.98 13.60 13.24 12.89 12.56 12.24 11.93 11.64 11.35 11.08 10.82 10.57 10.33 10.10 9.88 9.67 9.46 9.27 9.08 8.90 8.72 8.55 8.38 8.22 8.07 7.92 7.77 7.62 7.48 7.34

14.33 13.94 13.56 13.20 12.86 12.53 12.21 11.90 11.61 11.33 11.06 10.80 10.55 10.31 10.08 9.86 9.65 9.44 9.25 9.06 8.88 8.70 8.53 8.37 8.21 8.05 7.90 7.75 7.61 7.47 7.32

14.29 13.90 13.53 13.17 12.82 12.49 12.18 11.87 11.58 11.30 11.03 10.77 10.52 10.28 10.06 9.84 9.63 9.42 9.23 9.04 8.86 8.68 8.52 8.35 8.19 8.04 7.89 7.74 7.59 7.45 7.31

Source: American Society of Civil Engineering Committee on Sanitary Engineering Research, 1960.

STREAMS AND RIVERS STREAMS AND RIVERS

1.77

(b) at T 10°C DOsat 14.652 0.41022 10 0.0079910 102 0.000077774 103 11.27 (mg/L) (c) at T 20°C DOsat 14.652 0.41022 20 0.0079910 202 0.000077774 203 9.02 (mg/L) (d) at T 30°C DOsat 14.652 0.41022 30 0.0079910 302 0.000077774 303 7.44 (mg/L) The DO saturation concentrations generated by the formula must be corrected for differences in air pressure caused by air temperature changes and for elevation above the mean sea level (MSL). The correction factor can be calculated as follows: f

2116.8 (0.08 0.000115A)E 2116.8

(2.5)

where f correction factor for above MSL A air temperature, °C E elevation of the site, feet above MSL EXAMPLE 2: Find the correction factor of DOsat value for water at 620 ft above the MSL and air temperature of 25°C? What is DOsat at a water temperature of 20°C?

Solution: Step 1. Using Eq. (2.5) 2116.8 (0.08 0.000115A)E 2116.8 2116.8 (0.08 0.000115 25)620 2116.8 2116.8 47.8 2116.8 0.977

Step 2. Compute DOsat From Example 1, at T 20°C DOsat 9.02 mg/L With an elevation correction factor of 0.977 DOsat 9.02 mg/L 0.977 8.81 mg/L 5.2 Dissolved Oxygen Availability Most regulatory agencies have standards for minimum DO concentrations in surface waters to support indigenous fish species in surface waters. In Illinois, for example, the Illinois Pollution Control Board stipulates that dissolved oxygen shall not be less than 6.0 mg/L during at least 16 hours of any 24-hour period, nor less than 5.0 mg/L at any time (IEPA, 1990).

STREAMS AND RIVERS 1.78

CHAPTER 1.2

The availability of dissolved oxygen in a flowing stream is highly variable due to several factors. Daily and seasonal variations in DO levels have been reported. The diurnal variations in DO are primarily induced by algal productivity. Seasonal variations are attributable to changes in temperature that affect DO saturation values. The ability of a stream to absorb or reabsorb oxygen from the atmosphere is affected by flow factors such as water depth and turbulence, and it is expressed in terms of the reaeration coefficient. Factors that may represent significant sources of oxygen use or oxygen depletion are biochemical oxygen demand (BOD) and sediment oxygen demand (SOD). BOD, including carbonaceous BOD (CBOD) and nitrogenous BOD (NBOD), may be the product of both naturally occurring oxygen use in the decomposition of organic material and oxygen depletion in the stabilization of effluents discharged from wastewater treatment plants (WTPs). The significance of any of these factors depends upon the specific stream conditions. One or all of these factors may be considered in the evaluation of oxygen use and availability.

6 BIOCHEMICAL OXYGEN DEMAND ANALYSIS Laboratory analysis for organic matter in water and wastewater includes testing for biochemical oxygen demand, chemical oxygen demand (COD), total organic carbon (TOC), and total oxygen demand (TOD). The BOD test is a biochemical test involving the use of microorganisms. The COD test is a chemical test. The TOC and TOD tests are instrumental tests. The BOD determination is an empirical test that is widely used for measuring waste (loading to and from wastewater treatment plants), evaluating the organic removal efficiency of treatment processes, and assessing stream assimilative capacity. The BOD test measures (1) the molecular oxygen consumed during a specific incubation period for the biochemical degradation of organic matter (carbonaceous BOD); (2) oxygen used to oxidize inorganic material such as sulfide and ferrous iron; and (3) reduced forms of nitrogen (nitrogenous BOD) with an inhibitor (trichloromethylpyridine). If an inhibiting chemical is not used, the oxygen demand measured is the sum of carbonaceous and nitrogenous demands, so-called total BOD or ultimate BOD. The extent of oxidation of nitrogenous compounds during the 5-day incubation period depends upon the type and concentration of microorganisms that carry out biooxidation. The nitrifying bacteria usually are not present in raw or settleable primary sewage. These nitrifying organisms are present in sufficient numbers in biological (secondary) effluent. A secondary effluent can be used as ‘seeding’ material for an NBOD test of other samples. Inhibition of nitrification is required for a CBOD test for secondary effluent samples, for samples seeded with secondary effluent, and for samples of polluted waters. The result of the 5-day BOD test is recorded as carbonaceous biochemical oxygen demand, CBOD5, when inhibiting nitrogenous oxygen demand. When nitrification is not inhibited, the result is reported as BOD5 (with 5-day incubation at 20 C). The BOD test procedures can be found in Standard methods for the examination of water and wastewater (APHA, AWWA, and WEF, 1995). When the dilution water is seeded, oxygen uptake (consumed) is assumed to be the same as the uptake in the seeded blank. The difference between the sample BOD and the blank BOD, corrected for the amount of seed used in the sample, is the true BOD. Formulas for calculation of BOD are as follows (APHA, AWWA, and WEF, 1995): When dilution water is not seeded: D1 D2 P

(2.6)

(Di De ) (Bi Be ) f P

(2.7)

BOD, mg/L When dilution water is seeded: BOD, mg/L

where D1, Di DO of diluted sample immediately after preparation, mg/L D2, De DO of diluted sample after incubation at 20°C, mg/L P decimal volumetric fraction of sample used; mL of sample/300 mL (for Eq. (2.6)) Bi DO of seed control before incubation, mg/L

STREAMS AND RIVERS STREAMS AND RIVERS

1.79

Be DO of seed control after incubation, mg/L f ratio of seed in diluted sample to seed in seed control P percent seed in diluted sample/percent seed in seed control (for Eq. (2.7)) If seed material is added directly to the sample and to control bottles: f volume of seed in diluted sample/volume of seed in seed control EXAMPLE 1: For a BOD test, 75 mL of a river water sample is used in the 300 mL of BOD bottles without seeding with three duplications. The initial DO in three BOD bottles read 8.86, 8.88, and 8.83 mg/L, respectively. The DO levels after 5 days at 20°C incubation are 5.49, 5.65, and 5.53 mg/L, respectively. Find the 5-day BOD (BOD5) for the river water.

Solution: Step 1. Determine average DO uptake g(D1 D2) 3 [(8.86 5.49) (8.88 5.65) (8.83 5.53)] 3 3.30 (mg/L)

Step 2. Determine P P

75 0.25 300

Step 3. Compute BOD5 BOD5

3.30 mg/L x 13.2 mg/L P 0.25

The wastewater is diluted by a factor of 1/20 using seeded control water. DO levels in the sample and control bottles are measured at 1-day intervals. The results are shown in Table 2.3. One milliliter of seed material is added directly to diluted and to control bottles. Find daily BOD values.

EXAMPLE 2:

Solution: Step 1. Compute f and P 1 mL 1.0 1 mL 1 0.05 P 20 f

TABLE 2.3 Change in Dissolved Oxygen and Biochemical Oxygen Demand with Time Dissolved oxygen, mg/L Time, days 0 1 2 3 4 5 6 7

Diluted sample

Seeded control

7.98 5.05 4.13 3.42 2.95 2.60 2.32 2.11

8.25 8.18 8.12 8.07 8.03 7.99 7.96 7.93

BOD, mg/L — 57.2 74.4 87.6 96.2 102.4 107.4 111.0

STREAMS AND RIVERS 1.80

CHAPTER 1.2

Step 2. Find BODs using Eq. (2.7) Day 1: (Di De) (Bi Be)f P (7.98 5.05) (8.25 8.18)1 0.05 57.2 (mg/L)

BOD1

Similarly Day 2: (7.98 4.13) (8.25 8.12)1 0.05 74.4 (mg/L)

BOD2

For other days, BOD can be determined in the same manner. The results of BODs are also presented in the above table. It can be seen that BOD5 for this wastewater is 102.4 mg/L.

7 STREETER–PHELPS OXYGEN SAG FORMULA The method most widely used for assessing the oxygen resources in streams and rivers subjected to effluent discharges is the Streeter–Phelps oxygen sag formula that was developed for the use on the Ohio River in 1914. The well-known formula is defined as follows (Streeter and Phelps, 1925): Dt

K1La [eK1t eK2t] Da eK2t K2 K1

(2.8a)

k1La [10k1t 10k2t] Da10k2t k2 k1

(2.8b)

where Dt DO saturation deficit downstream, mg/L or lb (DOsatDOa) at time t t time of travel from upstream to downstream, days Da initial DO saturation deficit of upstream water, mg/L or lb La ultimate upstream biochemical oxygen demand (BOD), mg/L e base of natural logarithm, 2.7183 K1 deoxygenation coefficient to the base e, per day K2 reoxygenation coefficient to the base e, per day k1 deoxygenation coefficient to the base 10, per day k2 reoxygenation coefficient to the base 10, per day In the early days, K1or K2 and k1or k2 were used classically for values based on e and 10, respectively. Unfortunately, in recent years, many authors have mixed the usage of K and k. Readers should be aware of this. The logarithmic relationships between k and K are K1 2.3026 k1 and K2 2.3026 k2; or k1 0.4343 K1 and k2 0.4343 K2. The Streeter–Phelps oxygen sag equation is based on two assumptions: (1) at any instant the deoxygenation rate is directly proportional to the amount of oxidizable organic material present; and (2) the reoxygenation rate is directly proportional to the dissolved oxygen deficit. Mathematical expressions for assumptions (1) and (2) are dD K1(La Lt) dt

(2.9)

STREAMS AND RIVERS STREAMS AND RIVERS

1.81

and dD K2 D dt where

(2.10)

dD the net rate of change in the DO deficit, or the absolute change of DO deficit (D) over an dt increment of time dt due to stream waste assimilative capacity affected by deoxygenation coefficient K1 and due to an atmospheric exchange of oxygen at the air/water interface affected by the reaeration coefficient K2 La ultimate upstream BOD, mg/L Lt ultimate downstream BOD at any time t, mg/L

Combining the two above differential equations and integrating between the limits Da, the initial upstream sampling point, and t any time of flow below the initial point, yields the basic equation devised by Streeter and Phelps. This stimulated intensive research on BOD, reaction rates, and stream sanitation. The formula is a classic in sanitary engineering stream work. Its detailed analyses can be found in almost all general environmental engineering texts. Many modifications and adaptations of the basic equation have been devised and have been reported in the literature. Many researches have been carried out on BOD, K1, and K2 factors. Illustrations for oxygen sag formulas will be presented in the latter sections.

8 BOD MODELS AND K1 COMPUTATION Under aerobic conditions, organic matter, and some inorganics can be used by bacteria to create new cells, energy, carbon dioxide, and residue. The oxygen used to oxidize total organic material and all forms of nitrogen for 60 to 90 days is called the ultimate BOD (UBOD). It is common that measurements of oxygen consumed in a 5-day test period called 5-day BOD or BOD5 is practiced. The BOD progressive curve is shown in Fig. 2.2.

Carbonaceous 1st stage

Nitrification 2nd stage

BOD EXERTED, mg/L Lt La y TIME, days FIGURE 2.2

BOD progressive curve.

STREAMS AND RIVERS 1.82

CHAPTER 1.2

8.1 First-order Reaction Phelps law states that the rate of biochemical oxidation of organic matter is proportional to the remaining concentration of unoxidized substance. Phelps law can be expressed in differential form as follows (monomolecular or unimolecular chemical reaction):

dLt K1 Lt dt (2.11) dLt K1 d t Lt

by integration Lt dL t t 3 L Lt K1 3 0 dt1 a

Lt K1t or La Lt eK1t or La Lt La # eK1t or

log

Lt 0.434K1t k1t La Lt 10k1t La L L # 10k1t t

(2.12)

where Lt BOD remaining after time t days, mg/L La first-stage BOD, mg/L, per day K1 deoxygenation rate, based on e, K1 2.303 k1, per day k1 deoxygenation rate, based on 10, k1 0.4343 K1 (k1 0.1 at 20°C) e base of natural logarithm, 2.7183 Oxygen demand exerted up to time t, y, is a first-order reaction: y La Lt (see Fig. 2.2) y La(1 eK1t)

(2.13a)

or based on log10 y La(1 10k1t)

(2.13b)

When a delay occurs in oxygen uptake at the onset of a BOD test, a lag-time factor, t0 should be included and Eqs. (2.13a) and (2.13b) become y La[1 eK1(tt0)]

(2.14a)

y La[1 10k1(tt0)]

(2.14b)

For the Upper Illinois Waterway study, many of the total and nitrogenous BOD curves have an S-shaped configuration. The BOD in waters from pools often consists primarily of high-profile secondstage or nitrogenous BOD, and the onset of the exertion of this NBOD is often delayed 1 or 2 days. The delayed NBOD and the total BOD (TBOD) curves, dominated by the NBOD fraction, often exhibit an S-shaped configuration. The general mathematical formula used to simulate the S-shaped curve is (Butts et al., 1975) y La[1 eK1(tt0) ] m

where m is a power factor, and the other terms are as previously defined.

(2.15)

STREAMS AND RIVERS STREAMS AND RIVERS

1.83

Statistical results show that a power factor of 2.0 in Eq. (2.15) best represents the S-shaped BOD curve generated in the Lockport and Brandon Road areas of the waterway. Substituting m 2 in Eq. (2.15) yields y La[1 eK1(tt0) ] 2

(2.15a)

Given K1 0.25 per day, BOD5 6.85 mg/L, for a river water sample. Find La when t0 0 days and t0 2 days.

EXAMPLE 1:

Solution: Step 1. When t0 0 y La(1 eK1t) 6.85 La(1 e0.255) 6.85 1 0.286 9.60 (mg/L)

Another solution using k1: Since k1 0.4343 K1 0.4343 0.25 per day 0.109 per day y La(1 10k1t) 6.85 La(1 100.1095) 6.85 1 0.286 9.60 (mg/L)

Step 2. When t0 2 days, using Butts et al.’s equation [Eq. (2.15a)]: y La[1 eK1(tt0) ] y La K1(tt0)2 1e 6.85 2 1 e0.25(52) 6.85 1 e2.25 6.85 1 0.105 7.65 (mg/L) 2

EXAMPLE 2: Compute the portion of BOD remaining to the ultimate BOD (1 eK1t or 1 10k1t) for k1 0.10 (or K1 0.23).

Solution:

By Eq. (2.13b) y La(1 10k1t) y 1 10k1t La

1 eK1t

STREAMS AND RIVERS 1.84

CHAPTER 1.2

TABLE 2.4 Relationship between t and the Ultimate BOD (1 10k1t or 1 eK1t) t 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.0 2.5 3.0 3.5 4.0

1 10k1t or 1 eK1t

0.056 0.109 0.159 0.206 0.250 0.291 0.332 0.369 0.438 0.500 0.553 0.602

4.5 5.0 6.0 7.0 8.0 9.0 10.0 12.0 16.0 20.0 30.0

0.646 0.684 0.749 0.800 0.842 0.874 0.900 0.937 0.975 0.990 0.999 1.0

when t 0.25 days y 1 100.100.25 1 0.944 0.056 La Similar calculations can be performed as above. The relationship between t and (1 10k1t) is listed in Table 2.4. 8.2 Determination of Deoxygenation Rate and Ultimate BOD Biological decomposition of organic matter is a complex phenomenon. Laboratory BOD results do not necessarily fit actual stream conditions. BOD reaction rate is influenced by immediate demand, stream or river dynamic environment, nitrification, sludge deposit, and types and concentrations of microbes in the water. Therefore laboratory BOD analyses and stream surveys are generally conducted for raw and treated wastewaters and river water to determine BOD reaction rate. Many investigators have worked on developing and refining methods and formulas for use in evaluating the deoxygenation (K1) and reaeration (K2) constants and the ultimate BOD (La). There are several methods proposed to determine K1 values. Unfortunately, K1 values determined by different methods given by the same set of data have considerable variations. Reed–Theriault least-squares method published in 1927 (US Public Health Service, 1927) give the most consistent results, but it is time consuming and tedious. Computation using a digital computer was developed by Gannon and Downs (1964). In 1936, a simplified procedure, the so-called log-difference method of estimating the constants of the first-stage BOD curve, was presented by Fair (1936). The method is also mathematically sound, but is also difficult to solve. Thomas (1937) followed Fair et al. (1941a, 1941b) and developed the “slope” method, which, for many years, was the most used procedure for calculating the constants of the BOD curve. Later, Thomas (1950) presented a graphic method for BOD curve constants. In the same year, Moore et al. (1950) developed the “moment method” that was simple, reliable, and accurate to analyze BOD data; this soon became the most used technique for computing the BOD constants. Researchers found that K1 varied considerably for different sources of wastewaters and questioned the accepted postulate that the 5-day BOD is proportional to the strength of the sewage. Orford and Ingram (1953) discussed the monomolecular equation as being inaccurate and unscientific in its relation to BOD. They proposed that the BOD curve could be expressed as a logarithmic function. Tsivoglou (1958) proposed a ‘daily difference’ method of BOD data solved by a semi-graphical solution. A “rapid ratio” method can be solved using curves developed by Sheehy (1960), O’Connor (1966) modified least-squares method using BOD5.

STREAMS AND RIVERS STREAMS AND RIVERS

1.85

This book describes Thomas’s slope method, method of moments, logarithmic function, and rapid methods calculating K1 (or k1) and La. Slope method. The slope method (Thomas, 1937) gives the BOD constants via the least-squares treatment of the basic form of the first-order reaction equation or dy K1(La y) K1 La K1 y dt

(2.16)

where dy increase in BOD per unit time at time t K1 deoxygenation constant, per day La first stage ultimate BOD, mg/L y BOD exerted in time t, mg/L This differential equation (Eq. (2.16)) is linear between dy/dt and y. Let y dy/dt to be the rate of change BOD and n be the number of BOD measurements minus one. Two normal equations for finding K1 and La are na bgy gyr 0

(2.17)

ag y bg y2 g yy r 0

(2.18)

and

Solving Eqs. (2.17) and (2.18) yields values of a and b, from which K1 and La can be determined directly by following relations: K1 b

(2.19)

and La a/b

(2.20) 2

The calculations include first determinations of y, y y, y for each value of y. The summation of these gives the quantities of y, yy, and y2 which are used for the two normal equations. The values of the slopes are calculated from the given data of y and t as follows: (yi yi1) A ti dyi yri dt

ti1 ti ti1

B (yi1 yi) A t

ti1 ti1

ti ti1 ti i1

(2.21)

For the special case, when equal time increments ti1 ti t3 t2 t2 t1 t, y becomes dyi yi1 yi1 dt 2 t

yi1 yi1 ti1 ti1

(2.21a)

A minimum of six observations (n 6)of y and t are usually required to give consistent results. Equal time increments, BOD data at temperature of 20 C, t and y are shown in Table 2.5. Find K1 and La.

EXAMPLE 1:

Solution: Step 1. Calculate y, yy, and y2 Step 2. Determine a and b. Writing normal equations (Eqs. (2.17 and (2.18)), n 9 na bgy gyr 0 9a 865.8b 89.6 0 a 96.2b 9.96 0

(1)

STREAMS AND RIVERS 1.86

CHAPTER 1.2

TABLE 2.5 Calculations for y, yy, and y2 Values t, day (1) 0 1 2 3 4 5 6 7 8 9 10

y (2) 0 56.2 74.4 87.6 96.2 102.4 107.4 111.0 114.0 116.6 118.8 865.8†

y3 y1 t3 t1

y (3)

yy (4)

y2 (5)

37.2* 15.7 10.9 7.4 5.6 4.3 3.3 2.8 2.4

2090.64 1168.08 954.84 711.88 573.44 461.82 366.30 319.20 279.84

3158.44 5535.36 7673.76 9254.44 10485.76 11534.76 12321.00 12996.00 13595.56

89.6

6926.04

86555.08

74.4 0 2 0

*yr1 37.2 † Sum of first nine observations.

and agy bgy2 gy yr 0 865.8a 86,555b 6926 0 a 99.97b 8.0 0

(2)

Eq. (2) – Eq. (1) 3.77b 1.96 0 b 0.52 From Eq. (2) a 99.97(0.52) 8.0 0 a 59.97 Step 3. Calculate K1 and La with Eqs. (2.19) and (2.20) K1 b (0.52) 0.52 (per day) La a/b 59.97/(0.52) 115.3 (mg/L) EXAMPLE 2:

and La.

Unequal time increments, observed BOD data, t and y are given in Table 2.6. Find K1

Solution: Step 1. Calculate t, y, y, yy, and y2; then complete Table 2.6. From Eq. (2.21) (see Table 2.6) *yri

(yi yi1) A ti

ti1 ti ti1

ti ti1 ti i1

ti1 ti1

( yi1) A t B ( yi1) A t B

ti1

( ti1) ( ti1)

(28.8) A 0.4 B (27.4) A 0.6 B 61.47 0.4 0.6 0.6

yr1

B (yi1 yi) A t

0.4

STREAMS AND RIVERS STREAMS AND RIVERS

1.87

TABLE 2.6 Various t and y Values t

61.47*

1770.24

829.44

30.90

1736.75

3158.44

19.48

1276.00

4290.25

15.48

1238.48

6400.00

9.10

797.16

7673.76

7.40

711.88

9254.44

5.57

570.03

10485.76

3.55

394.05

12321.00

2.43

282.95

13595.56

155.37

8777.53

68008.65

0 0.4

0.4

28.8 28.8

0.6 1

27.4 56.2

0.5 1.5

9.3 65.5

0.7 2.2

14.5 80.0

0.8 3

7.6 87.6

1 4

8.6 96.2

1 5

6.2 102.4

2 7

8.6 111.0

2 9

5.6 116.6

5.6

122.2

Sum

744.3

*See text for calculation of this value.

Step 2. Compute a and b; while n 9 na bg y g yr 0 9a 744.3b 155.37 0 a 82.7b 17.26 0

(1)

ag y bg y2 g yyr 0 744.3a 68,008.65b 8777.53 0 a 91.37b 11.79 0

(2)

and

Eq. (2) Eq. (1) 8.67b 5.47 0 b 0.63 with Eq. (2) a 91.37(0.63) 11.79 0 a 69.35 Step 3. Determine K1 and La K1 b 0.63 (per day) La a/b 69.35/0.63 110.1 (mg/L)

STREAMS AND RIVERS 1.88

CHAPTER 1.2

Moment method. This method requires that BOD measurements must be a series of regularly spaced time intervals. Calculations are needed for the sum of the BOD values, y, accumulated to the end of a series of time intervals and the sum of the product of time and observed BOD values, ty, accumulated to the end of the time series. The rate constant K1 and the ultimate BOD La can then be easily read from a prepared graph by entering values of y/ ty on the appropriate scale. Treatments of BOD data with and without lag phase will be different. The authors (Moore et al., 1950) presented three graphs for 3-, 5-, and 7-day sequences (Figs. 2.3, 2.4, and 2.5) with daily intervals for BOD value without lag phase. There is another chart presented for a 5-day sequence with lag phase (Fig. 2.6).

FIGURE 2.3

y/L and y/ ty for various values of k1 in a 3-day sequence.

STREAMS AND RIVERS STREAMS AND RIVERS

FIGURE 2.4

EXAMPLE 1:

and La.

1.89

y/L and y/ ty for various values of k1 in a 5-day sequence.

Use the BOD (without lag phase) on Example 1 of Thomas’ slope method, find K1

Solution: Step 1. Calculate y and ty (see Table 2.7). Step 2. Compute y/ ty gy/gty 635.2/2786 0.228 Step 3. Find K1 and La On the 7-day time sequence graph (Fig. 2.5), enter the value 0.228 on the y/ ty scale, extend a horizontal line to the curve labeled y/ ty, and from this point follow a vertical line to the k1 scale. A value of k1 0.264. K1 2.3026 k1 2.3026 0.264 0.608 (per day)

STREAMS AND RIVERS 1.90

CHAPTER 1.2

FIGURE 2.5

y/L and y/ ty for various values of k1 in a 7-day sequence.

Extend the same vertical line to the curve labeled y/La, obtaining a value of 5.81. Since La gy/5.81 635.2/5.81 109.3 (mg/L) The technique of the moment method for analyzing a set of BOD data containing a lag phase is as follows: 1. Compute t, y, ty 2. Compute t 2 and t 2 y and take the sum of the values of each quantity as t 2 and t 2y 3. Compute the derived quantity: gty/gt gy/n g(t2y)/gt2 gy/n in which n is the number of observations 4. Enter the above quantity on appropriate curve to find k1 5. Project to other curves for values and solve equations for C and La. The BOD equation with log phase is expressed y L[1 10k1(tt0)]

(2.14b)

y La[1 C10

(2.14c)

k1t

in which t0 is the lag period and C 10 k1t0

]

STREAMS AND RIVERS STREAMS AND RIVERS

FIGURE 2.6

1.91

Curves for BOD computation with lag phase, based on a 5-day time sequence.

With a lag phase BOD data, BOD values are shown in Table 2.8. Find K1 and La and complete an equation of the curve of best fit for the BOD data.

EXAMPLE 2:

Solution: Step 1. Compute t, y, ty, t 2, and t 2y (Table 2.8) TABLE 2.7 Calculation of y and ty t

1 2 3 4 5 6 7

56.2 74.4 87.6 96.2 102.4 107.4 111.0

56.2 148.8 262.8 384.8 512.0 644.4 777.0

Sum

635.2

2786.0

STREAMS AND RIVERS 1.92

CHAPTER 1.2

TABLE 2.8 Computations of t, y, ty, t 2, and t 2y t

t 2y

1 2 3 4 5

12 74 101 126 146

12 148 303 504 730

1 4 9 16 25

12 296 909 2016 3650

459

1697

6883

Step 2. Compute some quantities gty/gt 1697/15 113.13 gy/n 459/5 91.8 gt2y/gt2 6883/55 125.14 and gty/gt gy/n 113.13 91.8 0.640 125.14 91.8 gt2y/gt2 gy/n Step 3. Enter Fig. 2.6 on the vertical axis labeled gty/gt gy/n gt2y/gt2 gy/n with the value 0.640, and proceed horizontally to the diagonal straight line. Extend a vertical line to the axis labeled k1 and read k1 0.173 per day (K1 0.173 per day 2.3026 0.398 per day). Extend the vertical line to curve labeled gty/gt gy/n CL and proceed horizontally from the intersection to the scale at the far right; read gty/gt gy/n 0.0878 CL Step 4. Find CL CL

113.13 91.8 242.94 0.0878

Step 5. Find La, La L for this case. Continue to the same vertical line to the curve labeled L gy/n CL and read horizontally on the inside right-hand scale L gy/n 0.353 CL Then L La 0.353(242.94) 91.8 177.6 (mg/L) Also

242.94 CL 1.368 L 177.6

STREAMS AND RIVERS STREAMS AND RIVERS

1.93

and lag period t0

1 log10 C k1

1 log10 1.368 0.173 0.787 (days)

Step 6. Write complete equation of the best fit for the data y 177.6 C 1 100.173(t0.787) D or y 177.6 C 1 e0.398(t0.787) D Logarithmic formula. Orford and Ingram (1953) reported that there is a relationship between the observed BOD from domestic sewage and the logarithm of the time of observation. If the BOD data are plotted against the logarithm of time, the resultant curve is approximately a straight line. The general equation is expressed as yt m log t b

(2.22)

where m is the slope of the line and b is a constant (the intercept). The general equation can be transformed by dividing each side by the 5-day BOD (or BOD5) intercept of the line to give yt b m s s log t s yt s M log t B

(2.22a)

yt s(M log t B)

(2.22b)

where s BOD5 intercept of the line M m/s, BOD rate parameter B b/s, BOD rate parameter For domestic sewage oxidation at 20 C, the straight line through the observed plotted points, when extrapolated to the log t axis, intercepts the log t axis at 0.333 days. The general equation is yt s(0.85 log t 0.41)

(2.23a)

where 0.85 and 0.41 are the BOD rate parameters for domestic sewage. For any observed BOD curve with different oxidation time, the above equation may generalize as yt S(0.85 log at 0.41)

(2.23b)

where S BOD intercept (y-axis) of the line at 5/a days 5-day BOD at the standardized domestic sewage oxidation rate, when a 1 strength factor a log t at x-axis intercept of normal domestic sewage BOD curve divided by the x-axis intercept of 0.333 xaxis intercept 1 for a standardized domestic sewage BOD5 curve

STREAMS AND RIVERS 1.94

CHAPTER 1.2

The oxidation rate dy 0.85S t K1(La yt) dt

(2.24a)

0.85S k1(La yt) 2.303t

(2.24b)

For the logarithmic formula method you need to determine the two constants S and a. Observed BOD data are plotted on semilogarithmic graph paper. Time in days is plotted on a logarithmic scale on the x-axis and percent of 5-day BOD on a regular scale on the y-axis. The straight line of best fix is drawn through the plotted points. The time value of the x-axis intercept is then read, x1 (for a standardized domestic sewage sample, this is 0.333). The a value can be calculated by a 0.333/x1. EXAMPLE:

sample?

At 20 C, S 95 mg/L and x1 0.222 days. What is the BOD equation and K1 for the

Solution: Step 1. Compute a a 0.333/x1 0.333/0.222 1.5 Step 2. Determine the simplified Eq. (2.23b) yt S(0.85 log at 0.41) 95(0.85 log 1.5t 0.41) Step 3. Compute the oxidation rate at t 5 days, using Eq. (2.24a) 0.85 95 mg/L dy 0.85S t dt 5 days 16.15 mg/(L # d) Step 4. Determine La when t 20 days, using Eq. (2.24b) from Step 2 La y20 95(0.85 log (1.5 20) 0.41) 158 (mg/L) Step 5. Compute K1 dy K1(La y5) dt 16.15 mg/(L # d) K1 (158 95) mg/L 0.256 per day Rapid methods. Sheehy (1960) developed two rapid methods for solving first-order BOD equations. The first method, using the ‘BOD slide rule,’ is applicable to the ratios of observations on whole day intervals, from 1 to 8 days, to the 5-day BOD. The second method, the graphical method, can be used with whole or fractional day BOD values.

STREAMS AND RIVERS STREAMS AND RIVERS

1.95

The BOD slide rule consists of scales A and B for multiplication and division. Scales C and D are the values of 1 10k1t and 10k1t plotted in relation to values of k1t on the B scale. From given BOD values, the k1 value is easily determined from the BOD slide rule. Unfortunately, the BOD slide rule is not on the market. The graphical method is based on the same principles as the BOD slide rule method. Two figures are used to determine k1 values directly. One figure contains k1 values based on the ratios of BOD at time t(BODt) to the 5-day BOD for t less than 5 days. The other figure contains k1 values based on the ratio of BODt to BOD5 at times t greater than 5 days. The value of k1 is determined easily from any one of these figures based on the ratio of BODt to BOD5. 8.3 Temperature Effect on K1 A general expression of the temperature effect on the deoxygenation coefficient (rate) is K1a u(TaTb) K1b

(2.25)

where K1a reaction rate at temperature Ta, per day K1b reaction rate at temperature Tb, per day u temperature coefficient On the basis of experimental results over the usual range of river temperature, u is accepted as 1.047. Therefore the BOD reaction rate at any T in Celsius (working temperature), deviated from 20 C, is K1(T) K1(20 C) 1.047(T20)

(2.26)

k1(T) k1(20 C) 1.047(T20)

(2.26a)

La(T) La(20 C)[1 0.02(T 20)]

(2.27)

La(T) La(20 C)(0.6 0.02T )

(2.28)

Thus

EXAMPLE 1: A river water sample has k1 0.10 (one base 10) and La 280 mg/L at 20 C. Find k1 and La at temperatures 14 C and 29 C.

Solution: Step 1. Using Eq. (2.26a) k1(T) k1(20 C) 1.047(T20) at 14 C k1(14 C) k1(20 C) 1.047(1420) 0.10 1.0476 0.076 (per day) at 29 C k1(29 C) k1(20 C) 1.047(2920) 0.10 1.0479 0.15 (per day)

STREAMS AND RIVERS 1.96

CHAPTER 1.2

Step 2. Find La(T ), using Eq. (2.28) La(T) La(20 C)(0.6 0.02T) at 14 C La(14 C) La(20 C)(0.6 0.02T ) 280(0.6 0.02 14) 246 (mg/L) at 29 C La(29 C) 280(0.6 0.02 29) 330 (mg/L) The ultimate BOD and K1 values found in the laboratory at 20 C have to be adjusted to river temperatures using the above formulas. Three types of BOD can be determined: i.e. total, carbonaceous, and nitrogenous. TBOD (uninhibited) and CBOD (inhibited with trichloromethylpyridine for nitrification) are measured directly, while NBOD can be computed by subtracting CBOD values from TBOD values for given time elements. EXAMPLE 2: Tables 2.9 and 2.10 show typical long-term (20-day) BOD data for the Upper Illinois Waterway downstream of Lockport (Butts and Shackleford, 1992). The graphical plots of these BOD progressive curves are presented in Figs. 2.7–2.9. Explain what are their unique characteristics.

TABLE 2.9 Biochemical Oxygen Demand at 20 C in the Upper Illinois Waterway Station sample: Date: pH: Temp:

Lockport 18 01/16/90 7.03 16.05 C

Station sample: Date: pH: Temp:

Lockport 36 09/26/90 6.98 19.10 C

Time, days

TBOD, mg/L

CBOD, mg/L

NBOD, mg/L

Time, days

TBOD, mg/L

CBOD, mg/L

NBOD, mg/L

0.89 1.88 2.87 3.87 4.87 5.62 6.58 7.62 8.57 9.62 10.94 11.73 12.58 13.62 14.67 15.67 16.58 17.81 18.80 19.90 20.60

1.09 1.99 2.74 3.39 4.60 5.48 7.41 9.87 12.62 16.45 21.55 25.21 29.66 34.19 35.68 36.43 37.08 37.45 37.64 38.19 38.56

0.98 1.73 2.06 2.25 3.08 3.45 4.17 4.83 5.28 6.15 6.62 6.96 7.26 7.90 8.14 8.38 8.75 8.92 9.16 9.39 9.55

0.11 0.25 0.68 1.14 1.52 2.04 3.24 5.04 7.34 10.30 14.94 18.25 22.40 26.29 27.55 28.05 28.33 28.53 28.47 28.79 29.01

0.72 1.69 2.65 3.65 4.72 5.80 6.75 7.79 8.67 9.73 10.69 11.74 12.66 13.69 14.66 15.65 16.64 17.76 17.67 19.66

0.19 0.66 0.99 1.38 1.81 2.74 3.51 4.17 4.65 5.59 6.22 6.95 7.28 7.79 7.79 8.36 8.56 8.80 8.80 8.92

0.18 0.18 0.92 1.25 1.25 1.62 2.09 2.53 2.92 2.92 3.08 3.08 3.12 3.56 3.56 3.56 3.57 3.82 3.82 3.87

1.01 0.48 0.07 0.13 0.56 1.12 1.43 1.64 1.74 2.67 3.14 3.88 4.16 4.23 4.23 4.80 4.99 4.99 4.98 5.05

STREAMS AND RIVERS STREAMS AND RIVERS

1.97

TABLE 2.10 Biochemical Oxygen Demand at 20 C in the Kankakee River Station sample: Date: pH: Temp:

Kankakee 11 08/13/90 8.39 23.70 C

Time, days

TBOD, mg/L

CBOD, mg/L

NBOD, mg/L

0.78 2.79 3.52 6.03 6.77 7.77 8.76 9.49 10.76 13.01 13.43 14.75 15.48 17.73 19.96

0.81 1.76 2.20 2.42 2.83 3.20 3.54 3.83 4.19 4.85 5.08 5.24 5.57 6.64 7.03

0.44 1.13 1.26 1.62 1.94 2.22 2.49 2.81 3.09 3.76 3.94 4.02 4.42 5.30 5.85

0.37 0.63 0.93 0.79 0.89 0.98 1.06 1.02 1.10 1.10 1.14 1.22 1.15 1.17 1.19

Solution: These three sets of data have several unique characteristics and considerations. Figure 2.7 demonstrates that S-shaped NBOD and TBOD curves exist at Lockport station (Illinois) that fit the mathematical formula represented by Eq. (2.8). These curves usually occur at Lockport during cold weather periods, but not always.

FIGURE 2.7

BOD progressive curves for Lockport 18, January 16, 1990 (Butts and Shackleford, 1992).

STREAMS AND RIVERS 1.98

CHAPTER 1.2

FIGURE 2.8

BOD progressive curves for Lockport 36, September 26, 1990 (Butts and Shackleford, 1992).

Figure 2.8 illustrates Lockport warm weather BOD progression curves. In comparisons with Figs. 2.7 and 2.8, there are extreme differences between cold and warm weather BOD curves. The January NBOD20 represents 75.3% of the TBOD20, whereas the September NBOD20 contains only 56.6% of the TBOD20. Furthermore, the September TBOD20 is only 23.3% as great as the January TBOD20 (see Table 2.9). Figure 2.9 shows most tributary (Kankakee River) BOD characteristics. The warm weather TBOD20 levels for the tributary often approach those observed at Lockport station (7.03 mg/L versus 8.92 mg/L), but the fraction of NBOD20 is much less (1.19 mg/L versus 5.05 mg/L) (Tables 2.9 and 2.10). Nevertheless, the TBOD20 loads coming from the tributaries are usually much less than those originating from Lockport since the tributary flows are normally much lower. 8.4 Second-order Reaction In many cases, researchers stated that a better fit of BOD data can be obtained by using a second order chemical reaction equation, i.e. the equation of a rotated rectangular hyperbola. A second-order chemical reaction is characterized by a rate of reaction dependent upon the concentration of two reactants. It is defined as (Young and Clark, 1965): dC KC2 (2.29) dt When applying this second-order reaction to BOD data, K is a constant; C becomes the initial substance concentration; La, minus the BOD; y, at any time, t; or

d(La y) K(La y)2 dt

rearranging d(La y) (La y)2

Kd t

(2.30)

STREAMS AND RIVERS STREAMS AND RIVERS

FIGURE 2.9

1.99

BOD progressive curves for Kankakee 10, August 13, 1990 (Butts and Shackleford, 1992).

Integrating yields yy

3 or

d(La y)

3 (La y)2

Kd t t0

1 1 K t La La y

(2.31)

Multiplying each side of the equation by La and rearranging, La KLat La y La y La KLa t La y y KLa t La y y KLa t(La y) y KL2a t KLa ty (1 KLa t)y KL2a t 1

KL2a t 1 KLa t t y 1 1 2 L t KL

(2.32)

or y

t a bt

(2.33)

(2.34)

STREAMS AND RIVERS 1.100

CHAPTER 1.2

where a is 1/KLa2 and b represents 1/La. The above equation is in the form of a second-order reaction equation for defining BOD data. The equation can be linearized in the form t y a bt

(2.35)

in which a and b can be solved by a least-squares analysis. The simultaneous equations for the leastsquares treatment are as follows: t a aa bt y b 0 t a aa bt y bt 0 gt at bgt y 0 gt2 agt bgt2 y 0 gt/y gt a ba t b a t b 0 gt2/y gt2 b a b 0 a ba gt gt To solve a and b using five data points, t 5, t 15, and t 2 55; then gt/y 15 b a b 0 5 5 gt2/y 55 a ba b a b 0 15 15 a ba

Solving for b a

gt2/y gt/y 15 55 bb c a b a bd 0 15 5 15 5 a

3gt 55 45 1 gt2 bb a y b 0 15 15 y 3gt gt2 10b y y

or 3gt gt2 b 0.10a y y b

(2.36)

Solving for a, by substituting b in the equation a

1 gt 15bb a 5 y

3gt gt2 1 gt c y 15 0.1a y y b d 5 1.5gt2 1 5.5gt a y y b 5

gt2 gt 1.1a y b 0.3a y b

(2.37)

STREAMS AND RIVERS STREAMS AND RIVERS

1.101

The velocity of reaction for the BOD curve is y/t. The initial reaction is 1/a (in mg/(L . d)) and is the maximum velocity of the BOD reaction denoted as ym. The authors (Young and Clark, 1965) claimed that a second-order equation has the same precision as a first-order equation at both 20 C and 35 C. Dougal and Baumann (1967) modified the second-order equation for BOD predication as y

t/bt t a a bt bt 1

a bt

1/b 1

La 1 Kr t

(2.38)

where La 1/b

(2.39)

Kr b/a

(2.40)

This formula has a simple rate constant and an ultimate value of BOD. It can also be transformed into a linear function for regression analysis, permitting the coefficients a and b to be determined easily. EXAMPLE: The laboratory BOD data for wastewater were 135, 198, 216, 235, and 248 mg/L at days 1, 2, 3, 4, and 5, respectively. Find K, La, and ym for this wastewater.

Solution: Step 1. Construct a table for basic calculations (Table 2.11). Step 2. Solve a and b using Eqs. (2.36) and (2.37) 3gt gt2 b 0.10a y y b 0.10(0.238167 3 0.068579) 0.003243 gt gt2 a 1.1a y b 0.3a y b 1.1 0.068579 0.3 0.238167 0.003987 Step 3. Calculate La, K, ym, and K using Eqs. (2.38)–(2.40).

La K ym Kr

1/b 1/0.003243 308 (mg/L) 1/aL2a 1/0.003987(308)2 0.00264 (per mg/(L # d)) 1/a 1/0.003987 mg/(L # d) 250.8 mg/(L # d) 10.5 mg/(L # h)or b/a 0.003243/0.003987 0.813 (per day)

STREAMS AND RIVERS 1.102

CHAPTER 1.2

TABLE 2.11 Data for Basic Calculations for Step 1 t2/y

t/y

1 2 3 4 5

135 198 216 235 248

0.007407 0.010101 0.013889 0.017021 0.020161

0.007407 0.020202 0.041667 0.068085 0.100806

0.068579

0.238167

Sum

9 DETERMINATION OF REAERATION RATE CONSTANT K2 9.1 Basic Conservation For a stream deficient in DO but without BOD load, the classical formula is dD (2.41) K2 D dt where dD/dt is the absolute change in DO deficit D over an increment of time dt due to an atmospheric exchange of oxygen at air/water interface; K2 is the reaeration coefficient, per day; and D is DO deficit, mg/L. Integrating the above equation from t1 to t2 gives D2 t2 dD 3 D D K2 3 t d t 1 1

Dt t dD 3 D D K2 3 0d t a

D2 K2(t2 t1) D1

D2 K2 t D1

Dt K2t Da Dt Da eK2t

ln D2 1

t The K2 values are needed to correct for river temperature according to the equation K2(@T) K2(@20)(1.02)T20

(2.42)

(2.43)

where K2(@T ) K2 value at any temperature T C and K2(@20) K2 value at 20 C. EXAMPLE 1: The DO deficits at upstream and downstream stations are 3.55 and 2.77 mg/L, respectively. The time of travel in this stream reach is 0.67 days. The mean water temperature is 26.5 C. What is the K2 value for 20 C?

Solution: Step 1. Determine K2 at 26.5 C D

K2(@26.5)

lnD2 1

2.77 b 0.67 3.55 ^ (0.248)/0.67 0.37 (per day) aln

STREAMS AND RIVERS STREAMS AND RIVERS

1.103

Step 2. Calculate K2(@20) K2(@T) K2(@20)(1.02)T20 therefore K2(@20) 0.37/(1.02)26.520 0.37/1.137 0.33 (per day)

9.2 From BOD and Oxygen Sag Constants In most stream survey studies involving oxygen sag equations (discussed later), the value of the reoxygenation constant (K2) is of utmost importance. Under different conditions, several methods for determining K2 are listed below: I. K2 may be computed from the oxygen sag equation, if all other parameters are known; however, data must be adequate to support the conclusions. A tria-and-error procedure is generally used. II. The amount of reaeration (rm ) in a reach (station A to station B) is equal to the BOD exerted (LaA LaB) plus oxygen deficiency from station A to station B, (DA DB ). The relationship can be expressed as rm (LaA LaB) (DA DB)

(2.44)

K2 rm/Dm

(2.45)

where rm is the amount of reaeration and Dm is the mean (average) deficiency. Given La for the upper and lower sampling stations are 24.6 and 15.8 mg/L, respectively; the DO concentration at these two stations are 5.35 and 5.83 mg/L, respectively; and the water temperature is 20 C; find the K2 value for the river reach.

EXAMPLE 2:

Solution: Step 1. Find DA, DB, and Dm at 20 C. DOsat at 20 C 9.02 mg/L (Table 2.2) DA 9.02 5.35 3.67 (mg/L) DB 9.02 5.83 3.19 (mg/L)

Dm (3.67 3.19)/2 3.43 (mg/L) Step 2. Calculate rm and K2 rm (LaA LaB) (DA DB) (24.6 15.8) (3.67 3.19) 9.28 (mg/L) K2 rm /Dm 9.28/3.43

2.71 (per day) III. For the case, where DO 0 mg/L for a short period of time and without anaerobic decomposition (O’Connor, 1958), K2Dmax K2Cs

(2.46)

STREAMS AND RIVERS 1.104

CHAPTER 1.2

where Dmax is the maximum deficit, mg/L, and Cs is the saturation DO concentration, mg/L. The maximum deficiency is equal to the DO saturation concentration and the oxygen transferred is oxygen utilized by organic matter. Organic matter utilized LaA LaB, during the time of travel t; therefore rate of exertion (LaA LaB)/t K2Dmax and (LaA LaB)/t K2Cs then K2 (LaA LaB)/Cs t

(2.47)

Given that the water temperature is 20 C; LaA and LaB are 18.3 and 13.7 mg/L, respectively; and the time of travel from station A to station B is 0.123 days. Compute K2.

EXAMPLE 3:

Solution:

At 20 C, Cs 9.02 mg/L K2 (LaA LaB )/Cs t (18.3 13.7)/(9.02 0.123) 4.15 (per day)

IV. At the critical point in the river (O’Connor, 1958) dD 0 dt

(2.48)

Kd Lc K2Dc

(2.49)

and K2 Kd Lc/Dc

(2.50)

where Kd is the deoxygenation rate in stream conditions, Lc is the first-stage ultimate BOD at critical point, and Dc is the DO deficit at the critical point. These will be discussed in a later section. V. Under steady-state conditions, at a sampling point (O’Connor, 1958): dD 0 dt

(2.48)

K2D Kd L

(2.51)

or at this point

The value of D can be obtained from a field measurement. The value of Kd L can be obtained by measuring oxygen uptake from a sample taken from a given point on a river. Thus, K2 can be computed from the above equation. 9.3 Empirical Formulas The factors in the Streeter–Phelps equation has stimulated much research on the reaeration rate coefficient K2. Considerable controversy exists as to the proper method of formulas to use. Several empirical and semi-empirical formulas have been developed to estimate K2, almost all of which relate stream velocity and water depth to K2, as first proposed by Streeter (1926). Three equations below are widely known and employed in stream studies:

STREAMS AND RIVERS STREAMS AND RIVERS

1.105

13.0V 0.5 H1.5

O’Connor and Dobbins (1958)

(2.52)

11.57V 0.969 H1.673

Churchill et al. (1962)

(2.53)

7.63V H1.33

Langbein and Durum (1967)

(2.54)

where K2 reaeration rate coefficient, per day V average velocity, ft/s (fps) H average water depth, ft On the basis of K2-related physical aspects of a stream, O’Connor (1958) and Eckenfelder and O’Connor (1959) proposed K2 formulas as follows: for an isotropic turbulence (deep stream) K2

(DL V )0.5 2.3 H1.5

(2.55)

for a nonisotropic stream K2

480DL0.5S 0.25 H1.25

where DL diffusivity of oxygen in water, ft2/d H average depth, ft S slope V velocity of flow, fps B 2(HS)

(2.56)

(nonisotropic if B 17; isotropic if B 17)

The O’Connor and Dobbins (1958) equation, (Eq. (2.52)), is based on a theory more general than the formula developed by several of the other investigators. The formula proposed by Churchill (Eq. (2.53)) appears to be more restrictive in use. The workers from the US Geological Survey (Langbein and Durum, 1967) have analyzed and summarized the work of several investigators and concluded that the velocities and cross-section information were the most applicable formulation of the reaeration factor. A stream has an average depth of 9.8 ft (2.99 m) and velocity of flow of 0.61 ft/s (0.20 m/s). What are the K2 values determined by the first three empirical formulas in this section (Eqs. (2.52)–(2.54))? What is K2 at 25 C of water temperature (use Eq. (2.54) only). EXAMPLE 4:

Solution: Step 1. Determine K2 at 20 C (a) by the O’Conner and Dobbins formula (Eq. (2.52)) K2

13.0 (0.61)0.5 13.0V 0.5 1.5 H (9.8)1.5

0.33 (per day) (b) by the Churchill et al. formula (Eq. (2.53)) K2

11.57 (0.61)0.969 0.16 (per day) (9.8)1.673

7.63 0.61 0.22 (per day) (9.8)1.33

STREAMS AND RIVERS 1.106

CHAPTER 1.2

Step 2. For a temperature of 25 C K2(@25) K2(@20)(1.02)T 20 0.22(1.02)2520 0.24 (per day) 9.4 Stationary Field Monitoring Procedure Larson et al. (1994) employed physical, biochemical, and biological factors to analyze the K2 value. Estimations of physical reaeration and the associated effects of algal photosynthetic oxygen production (primary productivity) are based on the schematic formulations as follows: Physical aeration ambient DO light chamber DO SOD Algal productivity: Gross light chamber DO dark chamber DO Net light chamber DO at end light chamber DO at beginning The amount of reaeration (REA) is computed with observed data from DataSonde as REA C2 C1 POP PAP TBOD SOD where

(2.57)

REA reaeration ( D2 D1 dD), mg/L C2 and C1 observed DO concentrations in mg/L at t2 and t1, respectively POP net periphytonic (attached algae) oxygen production for the time period (t2 t1 t), mg/L PAP net planktonic algae (suspended algae) oxygen production for the time period t2 t1, mg/L TBOD total biochemical oxygen demand (usage) for the time period t2 t1, mg/L SOD net sediment oxygen demand for the time period t2 t1, mg/L

The clear periphytonic chamber is not used if periphytonic productivity/respiration is deemed insignificant, i.e. POD 0. The combined effect of TBOD PAP is represented by the gross output of the light chamber, and the SOD is equal to the gross SOD chamber output less than dark chamber output. The DO deficit D in the reach is calculated as D

C1 C2 S1 S2 2 2

(2.58)

where S1 and S2 are the DO saturation concentration in mg/L at t1 and t2, respectively, for the average water temperature (T) in the reach. The DO saturation formula is given in the preview section (Eq. (2.4)). Since the Hydrolab’s DataSondes logged data at hourly intervals, the DO changes attributable to physical aeration (or deaeration) are available for small time frames, which permits the following modification of the basic natural reaeration equation to be used to calculate K2 values (Broeren et al., 1991) 24

a (Ri1 Ri) K2 a a

where

Si Si1 Ci Ci1 b 2 2

(2.59)

Ri an ith DO concentration from the physical aeration DO-used “mass diagram” curve Ri1 a DO concentration 1 hour later than Ri on the physical reaeration DO-used curve Si an ith DO saturation concentration

STREAMS AND RIVERS STREAMS AND RIVERS

1.107

Si1 a DO saturation concentration 1 hour later than Si Ci an ith-observed DO concentration Ci1 an observed DO concentration 1 hour later than Ci Note: All units are in mg/L. Thus, the reaeration rate can be computed by K2

REA D(t2 t1)

(2.60)

Both the algal productivity/respiration (P/R) rate and SOD are biologically associated factors which are normally expressed in terms of grams of oxygen per square meter per day (g/(m2 . day)). Conversion for these areal rates to mg/L of DO usage for use in computing physical aeration (REA) is accomplished using the following formula (Butts et al., 1975): U

3.28Gt H

(2.61)

where U DO usage in the river reach, mg/L G SOD or P/R rate, g/(m2 . day) t time of travel through the reach, days H average depth, ft Given that C1 6.85 mg/L, C2 7.33 mg/L, POP 0, at T 20 C, gross DO output in light chamber 6.88 mg/L, dark chamber DO output 5.55 mg/L, gross SOD chamber output 6.15 mg/L, and water temperature at beginning and end of field monitoring 24.4 and 24.6 C, respectively, t2 t1 2.50 days, calculate K2 at 20 C.

EXAMPLE 5:

Solution: Step 1. Determine REA TBOD PAP gross 6.88 mg/L 5.55 mg/L 1.33 mg/L SOD 6.15 mg/L 5.55 mg/L 0.60 mg/L From Eq. (2.57) REA C2 C1 POP (PAP TBOD) SOD 7.33 6.85 0 1.33 0.60 2.41 (mg/L) Step 2. Calculate S1, S2, and D S1 at T 24.4 C S1 14.652 0.41022(24.4) 0.007991(24.4)2 0.00007777(24.4)3 8.27 mg/L or from Table 2.2 S2 at T 24.6 C, use same formula S2 8.24 (mg/L) Note: The elevation correction factor is ignored, for simplicity. From Eq. (2.29) S1 S2 C1 C2 2 2 1 (8.27 8.24 6.85 7.33) 2 1.17 (mg/L)

STREAMS AND RIVERS 1.108

CHAPTER 1.2

Step 3. Compute K2 with Eq. (2.60) K2

REA D(t2 t1) 2.41 mg/L 1.17 mg/L 2.50 days

0.82 per day

10 SEDIMENT OXYGEN DEMAND Sediment oxygen demand (SOD) is a measure of the oxygen demand characteristics of the bottom sediment which affects the dissolved oxygen resources of the overlying water. To measure SOD, a bottom sample is especially designed to entrap and seal a known quantity of water at the river bottom. Changes in DO concentrations (approximately 2 h, or until the DO usage curve is clearly defined) in the entrapped water are recorded by a DO probe fastened in the sampler (Butts, 1974). The test temperature should be recorded and factored for temperature correction at a specific temperature. SOD curves can be plotted showing the accumulated DO used ( y-axis) versus elapsed time (x-axis). SOD curves resemble first-order carbonaceous BOD curves to a great extent; however, firstorder kinetics are not applicable to the Upper Illinois Waterway’s data. For the most part, SOD is caused by bacteria reaching an “unlimited” food supply. Consequently, the oxidation rates are linear in nature. The SOD rates, as taken from SOD curves, are in linear units of milligrams per liter per minute (mg/(L . min)) and can be converted into grams per square meter per day (g/(m2 . d)) for practical application. A bottle sampler SOD rate can be formulated as (Butts, 1974; Butts and Shackleford, 1992): SOD

1440 SV 103 A

(2.62)

where SOD sediment oxygen demand, g/(m2 . d) S slope of stabilized portion of the curve, mg/(L . min) V volume of sampler, L A bottom area of sampler, m2 EXAMPLE 1: An SOD sampler is a half-cylinder in shape (half-section of steel pipe). Its diameter and length are 116 ft (14 in) and 2.0 ft, respectively. (1) Determine the relationship of SOD and slope of linear portion of usage curve. (2) What is SOD in g/(m2 . d) for S 0.022 mg/(L . min)?

Solution: Step 1. Find area of the SOD sampler A 116 (ft) 2.0(ft) 0.0929 (m2/ft2) 0.217 m2 Step 2. Determine the volume of the sampler 1 V 2p r2l

1 2

3.14 (7/12 ft)2 (2.0 ft) (28.32 L/ft3)

30.27 L

STREAMS AND RIVERS STREAMS AND RIVERS

1.109

Step 3. Compute SOD with Eq. (2.62) SOD

1440 SV 103 A

1400 S 30.27 103 0.217 195 S ignoring the volume of two union connections

Note: If two union connections are considered, the total volume of water contained within the sampler is 31.11 L. Thus, the Illinois State Water Survey’s SOD formula is SOD 206.6 S Step 4. Compute SOD SOD 206.6 0.022 4.545 [g/(m2 . d)] 10.1 Relationship of Sediment Characteristics and SOD Based on extensive data collections by the Illinois State Water Survey, Butts (1974) proposed the relationship of SOD and the percentage of dried solids with volatile solids. The prediction equation is SOD 6.5(DS)0.46 (VS)0.38

(2.63)

where SOD sediment oxygen demand g/(m . d) DS percent dried solids of the decanted sample by weight VS percent volatile solids by weight 2

EXAMPLE 2:

Given DS 68% and VS 8%, predict the SOD value of this sediment.

Solution: SOD 6.5(DS)0.46(VS)0.38

6.5 (8)0.38 6.5 2.20 6.965 (68)0.46

2.05 [g/(m2 # d)] 10.2 SOD Versus DO An SOD–DO relationship has been developed from the data given by McDonnell and Hall (1969) for 25-cm deep sludge. The formula is SOD/SOD1 DO0.28

(2.64)

where SOD SOD at any DO level, g/(m2 . d) SOD1 SOD at a DO concentration of 1.0 mg/L, g/(m2 . d) DO dissolved oxygen concentration, mg/L This model can be used to estimate the SOD rate at various DO concentration in areas having very high benthic invertebrate populations.

STREAMS AND RIVERS 1.110

CHAPTER 1.2

EXAMPLE 3:

Given SOD1 2.4 g/(m2 . d), find SOD at DO concentrations of 5.0 and 8.0 mg/L.

Solution: Step 1. For DO 5.0 mg/L, with Eq. (2.64) SOD5

SOD1 DO0.28 2.4 50.28 2.4 1.57 3.77 (mg/L)

Step 2. For DO 8.0 mg/L SOD8 2.4 80.28 2.4 1.79 4.30 (mg/L) Undisturbed samples of river sediments are collected by the test laboratory to measure the oxygen uptake of the bottom sediment: the amount of oxygen consumed over the test period is calculated as a zero-order reaction (USEPA, 1997): dC SOD H dt

(2.65)

where dC/d t rate change of oxygen concentration, g O2/(m2 . d) SOD sediment oxygen demand, g O2/(m2 . d) H average river depth, m

11 ORGANIC SLUDGE DEPOSITS Solids in wastewaters may be in the forms of settleable, flocculation or coagulation of colloids, and suspended. When these solids are being carried in a river or stream, there is always a possibility that the velocity of flow will drop to some value at which sedimentation will occur. The limiting velocity at which deposition will occur is probably about 0.5–0.6 ft/s (15–18 cm/s). Deposition of organic solids results in a temporary reduction in the BOD load of the stream water. Almost as soon as organic matter is deposited, the deposits will start undergoing biological decomposition, which results in some reduction in the DO concentration of the water adjacent to the sediment material. As the deposited organic matter increases in volume, the rate of decomposition also increases. Ultimately, equilibrium will be established. Velz (1958) has shown that at equilibrium the rate of decomposition (exertion of a BOD) equals the rate of deposition. In order to properly evaluate the effects of wastewater on the oxygen resources of a stream, it may be necessary to account for the contribution made by solids being deposited or by sediments being scoured and carried into suspension. From field observations, Velz (1958) concluded that enriched sediments will deposit and accumulate at a stream velocity of 0.6 ft/s (18 cm/s) or less, and resuspension of sediments and scouring will occur at a flow velocity of 1.0–1.5 ft/s (30–45 cm/s). Velz (1958) has reported that the effect of sludge deposits can be expressed mathematically. The accumulation of sludge deposition (Ld) is Ld

Pd (1 10kt) 2.303k

where Ld accumulation of BOD in area of deposition, lb/d Pd BOD added, lb/d k rate of oxidation of deposit on base 10 t time of accumulation, days

(2.66)

STREAMS AND RIVERS STREAMS AND RIVERS

EXAMPLE:

Solution:

1.111

Given (A)k 0.03; t 2 days (B)k 0.03; t 30 days (C)k 0.03; t 50 days Find the relationship of Ld and Pd For A with Eq. (2.66) Pd (1 10kt) 2.303k Pd (1 100.032) 2.303 0.03 Pd (0.13) 0.0691 1.88Pd

This means that in 2 days, 188% of daily deposit (Pd) will have accumulated in the deposit area, and 13% of Pd will have been oxidized. The rate of utilization is about 6.9% per day of accumulated sludge BOD. Solution for B: Pd (1 100.0330) 0.0691 Pd (0.874) 0.0691 12.65Pd

In 30 days, 1265% of daily deposit will have accumulated and the rate of utilization is about 87.4% of daily deposit. Solution for C: Pd (1 100.0350) 0.0691 Pd (0.97) 0.0691 14.03Pd

In 50 days, 1403% of daily deposit will have accumulated and the rate of utilization is about 97% of daily deposit (98% in 55 days, not shown). This suggests that equilibrium is almost reached in approximately 50 days, if there is no disturbance from increased flow.

12 PHOTOSYNTHESIS AND RESPIRATION Processes of photosynthesis and respiration by aquatic plants such as phytoplankton (algae), periphyton, and rooted aquatic plants (macrophytes) could significantly affect the DO concentrations in the water column. Plant photosynthesis consumes nutrients and carbon dioxide under the light, and produces oxygen. During dark conditions, and during respiration, oxygen is used. The daily average oxygen production and reduction due to photosynthesis and respiration can be expressed in the QUALZE model as (USEPA, 1997): dC PR dt dC (a3 m a4 r)Ag dt

(2.67) (2.68)

STREAMS AND RIVERS 1.112

CHAPTER 1.2

where

dC/dt rate of change of oxygen concentration, mg O2/(L . d) P average gross photosynthesis production, mg O2/(L . d) R average respiration, mg O2/(L . d) a3 stoichiometric ratio of oxygen produced per unit of algae photosynthesis, mg/mg a4 stoichiometric ratio of oxygen uptake per unit of algae respired, mg/mg m algal growth rate coefficient, per day r algal respiration rate coefficient, per day Ag algal mass concentration, mg/L

13 NATURAL SELF-PURIFICATION IN STREAMS When decomposable organic waste is discharged into a stream, a series of physical, chemical, and biological reactions are initiated and thereafter the stream ultimately will be relieved of its pollutive burden. This process is so-called natural self-purification. A stream undergoing self-purification will exhibit continuously changing water quality characteristics throughout the reach of the stream. Dissolved oxygen concentrations in water are perhaps the most important factor in determining the overall effect of decomposable organic matters in a stream. It is also necessary to maintain mandated DO levels in a stream. Therefore the type and the degree of wastewater treatment necessary depend primarily on the condition and best usage of the receiving stream. Recently the US Environmental Protection Agency has revised the water quality model QUALZE for total maximum daily loads to rivers and streams (US EPA, 1997). Interested readers may refer to an excellent example of total maximum daily load analysis in Appendix B of the US EPA technical guidance manual. 13.1 Oxygen Sag Curve The dissolved oxygen balance in a stream which is receiving wastewater effluents can be formulated from a combination of the rate of oxygen utilization through BOD and oxygen transfer from the atmosphere into water. Many factors involved in this process are discussed in the previous sections. The oxygen sag curve (DO balance) is as a result of DO added minus DO removed. The oxygen balance curve or oxygen profile can be mathematically expressed (Streeter and Phelps, 1925) as previously discussed: Dt

k1La (10k1t 10k2t) Da 10k2t k2 k1

(2.8b)

where k1 and k2 are, respectively, deoxygenation and reoxygenation rates to the base 10 which are popularly used. Since k1 is determined under laboratory conditions, the rate of oxygen removed in a stream by oxidation may be different from that under laboratory conditions. Thus, a term kd is often substituted. Likewise, the rate of BOD removal in a stream may not equal the deoxygenation rate in a laboratory bottle and the oxidation rate in a stream, so the term kr is used to reflect this situation. Applying these modified terms, the oxygen sag equation becomes Dt

kd La (10kr t 10k2t) Da 10k2t k 2 kr

(2.69)

If deposition occurs, kr will be greater than k1; and if scour of sediment organic matter occurs, kr will be less than k1. Computation of organic waste-load capacity of streams or rivers may be carried out by using Eq. (2.69) (the following example), the Thomas method, the Churchill–Buckingham method, and other methods.

STREAMS AND RIVERS STREAMS AND RIVERS

1.113

EXAMPLE: Station 1 receives a secondary effluent from a city. BOD loading at station 2 is negligible. The results of two river samplings at stations 3 and 4 (temperature, flow rates, BOD5, and DO) are adopted from Nemerow (1963) and are shown below. Construct an oxygen sag curve between stations 3 and 4.

Station

Temp C

Flow Q

(1) 3

(2) 20.0 17.0

cfs (3) 60.1 54.0

mg/L (4) 36.00 10.35

lb/d (5) 11662 3012

Measure (6) 4.1 5.2

Saturated (7) 9.02 9.61

mg/L (8) 4.92 4.41

lb/d (9) 1594 1284

20.0 16.5

51.7 66.6

Mean 21.2 5.83

7337 5908 2093

2.6 2.8

9.02 9.71

6.42 6.91

1439 1789 2481

Mean

4000

BOD5

DO, mg/L

DO deficit

2135

1 3 2

3.6 miles v=1 mile/h 4

Solution: Step 1. Calculation for columns 5, 7, 8, and 9 Col. 5: lb/d 5.39 col. 3 col. 4 Col. 7: taken from Table 2.2 Col. 8: mg/L col. 7 col. 6 Col. 9: lb/d 5.39 col. 3 col. 8 Note: 5.39 lb/d 1 cfs 62.4 lb/ft3 86,400 s/d (1 mg/106 mg) Step 2. Determine ultimate BOD at stations 3 (La3 ) and 4 (La4). Under normal deoxygenation rates, a factor of 1.46 is used as a multiplier to convert BOD5 to the ultimate first-stage BOD. La3 7337 lb/d 1.46 10,712 lb/d La4 4000 lb/d 1.46 5840 lb/d Step 3. Compute deoxygenation rate kd 3.6 mi 0.15 day 1.0 mi/h 24 h/d La3 10,712 1 1 log log kd La4 0.15 day 5840

1.76 (per day)

STREAMS AND RIVERS 1.114

CHAPTER 1.2

Step 4. Compute average BOD load L, average DO deficit D, and the difference in DO deficit

D in the reach (stations 3 and 4) from above L 12(La3 La4) 12(10,712 5840) lb/d 8276 lb/d

1 2 (1439

2135) lb/d 1787 lb/d (1789 2481) (1594 1284) 1392 (lb/d)

Step 5. Compute reaeration rate k2 L

D D 2.303 tD 1392 8276 1.76 1787 2.303 0.15 1787 5.90 (per day)

k2 kd

f k2/kd 5.9/1.76 3.35 Step 6. Plot the DO sag curve Assuming the reaction rates kd(kd k1) and k2 remain constant in the reach (stations 3–4), the initial condition at station 3: La 10,712 lb/d Da 1439 lb/d Using Eq. (2.69) Dt

kd La (10kr t 10k2t) Da 10k2t k 2 kr

When t 0.015 1.76 10,712 (101.76 0.015 105.9 0.015) 1439 105.9 0.015 5.9 1.76 1745 (lb/d)

D0.015

D values at various t can be computed in the same manner. We get D values in lb/d as below. It was found that the critical point is around t 0.09 days. The profile of DO deficit is depicted as below (Fig. 2.10): D0 1439 D0.015 1745 D0.03 1960 D0.045 2104 D0.06 2192 D0.075 2235

D0.09 2245 D0.105 2227 D0.12 2190 D0.135 2137 D0.15 2073

Thomas method. Thomas (1948) developed a useful simplification of the Streeter–Phelps equation for evaluating stream waste—assimilative capacity. His method presumes the computation of the

STREAMS AND RIVERS STREAMS AND RIVERS

FIGURE 2.10

1.115

Profile of DO deficit.

stream deoxygenation coefficient (k1) and reoxygenation coefficient (k2) as defined in previous sections (also example). He developed a nomograph to calculate the DO deficit at any time, t (DO profile), downstream from a source of pollution load. The nomograph (not shown, the interested reader should refer to his article) is plotted as D/La versus k2 t for various ratios of k2/k1. In most practical applications, this can be solved only by a tedious trial-and-error procedure. Before the nomograph is used, k1, k2, Da, and La must be computed (can be done as in the previous example). By means of a straight edge, a straight line (isopleth) is drawn, connecting the value of Da/La at the left of the point representing the reaeration constant time-of-travel (k2 t) on the appropriate k2/k1 curve (a vertical line at k2t and intercept on the k2/k1 curve). The value Da /La is read at the intersection with the isopleth. Then, the value of the deficit at time t (Dt) is obtained by multiplying La with the interception value. Nemerow (1963) claimed that he had applied the Thomas nomograph in many practical cases and found it very convenient, accurate, and timesaving. He presented a practical problem-solving example for an industrial discharge to a stream. He illustrated three different DO profiles along a 38-mile river: i.e. (1) under current loading and flow conditions; (2) imposition of an industrial load at the upstream station under current condition; and (3) imposition of industrial load under 5-year low-stream conditions. Churchill–Buckingham method. Churchill and Buckingham (1956) found that the DO values in streams depend on only three variables: temperature, BOD, and stream flow. They used multiple linear correlation with three normal equations based on the principle of least squares: g x1y ax21 bx1x2 cx1x3

(2.70)

g x2y ax1x2 bx22 cx2 x3

(2.71)

g x3y ax1x3 bx2x3 cx23

(2.72)

y DO dropped in a reach, mg/L x1 BOD at sag, mg/L x2 temperature, C x3 flow, cfs (or MGD) a, b, and c constants

where

STREAMS AND RIVERS 1.116

CHAPTER 1.2

TABLE 2.12 DO Sag between Stations 2 and 4 Observed DO, mg/L Station 2

Station 4

DO drop, mg/L y

7.8 8.2 6.1 6.0 5.5 6.2

5.7 5.9 4.8 3.2 2.3 2.9

2.1 2.3 1.9 2.8 3.2 3.3

6.8 12.0 14.8 20.2 18.9 14.3

11.0 16.5 21.2 23.4 28.3 25.6

13.21 11.88 9.21 6.67 5.76 8.71

15.6 2.6

87.0 14.5

126.0 21.0

55.44 9.24

Sum Mean

BOD @ sag, mg/L x1

Temperature, C x2

Flow, 1000 cfs x3

By solving these three equations, the three constants a, b, and c can be obtained. Then, the DO drop is expressed as y ax1 bx2 cx3 d

(2.73)

This method eliminates the often questionable and always cumbersome procedure for determining k1, k2, kd, and kr. Nemerow (1963) reported that this method provides a good correlation if each stream sample is collected during maximum or minimum conditions of one of the three variables. Only six samples are required in a study to produce practical and dependable results. EXAMPLE: Data obtained from six different days’ stream surveys during medium- and low-flow periods showed that the DO sag occurred between stations 2 and 4. The results and computation are as shown in Table 2.12. Develop a multiple regression model for DO drop and BOD loading.

Solution: Step 1. Compute element for least-squares method (see Table 2.13). Step 2. Apply to three normal equations by using the values of the sum or average of each element. 229.84a 326.7b 122.81c 39.04 326.7a 474.42b 179.77c 56.98 122.81a 179.77b 92.33c 23.07 Solving these three equations by dividing by the coefficient of a: a 1.421b 0.534c 0.1699 a 1.452b 0.550c 0.1744 a 1.464b 0.752c 0.1879 TABLE 2.13 Elements for Least-Squares Method

Sum Mean

yx1

yx2

yx3

x21

x1x2

x1x3

x22

x2x3

x23

4.41 5.29 3.61 7.84 10.24 10.89

14.28 27.60 28.12 56.56 60.48 47.19

23.10 37.95 40.28 65.52 90.56 84.48

27.74 27.32 17.50 18.68 18.43 28.74

46.24 144.00 219.04 408.04 357.21 204.49

74.8 198.0 313.76 472.68 534.87 366.08

89.83 142.56 136.31 134.74 108.86 124.55

121.00 272.25 449.44 547.56 800.89 655.36

145.31 176.02 195.25 196.08 163.01 222.98

174.50 141.13 84.82 44.49 33.18 75.86

42.28 7.05

234.23 39.04

341.89 56.98

138.42 23.07

1379.02 229.84

1960.19 326.70

736.85 122.81

2846.50 474.42

1078.64 179.77

553.99 92.33

STREAMS AND RIVERS STREAMS AND RIVERS

1.117

Subtracting one equation from the other: 0.043b 0.218c 0.0180 0.012b 0.202c 0.0135 Dividing each equation by the coefficient of b: b 5.07c 0.4186 b 16.83c 1.125 Subtracting one equation from the other: 11.76c 0.7064 c 0.0601 Substituting c in the above equation: b 5.07 0.0601 0.4186 b 0.1139 Similarly a 0.1699 1.421 0.1139 0.534 0.0601 0.0241 Check on to the three normal equations with values of a, b, and c: d Y (ax1 bx2 cx3) 39.04 (229.84 0.0241 326.7 0.1139 122.81 0.0601) 0 The preceding computations yield the following DO drop: Y 0.0241x1 0.1139x2 0.0601x3 Step 3. Compute the DO drop using the above model: The predicted and observed DO drop is shown in Table 2.14. It can be seen from Table 2.14 that the predicted versus the observed DO drop values are reasonably close. Step 4. Compute the allowable BOD loading at the source of the pollution. The BOD equation can be derived from the same least-squares method by correlating the upstream BOD load (as x1) with the water temperature (as x2), flow rate (as x3) and resulting BOD (as z) at the TABLE 2.14 Predicted and Observed DO Drop Dissolved oxygen drop, mg/L Sampling date 1 2 3 4 5 6

Calculated

Observed

1.93 2.35 2.61 2.62 3.07 3.04

2.1 2.3 1.9 2.8 3.2 3.3

STREAMS AND RIVERS 1.118

CHAPTER 1.2

sag point in the stream. Similar to steps 1 and 2, the DO drop is replaced by BOD at station 4 (not shown). It will generate an equation for the allowable BOD load: z ax1 bx2 cx3 d Then, applying the percent of wastewater treatment reduction, the BOD in the discharge effluent is calculated as x1. Selecting the design temperature and low-flow data, one can then predict the BOD load (lb/d or mg/L) from the above equation. Also, applying z mg/L in the DO drop equation gives the predicted value for DO drop at the sag. 13.2 Determination of kr The value kr can be determined for a given reach of stream by determining the BOD of water samples from the upper (A) and lower (B) end of the section under consideration: log

L aB kr t L aA

(2.74)

L aB L aA 1 1 t log kr t log L aA L aB

(2.75a)

or 1 (2.75b) kr t (log L aA log L aB) While the formula calls for first-stage BOD values, any consistent BOD values will give satisfactory results for kr. The first-stage BOD values for river stations 3 and 4 are 34.6 and 24.8 mg/L, respectively. The time of travel between the two stations is 0.99 days. Find kr for the reach.

EXAMPLE 1:

Solution: 1 kr t (log LaA log LaB) 1 (log 34.6 log 24.8) 0.99 0.15 (per day for log base 10) Kr 2.3026 kr 0.345 per day for log base e EXAMPLE 2:

Given LaA 32.8 mg/L, DA 3.14 mg/L, t 1.25 day

LaB 24.6 mg/L DB 2.58 mg/L

Find k2 and kr Solution: Step 1. Determine k2 using Eq. (2.44) and Eq. (2.45) Reaeration rm (LaA LaB) (DA DB) (32.8 24.6) (3.14 2.58) 8.76 (mg/L)

STREAMS AND RIVERS STREAMS AND RIVERS

1.119

Mean deficit Dm (3.14 2.58)/2 2.86 rm 8.76 3.06 (per day) K2 Dm 2.86

3.06 1.33 (per day) 2.303

Step 2. Compute kr 1 kr t (log LaA log LaB) 1 (log 32.8 log 24.6) 1.25 0.10 (per day) 13.3 Critical Point on Oxygen Sag Curve In many cases, only the lowest point of the oxygen sag curve is of interest to engineers. The equation can be modified to give the critical value for DO deficiency (Dc) and the critical time (tc) downstream at the critical point. At the critical point of the curve, the rate of deoxygenation equals the rate of reoxygenation: dD K1 Lt K2 Dc 0 dt

(2.76)

thus K1Lt K2Dc Dc

K1 L K2 t

(2.77a)

Kd L K2 t

(2.77b)

where Lt BOD remaining. Use Kd if deoxygenation rate at critical point is different from K1. Since Lt La eKr tc then Dc

Kd (L eKr tc) K2 a

(2.78a)

kd (L 10kr tc) k2 a

(2.78b)

Let f

K2 k2 Kd kd

(2.79)

or Dc

1 (L eKr tc) f a

(2.80)

STREAMS AND RIVERS 1.120

CHAPTER 1.2

and from Thomas (1948) Da(k2 kr) k2 1 log c1 d k2 kr kr kd La

(2.81)

Da 1 log e f c1 ( f 1) d f La kr( f 1)

(2.82)

tc then tc

Substitute tc in Dc formula (Thomas, 1948): Da 0.418 k2 kr La Dc a b c1 a1 b d Dc kr k2 kr

(2.83)

or log La log Dc c1

Da 0.418 kr k2 a1 b d log a b Dc k2 kr kr

(2.84)

Thomas (1948) provided this formula which allows us to approximate La: the maximum BOD load that may be discharged into a stream without causing the DO concentration downstream to fall below a regulatory standard (violation). EXAMPLE 1: The following conditions are observed at station A. The water temperature is 24.3 C, with k1 0.16, k2 0.24, and kr 0.19. The stream flow is 880 cfs (ft3/s). DO and La for river water is 6.55 and 5.86 mg/L, respectively. The state requirement for minimum DO is 5.0 mg/L. How much additional BOD (Q 110, cfs, DO 2.22 mg/L) can be discharged into the stream and still maintain 5.0 mg/L DO at the flow stated?

Solution: Step 1. Calculate input data with total flow Q 880 110 990 cfs; at T 24.3 C Saturated DO from Table 1.2 DOS 8.29 mg/L After mixing 6.55 880 2.22 110 990 6.07 (mg/L)

DOa

Deficit at station A Da DOs DOa 8.29 mg/L 6.07 mg/L 2.22 mg/L Deficit at critical point Dc DOs DOmin 8.29 mg/L 5.00 mg/L 3.29 mg/L Rates kd kr 0.19 (per day) f

k2 0.24 1.5 k1 0.16

f 1 0.5

STREAMS AND RIVERS STREAMS AND RIVERS

1.121

Step 2. Assume various values of La and calculate resulting Dc. tc

Da 1 log e f c1 ( f 1) d f La kr ( f 1) 2.22 1 log e 1.5 c1 (0.5) df La 0.19(0.5)

10.53 log a1.5

1.665 b La

Let La 10 mg/L tc 10.53 log (1.5 0.1665) 1.32 (days) kd Dc (La 10kr tc) k2 0.19 (10 100.191.32) 0.24 0.792(10 0.561) 4.45 (mg/L) Similarly, we can develop a table La, mg/L

Tc, days

Dc, mg/L

10.00 9.00 8.00 7.00 6.40 6.00

1.32 1.25 1.17 1.06 0.90 0.92

4.45 4.13 3.80 3.48 3.29 3.17

Therefore, maximum La, max 6.40 mg/L Step 3. Determine effluent BOD load (Ye ) that can be added. BOD that can be added maximum load existing load 110Ye 6.40 990 5.86 880 Ye 10.72 mg/L This means that the first-stage BOD of the effluent should be less than 10.72 mg/L. EXAMPLE 2: Given: At the upper station A of the stream reach, under standard conditions with temperature of 20 C, BOD5 3800 lb/d, k1 0.14, k2 0.25, kr 0.24, and kd kr. The stream temperature is 25.8 C with a velocity 0.22 miles/h. The flow in the reach (A → B) is 435 cfs (including effluent) with a distance of 4.8 miles. DOA 6.78 mg/L, DOmin 6.00 mg/L. Find how much additional BOD can be added at station A and still maintain a satisfactory DO level at station B?

Step 1. Calculate La at T 25.8 C. y La(1 10k1 t) When t 5 days, T 20 C (using loading unit of lb/d): 3800 La(1 100.145) La 3800/0.80 4750 (lb/d) La(20)

STREAMS AND RIVERS 1.122

CHAPTER 1.2

Convert to T 25.8 C: La(T) La(20)[1 0.02(T 20)] La(25.8) 4750[1 0.02(25.8 20)] 5300 (lb/d) Step 2. Change all constants to 25.8 C basis. At 25.8 C, k1(T) k1(20) 1.047T20 k1(25.8) 0.14 1.047(25.820) 0.18 (per day) k2(T) k2(20) (1.02)T20 k2(25.8) 0.25 (1.02)25.820 0.28 (per day) kr(25.8) 0.24 1.047(25.820) 0.31 (per day) kd(25.8) 0.31 (per day) Step 3. Calculate allowable deficit at station B. At 25.8 C, DOsat 8.05 mg/L DOB 6.00 mg/L DB 8.05 mg/L 6.00 mg/L 2.05 mg/L DA 8.05 mg/L 6.78 mg/L 1.27 mg/L Step 4. Determine the time of travel (t) in the reach. t

4.8 miles distance 0.90 days V 0.22 mph 24 h/d

Step 5. Compute allowable La at 25.8 C using Eq. (2.69). Dt

kd La (10kr t 10k2 t) Da 10k2 t k2 kr

Here Dt DB and Da DA 0.31 La (100.31 0.9 100.28 0.9) 1.27 100.28 0.9 0.28 0.31 2.05 (10.33)La(0.526 0.560) 0.71 2.05

La 1.34/0.35 3.83 (mg/L) Allowable load: lb/day 5.39 La # Q 5.39(3.83 mg/L)(435 cfs) 8980 (lb/d)

STREAMS AND RIVERS STREAMS AND RIVERS

Step 6. Find load that can be added at station A. Added allowable existing 8980 5300 3680 (lb/d) Step 7. Convert answer back to 5-day 20 C BOD. 3680 La(20)[1 0.02(25.8 20)] La(20) 3297 (lb/d) EXAMPLE 3:

Given: The following data are obtained from upstream station A. k1 0.14@20 C

BOD5 load at station A 3240 lb/d

k2 0.31@20 C

Stream water temperature 26 C

kr 0.24@20 C

Flow u 188 cfs

kd kr

Velocity V 0.15 mph

DOA 5.82 mg/L

A B distance 1.26 miles

Allowable DOmin 5.0 mg/L Find: Determine the BOD load that can be discharged at downstream station B. Solution: Step 1. Compute DO deficits. At T 26 C, DOs 8.02 mg/L and Da at station A Da DOs DOA 8.02 mg/L 5.82 mg/L 2.20 mg/L Critical deficit at station B. Dc 8.02 DOmin 8.02 mg/L 5.0 mg/L 3.02 mg/L Step 2. Convert all constants to 26 C basis. At 26 C, k1(26) k1(20) 1.047(2620) 0.14 1.0476 0.18 (per day)

1.123

STREAMS AND RIVERS 1.124

CHAPTER 1.2

k2(26) k2(20) (1.02)2620 0.31 (1.02)6 0.34 (per day) kr(26) 0.24 1.0476 0.32 (per day) k2(26) 0.34 1.0625 0.32 kr(26) Step 3. Calculate allowable BOD loading at station B at 26 C. From Eq. (2.84): log LaB log Dc c1

Da 0.418 kr k2 a1 b d log a b Dc k2 kr kr

log LaB log 3.02 c1

LaB

0.32 2.2 0.418 a1 b d log 1.0625 0.34 0.32 3.02

0.480 [1 16 (0.272)0.418]0.0263 0.480 10.28 0.0263 0.750 5.52 (mg/L)

Allowable BOD loading at station B: lb/d 5.39 LaB (mg/L) # Q (cfs) 5.39 5.52 188 5594 Step 4. Compute ultimate BOD at station A at 26 C. LaA BOD5 loading/(1 10k1t) 3240/(1 100.145) 4050 (lb/d at 20 C) LaA(26) LaA[1 0.02 (26 20)] 4050(1 0.12) 4536 (lb/d) Step 5. Calculate BOD at 26 C from station A oxidized in 1.26 miles. Time of travel, t

1.26 miles 0.35 days 0.15 mph 24 h/d

yB LaA(1 10kr t) 4536(1 100.320.35) 4536 0.227 1030 (lb/d) Step 6. Calculate BOD remaining from station A at station B. Lt 4536 lb/d 1030 lb/d 3506 lb/d

STREAMS AND RIVERS STREAMS AND RIVERS

1.125

Step 7. Additional BOD load that can be added in stream at station B (x) x(26) allowable BOD Lt 5594 lb/d 3506 lb/d 2088 lb/d This is first-stage BOD at 26 C. At 20 C, x(26) x(20)[1 0.02 (26 20)] x(20) 2088/1.12 1864 (lb/d) For BOD5 at 20 C (y5) y5 x20(1 100.145) 1864(1 100.7) 1492 (lb/d) Given: Stations A and B are selected in the main stream, just below the confluences of tributaries. The flows are shown in the sketch below. Assume there is no significant increased flow between stations A and B, and the distance is 11.2 miles (18.0 km). The following information is derived from the laboratory results and stream field survey:

EXAMPLE 4:

k1 0.18 per day @20 C

stream water temperature 25 C

k2 0.36 per day @20 C

DO above station A 5.95 mg/L

kr 0.22 per day @20 C

La just above A 8870 lb/d @20 C

kd kr Velocity, V 0.487 mph

La added at A 678 lb/d @20 C DO deficit in the tributary above A 256 lb/d

11.4 MGD

98.6 MGD

DO = 7.25 mg/L A 8.9 MGD

11.2 miles 107.5 MGD

B 118.9 MGD

Find: Expected DO concentration just below station B. Solution: Step 1. Calculate total DO deficit just below station A at 25 C. DO deficit above station A DA 8.18 mg/L 5.95 mg/L 2.23 mg/L lb/d of DO deficit above A DA (mg/L) Q (MGD) 8.34 (lb/gal) 2.23 98.6 8.34 1834 DO deficit in tributary above A 256 lb/d Total DO deficit below A 1834 lb/d 256 lb/d 2090 lb/d

STREAMS AND RIVERS 1.126

CHAPTER 1.2

Step 2. Compute ultimate BOD loading at A at 25 C. At station A at 20 C, LaA 8870 lb/d 678 lb/d 9548 lb/d At station A at 25 C, LaA 9548[1 0.02(25 20)] 10,503 (lb/d) Step 3. Convert rate constants for 25 C. k1(25) k1(20) 1.047(2520) 0.18 1.258 0.23 (per day) k2(25) k2(20) 1.0242520 0.36 1.126 0.40 (per day) kr(25) kr(20) 1.0472520 0.22 1.258 0.28 (per day) kd(25) Step 4. Calculate DO deficit at station B from station A at 25 C. Time of travel t 11.2/(0.487 24) 0.958 day DB

kd La (10kr t 10k2t) Da 10k2 t k2 kr 0.28 10,503 (100.280.958 100.40.958) 2090 100.40.958 0.40 0.28

24,507(0.5392 0.4138) 2090 0.4138 3073 865 2208 (lb/d) Convert the DO deficit into concentration. DB amount, lb/d/(8.34 lb/gal flow, MGD) 2208/(8.34 107.5) 2.46 (mg/L) This is DO deficiency at station B from BOD loading at station A. Step 5. Calculate tributary loading above station B. DO deficit 8.18 mg/L 7.25 mg/L 0.93 mg/L Amount of deficit 0.93 8.34 11.4 88 (lb/d) Step 6. Compute total DO deficit just below station B. Total deficit DB 2208 lb/d 88 lb/d 2296 (lb/d) or

2296/(8.34 118.9) 2.23 (mg/L)

STREAMS AND RIVERS STREAMS AND RIVERS

1.127

Step 7. Determine DO concentration just below station B. DO 8.18 mg/L 2.23 mg/L 5.95 mg/L A treated wastewater effluent from a community of 108,000 persons is to be discharged into a river which is not receiving any other significant wastewater discharge. Normally, domestic wastewater flow averages 80 gal (300 L) per capita per day. The 7-day, 10-year low flow of the river is 78.64 cfs. The highest temperature of the river water during the critical flow period is 26.0 C. The wastewater treatment plant is designed to produce an average carbonaceous 5-day BOD of 7.8 mg/L; an ammonia–nitrogen concentration of 2.3 mg/L, and DO for 2.0 mg/L. Average DO concentration in the river upstream of the discharge is 6.80 mg/L. After the mixing of the effluent with the river water, the carbonaceous deoxygenation rate coefficient (KrC or KC) is estimated at 0.25 per day (base e) at 20 C and the nitrogenous deoxygenation coefficient (KrN or KN) is 0.66 per day at 20 C. The lag time (t0) is approximately 1.0 day. The river cross-section is fairly constant with mean width of 30 ft (10 m) and mean depth of 4.5 ft (1.5 m). Compute DO deficits against time t. EXAMPLE 5:

Solution: Step 1. Determine total flow downstream Q and V. Effluent flow Qe 80 108,000 gpd 8.64 106 gpd 1.54/106 cfs/gpd 13.31 cfs Upstream flow Qu 78.64 cfs Downstream flow Qd Qe Qu 13.31 cfs 78.64 cfs 91.95 cfs Velocity V Qd/A 91.95 ft3/s/(30 4.5)ft2 0.681 ft/s Step 2. Determine reaeration rate constant K2. The value of K2 can be determined by several methods as mentioned previously. From the available data, the method of O’Connor and Dobbins (1958), Eq. (2.52), is used at 20 C: K2 13.0V1/2H3/2 13.0(0.681)1/2(4.5)3/2 1.12 (per day) Step 3. Correct temperature factors for coefficients. K2(26) 1.12 per day 1.0242620 1.29 per day KC(26) 0.25 per day 1.0476 0.33 per day Zanoni (1967) proposed correction factors for nitrogenous KN in wastewater effluent at different temperatures as follows: KN(T) KN(20) 1.097T20 for 10 22 C

(2.85)

KN(T) KN(20) 0.877T22 for 22 30 C

(2.86)

and For 26 C, KN(26) 0.66 per day 0.8772622 0.39 per day

STREAMS AND RIVERS 1.128

CHAPTER 1.2

Step 4. Compute ultimate carbonaceous BOD (LaC). At 20 C, K1C 0.25 per day BOD5 7.8 mg/L LaC BOD5/(1 eK1C5) 7.8 mg/L/(1 e0.255) 10.93 mg/L At 26 C, using Eq. (2.28) LaC(26) 10.93 mg/L(0.6 0.02 26) 12.24 mg/L Step 5. Compute ultimate nitrogenous oxygen demand (LaN). Reduced nitrogen species (NH4, NO3, and NO2) can be oxidized aerobically by nitrifying bacteria which can utilize carbon compounds but always require nitrogen as an energy source. The two-step nitrification can be expressed as 3

NH 4 2 O2

Nitrosomonas NO 2 h

2H H2O

(2.87)

and 1 NO 2 2 O2

Nitrobacter h

NO 3

NH 4 2O2 h NO3 2H H2O

(2.88) (2.89)

Overall 2 moles of O2 are required for each mole of ammonia; i.e. the N:O2 ratio is 14:64 (or 1:4.57). Typical domestic wastewater contains 15–50 mg/L of total nitrogen. For this example, effluent NH3 2.3 mg/L as N; therefore, the ultimate nitrogenous BOD is LaN 2.3 mg/L 4.57 10.51 mg/L Step 6. Calculate DO deficit immediately downstream of the wastewater load. 13.31 2 78.64 6.8 13.31 78.64 6.10 (mg/L)

From Table 2.2, the DO saturation value at 26 C is 8.02 mg/L. The initial DO deficit Da is Da 8.02 mg/L 6.10 mg/L 1.92 mg/L Step 7. Compute Dt and DO values at various times t. Using Dt

KC LaC KN LaN (eKC t eK2t) Da eK2 t [eKN(tt0) eK2(tt0)] K2 KC K2 KN

and at 26 C K2 1.29 per day KC 0.33 per day KN 0.39 per day LaC 12.24 mg/L LaN 10.51 mg/L Da 1.92 mg/L t0 1.0 day

(2.90)

STREAMS AND RIVERS STREAMS AND RIVERS

1.129

For t 1.0 day: When t 0.1 day 0.33 12.24 0.330.1 (e e1.290.1) 1.92e1.290.1 1.29 0.33 4.2075(0.9675 0.8790) 1.6876 2.060 (mg/L) D0.2 2.17 mg/L and so on

D0.1

For t 1.0 day D1.1 4.2075(e0.331.1 e1.291.1) 1.92e1.291.1 0.39 10.51 0.39(1.11) Ce e1.29(1.11) D 1.29 0.39 4.2075(0.6956 0.2420) 0.4646 4.5543(0.9618 0.8790) 2.75 (mg/L) D1.2 3.04 mg/L and so on Step 8. Produce a table for DO sag. Table 2.15 gives the results of the above calculations for DO concentrations in the stream at various locations. V 0.681 ft/s 0.681 ft/s 3600 s/h 0.463 miles/h When t 0.1 days Distance 0.463 miles/h 24 h/d 0.1 day 1.11 miles

1 mile 5280 ft

TABLE 2.15 DO Concentrations at Various Locations t, days

Distance below outfall, miles

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.8 2.0 2.1 2.2

0.00 1.11 2.22 3.33 4.44 5.55 6.67 7.78 7.89 10.00 11.11 12.22 13.33 14.44 15.55 16.67 17.78 18.89 22.22 23.33 24.44

DO deficit, Expected DO* mg/L mg/L 1.92 2.06 2.17 2.26 2.32 2.37 2.40 2.41 2.42 2.41 2.40 2.75 3.04 3.27 3.45 3.59 3.69 3.81 3.83 3.81 3.79

6.10 5.96 5.85 5.74 5.70 5.64 5.62 5.61 5.60 5.61 5.62 5.27 4.98 4.75 4.57 4.43 4.33 4.21 4.19 4.21 4.23

*Expected DO Saturated DO (8.02 mg/L) DO deficit.

STREAMS AND RIVERS 1.130

CHAPTER 1.2

Step 9. Explanation From Table 2.15, it can be seen that minimum DO of 5.60 mg/L occurred at t 0.8 days due to carbonaceous demand. After this location, the stream starts recovery. However, after t 1.0 days, the stream DO is decreasing due to nitrogenous demand. At t 2.0, it is the critical location (22.22 miles below the outfall) with a critical DO deficit of 3.83 mg/L and the DO level in the water is 4.19 mg/L. 13.4 Simplified Oxygen Sag Computations The simplified oxygen sag computation was suggested by Le Bosquet and Tsivoglou (1950). As shown earlier at the critical point dD 0 dt Kd Lc K2 Dc

and

(2.48) (2.49)

Let the lb/day of first-stage BOD be C C (lb/day)

La (mg/L) Q (cfs) 62.38 (lb/cf) 86,400 (s/d) 1,000,000

5.39La Q or La

C1 C Q 5.39Q

(2.91)

where C1 constant If Da 0, then Dc is a function of La, or D LaC2

C3 C1 C2 Q Q

(2.92)

where C2, C3 constants DQ C3

(2.93)

Since Kd Lc K2 Dc Dc

C3 Kd L La C2 K2 c Q

(2.94)

Hence the minimum allowable DO concentration at the critical point (DOC) is the DO saturation value (S, or DOsat) minus Dc. It can be written as DOC S Dc 1 S C3 a b Q

(2.95) (2.95a)

From the above equation, there is a linear relationship between DOC and 1/Q. Therefore, a plot of DOC versus 1/Q will give a straight line, where S is the y-intercept and C3 is the slope of the line. Observed values from field work can be analyzed by the method of least squares to determine the degree of correlation.

STREAMS AND RIVERS STREAMS AND RIVERS

1.131

14 SOD OF DO USAGE The SOD portion of DO usage is calculated using the following formula: DOsod

3.28Gt H

(2.96)

where DOsod oxygen used per reach, mg/L G the SOD rate, g/(m2 . d) t the retention time for the reach H the average water depth in the reach, ft All biological rates, including BOD and SOD, have to be corrected for temperature using the basic Arrhenius formula: RT RA(uTA)

(2.97)

where RT biological oxygen usage rate at a temperature, T C RA biological oxygen usage rate at ambient or standard temperature (20 C as usual), A C u proportionality constant, 1.047 for river or stream EXAMPLE: A stream reach has an average depth of 6.6 ft. Its SOD rate was determined as 3.86 g/(m2 . day) at 24 C. The detention time for the reach is 0.645 days. What is the SOD portion of DO usage at the standard temperature of 20 C?

Solution: Step 1. Determine SOD rate G at 20 C. G24 G20(1.0472420) G20 G24/(1.047)4 3.86/1.2017 3.21 (g/(m2 # d)) Step 2. Calculate DOsod: 3.28Gt H 3.28 3.21 0.645 6.6 1.03 (mg/L)

DOsod

15 APPORTIONMENT OF STREAM USERS In the United States, effluent standards for wastewater treatment plants are generally set for BOD, total suspended solids (TSS), and ammonia-nitrogen (NH3-N) concentrations. Each plant often should meet standards of 10–10–12 (mg/L for BOD, TSS, and NH3-N, respectively), right or wrong. It is not like classically considering the maximum use of stream-assimilative capacity. However, in some parts of the world, maximum usage of stream natural self-purification capacity may be a valuable tool for cost saving of wastewater treatment. Thus, some concepts of apportionment of selfpurification capacity of a stream among different users are presented below. The permissible load equation can be used to compute the BOD loading at a critical point Lc, as previously stated (Eq. (2.84); Thomas, 1948): log Lc log Dc c1

Da 0.418 kr k2 a1 b d loga b Dc k2 kr kr

(2.98)

STREAMS AND RIVERS 1.132

CHAPTER 1.2

where Lc BOD load at the critical point Dc DO deficit at the critical point Da DO deficit at upstream pollution point kr river BOD removal rate to the base 10, per day k2 reoxygenation rate to the base 10, per day Lc also may be computed from limited BOD concentration and flow. Also, if city A and city B both have the adequate degree of treatment to meet the regulatory requirements, then Lc paLA bLB where

(2.99)

p part or fraction of LA remaining at river point B a fraction of LA discharged into the stream b fraction of LB discharged into the stream LA BOD load at point A, lb/d or mg/L LB BOD load at point B, lb/d or mg/L

B t

C t

If LA LB L Lc paL bL

(2.99a)

A numerical value of Lc is given by Eq. (2.98) and a necessary relationship between a, b, p, LA, and LB is set up by Eq. (2.99). The value p can be computed from log p kr t Since log

Lt kr t log fraction remaining La

Therefore log p kr t

(2.100)

or p 10kr t

(2.100a)

The apportionment factors a and b can be determined by the following methods. 15.1 Method 1 Assume: City A is further upstream than city B, with time-of-travel t given, by nature, a larger degree of stream purification capacity. The time from city B to the lake inlet is also t. Since y La(1 10kr t) y 1 10kr t 1 p La

(2.101)

The proportion removed by the stream of BOD added by city A is y 1 10kr t2 1 102kr t LA 1 p2

(2.102) (2.102a)

STREAMS AND RIVERS STREAMS AND RIVERS

The proportion removed by the stream of BOD added by city B is y 1 10kr t 1 p LB

1.133

(2.102b)

Then the percent amount removed BOD from cities A and B, given as PA and PB, respectively, are 1 p2 (1 p)(1 p) (1 p)(1 p) (1 p) (1 p2) (1 p) (1 p)(1 p) (1 p)(1 p 1) 1p PA 2p 1p (1 p) PB (1 p)(1 p 1) (1 p 2) (1 p)

1 2p

(2.103)

(2.104)

Calculate the degree of BOD removal proportion: PA 1p (1 p)/(2 p) a P b 1/(2 p) 1 B a (1 p)b

(2.105)

Solving Eqs. (2.99a) and (2.105) simultaneously for a and b: Lc pa L b L Lc pa b L Lc p(1 p)b b (p p2 1)b L Lc 1 b b a L 1 p p2

(2.106)

a (1 p)b

Lc 1p # L 1 p p2

(2.107)

The required wastewater treatment plant efficiencies for plants A and B are 1a and 1b, respectively. The ratio of cost for plant A to plant B is (1a) to (1b). Assume all conditions are as in Method 1. The following data are available: kr 0.16 per day; t 0.25 days; flow at point C, Qc 58 MGD; L 6910 lb/d; and regulation required Lc 10 mg/L. Determine the apportionment of plants A and B, required plant BOD removal efficiencies, and their cost ratio. EXAMPLE 1:

Solution: Step 1. Compute allowable Lc. Lc 10 mg/L 8.34 lb/d/MGD # mg/L 58 MGD 4837 lb/d

STREAMS AND RIVERS 1.134

CHAPTER 1.2

Step 2. Compute p p 10kr t 100.160.25 0.912 Step 3. Calculate a and b by Eqs. (2.106) and (2.107): a

Lc 1p 1 0.912 4837 a a b b L 1 p p2 6910 1 0.912 0.9122

0.70 0.697 0.49 Lc 1 1 b a b 0.70 L 1 p p2 2.744 0.255 Step 4. Determine percent BOD removal required For plant A: 1 a 1 0.49 0.51

i.e. 51% removal needed

For plant B: 1 b 1 0.255 0.745

i.e. 74.5% removal needed

Step 5. Calculate treatment cost ratio Cost, plant A 0.51 0.68 1a 1 1b 1 Cost, plant B 1.46 0.745 15.2 Method 2 Determine a permissible BOD load for each city as if it were the only city using the river or stream. In other words, calculate maximum BOD load LA for plant A so that dissolved oxygen concentration is maintained above 5 mg/L (most state requirements), assuming that plant B does not exist. Similarly, also calculate maximum BOD load LB for plant B under the same condition. It should be noted that LB may be larger than LA and this cannot be the case in Method 1. The BOD removed by the stream from that added at point A: referring to Eq. (2.102a) (1 p2)LA at point B: referring to Eq. (2.102b) (1 p )LB The percentage removed is at point A, PA

(1 p2)LA (1 p2)LA (1 p)LB

(2.108)

at point B, PB

(1 p)LB (1 p )LA (1 p)LB 2

(2.109)

STREAMS AND RIVERS STREAMS AND RIVERS

1.135

The degree of BOD removal is PA a PB b

(2.105)

Solve equations (2.99), (2.105), (2.108), and (2.109) for a and b. Also determine (1a) and (1b) for the required wastewater treatment efficiencies and cost ratios as (1a)/(1b). EXAMPLE 2:

All conditions and questions are the same as for Example 1, except use Method 2.

Solution: Step 1. Compute allowable LA and LB. For point B, using Eq. (1.100) log p log p

Lt kr t 0.16 0.25 0.04 LB

Lt 100.04 0.912 LB

LB Lt /0.912 (Lt Lc in this case) 4837/0.912 5304 (lb/d) For point A, Lc 100.160.252 100.08 0.8318 LA LA 4837 lb/d/0.8318 5815 lb/d Step 2. Calculate PA and PB. PA

(1 p2)LA (1 p2)LA (1 p)LB

(1 0.9122)5815 (1 0.9122)5815 (1 0.912)5304 978.4 978.4 466.8 0.677 PB 1 0.677 0.323

Step 3. Determine a and b. a/b PA/PB 0.677/0.323 2.1/1 a 2.1b Lc pa LA b LB 4837 0.912 (2.1b) 5815 b 5304 4837 11,137b 5304b b 4837/16,441 0.29 a 2.1 0.29 0.61

STREAMS AND RIVERS 1.136

CHAPTER 1.2

Step 4. Determine degree of BOD removal For plant A 1 a 1 0.61 0.39

i.e. 39% removal efficiency

1 b 1 0.29 0.71

i.e. 71% removal efficiency

For plant B

Step 5. Determine cost ratio 0.39 0.55 Plant A 1a 1 or 1b 0.71 1.82 1 Plant B 15.3 Method 3 The principle of this method is to load the river to the utmost so as to minimize the total amount of wastewater treatment. The proportion of PA and PB can be computed by either Method 1 or Method 2. The cost of treatment can be divided in proportion to population. There are two cases of initial BOD loading: Case 1. LA > L (L is the BOD load of plant A influent) • • • •

Let plant A discharge L without a treatment. Have plant B treat an amount equal to L Lc. City A is assessed for PA(L Lc). City B is assessed for PB(L Lc).

Case 2. LA < L • • • • •

Let plant A discharge LA and treat L LA. Have plant B treat an amount equal to (L Lc) pLA. Then the total treatment will remove (L LA) (L Lc pLA) 2L (1 p)LA Lc. City A is assessed for PA[2L (1 p)LA Lc]. City B is assessed for PB[2L (1 p)LA Lc].

EXAMPLE 3: Using data listed in Examples 1 and 2, determine the cost ratio assessed to plants A and B by Method 3.

Solution: Step 1. Select case Since LA 5815 lb/d L 6910 lb/d Case 2 will be applied Step 2. Determine amounts needed to be treated. For plant A: 6910 5815 1095 (lb/d treated) 5815 lb/d discharged without treatment For plant B: L Lc pLA 6910 4837 0.912 1095 1074 (lb/d)

STREAMS AND RIVERS STREAMS AND RIVERS

1.137

Total treatment 1095 1074 2169 (lb/d) Step 3. Compute cost assessments For Example 2, PA 0.677 PB 0.323 For plant A Cost 0.677 treatment cost of 2169 lb/d For plant B Cost 0.323 treatment cost of 2169 lb/d

16 VELZ REAERATION CURVE (A PRAGMATIC APPROACH) Previous sections discuss the conceptual approach of the relationships between BOD loading and DO in the stream. Another approach for determining the waste-assimilative capacity of a stream is a method first developed and used by Black and Phelps (1911) and later refined and used by Velz (1939). For this method, deoxygenation and reoxygenation computations are made separately, then added algebraically to obtain the net oxygen balance. This procedure can be expressed mathematically as DOnet DOa DOrea DOused

(2.110)

where DOnet dissolved oxygen at the end of a reach DOa initial dissolved oxygen at beginning of a reach DOrea dissolved oxygen absorbed from the atmosphere DOused dissolved oxygen consumed biologically All units are in lb/d or kg/d. 16.1 Dissolved Oxygen Used The dissolved oxygen used (BOD) can be computed with only field-observed dissolved oxygen concentrations and the Velz reoxygenation curve. From the basic equation, the dissolved oxygen used in pounds per day can be expressed as: DOused (DOa DOnet) DOrea (in lb/d) The terms DOa and DOnet are the observed field dissolved oxygen values at the beginning and end of a subreach, respectively, and DOrea is the reoxygenation in the subreach computed by the Velz curve. If the dissolved oxygen usage in the river approximates a first-order biological reaction, a plot of the summations of the DOused versus time-of-travel can be fitted to the equation: DOused La[1 exp(K1 t)]

(2.111)

and La and K1 are determined experimentally in the same manner as for the Streeter–Phelps method. The fitting of the computed values to the equations must be done by trial and error, and a digital computer is utilized to facilitate this operation.

STREAMS AND RIVERS 1.138

CHAPTER 1.2

16.2 Reaeration The reaeration term (DOrea) in the equation is the most difficult to determine. In their original work, Black and Phelps (1911) experimentally derived a reaeration equation based on principles of gas transfer of diffusion across a thin water layer in acquiescent or semiquiescent system. A modified but equivalent form of the original Black and Phelps equation is given by Gannon (1963): R 100 a

1 B0 e25K c e9K b(81.06)aeK b 100 9 25

(2.112)

where R percent of saturation of DO absorbed per mix B0 initial DO in percent of saturation K p2 am/4L2 in which m is the mix or exposure time in hours, L is the average depth in centimeters, and a is the diffusion coefficient used by Velz The diffusion coefficient a was determined by Velz (1947) to vary with temperature according to the expression: aT a20(1.1T20)

(2.113)

where aT diffusion coefficient at T C a20 diffusion coefficient at 20 C when the depth is in feet, a20 0.00153 when the depth is in centimeters, a20 1.42 Although the theory upon which this reaeration equation was developed is for quiescent conditions, it is still applicable to moving, and even turbulent, streams. This can be explained in either of two ways (Phelps, 1944): (1) Turbulence actually decreases the effective depth through which diffusions operate. Thus, in a turbulent stream, mixing brings layers of saturated water from the surface into intimate contact with other less oxygenated layers from below. The actual extent of such mixing is difficult to envision, but it clearly depends upon the frequency with which surface layers are thus transported. In this change-of-depth concept therefore there is a time element involved. (2) The other and more practical concept is a pure time effect. It is assumed that the actual existing conditions of turbulence can be replaced by an equivalent condition composed of successive periods of perfect quiescence between which there is instantaneous and complete mixing. The length of the quiescent period then becomes the time of exposure in the diffusion formula, and the total aeration per hour is the sum of the successive increments or, for practical purposes, the aeration period multiplied by the periods per hour.

Because the reaeration equation involves a series expansion in its solution, it is not readily solved by desk calculations. To facilitate the calculations, Velz (1939, 1947) published a slide-rule curve solution to the equation. This slide-rule curve is reprinted here as Fig. 2.11 (for its use, consult Velz, 1947). Velz’s curve has been verified as accurate by two independent computer checks, one by Gannon (1963) and the other by Butts (1963). Although the Velz curve is ingenious, it is somewhat cumbersome to use. Therefore, a nomograph (Butts and Schnepper, 1967) was developed for a quick, accurate desk solution of R at zero initial DO (B0 in the Black and Phelps equation). The nomograph, presented in Fig. 2.12, was constructed on the premise that the exposure or mix time is characterized by the equation M (13.94) loge(H) 7.45

(2.114)

STREAMS AND RIVERS STREAMS AND RIVERS

FIGURE 2.11

1.139

Standard reoxygeneration curve (Butts et al., 1973).

in which M is the mix time in minutes and H is the water depth in feet. This relationship was derived experimentally and reported by Gannon (1963) as valid for streams having average depths greater than 3 ft. For an initial DO greater than zero, the nomograph value is multiplied by (1 B0) expressed as a fraction because the relationship between B0 and the equation is linear. The use of the nomograph and the subsequent calculations required to compute the oxygen absorption are best explained by the example that follows: EXAMPLE: Given that H 12.3 ft; initial DO of stream 48% of saturation; T 16.2 C; DO load at saturation at 16.2 C 8800 lb; and time-of-travel in the stream reach 0.12 days. Find the mix time (M) in minutes; the percent DO absorbed per mix at time zero initial DO (R0); and the total amount of oxygen absorbed in the reach (DOrea).

Solution: Step 1. Determine M and R0. From Fig. 2.12, connect 12.3 ft on the depth-scale with 16.2 C on the temperature-scale. It can be read that: M 27.5 min on the mix time-scale, and R0 0.19% on the R0-scale Step 2. Compute DOrea Since R0 is the percent of the saturated DO absorbed per mix when the initial DO is at 100% deficit (zero DO), the oxygen absorbed per mix at 100% deficit will be the product of R0 and the DO load at saturation. For this example, lb/mix per 100% deficit is 8800 lb

0.19 16.72 (lb/mix per 100% deficit) 100

STREAMS AND RIVERS 1.140

CHAPTER 1.2

FIGURE 2.12

Oxygen obsorption nomograph (Butts and Schnepper, 1967).

STREAMS AND RIVERS STREAMS AND RIVERS

1.141

However, the actual DO absorbed (at 48% saturation) per mix is only 16.72 [1 (48/100)] 8.7 (lb/mix per 100% deficit) Because the amount absorbed is directly proportional to the deficit, the number of mixes per reach must be determined: t 0.12 days 0.12 days 1440 min/d 173 min

The time-of-travel

The mix time M 27.5 min Consequently, the number of mixes in the reach 173/27.5 6.28. Reaeration in the reach is therefore equal to the product of the DO absorbed per mix the number of mixes per reach. DOrea is computed according to the expression: DOrea a1 a1

R0 % DO saturation t ba b a b (DO saturation load) M 100 100

(2.115)

0.19 173 48 ba ba b(8800) 100 100 27.5

8.7 6.28 54.6 (lb)

17 STREAM DO MODEL (A PRAGMATIC APPROACH) Butts and his coworkers (Butts et al., 1970, 1973, 1974, 1975, 1981; Butts, 1974; Butts and Shackleford, 1992) of the Illinois State Water Survey expanded Velz’s oxygen balance equation. For the pragmatic approach to evaluate the BOD/DO relationship in a flowing stream, it is formulated as follows: DOn DOa DOu DOr DOx

(2.116)

where DOn net dissolved oxygen at the end of a stream reach DOa initial dissolved oxygen at the beginning of a reach DOu dissolved oxygen consumed biologically within a reach DOr dissolved oxygen derived from natural reaeration within a reach DOx dissolved oxygen derived from channel dams, tributaries, etc. Details of computation procedures for each component of the above equation have been outlined in several reports mentioned above. Basically the above equation uses the same basic concepts as employed in the application of the Streeter–Phelps expression and expands the Velz oxygen balance equation by adding DOx. The influences of dams and tributaries are discussed below. 17.1 Influence of a Dam Low head channel dams and large navigation dams are very important factors when assessing the oxygen balance in a stream. There are certain disadvantages from a water quality standpoint associated with such structures but their reaeration potential is significant during overflow, particularly if a low DO concentration exists in the upstream pooled waters. This source of DO replenishment must be considered.

STREAMS AND RIVERS 1.142

CHAPTER 1.2

Procedures for estimating reaeration at channel dams and weirs have been developed by researchers in England (Gameson, 1957; Gameson et al., 1958; Barrett et al., 1960; Grindrod, 1962). The methods are easily applied and give satisfactory results. Little has been done in developing similar procedures for large navigation and power dams. Preul and Holler (1969) investigated larger structures. Often in small sluggish streams more oxygen may be absorbed by water overflowing a channel dam than in a long reach between dams. However, if the same reach were free flowing, this might not be the case; i.e. if the dams were absent, the reaeration in the same stretch of river could conceivably be greater than that provided by overflow at a dam. If water is saturated with oxygen, no uptake occurs at the dam overflow. If it is supersaturated, oxygen will be lost during dam overflow. Water, at a given percent deficit, will gain oxygen. The basic channel dam reaeration formula takes the general form r 1 0.11qb(1 0.046T )h

(2.117)

where r dissolved oxygen deficit ratio at temperature T q water quality correction factor b weir correction factor T water temperature, C h height through which the water falls, ft The deficit dissolved oxygen ratio is defined by the expression r (Cs CA )/(Cs CB ) DA /DB

(2.118)

where CA dissolved oxygen concentration upstream of the dam, mg/L CB dissolved oxygen concentration downstream of the dam, mg/L Cs dissolved oxygen saturation concentration, mg/L DA dissolved oxygen deficit upstream of the dam, mg/L DB dissolved oxygen deficit downstream of the dam, mg/L Although Eqs. (2.117) and (2.118) are rather simplistic and do not include all potential parameters which could affect the reaeration of water overflowing a channel dam, they have been found to be quite reliable in predicting the change in oxygen content of water passing over a dam or weir (Barrett et al., 1960). The degree of accuracy in using the equations is dependent upon the estimate of factors q and b. For assigning values for q, three generalized classifications of water have been developed from field observations. They are q 1.25 for clean or slightly polluted water; q 1.0 for moderately polluted water; and q 0.8 for grossly polluted water. A slightly polluted water is one in which no noticeable deterioration of water quality exists from sewage discharges; a moderately polluted stream is one which receives a significant quantity of sewage effluent; and a grossly polluted stream is one in which noxious conditions exist. For estimating the value of b, the geometrical shape of the dam is taken into consideration. This factor is a function of the ratio of weir coefficients W of various geometrical designs to that of a free weir where W (r 1)/h

(2.119)

Weir coefficients have been established for a number of spillway types, and Gameson (1957) in his original work has suggested assigning b values as follows: Spillway type Free Step Slope (ogee) Sloping channel

b 1.0 1.3 0.58 0.17

STREAMS AND RIVERS STREAMS AND RIVERS

1.143

For special situations, engineering judgment is required. For example, a number of channel dams in Illinois are fitted with flashboards during the summer. In effect, this creates a free fall in combination with some other configuration. A value of b, say 0.75, could be used for a flashboard installation on top of an ogee spillway (Butts et al., 1973). This combination would certainly justify a value less than 1.0, that for a free weir, because the energy dissipation of the water flowing over the flashboards onto the curved ogee surface would not be as great as that for a flat surface such as usually exists below free and step weirs. Aeration at a spillway takes place in three phases: (1) during the fall; (2) at the apron from splashing; and (3) from the diffusion of oxygen due to entrained air bubbles. Gameson (1957) has found that little aeration occurs in the fall. Most DO uptake occurs at the apron; consequently, the greater the energy dissipation at the apron the greater the aeration. Therefore, if an ogee spillway is designed with an energy dissipator, such as a hydraulic jump, the value of b should increase accordingly. Given CA 3.85 mg/L, T 24.8 C, h 6.6 ft (2.0 m), q 0.9, and b 0.30 (because of hydraulic jump use 0.30). Find DO concentration below the dam CB.

EXAMPLE:

Solution: Step 1. Using dam reaeration equation to find r r 1 0.11qb(1 0.046T)h 1 0.11 0.9 0.30(1 0.046 24.8) 6.6 1.42 Step 2. Determine DO saturation value. From Table 2.2, or by calculation, for Cs at 24.8 C this is Cs 8.21 mg/L Step 3. Compute CB. Since r (Cs CA)/(Cs CB) therefore CB Cs (Cs CA)/r 8.21 (8.21 3.85)/1.42 8.21 3.07 5.14 (mg/L) 17.2 Influence of Tributaries Tributary sources of DO are often an important contribution in deriving a DO balance in stream waters. These sources may be tributary streams or outfalls of wastewater treatment plants. A tributary contribution of DOx can be computed based on mass balance basis of DO, ammonia, and BOD values from tributaries. The downstream effect of any DO input is determined by mass balance computations: in terms of pounds per day the tributary load can simply add to the mainstream load occurring above the confluence. The following two examples are used to demonstrate the influence of tributary sources of DO. Example 1 involves a tributary stream. Example 2 involves the design of an outfall structure to achieve a minimum DO at the point of discharge. EXAMPLE 1:

Given:

Tributary flow Tributary DO Mainstream flow Mainstream DO

Q1 123 cfs 6.7 mg/L Q2 448 cfs 5.2 mg/L

Find: DO concentration and DO load at the confluence.

STREAMS AND RIVERS 1.144

CHAPTER 1.2

Solution: Step 1. Determine DO concentration. Assuming there is a complete mix immediately below the confluence, the DO concentration downstream on the confluence can be determined by mass balance. DO

Q1 DO1 Q2 DO2 Q1 Q2

(123 6.7 448 5.2)/(123 448) 5.5 (mg/L) Step 2. Compute DO loadings. Since DO load (DO, mg/L) (Q, cfs) mg L/s cfs 28.32 b L cfs mg 86,400 s 1 lb b 28.32 DO Qa s day 454,000 mg DO Qa

5.39 DO Q lb/d Tributary load 5.39 6.7 123 lb/d or 5.39 lb/(d # mg/L # cfs) DO mg/L Q cfs 4442 lb/d Note: The factor 5.39 is the same as that in Eq. (2.91) Similarly, Mainstream load 5.39 5.2 448 12,557 (lb/d) Total DO load 4442 lb/d 12,557 lb/d 16,999 lb/d The difference in elevation between the outfall crest and the 7-day, 10-year low flow is 4.0 ft (1.22 m). Utilizing this head difference, determine if a free-falling two-step weir at the outfall will insure a minimum DO of 5.0 mg/L at the point of discharge. The flow of wastewater effluent and the stream are, respectively, 6.19 and 14.47 cfs (175 and 410 L/s, or 4 MGD and 10 MGD). DO for the effluent and stream are 1.85 and 6.40 mg/L, respectively. The temperature of the effluent is 18.9 C. The stream is less than moderately polluted. Determine the stream DO at the point of effluent discharge and the oxygen balance.

EXAMPLE 2:

Solution: Step 1. The water quality correction factor is selected as q 1.1 b Cs CA h

1.3 for free fall step-weir 9.23 mg/L, when T 18.9 C from Table 2.2 1.85 mg/L as effluent DO 4.0 ft

Step 2. Determine r with the channel dam reaeration model r 1 0.11qb(1 0.046T)h 1 0.11 1.1 1.3(1 0.046 18.9) 4 2.18

STREAMS AND RIVERS STREAMS AND RIVERS

1.145

Step 3. Compute CB r (Cs CA)/(Cs CB) CB Cs (Cs CA)/r 9.23 (9.23 1.85)/2.18 5.85 (mg/L) Note: In this case, CB is the effluent DO as it reaches the stream. Step 4. Compute the resultant DO in the stream assuming there is a complete mixing with the stream water, the resultant DO at the point of discharge is 5.85 6.19 6.40 14.47 6.19 14.47 6.23 (mg/L)

Step 5. Calculate oxygen mass balance DO load from effluent 5.39 DO Q 5.39 5.85 6.19 195 (lb/d) DO load in stream above the point of effluent discharge 5.39 DO Q 5.39 6.40 14.47 499 (lb/d) The total amount of oxygen below the point of discharge is 694 lb/d. 17.3 DO Used The term DOused (DOu) in Eq. (2.111) represents the oxygen consumed biologically within a stream reach. This term can be determined by three methods (Butts et al., 1973): (1) observed DO concentrations in conjunction with reaeration estimate; (2) bottle BOD and deoxygenation rate determinations of river water samples; and (3) long-term bottle BOD progression evaluation of a wastewater effluent. The method used will probably be dictated by the existing data or the resources available for collecting usable data. The term DOu includes DO usage due to carbonaceous and nitrogenous BOD and to SOD, as stated previously. The ratio of DO contribution by algal photosynthesis to DO consumption by algal respiration is assumed to be unity, although it can handle values greater or less than one when derived on a diurnal basis. For a series of stream reaches, each incremental DOu value is added and the accumulated sums, with the corresponding time-of-travel in the stream, are fitted to the first-order exponential expression y La(1 eKd t)

(2.120)

where y oxygen demand exerted (DOu) La ultimate oxygen demand, including carbonaceous and nitrogenous Kd in-stream deoxygenation coefficient to the base e, per day t time-of-travel, days Note: The coefficient Kd is comparable to the composite of the terms KC and KN previously defined in the discussion regarding the conceptual approach to waste-assimilative analysis.

STREAMS AND RIVERS 1.146

CHAPTER 1.2

The use of field DO values in estimating the waste-degradation characteristics of a stream has certain advantages: the need for laboratory BOD tests is eliminated which saves time and cost; the reliability of the results should be better since the measurement of dissolved oxygen is far more precise and accurate than the BOD test; also, stream measured DO concentrations take into account the in situ oxygen demand in the stream, which includes both dissolved and benthic demand, whereas laboratory BOD results generally reflect only the oxygen demand exerted by dissolved matter. Estimations of Kd and La. The data obtained from stream survey, a summation of DOu values and corresponding time-of-travel t at an observed water temperature, have to be computed along the reach of a stream. Nowadays, for determinations of Kd and La one can simplify the calculation by using a computer or a programmable calculator. However, the following example illustrates the use of the Thomas slope method for determining Kd and La.

17.4 Procedures of Pragmatic Approach The steps of the pragmatic approach involved in estimating the waste-assimilative capacity of a stream based upon a pragmatic approach can be summarized as follows (Butts et al., 1973): 1. Develop a full understanding of the stream length, its channel geometry, water stage and flow patterns, and the general hydrologic features of the watershed. 2. Determine the 7-day, 10-year flow of the stream and select a design water temperature. 3. Define the location of all dams and their physical features; define also the location of all tributary flows and relevant data regarding them. 4. Divide the stream into reaches consistent with significant changes in cross-sections and determine the volumes and average depth in each reach. 5. At the beginning and end of each reach, during low-flow conditions and summer temperatures, undertake a series of field determinations for at least water temperature and dissolved oxygen concentrations and, if desired, collect water samples for BOD determinations. 6. Compute the time-of-travel within each reach at stream flows observed during the time-of-sampling as well as that during 7-day, 10-year low flow. 7. From the observed DO values, flow, and time-of-travel compute DOa DOn, as demonstrated in Table 2.17 (see later). 8. Select DO saturation values from Table 2.2 for observed stream temperature conditions and compute the natural reaeration for each reach using Figs. 2.12 and 2.13 in conjunction with appropriate equations for finding the mix time M and the percent absorption at 100% deficit R0. Keep in mind the need to make adjustments in accordance with weir and mass balance formulas where dams and tributaries are encountered. 9. Calculate, by summation, the DOu for each reach as demonstrated in Table 2.18 (see later). 10. From an array of the DOu versus t data, determine La and Kd, preferably by the methods of Reed–Theriault, steepest descent, or least squares. For a graphical solution, the Thomas slope method is satisfactory. Adjust the values for La and Kd for the selected design water temperature by the use of Eqs. (2.26) and (2.28). 11. Apply the removal efficiency anticipated to the computed ultimate oxygen demand La and develop the required expression DOu La(1 eKt ). 12. From the values developed in step 6 for time-of-travel and depth at 7-day, 10-year low flow, use the observed DO just upstream of the discharge point (Da at the beginning of the reach) as a starting-point, and follow the computation. 13. Note whether or not the removal efficiency selected will permit predicted DO concentrations in the stream compatible with water quality standards.

STREAMS AND RIVERS STREAMS AND RIVERS

FIGURE 2.13

1.147

Modified Velz curve (Butts et al., 1973).

EXAMPLE: A long reach of a moderately sized river is investigated. There is an overloaded secondary wastewater treatment plant discharging its effluent immediately above sampling station 1 (upstream). The plant, correctly, has BOD removal efficiency of 65%. A total of 12 sampling stations are established in reaches of the study river. Downstream stations 3 and 4 are above and below an ogee channel dam, respectively. Stations 10 and 11 are immediately above and below the confluence of the main stream and a tributary which has relatively clean water. The 7-day, 10-year low flow of the main receiving stream and the tributary are 660 and 96 cfs, respectively. The design water temperature is selected as 27 C. A stream field survey for flow, water temperature, DO, average depth, and time-of-travel is conducted during summer low-flow conditions. During the field survey, the receiving stream in the vicinity of the effluent discharge was 822 cfs and the tributary flow was 111 cfs. The DO values are 5.96 and 6.2 mg/L, respectively. Data obtained from field measurements and subsequent computations for DO values are presented in Tables 2.16, 2.17, and 2.18. The DO levels in the stream are not less than 5.0 mg/L, as required.

Question: Determine the DO concentrations along the reach of the stream at design temperature and design flow conditions, if the efficiency of sewage treatment plant is updated to at least 90% BOD removal. Answer: The solution is approached in two sessions: (1) The deoxygenation rate coefficients Kd and La should be determined under the existing conditions; this involves steps 1 to 5. (2) A predictive profile of DO concentrations will be calculated under 7-day 10-year low flow and 27 C design conditions (steps 6 and 7).

STREAMS AND RIVERS 1.148

CHAPTER 1.2

TABLE 2.16 Field Data and DO Computations DOa observed†

Flow, cfs Station 1 2 3 Dam 4 5 6 7 8 9 10 Tributary 11 12 ∗ †

DO averaged

at station

mean

mg/L

lb/d∗

DOa – DOn, lb/day

mg/L

lb/d

822 830 836

826 833

5.96 3.44 2.40

26406 15390 10814

11016 4575

4.70 2.92

20925 13110

836 844 855 871 880 880 888

840 849 863 875 880 884

3.96 3.90 3.70 3.60 3.90 4.04 4.44

17844 17741 17051 16900 18498 19163 21251

103 690 151 1598 665 2288

3.93 3.80 3.65 3.75 3.97 4.27

17793 17389 16978 17686 18831 20202

999 1010

1005

5.18 5.28

27892 28743

6641 851

4.81 5.23

24474 28330

DO (lb/d) 5.39 flow (cfs) DO (mg/L). DOa at the end of a reach is the DOn of that reach.

Solution: Step 1. Compute DO values. In Table 2.16, data are obtained from field work and subsequent DO computations for other purposes. The term DOa DOn (lb/d) will be used in the equation for Table 2.18. A plot of DO measured and time-of-travel t can be made (not shown). It suggests that a profound DO sag exists in the stream below the major pollution source of effluent discharge. Step 2. Compute the natural reaeration, DOr. In Table 2.17, values of saturated DO (DOs) at the specified water temperature are obtained from Table 2.2. The values of M and R0 are taken from Figs. 2.12 and 2.13 on the basis of the physical dimension of the stream listed in Table 2.16 or calculated from Eq. (2.114) (M 13.94 ln H 745). Step 3. Calculate DO used, DOu. The DOU is calculated from equation DOu (DOa DOn) DOr DOx. Without consideration of DOx, the values of DOu for each reach of the stream are given in Table 2.18. It should be noted that DOx does not apply yet. Step 4. Determine Kd(23) and Ld(23) for T 23 C (during the field survey). Using the Thomas (1950) slope method, the factor ( t/ DOu)1/3 is plotted on arithmetic graph paper against t as shown in Fig. 2.14. A line of best fit is then drawn, often neglecting the first two points. The ordinate intercept b is read as 0.0338; and the slope of the line S is computed as 0.00176 [i.e. (0.0426 0.0338)/5]. Then determine Kd and La at 23 C. Kd(23) 6S/b 6 0.00176/0.0338 0.31 (per day) La(23) 1/(Kd # b3) 1/(0.31 0.03383) 83,500 (lb/d) The curve is expressed as y 83,500(1 e0.31t)

†

R0 100

4.70 2.92 3.93 3.80 3.65 3.75 3.97 4.24 4.81 5.23

8.50 8.49 8.48 8.47 8.46 8.46 8.47 8.48

Mean Da†, mg/L

8.53 8.51

Mean Ds, mg/L

4.4 11.2 15.5 17.6 18.2 18.6 19.9

7.2

18.8 19.4

M, min

2.9 3.8 5.2 6.1 6.3 6.5

6.6 6.8

Depth h, ft

656.6

396.0 792.0 1591.2 1630.0 1118.9 911.5

846.7 1201.0

t, min

DOa DOs

0.538 0.552 0.57 0.557 0.531 0.499 0.432 0.383

0.449 0.657

tr b a5.39Q DOs b and is determined as the same manner of example in Section 16.2. M

8.53 8.52 8.50 8.50 8.50 8.48 8.48 8.46 8.46 8.45 8.48 8.48

DOs mg/L

Mean Da is from Table 2.16.

DOa

DOr a1 DOs

22.8 22.9 23.0 23.0 23.0 23.1 23.1 23.2 23.2 23.3 23.1 23.1

1 2 3 4 5 6 7 8 9 10 11 12

Temp., C

Computations of Natural Reaeration

Station

TABLE 2.17

0.0040

0.0046 0.0052 0.0049 0.0045 0.0044 0.0043

0.0042 0.0041

Ro 100

33.0

90.0 70.7 102.7 92.6 61.5 49.0

45.0 61.9

tr M

45936

38485 38851 39445 39946 40127 40310

37977 38209

5.39Q DOs lb/d

2324

8566 7892 11301 9277 5761 4237

3225 6370

DOr,∗ lb/d

STREAMS AND RIVERS

1.149

STREAMS AND RIVERS 1.150

CHAPTER 1.2

TABLE 2.18 Calculation of DO Used Station 1 2 3 4 5 6 7 8 9 10 11 12

t, days

DOa – DOn, lb/d

DOr, lb/d

DOu,* lb/d

DOu, lb/d

0.588 0.834

0.588 1.422

11016 4575

3225 6370

14241 10945

0.275 0.550 1.105 1.132 0.777 0.633

1.697 2.247 3.352 4.484 5.261 5.894

103 690 151 1598 665 2888

8566 7892 11301 9277 5761 4237

8669 8582 11452 7679 5096 1349

0.456

6.350

851

2324

1473

14241 25186 25186 33855 42437 53889 61568 66664 68013 68013 69486

( t/ DOu)1/3 0.03456 0.03836 0.03687 0.03755 0.03962 0.04176 0.04289 0.04425 0.04504

*DOu DOa DOn DOr

0.046

0.044

0.042

(Σt/ΣDOu)1/3

0.040

0.038

0.036

0.034

0.032

0.030 0 FIGURE 2.14

3 4 Time of Travel, Σt, days

Plot of ( t/ DOu)1/3 versus time of travel.

STREAMS AND RIVERS STREAMS AND RIVERS

1.151

Step 5. Compute Kd(27) and La(27) for design temperature 27 C. Using Eq. (2.26) Kd(27) Kd(20) 1.0472720 Kd(23) Kd(20) 1.0472320 Kd(27) Kd(23) 1.0477/1.0473 0.31 1.3792/1.1477 0.37 (per day) Similarly, using Eq. (2.28) La(T) La(20) (0.6 0.02T) La(27) La(23) (0.6 0.02 27)/(0.6 0.02 23) 83,500 1.14/1.06 89,800 (lb/d) Next step goes to design phase. Step 6. Calculate the design BOD loading after plant improvement. Since the current overloaded wastewater treatment plant only removes 65% of the incoming BOD load, the raw wastewater BOD load is 89,800 lb/d /(1 0.65) 256,600 lb/d The expanded (updated) activated sludge process may be expected to remove 90% to 95% of the BOD load from the raw wastewater. For the safe side, 90% removal is selected for the design purpose. Therefore the design load La at 27 C to the stream will be 25,660 lb/d (256,600 0.1). Step 7. Gather new input data for DO profile. In order to develop the predictive profile for DO concentration in the receiving stream after the expanded secondary wastewater treatment plant is functioning, the following design factors are obtained from field survey data and the results computed: Stream flow 660 cfs At the dam q 1.1 Tributary flow 96 cfs b 0.58 Tributary DO 6.2 mg/L h 6.1 ft DO at station 1 5.90 mg/L Kd(27) 0.37 per day Water temperature 27 C La(27) 25,660 lb/d H and t are available from cross-section data Step 8. Perform DO profile computations. Computations are essentially the same as previous steps with some minor modifications (Table 2.19). For columns 9, 10, 13, 19, 20, and 21, calculate iteratively from station by station from upstream downward. The other columns can be calculated independently. At station 4 (below the dam), the dam aeration ratio is given as Eq. (1.117) and Eq. (1.118) r 1 0.11qb(1 0.046 T )h 1 0.11 1.1 0.58(1 0.046 27) 6.1 1.960 r (Cs CA)/(Cs CB) 1.96 (7.87 5.09)/(7.87 CB) CB 2.78/1.96 7.87 6.45 (mg/L) This DO value of 6.41 mg/L is inserted into column 21 at station 4. The corresponding DOn (in lb/d) is computed by multiplying 5.39Q to obtain 23,148 lb/d, which is inserted into column 20 at station 4.

STREAMS AND RIVERS 1.152

CHAPTER 1.2

At station 11, immediately below the tributary, DO contribution is 3208 lb/d (i.e. 5.39 6.2 96). This value plus 28,068 lb/d (31,276 lb/d) is inserted into column 20. Then computations continue with iterative procedures for station 12. It should be noted that the parameters H, M, t, t/M, and R/100 can be computed earlier (i.e. in earlier columns). Therefore, the other DO computations will be in the latter part of Table 2.19. It seems easier for observation. Also, these steps of computation can be easily programmed using a computer. The results of computations for solving equation, DOn DOa DOu DOr DOx are given in Table 2.19, and the DO concentrations in column 21 of Table 2.19 are the predictive DO values for 7-day, 10-year low flow with secondary wastewater treatment. All DO values are above the regulatory standard of 5.0 mg/L, owing to expanded wastewater treatment and dam aeration. TABLE 2.19 DO Profile Computation at 27 C and a 7-Day, 10-Year Low Flow (1)

(2)

(3)

(4)

(5)

(6) 1 eKd t

(7) Total DOu, lb/d

(8) Subreach DOu, lb/d

at station

mean

t, days

t days

660 666 670

663 668

0.750 1.076

0.750 1.826

0.242 0.491

0 6218 12603

6218 6385

672 678 689 699 703 706

0.345 0.626 1.313 1.370 0.938 0.845

2.171 2.797 4.110 5.48 6.418 7.263

0.551 0.645 0.781 0.868 0.907 0.932

14168 16544 20052 22282 23272 23913

1565 2376 3508 2230 990 941

806

0.608

7.871

0.946

24265

351

Flow, cfs Station 1 2 3 Dam 4 5 6 7 8 9 10 Tributary 11 12

670 674 682 696 702 704 708 96 804 810

TABLE 2.19 (contd.) (9)

Station 1 2 3 Dam 4 5 6 7 8 9 10 Tributary 11 12

DOa – DOu, lb/d

(10) Mean DOa, lb/d

(11)

(12)

mg/L

lb/d

14866 11977

17975 15170

7.87 7.87

21084 28124 28336

21584 23895 22908 24556 26367 27093

22366 25083 24662 25671 26862 27404

7.87 7.87 7.87 7.87 7.87 7.87

30924

31100

7.87

(13)

(14) Mean depth H, ft

(15)

0.361 0.465

5.9 6.1

17.4 17.5

28506 28760 29227 29651 29821 29948

0.215 0.128 0.165 0.134 0.099 0.085

2.5 3.4 4.6 5.5 5.7 6.0

4.1 9.6 13.8 16.2 16.6 17.4

34190

0.090

6.5

18.6

DOs 1

DOa DOs

M, min

STREAMS AND RIVERS STREAMS AND RIVERS

1.153

TABLE 2.19 (contd.) (16) t 1440t, min

Station 1 2 3 Dam 4 5 6 7 8 9 10 Tributary 11 12 Notes: Col 7 Col 8n Col 9n Col 10n Col 11 Col 12 Col 15 Col 18 Col 19 Col 20 Col 21 trib.

(17)

(18)

(19)

(20) (21) DOn (or DOa)

t/M

R0/100

DOr, lb/d

lb/d 5.90 5.14 5.09

Remark

1080 1549

62.1 88.5

0.00555 0.00545

3696 6353

21084 18362 18330

497 901 1891 1973 1351 1217

121.2 93.9 137.0 121.8 81.4 69.9

0.00630 0.00730 0.00620 0.00578 0.00568 0.00550

4687 2521 3878 2802 1367 975

23148 26271 26416 26786 27357 27734 28067

6.41 7.25 7.23 7.21 7.26 7.32 7.38

876

47.1

0.00521

788

31276 31682

7.29

Above dam Below dam

Above trib. Below trib.

La Col 6 25,660 Col 6 Col 7n – Col 7n–1 Col 20n–1 – Col 8n (Col 20n–1 Col 9n)/2 DOS at T of 27 C 5.39 Col 3 Col 11 Calculated or from monogram Fig. 2.12 Calculated or from monogram Fig. 2.12 Col 12 Col 13 Col 17 Col18 Initial value is 663 5.90 5.39 and thereafter Col 9 Col 19 Col 20/Col 3/5.39 Tributary

18 BIOLOGICAL FACTORS 18.1 Algae Algae are most commonly used to assess the extent that primary productivity (algal activity) affects the DO resources of surface waters. Many factors affect the distribution, density, and species composition of algae in natural waters. These include the physical characteristics of the water, length of storage, temperature, chemical composition, in situ reproduction and elimination, floods, nutrients, human activities, trace elements, and seasonal cycles. Of the many methods suggested for defining the structure of a biological community, the most widely used procedure has been the diversity index. A biological diversity index provides a means of evaluating the richness of species within a biological community using a mathematical computation. Although different formulas have been used, the one determined by the information theory formula, the Shannon–Weiner diversity index, is widely used for algal communities or communities of other organisms. The formula is (Shannon and Weaver, 1949): m

D a Pilog2 Pi i1

where

D diversity index Pi ni /N ni the number (density) of the ith genera or species N total number (density) of all organisms of the sample

(2.121)

STREAMS AND RIVERS 1.154

CHAPTER 1.2

i 1, 2, . . . , m m the number of genera or species log2 Pi 1.44 ln Pi or m

D 1.44 a (ni/N) ln(ni/N)

(2.122)

The index D has a minimum value, when a community consisting solely of one (m 1) species has no diversity or richness and takes on a value of unity. As the number of species increases and as long as each species is relatively equal in number, the diversity index increases numerically. It reaches a maximum value when m N. EXAMPLE: Seven species of algae are present in a stream water sample. The numbers of seven species per mL are 32, 688, 138, 98, 1320, 424, and 248, respectively. Compute the diversity index of this algal community.

Solution:

Using formula (Eq. (2.122)) m

D 1.44 a (ni/N) ln(ni/N) i1

and construct a table of computations (Table 2.20).

Answer: D 2.15

18.2 Indicator Bacteria Pathogenic bacteria, pathogenic protozoan cysts, and viruses have been isolated from wastewaters and natural waters. The sources of these pathogens are the feces of humans and of wild and domestic animals. Identification and enumeration of these disease-causing organisms in water and wastewater are not recommended because no single technique is currently available to isolate and identify all the pathogens. In fact, concentrations of these pathogens are generally low in water and wastewater. In addition, the methods for identification and enumeration of pathogens are labor-intensive and expensive. Instead of direct isolation and enumeration of pathogens, total coliform (TC) has long been used as an indicator of pathogen contamination of a water that poses a public health risk. Fecal coliform (FC), which is more fecal-specific, has been adopted as a standard indicator of contamination in natural waters in Illinois, Indiana, and many other states. Both TC and FC are used in standards for drinking-water and natural waters. Fecal streptococcus (FS) is used as a pollution indicator in Europe. FC/FS ratios have been employed for identifying pollution sources in the United States. Fecal streptococci are present in the intestines of warm-blooded animals and of insects, and they are present in the environment (water, soil, and vegetation) for long periods of time. Escherichia coli bacteria have also been used as an indicator. TABLE 2.20 Values for Eq. (2.122) i 1 2 3 4 5 6 7

ni 32 688 138 98 1320 424 248 2948(N )

ni /N

1.44 ln (ni /N)

1.44(ni /N)ln (ni /N)

0.0109 0.2334 0.0468 0.0332 0.4478 0.1438 0.0841

6.513 2.095 4.409 4.902 1.157 2.792 3.565

0.071 0.489 0.206 0.163 0.518 0.402 0.300

1.0000

2.149

STREAMS AND RIVERS STREAMS AND RIVERS

1.155

Calculation of bacterial density. The determination of indicator bacteria, total coliform, fecal coliform, and fecal streptococcus or enterococcus is very important for assessing the quality of natural waters, drinking-waters, and wastewaters. The procedures for enumeration of these organisms are presented elsewhere (APHA et al., 1995). Examinations of indicator bacterial density in water and in wastewater are generally performed by using a series of four-decimal dilutions per sample, with 3 to 10, usually five, tubes for each dilution. Various special broths are used for the presumptive, confirmation, and complete tests for each of the bacteria TC, FC, and FS, at specified incubation temperatures and periods. MPN method. Coliform density is estimated in terms of the most probable number (MPN). The multiple-tube fermentation procedure is often called an MPN procedure. The MPN values for a variety of inoculation series are listed in Table 2.21. These values are based on a series of five tubes for three dilutions. MPN values are commonly determined using Table 2.21 and are expressed as MPN/100 mL. Only two significant figures are used for the reporting purpose. Other methods are based on five 20 mL or ten 10 mL of the water sample used. Table 2.21 presents MPN index for combinations of positive and negative results when five 10-mL, five 1-mL, and five 0.1-mL volumes of sample are inoculated. When the series of decimal dilution is

TABLE 2.21 MPN Index for Various Combinations of Positive Results when Five Tubes are Used per Dilution (10 mL, 1.0 mL, 0.1 mL) Combination of positives

MPN index 100 mL

Combination of positives

0-0-0 0-0-1 0-1-0 0-2-0

6.0

REFERENCES American Public Health Association (APHA), American Water Works Association (AWWA), and Water Environment Federation (WEF). 1995. Standard methods for the examination of water and wastewater, 19th edn. Washington, DC: American Public Health Association. American Society of Civil Engineering Committee on Sanitary Engineering Research. 1960. Solubility of atmospheric oxygen in water. J. Sanitary Eng Div. 86(7): 41–53. Barrett, M. J., Gameson, A. L. H. and Ogden, C. G. 1960. Aeration studies at four weir systems. Water and Water Eng. 64: 407.

STREAMS AND RIVERS STREAMS AND RIVERS

1.163

Black, W. M. and Phelps, E. B. 1911. Location of sewer outlets and discharge of sewage in New York Harbor. New York City Board of Estimate, March 23. Broeren, S. M., Butts, T. A. and Singh, K. P. 1991. Incorporation of dissolved oxygen in aquatic habitat assessment for the upper Sangamon River. Contract Report 513. Champaign: Illinois State Water Survey. Butts, T. A. 1963. Dissolved oxygen profile of Iowa River. Masters Thesis. Iowa City: University of Iowa. Butts, T. A. 1974. Measurement of sediment oxygen demand characteristics of the Upper Illinois Waterway. Report of Investigation 76 Urbana: Illinois State Water Survey. Butts, T. A. and Schnepper, D. H. 1967. Oxygen absorption in streams. Water and Sewage Works, 114(10): 385–386. Butts, T. A. and Shackleford, D. B. 1992. Reduction in peak flows and improvements in water quality in the Illinois Waterway downstream of Lockport due to implementation of Phase I and II of TARP, Vol. 2: Water Quality. Contract Report 526. Urbana: Illinois State Water Survey. Butts, T. A., Evans, R. L. and Stall, J. B. 1974. A waste allocation study of selected streams in Illinois. Contract Report prepared for Illinois EPA. Urbana: Illinois State Water Survey. Butts, T. A., Evans, R. L. and Lin, S. D. 1975. Water quality features of the Upper Illinois Waterway. Report of Investigation 79. Urbana: Illinois State Water Survey. Butts, T. A., Kothandaraman, V. and Evans, R. L. 1973. Practical considerations for assessing waste assimilative capacity of Illinois streams. Circular 110. Urbana: Illinois State Water Survey. Butts, T. A., Roseboom, D., Hill, T., Lin, S. D., Beuscher, D., Twait, R. and Evans, R. L. 1981. Water quality assessment and waste assimilative analysis of the LaGrange Poole, Illinois River. Contract Report 260. Urbana: Illinois State Water Survey. Butts, T. A., Schnepper, D. H. and Evans, R. L. 1970. Dissolved oxygen resources and waste assimilative capacity of a LaGrange pool, Illinois River. Report of Investigation 64. Urbana: Illinois State Water Survey. Churchill, M. A. and Buckingham, R. A. 1956. Statistical method for anlaysis of stream purification capacity. Sewage and Industrial Wastes 28(4): 517–537. Churchill, M. A., Elmore, R. L. and Buckingham, R. A. 1962. The prediction of stream reaeration rates. J. Sanitary Eng. Div. 88(7): 1–46. Dougal, M. D. and Baumann, E. R. 1967. Mathematical models for expressing the biochemical oxygen demand in water quality studies. Proc. 3rd Ann. American Water Resources Conference, pp. 242–253. Eckenfelder, W. and O’Conner, D. J. 1959. Stream analysis biooxidation of organic wastes—theory and design. Civil Engineering Department, Manhattan College, New York. Fair, G. M. 1936. The log-difference method of estimating the constants of the first stage biochemical oxygendemand curve. Sewage Works J. 8(3): 430. Fair, G. M., Moore, W. E. and Thomas, Jr., H. A. 1941a. The natural purification of river muds and pollutional sediments. Sewage Works J. 13(2): 270–307. Fair, G. M., Moore, W. E. and Thomas, Jr., H. A. 1941b. The natural purification of river muds and pollutional sediments. Sewage Works J. 13(4): 756–778. Gameson, A. L. H. 1957. Weirs and the aeration of rivers. J. Inst. Water Eng. 11(6): 477. Gameson, A. L. H., Vandyke, K. G. and Ogden, C. G. 1958. The effect of temperature on aeration at weirs. Water and Water Eng. 62: 489. Gannon, J. J. 1963. River BOD abnormalities. Final Report USPHS Grants RG-6905, WP-187(C1) and WP-187(C2). Ann Arbor, Michigan: University of Michigan School of Public Health. Gannon, J. J. and Downs, T. D. 1964. Professional Paper Department of Environmental Health. Ann Arbor, Michigan: University of Michigan. Geldreich, E. E. 1967. Fecal coliform concepts in stream pollution. Water and Sewage Works 114(11): R98–R110. Grindrod, J. 1962. British research on aeration at weirs. Water and Sewage Works 109(10): 395. Illinois Environmental Protection Agency. 1987. Quality Assurance and Field Method Manual. Springfield, Illinois: IEPA. Illinois Environmental Protection Agency. 1990. Title 35: Environmental Protection, Subtitle C: Water pollution, Chapter I: Pollution Control Board. Springfield, Illinois: IEPA. Kittrel, F. W. and Furfari, S. A. 1963. Observation of coliform bacteria in streams. J. Water Pollut. Control Fed. 35: 1361.

STREAMS AND RIVERS 1.164

CHAPTER 1.2

Langbein, W. B. and Durum, W. H. 1967. Aeration capacity of streams. Circular 542. Washington, DC: US Geological Survey, 67. Larson, R. S., Butts, T. A. and Singh, K. P. 1994. Water quality and habitat suitability assessment: Sangamon River between Decatur and Petersburg. Contract Report 571, Champaign: Illinois State Water Survey. Le Bosquet, M. and Tsivoglou, E. C. 1950. Simplified dissolved oxygen computations. Sewage and Industrial Wastes 22: 1054–1061. McDonnell, A. J. and Hall, S. D. 1969. Effect of environmental factors on benthal oxygen uptake. J. Water Pollut. Control Fed. Research Supplement, Part 2 41: R353–R363. Moore, E. W., Thomas, H. A. and Snow, W. B. 1950. Simplified method for analysis of BOD data. Sewage and Industrial Wastes 22(10): 1343. Nemerow, N. L. 1963. Theories and practices of industrial waste treatment. Reading, Massachusetts: AddisonWesley. O’Connor, D. J. 1958. The measurement and calculation of stream reaeration ratio. In: Oxygen relation in streams. Washington, DC: US Department of Health, Education, and Welfare, pp. 35–42. O’Connor, D. J. 1966. Stream and estuarine analysis. Summer Institute in Water Pollution Control, Training Manual. New York: Manhattan College. O’Connor, D. J. and Dobbins, W. E. 1958. The mechanics of reaeration in natural streams. Trans. Am. Soc. Civil Engineers 123: 641–666. Orford, H. E. and Ingram, W. T. 1953. Deoxygenation of sewage. Sewage and Industrial Waste 25(4): 419–434. Perry, R. H. 1959. Engineering manual, 2nd edn. New York: McGraw-Hill. Phelps, E. B. 1944. Stream sanitation. New York: John Wiley. Preul, H. C. and Holler, A. G. 1969. Reaeration through low days in the Ohio river. Proceedings of the 24th Purdue University Industrial Waste Conference. Lafayette, Indiana: Purdue University. Shannon, C. E. and Weaver, W. 1949. The mathematical theory of communication. Urbana: University of Illinois Press, 125 pp. Sheehy, J. P. 1960. Rapid methods for solving monomolecular equations. J. Water Pollut. Control Fed. 32(6): 646–652. Streeter, H. W. 1926. The rate of atmospheric reaeration of sewage polluted streams. Trans. Amer. Soc. Civil Eng. 89: 1351–1364. Streeter, H. W. and Phelps, E. B. 1925. A study of the pollution and natural purification of the Ohio River. Cincinatti: US Public Health Service, Bulletin No. 146. Thomas, H. A., Jr. 1937. The ‘slope’ method of evaluating the constants of the first-stage biochemical oxygen demand curve. Sewage Works J. 9(3): 425. Thomas, H. A. 1948. Pollution load capacity of streams. Water and Sewage Works 95(11): 409. Thomas, H. A., Jr. 1950. Graphical determination of BOD curve constants. Water and Sewage Works 97(3): 123–124. Tsivoglou, E. C. 1958. Oxygen relationships in streams. Technical Bulletin W-58-2. US Environmental Protection Agency. 1997. Technical guidance manual for developing total maximum daily loads, book 2: streams and rivers. Part 1: Biological oxygen demand/dissolved oxygen and nutrients/eutrophication. EPA 823-B-97-002. Washington, DC: US EPA. US Public Health Service. 1927. The oxygen demand of polluted waters. Public Health Bulletin No. 172. Washington, DC: US Public Health Service. Velz, C. J. 1958. Significance of organic sludge deposits. In: Taft, R. A. (ed.), Oxygen relationships in streams. Sanitary Engineering Center Technical Report W58-2. Cincinnati, Ohio: US Department of Health, Education, and Welfare. Velz, C. J. 1939. Deoxygenation and reoxygenation. Trans. Amer. Soc. Civil Eng. 104: 560–572. Wetzel, R. G. 1975. Limnology. Philadelphia, Pennsylvania: Saunders. Young, J. C. and Clark, J. W. 1965. Second order equation for BOD. J. Sanitary Eng. Div., ASCE 91(SA): 43–57. Zanoni, A. E. 1967. Effluent deoxygenation at different temperatures. Civil Engineering Department Report 100-5A. Milwaukee, Wisconsin: Marquette University. \

Source: HANDBOOK OF ENVIRONMENTAL ENGINEERING CALCULATIONS

CHAPTER 1.3

LAKES AND RESERVOIRS Shun Dar Lin

1 LAKES AND IMPOUNDMENT IMPAIRMENTS 1.165 2 LAKE MORPHOMETRY

1.166

3 WATER QUALITY MODELS

1.169

4 EVAPORATION 1.170 4.1 Water Budget Method 1.170 4.2 Energy Budget Method 1.171 5 THE CLEAN LAKES PROGRAM 5.1 Types of Funds 1.175 5.2 Eligibility for Financial Assistance 1.175

1.175

5.3 State Lake Classification Survey 1.175 5.4 Phase 1: Diagnostic/Feasibility Study 1.177 5.5 Phase 2: Implementation 1.180 5.6 Phase 3: Post-Restoration Monitoring 1.180 5.7 Watershed Protection Approach 1.180 5.8 In-lake Monitoring 1.181 5.9 Trophic State Index 1.188 5.10 Lake Use Support Analysis 1.191 5.11 Lake Budgets 1.197 5.12 Soil Loss Rate 1.201 REFERENCES

1.203

This chapter includes mainly lake morphometry, evaporation, and the Clean Lakes Program (CLP). Since most lake management programs in the United States are based on the CLP, the CLP is discussed in detail. Regulatory requirements and standardization of research and application are provided with a focus on the Phase 1, diagnostic/feasibility study.

LAKES AND IMPOUNDMENT IMPAIRMENTS Lakes are extremely complex systems whose conditions are a function of physical, chemical, and biological (the presence and predominance of the various plants and organisms found in the lake) factors. Lakes inherently function as traps or sinks for pollutants from tributaries watershed (drainage basins) or from precipitation and atmospheric deposits. Like streams, lakes are most often impaired by agricultural activities (main sources in the United States), hydrologic/habitat modification (stream channelization), and point pollution sources. These activities contribute to nutrient and sediment loads, suspended solids, and organic matter, and subsequently cause overgrowth of aquatic plants. The resulting decline in water quality limits recreation, impairs other beneficial uses, and shortens the expected life span of a lake. Common lake problems are eutrophication, siltation, shoreline erosion, algal bloom, bad taste and/or odor, excessive growth of aquatic vegetation, toxic chemicals, and bacterial contamination. Eutrophication, or aging, the process by which a lake becomes enriched with nutrients, is caused primarily by point and nonpoint pollution sources from human activities. Some man-made lakes

1.165

LAKES AND RESERVOIRS 1.166

CHAPTER 1.3

and impoundments may be eutrophic from their birth. These problems impact esthetic and practical uses of the lake. For example, the growth of planktonic algae in water-supply impoundments may cause tase and odor problems, shortened filter runs, increased chlorine demand, increased turbidity, and, for some facilities, increased trihalomethane precursors. The effects ultimately lead to increased water treatment costs and, in some instances, even to abandonment of the lake as a public watersupply source. Lakes and reservoirs are sensitive to pollution inputs because they flush out their contents relatively slowly. Even under natural conditions, lakes and reservoirs undergo eutrophication, an aging process caused by the inputs of organic matters and siltation.

LAKE MORPHOMETRY Lake morphometric data can be calculated from either a recent hydrographic map or a preimpoundment topographic map of the basin. In general, pre-impoundment maps may be obtainable from the design engineering firm or local health or environmental government agencies. If the map is too old and there is evidence of significant siltation, sections of the lake, such as the upper end and coves with inflowing streams, will need to be remapped. In some cases it is necessary to create a new lake map. The procedures include the following: 1. The outline of the shoreline is drawn either from aerial photographs or from United States Geological Survey (USGS) 7.5-minute topographic maps. 2. The water depth for transacts between known points on the shoreline of the lake is measured by a graphing sonar. 3. The sonar strip chart of the transacts is interpreted and drawn on to an enlarged copy of the lake outline. 4. Contours may be drawn by hand or with a computer on the map. 5. All maps are then either digitized or scanned and entered into a geographic information system (GIS). 6. Coverages are then converted into sea level elevations by locating the map on a 7.5-minute USGS topographic map and digitizing reference points both on the map and on the quadrangle for known sea level elevations. 7. Depths are assigned to the contour. The GIS gives the length of each contour and areas between adjacent contour lines. From this data, surface area, maximum depth of the lake, and shoreline length can be computed. Lake volume (V), shoreline development index (SDI) or shoreline configuration ratio, and mean depth ( D ) can be calculated by the formula in Wetzel (1975): 1. Volume n

h V a sAi Ai1 2Ai Ai1d 3 i0 where V volume, ft3, acre⋅ft, or m3 h depth of the stratum, ft or m i number of depth stratum Ai area at depth i, ft2, acre, or m2

(3.1)

LAKES AND RESERVOIRS LAKES AND RESERVOIRS

1.167

2. Shoreline development index SDI

L 2 2p A0

(3.2)

where L length of shoreline, miles or m A0 surface area of lake, acre, ft2, or m2 3. Mean depth D

V A0

(3.3)

where D mean depth, ft or m V volume of lake, ft3, acreft or m3 A0 surface area, ft2, acre or m2 Other morphometric information also can be calculated by the following formulas: 4. Bottom slope S

D Dm

(3.4)

where S bottom slope D mean depth, ft or m Dm maximum depth, ft or m 5. Volume development ratio, Vd (Cole, 1979) Vd 3

D Dm

(3.5)

6. Water retention time RT

storage capacity, acre # ft or m3 annual runoff, acre # ft/year or m3/year

(3.6)

where RT retention time, year 7. Ratio of drainage area to lake capacity R R

drainage area, acre or m2 storage capacity, acre # ft or m3

(3.7)

EXAMPLE: A reservoir has a shoreline length of 9.80 miles. Its surface area is 568 acres. Its maximum depth is 10.0 ft. The areas for each foot depth are 480, 422, 334, 276, 205, 143, 111, 79, 30, and 1 acres. Annual rainfall is 38.6 inches. The watershed drainage is 11,620 acres. Calculate morphometric data with the formulas described above.

LAKES AND RESERVOIRS 1.168

CHAPTER 1.3

Solution:

Compute the following parameters:

1. The volume of the lake 10

h V a (Ai Ai1 2Ai Ai1) i0 3

1 [(568 480 2568 480) (480 422 2480 422) 3 (422 324 2422 324) c (30 1 230 1)]

1 [7062] 3

2354 (acre # ft) 2,902,000 m3 2. Shoreline development index or shoreline configuration ratio A0 568 acres 568 acres SDI

1 sq miles 0.8875 sq miles 640 acres

L 2 2p A0 9.80 miles 2 23.14 0.8875 sq miles 9.80 3.34

2.93 3. Mean depth D

V A0

2354 acre # ft 568 acres 4.14 ft

4. Bottom slope S

4.13 ft D 0.41 Dm 10.0 ft

5. Volume development ratio

Vd 3

D Dm

3 0.41 1.23 (or 1.23:1)

LAKES AND RESERVOIRS LAKES AND RESERVOIRS

1.169

6. Water retention time Storage capacity V 2354 acre # ft Annual runoff 38.6 inches/year 11,620 acres 38.6 inches/year

1 ft 11,620 acres 12 inches

37,378 acre # ft/year RT

storage capacity annual runoff 2354 acre # ft 37,378 acre # ft/year

0.063 years 7. Ratio of drainage area to lake capacity R

drainage area storage capacity

11,620 acres 2354 acre # ft

4.94 acres 1 acre # ft

4.94 acre/acre # ft

WATER QUALITY MODELS Lakes and reservoirs are usually multipurpose, serving municipal and industrial water supplies, recreation, hydroelectric power, flood control, irrigation, drainage, and/or agriculture, due to the importance of protecting these natural resources. Water quality involves the physical, chemical, and biological integrity of water resources. Water quality standards promulgated by the regulatory agencies define the water quality goals for protection of water resources in watershed management. Modeling the water quality in lakes and reservoirs is very different from that in rivers (Chapter 1.2) or estuaries. A variety of models are based on some of physical, chemical, and biological parameters and/or combinations. Physical models deal with temperature, dissolved oxygen (DO), energy budget diffusion, mixing, vertical and horizontal aspects, seasonal cycles, and meteorological data setting. Chemical models involve mass balance and toxic substances. Biological models deal with nutrients and the food chain, biological growth, perdition, oxygen balance, and tropical conditions, etc. A detailed modeling of lake water quality is given in a book by Chapra and Reckhow (1983). Clark et al. (1977), Tchobanoglous and Schroeder (1985), and James (1993) also present modeling water quality in lakes and reservoirs. However, most of those models are not used for lake management practices. For construction of a new lake or reservoir, it is required that the owner submit the prediction of water quality in the future new lake to the US Army Corps of Engineers. For example, Borah et al. (1997) used the US Army Corps of Engineers’ HEC-5Q model (US Army Corps of Engineers, 1986, 1989) for a central Illinois lake and two proposed new lakes. The model was calibrated and verified on the existing water supply lake used monitored data in the draft years of 1986 and 1988. Eight

LAKES AND RESERVOIRS 1.170

CHAPTER 1.3

water quality constituents were simulated. Calibration, verification, and predictions were made only for water temperature, DO, nitrate nitrogen, and phosphate phosphorus. The calibrated and verified model was used to predict water surface elevations and constituent concentrations in the three lakes, individually and in combination, with meteorological and estimated flow data for a 20-month period from May 1953 through December 1954, the most severe drought of record. The HEC-5Q model provided a useful tool in the water quality evaluation of the three lakes. The HEC-5Q model was developed by the Hydraulic Engineering Center in Davis, California, for simulation of flood control and conservation systems with water quality analysis. The details of these models are beyond the scope of this book. The major portion of this chapter will cover evaporation and lake management programs in the Clean Lakes Program. Because these subjects are general, they are not usually included in environmental engineering textbooks.

EVAPORATION Evaporation converts water in its liquid or solid state into water vapor which mixes with the atmosphere. The rate of evaporation is controlled by the availability of energy at the evaporating surface, and the ease with which water vapor can diffuse into the atmosphere. Shuttleworth (1993) presented a detailed description with modeling of evaporation and transpiration from soil surfaces and crops. Modelings include fundamental and empirical equations. Measurements of evaporation and methods for estimating evaporation are also given. Evaporation of water from a lake surface uses energy provided by the sun to heat the water. The rate of evaporation is controlled primarily by water temperature, air temperature, and the level of moisture saturation in the air. Knowledge of evaporative processes is important in understanding how water losses through evaporation from a lake or reservoir are determined. Evaporation increases the storage requirement and decreases the yield of lakes and reservoirs. Determination of evaporation from a lake surface can be modeled by the water budget method, the mass transfer method, and the energy budget method (Robert and Stall, 1967). Many empirical equations have been proposed elsewhere (Fair et al., 1966, Linsley and Franzinni, 1964).

4.1

Water Budget Method The water budget method for lake evaporation depends on an accurate measurement of the inflow and outflow of the lake. The simple mathematical calculation shows that the change in storage equals the input minus output. It is expressed as ΔS P R GI GO E T O

(3.8)

where ΔS change in lake storage, mm (or in) P precipitation, mm R surface runoff or inflow, mm GI groundwater inflow, mm GO groundwater outflow, mm E evaporation, mm T transpiration, mm O surface water release, mm For the case of a lake with little vegetation and negligible groundwater inflow and outflow, lake evaporation can be estimated by E P R O ΔS

(3.9)

The water budget method has been used successfully to estimate lake evaporation in some areas.

LAKES AND RESERVOIRS LAKES AND RESERVOIRS

4.2

1.171

Energy Budget Method The principal elements in energy budget of the lake evaporation are shown in Fig. 3.1. The law of conservation of energy suggests that the total energy reaching the lake must be equal to the total energy leaving the lake plus the increase in internal energy of the lake. The energy budget in Fig. 3.1 can be expressed as (US Geological Survey, 1954) Qe Qs Qr Qb Qh Q Qv

(3.10)

where Qe energy available for evaporation Qs solar radiation energy Qr reflected solar radiation Qb net long-wave radiation Qh energy transferred from the lake to the atmosphere Q increase in energy stored in the lake Qv energy transferred into or from the lake bed The energy budget shown in Fig. 3.1 can be applied to evaporation from a class A pan (illustrated in Fig. 3.2). The pan energy budget (Fig. 3.2) contains the same elements as the lake budget (Fig. 3.1) except that heat is lost through the side and the bottom of the pan. Equation (3.10) is also applied to pan energy budget calculation. A pioneering and comprehensive research project centered on techniques for measuring evaporation was conducted at Lake Hefner, Oklahoma, in 1950–51 with the cooperation of five federal agencies (US Geological Survey, 1954). The energy budget data from a Weather Bureau class A evaporation pan was used to evaluate the study. Growing out of that project was the US Weather Bureau nomograph, a four-gradient diagram (Fig. 3.3) for general use for lake evaporation (Kohler et al., 1959). Subsequently, the formula for a mathematical solution of the nomograph was adapted

FIGURE 3.1 Elements of energy budget method for determining lake evaporation (Robert and Stall, 1967).

LAKES AND RESERVOIRS 1.172

CHAPTER 1.3

FIGURE 3.2

FIGURE 3.3

Energy budget method applied to class A pan (Robert and Stall, 1967).

Lake evaporation nomograph (Kohler et al., 1959).

LAKES AND RESERVOIRS LAKES AND RESERVOIRS

1.173

from computer use (Lamoreux, 1962). These procedures allow lake and pan evaporation to be computed from four items of climate data, i.e. air temperature, dew point temperature, wind movement, and solar radiation. Although the nomograph (Fig. 3.3) was developed for the use with daily values, it is also usable with monthly average values and gives reasonably reliable results with annual average values. The four-quadra diagram may be used to determine a lak evaporation rate is illustrated in Figure 3.3. For example, at the upper left quadrant, a horizontal line passed from the mean daily air temperature 81F to the first intersection at the solar radiation 600-langleys curve, and project toward the right to intersection of the 55F mean daily Dew-point curve. At the stop, a vertical line runs downward to the line of 120-mile per day wind movement, from where a horizontal line projects toward to the left to interset a vertical line from the first intersection. The average daily lake evaporation, 0.27 in, is read at this point. Since long-term climate data are mostly available, these techniques permit extended and refined evaporation determination. The following two equations for evaporation can be computed by computer (Lamoreux, 1962) as exemplified by the Weather Bureau computational procedures for the United States. For pan evaporation: Ep {exp[(Ta 212)(0.1024 0.01066 ln R)] 0.0001 0.025(es ea)0.88 (0.37 0.0041 Up)} {0.025 (Ta 398.36)2 4.7988 1010 exp[7482.6/(Ta 398.36)]}1

(3.11)

For lake evaporation, the expression is EL {exp[(Ta 212)(0.1024 0.01066 ln R)] 0.0001 0.0105(es ea)0.88 (0.37 0.0041Up)} {0.015 (Ta 398.36)2 6.8554 1010

(3.12)

exp[7482.6/(Ta 398.36)]}1 where Ep pan evaporation, in EL lake evaporation, in Ta air temperature, F ea vapor pressure, in of mercury at temperature Ta es vapor pressure, in of mercury at temperature Td Td dew point temperature, F R solar radiation, langleys per day Up wind movement, miles per day Pan evaporation is used widely to estimate lake evaporation in the United States. The lake evaporation EL is usually computed for yearly time periods. Its relationship with the pan evaporation Ep can be expressed as EL pc Ep

(3.13)

where pc is the pan coefficient. The pan coefficient on an annual basis has been reported as 0.65 to 0.82 by Kohler et al. (1955). Robert and Stall (1967) used Eqs. (3.9) and (3.10) for 17 stations in and near Illinois over the MayOctober season for a 16-year period to develop the general magnitudes and variability of lake evaporation in Illinois. Evaporation loss is a major factor in the net yield of a lake or reservoir. The net evaporation loss from a lake is the difference between a maximum expected gross lake evaporation and the minimum expected precipitation on the lake surface for various recurrence intervals and for critical periods having various durations. This approach assumes maximum evaporation and minimum precipitation would occur simultaneously. The following example illustrates the calculation of net draft rate for a drought recurrence interval of 40 years.

LAKES AND RESERVOIRS 1.174

CHAPTER 1.3

EXAMPLE: Calculate the net yield or net draft for 40-year recurrence interval for a reservoir (333 acres, 135 ha). Given: drainage area 15.7 miles2, unit RC 2.03 in/miles2, draft 1.83 MGD, E 31 in/acre.

Solution: Step 1. Compute reservoir capacity (RC) in million gallons (Mgal) RC 2.03 in 15.7 miles2 2.03 in (1 ft/12 in) 15.7 miles2 (640 acres/miles2) 1700 acreft 1700 acreft 43,560 ft2 acre 7.48 gal/ft3 554 Mgal 2,097,000 m3 Note: 1 Mgal 3785 m3 Step 2. Compute the total gross draft 16 months 365 days/12 16 486 days Draft 1.83 MGD 486 days 888 Mgal Step 3. Compute the inflow to the reservoir (total gross draft minus RC) Inflow draft RC 888 Mgal 554 Mgal 334 Mgal Step 4. Compute the evaporation loss E Effective evaporation surface area of the lake is 333 acres 0.65 216 acres E 31 in (1 ft/12 in) 216 acres 558 acreft 182 Mgal Step 5. Compute the net usable reservoir capacity (the total reservoir minus the evaporation loss) RC E 554 Mgal 182 Mgal 372 Mgal Step 6. Compute the total net draft which the reservoir can furnish (the net usable reservoir capacity plus the inflow) Net draft 372 Mgal 334 Mgal 706 Mgal Step 7. Compute the net draft rate, or the net yield, which the reservoir can furnish (the total net draft divided by total days in the critical period) Net draft rate 706 Mgal/486 days 1.45 MGD 5490 m3/d Note: 1 MGD 3785 m3/d

LAKES AND RESERVOIRS LAKES AND RESERVOIRS

1.175

THE CLEAN LAKES PROGRAM In the 1960s, many municipal wastewater treatment plants were constructed or upgraded (point source pollution control) in the United States. However, the nation’s water quality in lakes and reservoirs has not been improved due to nonpoint sources of pollution. In 1972, the US Congress amended Section 314 of the Federal Water Pollution Control Act (Public Law 92-500) to control the nation’s pollution sources and to restore its freshwater lakes. In 1977, the Clean Lakes Program (CLP) was established pursuant to Section 314 of the Clean Water Act. Under the CLP, publicly owned lakes can apply and receive financial assistance from the US Environmental Protection Agency (USEPA) to conduct diagnostic studies of the lakes and to develop feasible pollution control measures and water-quality enhancement techniques for lakes. This is the so-called Phase 1 diagnostic/feasibility (D/F) study. The objectives of the CLP are to (USEPA, 1980): 1. classify publicly owned freshwater lakes according to trophic conditions; 2. conduct diagnostic studies of specific publicly owned lakes, and develop feasible pollution control and restoration programs for them; and 3. implement lake restoration and pollution control projects.

5.1

Types of Funds The CLP operates through four types of financial assistance by cooperative agreements. They are State lake classification survey Phase 1—diagnostic/feasibility study Phase 2—implementation Phase 3—post-restoration monitoring Funding is typically provided yearly for the state lake classification survey, but on a long-term basis for Phases 1, 2, and 3. Phase 1 funds are used to investigate the existing or potential causes of decline in the water quality of a publicly owned lake; to evaluate possible solutions to existing or anticipated pollution problems; and to develop and recommend the most ‘feasible’ courses of action to restore or preserve the quality of the lake. Activities typically associated with sample collection, sample analyses, purchase of needed equipment, information gathering, and report development are eligible for reimbursem*nt. During Phase 1, total project costs cannot exceed $100,000. Fifty percent of this funding will come from the CLP, while the rest must come from non-federal or local sources.

5.2

Eligibility for Financial Assistance Only states are eligible for CLP financial assitance, and cooperative agreements will be awarded to state agencies. A state may in turn make funds available to a sub-state agency or agencies for all or any portion of a specific project. The D/F study is generally carried out by a contracted organization, such as research institutes or consulting engineers. For nearly two decades, 46 percent of the US lake acres were assessed for their water qualities and impairments (USEPA, 1994). Since the early 1990s, federal funds for CLP were terminated due to the budget cuts. Some states, such as Illinois, continued the CLP with state funds.

5.3

State Lake Classification Survey In general, most states use state lake classification survey funding to operate three types of lake survey activities. These include the volunteer lake monitoring program (VLMP), the ambient lake monitoring

LAKES AND RESERVOIRS 1.176

CHAPTER 1.3

program (ALMP), and lake water quality assessment (LWQA) grant. All three programs are partially supported by a Section 314 Federal CPL LWQA grant. State funds matched equally with federal grant funds are used to improve the quantity and quality of lake information reported in the annual 305(b) report to the US Congress. Volunteer lake monitoring program. The VLMP is a statewide cooperative program which volunteers to monitor lake conditions twice a month from May through October and transmit the collected data to the state agency (EPA or similar agency) for analysis and report preparation. In Illinois, the VLMP was initiated by Illinois Environmental Protection Agency (IEPA) in 1981 (IEPA, 1984). The volunteers can be personnel employed by the lake owner (a water treatment plant, Department of Conservation or other state agency, city, lake association, industry, etc.), or they can be local citizens. (The lakes monitored are not limited to publicly owned lakes.) Volunteers must have a boat and an anchor in order to perform the sampling. Volunteers receive a report prepared by the state EPA, which evaluates their sampling results. Volunteers measure Secchi disc transparencies and total depths at three sites in the lake. For reservoirs, site 1 is generally located at the deepest spot (near the dam); site 2 is at midlake or in a major arm; and site 3 is in the headwater, a major arm, or the tributary confluence. The data are recorded on standard forms. In addition, the volunteers also complete a field observation form each time the lake is sampled and record the number that best describes lake conditions during sampling for each site on the lake. Observations include color of the water, the amount of sediment suspended in the water, visible suspended algae, submerged or floating aquatic weeds, weeds near the shore, miscellaneous substances, odors, cloudiness, precipitation, waves, air temperature at the lake, lake water levels, recreational usage, and lake management done since the last sampling (IEPA, 1983). Ambient lake monitoring program. The ALMP is also a statewide regular water quality monitoring program. Water quality samples are collected and analyzed annually by IEPA personnel on selected lakes throughout the state. The major objectives of the ALMP are to (IEPA, 1992): • characterize and define trends in the condition of significant lakes in the state; • diagnose lake problems, determine causes/sources of problems, and provide a basis for identifying alternative solutions; • evaluate progress and success of pollution control/restoration programs; • judge effectiveness of applied protection/management measures and determine applicability and transferability to other lakes; • revise and update the lake classification system; • meet the requirements of Section 314 CLP regulation and/or grant agreements. The ALMP provides a much more comprehensive set of chemical analyses on the collected water samples than does the VLMP. In Illinois, the ALMP was initiated in 1977. Water samples are analyzed for temperature, dissolved oxygen, Secchi disc transparency, alkalinity, conductivity, turbidity, total and volatile suspended solids, dissolved and total phosphorus, and nitrogen (ammonia, nitrite/nitrate, and total kjeldahl). In addition, 9 metals and 11 organics analyses are performed on water and sediment samples. Lake water quality assessment. The LWQA grant is focused on non-routinely monitored lakes to gather basic data on in-lake water quality and sediment quality. This information increases the efficiency, effectiveness, and quality of the state EPA’s lake data management and 305(b) report. LWQA fieldwork involves two major tasks: namely, collection of lake assessment information and in-lake water and sediment sampling.

LAKES AND RESERVOIRS LAKES AND RESERVOIRS

1.177

Lake assessment information collected includes lake identification and location, morphometric data, public access, designated uses and impairments, lake and shoreline usages, watershed drainage area usage, water quality and problems, status of fisheries, causes and sources of impairment (if any), past lake protection and management techniques, and lake maps. In general, only one sampling trip per lake is made in order to evaluate a maximum number of lakes with the limited resources. The water and sediment samples are collected in summer by contracted agencies and analyzed by the state EPA laboratory. Chemical analyses for water and sediment samples are the same as in the ALMP. 5.4

Phase 1: Diagnostic/Feasibility Study A diagnostic/feasibility study is a two-part study to determine a lake’s current condition and to develop plans for its restoration and management. Protocol for the D/F study of a lake (Lin, 1994) is a digest from the CLP Guidance Manual (USEPA, 1980). For a Phase 1 D/F study, the following activities should be carried out and completed within three years (USEPA, 1980): 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

development of a detailed work plan; study of the natural characteristics of the lake and watershed; study of social, economic, and recreational characteristics of the lake and watershed; lake monitoring; watershed monitoring; data analysis; development and evaluation of restoration alternatives; selection and further development of watershed management plans; projection of benefits; environmental evaluation; public participation; public hearings (when appropriate); report production.

Activities 1 through 6 are part of the diagnostic study; the other activities pertain to the feasibility study. Diagnostic study. The diagnostic study is devoted to data gathering and analysis. It involves collecting sufficient limnological, morphological, demographic, and socioeconomic information about the lake and its watershed. The study of lake and watershed natural characteristics, and most of the social, economic, and recreational information of the lake region can often be accomplished by obtaining and analyzing secondary data; i.e. data already available from other sources. Baseline limnological data include a review of historical data and 1 year of current limnological data. Lake monitoring is expensive. Monthly and bimonthly samples are required for assessing physical and chemical characteristics. At least three sites are chosen for each lake, as mentioned in the discussion of the ALMP (Sec. 5.3). In addition, biological parameters, such as chlorophylls, phytoplankton, zooplankton, aquatic macrophytes, indicator bacteria, benthic macroinvertebrate, and fish surveys, must be assessed at different specified frequencies. At least one surficial and/or core sediment sample and water sample at each site must be collected for heavy metals and organic compounds analyses. Water quality analyses of tributaries and ground water may also be included, depending on each lake’s situation.

LAKES AND RESERVOIRS 1.178

CHAPTER 1.3

Watershed monitoring is done by land-use stream monitoring manually or automatically to determine the nutrient, sediment, and hydraulic budget for a lake. Primary and secondary data collected during the diagnostic study are analyzed to provide the basic information for the feasibility portion of the Phase 1 study. Data analysis involves: 1. 2. 3. 4. 5. 6. 7.

inventory of point source pollutant discharges; watershed land use and nonpoint nutrient/solids loading; analyses of lake data—water and sediment quality; analyses of stream and ground-water data; calculating the hydrologic budget of the watershed and lake; calculating the nutrient budget of the lake; assessing biological resources and ecological relationships—lake fauna, terrestrial vegetation, and animal life; 8. determining the loading reductions necessary to achieve water quality objectives. The following analyses of lake data are typically required: (1) identification of the limiting nutrient based on the ratio of total nitrogen to total phosphorus; (2) determination of the trophic state index based on total phosphorus, chlorophyll a, Secchi depth, and primary productivity; (3) calculation of the fecal coliform to fecal streptococcus ratios to identify causes of pollution in the watershed; and (4) evaluation of the sediment quality for the purpose of dredging operations. A description of the biological resources and ecological relationships is included in the diagnostic study based on information gathered from secondary sources. This description generally covers lake flora (terrestrial vegetation, such as forest, prairie, and marsh) and fauna (mammals, reptiles and amphibians, and birds). Feasibility study. The feasibility study involves developing alternative management programs based on the results of the diagnostic study. First, existing lake quality problems and their causes should be identified and analyzed. Problems include turbid water, eroding shorelines, sedimentation and shallow water depths, low water levels, low dissolved oxygen levels, excessive nutrients, algal bloom, unbalanced aquatic vegetative growth (excessive growth of macrophytes), non-native exotic species, degraded or unbalanced fishery and aquatic communities, bad taste and odor, acidity, toxic chemicals, agricultural runoff, bacterial contamination, poor lake esthetics, user conflicts, and negative human impacts. Once the lake problems have been defined, a preliminary list of corrective alternatives needs to be established and discussed. Commonly adopted pollution control and lake restoration techniques can be found in the literature (USEPA, 1980, 1990). The corrective measures include in-lake restoration and/or best management practices in the watershed. Each alternative must be evaluated in terms of cost; reliability; technical feasibility; energy consumption; and social, economic, and environmental impacts. For each feasible alternative, a detailed description of measures to be taken, a quantitative analysis of pollution control effectiveness, and expected lake water quality improvement must be provided. There are usually many good methods of achieving specific objectives or benefits. The Phase 1 report should document the various alternatives by describing their relative strengths and weaknesses and showing how the one chosen is superior. Final selection should be discussed in a public meeting. Once an alternative has been selected, work can proceed on developing other program details required as part of Phase 1. These include developing the Phase 2 monitoring program, schedule and budget, and source of non-federal funds; defining the relationship of Phase 2 plans to other programs; developing an operation and maintenance plan; and obtaining required permits. In other words, the feasibility study portion of the Phase 1 report actually constitutes the proposal for Phase 2 study. The report also must include the projection of the project benefits, environmental evaluations, and public participation.

LAKES AND RESERVOIRS LAKES AND RESERVOIRS

1.179

Final report. The final report will have to follow strictly the format and protocol stipulated in the Clean Lakes Program Guidance Manual. Requirements for diagnostic feasibility studies and environmental evaluations are at a minimum, but not limited, as follows: 1. Lake identification and location 2. Geologic and soils description of drainage basins a. Geologic description b. Ground-water hydrology c. Topography d. Soils 3. Description of public access 4. Description of size and economic structure of potential user population 5. Summary of historical lake uses 6. Population segments adversely affected by lake degradation 7. Comparison of lake uses to uses of other lakes in the region 8. Inventory of point-source pollution discharges 9. Land uses and nonpoint pollutant loadings 10. Baseline and current limnological data a. Summary analysis and discussion of historical baseline limnological data b. Presentation, analysis, and discussion of one year of current limnological data c. Tropic condition of lake d. Limiting algal nutrient e. Hydraulic budget for lake f. Nutrient sediment budget 11. Biological resources and ecological relationships 12. Pollution control and restoration procedures 13. Benefits expected from restoration 14. Phase II monitoring program 15. Schedule and budget 16. Source of matching funds 17. Relationship to other pollution control programs 18. Public participation summary 19. Operation and maintenance plan 20. Copies of permits or pending applications 21. Environmental evaluation a. Displacement of people b. Defacement of residential area c. Changes in land use patterns d. Impacts on prime agricultural land e. Impacts on parkland, other public land and scenic resources f. Impacts on historic, architectural, archaeological, or cultural resources g. Long range increases in energy demand h. Changes in ambient air quality or noise levels i. Adverse effects of chemical treatments j. Compliance with executive order 11988 on flood plain management k. Dredging and other channel bed or shoreline modifications l. Adverse effects on wetlands and related resources m. Feasible alternatives to proposed project n. Other necessary mitigative measures requirements

LAKES AND RESERVOIRS 1.180

5.5

CHAPTER 1.3

Phase 2: Implementation A Phase 1 D/F study identifies the major problems in a lake and recommends a management plan. Under the CLP, if funds are available, the plan can then be implemented and intensive monitoring of the lake and tributaries conducted either by the state EPA or a contracted agency. The physical, chemical, and biological parameters monitored and the frequencies of monitoring are similar to that for Phase 1 D/F study (or more specific, depending on the objectives). One year of water quality monitoring is required for post-implementation study. Non-federal cost-share (50%) funds are usually provided by state and local landowners. A comprehensive monitoring program should be conducted to obtain post-implementation data for comparison with pre-implementation (Phase 1) data. The implementation program may include keeping point- and nonpoint-source pollutants from entering a lake, implementing in-lake restoration measures to improve lake water quality, a monitoring program, environmental impact and cost evaluation, and public participation.

5.6

Phase 3: Post-Restoration Monitoring The CLP has recently begun setting up a system for managing and evaluating lake project data, designed to derive benefit from past projects. Quantitative scientific data on long-term project effectiveness will become increasingly available through post-restoration monitoring studies being conducted under Phase 3 CLP grants instituted in 1989. Several years after Phase 2 implementation, a lake may be selected for Phase 3 post-restoration monitoring with matching funds. For Phase 3 monitoring, the water quality parameters monitored and the frequency of sampling in the lake and tributaries are similar to that for Phases 1 and 2. However, the period of monitoring for Phase 3 is generally 3 years with a total project period of 4 to 5 years. The purpose is to build upon an extensive database of information gathered under Phases 1 and 2 investigations. Data gathered from Phases 1 are compared with Phases 2 and 3 databases to determine the longterm effectiveness of watershed protection measures and in-lake management techniques implemented during and since Phase 2 completion.

5.7

Watershed Protection Approach The watershed protection approach (WPA) is an integrated strategy for more effective restoration and protection of aquatic ecosystems and human health; i.e. drinking water supplies and fish consumption (USEPA, 1993). The WPA focuses on hydrologically defined drainage basins (‘watersheds’) rather than on areas arbitrarily defined by political boundaries. Local decisions on the scale of geographic units consider many factors including the ecological structure of the basin, the hydrologic factors of groundwaters, the economic lake uses, the type and scope of pollution problems, and the level of resources available for protection and restoration projects (USEPA, 1991, 1993). The WPA has three major principles (USEPA, 1991, 1993): 1. problem identification—identify the primary risk to human health and ecosystem within the watershed; 2. stakeholder involvement—involve all parties most likely to be concerned or most able to take action for solution; 3. integrated actions—take corrective actions in a manner that provides solutions and evaluates results. The WPA is not a new program, and any watershed planning is not mandated by federal law. It is a flexible framework to achieve maximum efficiency and effect by collaborative activities. Everyone—individual citizens, the public and private sectors—can benefit from a WPA. The USEPA’s goal for the WPA is to maintain and improve the health and integrity of aquatic ecosystems using comprehensive approaches that focus resources on the major problems facing these systems within the watershed (USEPA, 1993).

LAKES AND RESERVOIRS LAKES AND RESERVOIRS

1.181

For more than two decades, the CLP has emphasized using the watershed protection approach. A long-standing program policy gives greater consideration to applicants who propose restoration and protection techniques that control pollutants at the source through watershed-wide management rather than dealing with symptoms in the lake. 5.8

In-lake Monitoring In Phase 1, 2, and 3 studies, in-lake water quality parameters that need to be monitored generally include water temperature, dissolved oxygen, turbidity, Secchi transparency, solids (total suspended and volatile), conductivity, pH, alkalinity, nitrogen (ammonia, nitrite/nitrate, and total kjeldahl), phosphorus (total and dissolved), total and fecal coliforms, fecal streptococci, algae, macrophytes, and macroinvertebrates. Except for the last two parameters, samples are collected bimonthly (April–October) and monthly (other months) for at least 1 year. At a minimum, three stations (at the deepest, the middle, and upper locations of the lake) are generally monitored (Lin et al., 1996, 1998). Lake water quality standards and criteria. The comprehensive information and data collected during Phase 1, 2, or 3 study are evaluated with state water quality standards and state lake assessment criteria. Most states in the United States have similar state standards and assessment criteria. For example, Illinois generally use the water quality standards applied to lake water listed in Table 3.1; and Illinois EPA’s lake assessment criteria are presented in Table 3.2. The data obtained from Phase 1, 2, and 3 of the study can be plotted from temporal variations for each water quality parameter, as shown in the example given in Fig. 3.4. Similarly, the historical data also can be plotted in the same manner for comparison purposes. Indicator bacteria (total and fecal coliforms, and fecal streptococcus) densities in lakes and reservoirs should be evaluated in the same manner as that given for rivers and streams in Chapter 1.2. The density and moving geometric mean of FC are usually stipulated by the state’s bacterial quality standards. The ratio of FC/FS is used to determine the source of pollution. Temperature and dissolved oxygen. Water temperature and dissolved oxygen in lake waters are usually measured at 1 ft (30 cm) or 2 ft (60 cm) intervals. The obtained data is used to calculate values of the percent DO saturation using Eq. (2.4) or from Table 2.1. The plotted data is shown in Fig. 3.5: observed vertical DO and temperature profiles on selected dates at a station. Data from long-term observations can also be depicted in Fig. 3.6 which shows the DO isopleths and isothermal plots for a near-bottom (2 ft above the bottom station).

TABLE 3.1 Illinois General Use Water Quality Standards 5 mg/L at any time 6 mg/L at 16 h/24 h Temperature: 30C Total dissolved solids: 100 mg/L (conductivity: 1700 –1/cm) Chloride: 500 mg/L for protection of aquatic life pH: 6.5–9.0, except for natural causes (alkalinity in Illinois lakes 20–200 mg/L as CaCO3) NH3-N:

1.5 mg/L at 20C (68F) and pH of 8.0

15 mg/L under no conditions NO3-N:

10 mg/L in all waters

1.0 mg/L in public water supply Total P:

0.05 mg/L in any lake 8.1 ha (20 acres)

1. Dissolved oxygen: 2. 3. 4. 5. 6. 7. 8.

Source: IEPA, 1987.

LAKES AND RESERVOIRS TABLE 3.2 Illinois EPA Lake Assessment Criteria Parameter

Minimal

Secchi depth TSS, mg/L Turbidity, NTU COD, mg/L

Slight

Moderate

High

Indication

49–79 5–15 3–7 10–20

18–48 15–25 7–15 20–30

18 25 15 30

lake use impairment lake use impairment suspended sediment organic enrichment

Source: IEPA, 1987.

FIGURE 3.4

1.182

Temporal variations of surface water characteristics at RHA-2 (Lin et al., 1996).

LAKES AND RESERVOIRS LAKES AND RESERVOIRS

FIGURE 3.5

Temperature and dissolved oxygen profiles at station 2 (Lin and Raman, 1997).

1.183

FIGURE 3.6

Isothermal and iso-dissolved oxygen plots for the deep stations (Lin et al., 1996).

LAKES AND RESERVOIRS

1.184

LAKES AND RESERVOIRS LAKES AND RESERVOIRS

1.185

Algae. Most algae are microscopic, free-floating plants. Some are filamentous and attached ones. Algae are generally classified into four major types: blue-greens, greens, diatoms, and flagellates. Through their photosynthesis processes, algae use energy from sunlight and carbon dioxide from bicarbonate sources, converting it to organic matter and oxygen (Fig. 3.7). The removal of carbon dioxide from the water results in an increase in pH and a decrease in alkalinity. Algae are important sources of dissolved oxygen in water. Phytoplanktonic algae form at the base of the aquatic food web and provide the primary source of food for fish and other aquatic insects and animals. Excessive growth (blooms) of algae may cause problems such as bad taste and odor, increased color and turbidity, decreased filter run at a water treatment plant, unsightly surface sc*ms and esthetic problems, and even oxygen depletion after die-off. Blue-green algae tend to cause the worst problems. To prevent such proliferation (upset the ecological balance), lake and watershed managers often try to reduce the amounts of nutrients entering lake waters, which act like fertilizers in promoting algal growth. Copper sulfate is frequently used for algal control, and sometimes it is applied on a routine basis during summer months, whether or not it is actually needed. One should try to identify the causes of the particular problems then take corrective measures selectively rather than on a shot-in-the-dark basis. Copper sulfate (CuSO4 ⋅ 5H2O) was applied in an Illinois impoundment at an average rate of 22 pounds per acre (lb/a); in some lakes, as much as 80 lb/a was used (Illinois State Water Survey, 1989). Frequently, much more copper sulfate is applied than necessary. Researchers have shown that 5.4 pounds of copper sulfate per acre of lake surface (6.0 kg/ha) is sufficient to control problemcausing blue-green algae in waters with high alkalinity (> 40 mg/L as CaCO3). The amount is equivalent to the rate of 1 mg/L of copper sulfate for the top 2 ft (60 cm) of the lake surface (Illinois State Water Survey, 1989). The literature suggests that a concentration of 0.050.10 mg/L as Cu2 is effective in controlling blue-green algae in pure cultures under laboratory conditions.

FIGURE 3.7

Photosynthesis and its chemical processes (Illinois State Water Survey, 1989).

LAKES AND RESERVOIRS 1.186

CHAPTER 1.3

EXAMPLE 1:

What is the equivalent concentration of Cu2+ of 1 mg/L of copper sulfate in water?

Solution: MW of CuSO4 # 5H2O 63.5 32 16 4 5(2 16) 249.5 2

MW of Cu

63.5

Cu2, mg/L 63.5 0.255 CuSO4 # 5H2O, mg/L 249.5 Cu2 0.255 1 mg/L 0.255 mg/L Since 0.050.10 mg/L of Cu2+ is needed to control blue-green algae, what is the theoretical concentration expressed as CuSO45H2O? EXAMPLE 2:

Solution: Copper sulfate 0.05 mg/L

249.5 63.5

> 0.20 mg/L Answer: This is equivalent to 0.200.4 mg/L as CuSO45H2O. Note: for field application, however, a concentration of 1.0 mg/L as CuSO45H2O is generally suggested. For a lake with 50 acres (20.2 ha), 270 lb (122 kg) of copper sulfate is applied. Compute the application rate in mg/L on the basis of the top 2 ft of lake surface.

EXAMPLE 3:

Solution: Step 1. Compute the volume (V) of the top 2 ft V 2 ft 50 acre 2 ft 0.3048 m/ft 50 acre 4047 m2/acre 123.350 m3 123.3 106 L Step 2. Convert weight (W) in pounds to milligrams W 270 lb 453,600 mg/lb 122.5 106 mg Step 3. Compute copper sulfate application rate Rate

122.5 106 mg W V 123.3 106 L

0.994 mg/L > 1.0 mg/L

LAKES AND RESERVOIRS LAKES AND RESERVOIRS

1.187

Secchi disc transparency. The Secchi disc is named after its Italian inventor Pietro Angelo Secchi, and is a black and white round plate used to measure water clarity. Values of Secchi disc transparency are used to classify a lake’s trophic state. Secchi disc visibility is a measure of a lake’s water transparency, which suggests the depth of light penetration into a body of water (its ability to allow sunlight to penetrate). Even though Secchi disc transparency is not an actual quantitative indication of light transmission, it provides an index for comparing similar bodies of water or the same body of water at different times. Since changes in water color and turbidity in deep lakes are generally caused by aquatic flora and fauna, transparency is related to these entities. The euphotic zone or region of a lake where enough sunlight penetrates to allow photosynthetic production of oxygen by algae and aquatic plants is taken as two to three times the Secchi disc depth (USEPA, 1980). Suspended algae, microscopic aquatic animals, suspended matter (silt, clay, and organic matter), and water color are factors that interfere with light penetration into the water column and reduce Secchi disc transparency. Combined with other field observations, Secchi disc readings may furnish information on (1) suitable habitat for fish and other aquatic life; (2) the lake’s water quality and esthetics; (3) the state of the lake’s nutrient enrichment; and (4) problems with and potential solutions for the lake’s water quality and recreational use impairment. Phosphorus. The term total phosphorus (TP) represents all forms of phosphorus in water, both particulate and dissolved forms, and includes three chemical types: reactive, acid-hydrolyzed, and organic. Dissolved phosphorus (DP) is the soluble form of TP (filterable through a 0.45 m filter). Phosphorus as phosphate may occur in surface water or groundwater as a result of leaching from minerals or ores, natural processes of degradation, or agricultural drainage. Phosphorus is an essential nutrient for plant and animal growth and, like nitrogen, it passes through cycles of decomposition and photosynthesis. Because phosphorus is essential to the plant growth process, it has become the focus of attention in the entire eutrophication issue. With phosphorus being singled out as probably the most limiting nutrient and the one most easily controlled by removal techniques, various facets of phosphorus chemistry and biology have been extensively studied in the natural environment. Any condition which approaches or exceeds the limits of tolerance is said to be a limiting condition or a limiting factor. In any ecosystem, the two aspects of interest for phosphorus dynamics are phosphorus concentration and phophorus flux (concentration flow rate) as functions of time and distance. The concentration alone indicates the possible limitation that this nutrient can place on vegetative growth in the water. Phosphorus flux is a measure of the phosphorus transport rate at any point in flowing water. Unlike nitrate-nitrogen, phosphorus applied to the land as a fertilizer is held tightly to the soil. Most of the phosphorus carried into streams and lakes from runoff over cropland will be in the particulate form adsorbed to soil particles. On the other hand, the major portion of phosphate-phosphorus emitted from municipal sewer systems is in a dissolved form. This is also true of phosphorus generated from anaerobic degradation of organic matter in the lake bottom. Consequently, the form of phosphorus, namely particulate or dissolved, is indicative of its source to a certain extent. Other sources of dissolved phosphorus in the lake water may include the decomposition of aquatic plants and animals. Dissolved phosphorus is readily available for algae and macrophyte growth. However, the DP concentration can vary widely over short periods of time as plants take up and release this nutrient. Therefore, TP in lake water is the more commonly used indicator of a lake’s nutrient status. From his experience with Wisconsin lakes, Sawyer (1952) concluded that aquatic blooms are likely to develop in lakes during summer months when concentrations of inorganic nitrogen and inorganic phosphorus exceed 0.3 and 0.01 mg/L, respectively. These critical levels for nitrogen and phosphorus concentrations have been accepted and widely quoted in scientific literature. To prevent biological nuisance, the IEPA (1990) stipulates, ‘Phosphorus as P shall not exceed a concentration of 0.05 mg/L in any reservoir or lake with a surface area of 8.1 hectares (20 acres) or more or in any stream at the point where it enters any reservoir or lake.’ Chlorophyll. All green plants contain chlorophyll a, which constitutes approximately 1 to 2 percent of the dry weight of planktonic algae (APHA et al., 1992). Other pigments that occur in phytoplankton include chlorophyll b and c, xanthophylls, phycobilius, and carotenes. The important

LAKES AND RESERVOIRS 1.188

CHAPTER 1.3

chlorophyll degration products in water are the chlorophyllides, pheophorbides, and pheophytines. The concentration of photosynthetic pigments is used extensively to estimate phytoplanktonic biomass. The presence or absence of the various photosynthetic pigments is used, among other features, to identify the major algal groups present in the water body. Chlorophyll a is a primary photosynthetic pigment in all oxygen-evolving photosynthetic organisms. Extraction and quantification of chlorophyll a can be used to estimate biomass or the standing crop of planktonic algae present in a body of water. Other algae pigments, particularly chlorophyll b and c, can give information on the type of algae present. Blue-green algae (Cyanophyta) contain only chlorophyll a, while both the green algae (Chlorophyta) and the euglenoids (Euglenophyta) contain chlorophyll a and c. Chlorophyll a and c are also present in the diatoms, yellow-green and yellow-brown algae (Chrysophyta), as well as dinoflagellates (Pyrrhophyta). These accessory pigments can be used to identify the types of algae present in a lake. Pheophytin a results from the breakdown of chlorophyll a, and a large amount indicates a stressed algal population or a recent algal die-off. Because direct microscopic examination of water samples is used to identify and enumerate the type and concentrations of algae present in the water samples, the indirect method (chlorophyll analyses) of making such assessments may not be employed. Nutrient in lake water will impact the aquatic community in the surface water during summer. Dillon and Rigler (1974) used data from North American lakes to drive the relationship between spring phosphorus and chlorophyll a concentration in summer as follows: CHL 0.0731 (TP)1.449

(3.14)

where CHL summer average chlorophyll a concentration at lake surface water, mg/m TP spring average total phosophorus concentration at lake surface water, mg/m3 3

EXAMPLE: Estimate the average chlorophyll a concentration in a North American lake during the summer if the average spring total phosphorus is 0.108 mg/L.

Solution: TP 0.108 mg/L 0.108 mg/L 1000 L/m3 108 mg/m3 Estimate summer CHL using Eq. (3.14) CHL 0.0731 (TP)1.449 0.0731 (108)1.449 mg/m3 64.6 mg/m3 5.9

Trophic State Index Eutrophication is a normal process that affects every body of water from its time of formation (Walker, 1981a, 1981b). As a lake ages, the degree of enrichment from nutrient materials increases. In general, the lake traps a portion of the nutrients originating in the surrounding drainage basin. Precipitation, dry fallout, and groundwater inflow are the other contributing sources. A wide variety of indices of lake trophic conditions have been proposed in the literature. These indices have been based on Secchi disc transparency; nutrient concentrations; hypolimnetic oxygen depletion; and biological parameters, including chlorophyll a, species abundance, and diversity. In its Clean Lake Program Guidance Manual, the USEPA (1980) suggests the use of four parameters as trophic indicators: Secchi disc transparency, chlorophyll a, surface water total phosphorus, and total organic carbon. In addition, the lake trophic state index (TSI) developed by Carlson (1977) on the basis of Secchi disc transparency, chlorophyll a, and surface water total phosphorus can be used to calculate a lake’s

LAKES AND RESERVOIRS LAKES AND RESERVOIRS

1.189

trophic state. The TSI can be calculated from Secchi disc transparency (SD) in meters (m), chlorophyll a (CHL) in micrograms per liter (g/L), and total phosphorus (TP) in g/L as follows: on the basis of SD, TSI 60 14.4 ln (SD) on the basis of CHL, TSI 9.81 ln (CHL) 30.6 on the basis of TP, TSI 14.42 ln (TP) 4.15

(3.15) (3.16) (3.17)

The index is based on the amount of algal biomass in surface water, using a scale of 0 to 100. Each increment of 10 in the TSI represents a theoretical doubling of biomass in the lake. The advantages and disadvantages of using the TSI were discussed by Hudson et al. (1992). The accuracy of Carlson’s index is often diminished by water coloration or suspended solids other than algae. Applying TSI classification to lakes that are dominated by rooted aquatic plants may indicate less eutrophication than actually exists. Lakes are generally classified by limnologists into one of three trophic states: oligotrophic, mesotrophic, or eutrophic (Table 3.4). Oligotrophic lakes are known for their clean and cold waters and lack of aquatic weeds or algae, due to low nutrient levels. There are few oligotrophic lakes in the Midwest. At the other extreme, eutrophic lakes are high in nutrient levels and are likely to be very productive in terms of weed growth and algal blooms. Eutrophic lakes can support large fish populations, but the fish tend to be rougher species that can better tolerate depleted levels of DO. Mesotrophic lakes are in an intermediate stage between oligotrophic and eutrophic. The great majority of Midwestern lakes are eutrophic. A hypereutrophic lake is one that has undergone extreme eutrophication to the point of having developed undesirable esthetic qualities (e.g. odors, algal mats, and fish kills) and water-use limitations (e.g. extremely dense growths of vegetation). The natural aging process causes all lakes to progress to the eutrophic condition over time, but this eutrophication process can be accelerated by certain land uses in the contributing watershed, streams, and from unprotected areas (e.g. agricultural activities, application of lawn fertilizers, and erosion from construction sites, streams, and from unprotected areas). Given enough time, a lake will grow shallower and will eventually fill in with trapped sediments and decayed organic matter, such that it becomes a shallow marsh or emergent wetland. EXAMPLE 1: Lake monitoring data shows that the Secchi disc transparency is 77 inches; total phosphorus, 31 g/L; and chlorophyll a 3.4 g/L. Calculate TSI values using Eqs. (3.15)(3.17).

Solution: Step 1. Using Eq. (3.15) SD 77 in 2.54 cm/in 0.01 m/cm 1.956 m TSI 60 14.4 ln (SD) 60 14.4 ln (1.956) 50.3 Step 2. Using Eq. (3.16) TSI 9.81 ln (CHL) 30.6 9.81 ln (3.4) 30.6 42.6 Step 3. Using Eq. (3.17) TSI 14.42 ln (TP) 4.15 14.42 ln (31) 4.15 53.7 EXAMPLE 2: One-year lake monitoring data for a Secchi disc transparency, total phosphorus, and chlorophyll a in a southern Illinois lake (at the deepest station) are listed in Table 3.3. Determine the trophic condition of the lake.

LAKES AND RESERVOIRS 1.190

CHAPTER 1.3

TABLE 3.3 Trophic State Index and Trophic State of an Illinois Lake Secchi disc trans Date 10/26/95 11/20/95 12/12/95 1/8/96 2/14/96 3/11/96 4/15/96 5/21/96 6/6/96 6/18/96 7/1/96 7/16/96 8/6/96 8/20/96 9/6/96 9/23/96 10/23/96

Total phosphorus

Chlorophyll a

TSI

g/L

TSI

g/L

TSI

77 104 133 98 120 48 43 52 51 60 60 42 72 48 54 69 42

50.3 46.0 42.5 46.9 44.0 57.1 58.7 56.0 56.3 53.9 53.9 59.1 51.3 57.1 55.4 51.9 59.1

31 33 21 29 26 22 49 37 36 30 21 38 19 20 32 22 42

53.7 54.6 48.1 52.7 51.1 48.7 60.3 56.2 55.8 53.2 48.1 56.6 46.6 47.3 54.1 48.9 58.0

3.4 4.7 3

42.6 45.8 41.4

2.1 6.2 10.4 12.4 7.1 14.6 7.7 12.4 6 1.7 9.8 11.3

37.9 48.5 53.6 55.3 49.8 56.9 50.6 55.3 48.2 35.8 53.0 54.4

Mean Trophic state

52.9 eutrophic

Overall mean Overall trophic state

51.4 eutrophic

52.6 eutrophic

48.6 mesotrophic

Source: Bogner et al. (1997).

Solution: Step 1. Calculate TSI values: As example 1, the TSI values for the lake are calculated using Eqs. (3.15)(3.17) based on SD, TP, and chlorophyll a concentration of each sample. The TSI values are included in Table 3.3. Step 2. Determine trophic state: a. b. c. d.

Calculate average TSI based on each parameter. Classify trophic state based on the average TSI and the criteria listed in Table 3.4. Calculate overall mean TSI for the lake. Determine overall trophic state. For this example, the lake is classified as eutrophic. The classifications are slightly different if based on each of the three water quality parameters.

TABLE 3.4 Quantitative Definition of a Lake Trophic State Secchi disc transparency Trophic state Oligotrophic Mesotrophic Eutrophic Hypereutrophic

Chlorophyll a (g/L)

157 79–157 20–79

4.0 2.0–4.0 0.5–2.0

0.5

2.6 2.6–7.2 7.2–55.5 55.5

Total phosphorus, lake surface (g/L)

TSI

12 12–24 24–96 96

40 40–50 50–70 70

LAKES AND RESERVOIRS LAKES AND RESERVOIRS

5.10

1.191

Lake Use Support Analysis Definition. An analysis of a lake’s use support can be carried out employing a methodology developed by the IEPA (1994). The degree of use support identified for each designated use indicates the ability of the lake to (1) support a variety of high-quality recreational activities, such as boating, sport fishing, swimming, and esthetic enjoyment; (2) support healthy aquatic life and sport fish populations; and (3) provide adequate, long-term quality and quantity of water for public or industrial water supply (if applicable). Determination of a lake’s use support is based upon the state’s water quality standards as described in Subtitle C of Title 35 of the State of Illinois Administrative Code (IEPA, 1990). Each of four established use-designation categories (including general use, public and food processing water supply, Lake Michigan, and secondary contact and indigenous aquatic life) has a specific set of water quality standards. For the lake uses assessment, the general use standards—primarily the 0.05 mg/L TP standard—shall be used. The TP standard has been established for the protection of aquatic life, primary-contact (e.g. swimming) and secondary-contact (e.g. boating) recreation, agriculture, and industrial uses. In addition, lake-use support is based in part on the amount of sediment, macrophytes, and algae in the lake and how these might impair designated lake uses. The following is a summary of the various classifications of use impairment: • Full full support of designated uses, with minimal impairment. • Full/threatened full support of designated uses, with indications of declining water quality or evidence of existing use impairment. • Partial/minor partial support of designated uses, with slight impairment. • Partial/moderate partial support of designated uses, with moderate impairment; only partial is used after the year of 2000. • Nonsupport no support of designated uses, with severe impairment. Lakes that fully support designated uses may still exhibit some impairment, or have slight-tomoderate amounts of sediment, macrophytes, or algae in a portion of the lake (e.g. headwaters or shoreline); however, most of the lake acreage shows minimal impairment of the aquatic community and uses. It is important to emphasize that if a lake is rated as not fully supporting designated uses, it does not necessarily mean that the lake cannot be used for those purposes or that a health hazard exists. Rather, it indicates impairment in the ability of significant portions of the lake waters to support either a variety of quality recreational experiences or a balanced sport fishery. Since most lakes are multiple-use water bodies, a lake can fully support one designated use (e.g. aquatic life) but exhibit impairment of another (e.g. swimming). Lakes that partially support designated uses have a designated use that is slightly to moderately impaired in a portion of the lake (e.g. swimming impaired by excessive aquatic macrophytes or algae, or boating impaired by sediment accumulation). So-called nonsupport lakes have a designated use that is severely impaired in a substantial portion of the lake (e.g. a large portion of the lake has so much sediment that boat ramps are virtually inaccessible, boating is nearly impossible, and fisheries are degraded). However, in other parts of the same nonsupport lake (e.g. near a dam), the identical use may be supported. Again, nonsupport does not necessarily mean that a lake cannot support any uses, that it is a public health hazard, or that its use is prohibited. Lake-use support and level of attainment shall be determined for aquatic life, recreation, swimming, fish consumption, drinking water supply, secondary contact, and overall lake use, using methodologies described in the IEPA’s Illinois Water Quality Report 19941995 (IEPA, 1996). The primary criterion in the aquatic life use assessment is an aquatic life use impairment index (ALI), while in the recreation use assessment the primary criterion is a recreation use impairment index (RUI). While both indices combine ratings for TSI (Carlson, 1977) and degree of use impairment from sediment and aquatic macrophytes, each index is specifically designed for the assessed use. ALI and RUI relate directly to the TP standard of 0.05 mg/L. If a lake water sample is found to have a TP concentration at or below the standard, the lake is given a ‘full support’ designation. The aquatic life use rating reflects the degree of attainment of the ‘fishable goal’ of the Clean Water Act, whereas the recreation use rating reflects the degree to which pleasure boating, canoeing, and esthetic enjoyment may be obtained at an individual lake.

LAKES AND RESERVOIRS 1.192

CHAPTER 1.3

The assessment of swimming use for primary-contact recreation was based on available data using two criteria: (1) Secchi disc transparency depth data and (2) Carlson’s TSI. The swimming use rating reflects the degree of attainment of the ‘swimmable goal’ of the Clean Water Act. If a lake is rated ‘nonsupport’ for swimming, it does not mean that the lake cannot be used or that health hazards exist. It indicates that swimming may be less desirable than at those lakes assessed as fully or partially supporting swimming. Finally, in addition to assessing individual aquatic life, recreation, and swimming uses, and drinking water supply, the overall use support of the lake is also assessed. The overall use support methodology aggregates the use support attained for each of the individual lake uses assessed. Values assigned to each use-support attainment category are summed and averaged, and then used to assign an overall lake-use attainment value for the lake. Designated uses assessment. Multiple lakes designated are assessed for aquatic life, recreation, drinking water supply, swimming, fish consumption, and overall use. Specific criteria for determining attainment of these designated lake uses are described below. The degree of use support attainment is described as full, full/threatened, partial/minor impairment, partial/moderate impairment, or nonsupport. Aquatic life. An aquatic life use impairment index (ALI) which combines ratings for trophic state index and the amount of use impairments from aquatic macrophytes and sediment is used as the primary criteria for assessing aquatic life lake use (Table 3.5). The higher the ALI number, the more impaired the lake. Specific critiera used for each level of aquatic life use support attainment are presented in Table 3.6. Recreation. A recreation use impairment index (RUI), which combines TSI and the amount of use impairments from aquatic life and from sediment is utilized as the primary criteria for assessing recreation lake use (Table 3.7). Lake uses include pleasure boating, canoeing, skiing, sailing, esthetic enjoyment, and fishing. The higher the RUI number, the more impaired the lake. Specific criteria used for each level of attaining recreation use support are listed in Table 3.8. Swimming. The assessment criteria for swimming use is based primarily on the Secchi disc transparency depth and on the fecal coliform (FC) density—percent that exceed the 200 FC/100 mL standard. If FC data are not available, a TSI is calculated and used to make the assessment. The degree of swimming use support attainment is presented in Table 3.9. Drinking water supply. Drinking water supply use assessment for a lake is determined on the basis of water supply advisories or closure issued through state regulatory public water supply TABLE 3.5 Aquatic Life Use Impairment Index (ALI) Evaluation factor

Parameter

Weighting criteria TSI 60 60 TSI 85 85 TSI 90 90 TSI

Mean trophic state index (Carson, 1977)

Mean TSI value between 30 and 100

a. b. c. d.

Macrophyte impairment

Percent of lake surface area covered by weeds, or amount of weeds recorded on form

a. 15 % 40; or minimal (1) b. 10 % 15 & 40 % 50; or slight (2) c. 5 % 10 & 50 % 70; or moderate (3) d. % 5 & 70 %; or substantial (4)

Sediment impairment

Concentration of nonvolatile suspended solids (NVSS); or amount of sediment value reported on form

Source: Modified from Illinois EPA (1996).

a. b. c. d.

NVSS 12; or minimal (1) 12 NVSS 15; or slight (2) 15 NVSS 20; or moderate (3) 20 NVSS; or substantial (4)

Points a. b. c. d.

40 50 60 70

a. b.

0 5

a. b. c. d.

0 5 10 15

LAKES AND RESERVOIRS LAKES AND RESERVOIRS

1.193

TABLE 3.6 Assessment Criteria for Aquatic Life and Overall Use in Illinois Lakes Degree of use support

Criteria

Full

a. Total ALI points are 75 b. Direct field observations of minimal aquatic life impairment

Full/threatened

a. Total ALI point are 75 and evidence of a declined water quality trend exists b. Specific knowledge of existing or potential threats to aquatic life impairment

Partial/minor

a. 75 total ALI points 85 b. Direct field observations of slight aquatic life impairment

Partial/moderate

a. 85 total ALI points 95 b. Direct field observations of moderate aquatic life impairment

Nonsupport

a. Total ALI points 95 b. Direct field observation of substantial aquatic life impairment

Source: Modified from Illinois EPA (1996).

TABLE 3.7 Recreation Use Impairment Index (RUI) Evaluation factor

Parameter

Weighting criteria

Points

Mean trophic state index (Carlson, 1977)

Mean TSI value (30–110)

Actual TSI value

Macrophyte impairment

Percent of lake surface area covered by weeds; or amount of weeds value reported on form

a. b. c. d.

% 5; or minimal (1) 5 % 15; or slight (2) 15 % 25; or moderate (3) 25 %; or substantial (4)

a. b. c. d.

0 5 10 15

Sediment impairment

Concentration of nonvolatile suspended solids (NVSS); or amount of sediment value reported on form

a. b. c. d.

NVSS % 3; or minimal (1) 3 NVSS 7; or slight (2) 7 NVSS 15; or moderate (3) 15 %; or substantial (4)

a. b. c. d.

0 5 10 15

Source: Modified from Illinois EPA (1996).

TABLE 3.8 Assessment Criteria for Recreation Use in Illinois Lakes Degree of use support

Criteria

Full

a. Total RUI points are less than 60 b. Direct field observation of minor recreation impairment

Full/threatened

a. Total RUI points are 60, and evidence of a decline in water quality trend exists b. Specific knowledge or potential threats to recreation impairment

Partial/minor moderate

a. 60 total RUI points 75 b. Direct field observation of slight impairment

Partial/moderate

a. 75 total RUI points 90 b. Direct field observation of moderate recreation impairment

Nonsupport

a. Total RUI points 90 b. Direct field observation of substantial recreation impairment

Source: Illinois EPA (1996).

LAKES AND RESERVOIRS 1.194

CHAPTER 1.3

TABLE 3.9 Assessment Criteria for Swimming Use in Illinois Lakes Degree of use support

Criteria

Full

a. No Secchi depths are 24 in b. 10% fecal coliforms (FC) samples exceed the standard c. TSI 50

Partial/minor

a. 50% of Secchi depths were 24 in b. 10% FC 25% exceed the standard c. TSI 65

Partial/moderate

a. 50–100% of Secchi depths were 24 in b. 10% FC sample 25% exceed the standard TSI 75

Nonsupport

a. 100% of Secchi depths were 24 in b. 25% FC sample exceed the standard c. TSI 75

Source: Modified from Illinois EPA (1996).

programs. For example, in Illinois, the primary criteria used is the length of time (greater than 30 days) nitrate and/or atrazine concentration exceeding the public water supply standards of 10 mg/L and 3 g/L, respectively. Other problems which affect the quality of finished waters, such as chemical or oil spills or severe taste and odor problems requiring immediate attention, are also included in assessing use support. Specific criteria used for assessing the drinking water supply and the degree of use support are shown in Table 3.10. Fish consumption. The assessment of fish consumption use is based on fish tissue data and resulting sport fish advisories generated by the state fish contaminant monitoring program. The degree of fish consumption use support attainment can be found in Table 3.11. Overall use. After assessing individual lake uses, the overall use support of a lake can be determined. The overall use support methodology aggregates the use support attained for each of the individual lake uses assessed (i.e. aquatic life, recreation, swimming, drinking water supply, and fish consumption). The aggregation is achieved by averaging individual use attainments for a lake. For instance, individual uses meeting full, full/threatened, partial/minor, partial/moderate, or nonsupport are assigned values from five (5) to one (1), respectively. The values assigned to each individual use are subsequently summed and averaged. The average value is rounded down to the next whole number, which is then applied to assign an overall lake use attainment. Full support attainment is assigned to an average value of 5; full/threatened, 4; partial/minor, 3; partial/moderate, 2; and nonsupport, 1.

TABLE 3.10 Assessment Criteria for Drinking Water Supply in Illinois Lakes Degree of use support

Criteria

Full

No drinking water closures or advisories in effect during reporting period; no treatment necessary beyond ‘reasonable levels’ (copper sulfate may occasionally be applied for algae/taste and odor control).

Partial/minor

One or more drinking water supply advisory(ies) lasting 30 days or less; or problems not requiring closure or advisories but adversely affecting treatment costs and quality of treated water, such as taste and odor problems, color, excessive turbidity, high dissolved solids, pollutants requiring activated carbon filters.

Partial/moderate

One or more drinking water supply advisories lasting more than 30 days per year.

Nonsupport

One or more drinking water supply closures per year.

Source: Illinois EPA (1996).

LAKES AND RESERVOIRS LAKES AND RESERVOIRS

1.195

TABLE 3.11 Assessment Criteria for Fish Consumption Use in Illinois Lakes Degree of use support

Criteria

Full

No fish advisories or bans are in effect.

Partial/moderate

Restricted consumption fish advisory or ban in effect for general population or a subpopulation that could be at potentially greater risk (pregnant women, children). Restricted consumption is defined as limits on the number of meals or size of meals consumed per unit time for one or more fish species. In Illinois, this is equivalent to a Group II advisory.

Nonsupport

No consumption fish advisory or ban in effect for general population for one or more fish species; commercial fishing ban in effect. In Illinois this is equivalent to a Group III advisory.

Source: Illinois EPA (1996).

EXAMPLE: The mean TSI is determined by averaging 18 months’ SD-TSI, TP-TSI, and CHL-TSI

values at station 1-surface (deepest station) and station 2-surface (mid-lake station) of a central Illinois lake, for values of 54.3 and 58.9, respectively. All secchi disc reading were greater than 24 inches. The mean of nonvolatile suspended solids concentrations observed during 1996–97 at stations 1-surface and 2-surface are 5 and 7 mg/L, respectively. Estimated macrophyte impairment and other observed data are given in Table 3.12. Determine support of designated uses in the lake based on Illinois lake-use support assessment criteria. TABLE 3.12 Use Support Assessment for Otter Lake, 1996–97 Station 1

I. Aquatic life use 1. Mean trophic state index 2. Macrophyte impairment 3. Mean NVSS Total points: Criteria points: Use support:

II. Recreation use 1. Mean trophic state index 2. Macrophyte impairment 3. Mean NVSS Total points: Criteria points: Use support:

ALI points*

value

ALI points

54.3

5% 5 mg/L

40 10 0 50

75 Full

58.9

5% 7 mg/L

40 10 5 55

75 Full

Value

RUI points*

Value

RUI points

54.3

5% 5 mg/L

54 0 5 59

60 Full

58.9

5% 7 mg/L

59 0 5 64 65 R 75 Partial

Value III. Swimming use 1. Secchi depth 24 in 2. Fecal coliform 200/100 mL 3. Mean trophic state index Use support:

0% 0% 54.3

IV. Drinking water supply V. Overall use Use support:

Station 2

value

Degree of use support Full Full Full Full

Value 0% 0% 58.9

Full 5

Degree of use support Full Full Full Full Full

4 Full

*ALI, aquatic life use impairment index; RUI, recreation use impairment index.

Full/threatened

LAKES AND RESERVOIRS 1.196

CHAPTER 1.3

Solution: Solve for station 1 (Steps 15); then station 2 (Step 6) can be solved in the same manner. Step 1. Assess aquatic life use 1. Find ALI points for TSI Since mean TSI 54.3, TSI points 40 is obtained from Table 3.5, section 1. 2. Determine macrophyte impairment (MI) of ALI points The estimated macrophyte impairment is 5% of the lake surface area. From section 2 of Table 3.5, the MI point is 10. 3. Determine mean nonvolatile suspended solids (NVSS) of ALI point The NVSS 5 mg/L. From section 3 of Table 3.5, we read NVSS point 0. 4. Calculate total ALI points Total ALI points TSI points MI points NVSS points 40 10 0 50 5. Determine the degree of aquatic life index use support Since ALI points 50 (i.e. 75) of the critical value, and there are potential threats to aquatic life impairment, the degree of use support is full from Table 3.6. Step 2. Assess recreation use 1. Determine the mean TSI of RUI point Since mean TSI value is 54.3, then RUI points for TSI 54, from section 1 of Table 3.7. 2. Determine macrophyte impairment of RUI point Since 5% of the lake’s surface area is covered by weeds, from section 2 of Table 3.7, the RUI point for MI 0. 3. Determine sediment impairment The mean NVSS is 5 mg/L. From section 3 of Table 3.7, for sediment impairment (NVSS), RUI 5. 4. Compute total RUI points Total RUI points TSI value MI points NVSS points 54 0 5 59 5. Determine the degree of recreation use support Total RUI points 60, therefore the degree of recreation use is considered as Full from Table 3.8. Step 3. Assess swimming use 1. Based on Secchi disc transparency Since no Secchi depth is less than 24 in, from Table 3.9, the degree of use support is classified as Full. 2. Fecal coliform criteria No fecal coliform density is determined. 3. Based on trophic state index Mean TSI 54.3. From Table 3.9, it can be classified as Full. 4. Determine swimming use support Based on the above use analysis, it is assessed as Full for swimming use support.

LAKES AND RESERVOIRS LAKES AND RESERVOIRS

1.197

Step 4. Assess drinking water supply use There was no drinking water supply closure or advisories during the 1996–97 study period. Also no chemicals were applied to the lake. It is considered as Full use support for drinking water supply. Step 5. Assess overall use for station 1 On the basis of overall use criteria, the assigned score values for each individual use are as follows: Aquatic life use Recreation use Swimming use Drinking water use Total:

5 5 5 5 20

The average value is 5. Then the overall use for station 1 is attained as Full (Table 3.12). Step 6. Assess for station 2. Most assessments are as for station 1 except the following: 1. Recreation use: From Table 3.12, we obtain RUI points 64 which is in the critical range 60 RUI 75; therefore, the degree of research on use is considered as partial/minor from Table 3.8. 2. The assigned scores for overall use are Aquatic life use Recreation use Swimming use Drinking water supply Total

5 3 5 5 18

The average is 4.5, which is rounded doun to the next whole number, 4. Then, the overall use for station 2 is attained as full/threatened.

5.11

Lake Budgets Calculation of lake budgets (fluxes) usually includes the hydrologic budget, nutrients (nitrogen and phosphorus) budgets, and the sediment budget. These data should be generated in Phases I, II, and III studies. Hydrologic budget. A lake’s hydrologic budget is normally quantified on the basis of the height of water spread over the entire lake surface area entering or leaving the lake in a year period. The hydrologic budget of a lake is determined by the general formula: Storage change inflows outflows

(3.8)

S P I U E O R

(3.8a)

(Eqs. (3.8) and (3.8a) are essentially the same.) In general, inflow to the lake includes direct precipitation (P), watershed surface runoff (I), subsurface groundwater inflow through lake bottom (U), and pumped input if any. Outflows include lake

LAKES AND RESERVOIRS 1.198

CHAPTER 1.3

surface evaporation (E), discharge through surface outlet (O), outflow through lake bottom (groundwater recharge, R), and pumped outflow for water supply use, if any. The storage term is positive if the water level increases in the period and negative if it decreases. The unit of the storage can be, simply, in (or mm) or acre-foot (or ha-cm). The hydrologic budget is used to compute nutrient budgets and the sediment budget and in selecting and designing pollution control and lake restoration alternatives. EXAMPLE: Table 3.13 illustrates the hydrologic analysis from Vienna Correctional Center Lake, a water-supply lake (Bogner et al., 1997). Date needed for evaluating various parameters (Eq. (3.8)) to develop a hydrologic budget for the lake were collected for a 1-year period (October 1995 to September 1996). Table 3.13 presents monthly and annual results of this monitoring. Data sources and methods are explained as below:

Solution: Step 1. Determine lake storage change Lake storage change was determined on the basis of direct measurement of the lake level during the study period. Lake-level data were collected by automatic water-level recorder at 15-min intervals and recorded at 6-h intervals or less. On the basis of water-level record frequency, changes in storage were estimated by multiplying the periodic change in lake storage from the water-level recorder by the lake surface area (A) to determine net inflow or outflow volume. Step 2. Compute direct precipitation Direct precipitation was obtained from the precipitation record at the University of Illinois’ Dixon Spring Experiment Station. The volume of direct precipitation input to the lake was determined by multiplying the precipitation depth by A.

TABLE 3.13 Summary of Hydrologic Analysis for Vienna Correction Center Lake, October 1995–September 1996 Storage change (acreft)

Direct precipitation (acreft)

Groundwater inflow outflow(–) (acreft)

1995 October November December

–56 –14 –13

9 16 14

7 –13 –2

3 40 31

14 7 4

0 0 0

61 49 52

1996 January February March April May June July August September

40 –11 153 74 –8 –37 –55 –76 –36

19 4 27 34 33 21 30 3 34

18 24 36 31 24 –3 –6 4 –18

63 19 154 221 262 76 11 4 27

4 6 12 20 28 30 33 29 21

0 0 0 142 244 42 0 0 0

55 51 52 50 55 59 58 59 58

Annual

–39

244

102

911

208

428

658

Date

Surface inflow (acreft)

Monthly evaporation (acreft)

Spillway discharge (acreft)

Water supply withdrawal (acreft)

LAKES AND RESERVOIRS LAKES AND RESERVOIRS

1.199

Step 3. Compute spillway discharge Q The general spillway rating equation was used as below: Q 3.1 LH 1.5 3.1 (120)(H1.5) where L spillway length, ft H the height of water level exceeding the spillway Step 4. Estimate evaporation Evaporation was estimated using average monthly values for Carbondale in Lake Evaporation in Illinois (Robert and Stall, 1967). Monthly evaporation rates were reduced to daily rates by computing an average daily value for each month. The daily lake surface evaporation volume was determined for the study period by multiplying the daily average evaporation depth by A. Step 5. Obtain water-supply withdraw rate The daily water-supply withdraw rates were taken directly from the monthly reports of the water treatment plant. Step 6. Estimate groundwater inflow and surface inflow Groundwater inflow and surface water inflow could not be estimated from direct measurements and instead were determined on the basis of a series of sorting steps: a. If the daily spillway discharge was zero and no rainfall occurred during the preceding 3 days, all inflow was attributed to seepage from the groundwater system. b. If there was precipitation during the preceding 3-day period, or if there was discharge over the spillway, the groundwater input was not determined in step 6a. In this case, a moving average was used for the groundwater parameter based on the step 6a values determined for the preceding 5-day and following 10-day periods. c. If the daily balance indicated an outflow from the lake, it was attributed to seepage into the groundwater inflow/outflow to the lake; the surface water inflow to the lake was determined to be any remaining inflow volume needed to achieve a daily balance. Step 7. Summary Table 3.14 summarizes the hydrologic budget for the 1-year monitoring period. The inflows and outflows listed in Table 3.14 are accurate within the limits of the analysis. During the study period, 18.9, TABLE 3.14 Annual Summary of the Hydrologic Budget for Vienna Correction Center Lake, October 1995–September 1996 Source Storage change Direct precipitation Surface inflow Groundwater inflow Spillway discharge Evaporation Water supply withdrawal Totals

Inflow volume (acreft)

Outflow volume (acreft)

37.1 243.8 910.2 100.6

Inflow(%)

426.9 207.8 656.9 1291.7

Note: Blank spaces—not applicable, 1 acreft = 1233 m3.

Outflow(%)

2.9 18.9 70.4 7.8

1291.7

33.0 16.1 50.9 100.0

100.0

LAKES AND RESERVOIRS 1.200

CHAPTER 1.3

70.4, 7.8, and 2.9% of the inflow volume to the lake were, respectively, direct precipitation on the lake surface, watershed surface runoff, groundwater inflow, and decrease in storage. Outflow volume was 59.0%, water-supply withdrawal, 33.0% spillway overflow, and 16.1% evaporation. Nutrient and sediment budgets. Although nitrogen and phosphorus are not the only nutrients required for algal growth, they are generally considered to be the two main nutrients involved in lake eutrophication. Despite the controversy over the role of carbon as a limiting nutrient, the vast majority of researchers regard phosphorus as the most frequently limiting nutrient in lakes. Several factors have complicated attempts to quantify the relationship between lake trophic status and measured concentrations of nutrients in lake waters. For example, measured inorganic nutrient concentrations do not denote nutrient availability but merely represent what is left over by the lake production process. A certain fraction of the nutrients (particularly phosphorus) becomes refractory while passing through successive biological cycles. In addition, numerous morphometric and chemical factors affect the availability of nutrients in lakes. Factors such as mean depth, basin shape, and detention time affect the amount of nutrients a lake can absorb without creating nuisance conditions. Nutrient budget calculations represent the first step in quantifying the dependence of lake water quality on the nutrient supply. It is often essential to quantify nutrients from various sources for effective management and eutrophication control. A potential source of nitrogen and phosphorus for lakes is watershed drainage, which can include agricultural runoff, urban runoff, swamp and forest runoff, domestic and industrial waste discharges, septic tank discharges from lakeshore developments, precipitation on the lake surface, dry fallout (i.e. leaves, dust, seeds, and pollen), groundwater influxes, nitrogen fixation, sediment recycling, and aquatic bird and animal wastes. Potential sink can include outlet losses, fish catches, aquatic plant removal, denitrification, groundwater recharge, and sediment losses. The sources of nutrients considered for a lake are tributary inputs from both gaged and ungaged streams, direct precipitation on the lake surface, and internal nutrient recycling from bottom sediments under anaerobic conditions. The discharge of nutrients from the lake through spillway is the only readily quantifiable sink. The flow weighted-average method of computing nutrient transport by the tributary are generally used in estimating the suspended sediments, phosphorus, and nitrogen loads delivered by a tributary during normal flow conditions. Each individual measurement of nitrogen and phosphorus concentrations in a tributary sample is used with the mean flow values for the period represented by that sample to compute the nutrient transport for the given period. The total amount of any specific nutrient transported by the tributary is given by the expression (Kothandaraman and Evans, 1983; Lin and Raman, 1997): T lb 5.394 qicini

(3.18a)

T kg 2.446 qicini

(3.18b)

where T total amount of nutrient (nitrogen or phosphorus) or TSS, lb or kg qi average daily flow in cfs for the period represented by the ith sample ci concentration of nutrient, mg/L ni number of days in the period represented by the ith sample A similar algorithm with appropriate constant (0.0255) can be used for determining the sediment and nutrient transport during storm events. For each storm event, ni is the interval of time represented by the ith sample and qi is the instantaneous flow in cfs for the period represented by the ith sample. The summation is carried out for all the samples collected in a tributary during each storm event. An automatic sampler can be used to take storm water samples. The level of nutrient or sediment input is expressed either as a concentration (mg/L) of pollutant or as mass loading per unit of land area per unit time (kg/ha-yr). There is no single correct way to express the quantity of nutrient input to a lake. To analyze nutrient inputs to a lake, the appropriate averaging time is usually 1 year, since the approximate hydraulic residence time in a lake is of this order of magnitude and the concentration value is a long-term average.

LAKES AND RESERVOIRS LAKES AND RESERVOIRS

5.12

1.201

Soil loss rate Specific soil loss rate from the watershed (for agricultural land, large construction sites, and other land uses of open land) can be estimated through the universal soil loss equation or USLE (Wischmeier and Smith, 1965): A R LS K C P

(3.19)

where A average soil loss rate, tons/(acre · year) R rainfall factor L slope length factor, ft S slope steepness factor, % LS length-slope factor or topographic factor (not L times S) K soil erodibility factor C cropping and management factor P conservation practices factor The slope steepness, slope length, cropping factors, and erodibility of each soil type cropping factor can be determined for various land uses in consultation with the local (county) USDA Soil and Water Conservation Service. The R P factor value is assigned as 135 for Illinois agricultural cropland and 180 for other land uses. Wischmeier and Smith (1965) and Wischmeier et al. (1971) are useful references. The value of P is one for some CRP (Conservation Reserve Program) implementation after 1985. Based on the soil information compiled in the watershed or subwatersheds, the soil loss rates can be computed. The soil loss for each soil type for each subwatershed is obtained by multiplying the rate and soil acreage. The total soil loss for the watershed is the sum of soil loss in all subwatersheds expressed as tons per year. Excluding the lake surface area, the mean soil erosion rate for the watershed is estimated in terms of tons/(acre·year). The USLE accounts for a series of factors that are the most significant influences on the erosion of soil by precipitation. The USLE is a calculation of in-field soil losses and does not account for deposition from the field to the stream or lake. Redeposition within the field or drainage system is accounted for by a sediment delivery ratio that defines the proportion of the upstream soil losses that actually pass through the stream then to the lake. The RUSLE (Revised Universal Soil Loss Equation) is an updated version of the USLE and was adopted recently by the USDA and many states. The original USLE has been retained in RUSLE; however it has been put into a computer program to facilitate calculations, the technology for factor evaluation has been altered, and new data has been introduced to evaluate each factor under more specific conditions. There are three tables for LS value determinations. For the following example, taking the sum of annual loss for each soil type, the total annual soil loss in the Nashville City Reservoir watershed (984.3 acres) is 2.953 tons (Table 3.16). This analysis shows that the average annual soil loss in the watershed in 3.5 tons per acre per year. EXAMPLE: Estimate the watershed soil loss rate for the Nashville City Reservoir watershed in Washington County, Illinois (southern). Use USLE instead of RUSLE, since new factors are not yet available. The following values are given for each of these factors for Nashville City Reservoir:

• The rainfall factor (R) is set to 200, the value applicable for southern Illinois. • The soil classifications for the watershed and related information are presented in Table 3.15. • The soil erodibility factor (K), the slope length (L), slope steepness (S) and the cropping factor (C) are provided in Table 3.16 by the Washington County Soil and Water Conservation District. • The conservation factor (P) is one. Solution: (Assistance in developing these estimates can be obtained through the county conservation district or directly from the Soil and Water Conservation District of US Department of Agriculture. Both are in the same office).

LAKES AND RESERVOIRS 1.202

CHAPTER 1.3

TABLE 3.15 Soil Classifications in the Nashville City Reservoir Watershed Soil type 2 3A 3B 5C2 5C3 5D3 8D2 12 13A 13B 13B2 14B 48 912A 912B2 991 1334 3415 Water

Soil name

Slope

Area (acres)

Cisne silt loam Hoyleton silt loam Hoyleton silt loam Blair silt loam, eroded Blair silt loam, severely eroded Blair silt loam, severely eroded Hickory silt loam Wynoose silt loam Bluford silt loam Bluford silt loam Bluford silt loam, eroded Ava silt loam Ebbert silt loam Dartstadt-Hoyleton complex Dartstadt-Hoyleton complex, eroded Huey-Cisne complex Birds silt loam Orian silt loam, wet

– 0–2 2–5 5–10 5–10 10–15 10–15 – 0–2 2–5 2–5 2–5 – 0–2 2–5 – – –

38.0 14.9 11.9 36.1 31.0 9.8 6.2 186.0 208.4 73.7 17.5 14.2 7.5 61.4 164.7 32.6 10.9 17.5 42.0

Total

984.3

Note: Means flat, using 0.5% for calculation. Sources: USDA, 1998; J. Quinn, personal communication, 2000.

TABLE 3.16 Summary of Soil Losses in the Nashville City Reservoir Watershed Due to Precipitation Runoff Soil type

Areal coverage (acres)

2 3A 3B 5C2 5C3 5D3 8D2 12 13A 13B 13B2 14B 48 912A 912B2 991 1334 3415 Water

38.0 14.9 11.9 36.1 31.0 9.8 6.2 186.0 208.4 73.7 17.5 14.2 7.5 61.4 164.7 32.6 10.9 17.5 42.0

Total Average

984.3

K 0.37 0.32 0.32 0.37 0.37 0.37 0.37 0.43 0.43 0.43 0.43 0.43 0.32 0.37 0.37 0.43 0.43 0.37

L(ft)

S(%)

Annual soil loss rate (tons/acre)

175 175 125 100 100 75 75 175 175 125 100 125 175 175 125 175 175 175

0.5 1.1 3.5 7.5 7.0 12.5 12.5 0.5 1.1 3.5 4.0 3.5 0.5 1.1 3.5 0.5 0.5 0.5

0.11 0.24 0.37 0.91 0.82 1.67 1.67 0.11 0.24 0.37 0.40 0.37 0.11 0.24 0.37 0.11 0.11 0.11

0.37 0.20 0.13 0.06 0.05 0.02 0.04 0.32 0.15 0.09 0.09 0.13 0.71 0.17 0.11 0.21 0.57 0.61

3.0 3.1 3.1 4.0 3.0 2.5 4.9 3.0 3.1 2.9 3.1 4.1 5.0 3.0 3.0 2.0 5.4 5.0

Total annual loss (tons) 114 46 37 146 94 24 31 563 645 211 54 59 37 185 496 65 59 87 2953

0.38

Note: Blank spaces—not applicable. Source: Lin, 2001.

3.5

LAKES AND RESERVOIRS LAKES AND RESERVOIRS

1.203

Step 1. Determine watershed boundary The watershed boundary of the lake can be determined from a 7.5-minute US Geological Survey topographic map and digitized into a GIS coverage. Step 2. Obtain the values for, K, L, S, and C Obtain “Soil Survey of Washington County, Illinois” from the USDA office. The soil types, values of K and S, and soil classifications, etc. are listed in the book. Values for L and C can be obtained by consultation with USDA or the county’s Soil and Water Conservation District personnel. These values for each type of soil are presented in Table 3.16. Step 3. Measure the area coverage for each type of soil The area coverage for each type of soil in the watershed can be directly measured from Soil Survey of Washington County, Illinois. The area coverage for each type of soil is listed in Table 3.15. Step 4. Find LS value The LS value can be determined from tables of topographic factor based on L, S, R, and land use; and then listed in Table 3.16 for calculations. Step 5. Calculate annual soil loss rate for a type of soil For soil type 2: Using Eq. (3.19), A R K LS C P 200 0.37 0.11 0.37 1.0 3.0 (tons/acre/year) (see Table 3.16) Step 6. Calculate total annual soil loss for each type of soils The total annual soil loss for each type of soil is calculated from annual soil loss rates for that soil type multiplying by the area coverage, such as For soil type 2: Total soil loss 3.0 tons/acre/year 38 acres 114 tons/year Step 7. Calculate the total annual soil loss (2953 tons, from Table 3.16) in the watershed

REFERENCES American Public Health Association, American Water Works Association and Water Environment Federation. 1992. Standard methods for the examination of water and wastewater, 18th edn. Washington, DC: APHA. Bogner, W. C., Lin, S. D., Hullinger, D. L. and Raman, R. K. 1997. Diagnostic—feasibility study of Vienna Correctional Center Lake Johnson County, Illinois. Contract report 619. Champaign: Illinois State Water Survey. Borah, D. K., Raman, R. K., Lin, S. D., Knapp, H. V. and Soong, T. W. D. 1997. Water quality evaluations for Lake Springfield and proposed Hunter Lake and proposed Lick Creek Reservoir. Contract Report 621. Champaign: Illinois State Water Survey. Carlson, R. E. 1977. A trophic state index for lakes. Limnology and Oceanography 22(2): 361369. Chapra, S. C. and Reckhow, K. H. 1983. Engineering approaches for lake management. Vol. 2: Mechanistic modeling. Woborn, Massachusetts: Butterworth. Clark, J. W., Viessman, Jr., W. and Hammer, M. J. 1977. Water supply and pollution control, 3rd edn. New York: Dun-Donnelley. Cole, G. A. 1979. Textbook of limnology, 2nd edn. St. Louis, Missouri: Mosby.

LAKES AND RESERVOIRS 1.204

CHAPTER 1.3

Dillon, P. J. and Rigler, F. H. 1974. The phosphorus-chlorophyll relationship in lakes. Limnology Oceanography 19(5): 767773. Fair, G. M., Geyer, J. C. and Okun, D. A. 1966. Water and wastewater engineering, Vol. 1: Water supply and wastewater removal. New York: John Wiley. Hudson, H. L., Kirschner, R. J. and Clark, J. J. 1992. Clean Lakes Program, Phase 1: Diagnostic/feasibility study of McCullom Lake, McHenry County, Illinois. Chicago: Northeastern Illinois Planning Commission. Illinois Environmental Protection Agency. 1983. Illinois Water Quality Report 19901991. IEPA/WPC/92-055. Springfield, Illinois: IEPA. Illinois Environmental Protection Agency. 1984. Volunteer Lake Monitoring Program—Report for 1983 Wolf Lake/Cook Co. IEPA/WPC/84-015. Springfield, Illinois: IEPA. Illinois Environmental Protection Agency. 1990. Title 35: Environmental protection, Subtitle C: Water pollution. State of Illinois, Rules and Regulations, Springfield, Illinois: IEPA. Illinois Environmental Protection Agency. 1992. Illinois Water Quality Report 19901991. IEPA/WPC/92-055. Springfield, Illinois: IEPA. Illinois Environmental Protection Agency. 1996. Illinois Water Quality Report 19941995. Vol. 1. Springfield, Illinois: Bureau of Water, IEPA. Illinois State Water Survey. 1989. Using copper sulfate to control algae in water supply impoundment. Misc. Publ. 111. Champaign: Illinois State Water Survey. James, A. 1993. Modeling water quality in lakes and reservoirs. In: James, A. (ed.), An introduction to water quality modeling, 2nd edn. Chichester: John Wiley. Kohler, M. A., Nordenson, T. J. and Fox, W. E. 1955. Evaporation from pan and lakes. US Weather Bureau Research Paper 38. Washington, DC. Kohler, M. A., Nordenson, T. J. and Baker, D. R. 1959. Evaporation maps for the United States. US Weather Bureau Technical Paper 37, Washington, DC. Kothandaraman, V. and Evans, R. L. 1983. Diagnostic-feasibility study of Lake Le-Aqua-Na., Contract Report 313. Champaign: Illinois State Water Survey. Lamoreux, W. M. 1962. Modern evaporation formulae adapted to computer use. Monthly Weather Review, January. Lin, S. D. 1994. Protocol for diagnostic/feasibility study of a lake. Proc. Aquatech Asia ’94, November 2224, 1994, Singapore, pp. 16576. Lin, S. D. and Raman, R. K. 1997. Phase III, Post-restoration monitoring of Lake Le-Aqua-na. Contract Report 610. Champaign: Illinois State Water Survey. Lin, S. D. et al. 1996. Diagnostic feasibility study of Wolf Lake, Cook County, Illinois, and Lake County, Indiana. Contract Report 604. Champaign: Illinois State Water Survey. Lin, S. D., Bogner, W. C. and Raman, R. K. 1998. Diagnostic-feasibility study of Otter Lake, Illinois. Contract Report (draft). Champaign: Illinois State Water Survey. Linsley, R. K. and Franzinni, J. B. 1964. Water resources engineering. New York: McGraw-Hill. Reckhow, K. H. and Chapra, S. C. 1983. Engineering approaches for lake management. Vol. 1: Data analysis and empirical modeling. Woburn, Massachusetts: Butterworth. Robert, W. J. and Stall, J. B. 1967. Lake evaporation in Illinois. Report of Investigation 57. Urbana: Illinois State Water Survey. Sawyer, C. N. 1952. Some aspects of phosphate in relation to lake fertilization. Sewage and Industrial Wastes 24(6): 768776. Shuttleworth, W. J. 1993. Evaporation. In: Maidment, D. R. (ed.) Handbook of hydrology. New York: McGraw-Hill. Tchobanoglous, G. and Schroeder, E. D. 1985. Water quality. Reading, Massachusetts: Addison-Wesley. US Environmental Protection Agency. 1980. Clean lakes program guidance manual. USEPA-440/5-81-003. Washington, DC: USEPA. US Environmental Protection Agency. 1990. The lake and reservoir restoration guidance manual, 2nd edn. USEPA-440/4-90-006. Washington, DC: USEPA. US Environmental Protection Agency. 1991. The watershed protection approach—an overview. USEPA/503/ 9-92/002. Washington, DC: USEPA.

LAKES AND RESERVOIRS LAKES AND RESERVOIRS

1.205

US Environmental Protection Agency. 1993. The watershed protection approach—annual report 1992. USEPA840-S-93-001. Washington, DC: USEPA. US Environmental Protection Agency. 1994. The quality of our nation’s water: 1992. USEPA841-S-94-002. Washington, DC: USEPA. US Geological Survey. 1954. Water loss investigations: Lake Hefner Studies. Technical Report, Professional paper 269. US Army Corps of Engineers. 1986. HEC-5: simulation of flood control and conservation systems: Appendix on water quality analysis. Davis, California: Hydrologic Engineering Center, US Army Corps of Engineers. US Army Corps of Engineers. 1989. HEC-5: Simulation of flood control and conservation systems: Exhibit 8 of user’s manual: input description. Davis, California: Hydrologic Engineering Center, US Army Corps of Engineers. Walker, Jr., W. W. 1981a. Empirical methods for predicting eutrophication in impoundments. Part 1. Phase I: data base development. Environmental Engineers. Tech. Report E-81-9, Concord, Massachusetts. Walker, Jr., W. W. 1981b. Empirical methods for predicting eutrophication in impoundments. Part 2. Phase II: model testing. Environmental Engineers Tech. Report E-81-9, Concord, Massachusetts. Wetzel, R. G. 1975. Limnology. Philadelphia, Pennsylvania: Saunders. Wischmeier, W. H. and Smith, D. D. 1965. Predicting rainfall—erosion losses from cropland east of the Rocky Mountains. Agriculture Handbook 282. Washington, DC: US Department of Agriculture. Wischmeier, W. H., Johnson, C. B. and Cross, B. V. 1971. A soil-erodibility monograph for farmland and construction sites. J. Soil and Water Conservation, 26(5), 18993.

LAKES AND RESERVOIRS

Source: HANDBOOK OF ENVIRONMENTAL ENGINEERING CALCULATIONS

CHAPTER 1.4

GROUNDWATER Shun Dar Lin

1 DEFINITION 1.207 1.1 Groundwater and Aquifer 1.2 Zones of Influence and Capture 1.209 1.3 Wells 1.211

1.207

2 HYDROGEOLOGIC PARAMETERS 2.1 Aquifer Porosity 1.211 2.2 Storativity 1.212 2.3 Transmissivity 1.212 2.4 Flow Nets 1.213 2.5 Darcy’s Law 1.214 2.6 Permeability 1.217 2.7 Specific Capacity 1.220 3 STEADY FLOWS IN AQUIFERS 4 ANISOTROPIC AQUIFERS

1 1.1

1.220

1.221

5 UNSTEADY (NONEQUILIBRIUM) FLOWS 1.222

1.211

5.1 5.2 5.3 5.4

Theis Method 1.222 Cooper–Jacob Method 1.225 Distance-drawdown Method 1.227 Slug Tests 1.229

6 GROUNDWATER CONTAMINATION 1.233 6.1 Sources of Contamination 1.233 6.2 Contaminant Transport Pathways 1.233 6.3 Underground Storage Tank 1.235 6.4 Groundwater Treatment 1.236 7 SETBACK ZONES 1.236 7.1 Lateral Area of Influence 1.237 7.2 Determination of Lateral Radius of Influence 1.238 7.3 The Use of Volumetric Flow Equation 1.240 REFERENCES

1.241

DEFINITION Groundwater and Aquifer Groundwater is subsurface water which occurs beneath the earth’s surface. In a hydraulic water cycle, groundwater comes from surface waters (precipitation, lake, reservoir, river, sea, etc.) and percolates into the ground beneath the water table. The groundwater table is the surface of the groundwater exposed to an atmospheric pressure beneath the ground surface (the surface of the saturated zone). A water table may fluctuate in elevation. An aquifer is an underground water-saturated stratum or formation that can yield usable amounts of water to a well. There are two different types of aquifers based on physical characteristics. If the saturated zone is sandwiched between layers of impermeable material and the groundwater is under pressure, it is called a confined aquifer (Fig. 4.1). If there is no impermeable layer immediately above the saturated zone, that is called an unconfined aquifer. In an unconfined aquifer, the top of the saturated zone is the water table defined as above. Aquifers are replenished by water infiltrated through the earth above from the upland area. The area replenishing groundwater is called the recharge area. In reverse, if the groundwater flows to the lower land area, such as lakes, streams, or wetlands, it is called discharge.

1.207

GROUNDWATER 1.208

CHAPTER 1.4

FIGURE 4.1 Drawdown, cone of depression, and radius of influence in unconfined and confined aquifers (Illinois EPA, 1995).

GROUNDWATER GROUNDWATER

1.209

On the earth, approximately 3 percent of the total water is fresh water. Of this, groundwater comprises 95 percent, surface water 3.5 percent, and soil moisture 1.5 percent. Out of all the fresh water on earth, only 0.36 percent is readily available to use (Leopold, 1974). Groundwater is an important source of water supply. Fifty-three percent of the population of the United States receives its water supply from groundwater or underground sources (USEPA, 1994). Groundwater is also a major source of industrial uses (cooling, water supply, etc.) and agricultural uses (irrigation and livestock). The quantity of groundwater available is an important value. The socalled safe yield of an aquifer is the practicable rate of withdrawing water from it perennially. Such a safe amount does not exist, however. The quantity of groundwater is also affected by water engineering. For decades and centuries, through improper disposal of wastes (solid, liquid, and gaseous) to the environment and subsurface areas, many groundwaters have become contaminated. Major sources of contaminants are possibly from landfill leachate, industrial wastes, agricultural chemicals, wastewater effluents, oil and gasoline (underground tanks, animal wastes, acid-mine drainage, road salts, hazardous wastes spillage, household and land chemicals, etc.). Illegal dumpings of wastes and toxic chemicals to waterways and to lands are major environmental problems in many countries. Efforts to protect the quantity and quality of groundwater have been made by cooperation between all government agencies, interested parties, and researchers. Most states are responsible for research, education, establishment of minimum setback zones for public and private water supply wells, and contamination survey and remediation. 1.2

Zones of Influence and Capture The withdrawal of groundwater by a pumping well causes a lowering of the water level. Referring to Fig. 4.2, the difference between water levels during non-pumping and pumping is called drawdown. The pattern of drawdown around a single pumping well resembles a cone. The area affected by the pumping well is called the cone of depression, the radius of influence, or lateral area of influence (LAI). Within the LAI, the flow velocity continuously increases as it flows toward the well due to gradually increased slope or hydraulic gradient. As a well pumps groundwater to the surface, the groundwater withdrawn from around the well is replaced by water stored within the aquifer. All water overlaid on the non-pumping potentiometric surface will eventually be pulled into the well. This area of water entering the area of influence of the well is called the zone of capture (ZOC), zone of contribution, or capture zone (Fig. 4.2). The ZOC generally extends upgradient from the pumping well to the edge of the aquifer or to a groundwater divide. The zone of capture is usually asymmetrical. It is important to identify the ZOC because any pollution will be drawn toward the well, subsequently contaminating the water supply. A zone of capture is usually referred to the time of travel as a time-related capture zone. For instance, a “ten-year” time-related capture zone is the area within which the water at the edge of the zone will reach the well within ten years. The state primacy has established setback zones for wells which will be discussed later. EXAMPLE 1: Groundwater flows into a well at 4 in/d (10 cm/d) and is 365 feet (111 m) from the well. Estimate the capture time.

Solution: Time 365 ft/(4/12 ft/d) 1095 days or

3-year time related capture zone

Calculate the distance from a well to the edge of the 3-year capture zone if the groundwater flow is 2 ft/d (0.61 m/d).

EXAMPLE 2:

GROUNDWATER 1.210

CHAPTER 1.4

FIGURE 4.2 Relationship between the cone of depression and the zone of capture within a regional flow field (Illinois EPA, 1995).

GROUNDWATER GROUNDWATER

1.211

Solution: Distance 2 ft/d × 3 years × 365 d/year 2190 ft 668 m

or 1.3

Wells Wells are classified according to their uses by the US Environmental Protection Agency (EPA). A public well is defined as having a minimum of 15 service connections or serving 25 persons at least 60 days per year. Community public wells serve residents the year round. Noncommunity public wells serve nonresidential populations at places such as schools, factories, hotels, restaurants, and campgrounds. Wells that do not meet the definition of public wells are classified as either semiprivate or private. Semi-private wells serve more than one single-family dwelling, but fewer than 15 connections or serving 25 persons. Private wells serve an owner-occupied single-family dwelling. Wells are also classified into types based on construction. The most common types of wells are drilled and bored dependent on the aquifer to be tapped, and the needs and economic conditions of the users (Babbitt et al., 1959; Forest and Olshansky, 1993).

HYDROGEOLOGIC PARAMETERS There are three critical aquifer parameters: porosity, specific yield (or storativity for confined aquifers), and hydraulic conductivity (including anisotropy). These parameters required relatively sophisticated field and laboratory procedures for accurate measurement. Porosity and specific yield/storativity express the aquifer storage properties. Hydraulic conductivity (permeability) and transmissivity describe the groundwater transmitting properties.

2.1

Aquifer Porosity The porosity of soil or fissured rock is defined as the ratio of void volume to total volume (Wanielista, 1990; Bedient and Huber, 1992): n

Vv (V Vs) Vs 1 V V V

(4.1)

where n porosity Vv volume of voids within the soil V total volume of sample (soil) Vs volume of the solids within the soil dry weight of sample/specific weight The ratio of the voids to the solids within the soil is called the void ratio e expressed as e Vv /Vs Then the relationship between void ratio and porosity is n e 1n or

e 1e

(4.2)

(4.3) (4.4)

GROUNDWATER 1.212

CHAPTER 1.4

The porosity may range from a small fraction to about 0.90. Typical values of porosity are 0.2 to 0.4 for sands and gravels depending on the grain size, size of distribution, and the degree of compaction; 0.1 to 0.2 for sandstone; and 0.01 to 0.1 for shale and limestone depending on the texture and size of the fissures (Hammer, 1986). When groundwater withdraws from an aquifer and the water table is lowered, some water is still retained in the voids. This is called the specific retention. The quantity drained out is called the specific yield. The specific yield for alluvial sand and gravel is of the order of 90 to 95 percent. EXAMPLE: If the porosity of sands and gravels in an aquifer is 0.38 and the specific yield is 92 percent, how much water can be drained per cubic meter of aquifer?

Solution: Volume 0.38 × 0.92 × 1 m3 0.35 m3 2.2

Storativity The term storativity (S) is the quantity of water that an aquifer will release from storage or take into storage per unit of its surface area per unit change in land. In unconfined aquifers, the storativity is in practice equal to the specific yield. For confined aquifers, storability is between 0.005 and 0.00005, with leaky confined aquifers falling in the high end of this range (USEPA, 1994). The smaller storativity of confined aquifers, the larger the pressure change throughout a wide area to obtain a sufficient supply from a well. However, this is not the case for unconfined aquifers due to gravity drainage.

2.3

Transmissivity Transmissivity describes the capacity of an aquifer to transmit a water. It is the product of hydraulic conductivity (permeability) and the aquifer’s saturated thickness: T Kb

(4.5)

where T transmissivity of an aquifer, gpd/ft or m3/(d m) K permeability, gpd/ft2 or m3/(d m2) b thickness of aquifer, ft or m A rough estimation of T is by multiplying specific capacity by 2000 (USEPA, 1994). EXAMPLE: If the aquifer’s thickness is 50 ft, estimate the permeability of the aquifer using data in the example of the specific capacity (in Sec. 2.7).

Solution: T 2000 × specific capacity 2000 × 15 gpm/ft 30,000 gpm/ft Rearranging Eq. (4.5) K T/b (30,000 gpm/ft)/50 ft 600 gpm/ft2

GROUNDWATER GROUNDWATER

2.4

1.213

Flow Nets Many groundwater systems are two or three dimensional. Darcy’s law was first derived in a one dimensional equation. Using Darcy’s law can establish a set of streamlines and equipotential lines to develop a two dimensional flow net. The details of this concept are discussed elsewhere in the text (Bedient et al., 1994). A flow net is constructed by flow lines that intersect the equipotential lines or contour lines at a right angle. Equipotential lines are developed based on the observed water levels in wells penetrating an isotropic aquifer. Flow lines are then drawn orthogonally to indicate the flow direction. Referring to Fig. 4.3, the horizontal flow within a segment in a flow net can be determined by the following equation (USEPA, 1994): qa TaHa Wa /La where

(4.6)

qa groundwater flow in segment A, m3/d or ft3/d Ta transmissivity in segment A, m3/d or ft3/d Ha drop in groundwater level across segment A, m or ft Wa average width of segment A, m or ft La average length of segment A, m or ft

The flow in the next segment, B, is similarly computed as Eq. (4.7) qb TbHbWb/Lb

(4.7)

Assuming that there is no flow added between segments A and B by recharge (or that recharge is insignificant), then qb qa or

TbHbWb/Lb TaHaWa/La

solving Tb computation of which allows Tb from Ta Tb Ta(LbHaWa/LaHbWb)

a a

FIGURE 4.3

Distribution of transmissivity in flow nets.

(4.8)

GROUNDWATER 1.214

CHAPTER 1.4

Measurement or estimation of transmissivity (T) for one segment allows the computation of variations in T upgradient and downgradient. If variations in aquifer thickness are known, or can be estimated for different segments, variation in hydraulic conductivity can also be calculated as K T/b

(4.9)

where K hydraulic conductivity, m/d or ft/d T transmissivity, m2/d or ft2/d b aquifer thickness, m or ft Eq. (4.9) is essentially the same as Eq. (4.5). 2.5

Darcy’s Law The flow movement of water through the ground is entirely different from the flow in pipes and in an open channel. The flow of fluids through porous materials is governed by Darcy’s law. It states that the flow velocity of fluid through a porous medium is proportional to the hydraulic gradient (referring to Fig. 4.4): Ki (h1 z1) (h2 z2) yK L

where

(4.10a) (4.10b)

y Darcy velocity of flow, mm/s or ft/s K hydraulic conductivity of medium or coefficient of permeability, mm/s or ft/s i hydraulic gradient, mm or ft/ft h1, h2 pressure heads at points 1 and 2, m or ft z1, z2 elevation heads at points 1 and 2, m or ft L distance between points (piezometers) 1 and 2, m or ft

The pore velocity p is equal to the Darcy velocity divided by porosity as follows: p /n

FIGURE 4.4

Groundwater flow (one dimensional).

(4.11)

GROUNDWATER GROUNDWATER

1.215

Darcy’s law is applied only in the laminar flow region. Groundwater flow may be considered as laminar when the Reynolds number is less than unity (Rose, 1949). The Reynolds number can be expressed as VD R v

(4.12)

where R Reynolds number V velocity, m/s or ft/s D mean grain diameter, mm or in v kinematic viscosity, m2/d or ft2/d Determine the Reynolds number when the groundwater temperature is 10°C (from Table 5.1a, v 1.31 × 10−6 m2/s); the velocity of flow is 0.6 m/d (2 ft/d); and the mean grain diameter is 2.0 mm (0.079 in). EXAMPLE 1:

Solution: V 0.6 m/d (0.6 m/d)/(86,400 s/d) 6.94 106 m/s D 2 mm 0.002 m 6.94 106 m/s 0.002 m VD R n 1.31 106 m2/s 0.011 Note: it is a laminar flow (R 1) If the mean grain diameter is 0.12 in (3.0 mm), its porosity is 40%, and the groundwater temperature is 50°F (10°C). Determine the Reynolds number for a flow 5 ft (1.52 m) from the centerline of the well with a 4400 gpm in a confined aquifer depth of 3.3 ft (1 m) thick.

EXAMPLE 2:

Solution: Step 1. Find flow velocity (V) Q Q n r h Since

4400 gpm 0.002228 cfs/gpm 9.80 cfs 0.4 5 ft 3.3 ft

Q nAV n(2prh) V V Q/n(2prh) 9.8 cfs/(0.4 2 3.14 5 ft 3.3 ft) 0.236 fps

Step 2. Compute R v 1.41 105 ft2/s (see Table 5.1b at 50 F) D 0.12 in 0.12 in/(12 in/ft) 0.01 ft 0.236 fps 0.01 ft VD R v 1.41 105 ft2/s 167

GROUNDWATER 1.216

CHAPTER 1.4

EXAMPLE 3: The slope of a groundwater table is 3.6 m per 1000 m. The coefficient of permeability of coarse sand is 0.51 cm/s (0.2 in/s). Estimate the flow velocity and the discharge rate through this aquifer of coarse sand 430 m (1410 ft) wide and 22 m (72 ft) thick.

Solution: Step 1. Determine the velocity of flow, , using Eq. (4.10a) i 3.6 m/1000 m 0.0036 y Ki 0.51 cm/s(0.0036) 0.00184 cm/s (86,400 s/d)(0.01 m/cm) 1.59 m/d 5.21 ft/d Step 2. Compute discharge Q yA 1.59 m/d 430 m 22 m 15,040 m3/d 15,040 m3/d 264.17 gal/m3 3.97 MGD (million gallons per day) If the difference in water level between two wells 1.6 miles (2.57 km) apart is 36 ft (11 m), and the hydraulic conductivity of the media is 400 gpd/ft2 (16,300 L/(d m2)). The depth of the media (aquifer) is 39 ft (12 m). Estimate the quantity of groundwater flow moving through the cross-section of the aquifer. EXAMPLE 4:

Solution: i 36 ft/1.6 miles 36 ft/(1.6 miles 5280 ft/miles) 0.00426 A 1.6 5280 ft 39 ft 329,500 ft2 then

Q KiA 400 gpd/ft2 0.00426 329,500 ft2 561,400 gpd 0.561 MGD 2123 m3/d

EXAMPLE 5: If the water moves from the upper to the lower lake through the ground. The following data is given:

difference in elevation h 25 m (82 ft) length of flow path L 1500 m (4920 ft) cross-sectional area of flow A 120 m2 (1290 ft2) hydraulic conductivity K 0.15 cm/d porosity of media n 0.25

GROUNDWATER GROUNDWATER

1.217

Estimate the time of flow between the two lakes. Solution: Step 1. Determine the Darcy velocity y Ki K h/L 0.0015 m/s (25 m/1500 m) 2.5 105 m/s Step 2. Calculate pore velocity yp y/n (2.5 105 m/s)/0.25 1.0 104 m/s Step 3. Compute the time of travel t t L/yp 1500 m/(1 104 m/s) 1.5 107s (1 d/86,400 s) 173.6 days 2.6

Permeability The terms permeability (P) and hydraulic conductivity (K) are often used interchangeably. Both are measurements of water moving through the soil or an aquifer under saturated conditions. The hydraulic conductivity, defined by Nielsen (1991), is the quantity of water that will flow through a unit cross-sectional area of a porous media per unit of time under a hydraulic gradient of 1.0 (measured at right angles to the direction of flow) at a specified temperature. Laboratory measurement of permeability. Permeability can be determined using permeameters in the laboratory. Rearranging Eq. (4.5) and Q A, for constant head permeameter, the permeability is K

LQ HpR2

(4.13)

where K permeability, m/d or ft/d L height of sample (media), m or ft Q flow rate at outlet, m3/d or ft3/d H head loss, m or ft R radius of sample column The permeability measure from the falling-head permeameter is h1 L r 2 K a t b a b ln a b R h2

(4.14)

where r radius of standpipe, m or ft h1 height of water column at beginning, m or ft h2 height of the water column at the end, m or ft t time interval between beginning and end, d other parameters are the same as Eq. (4.13) Groundwater flows through permeable materials; such as sand, gravel, and sandstone; and is blocked by less permeable material, such as clay. Few materials are completely impermeable in nature. Even solid bedrock has fine cracks, so groundwater can flow through. Groundwater recharge occurs when surface water infiltrates the soil faster than it is evaporated, used by plants, or stored as soil moisture.

GROUNDWATER 1.218

CHAPTER 1.4

Field measurement of permeability. Ideal steady-state flow of groundwater is under the conditions of uniform pump withdrawal, a stable drawdown curve, laminar and horizontal uniform flow, a flow velocity proportional to the tangent of the hydraulic gradient, and a hom*ogeneous aquifer. Assuming these ideal conditions, the well flow is a function of the coefficient of permeability, the shape of the drawdown curve, and the thickness of the aquifer. For an unconfined aquifer the well discharge can be expressed as an equilibrium equation (Steel and McGhee, 1979; Hammer and Mackichan, 1981): Q pK

H 2 h2w ln (r/rw)

(4.15)

where Q well discharge, L/s or gpm p 3.14 K coefficient of permeability, mm/s or fps H saturated thickness of aquifer before pumping, m or ft (see Fig. 4.1) hw depth of water in the well while pumping, m or ft h well losses in Fig. 4.1 r radius of influence, m or ft rw radius of well, m or ft Also under ideal conditions, the well discharge from a confined aquifer can be calculated as Q 2pKm

H hw ln (r/rw)

(4.16)

where m is the thickness of the aquifer, m or ft. Other parameters are the same as Eq. (4.15). Values of Q, H, and r may be assumed or measured from field well tests, with two observation wells, often establishing a steady-state condition for continuous pumping for a long period. The coefficient of permeability can be calculated by rearranging Eqs. (4.14) and (4.15). The K value of an unconfined aquifer is also computed by the equation: K

Q ln(r2/r1) p (h22 h21)

(4.17)

and for a confined aquifer: K

Q ln (r2/r1) 2mp (h2 h1)

(4.18)

where h1, h2 depth of water in observation wells 1 and 2, m or ft r1, r2 centerline distance from the well and observation wells 1 and 2, respectively, m or ft A well is pumped to equilibrium at 4600 gpm (0.29 m3/s) in an unconfined aquifer. The drawdown in the observation well at 100 ft (30.5 m) away from the pumped well is 10.5 ft (3.2 m) and at 500 ft (152 m) away is 2.8 ft (0.85 m). The water table is 50.5 ft (15.4 m). Determine the coefficient of permeability.

EXAMPLE 1:

Solution: h1 50.5 10.5 40.0 (ft) h2 50.5 2.8 47.7 (ft) r1 100 ft r2 500 ft Q 4600 gpm 4600 gpm 0.002228 cfs/gpm 10.25 cfs

GROUNDWATER GROUNDWATER

1.219

Using Eq. (4.17) K

Q ln (r2/r1) p (h22 h21)

(10.25 cfs) ln(500 ft/100 ft) 3.14 [(47.7 ft)2 (40 ft)2] 0.00778 ft/s

0.00237 m/s

Referring to Fig. 4.1, a well with a diameter of 0.46 m (1.5 ft) in a confined aquifer which has a uniform thickness of 16.5 m (54.1 ft). The depth of the top impermeable bed to the ground surface is 45.7 m (150 ft). Field pumping tests are carried out with two observation wells to determine the coefficient of permeability of the aquifer. The distances between the test well and observation wells 1 and 2 are 10.0 and 30.2 m (32.8 and 99.0 ft), respectively. Before pumping, the initial piezometric surface in the test well and the observation wells are 10.4 m (34.1 ft) below the ground surface. After pumping at a discharge rate of 0.29 m3/s (4600 gpm) for a few days, the water levels in the wells are stabilized with the following drawdowns: 8.6 m (28.2 ft) in the test well, 5.5 m (18.0 ft) in the observation well 1, and 3.2 m (10.5 ft) in the observation well 2. Compute (a) the coefficient of permeability of the aquifer and (b) the well discharge with the drawdown in the well 10 m (32.8 ft) above the impermeable bed if the radius of influence (r) is 246 m (807 ft) and ignoring head losses.

EXAMPLE 2:

Solution: Step 1. Let a datum be the top of the aquifer, then H 45.7 m 10.4 m 35.3 m hw H 8.6 m 35.3 m 8.6 m 26.7 m h1 35.3 m 5.5 m 29.8 m h2 35.3 m 3.2 m 32.1 m m 16.5 m Using Eq. (4.18) K

Q ln (r2/r1) 2pm (h2 h1) 0.29 m3/s ln (30.2 m/10 m) 2 3.14 1.65 m (32.1 m 29.8 m)

0.00134 m/s 0.00441 ft/s

Step 2. Estimate well discharge hw 10 m, H 35.3 m,

rw 0.46 m r 246 m

Using Eq. (4.16) H hw ln(r/rw) 2 3.14 0.00134 m/s 16.5 m (35.3 10) m/ln(264/0.46)

Q 2pKm

0.553 m3/s or

8765 gpm

GROUNDWATER 1.220

2.7

CHAPTER 1.4

Specific Capacity The permeability can be roughly estimated by a simple field well test. The difference between the static water level prior to any pumping and the level to which the water drops during pumping is called drawdown (Fig. 4.2). The discharge (pumping) rate divided by the drawdown is the specific capacity. The specific capacity gives the quantity of water produced from the well per unit depth (ft or m) of drawdown. It is calculated by Specific capacity Q/wd

(4.19)

where Q discharge rate, gpm or m3/s wd well drawdown, ft or m EXAMPLE: The static water elevation is at 572 ft (174.3 m) before pumping. After a prolonged normal well pumping rate of 120 gpm (7.6 L/s), the water level is at 564 ft (171.9 m). Calculate the specific capacity of the well.

Solution: Specific capacity Q/wd 120 gpm/(572 564) ft 15 gpm/ft Specific capacity (7.6 L/s)/(174.3 171.9) m

3.17 L/s m

STEADY FLOWS IN AQUIFERS Referring to Fig. 4.3, if z1 z2, for an unconfined aquifer, Q KA dh/dL

(4.20)

and let the unit width flow be q, then q Kh dh/dL qdL Kh dh

(4.21)

by integration: L

q 3 dL K 3 hdh h 0 2

qL (K/2)(h21 h22) K(h21 h22) q 2L

(4.22)

This is the so-called Dupuit equation. For a confined aquifer, it is a linear equation q where

KD(h1 h2) L

q unit width flow, m /d or ft /d K coeffcient of permeability, m/d or ft/d h1,h2 piezometric head at locations 1 and 2, m or ft L length of aquifer between piezometric measurements, m or ft D thickness of aquifer, m or ft 2

(4.23)

GROUNDWATER GROUNDWATER

1.221

EXAMPLE: Two rivers are located 1800 m (5900 ft) apart and fully penetrate an aquifer. The water elevation of the rivers are 48.5 m (159 ft) and 45.6 m (150 ft) above the impermeable bed. The hydraulic conductivity of the aquifer is 0.57 m/d. Estimate the daily discharge per meter of width between the two rivers, neglecting recharge.

Solution: This case can be considered as an unconfined aquifer: K 0.57 m/d h1 48.5 m h2 45.6 m L 1800 m Using Eq. (4.22) the Dupuit equation: q

K(h21 h22) 0.57 m/d[(48.5 m)2 (45.6 m)2] 2L 2 1800 m

0.0432 m2/d

ANISOTROPIC AQUIFERS Most real geologic formations tend to have more than one direction for the movement of water due to the nature of the material and its orientation. Sometimes a soil formation may have a hydraulic conductivity (permeability) in the horizontal direction, Kx, radically different from that in the vertical direction, Kz. This phenomenon (Kx Kz), is called anisotropy. When hydraulic conductivities are the same in all directions, (Kx Kz), the aquifer is called isotropic. In typical alluvial deposits, Kx is greater than Kz. For a two-layered aquifer of different hydraulic conductivities and different thicknesses, applying Darcy’s law to horizontal flow can be expressed as Kx

K1z1 K2z2 z1 z2

(4.25)

or, in general form Kx

a Ki zi a zi

(4.26)

where Ki hydraulic conductivity in layer i, mm/s or fps zi aquifer thickness of layer i, m or ft For a vertical groundwater flow through two layers, let qz be the flow per unit horizontal area in each layer. The following relationship exists: dh1 dh2 a

z1 z2 bqz K1 K2

(4.27)

Since (dh1 dh2)Kz (z1 z2)qz then dh1 dh2 a

z1 z2 bqz Kz

(4.28)

GROUNDWATER 1.222

CHAPTER 1.4

where Kz is the hydraulic conductivity for the entire aquifer. Comparison of Eqs. (4.27) and (4.28) yields z1 z2 z1 z2 Kz K1 K2 Kz

z1 z2 z1/K1 z2/K2

(4.29)

or in general form Kz

a zi

a zi /Ki The ratios of Kx to Kz for alluvium are usually between 2 and 10.

(4.30)

UNSTEADY (NONEQUILIBRIUM) FLOWS Equilibrium equations described in the previous sections usually overestimate hydraulic conductivity and transmissivity. In practical situations, equilibrium usually takes a long time to reach. Theis (1935) originated the equation relations for the flow of ground water into wells and was then improved on by other investigators (Jacob, 1940, 1947; Wenzel 1942; Cooper and Jacob, 1946). Three mathematical/graphical methods are commonly used for estimations of transmissivity and storativity for nonequilibrium flow conditions. They are the Theis method, the Cooper and Jacob (straight-line) method, and the distance-drawdown method.

5.1

Theis Method The nonequilibrium equation proposed by Theis (1935) for the ideal aquifer is d

Q ` eu Q W(u) du 4pT 3 u 4pT

(4.31)

and

r2S (4.32) 4Tt where d drawdown at a point in the vicinity of a well pumped at a constant rate, ft or m Q discharge of the well, gpm or m3/s T transmissibility, ft2/s or m2/s r distance from pumped well to the observation well, ft or m S coefficient of storage of aquifer t time of the well pumped, min W(u) well function of u u

The integral of the Theis equation is written as W(u), and is the exponential integral (or well function) which can be expanded as a series: u3 u4 u2 c 2 2! 3 3! 4 4! un 0.5772 ln u u u2/4 u3/18 u4/96 c (1)n1 n n!

W(u) 0.5772 ln u u

(4.33a) (4.33b)

Values of W(u) for various values of u are listed in Appendix B, which is a complete table by Wenzel (1942) and modified from Illinois EPA (1990).

GROUNDWATER GROUNDWATER

1.223

If the coefficient of transmissibility T and the coefficient of storage S are known, the drawdown d can be calculated for any time and at any point on the cone of depression including the pumped well. Obtaining these coefficients would be extremely laborious and is seldom completely satisfied for field conditions. The complete solution of the Theis equation requires a graphical method of two equations (Eqs. (4.31) and (4.32)) with four unknowns. Rearranging as: d

Q W(u) 4pT

(4.31)

and 4T r2 (4.34) t S u Theis (1935) first suggested plotting W(u) on log-log paper, called a type curve. The values of S and T may be determined from a series of drawdown observations on a well with known times. Also, prepare another plot, of values of d against r2/t on transparent log-log paper, with the same scale as the other figure. The two plots (Fig. 4.5) are superimposed so that a match point can be obtained in the region of which the curves nearly coincide when their coordinate axes are parallel. The coordinates of the match point are marked on both curves. Thus, values of u, W(u), d, and r2/t can be obtained. Substituting these values into Eqs. (4.31) and (4.32), values of T and S can be calculated.

FIGURE 4.5

Observed data and Theis type curve.

GROUNDWATER 1.224

CHAPTER 1.4

An artesian well is pumped at a rate of 0.055 m3/s for 60 h. Observations of drawdown are recorded and listed below as a function of time at an observation hole 90 m away. Estimate the transmissivity and storativity using the Theis method.

EXAMPLE:

Solution: Time t, min

Drawdown d, m

1 2 3 4 5 6 10 15 20 30 40 50 60 80 90 100 200 300 600 900

0.12 0.21 0.33 0.39 0.46 0.5 0.61 0.75 0.83 0.92 1.02 1.07 1.11 1.2 1.24 1.28 1.4 1.55 1.73 1.9

r2/t, m2/min 8100 4050 2700 2025 1620 1350 810 540 405 270 203 162 135 101 90 81 40.5 27 13.5 10

Step 1. Calculate r2/t and construct a table with t and d. Step 2. Plot the observed data r2/t versus d on log-log (transparency) paper in Fig. 4.5. Step 3. Plot a Theis type curve u versus W(u) on log-log paper (Fig. 4.5) using the data listed in Appendix B (Illinois EPA, 1988). Step 4. Select a match point on the superimposed plot of the observed data on the type curve. Step 5. From the plots, the coordinates of the match points on the two curves are Observed data: r2/t 320 m2/min d 0.90 m Type curve: W(u) 3.5 u 0.018 Step 6. Substitute the above values to estimate T and S. Using Eq. (4.31) Q W(u) 4pT QW(u) 0.055 m3/s 3.5 T 4p 0.90 m 4pd 0.017 m2/s d

Using Eq. (4.34) 4Tu r2 t S

GROUNDWATER GROUNDWATER

1.225

4 0.017 m2/s 0.018 4Tu r2/t 320 m2/min (1 min/60 s)

2.30 104 5.2

Cooper–Jacob Method Cooper and Jacob (1946) modified the nonequilibrium equation. It is noted that the parameter u in Eq. (4.32) becomes very small for large values of t and small values of r. The infinite series for small u, W(u), can be approximated by W(u) 0.5772 ln u 0.5772 ln

r2S 4Tt

(4.35)

Then Q Q r2S W(u) a0.5772 ln b 4pT 4pT 4Tt Further rearrangement and conversion to decimal logarithms yields d

(4.36)

2.303Q 2.25Tt (4.37) log 2 4pT rS Therefore, the drawdown is to be a linear function of log t. A plot of d versus the logarithm of t forms a straight line with the slope Q/4pT and an intercept at d 0, when t t0, yields d

2.25Tt0 2.3Q log 4pT r2S

(4.38)

Since log(1) 0 therefore 1 S

2.25Tt0 r2S 2.25Tt0

(4.39)

If the slope is measured over one log cycle of time, the slope will equal the change in drawdown d, and Eq. (4.37) becomes d

2.303Q 4pT

then T

2.303Q 4pd

(4.40)

The Cooper and Jacob modified method solves for S and T when values of u are less than 0.01. The method is not applicable to periods immediately after pumping starts. Generally, 12 h or more of pumping are required. EXAMPLE: Using the given data in the above example (by the Theis method) with the Cooper and Jacob method, estimate the transmissivity and storativity of a confined aquifer.

Solution: Step 1. Determine t0 and d. Values of drawdown (d) and time (t) are plotted on semilog paper with the t in the logarithmic scale as shown in Fig. 4.6. A best fit straight line is drawn through the observed data. The intercept

GROUNDWATER 1.226

CHAPTER 1.4

FIGURE 4.6

Relationship of drawdown and the time of pumping.

of the t axis is 0.98 min. The slope of the line d is measured over 1 log cycle of t from the figure. We obtain: t0 0.98 min 58.8 s and, for a cycle (t 10 to 100 min) d 0.62 m Step 2. Compute T and S Using Eq. (4.40) 2.303Q 2.303 0.055 m3/s 4 3.14 0.62 m 4pd

0.016 m2/s Using Eq. (4.39) S

2.25Tt0 2

2.25 0.016 m2/s 58.8 s (90 m)2

2.61 104

GROUNDWATER 1.227

GROUNDWATER

5.3

Distance-drawdown Method The distance-drawdown method is a modification of the Cooper and Jacob method and is applied to obtain quick information about the aquifer characteristics while the pumping test is in progress. The method needs simultaneous observations of drawdown in three or more observation wells. The aquifer properties can be determined from pumping tests by the following equations (Watson and Burnett, 1993): 0.366Q for SI units (h0 h)

(4.41)

528Q for U.S. customary units (h0 h)

(4.42)

T T and

where

2.25Tt for SI units r20

(4.43)

Tt U.S. customary units 4790r20

(4.44)

T transmissivity, m2/d or gpd/ft Q normal discharge rate, m3/d or gpm (h0 h) drawdown per log cycle of distance, m or ft S storativity, unitless t time since pumping when the simultaneous readings are taken in all observation wells, d or min r0 intercept of the straight-line plot with the zero-drawdown axis, m or ft

The distance-drawdown method involves the following procedures: 1. Plot the distance and drawdown data on semi-log paper; drawdown on the arithmetic scale and the distance on the logarithmic scale. 2. Read the drawdown per log cycle in the same manner for the Cooper and Jacob method: this gives the value of (h0 h). 3. Draw a best-fit straight line. 4. Extend the line to the zero-drawdown and read the value of the intercept, r0. EXAMPLE: A water supply well is pumping at a constant discharge rate of 1000 m3/d (11,000 gpm). It happens that there are five observation wells available. After pumping for 3 h, the drawdown at each observation well is recorded as below. Estimate transmissivity and storativity of the aquifer using the distance-drawdown method.

Distance m

Drawdown, m

3 7.6 20 50 70

10 25 66 164 230

3.22 2.21 1.42 0.63 0.28

Solution: Step 1. Plot the distance-drawdown data on semi-log paper as shown in Fig. 4.7. Step 2. Draw a best-fit straight line over the observed data and extend the line to the X axis.

GROUNDWATER 1.228

CHAPTER 1.4

FIGURE 4.7

Plot of observed distance-drawdown data.

Step 3. Read the drawdown value for one log cycle. From Fig. 4.7, for the distance for a cycle from 4 to 40 m, the value of the drawdown, then (h0 h) is read as 2.0 (2.8 0.8) m. Step 4. Read the intercept on the X axis for r0, r0 93 m. Step 5. Determine T and S by Eqs. (4.41) and (4.43). T

0.366Q 0.366 1000 m3/d 2.0 m (h0 h)

183 m2/d

GROUNDWATER GROUNDWATER

1.229

Time of pumping: t 3 h/24 h/d 0.125 days S

2.25 183 m2/d 0.125 day 2.25Tt 2 r0 (93 m)2

0.006 5.4

Slug Tests In the preceding sections, the transmissivity T and storativity S of the aquifer, and permeability K of the soil are determined by boring one or two more observation wells. Slug tests use only a single well for the determination of those values by careful evaluation of the drawdown curve and information of screen geometry. The tests involve either raising or lowering the water level in the well and measuring the return to a static water level as a function of time. A typical test procedure requires introducing an object to record the volume (the slug) of the well. The Hvorslev (1951) method using a piezometer in an confined aquifer is widely used in practice due to it being quick and inexpensive. The procedures of conducting and analyzing the Hvorslev test are as follows: 1. 2. 3. 4.

Record the instantaneously raised (or lowered) water level to the static water level as H0. Read subsequently changing water levels with time as h. Thus h H0 at t 0. Measure the final raised head as H at infinite time. There is a relationship as Hh et/T0 H H0

(4.45)

where pr2 Fk Hvorslev-defined basic time lag

(4.46)

F shape factor 5. Calculate ratios (H h)/(H H0 ) as recovery. 6. Plot on semi-logarithmic paper (H h)/(H H0 ) on the logarithmic scale versus time on the arithmetic scale. 7. Find T0 at recovery equals 0.37 (37 percent of the initial change caused by the slug). 8. For piezometer intake length divided by radius (L/R) greater than 8, Hvorslev evaluated the shape factor F and proposed an equation for hydraulic conductivity K as K

r2 ln (L/R) 2LT0

(4.47)

where K hydraulic conductivity, m/d or ft/d r radius of the well casing, cm or in R radius of the well screen, cm or in L length of the well screen, cm or in T0 time required for water level to reach 37 percent of the initial change, s Other slug test methods have been developed for confined aquifers (Cooper et al., 1967; Papadopoulous et al., 1973; Bouwer and Rice, 1976). These methods are similar to Theis’s type curves in that a curve-matching method is used to determine T and S for a given aquifer. A family of

GROUNDWATER 1.230

CHAPTER 1.4

type curves Ht /H0 versus Tr/r c2 were published for five values of the variable , defined as (r2s/r2c) S, to estimate transmissivity, storativity, and hydraulic conductivity. The Bouwer and Rice (1976) slug test method is most commonly used for estimating hydraulic conductivity in groundwater. Although the method was originally developed for unconfined aquifers, it can also be applied for confined or stratified aquifers if the top of the screen is some distance below the upper confined layer. The following formula is used to compute hydraulic conductivity: K where

r2 ln (R/rw) 1 y0 t ln yt 2L

(4.48)

K hydraulic conductivity, cm/s r radius of casing, cm y0 , yt vertical difference in water levels between inside and outside the well at time t 0, and t t, m R effective radius distance over which y is dissipated, cm rw radius distance of undisturbed portion of aquifer from well centerline (usually r plus thickness of gravel). L length of screen, m t time, s

h H

EXAMPLE 1: The internal diameters of the L well casing and well screen are 10 cm (4 in) and 15 cm (6 in), respectively. The length of the well screen is 2 m (6.6 ft). The static water R level measured from the top of the casing is 2.50 m (8.2 ft). A slug test is conducted and pumped FIGURE 4.8 Hvorslev slug test. to lower the water level to 3.05 m (10 ft). The time-drawdown h in the unconfined aquifer is recorded every 3 s as shown in the following table. Determine the hydraulic conductivity of the aquifer by the Hvorslev method.

Solution: Step 1. Calculate (h H0)/(H H0) Given: H0 2.50 m, H 3.05 m Then H H0 3.05 m 2.50 m 0.55 m Time t, s

h, m

h H0, m

(h H0)/0.55

0 3 6 9 12 15 18 21 24 27 30

3.05 2.96 2.89 2.82 2.78 2.73 2.69 2.65 2.62 2.61 2.59

0.55 0.46 0.39 0.32 0.28 0.23 0.19 0.15 0.12 0.10 0.09

1.00 0.84 0.71 0.58 0.51 0.42 0.34 0.27 0.22 0.18 0.16

GROUNDWATER GROUNDWATER

FIGURE 4.9

1.231

Plot of Hvorslev slug test results.

Step 2. Plot t versus (h H0)/(H H0) on semi-log paper as shown in Fig. 4.9, and draw a bestfit straight line. Step 3. Find T0. From Fig. 4.9, read 0.37 on the (h H0)/(H H0) scale and note the time for the water level to reach 37 percent of the initial change T0 caused by the slug. This is expressed as T0. In this case T0 16.2 s. Step 4. Determine the L/R ratio R 15 cm/2 7.5 cm L/R 200 cm/7.5 cm 26.7 8 Thus Eq. (4.47) can be applied. Step 5. Find K by Eq. (4.47) r 10 cm/2 5 cm K

r2ln(L/R) 2LT0

(5 cm)2 ln 26.7/(2 200 cm 16.2 s) 0.0127 cm/s 0.0127 cm/s (1 m/100 cm) 86,400 s/d 11.0 m/d 36 ft/d

GROUNDWATER 1.232

CHAPTER 1.4

EXAMPLE 2: A screened, cased well penetrates a confined aquifer with gravel pack 3.0 cm thickness around the well. The radius of casing is 5.0 cm and the screen is 1.2 m long. A slug of water is injected and water level raised by m. The effective radial distance over which y is dissipated is 12 cm. Estimate hydraulic conductivity for the aquifer. The change of water level with time is as follows:

t, s

yt, cm

t, s

yt, cm

1 2 3 4 5 6 8

30 24 17 14 12 9.6 5.5

10 13 16 20 30 40

4.0 2.0 1.1 0.6 0.2 0.1

Solution: Step 1. Plot values of y versus t on semi-log paper as shown in Fig. 4.10. Draw a best-fit straight line. The line from y0 36 cm to yt 0.1 cm covers 2.5 log cycles. The time increment between the two points is 26 s. Step 2. Determine K by Eq. (4.48), the Bouwer and Rice equation: r 5 cm R 12 cm

FIGURE 4.10

Plot of yt versus t.

GROUNDWATER GROUNDWATER

1.233

rw 5 cm 3 cm 8 cm K

r2 ln (R/rw) 1 y0 t ln yt 2L (5 cm)2 ln(12 cm/8 cm) 1 36 cm ln 2(120 cm) 26 s 0.1 cm

9.56 103 cm/s

6 6.1

GROUNDWATER CONTAMINATION Sources of Contamination There are various sources of groundwater contamination and various types of contaminants. Underground storage tanks, agricultural activity, municipal landfills, abandoned hazardous waste sites, and septic tanks are the major threats to groundwater. Other sources may be from industrial spill, injection wells, land application, illegal dumps (big problems in many countries), road salt, saltwater intrusion, oil and gas wells, mining and mine drainage, municipal wastewater effluents, surface impounded waste material stockpiles, pipelines, radioactive waste disposal, and transportation accidents. Large quantities of organic compounds are manufactured and used by industries, agriculture, and municipalities. These man-made organic compounds are of most concern. The inorganic compounds occur in nature and may come from natural sources as well as human activities. Metals from mining, industries, agriculture, and fossil fuels also may cause groundwater contamination. Types of contaminants are classified by chemical group. They are metals (arsenic, lead, mercury, etc.), volatile organic compounds (VOCs, gasoline, oil, paint thinner), pesticides and herbicides, and radionuclides (radium, radon). The most frequently reported groundwater contaminants are the VOCs (Voelker, 1984; Rehfeldt et al., 1992). Volatile organic compounds are made of carbon, hydrogen, and oxygen and have a vapor pressure less than one atmosphere. Compounds such as gasoline or dry cleaning fluid evaporate when left open to the air. They are easily dissolved in water. There are numerous incidences of VOCs contamination caused by leaking underground storage tanks. Groundwater contaminated by VOCs poses cancer risks to humans either by ingestion of drinking water or inhalation.

6.2

Contaminant Transport Pathways The major contaminant transport mechanisms in groundwater are advection, diffusion, dispersion, adsorption, chemical reaction, and biodegradation. Advection is the movement of contaminant(s) with the flowing groundwater at the seepage velocity in the pore space and is expressed as Darcy’s law: K dh yx n L

(4.49)

where x seepage velocity, m/s or pps K hydraulic conductivity, m/s or fps n porosity dh pressure head, m or ft L distance, m or ft Eq. (4.49) is similar to Eq. (4.10b). Diffusion is a molecular-scale mass transport process that moves solutes from an area of higher concentration to an area of lower concentration. Diffusion is expressed by Fick’s law: Fx Dd

dc dx

(4.50)

GROUNDWATER 1.234

CHAPTER 1.4

where

Fx mass flux, mg/m2 s or lb/ft2 s Dd diffusion coefficient, m2/s or ft2/s dc/dx concentration gradient, mg/(m3 m) or (lb/ft3 ft)

Diffusive transport can occur at low or zero flow velocities. In a tight soil or clay, typical values of Dd range from 1 to 2 × 109 m2/s at 25°C (Bedient et al. 1994). However, typical dispersion coefficients in groundwater are several orders of magnitude greater than that in clay. Dispersion is a mixing process caused by velocity variations in porous media. Mass transport due to dispersion can occur parallel and normal to the direction of flow with two dimensional spreading. Sorption is the interaction of a contaminant with a solid. It can be divided into adsorption and absorption. An excess concentration of contaminants at the surfaces of solids is called adsorption. Absorption refers to the penetration of the contaminants into the solids. Biodegradation is a biochemical process that transforms contaminants (certain organics) into simple carbon dioxide and water by microorganisms. It can occur in aerobic and anaerobic conditions. Anaerobic biodegradation may include fermentation, denitrification, iron reduction, sulfate reduction, and methane production. Excellent and complete coverage of contaminant transport mechanisms is presented by Bedient et al. (1994). Theories and examples are covered for mass transport, transport in groundwater by advection, diffusion, dispersion, sorption, chemical reaction, and by biodegradation. Some example problems of contaminant transport are also given by Tchobanoglous and Schroeder (1985). Mathematical models that analyze complex contaminant pathways in groundwater are also discussed elsewhere (Canter and Knox, 1986; Willis and Yeh, 1987; Canter et al., 1988; Mackay and Riley, 1993; Smith and Wheatcraft, 1993; Watson and Burnett, 1993; James, 1993; Gupta, 1997). The transport of contaminants in groundwater involves adsorption, advection, diffusion, dispersion, interface mass transfer, biochemical transformations, and chemical reactions. On the basis of mass balance, the general equation describing the transport of a dissolved contaminant through an isotropic aquifer under steady-state flow conditions can be mathematically expressed as (Gupta 1997) rb rb 's 'c r ck, Sks w 't 't (4.51) '2c '2c '2c 'c 'c 'c Dx 2 Dy 2 Dz 2 Vx Vy Vz 'x 'y 'z 'x 'y 'z where

c solute (contaminant in liquid phase) concentration, g/m3 S concentration in solid phase as mass of contaminant per unit mass of dry soil, g/g t time, d b bulk density of soil, kg/m3x effective porosity k,, ks first-order decay rate in the liquid and soil phases, respectively, d−1 x, y, z Cartesian coordinates, m Dx, Dy, Dz directional hydrodynamic dispersion coefficients, m2/d Vx, Vy, Vz directional seepage velocity components, m/d

There are two unknowns (c and S) in Eq. (4.51). Assuming a linear adsorption isotherm of the form S Kd c

(4.52)

where Kd distribution coefficient due to chemical reactions and biological degradation and substituting Eq. (4.52) into Eq. (4.51), we obtain R

'c '2c '2c '2c 'c 'c 'c kc Dx 2 Dy 2 Dz 2 Vx Vy Vz 't 'x 'y 'z 'x 'y 'z

where R 1 kd(b) retardation factor which slows the movement of solute due to adsorption

(4.53)

GROUNDWATER GROUNDWATER

1.235

k k1 + ks Kd(b) overall first-order decay rate, d−1 The general equation under steady-state flow conditions in the x direction, we modify from Eq. (4.53). 6.3

Underground Storage Tank There are 3 to 5 million underground storage tanks (UST) in the United States. It is estimated that 3 to 10 percent of these tanks and their associated piping systems may be leaking (USEPA 1987). The majority of UST contain petroleum products (gasoline and other petroleum products). When the UST leaks and is left unattended, and subsequently contaminates subsurface soils, surface and groundwater monitoring and corrective actions are required. The migration or transport pathways of contaminants from the UST depend on the quantity released, the physical properties of the contaminant, and characteristics of the soil particles. When a liquid contaminant is leaked from a UST below the ground surface, it percolates downward to the unconfined groundwater surface. If the soil characteristics and contaminant properties are known, it can be estimated whether the contaminant will reach the groundwater. For hydrocarbons in the unsaturated (vadose) zone it can be estimated by the equation (USEPA 1987) D

Rv V A

(4.54) where D maximum depth of penetration, m Rv a coefficient of retention capacity of the soil and the viscosity of the product (contaminant) V volume of infiltrating hydrocarbon, m3 A area of spill, m2 The typical values for Rv are as follows (USEPA 1987): Soil Coarse gravel Gravel to coarse sand Coarse to medium sand Medium to fine sand Fine sand to silt

Gasoline

Kerosene

400 250 130 80 50

200 125 66 40 25

Light fuel oil 100 62 33 20 12

Retention (attenuative) capacities for hydrocarbons vary approximately from 5 L/m3 in a coarse gravel to more than 40 L/m3 in silts. Leaked gasoline can travel 25 ft (7.6 m) through unsaturated, permeable, alluvial, or glacial sediments in a few hours, or at most a few days, which is extremely site specific. The major movement of nonaqueous-phase liquids that are less dense than water (such as gasoline and other petroleum products) in the capillary zone (between unsaturated and saturated zones) is lateral. The plume will increase thickness (vertical plane) and width depending on leakage rates and the site’s physical conditions. The characteristic shape of the flow is the so-called “oil package.’’ Subsequently it will plug the pores of the soil or be diluted and may be washed out into the water table. The transport of miscible or dissolved substances in saturated zones follows the general direction of groundwater flow. The transport pathways can be applied to the models (laws) of advection and dispersion. The dispersion may include molecular diffusion, microscopic dispersion, and macroscopic dispersion. Immiscible substances with a specific gravity of less than 1.0 (lighter than water) are usually found only in the shallow part of the saturated zone. The transport rate depends on the groundwater gradient and the viscosity of the substance.

GROUNDWATER 1.236

CHAPTER 1.4

Immiscible substances with a specific gravity of more than 1.0 (denser than water) move downward through the saturated zone. A dense immiscible substance poses a greater danger in terms of migration potential than less dense substances due to its deeper penetration into the saturated zone. When the quantity of released contaminant exceeds the retention capacity of the unsaturated and saturated zones, the denser nonaqueous-phase liquid continues its downward migration until it reaches an impermeable boundary. A liquid substance leaking from a UST enters the vapor phase in the unsaturated zone according to its specific vapor pressure. The higher the vapor pressure of the substance, the more it evaporates. The contaminant in the vapor phase moves by advection and by diffusion. Vapor moves primarily in a horizontal direction depending on the slope of the water table and the location of the impermeable bedrock. If the vapor, less dense than air, migrates in a vertical direction, it may accumulate in sewer lines, basem*nts, and such areas. 6.4

Groundwater Treatment Once a leaking UST is observed, corrective action should be carried out, such as tank removal, abandonment, rehabilitation, removal/excavation of soil and sediment, onsite and/or offsite treatment and disposal of contaminants, product and groundwater recovery, and groundwater treatment, etc. Selection of groundwater treatment depends on the contaminants to be removed. Gasoline and volatile organic compounds can be removed by air stripping and stream stripping processes. Activated carbon adsorption, biological treatment, and granular media filtration can be used for removal of gasoline and other organics. Nonvolatile organics are removable by oxidation/reduction processes. Inorganic chemicals can be treated by coagulation/sedimentation, neutralization, dissolved air floatation, granular media filtration, ion exchange, resin adsorption, and reverse osmosis. These treatment processes are discussed in detail in Chapter 1.6, Public Water Supply and in Chapter 1.7, Wastewater.

SETBACK ZONES Section 1428 of the Safe Drinking Water Act (SDWA) requires each state to submit a wellhead protection program to the USEPA. For example, Illinois EPA promulgated the Illinois Groundwater Protection Act (IGPA) in 1991. The Act assigned the responsibilities of existing state agencies and established groundwater classifications. On the basis of classification, different water quality standards, monitoring, and remedial requirements can be applied. Classifications are based on the PCB levels which may be associated with hazardous wastes. Groundwater used as a public water supply source is called potable groundwater. This requires the highest degree of protection with the most stringent standards. The groundwater quality standards for potable groundwater are generally equal to the USEPA’s maximum contamination levels applicable at-the-tap pursuant to the SDWA. The rationale is that potable groundwater should be safe for drinking water supply without treatment. The state primacy agency establishes a comprehensive program for the protection of groundwater. Through inter-agency cooperation, local groundwater protection programs can help to prevent unexpected and costly water supply systems. An Illinois community experienced a leaking gasoline underground tank, operated by the city-owned garage. It contaminated one well and threatened to contaminate the entire well field. The city has spent more than $300,000 in an attempt to replace the water supply. Some parts of the groundwater protection programs, such as minimum and maximum setback zones for wellhead protection, are used to protect public and private drinking water supplies from potential sources of groundwater contamination. Each community well must have a setback zone to restrict land use near the well. The setback zone provides a buffer between the well and potential contamination sources and routes. It will give time for cleanup efforts of contaminated groundwater or to obtain an alternative water supply source before the existing groundwater source becomes unfit for use. A minimum setback zone is mandatory for each public well. Siting for new potential primary or secondary pollution sources or potential routes is prohibited within the setback zone. In Illinois, the

GROUNDWATER GROUNDWATER

FIGURE 4.11

1.237

Maximum and minimum setback zones.

minimum setback zone is a 200 ft (61 m) radius area around the wellhead for every water supply well. For some vulnerable aquifers, the zone may be 400 ft in radius. The maximum setback zone is a second level of protection from pollution. It prohibits the siting of new potential primacy pollution sources within the area outside the minimum setback zone up to 1000 ft (305 m) from a wellhead in Illinois (Fig. 4.11). Maximum setback zones allow the well owners, county or municipal government, and state to regulate land use beyond the minimum setback zone. The establishment of a maximum zone is voluntary. A request to determine the technical adequacy of a maximum setback zone determination must first be submitted to the state by a municipality or county. Counties and municipalities served by community water supply wells are empowered to enact maximum setback zone ordinances. If the community water supply wells are investor or privately owned, a county or municipality served by that well can submit an application on the behalf of the owner. 7.1

Lateral Area of Influence As described in the previous section, the lateral radius of influence (LRI) is the horizontal distance from the center of the well to the outer limit of the cone of depression (Figs. 4.1 and 4.12). It is the distance from the well to where there is no draw of groundwater (no reduction in water level). The lateral area of influence (LAI) outlines the extent of the cone of depression on the land surface as shown in the hatched area in Fig. 4.12. The lateral area of influence in a confined aquifer is generally 4000 times greater than the LAI in an unconfined aquifer (Illinois EPA, 1990). If a pollutant is introduced within the LAI, it will reach the well faster than other water replenishing the well. The slope of the water table steepens toward the well within the LAI. Therefore, it is extremely important to identify and protect the lateral area of influence. The LAI is used to establish a maximum setback zone.

GROUNDWATER 1.238

CHAPTER 1.4

FIGURE 4.12 Lateral area of influence and the cone of depression under normal operation conditions (Illinois EPA, 1990).

7.2

Determination of Lateral Radius of Influence There are three types of determination methods for the lateral radius of influence. They are the direct measurement method, the use of the Theis equation or volumetric flow equation, and the use of a curve-matching technique with the Theis equation. The third method involves the interpretation of pump test data from observation wells within the minimum setback zone using a curve-matching technique and the Theis equation to determine the transmissivity and storativity of the aquifer. Once these aquifer constants have been obtained, the Theis equation is then used to compute the lateral radius of influence of the well. It should be noted that when the observation well or piezometer measurement was made outside of the minimum setback zone, the determination could have been conducted by the direct measurement method. Volumetric flow equations adopted by the Illinois EPA (1990) are as follows. For unconfined unconsolidated or unconfined nonfractured bedrock aquifers, the lateral radius of influence can be calculated by r

Qt 4524nH B

where r radius of influences, ft Q daily well flow under normal operational conditions, cfs t time that the well is pumped under normal operational conditions, min H open interval or length of well screen, ft n aquifer porosity, use following data unless more information is available:

(4.55)

GROUNDWATER GROUNDWATER

Sand Gravel Sand and gravel Sandstone Limestone primary dolomites Secondary dolomites

1.239

0.21 0.19 0.15 0.06 0.18 0.18

If pump test data is available for an unconfined/confined unconsolidated or nonfractured bedrock aquifer, the LRI can be calculated by r

uTt B 2693 S

(4.56)

where r radius of influence, ft T aquifer transmissivity, gpd/ft t time that well is pumped under normal operational conditions, min S aquifer storativity or specific yield, dimensionless u a dimensionless parameter related to the well function W(u) and W(u) where

T(h0 h) 114.6 Q

(4.57)

W(u) well function h0 h drawdown in the piezometer or observation well, ft Q production well pumping rate under normal operational conditions, gpm

Direct measurement method. The direct measurement method involves direct measurement of drawdown in an observation well piezometer or in an another production well. Both the observation well and another production well should be located beyond the minimum setback zone of the production well and also be within the lateral area of influence. EXAMPLE: A steel tape is used to measure the water level of the observation well located 700 ft from a drinking water well. The pump test result shows that under the normal operational conditions, 0.58 ft of drawdown compared to the original nonpumping water level is recorded. Does this make the well eligible for applying maximum setback zone?

Answer: Yes. A municipality or county may qualify to apply for a maximum setback zone protection when the water level measurements show that the drawdown is greater than 0.01 ft and the observation well is located beyond (700 ft) the minimum setback zone (200 to 400 ft). It is also not necessary to estimate the lateral radius of influence of the drinking water well. Theis equation methods. The Theis equation (Eqs. (4.56) and (4.57)) can be used to estimate the lateral radius of influence of a well if the aquifer parameters, such as transmissivity, storativity, and hydraulic conductivity, are available or can be determined. The Theis equation is more useful if the observation well is located within the minimum setback zone when the direct measurement method can not be applied. Equations (4.52) and (4.53) can be used if pump test data is available for an unconfined/confined, unconsolidated or nonfractured bedrock aquifer (Illinois EPA, 1990). EXAMPLE: An aquifer test was conducted in a central Illinois county by the Illinois State Water Survey (Walton 1992). The drawdown data from one of the observations located 22 ft from the production well was analyzed to determine the average aquifer constants as follows:

GROUNDWATER 1.240

CHAPTER 1.4

Transmissivity: T 340,000 gpd/ft of drawdown Storativity: S 0.09 The pumping rate under normal operational conditions was 1100 gpm. The volume of the daily pumpage was 1,265,000 gal. Is this aquifer eligible for a maximum setback zone protection? Solution: Step 1. Determine the time of pumping, t 1,265,000 gal y Q 1100 gal/min

1150 min Step 2. The criteria used to represent a minimum level of drawdown (h0 h) is assumed to be h0 h 0.01 ft Step 3. Calculate W(u) using Eq. (4.57): T(h0 h) 114.6 Q

W(u)

340,000 gpm/ft 0.01 ft 114.6 1100 gpm

0.027 Step 4. Find u associated with a W(u) 0.027 in Appendix B: u 2.43 Step 5. Compute the lateral radius of influence r using Eq. (4.56): r

uTt B 2693 S 2.43 340,000 gpd/ft 1150 min 2693 0.09 B

1980 ft Answer: Yes. The level of drawdown was estimated to be 0.01 ft at a distance of 1980 ft from the drinking water wellhead. Since r is greater than the minimum setback distance of 400 ft from the well head, the well is eligible for a maximum setback zone. 7.3

The Use of Volumetric Flow Equation The volumetric flow equation is a cost-effective estimation for the radius of influence and is used for wells that pump continuously. It is an alternative procedure where aquifer constants are not available and where the Theis equation cannot be used. The volumetric flow equation (Eq. (4.55)) may be utilized for wells in unconfined unconsolidated or unconfined nonfractured bedrock aquifers. EXAMPLE: A well in Peoria, Illinois (Illinois EPA, 1995), was constructed in a sand and gravel deposit which extends from 40 to 130 ft below the land surface. The static water level in the aquifer is 65 ft below the land surface. Because the static water level is below the top of the aquifer, this

GROUNDWATER GROUNDWATER

1.241

would be considered an unconfined aquifer. The length of the well screen is 40 ft. Well operation records indicate that the pump is operated continuously for 30 days every other month. The normal discharge rate is 1600 gpm (0.101 m3/s). Is the well eligible for a maximum setback zone protection? Solution: Step 1. Convert discharge rate from gpm to ft3/d: Q 1600 gpm gal 1440 min 1 ft3 1600 min 7.48 gal 1 day 308,021 ft3/d 8723 m3/d Step 2. Calculate t in min: t 30 day 30 day × 1440 min/d 43,200 min Step 3. Find the aquifer porosity n: n 0.15 for sand and gravel Step 4. Calculate r using Eq. (4.55): H 40 ft r

Qt B 4524 nH 308,021 ft3/d 43,200 min B 4524 0.15 40 ft

700 ft 213.4 m Yes, the well is eligible for maximum setback zone. Note: The volumetric flow equation is very simple and economical.

REFERENCES Babbitt, H. E., Doland, J. J. and Cleasby, J. L. 1959. Water supply engineering. 6th edn. New York: McGraw-Hill. Bedient, P. B. and Huber, W. C. 1992. Hydrology and floodplain analysis. 2nd edn. Reading, Massachusetts: Addison-Wesley. Bedient, P. B., Rifai, H. S. and Newell, C. J. 1994. Ground water contamination transport and remediation. Englewood Cliffs, New Jersey: Prentice Hall. Bouwer, H. and Rice, R. C. 1976. A slug test for determining hydraulic conductivity of unconfined aquifers with completely or partially penetrating wills. Water Resources Res, 12: 423–428. Canter, L. W. and Knox, R. C. 1986. Ground water pollution control. Chelsea, Michigan: Lewis. Canter, L. W., Knox, R. C. and Fairchild, D. M. 1988. Gound water quality protection. Chelsea, Michigan: Lewis. Cooper, H. H., Jr., Bredehoeft, J. D. and Papadopoulos, I. S. 1967. Response of a finite-diameter well to an instantaneous charge of water. Water Resources Res, 3: 263–269. Cooper, H. H. Jr. and Jacob, C. E. 1946. A generalized graphical method for evaluating formation constants and summarizing well field history. Trans. Amer. Geophys. Union, 27: 526–34.

GROUNDWATER 1.242

CHAPTER 1.4

Forrest, C. W. and Olshansky, R. 1993. Groundwater protection by local government. University of Illinois at Urbana-Champaign. Illinois. Gupta, A. D. 1997. Groundwater and the environment. In: Biswas, A. K. (ed.) Water resources. New York: McGraw-Hill. Hammer, M. J. 1986. Water and wastewater technology. New York: John Wiley. Hammer, M. J. and Mackichan, K. A. 1981. Hydrology and quality of water resources. New York: John Wiley. Hvorslev, M. J. 1951. Time lag and soil permeability in groundwater observations. U.S. Army Corps of Engineers Waterways Experiment Station, Bulletin 36, Vicksburg, Mississippi. Illinois Environmental Protection Agency (IEPA). 1987. Quality assurance and field method manual. Springfield: IEPA. Illinois Environmental Protection Agency (IEPA). 1988. A primer regarding certain provisions of the Illinois groundwater protection act. Springfield: IEPA. Illinois Environmental Protection Agency (IEPA). 1990. Maximum setback zone workbooks. Springfield: IEPA. Illinois Environmental Protection Agency (IEPA). 1994. Illinois water quality report 1992–1993. Springfield: IEPA. Illinois Environmental Protection Agency (IEPA). 1995. Guidance document for groundwater protection needs assessments. Springfield: IEPA. Illinois Environmental Protection Agency (IEPA). 1996. Illinois Water Quality Report 1994–1995. Vol. I. Springfield, Illinois: Bureau of Water, IEPA. Jacob, C. E. 1947. Drawdown test to determine effective radius of artesian well. Trans. Amer. Soc. Civil Engs., 112(5): 1047–1070. James, A. (ed.) 1993. Introduction to water quality modelling. 2nd edn. Chichester: John Wiley. Leopold, L. B. 1974. Water: A primer. New York: W. H. Freeman. Mackay, R. and Riley, M. S. 1993. Groundwater quality modeling. In: James, A. (ed.) An introduction to water quality modeling. 2nd edn. New York: John Wiley. Nielsen, D. M. (ed.). 1991. Practical handbook of groundwater monitoring. Chelsea, Michigan: Lewis. Papadopoulous, I. S., Bredehoeft, J. D. and Cooper, H. H. 1973. On the analysis of slug test data. Water Resources Research, 9(4): 1087–1089. Rehfeldt, K. R., Raman, R. K., Lin, S. D. and Broms, R. E. 1992. Assessment of the proposed discharge of ground water to surface waters of the American Bottoms area of southwestern Illinois. Contract report 539. Champaign: Illinois State Water Survey. Rose, H. E. 1949. On the resistance coefficient-Reynolds number relationship for fluid flow through a bed of granular materials. Proc. Inst. Mech. Engrs. (London) 154, 160. Smith, L. and Wheatcraft, S. W. 1993. In: Maidment, D. R. (ed.) Handbook of hydrology, New York: McGraw-Hill. Steel, E. W. and McGhee, T. J. 1979. Water supply and sewerage. 5th edn. New York: McGraw-Hill. Tchobanoglous, G. and Schroeder, E. D. 1985. Water quality, characteristics, modeling, modification. Reading, Massachusetts: Addison-Wesley. Theis, C. V. 1935. The relation between the lowering of the piezometric surface and the rate and duration of discharge of a well using ground storage. Trans. Am. Geophys. Union 16: 519–524. US Environmental Protection Agency (EPA). 1987. Underground storage tank corrective action technologies. EPA/625/6-87-015. Cincinnati, Ohio: USEPA. US Environmental Protection Agency (EPA). 1994. Handbook: Ground water and wellhead protection, EPA/625/R-94/001. Washington, DC: USEPA. Voelker, D. C. 1984. Quality of water in the alluvial aquifer, American Bottoms, East St. Louis, Illinois. US Geological Survey Water Resources Investigation Report 84-4180. Chicago, Illinois. Walton, W. C. 1962. Selected analytical methods for well and aquifer evaluation. Illinois State Water Survey. Bulletin 49, Urbana, Illinois. Wanielista, M. 1990. Hydrology and water quality control. New York: John Wiley. Watson, I., and Burnett, A. D. 1993. Hydrology: An environmental approach. Cambridge, Ft. Lauderdale, Florida: Buchanan Books. Wenzel, L. K. 1942. Methods for determining the permeability of water-bearing materials. US Geological Survey. Water Supply Paper 887, p. 88. Willis. R., and Yeh, W. W-G. 1987. Groundwater systems planning & management. Englewood Cliffs, New Jersey: Prentice-Hall.

Source: HANDBOOK OF ENVIRONMENTAL ENGINEERING CALCULATIONS

CHAPTER 1.5

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS Shun Dar Lin

1 DEFINITIONS AND FLUID PROPERTIES 1.243 1.1 Weight and Mass 1.243 1.2 Specific Weight 1.244 1.3 Pressure 1.244 1.4 Viscosity of Water 1.246 1.5 Perfect Gas 1.249 2 WATER FLOW IN PIPES 1.250 2.1 Fluid Pressure 1.250 2.2 Head 1.250 2.3 Pipeline Systems 1.265 2.4 Distribution Networks 1.269 2.5 Sludge Flow 1.275 2.6 Dividing-Flow Manifolds and Multiport Diffusers 1.275 3 PUMPS 1.275 3.1 Types of Pump 1.275 3.2 Pump Performance 1.275 3.3 Cost of Pumping 1.279

4 WATER FLOW IN OPEN CHANNELS 1.281 4.1 Che’zy Equation for Uniform Flow 1.281 4.2 Manning Equation for Uniform Flow 1.282 4.3 Partially Filled Conduit 1.284 4.4 Self-cleansing Velocity 1.289 4.5 Specific Energy 1.291 4.6 Critical Depth 1.292 4.7 Hydraulic Jump 1.294 5 FLOW MEASUREMENTS 1.297 5.1 Velocity Measurement in Open Channel 1.297 5.2 Velocity Measurement in Pipe Flow 1.297 5.3 Discharge Measurement of Water Flow in Pipes 1.298 5.4 Discharge Measurements 1.301 REFERENCES 1.310

1 DEFINITIONS AND FLUID PROPERTIES 1.1 Weight and Mass The weight (W) of an object, in the International System of Units (SI), is defined as the product of its mass (m, in grams, kilograms, etc.) and the gravitational acceleration (g 9.81 m/s2 on the earth’s surface) by Newton’s second law of motion: F ma. The weight is expressed as W mg

(5.1a)

The unit of weight is kg m/s and is usually expressed as newton (N). In the SI system, one newton is defined as the force needed to accelerate 1 kg of mass at a rate of 1 m/s2. Therefore 2

1 N 1 kg # m/s2 1 103 g # 102 cm/s2 105 g # cm/s2 105 dynes 1.243

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.244

CHAPTER 1.5

In the U.S. customary system, mass is expressed in slugs. One slug is defined as the mass of an object which needs one pound of force to accelerate to one ft/s2, i.e. m EXAMPLE:

W (lb) w g (in slugs) g (ft/s2)

(5.1b)

What is the value of the gravitational acceleration (g) in the U.S. customary system?

Solution: g 9.81 m/s2 9.81 m/s2 3.28 ft/m 32.174 ft/s2 Note: This is commonly used as g 32.2 ft/s2. 1.2 Specific Weight The specific weight (weight per unit volume) of a fluid such as water, , is defined by the product of the density () and the gravitational acceleration (g), i.e. g rg (in kg/m3 # m/s2) rg (in N/m3)

(5.2)

Water at 4C reaches its maximum density of 1000 kg/m3 or 1.000 g/cm3. The ratio of the specific weight of any liquid to that of water at 4C is called the specific gravity of that liquid. EXAMPLE:

What is the unit weight of water at 4C in terms of N/m3 and dyn/cm3?

Solution: Step 1. In N/m3, 1000 kg/m3 g rg 1000 kg/m3 9.81 m/s2 9810 N/m3 Step 2. In dyn/cm3, 1.000 g/cm3 g 1 g/cm3 981 cm/s2 981 (g # cm/s2)/cm3 981 dyn/cm3 1.3 Pressure Pressure (P) is the force (F) applied to or distributed over a surface area (A) as a measure of force per unit area: P

F A

(5.3a)

In the SI system, the unit of pressure can be expressed as barye, bar, N/m2, or pascal. One barye equals one dyne per square centimeter (dyn/cm2). The bar is one megabarye, 106 dynes per square centimeter (106 barye). One pascal equals one newton per square meter (N/m2). In the U.S. customary system the unit of pressure is expressed as lb/in2 (psi) or lb/ft2 etc.

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.245

Pressure is also a measure of the height (h) of the column of mercury, water, or other liquid which it supports. The pressure at the liquid surface is atmospheric pressure (Pa). The pressure at any point in the liquid is the absolute pressure (Pab) at that point. The absolute pressure is measured with respect to zero pressure. Thus Pab can be written as Pab gh Pa

(5.3b)

where specific weight of water or other liquid. In engineering work gage pressure is more commonly used. The gage pressure scale is designed on the basis that atmospheric pressure (Pa) is zero. Therefore the gage pressure becomes: P gh

(5.3c)

P h g

(5.3d)

and pressure head

According to Pascal’s law, the pressure exerted at any point on a confined liquid is transmitted undiminished in all directions. EXAMPLE: Under normal conditions, the atmospheric pressure at sea level is approximately 760 mm height of mercury. Convert it in terms of m of water, N/m2, pascals, bars, and psi.

Solution: Step 1. Solve for m of water Let 1 and 2 be specific weights of water and mercury, respectively, and h1 and h2 be column heights of water and mercury, respectively; then approximately g1 1.00,

g2 13.5936

From Eq. (5.3c): P g1h1 g2h2 g2 13.5936 h1 g h2 760 mm 10,331 mm 1 1 10.33 m (of water) Step 2. Solve for N/m2 Since 1 9810 N/m3 P g1h1 9810 N/m3 10.33 m 101,337 N/m2 1.013 105 N/m2 Step 3. Solve for pascals Since 1 pascal 1 N/m2, from step 2: P 1.013 105 N/m2 1.013 105 Pa Note: 1 atmosphere approximately equals 105 N/m2 or 105 pascals.

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.246

CHAPTER 1.5

Step 4. Solve for bars Since approximately 1 981 dyn/cm3 P g1h1 981 dyn/cm3 1033 cm 1,013,400 dyn/cm2 1.013 106 dyn/cm2

1 bar 1 106 dyn/cm2

1.013 bars Note: 1 atmosphere pressure is approximately equivalent to 1 bar. Step 5. Solve for lb/in2 (psi) From step 1, h1 10.33 m 10.33 m 3.28 ft/m 33.88 ft of water P 33.88 ft 1 (lb/in2)/2.31 ft 14.7 lb/in2 1.4 Viscosity of Water All liquids possess a definite resistance to change of their forms and many solids show a gradual yielding to force (shear stress) tending to change their forms. Newton’s law of viscosity states that, for a given rate of angular deformation of liquid, the shear stress is directly proportional to the viscosity. It can be expressed as where

du dy

(5.4a)

τ shear stress μ proportionality factor, viscosity du/dy velocity gradient u angular velocity y depth of fluid

The angular velocity and shear stress change with y. The viscosity can be expressed as absolute (dynamic) viscosity or kinematic viscosity. From Eq. (5.4a), it may be rewritten as (5.4b) du/dy The viscosity is frequently referred to absolute viscosity, or mass per unit length and time. In the SI system, the absolute viscosity is expressed as poise or in dyne-second per square centimter. One poise is defined as the tangent shear force () per unit area (dyn/cm2) required to maintain unit difference in velocity (1 cm/s) between two parallel planes separated by 1 cm of fluid. It can be written as 1 poise 1 dyne # s/cm2 1 dyne # s/cm2 1 g/cm # s

1 g # cm/s2 1 dyne

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.247

or 1 poise 1 dyne # s/cm2 1 dyne # s/cm2

1N 1 m2 ^ 105dyn 104 cm2

0.1 N # s/m2 also 1 poise 100 centipoise The usual measure of absolute viscosity of fluid and gas is the centipoise. The viscosity of fluids is temperature dependent. The viscosity of water at room temperature (20C) is one centipoise. The viscosity of air at 20C is approximately 0.018 centipoise. The U.S. customary system unit of viscosity is 1 lb s/ft2 or 1 slug ft s. EXAMPLE 1:

How much in U.S. customary units is 1 poise absolute viscosity?

Solution: 1 poise 0.1 N # s/m2

(3.28 ft)2 1 lb # s/m2^ 4.448 N m2 0.02248 lb # s/10.758 ft2 0.00209 lb # s/ft2 0.1 N

On the other hand: 1 lb # s/ft2 479 poise 47.9 N # s/m2 The kinematic viscosity is the ratio of viscosity (absolute) to mass density . It can be written as /

(5.5)

The dimension of is length squared per unit time. In the SI system, the unit of kinematic viscosity is the stoke. One stoke is defined as 1 cm2/s. However, the standard measure is the centistoke ( 10 2 stoke or 10 2 cm2/s). Kinematic viscosity offers many applications, such as the Reynolds number R VD/ . For water, the absolute viscosity and kinematic viscosity are essentially the same, especially when the temperature is less than 10C. The properties of water in SI units and U.S. customary units are respectively given in Tables 5.1a and 5.1b. EXAMPLE 2: At 21C, water has an absolute viscosity of 0.00982 poise and a specific gravity of 0.998. Compute (a) the absolute (Ns/m2) and kinematic viscosity in SI units and (b) the same quantities in U.S. customary units.

Solution: Step 1. For (a), at 21C (69.8F) 0.00982 poise 0.00982 poise 0.000982 N # s/m2

0.1 N # s/m2 1 poise

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.248

CHAPTER 1.5

0.000982, N # s/m2

1 kg # m/s2 1N

0.000982 kg/m # s 0.000982 kg/m # s m n r 0.998 1000 kg/m3 0.984 10 6 m2/s

TABLE 5.1a Physical Properties of Water—SI Units Temperature T, C 0 4 10 15 21 27 38 93

Specific gravity 0.9999 1.0000 0.9997 0.9990 0.9980 0.9966 0.9931 0.9630

Specific weight , N/m3 9805 9806 9804 9798 9787 9774 9739 9444

Absolute viscosity , N s/m2 0.00179 0.00157 0.00131 0.00113 0.00098 0.00086 0.00068 0.00030

Kinematic viscosity , m2/s

Surface tension , N/m2

Vapor pressure Pv, N/m2

1.795 10 6 1.568 10 6 1.310 10 6 1.131 10 6 0.984 10 6 0.864 10 6 0.687 10 6 0.371 10 6

0.0756 0.0750 0.0743 0.0735 0.0727 0.0718 0.0700 0.0601

608 809 1226 1762 2504 3495 6512 79,002

Source: Brater et al. (1996).

TABLE 5.1b Physical Properties of Water—U.S. customary Units

Temperature T, F 32 40 50 60 70 80 90 100 120 140 160 180 200 212

Density , slug/ft3

Specific weight , lb/ft3

Absolute viscosity 105, lb s/ft2

Kinematic viscosity 105, ft2/s

1940 1938 1936 1934 1931 1927 1923 1918 1908 1896 1890 1883 1868 1860

62.42 62.43 62.41 62.37 62.30 62.22 62.11 62.00 61.71 61.38 61.00 60.58 60.12 59.83

3.746 3.229 2.735 2.359 2.050 1.799 1.595 1.424 1.168 0.981 0.838 0.726 0.637 0.593

1.931 1.664 1.410 1.217 1.059 0.930 0.826 0.739 0.609 0.514 0.442 0.385 0.341 0.319

Source: Benefield et al. (1984) and Metcalf & Eddy, Inc. (1972).

Surface tension , lb/ft

Vapor pressure Pv, psia

0.00518 0.00514 0.00509 0.00504 0.00498 0.00492 0.00486 0.00480

0.09 0.12 0.18 0.26 0.36 0.51 0.70 0.95

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.249

Step 2. For (b), from Step 1 and Example 1 0.000982 N s/m2

1 slug/ft s 47.9 N s/m2

2.05 10 5 slug/ft s n 0.984 10 6 m2/s a

3.28 ft 2 b 1m

n 1.059 10 5 ft2/s 1.5 Perfect Gas A perfect gas is a gas that satisfies the perfect gas laws, such as the Boyle–Mariotte law and the Charles–Gay-Lussac law. It has internal energy as a function of temperature only. It also has specific heats with values independent of temperature. The normal volume of a perfect gas is 22.4136 liters/mol (commonly quoted as 22.4 L/mol). Boyle’s law states that at a constant temperature the volume of a given quantity of any gas varies inversely as the pressure applied to the gas. For a perfect gas, changing from pressure P1 and volume V1 to P2 and V2 at constant temperature, the following law exists: P1V1 P2V2

(5.6)

According to the Boyle–Mariotte law for perfect gases, the product of pressure P and volume V is constant in an isothermal process. PV nRT

(5.7)

or P

1 RT rRT V

where P pressure, pascal (Pa) or lb/ft2 V volume, m3 or ft3 n number of moles R gas constant T absolute temperature, K C 273, or R F 459.6 density, kg/m3 or slug/ft3 On a mole (M) basis, a pound mole (or kg mole) is the number of pounds (or kg) mass of gas equal to its molecular weight. The product MR is called the universal gas constant and depends on the units used. It can be MR 1545 ft lb/lb R The gas constant R is determined as 1545 ft lb/lb R M

(5.8a)

1545 32.2 ft lb/slug R M

(5.8b)

8312 n N/kg K M

(5.8c)

R or, in slug units R For SI units R

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.250

CHAPTER 1.5

EXAMPLE: For carbon dioxide with a molecular weight of 44 at a pressure of 12.0 psia (pounds per square inch absolute) and at a temperature of 70F (21C), compute R and its density.

Solution: Step 1. Determine R from Eq. (5.8b) 1545 32.2 ft lb/slug R M 1545 32.2 44

1130 ft lb/slug R Step 2. Determine density r

12 lb/in2 144 in2/ft2 P RT (1130 ft lb/slugR) (460 70R)

0.00289 slug/ft3 Based on empirical generation at constant pressure in a gaseous system (perfect gas), when the temperature varies the volume of gas will vary approximately in the same proportion. If the volume is exactly one mole of gas at 0C (273.15 K) and at atmosphere pressure, then, for ideal gases, these are the so-called standard temperature and pressure, STP. The volume of ideal gas at STP can be calculated from Eq. (5.7) (with R 0.08206 L atm/K mol): V

(1 mole) (0.08206 L ? atm/mol ? K) (273.15 K) nRT P 1 atm

22.41 liters. Thus, according to Avogadro’s hypothesis and the ideal-gas equation, one mole of any gas will occupy 22.41 L at STP.

2 WATER FLOW IN PIPES 2.1 Fluid Pressure Water and wastewater professionals frequently encounter some fundamentals of hydraulics, such as pressure, static head, pump head, velocity of flow, and discharge rate. The total force acting on a certain entire space, commonly expressed as the force acting on unit area, is called intensity of pressure, or simply pressure, P. The U.S. customary system of Units generally uses the pound per square inch (psi) for unit pressure. This quantity is also rather loosely referred to simply as pounds pressure; i.e., “20 pounds pressure’’ means 20 psi of pressure. The International System of Units uses the kg/cm2 (pascal) or g/cm2. To be technically correct, the pressure is so many pounds or kilograms more than that exerted by the atmosphere (760 mm of mercury). However, the atmospheric pressure is ignored in most cases, since it is applied to everything and acts uniformly in all directions. 2.2 Head The term head is frequently used, such as in energy head, velocity head, pressure head, elevation head, friction head, pump head, and loss of head (head loss). All heads can be expressed in the dimension of length, i.e. ft lb/lb ft, or m kg/kg m, etc.

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.251

The pump head equals the ft lb (m kg) of energy put into each pound (kg) of water passing through the pump. This will be discussed later in the section on pumps. Pressure drop causes loss of head and may be due to change of velocity, change of elevation, or to friction loss. Hydraulic head loss may occur at lateral entrances and is caused by hydraulic components such as valves, bends, control points, sharp crested weirs, and orifices. These types of head loss have been extensively discussed in textbooks and handbooks of hydraulics. Bend losses and head losses due to dividing and combining flows are discussed in detail by James M. Montgomery, Consulting Engineers, Inc. (1985). Velocity head. The kinetic energy (KE) of water with mass m is its capacity to do work by reason 1 of its velocity V and mass and is expressed as 2 mV 2. For a pipe with mean flow velocity V (m/s or ft/s) and pipe cross-sectional area A (cm2 or sq. in), the total mass of water flowing through the cross section in unit time is m VA, where is the fluid density. Thus the total kinetic energy for a pipe flow is 1 1 1 KE 2mV 2 2(rVA)V 2 2rAV 3

(5.9)

The total weight of fluid W m g AVg, where g is the gravitational acceleration. It is commonly expressed in terms of energy in a unit weight of fluid. The kinetic energy in unit weight of fluid is 1 3 V2 KE 2 rAV W rgAV 2g

(5.10)

This is the so-called velocity head, i.e. the height of the fluid column. Twenty-two pounds of water are moving at a velocity of 2 ft/s. What are the kinetic energy and velocity head?

EXAMPLE 1:

Solution: Step 1. KE 12mV 2

1 2

22 lb (2 ft/s)2

44 lb ft2/s2 44 ft lb ft/s2 44 slug ft, (since 1 slug 1 lb ft/s2) Step 2. hv

(2 ft/s)2 V2 2g 2 32.2 ft/s2

0.062 ft Ten kilograms of water are moving with a velocity of 0.61 m/s. What are the kinetic energy and velocity head?

EXAMPLE 2:

Solution: Step 1. KE 12 mV 2 12 (10 kg)(0.61 m/s)2 1.86 kg m2/s2 1.86 m kg m/s2 1.86 N m, since 1 N 1 kg m/s2

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.252

CHAPTER 1.5

Step 2. hy

(0.61 m/s)2 V2 2g 2 9.81 m/s2

0.019 m 0.019 m 3.28 ft/m 0.062 ft Note: Examples 1 and 2 are essentially the same. Pressure head. The pressure energy (PE) is a measure of work done by the pressure force on the fluid mass and is expressed as PE pAV

(5.11)

where p pressure at a cross section A pipe cross-sectional area, cm2 or in2 V mean velocity The pressure head (h) is the pressure energy in unit weight of fluid. The pressure is expressed in terms of the height of the fluid column h. The pressure head is pressure p p pAV PE (5.12) rg g sp. wt. W rgAV where is the weight per unit volume of fluid or its specific weight in N/m3 or lb/ft3. The general expression of unit pressure is h

p gh

(5.13)

and the pressure head is hp p/g in ft or m. EXAMPLE:

At the bottom of a water storage tank, the pressure is 31.2 lb/in2. What is the pressure head?

Solution: Step 1. Convert the pressure to lb/ft2 p 31.2 lb/in2 31.2 lb/in2 144 in2/ft2 31.2 144 lb/ft2 Step 2. Determine hp; 62.4 lb/ft3 for water p 31.2 144 lb/ft2 hp g 62.4 lb/ft3 72 ft Elevation head. The elevation energy (EE) of a fluid mass is simply the weight multiplied by the height above a reference plane. It is the work to raise this mass W to elevation h and can be written as EE Wz

(5.13a)

The elevation head, z, is EE divided by the total weight W of fluid: Elevation head

Wz EE z W W

(5.13b)

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.253

EXAMPLE: (1) Ten kilograms and (2) one pound of water are at 50 feet above the earth’s surface. What are their elevation heads?

Solution: Step 1.

For 10 kg he

10 kg 50 ft Wh W 10 kg

50 ft Step 2. For 1 lb he

Wh 1 lb 50 ft 50 ft W 1 lb

Both have 50 feet of head. Bernoulli equation. The total energy (H) at a particular section of water in a pipe is the algebraic sum of the kinetic head, pressure head, and elevation head. For sections 1 and 2, they can be expressed as H1

V12 p1 g z1 2g

(5.14)

V 22 p2 g z2 2g

(5.15)

and

If water is flowing from section 1 to section 2, friction loss (hf ) is the major loss. The energy relationship between the two sections can be expressed as the Bernoulli equation p2 p2 V12 V 22 g z1 g z2 hf 2g 2g

(5.16)

where hf is the friction head. This is also called the continuity equation. EXAMPLE: A nozzle of 12 cm diameter is located near the bottom of a storage water tank. The water surface is 3.05 m (10 ft) above the nozzle. Determine (a) the velocity of the effluent from the nozzle and (b) the discharge.

Solution: Step 1. (a) Determine V2 Let point 1 be at the water surface and point 2 at the center of the nozzle. By the Bernoulli equation p1 p2 V12 V 22 g z1 g z2 2g 2g Since p1 p2 0, and V 21 0 V 22 z1 z2 H 3.05 m 2g V2 22gH 22 9.806 3.05 7.73 m/s

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.254

CHAPTER 1.5

Step 2. Solve for (b) flow Q A2V2 pr2V2 3.14 (0.06 m)2 7.73 m/s 0.087 m3/s Friction head. Friction head (hf ) equals the loss of energy by each unit weight of water or other liquid through frction in the length of a pipe, in which the energy is converted into heat. The values of friction heads are usually obtained from the manufacturer’s tables. Darcy–Weisback equation. The Darcy–Weisback formula can be calculated from the friction head: L V2 hf f a b D 2g

(5.17)

where hf head of friction loss, cm or ft f friction factor, dimensionless L length of pipeline, cm or ft D2 diameter of pipe, cm or ft V 2g velocity head, cm or ft V average velocity of flow, cm/s or ft/s EXAMPLE 1:

Rewrite the Darcy–Weisback formula for hf in terms of flow rate Q instead of velocity V.

Solution: Step 1. Convert V to Q. In a circular pipe, A p(D/2)2 pD2/4 0.785 D2 The volumetric flow rate may be expressed in terms of velocity and area (A) as Q VA then V

Q 4Q/pD2 A

V 2 16Q2/p2D4 Step 2. Substitute in the formula 16Q2 L L V2 hf f a b fa b D 2g D 2gp2D4 fa

2 L 8Q b 2 5 p g D

(5.18)

The head of loss due to friction may be determined in three ways: 1. The friction factor f depends upon the velocity of flow and the diameter of the pipe. The value of f may be obtained from a table in many hydraulic textbooks and handbooks, or from the Moody chart (Fig. 5.1) of Reynolds number R versus f for various grades and sizes of pipe. R DV/ , where is the kinematic viscosity of the fluid.

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.255

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.256

CHAPTER 1.5

TABLE 5.2 Values of Equivalent Roughness e for New Commercial Pipes Type of pipe

e, in

Asphalted cast iron Cast iron Concrete Drawn tubing Galvanized iron PVC Riveted steel Steel and wrought iron Wood stave

0.0048 0.0102 0.01–0.1 0.00006 0.006 0.00084 0.04–0.4 0.0018 0.007–0.04

Source: Benefield et al. (1984).

2. The friction loss of head hf per 1000 ft of pipe may be determined from the Hazen–Williams formula for a pipe flow. A constant C for a particular pipe, diameter of pipe, and either the velocity or the quantity of flow should be known. A nomograph chart is most commonly used for solution by the Hazen–Williams formula. 3. Another empirical formula, the Manning equation, is also a popular formula for determining head loss due to free flow. The friction factor f is a function of Reynolds number R and the relative roughness of the pipe wall e/D. The value of e (Table 5.2), the roughness of the pipe wall (equivalent roughness), is usually determined from experiment. For laminar pipe flow (R 2000) f is independent of surface roughness of the pipe. The f value can be determined from f 64/R

(5.19)

When R 2000 then the relative roughness will affect the f value. The e/D values can be found from the manufacturer or any textbook. The relationship between f, R, and e/D are summarized in graphical expression as the Moody diagram (Fig. 5.1). A pumping station has three pumps with 1 MGD, 2 MGD, and 4 MGD capacities. Each pumps water from a river at an elevation of 588 ft above sea level to a reservoir at an elevation of 636 ft, through a cast iron pipe of 24 in diameter and 2600 ft long. The Reynolds number is 1600. Calculate the total effective head supplied by each pump and any combinations of pumping.

EXAMPLE 2:

Solution: Step 1. Write Bernoulli’s equation Let stations 1 and 2 be river pumping site and discharge site at reservoir, respectively; hp is effective head applied by pump in feet. V 22 p1 p2 V 12 g z1 hp g z2 hf 2g 2g Since V1 V2, p1 is not given (usually negative), assumed zero p2 zero, to atmosphere

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

then z1 hp z2 hf hp (z2 z1) hf (636 588) hf 48 hf Step 2. Compute flow velocity (V) for the 1 MGD pump A pr2 3.14 (1 ft)2 3.14 ft2 Q 1 MGD

106 gal 0.1337 ft3/gal 1 day 24 60 60 s/day

1.547 ft3/s (cfs) V

Q 1.547 cfs A 3.14 ft2

0.493 ft/s Step 3. Find hf and hp for 1 MGD pump From Eqs. (5.19) and (5.17), f 64/R 64/1600 0.04 hf f a

L V2 b D 2g

0.04a

2600 (0.493)2 b 2 2 32.2

0.20 ft hp 48 ft 0.2 ft 48.2 ft Step 4. Similar calculations for the 2-MGD pump V 0.493 ft/s 2 0.986 ft/s hf 0.2 ft 22 0.8 ft hp 48 ft 0.8 ft 48.8 ft Step 5. For the 4 MGD pump V 0.493 ft/s 4 1.972 ft/s hf 0.2 ft 42 3.2 ft hp 48 ft 3.2 ft 51.2 ft Step 6. For the 3 MGD: using the 1 and 2 MGD capacity pumps hf (0.2 0.8) ft 1.0 ft hp (48.2 48.8) ft 97 ft

1.257

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.258

CHAPTER 1.5

Step 7. For the 5 MGD: using the 1 and 4 MGD capacity pumps hf (0.2 3.2) ft 3.4 ft hp (48.2 51.2) ft 99.4 ft Step 8. For the 6 MGD: using the 2 and 4 MGD capacity pumps hf (0.8 3.2) ft 4.0 ft hp (48.8 51.2) ft 100.0 ft Step 9. For the 7 MGD: using all pumps simultaneously hf (0.2 0.8 3.2) ft 4.2 ft hp (48.2 48.8 51.2) ft 148.2 ft Hazen–Williams equation. Due to the difficulty of using the Darcy–Weisback equation for pipe flow, engineers continue to make use of an exponential equation with empirical methods for determining friction losses in Pipe flows. Among these the empirical formula of the Hazen–Williams equation is most widely used to express flow relations in pressure conduits, while the Manning equation is used for flow relations in free-flow conduits and in pipes partially full. The Hazen–Williams equation, originally developed for the British measurement system, has the following form: V 1.318 CR0.63S0.54

(5.20a)

where V average velocity of pipe flow, ft/s C coefficient of roughness (see Table 5.3) R hydraulic radius, ft S slope of the energy gradient line or head loss per unit length of the pipe (S hf /L) The hydraulic radius R is defined as the water cross-sectional area A divided by the wetted perimeter P. For a circular pipe, if D is the diameter of the pipe, then R is R

pD2/4 D A P pD 4

(5.21)

The Hazen–Williams equation was developed for water flow in large pipes (D 2 in, 5 cm) with a moderate range of velocity (V 10 ft/s or 3 m/s). The coefficient of roughness C values range from 140 for very smooth (new), straight pipe to 90 or 80 for old, unlined tuberculated pipe. The value of 100 is used for the average conditioned pipe. It is not a function of the flow condition (i.e. Reynolds number R). The major limitation of this equation is that the viscosity and temperature are considered. A nomograph (Fig. 5.2) can be used to solve the Hazen–Williams equation. In the SI system, the Hazen–Williams equation is written as V 0.85 C R0.63S0.54

(5.20b)

The units are R in m, S in m/1000 m, V in m/s, and D in m, and discharge Q in m3/s (from chart). Manning equation. Another popular empirical equation is the Manning equation, developed by the Irish engineer Robert Manning in 1889, and employed extensively for open channel flows. It is also commonly used for pipe free-flow. The Manning equation is V

1.486 2/3 1/2 n R S

where V average velocity of flow, ft/s n Manning’s coefficient of roughness (see Table 5.4) R hydraulic radius (ft) (same as in the Hazen–Williams equation) S slope of the hydraulic gradient, ft/ft

(5.22a)

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.259

TABLE 5.3 Hazen–Williams Coefficient of Roughness C for Various Types of Pipe Pipe material

C value

Brass Brick sewer Cast iron tar coated new, unlined cement lined uncertain Cement–asbestos Concrete Copper Fire hose (rubber lined) Galvanized iron Glass Lead Plastic Steel coal-tar enamel lined corrugated new unlined riveted Tin Vitrified clay Wood stave

130–140 100 130 130 130–150 60–110 140 130–140 130–140 135 120 140 130–140 140–150 145–150 60 140–150 110 130 110–140 110–120

Sources: Perry (1967), Hwang (1981), and Benefield et al. (1984).

This equation can be easily solved with the nomograph shown in Fig. 5.3. Figure 5.4 is used for partially filled circular pipes with varying Manning’s n and depth. The Manning equation for SI units is 1 V n R2/3 S1/2

(5.22b)

where V mean velocity, m/s n same as above R hydraulic radius, m S slope of the hydraulic gradient, m/m Assume the sewer line grade gives a sewer velocity of 0.6 m/s with a half full sewer flow. The slope of the line is 0.0081. What is the diameter of the uncoated cast iron sewer line?

EXAMPLE 1:

Solution: Step 1. Using the Manning equation to solve R 1 V n R2/3 S1/2 R2/3 V n/S1/2

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.260

CHAPTER 1.5

FIGURE 5.2

Nomograph for Hazen–Williams formula.

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.261

TABLE 5.4 Manning’s Roughness Coefficient n for Various Types of Pipe Type of pipe

Brick open channel pipe with cement mortar brass copper or glass pipe Cast iron pipe uncoated tuberculated Cement mortar surface Concrete open channel pipe Common clay drain tile Fiberglass Galvanized iron Gravel open channel Plastic pipe (smooth) Rock open channel Steel pipe Vitrified clay pipes liner plates Wood, laminated Wood stave Wrought iron

0.014–0.017 0.012–0.017 0.009–0.013 0.013 0.015–0.035 0.011–0.015 0.013–0.022 0.010–0.015 0.011–0.017 0.013 0.012–0.017 0.014–0.033 0.011–0.015 0.035–0.045 0.011 0.011–0.015 0.017–0.017 0.015–0.017 0.010–0.013 0.012–0.017

Sources: Perry (1967), Hwang (1981), and ASCE & WEF (1992).

From Table 5.4, n 0.013 R2/3 V n/S1/2 0.6 0.013/0.00811/2 0.0864 R 0.0254 Step 2. Calculate diameter (D) of sewer line using (Eq. (5.21)) D 4 R 4 0.0254 0.102 (m) 4 (in) EXAMPLE 2: Determine the energy loss over 1600 ft in a new 24-in cast iron pipe when the water temperature is 60F and the flow rate is 6.25 cfs, using (1) the Darcy–Weisback equation, (2) the Hazen–Williams equation, and (3) the Manning equation.

Solution: Step 1. By the Darcy–Weisback equation From Table 5.2, e 0.0102 in (a) Determine e/D, D 24 in 2 ft 0.0102 e 0.000425 D 24

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.262

CHAPTER 1.5

FIGURE 5.3

Nomograph for Manning formula.

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.263

FIGURE 5.4 Hydraulic elements graph for circular sewers (Metcalf and Eddy, Inc., Wastewater Engineering: Collection and Pumping of Wastewater, Copyright, 1990, McGraw-Hill, New York, reproduced with permission of McGraw-Hill ).

(b) Compute velocity V, r 1 ft V

Q 6.25 ft3/s A p (1 ft)2

1.99 ft/s (c) Compute Reynolds number R From Table 5.1b, 1.217 10 5 ft2/s for 60F 1.99 ft/s 2 ft VD R n 1.217 10 5 ft2/s 3.27 105

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.264

CHAPTER 1.5

(d) Determine f value from the Moody diagram, (Fig. 5.1) f 0.018 (e) Compute loss of head hf 1600 (1.99)2 L V2 0.018 a b hf f a b D 2g 2 2 32.2 0.885 (ft) Step 2. By the Hazen–Williams equation (a) Find C From Table 5.3, for new unlined cast pipe C 130 (b) Solve S Eq. 5.21: R D/4 2/4 0.5 Eq. 5.20a: V 1.318 CR0.63 S0.54 Rearranging, 1.852

V S c d 1.318CR0.63

1.852

1.99 c d 1.318 130 (0.5)0.63 0.000586 (ft/ft) (c) Compute hf hf S L 0.000586 ft/ft 1600 ft 0.94 ft Step 3. By the Manning equation (a) Find n From Table 5.4, for unlined cast iron pipe n 0.013 (b) Determine slope of the energy line S From Eq. (5.21), R D/4 2/4 0.5 1.486 From Eq. (5.22a), V n R2/3 S1/2 Rearranging S a

2 1.99 0.013 Vn b c d 1.486 R2/3 1.486 (0.5)2/3

0.000764 (ft/ft) (c) Compute hf hf S L 0.000764 ft/ft 1600 ft 1.22 ft

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.265

The drainage area of the watershed to a storm sewer is 10,000 acres (4047 ha). Thirtyeight percent of the drainage area has a maximum runoff of 0.6 ft3/s acre (0.0069 m3/s ha), and the rest of the area has a runoff of 0.4 ft3/s acre (0.0046 m3/s ha). Determine the size of pipe needed to carry the storm flow with a grade of 0.12 percent and n 0.011. EXAMPLE 3:

Solution: Step 1. Calculate the total runoff Q Q 0.6 cfs/acre 0.38 10,000 acre 0.4 ft3/(s acre) 0.62 10,000 acre 4760 ft3/s Step 2. Determine diameter D of the pipe by the Manning formula A

p 2 D 0.785 D2 4

R D/4 1.486 Q AV A n R2/3S1/2 4760 0.785 D2

1.486 D 2/3 a b (0.0012)1/2 0.011 4

3265 D8/3 D 20.78 (ft) Minor head losses due to hydraulic devices such as sharp-crested orifices, weirs, valves, bend, construction, enlargement, discharge, branching, etc. have been extensively covered in textbooks and handbooks elsewhere. Basically they can be calculated from empirical formulas. 2.3 Pipeline Systems There are three major types of compound piping system: pipes connected in series, in parallel, and branching. Pipes in parallel occur in a pipe network when two or more paths are available for water flowing between two points. Pipe branching occurs when water can flow to or from a junction of three or more pipes from independent outlets or sources. On some occasions, two or more sizes of pipes may connect in series between two reservoirs. Figure 5.5 illustrates all three cases. The flow through a pipeline consisting of three or more pipes connected in various ways can be analyzed on the basis of head loss concept with two basic conditions. First, at a junction, water flows in and out should be the same. Second, all pipes meeting at the junction have the same water pressure at the junction. The general procedures are to apply the Bernoulli energy equation to determine the energy line and hydraulic gradient mainly from friction head loss equations. In summary, using Fig. 5.5: 1. For pipelines in series Energy: H hf1 hf2 Continuity: Q Q1 Q2 Method: (a) Assume hf (b) Calculate Q1 and Q2 (c) If Q1 Q2, the solution is correct (d) If Q1 2 Q2, repeat step (a)

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.266

CHAPTER 1.5

FIGURE 5.5

Pipeline system for two reservoirs.

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.267

2. For pipelines in parallel Energy: H hf1 hf2 Continuity: Q Q1 Q2 Method: Parallel problem can be solved directly for Q1 and Q2 3. For branched pipelines Energy: hf1 z1 HJ hf2 z2 HJ hf3 HJ z3 where HJ is the energy head at junction J. Continuity: Q3 Q1 Q2 Method: (a) Assume HJ (b) Calculate Q1, Q2, and Q3 (c) If Q1 Q2 Q3, the solution is correct (d) If Q1 Q2 2 Q3, repeat step (a) Pipelines in series EXAMPLE: As shown in Fig. 5.5a, two concrete pipes are connected in series between two reservoirs A and B. The diameters of the upstream and downstream pipes are 0.6 m (D1 2 ft) and 0.45 m (D2 1.5 ft); their lengths are 300 m (1000 ft) and 150 m (500 ft), respectively. The flow rate of 15C water from reservoir A to reservoir B is 0.4 m3/s. Also given are coefficient of entrance Ke 0.5, coefficient of contraction Kc 0.13, and coefficient of discharge Kd 1.0. Determine the elevation of the water surface of reservoir B when the elevation of the water surface of reservoir A is 100 m.

Solution: Step 1. Find the relative roughness e/D from Fig. 5.1 for a circular concrete pipe, assume e 0.003 ft 0.0009 m e/D1 0.0009/0.6 0.0015 e/D2 0.0009/0.45 0.0020 Step 2. Determine velocities (V) and Reynolds number (R) V1

Q 0.4 m3/s p 1.41 m/s 2 A1 4 (0.6 m)

Q 0.4 p 2.52 m/s 2 A2 4 (0.45)

From Table 5.1a, 1.131 10 6 m2/s for T 15C R1

V1D1 1.41 m/s 0.6 m 5 n 1.131 10 6 m2/s 7.48 10

V2D2 2.51 0.45 5 n 1.131 10 6 9.99 10

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.268

CHAPTER 1.5

Step 3. Find f value from Moody chart (Fig. 5.1) corresponding to R and e/D values f1 0.022 f2 0.024 Step 4. Determine total energy loss for water flow from A to B he Ke hf1 f1

2 V 12 L1 V 21 300 V 1 0.022 11 D1 2g 2g 0.6 2g

h c Kc hf2 f2

V 12 V 12 0.5 2g 2g

V 22 V 22 0.13 2g 2g

2 V 22 L2 V 22 150 V 2 0.024 8 D2 2g 2g 0.45 2g

hd Kd

V 22 V 22 1 2g 2g

Step 5. Calculate total energy head H H he hf1 hc hf2 hd (0.5 11) 11.5

V 12 V 22 (0.13 8 1) 2g 2g

(1.41)2 (2.52)2 9.13 2 9.81 2 9.81

4.12 (m) Step 6. Calculate the elevation of the surface of reservoir B Ele. 100 m H 100 m 4.12 m 95.88 (m) Pipelines in parallel EXAMPLE: Circular pipelines 1, 2, and 3, each 1000 ft (300 m) long and of 6, 8, and 12 in (0.3 m) diameter, respectively, carry water from reservoir A and join pipeline 4 to reservoir B. Pipeline 4 is 2 ft (0.6 m) in diameter and 500 ft (150 m) long. Determine the percentages of flow passing pipelines 1, 2, and 3 using the Hazen–Williams equation. The C value for pipeline 4 is 110 and, for the other three, 100.

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.269

Solution: Step 1. Calculate velocity in pipelines 1, 2, and 3. Since no elevation and flow are given, any slope of the energy line is OK. Assuming the head loss in the three smaller pipelines is 10 ft per 1000 ft of pipeline: S 10/1000 0.01 For a circular pipe, the hydraulic radius is R

D 4

then, from the Hazen–Williams equation, Eq. (5.20a) V1 1.318 C R0.63 S0.54 1.318 100 a 10.96 a

0.5 0.63 b (0.01)0.54 4

0.5 0.63 b 4

2.96 (ft/s) V2 10.96a

0.667 0.63 b 3.54 (ft/s) 4

V3 10.96( 14)0.63 4.58 (ft/s) Step 2. Determine the total flow Q4 Q1 A1V1

p 2 D V 0.785(0.5)2 2.96 4 1 1

0.58 (ft3/s) Q2 A2V2 0.785(0.667)2 3.54 1.24 (ft3/s) Q3 A3V3 0.785(1)2 4.58 3.60 (ft3/s) Q4 Q1 Q2 Q3 0.58 1.24 3.60 5.42 (ft3/s) Step 3. Determine percentage of flow in pipes 1, 2, and 3 For

pipe 1 (Q1Q4) 100% 0.58 1005.42 10.7% pipe 2 1.24 100%5.42 22.9% pipe 3 3.60 100%/5.42 66.4%

2.4 Distribution Networks Most waterline or sewer distribution networks are complexes of looping and branching pipelines. The solution methods described for the analysis of pipelines in series, parallel, and branched systems are not suitable for the more complex cases of networks. A trial and error procedure should

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.270

CHAPTER 1.5

be used. The most widely used is the loop method originally proposed by Hardy Cross in 1936. Another method, the nodal method, was proposed by Cornish in 1939 (Chadwick and Morfett, 1986). For analysis flows in a network of pipes, the following conditions must be satisfied: 1. The algebraic sum of the pressure drops around each circuit must be zero. 2. Continuity must be satisfied at all junctions; i.e., inflow equals outflow at each junction. 3. Energy loss must be the same for all paths of water. Using the continuity equation at a node, this gives m

a qi 0 i1

(5.23)

where m is the number of pipes joined at the node. The sign conventionally used for flow into a joint is positive, and outflow is negative. Applying the energy equation to a loop, we get n

a hfi 0 i1

(5.24)

where n is the number of pipes in a loop. The sign for flow (qi) and head loss (hfi) is conventionally positive when clockwise. Since friction loss is a function of flow hfi (q1)

(5.25)

Equations (5.23) and (5.24) will generate a set of simultaneous nonlinear equations. An iterative solution is needed. Method of equivalent pipes. In a complex pipe system, small loops within the system are replaced by single hydraulically equivalent pipes. This method can also be used for determination of diameter and length of a replacement pipe; i.e., one that will produce the same head loss as the old one.

FIGURE 5.6

Two types of pipe network.

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.271

EXAMPLE 1: Assume QAB 0.042 m3/s (1.5 ft3/s), QBC 0.028 m3/s (1.0 ft3/s), and a Hazen–Williams coefficient C 100. Determine an equivalent pipe for the network of the following figure.

Solution: Step 1. For line BCE, Q 1 ft3/s. From the Hazen–Williams chart (Fig. 5.2) with C 100, to obtain loss of head in ft per 1000 ft, then determine the total head loss for the pipe proportionally (a) Pipe BC, 4000 ft, 14 in 0.44 hBC 4000 ft 1.76 ft 1000 (b) Pipe CE, 3000 ft, 10 in hCE

2.40 3000 ft 7.20 ft 1000

8.96 3733 ft 2.4

Step 2. For line BDE, Q 1.5 1.0 0.5 ft3/s or 0.014 m3/s (a) Pipe BD, 2500 ft, 12 in 0.29 hBD 2500 ft 0.725 ft 1000 (b) Pipe DE, 5000 ft, 8 in hDE

1.80 5000 ft 9.0 ft 1000

9.725 5403 ft 1.80

Step 3. Determine equivalent line BE from results of Steps 1 and 2, assuming hBE 8.96 ft (a) Line BCE, 3733 ft, 10 in S

8.96 2.4/1000 24% 3733

QBCD 1.0 ft3/s

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.272

CHAPTER 1.5

(b) Line BDE, 5403 ft, 8 in S

8.96 1.66/1000 5403

QBDE 0.475 ft3/s 0.0134 m3/s

From Fig. 5.2 (c) Total

QBE 1 ft3/s 0.475 ft3/s 1.475 ft3/s 0.0417 m3/s (d) Using Q 1.475 ft3/s, select 12-in pipe to find equivalent length From chart S 1.95 Since assumed total head loss HBE 8.96 L

8.96 1000 < 4600 ft 1.95

Answer: Equivalent length of 4600 ft of 12-in pipe. EXAMPLE 2: The sewer pipeline system below shows pipe diameter sizes (), grades, pipe section numbers, and flow direction. Assume there is no surcharge and full flow in each of sections 1, 2, 3, and 4. All pipes are fiberglass with n 0.013. A, B, C, and D represent inspection holes. Determine

(a) the flow rate and minimum commercial pipe size for section AB; (b) the discharge, sewage depth, and velocity in section BC; and (c) the slope required to maintain full flow in section CD.

Solution: Step 1. Compute cross-sectional areas and hydraulic radius Section 1 2 3 4 BC CD

Diameter, D (ft)

A (ft)

R D/4(ft)

1.167 0.667 0.833 1.0 1.5 1.667

1.068 0.349 0.545 0.785 1.766 2.180

0.292 0.167 0.208 0.250 0.375 0.417

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

Step 2. Compute flows with Manning formula 1.486 2/3 1/2 n R S 1.486 (0.292)2/3(0.0036)1/2 Q1 1.068 0.013 3.22 (ft3/s) QA

Q2 0.349 114.3 (0.167)2/3 (0.0025)1/2 0.60 (ft3/s) Q AB Q1 Q2 3.22 0.60 3.82 (ft3/s) Step 3. Compute DAB D 2/3 QAB 3.82 0.785 D2 114.3a b (0.0025)1/2 4 D8/3 2.146 D 1.33 (ft) 15.9 in Use 18-in commercially available pipe for answer (a). Step 4. Determine Q3, QBC, QBCQf Q3 0.545 114.3 (0.208)2/3(0.0049)1/2 1.53 (ft3/s) QBC QAB Q3 3.82 ft3/s 1.53 ft3/s 5.35 ft3/s At section BC: velocity for full flow Vf 114.3 (0.375)2/3 (0.0036)1/2 3.57 (ft/s) Qf AVf 1.766 3.57 6.30 (ft3/s) Percentage of actual flow QBC 5.35 ft3/s 0.85 Qf 6.30 ft3/s Step 5. Determine sewage depth d and V for section BC from Fig. 5.4 (page 1.285) We obtain d/D 0.70 Then d 0.70 18 in 12.6 in 1.05 ft and V/Vf 1.13

1.273

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.274

CHAPTER 1.5

Then V 1.13 Vf 1.13 3.57 ft/s 4.03 ft/s Step 6. Determine slope S of section CD Q4 0.785 114.3 (0.25)2/3(0.0025)1/2 1.78 (ft3/s) QCD Q4 QBC 1.78 ft3/s 5.35 ft3/s 7.13 ft3/s Section CD if flowing full: 7.13 2.18 114.3 (0.417)2/3 S1/2 S1/2 0.05127 S 0.00263 Hardy–Cross method. The Hardy–Cross method (1936) of network analysis is a loop method which eliminates the head losses from Eqs. (5.24) and (5.25) and generates a set of discharge equations. The basics of the method are: 1. 2. 3. 4. 5.

Assume a value for Qi for each pipe to satisfy Qi 0 Compute friction losses hfi from Qi; find S from Hazen–Williams equation If hfi 0, the solution is correct If fi 2 0, apply a correction factor Q to all Qi, then repeat step 1. A reasonable value of Q is given by Chadwick and Morfett (1986) Q

ghfi 2ghfi /Qi

(5.26)

This trial and error procedure solved by digital computer program is available in many textbooks on hydraulics (Hwang 1981, Streeter and Wylie 1975). The nodal method. The basic concept of the nodal method consists of the elimination of discharges from Eqs. (5.23) and (5.25) to generate a set of head loss equations. This method may be used for loops or branches when the external heads are known and the heads within the networks are needed. The procedure of the nodal method is as follows: 1. 2. 3. 4. 5.

Assume values of the head loss Hj at each junction. Compute Qi from Hj. If Qi 0, the solution is correct. If Qi 2 0, adjust a correction factor H to Hj, then repeat step 2. The head correction factor is

2gQi gQi /hfi

(5.27)

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.275

2.5 Sludge Flow To estimate every loss in a pipe carrying sludge, the Hazen–Williams equation with a modified C value and a graphic method based on field experience are commonly used. The modified C values for various total solids contents are as follows (Brisbin 1957) Total solids (%) C

8.5

100

2.6 Dividing-Flow Manifolds and Multiport Diffusers The related subjects of dividing-flow manifolds and multiport diffusers are discussed in detail by Benefield et al. (1984). They present basic theories and excellent design examples for these two subjects. In addition, there is a design example of hydraulic analysis for all unit processes of a wastewater treatment plant.

3 PUMPS 3.1 Types of Pump The centrifugal pump and the displacement pump are most commonly used for water and wastewater works. Centrifugal pumps have a rotating impeller which imparts energy to the water. Displacement pumps are often of the reciprocating type in which a piston draws water or slurry into a closed chamber and then expels it under pressure. Reciprocating pumps are widely used to transport sludge in wastewater treatment plants. Air-lift pumps, jet pumps, and hydraulic rams are also used in special applications. 3.2 Pump Performance The Bernoulli equation may be applied to determine the total dynamic head on the pump. The energy equation expressing the head between the suction (s) and discharge (d) nozzles of the pumps is as follows: Pd Ps Vd2 Vs2 H g zd a g zs b 2g 2g where

H total dynamic head, m or ft Pd, Ps gage pressure at discharge and suction, respectively, N/m2 or lb/in2 specific weight of water, N/m3 or lb/ft3 Vd, Vs velocity in discharge and suction nozzles, respectively, m/s or ft/s g gravitational acceleration, 9.81 m/s2 or 32.2 ft/s2 zd, zs elevation of discharge and suction gage above the datum, m or ft

The power Pw required to pump water is a function of the flow Q and the total head H and can be written as Pw jQH where Pw water power, kW (m m3/min) or hp (ft gal/min) j constant, at 20C 0.163 (for SI units)

(5.28)

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.276

CHAPTER 1.5

2.525 10 4 (for US customary units) Q discharge, m3/min or gpm H total head, m or ft EXAMPLE 1: Calculate the water power for a pump system to deliver 3.785 m3/min (1000 gpm) against a total system head of 30.48 m (100 ft) at a temperature of 20C.

Solution: Pw jQH 0.163 3.785 m3/min 30.48 m 18.8 kW 25.3 hp

The power output of a pump is the work done to lift the water to a higher elevation per unit time. The theoretical power output or horsepower (hp) required is a function of a known discharge and total pump lift. For US customery units, it may be written as Po (in hp) For SI units: where

QgH 550

Po (in N m/s or watt) QgH

(5.28a) (5.28b)

Q discharge, ft3/s (cfs) or m3/s specific weight of water or liquid, lb/ft3 or N/m3 H total lift, ft or m 550 conversion factor from ft lb/s to hp

EXAMPLE 2:

Determine how many watts equal one horsepower. 1 hp 550 ft lb/s 550 ft 0.305 m/ft 1 lb 4.448 N/(lb s) 746 N m/s ( joule/s watt) 746 watts (W) or 0.746 kilowatts (kW)

Conversely, 1 kW 1.341 hp The actual horsepower needed is determined by dividing the theoretical horsepower by the efficiency of the pump and driving unit. The efficiencies for centrifugal pumps normally range from 50 to 85 percent. The efficiency increases with the size and capacity of the pump. The design of a pump should consider total dynamic head which includes differences in elevations. The horsepower which should be delivered to a pump is determined by dividing Eq. (5.28b) by the pump efficiency, as follows: gQH (5.28c) hp 550 efficiency The efficiency of a pump (ep ) is defined as the ratio of the power output (Po QH/550) to the input power of the pump (Pi w ). It can be written as: for U.S. customary units ep

Po gQH Pi hp 550

(5.28d)

gQH gQH vt Pi

(5.28e)

for SI units

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

where

1.277

ep efficiency of the pump, % specific weight of water, kN/m3 or lb/ft3 Q capacity, m3/s or ft3/s hp brake horsepower 550 conversion factor for horsepower to ft lb/s w angular velocity of the turbo-hydraulic pump torque applied to the pump by a motor

The efficiency of the motor (em) is defined as the ratio of the power applied to the pump by the motor (Pi ) to the power input to the motor (Pm ), i.e. em

Pi Pm

(5.29)

The overall efficiency of a pump system (e) is combined as e ePem

Po Po Pi Pi Pm Pm

(5.30)

EXAMPLE 3: A water treatment plant pumps its raw water from a reservoir next to the plant. The intake is 12 ft below the lake water surface at elevation 588 ft. The lake water is pumped to the plant influent at elevation 611 ft. Assume the suction head loss for the pump is 10 ft and the loss of head in the discharge line is 7 ft. The overall pump effiency is 72 percent. The plant serves 44,000 persons. The average water consumption is 200 gallons per capita per day (gpcpd). Compute the horsepower output of the motor.

Solution: Step 1. Determine discharge Q Q 44,000 200 gpcpd 8.8 Mgal/day (MGD) 8.8 Mgal/day 1.547 (ft3/s)/(Mgal/day) 13.6 ft3/s Step 2. Calculate effective head H H (611 588) 10 7 40 (ft) Step 3. Compute overall horsepower output Using Eq. (5.28c), hp

QgH 13.6 ft3/s 62.4 lb/ft3 40 ft 550ep 550 ft lb/hp 0.72

85.7 hp EXAMPLE 4: A water pump discharges at a rate of 0.438 m 3/s (10 Mgal/day). The diameters of suction and discharge nozzles are 35 cm (14 in) and 30 cm (12 in), respectively. The reading of the suction gage located 0.3 m (1 ft) above the pump centerline is 11 kN/m2 (1.6 lb/in2). The reading of the discharge gage located at the pump centerline is 117 kN/m2 (17.0 lb/in2). Assume the pump efficiency is 80 percent and the motor efficiency is 93 percent; water temperature is 13C. Find (a) the power input needed by the pump and (b) the power input to the motor.

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.278

CHAPTER 1.5

Solution: Step 1. Write the energy equation Pd Ps Vd2 Vs2 zd b a g zs b H ag 2g 2g Step 2. Calculate each term in the above equation At T 13C, 9800 N/m3, refer to table 5.1a Pd 117,000 N/m2 g 9800 N/m3 11.94 m Qd 0.438 m3/s Vd 6.20 m/s Ad (p/4)(0.30 m)2 V d2 (6.20 m/s)2 1.96 m 2g 2(9.81 m/s2) Let zd 0 be the datum at the pump centerline Ps 11,000 N/m2 1.12 m g 9800 N/m3 Vs

0.438 m3/s 4.55 m/s (p/4)(0.35 m)2

V s2 (4.55 m/s)2 1.06 m 2g 2(9.81 m/s2) zs 0.30 m Step 3. Calculate the total head H H [11.94 1.96 0 (1.12 1.06 0.30)] m 11.42 m or 34.47 ft Step 4. Compute power input Pi by Eq. (5.28e) for question (a) Pq gQH (9.8 kN)(0.438 m3/s)(11.42 m) Pi e e 0.80 p p 61.27 kW 61.27 kW

1.341 hp 1 kW

82.2 hp Step 5. Compute power input to the motor (Pm ) for question (b) Pi 61.27 kW Pm e 0.93 m 65.88 kW or

88.4 hp

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.279

3.3 Cost of Pumping The cost of pumping through a pipeline is a function of head loss, flow rate, power cost, and the total efficiency of the pump system. It can be expressed as (Ductile Iron Pipe Research Association, 1997) CP 1.65 HQ$/E

(5.31)

where CP pumping cost, $/(yr 1000 ft) (based on 24 h/day operation) H head loss, ft/1000 ft Q flow, gal/min $ unit cost of electricity, $/kWh E total efficiency of pump system, % Velocity is related to flow by the following equation V

Q 2.448d 2

(5.32a)

where V velocity, ft/s Q flow, gal/min d actual inside diameter, in Head loss is determined by the Hazen–Williams formula H 1000a

1.852 V b 0.63 0.115 C d

(5.32b)

where C a coefficient, and other symbols are as above. EXAMPLE: Water is pumped at a rate of 6300 gal/min (0.40 m3/s) through a 24-in pipeline of 10,000 ft (3048 m) length. The actual inside diameters for ductile iron pipe and PVC pipe are respectively 24.95 and 22.76 in. Assume the unit power cost is $0.058/kWh; the total efficiency of pump system is 75%; the pump is operated 24 hours per day. Estimate: (a) the cost of pumping for each kind of pipeline, (b) the present value of the difference of pumping cost, assuming 50-year design pipe life (n), 6.6 percent annual rate (r) of return on the initial investment, and 3.5 percent annual inflation rate of power costs.

Solution: Step 1. Compute the velocity of flow for each pipeline. Let Vd and Vp represent velocity for ductile and PVC pipe respectively, using Eq. (5.32a) Vd

6300 gal/min Q 2.448d2 2.448(24.95 in)2

4.13 ft/s 6300 Vp 2.448(22.76)2 4.96 ft/s

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.280

CHAPTER 1.5

Step 2. Compute head losses (Hd and Hp ) for each pipe flow. Coefficients C for ductile and PVC pipes are 140 and 150, respectively. Hd 1000a 1000a

Vd 0.115Cd

0.63

1.852

1.852 4.13 b 0.63 0.115 140 24.95

1.89 (ft/1000 ft) Hp 1000a

1.852 4.96 b 0.115 150 22.760.63

2.59 (ft/1000 ft) Step 3. Compute the costs of pumping, CPd and CPp CPd 1.65 Hd Q $/E 1.65 1.89 6300 0.058/0.75 1519 ($/1000 ft/yr) CPp 1.65 2.59 6300 0.058/0.75 2082 ($/1000 ft/yr) Step 4. Compute the difference of total cost for 10,000 ft annually (A) A (2082 1519)$/(1000 ft/yr) (10,000 ft) 5630 $/yr Step 5. Compute the present worth (PW) of A adjusting for inflation using the appropriate equation below When g r PW An

(5.33a)

When g 2 r PW A c i

(1 i)n 1 d i(1 i)n

r g 1g

where PW present worth of annual difference in pumping cost, $ A annual difference in pumping cost, $ i effective annual investment rate accounting for inflation, % n design life of pipe, yr g inflation (growth) rate of power cost, % r annual rate of return on the initial investment, %

(5.33b) (5.33c)

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.281

In this example, g 2 n i

r g 0.066 0.035 0.030 1g 1 0.035

PW A c

(1 i)n 1 (1 0.03)50 1 d n d 5630 c i(1 i) 0.03(1 0.03)50

144,855 ($)

4 WATER FLOW IN OPEN CHANNELS 4.1 Che’zy Equation for Uniform Flow In 1769, the French engineer Antoine Che’zy proposed an equation for uniform open channel flow in which the average velocity is a function of the hydraulic radius and the energy gradient. It can be written as V C 2RS

(5.34)

where V average velocity, ft/s C Che’zy discharge coefficient, ft1/2/s R hydraulic radius, ft cross-sectional area A, (ft2) divided by the wetted perimeter P (ft) A/P S energy gradient (slope of the bed, slope of surface water for uniform flow) head loss (hf) over the length of channel (L) divided by L hf /L The value of C can be determined from C

8g Bf

(5.35a)

where g gravity constant, ft/s f Darcy–Weisback friction factor EXAMPLE: A trapezoidal open channel has a bottom width of 20 ft and side slopes of inclination 1:1.5. Its friction factor f 0.056. The depth of the channel is 4.0 ft. The channel slope is 0.025. Compute the flow rate of the channel.

Solution: Step 1. Determine C value from Eq. (5.34)

8g 8 32.2 B f B 0.056

67.8 (ft1/2/s)

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.282

CHAPTER 1.5

Step 2. Compute A, P, and R Width of water surface 20 ft 2 6 ft 32 ft A 12(20 32) ft 4 ft 104 ft2 P 2 ft 242 62 20 ft 34.42 ft R A/P 104 ft2/34.42 ft 3.02 ft Step 3. Determine the flow rate Q Q AV AC 2RS 104 ft2 67.8 ft1/2/s 23.02 ft 0.025 1937 ft3/s 4.2 Manning Equation for Uniform Flow As in the previous section, Eq. (5.22) was proposed by Robert Manning in 1889. The well-known Manning formula for uniform flow in open channel of non-pressure pipe is V

1.486 2/3 1/2 (for U.S. customary) n R S

1 V n R2/3S1/2

(for SI units)

(5.22a) (5.22b)

The flow rate (discharge) Q can be determined by 1.486 Q AV A n R2/3S1/2

(5.36)

All symbols are the same as in the Che’zy equation. The Manning roughness coefficient (n) is related to the Darcy–Weisback friction factor as follows: n 0.093 f 1/2R1/6

(5.37)

1/3

Manning also derived the relationship of n (in s/ft ) to the Che’zy coefficient C by the equation 1 C n R1/6

(5.35b)

Typical values of n for various types of channel surface are shown in Table 5.4. A 10-ft wide (w) rectangular source-water channel has a flow rate of 980 ft3/s at a uniform depth (d) of 3.3 ft. Assume n 0.016 s/ft1/3 for concrete. (a) Compute the slope of the channel. (b) Determine the discharge if the normal depth of the water is 4.5 ft. EXAMPLE 1:

Solution: Step 1. Determine A, P, and R for question (a) A wd 10 ft 3.3 ft 33 ft2 P 2d w 2 3.3 ft 10 ft 16.6 ft R A/P 33 ft2/16.6 ft 1.988 ft

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.283

Step 2. Solve S by the Manning formula 1.486 Q A n R2/3S1/2 Rewrite S a

2 2 Qn 980 cfs 0.016 s/ft1/3 b a b 0.041 1.486AR2/3 1.486 33 ft2 (1.988 ft)2/3

Answer for (a), S 0.041 Step 3. For (b), determine new A, P, and R A wd 10 ft 4.5 ft 45 ft2 P 2d w 2 (4.5 ft) 10 ft 19 ft R A/P 45 ft2/19 ft 2.368 ft Step 4. Calculate Q for answer of question (b) 1.486 Q A n R2/3S1/2 45

1.486 (2.368)2/3(0.041)1/2 0.016

1500 (ft3/s) EXAMPLE 2: A rock trapezoidal channel has bottom width 5 ft (1.5 m), water depth 3 ft (0.9 m), side slope 2:1, n 0.044, 5 ft wide. The channel bottom has 0.16% grade. Two equal-size concrete pipes will carry the flow downstream. Determine the size of pipes for the same grade and velocity.

Solution: Step 1. Determine the flow Q

58 ft 3 ft 2

19.5 ft2 R 5 ft 2 21.52 32 ft 11.71 ft

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.284

CHAPTER 1.5

By the Manning formula, Eq. (5.35) 1.486 Q A n R2/3S1/2 1.486 19.5 (11.71)2/3(0.0016)1/2 0.044 135.8 (ft3/s) Step 2. Determine diameter of a pipe D For a circular pipe flowing full, a pipe carries one-half of the total flow R

D 4

135.8 1.486 D 2/3 pD2 a b (0.0016)1/2 2 4 0.013 4 67.9 1.424 D8/3 D 4.26 (ft) 51.1 in Note: Use 48-in pipe, although the answer is slightly over 48 in. Step 3. Check velocity of flow Cross-sectional area of pipe Ap Ap

p(4 ft)2 12.56 ft2 4

Q 67.9 ft3/s AP 12.56 ft2 5.4 ft/s

This velocity is between 2 and 10 ft/s; it is thus suitable for a storm sewer. 4.3 Partially Filled Conduit The conditions of partially filled conduit are frequently encountered in environmental engineering, particularly in the case of sewer lines. In a conduit flowing partly full, the fluid is at atmospheric pressure and the flow is the same as in an open channel. The Manning equation (Eq. (5.22)) is applied. A schematic pipe cross section is shown in Fig. 5.7. The angle , flow area A, wetted perimeter P and hydraulic radius R can be determined by the following equations:

FIGURE 5.7 Flow in partially filled circular pipe.

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.285

For angle cos

u BC d 2d r d r 1 r 1

D 2 AC u 2 cos 1 a1

2d b D

(5.38a) (5.38b)

Area of triangle ABC, a a

u u 1 1 sin u 1 AB BC r sin r cos r2 2 2 2 2 2 2 1 D2 sin u 2 4 2

Flow area A A A

pD2 u

2a 4 360 pD2 u 1 D2 sin u

2a b 4 360 2 4 2 D2 pu sin u

a b 4 360 2

(5.39)

u 360

(5.40)

For wetted perimeter P P pD For hydraulic radius R R A/P Thus we can mathematically calculate the flow area A, the wetted perimeter P, and the hydraulic radius R. In practice, for a circular conduit, a chart is generally used which is available in hydraulic textbooks and handbooks (Chow, 1959; Morris and Wiggert, 1972; Zipparo and Hasen, 1993; Horvath, 1994). Assume a 24-in diameter sewer concrete pipe (n 0.012) is placed on a slope of 2.5 in 1000. The depth of the sewer flow is 10 in. What is the average velocity and the discharge? Will this grade produce a self-cleansing velocity for the sanitary sewer?

EXAMPLE 1:

Solution:

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.286

CHAPTER 1.5

Step 1. Determine from Eq. (5.38), referring to Fig. 5.7 and to the figure above u 2 10 2d 1

D 2 24 0.1667 u cos 1(0.1667) 80.4 2 160.8

cos

Step 2. Compute flow area A from Eq. (5.39) D 24 in 2 ft A

sin u u D2

ap b 4 2 360 160.8 1 22

sin 160.8b a3.14 4 2 360

1.40 0.16 1.24 (ft2) Step 3. Compute P from Eq. (5.40) P pD

u 160.8 3.14 2 ft 360 360

2.805 ft Step 4. Calculate R R A/P 1.24 ft2/2.805 ft 0.442 ft Step 5. Determine V from Eq. (5.22a) V

1.486 2/3 1/2 n R S 1.486 (0.442)2/3(2.5/1000)1/2 0.012

123.83 0.58 0.05 3.591 (ft/s) Step 6. Determine Q Q AV 1.24 ft2 3.591 ft/s 4.453 ft3/s Step 7. It will produce self-cleaning since V 2.0 ft/s.

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.287

A concrete circular sewer has a slope of 1 m in 400 m. (a) What diameter is required to carry 0.1 m3/s (3.5 ft3/s) when flowing six-tenths full? (b) What is the velocity of flow? (c) Is this a self-cleansing velocity?

EXAMPLE 2:

Solution: These questions are frequently encountered in sewer design engineering. In the partially filled conduit, the wastewater is at atmospheric pressure and the flow is the same as in an open channel, which can be determined with the Manning equation (Eq. (5.22)). Since this is inconvenient for mathematical calculation, a chart (Fig. 5.4) is commonly used for calculating area A, hydraulic radius R, and flow Q for actual values (partly filled), as opposed to full-flow values. Step 1. Find full flow rate Qf (subscript “f’’ is for full flow) Given:

d 0.60 D From Fig. 5.4 Q 0.68 Qf Q 0.1 m3/s 0.68 0.68 0.147 m3/s

Step 2. Determine diameter of pipe D from the Manning formula For full flow p Af D2 4 D Rf 4 For concrete n 0.013 Slope S 1/400 0.0025 From Eq. (5.22b): 1 V n R2/3 S1/2 1 Q AV A n R2/3 S1/2

p 2 D 2/3 1 D a b (0.0025)1/2 4 0.013 4

0.147 1.20 D8/3

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.288

CHAPTER 1.5

Answer (a) D 0.455 (m) 1.5 ft Step 3. Calculate Vf Vf

0.455 2/3 1 a b (0.0025)1/2 0.013 4

0.903 (m/s) 2.96 ft/s From Fig. 5.4 V 1.07 Vf Answer (b) V 1.07 Vf 1.07 0.903 m/s 0.966 m/s 3.17 ft/s Step 4. For answer (c) Since V 3.27 ft/s 2.0 ft/s it will provide self-cleansing for the sanitary sewer. A 12-in (0.3 m) sewer line is laid on a slope of 0.0036 with n value of 0.012 and flow rate of 2.0 ft3/s (0.057 m3/s). What are the depth of flow and velocity?

EXAMPLE 3:

Solution: Step 1. Calculate flow rate in full, Qf From Eq. (5.22a) D 1 ft Qf AV

p 2 1.486 D 2/3 D a b (0.0036)1/2 4 0.012 4

0.785 123.8 0.397 0.06 2.32 (ft3/s) Vf 2.96 ft/s Step 2. Determine depth of flow d Q 2.0 0.862 Qf 2.32

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.289

From Fig. 5.4 d 0.72 D d 0.72 1 ft 0.72 ft Step 3. Determine velocity of flow V From chart (Fig. 5.4) V 1.13 Vf V 1.13 2.96 ft/s 3.34 ft/s or

1.0 m/s

4.4 Self-cleansing Velocity The settling of suspended mater in sanitary sewers is of great concern to environmental engineers. If the flow velocity and turbulent motion are sufficient, this may prevent deposition and resuspend the sediment and move it along with the flow. The velocity sufficient to prevent deposits in a sewer is called the self-cleansing velocity. The self-cleansing velocity in a pipe flowing full is (ASCE and WEF 1992): for SI units R1/6 V n [B(s 1)Dp]1/2 1/2 8B c g(s 1)Dp d f

(5.41a)

for U.S. customary units V

1.486R1/6 [B(s 1)Dp]1/2 n

1/2 8B g(s 1)Dp d f

where V velocity, m/s or ft/s R hydraulic radius, m or ft n Manning’s coefficient of roughness B dimensionless constant 0.04 to start motion 0.8 for adequate self-cleansing s specific gravity of the particle Dp diameter of the particle f friction factor, dimensionless g gravitational acceleration 9.81 m/s2 or 32.2 ft/s2

(5.41b)

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.290

CHAPTER 1.5

Sewers flowing between 50 and 80 percent full need not be placed on steeper grades to be as self-cleansing as sewers flowing full. The reason is that velocity and discharge are functions of attractive force which depends on the friction coefficient and flow velocity (Fair et al. 1966). Figure 5.8 presents the hydraulic elements of circular sewers that possess equal self-cleansing effect. Using Fig. 5.8, the slope for a given degree of self-cleansing of partly full pipes can be determined. Applying Eq. (5.41) with the Manning equation (Eq. (5.22)) or Fig. 5.3, a pipe to carry a design full discharge Qf at a velocity Vf that moves a particle of size Dp can be selected. This same particle will be moved by a lesser flow rate between Qf and some lower discharge Qs. Figure 5.8 suggests that any flow ratio Q/Qf that causes the depth ratio d/D to be larger than 0.5 requires no increase in slope because Ss is less than Sf. For smaller flows, the invert slope must be increased to Ss to avoid a decrease in self-cleansing. EXAMPLE: A 10 in (25 cm) sewer is to discharge 0.353 ft3/s (0.01 m3/s) at a self-cleansing velocity. When the sewer is flowing full, its velocity is 3 ft/s (0.9 m/s). Determine the depth and velocity of flow and the required sewer line slope. Assume N n 0.013.

FIGURE 5.8

Hydraulic elements of circular sewers that possess equal self-cleansing properties at all depths.

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.291

Solution: Step 1. Determine the flow Qf and slope Sf during full flow Qf

pr2 V 0.785(10/12 ft)2 3.0 ft/s 4 f

1.64 ft3/s Using the Manning formula Qf 1.64

1.486 2/3 1/2 n R Sf 1.486 0.833 2/3 1/2 a b Sf 0.013 4

Rearranging Sf1/2 0.04084 Sf 0.00167 1.67% Step 2. Determine depth d, velocity Vs, and slope S for self-cleansing From Fig. 5.8 For N n and Qs/Qf 0.353/1.64 0.215, we obtain d/D 0.28 Vs/Vf 0.95 and S/Sf 1.50 Then d 0.33D 0.28 10 in 2.8 in Vs 0.95 V 0.95 3 ft/s 2.85 ft/s S 1.5 Sf 1.5 1.67% 2.50% 4.5 Specific Energy For a channel with small slope (Fig. 5.9) the total energy head at any section may be generally expressed by the general Bernoulli equation as E

V2 P g z 2g

(5.15)

where z is the elevation of the bed. For any stream line in the section, P/ z D (the water depth at the section). When the channel bottom is chosen as the datum (z 0), the total head or total energy E is called the specific energy (He). The specific energy at any section in an open channel is equal to the sum of the velocity head (kinetic energy) and water depth (potential energy) at the section. It is written as He

V2 D 2g

(5.42a)

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.292

CHAPTER 1.5

Energy gradient

V2/2g Water surface

D Channel bed

Z Datum FIGURE 5.9

Energy of open channel flow.

Since V Q/A flow/area, then He

Q2 D 2gA2

(5.42b)

For flow rate in a rectangular channel, area A is A WD

(5.43)

where W is the width of the channel. 4.6 Critical Depth Given the discharge (Q) and the cross-sectional area (A) at a particular section, Eq. (5.42a) may be rewritten for specific energy expressed in terms of discharge He

Q2 D 2gA2

(5.42b)

If the discharge is held constant, specific energy varies with A and D. At the critical state the specific energy of the flow is at a minimum value. The velocity and depth associated with the critical state are called the critical velocity and critical depth, respectively. At the critical state, the first derivative of the specific energy with respect to the water depth should be zero: Q2 dE d a Db 0 dD dD 2gA2 Q2 dA dE

10 dD gA3 dD Since near the free surface dA WdD

(5.44)

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.293

V2/2g

W FIGURE 5.10

Channel cross-section.

then dA dD where W is the top width of the channel section. For open channel flow, Dm A/W is known as the hydraulic depth (or mean depth, Dm). Hence for nonrectangular channels W

Q2W 10 gA3

Q2W gA3

(5.45) (5.45a)

Then the general expression for flow at the critical depth is Q 2gA3/W A 2gA/W Q A 2gDm

(5.46)

and the critical velocity Vc is Vc 2gA/W 2gDm Q2 Dm 2 gA Dm Hm Dc 2

(5.47) (5.48) (5.49)

where Hm is the minimum specific energy. For the special case of a rectangular channel, the critical depth Dc a

q2 1/3 Q2 1/3 b agb 2 gW

(5.50)

where q discharge per unit width of the channel, m3/s/m. For rectangular channels, the hydraulic depth is equal to the depth of the flow. Eq. (5.45) can be simplified as Q2 W 1 A2 Ag V2 1 gD V 2gD

(5.51)

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.294

CHAPTER 1.5

The quantity V/ !gD is dimensionless. It is the so-called Froude number, F. The Froude number is a very important factor for open channel flow. It relates three states of flow as follows: F 1 1

Velocity

V !gD V !gD V !gD

Flow state critical flow (V speed of surface wave) subcritical (V speed of surface wave) supercritical (V speed of surface wave)

EXAMPLE: Determine the critical depth of a channel when the water is flowing in a trapezoidal channel with a base width of 1.5 m (5.0 ft) and side slope of 2 on 1. The flow is 4.0 m3/s (63,400 gpm).

Solution:

Let h the critical depth of the channel

Then base 1.5 m W 1.5 m 2 (1/2 h) m (1.5 h) m A 12 [1.5 m (1.5 h) m] h m (1.5 h 0.5 h2) m2 Q 4.0 m3/s2 g 9.806 m/s2 Solve h using Eq. (4.45a) Q2W gA3 (4.0)2 (1.5 h) 9.806 (1.5h 0.5h2)3 (1.5h 0.5h2)3/ (1.5 h) 1.63 by trial h 0.818 (m) 2.68 ft

4.7 Hydraulic Jump When water flows in an open channel, an abrupt decrease in velocity may occur due to a sudden increase of water depth in the downstream direction. This is called hydraulic jump and may be a natural phenomenon. It occurs when the depth in an open channel increases from a depth less than the critical depth Dc to one greater than Dc. Under these conditions, the water must pass through a hydraulic jump, as shown in Fig. 5.11.

ΔE

V22/2g D2

V12/2g D2 V12

FIGURE 5.11

P1 Hydraulic jump.

Dc D1

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.295

The balance between the hydraulic forces P1 and P2, represented by the two triangles and the momentum flux through sections 1 and 2, per unit width of the channel, can be expressed as P1 P2 q (V2 V1)

(5.52)

where q is the flow rate per unit width of the channel. Substituting the following quantities in the above equation g g P1 D21; P2 D22 2 2 V1

q q ; V2 D1 D2

we get q q g 2 (D D22) r q a

b D2 D1 2 1 D1 D 2 r 1 (D D2)(D1 D2) g q2 a b D1D2 2 1 since r 1 gg then D1 D2 q2 b g D1D2 a 2

(5.53)

This equation may be rearranged into a more convenient form as follows: 1/2 D2 1 1 a 2F 12b

D1 4 2

(5.54)

where F1 is the upstream Froude number and is expressed by F1

V1 2gD1

(5.55)

If the hydraulic jump is to occur, F1 must be greater than one; i.e. the upstream flow must be supercritical. The energy dissipated (E) in a hydraulic jump in a rectangular channel may be calculated by the equation E E1 E2

(D2 D1)3 4D1D2

(5.56)

EXAMPLE 1: A rectangular channel, 4 m wide and with 0.5 m water depth, is discharging 12 m3/s flow before entering a hydraulic jump. Determine the critical depth and the downstream water depth.

Solution: Step 1. Calculate discharge per unit width q q

Q 12 m3/s 3 (m3/m # s or m2/s) W 4m

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.296

CHAPTER 1.5

Step 2. Calculate critical depth Dc from Eq. (5.50) q2 1/3 3 m2/s 3 m2/s 1/3 Dc a g b a b 9.81 m/s2 0.972 m Step 3. Determine upstream velocity V1 V1

q 3 m2/s 6.0 m/s D1 0.5 m

Step 4. Determine the upstream Froude number F1

V1 2gD1

6.0 29.81 0.5

2.71

Step 5. Compute downstream water depth D2 from Eq. (5.54) 1/2 1 1 D2 D1 c a 2F 12b d 4 2 1/2 1 1 0.5 c a 2 2.712 b d 4 2

1.68 (m) EXAMPLE 2: A rectangular channel 9 ft (3 m) wide carries 355 ft3/s (10.0 m3/s) of water with a water depth of 2 ft (0.6 m). Is a hydraulic jump possible? If so, what will be the depth of water after the jump and what will be the horsepower loss through the jump?

Solution: Step 1. Compute the average velocity in the channel V1

Q 355 ft3/s A1 9 ft 2 ft

19.72 ft/s Step 2. Compute F1 from Eq. (5.55) F1

2gD1 2.46

19.72 ft/s 232.2 ft/s2 2 ft

Since F1 1, the flow is supercritical and a hydraulic jump is possible. Step 3. Compute the depth D2 after the hydraulic jump D2 D1[(0.25 2F 21)1/2 0.5] 2[(0.25 2 2.462)1/2 0.5] 6.03 ft

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.297

Step 4. Compute velocity V2 after jump V2

Q 355 ft3/s A2 9 ft 6.03 ft

6.54 ft/s Step 5. Compute total energy loss E (or h) E1 D1

V12 19.722 2 8.04 (ft) 2g 2 32.2

E2 D2

V 22 6.542 6.03 6.69 (ft) 2g 2 32.2

E E1 E2 8.04 ft 6.69 ft 1.35 ft Step 6. Compute horsepower (hp) loss Loss

EgQ 1.35 ft 62.4 lb/ft3 355 ft3/s 550 550 ft lb/hp

54.4 hp

5 FLOW MEASUREMENTS Flow can be measured by velocity methods and direct discharge methods. The measurement flow velocity can be carried out by a current meter, Pitot tube, U-tube, dye study, or salt velocity. Discharge is the product of measured mean velocity and cross-sectional area. Direct discharge methods include volumetric gravimeter, Venturi meter, pipe orifice meter, standardized nozzle meter, weirs, orifices, gates, Parshall flumes, etc. Detail flow measurements in orifices, gates, tubes, weirs, pipes, and in open channels are discussed by Brater et al. (1996). 5.1 Velocity Measurement in Open Channel The mean velocity of a stream or a channel can be measured with a current meter. A variety of current meters is commercially available. An example of discharge calculation with known mean velocity in sub-cross section is presented in Chapter 1.2. 5.2 Velocity Measurement in Pipe Flow Pitot tube. A Pitot tube is bent to measure velocity due to the pressure difference between the two sides of the tube in a flow system. The flow velocity can be determined from V 22gh

(5.57)

where V velocity, m/s or ft/s g gravitational acceleration, 9.81 m/s2 or 32.2 ft/s2 h height of the fluid column in the manometer or a different height of immersible liquid such as mercury, m or ft

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.298

CHAPTER 1.5

EXAMPLE: The height difference of the Pitot tube is 5.1 cm (2 in). The specific weight of the indicator fluid (mercury) is 13.55. What is the flow velocity of the water?

Solution 1:

For SI units

Step 1. Determine the water column equivalent to h h 5.1 cm 13.55 69.1 cm 0.691 m of water Step 2. Determine velocity V 22gh 22 9.81 m/s2 0.691 m 3.68 m/s Solution 2:

For US customary units

Step 1. h 2/12 ft 13.55 2.258 ft Step 2. V 22gh 22 32.2 ft/s2 2.258 ft 12.06 ft/s 12.06 ft/s 0.304 m/ft 3.68 m/s 5.3 Discharge Measurement of Water Flow in Pipes Direct collection of volume (or weight) of water discharged from a pipe divided by time of collection is the simplest and most reliable method. However, in most cases it cannot be done by this method. A change of pressure head is related to a change in flow velocity caused by a sudden change of pipe cross-section geometry. Venturi meters, nozzle meters, and orifice meters use this concept. Venturi meter. A Venturi meter is a machine-cased section of pipe with a narrow throat. The device, in a short cylindrical section, consists of an entrance cone and a diffuser cone which expands to full pipe diameter. Two piezometric openings are installed at the entrance (section 1) and at the throat (section 2). When the water passes through the throat, the velocity increases and the pressure decreases. The decrease of pressure is directly related to the flow. Using the Bernoulli equation at sections 1 and 2, neglecting friction head loss, it can be seen that V 12 V 22 P1 P2 g z1 g z2 2g 2g For continuity flow between sections 1 and 2, Q1 Q2 A1V1 A2V2 Solving the above two equations, we get Q

A1A2 22g[(h1 h2) (z1 z2)] 2A21 A22

(5.16)

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

A1A2 22g(H Z ) 2A21 A22

CdA1 22g(H Z ) where

1.299

(5.58a)

Q discharge, m3/s or ft3/s A1, A2 cross-sectional areas at pipe and throat, respectively, m2 or ft2 g gravity acceleration, 9.81 m/s2 or 32.2 ft/s2 H h1 h2 pressure drop in Venturi tube, m or ft Z z1 z2 difference of elevation head, m or ft Cd coefficient of discharge

For a Venturi meter installed in a horizontal position, Z 0 Q

A1A2 22gH 2A21 A22

Cd A1 22gH

(5.58b)

where Cd

A2 2A21 A22

(5.59)

A1 2 a b 1 Å A2

EXAMPLE: A 6-cm throat Venturi meter is installed in an 18-cm diameter horizontal water pipe. The reading of the differential manometer is 18.6 cm of mercury column (sp gr 13.55). What is the flow rate in the pipe?

Solution: Step 1. Determine A1/A2 A1 p(18/2)2 254.4 cm2 0.2544 m2 A2 p(6/2)2 28.3 cm2 0.0283 m2 A1 254.4 8.99 A2 28.3 Step 2. Calculate Cd A1 2 Cd 1/ a b 1 1/ 2(8.99)2 1 Å A2 0.112

Step 3. Calculate H H gy 13.55

18.6 m 100

2.52 m Step 4. Calculate Q Q Cd A1 22gH 0.112 0.2544 22 9.81 2.52 0.20 (m3/s)

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.300

CHAPTER 1.5

Nozzle meter and orifice meter. The nozzle meter and the orifice meter are based on the same principles as the Venturi meter. However, nozzle and orifice meters encounter a significant loss of head due to a nozzle or orifice installed in the pipe. The coefficient (Cv) for the nozzle meter or orifice meter is added to the discharge equation for the Venturi meter, and needs to be determined by on-site calibration. The discharge for the nozzle meter and orifice meter is Q CvCdA1 22g(H Z )

(5.60)

Q CvCd A1 22gH

(5.61)

or, for horizontal installation

The nozzle geometry has been standardized by the American Society of Mechanical Engineers (ASME) and the International Standards Association. The nozzle coefficient Cv is a function of Reynolds number (R V2 d2/ ) and ratio of diameters (d2 /d1). A nomograph (Fig. 5.12) is available. Values of Cv range from 0.96 to 0.995 for R ranging from 5 104 to 5 106. EXAMPLE: Determine the discharge of a 30-cm diameter water pipe. An ASME nozzle of 12-cm throat diameter is installed. The attached differential manometer reads 24.6 cm of mercury column. The water temperature in the pipe is 20 C.

Solution: Step 1. Determine ratio A1 : A2 p (30 cm)2 707 cm2 0.0707 m2 4 p A2 (12 cm)2 113 cm2 0.0113 m2 4 A1 302 2 6.25 A2 12 A1

FIGURE 5.12.

ASME nozzle coefficients.

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.301

Step 2. Calculate Cd Cd 1/ 2(A1/A2)2 1 1/ 2(6.25)2 1 0.162 Step 3. Calculate H g 13.55 at 20C H gy 13.55 24.6 cm 333 cm 3.33 m Step 4. Estimate discharge Q Assuming Cv 0.98 Q CvCd A1 22gH 0.98 0.162 0.0707 22 9.81 3.33 0.091 (m3/s) This value needs to be verified by checking the corresponding Reynolds number R of the nozzle. Step 5. Calculate R V2 Q/A2 0.091 m3/s/0.0113 m2 8.053 m/s The kinematic viscosity at 20C is (Table 5.1a) n 1.007 10 6 m2/s then V2d2 8.053 m/s 0.12 m n 1.007 10 6 m2/s 9.6 105

Using this R value, with d1/d2 0.4, the chart in Fig. 5.12 gives Cv 0.977 Step 6. Correct Q value 0.977 0.091 m3/s 0.98 0.0907 m3/s

5.4 Discharge Measurements Orifices. The discharge of an orifice is generally expressed as Q Cd A 22gH where Q flow rate Cd coefficient of discharge A area of orifice g gravity acceleration H water height above center of orifice All units are as Eq. (5.58a)

(5.62)

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.302

CHAPTER 1.5

EXAMPLE: The water levels upstream and downstream are 4.0 and 1.2 m, respectively. The rectangular orifice has a sharp-edged opening 1 m high and 1.2 m wide. Calculate the discharge, assuming Cd 0.60.

Solution: Step 1. Determine area A A 1 m 1.2 m 1.2 m2 Step 2. Determine discharge Q Q Cd A 22g(h1 h2) 0.60 1.2 m2 22 9.81 m/s2 (4 1.2) m 5.34 m3/s Weirs.

The general discharge equation for a rectangular, horizontal weir is Q CdLH3/2

(5.63a)

where Q flow rate, ft3/s or m3/s L weir length, ft or m H head on weir, ft or m the water surface above the weir crest Cd coefficient of discharge In the US customary system, let y be the weir height in feet. H Cd 3.22 0.40 y

(US customary units)

(5.64a)

H Cd 1.78 0.24 y

(SI units)

(5.64b)

The theoretical equation for a rectangular streamlined weir is (ASCE & WEF, 1992) 2 3 2 gH Q L 3 B3

(5.63b)

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.303

where g is the gravitational acceleration rate. This formula reduces to

Q 3.09 L H 3/2 (for US customary units system)

(5.63c)

Q 1.705 L H

(5.63d)

3/2

(for SI units)

EXAMPLE 1: A rectangular flow control tank has an outflow rectangular weir 1.5 m (5 ft) in length. The inflow to the box is 0.283 m3/s (10 ft3/s). The crest of the weir is located 1.2 m above the bottom of the tank. Find the depth of water in the tank.

Solution: Step 1. Determine head on weir H Using Eq. 5.63d Q 1.705 L H 3/2 0.283 1.705(1.5) H 3/2 H 3/2 0.111 H 0.23 (m) Step 2. Calculate water depth D D 1.2 m H 1.2 m 0.23 m 1.43 m EXAMPLE 2: In a rectangular channel 4 m (13 ft) high and 12 m (40 ft) wide, a sharp-edged rectangular weir 1.2 m (4 ft) high without end contraction will be installed. The flow of the channel is 0.34 m3/s (12 ft3/s). Determine the length of the weir to keep the head on the weir 0.15 m (0.5 ft).

Solution: Step 1. Compute velocity of approach at channel, V V

Q 0.34 m3/s 0.007 m/s (negligible) A 12 4

Head due to velocity of approach h

V2 2g

(negligible)

Step 2. Use weir formula (Eq. (5.63d)) (a) Without velocity of approach Q 1.705 L H3/2 0.34 1.705 L H3/2 0.34 1.705 L (0.15)1.5 L 3.43 (m) 11.2 ft (b) Including velocity of approach Q 1.705 L [(0.15 h)1.5 h1.5] L 3.4 m

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.304

CHAPTER 1.5

For freely discharging rectangular weirs (sharp-crested weirs), the Francis equation is most commonly used to determine the flow rate. The Francis equation is Q 3.33 LH 3/2

(5.63e)

where Q flow rate, ft3/s L weir length, ft H head on weir, ft the water surface above the weir crest For constracted rectangular weirs Q 3.33 (L 0.1 nH) H 3/2 (US customary units)

(5.63f)

Q 1.705 (L 0.1 nH) H

(5.63g)

1/2

(SI units)

where n number of end constractions. EXAMPLE 3: Conditions are the same as in Example 2, except that the weir has two end constractions. Compute the width of the weir.

Solution: Step 1. For constracted weir, without velocity of approach Q 1.705 (L 0.2 H) H 3/2 0.34 1.705 (L 0.2 0.15) (0.15)1.5 L 3.40 (m) 11.1 ft Step 2. Discharge including velocity of approach Q 1.705 (L 0.2H)[(H h)3/2 h3/2] Since velocity head is negligible L 3.4 m EXAMPLE 4:

Estimate the discharge of a weir 4 ft long where the head on the weir is 3 in.

Solution: By the Francis equation Q 3.33 L H3/2/s 3.33 4 (3/12)3/2 1.66 (ft3/s) For the triangular weir and V-notch weir, the flow is expressed as u Q Cd atan bH5/2 2

(5.65)

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.305

where weir angle Cd discharge coefficient, calibrated in place (different values have been proposed) 1.34 for SI units 2.50 for US customary units The triangular weir is commonly used for measuring small flow rates. Several different notch angles are available. However, the 90 V-notch weir is the one most commonly used. The discharge for 90 V-notch with free flow is Q 2.5 H 2.5

(5.66)

where Q discharge, ft3/s H head on the weir, ft Detailed discussion of flow equations for SI units for various types of weir is presented by Brater et al. (1996). EXAMPLE 5: A rectangular control tank has an outflow rectangular weir 8 ft (2.4 m) in length. The crest of the weir is 5 ft (1.5 m) above the tank bottom. The inflow from a pipe to the tank is 10 ft3/s (0.283 m3/s). Estimate the water depth in the tank, using the Francis equation.

Solution: Q 3.33 LH3/2 Rewrite equation for H H a

2/3 2/3 Q 10 b a b 3.33 L 3.33 8

0.52 (ft) EXAMPLE 6: A rectangular channel 6 ft (1.8 m) wide has a sharp-crested weir across its whole width. The weir height is 4 ft (1.2 m) and the head is 1 ft (0.3 m). Determine the discharge.

Solution: Step 1. Determine Cd using Eq. (5.64) H 1 Cd 3.22 0.40 y 3.22 0.40 4 3.32 Step 2. Compute discharge Q Q CdL H3/2 3.32 6 (1)3/2 19.92 (ft3/s) A circular sedimentation basin has a diameter of 53 ft (16 m). The inflow from the center of the basin is 10 MGD (0.438 m3/s). A circular effluent weir with 90 V-notches located at 0.5 ft (0.15 m) intervals is installed 1.5 ft (0.45 m) inside the basin wall. Determine the water depth on each notch and the elevation of the bottom of the V-notch if the water surface of the basin is at 560.00 ft above mean sea level (MSL). EXAMPLE 7:

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.306

CHAPTER 1.5

Solution: Step 1. Determine the number of V-notches Diameter of the weir d 53 ft 2 1.5 ft 50 ft Weir length l pd 3.14 50 ft 157 ft No. of V-notches 2 notches ft 314 notches

n 157 ft

Step 2. Compute the discharge per notch 10,000,000 gal/day 31,847 (gal/day)/notch 314 notches 1 day gal/day 1 ft3 31,847 notch 7.48 gal 86,400 s

0.0493(ft3/h)/notch Step 3. Compute the head of each notch from Eq. (5.66) Q 2.5 H 2.5 H (Q/2.5)1/2.5 (Q/2.5)0.4 (0.0493/2.5)0.4 0.21 (ft) Step 4. Determine the elevation of the bottom of the V-notch Elevation 560.00 ft 0.21 ft 559.79 ft MSL Parshall flume. The Parshall flume was developed by R. L. Parshall in 1920 for the British measurement system. It is widely used for measuring the flow rate of an open channel. It consists of a converging section, a throat section, and a diverging section. The bottom of the throat section is inclined downward and the bottom of the diverging section is inclined upward. The dimensions of the Parshall flume are specified in the U.S. customary system, not the metric system. The geometry creates a critical depth to occur near the beginning of the throat section and also generates a back water curve that allows the depth Ha to be measured from observation well a. There is a second measuring point Hb located at the downstream end of the throat section. The ratio of Hb to Ha is defined as the submergence of the flume. Discharge is a function of Ha. The discharge equation of a Parshall flume is determined by its throat width, which ranges from 3 in to 50 ft. The relationship between the empirical discharge and the gage reading Ha for various sizes of flume is given below: Throat width W, ft

Discharge equation, ft3/s

3 in (0.25 ft) 6 in (0.5 ft) 9 in (0.75 ft) 1–8 ft 10–50 ft

Q Q Q Q Q (3.6875 W 2.5) H 1.6 a

Source: R. L. Parshall (1926).

0.992 H a1.547 2.06 H 1.58 a 3.07 Ha1.53 0.26 1.522W 4 W Ha

Flow capacity, ft3/s 0.03–1.9 0.05–3.9 0.09–8.9 up to 140

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.307

FIGURE 5.13 Diagram for determining rate of submerged flow through a 6-in Parshall flume (U.S. Department of the Interior, 1997).

When the ratio of Ha to Hb exceeds the following values: 0.5 for flumes of 1, 2, and 3 in width, 0.6 for flumes of 6 and 9 in width, 0.7 for flumes of 1 to 8 ft width, 0.8 for flumes of 8 to 50 ft width. The flume is said to be submerged. When a flume is submerged, the actual discharge is less than that determined by the above equations. The diagram presented in Figs. 5.13 and 5.14 can be used to determine discharges for submerged Parshall flumes of 6 and 9 in width, respectively. Figure 5.15 shows the discharge correction for flumes with 1 to 8 ft throat width and with various percentages of submergence. The correction for the 1-ft flume can be applicable to larger flumes by multiplying by a correction factor of 1.0, 1.4, 1.8, 2.4, 3.1, 4.3, and 5.4 for flume sizes of 1, 1.5, 2, 3, 4, 6, and 8 ft width, respectively. Figure 5.16 is used for determining the correction to be subtracted from the free-flow value for a 10-ft Parshall flume. For larger sizes, the correction equals the value from the diagram multiplied by the size factor M. Figs. 5.13–5.16 are improved curves from earlier developments (Parshall 1932, 1950). Calculate the discharge through a 9-in Parshall flume for the gage reading H a of 1.1 ft when (a) the flume has free flow, and (b) the flume is operating at 72 percent of submergence. EXAMPLE 1:

Solution: Step 1. Calculate Q for free flow for (a) 3.07 (1.1)1.53 Q 3.07 H1.53 a 3.55 (ft3/s)

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.308

CHAPTER 1.5

FIGURE 5.14 1997).

Diagram for determining rate of submerged flow through a 9-in Parshall flume (U.S. Department of the Interior,

FIGURE 5.15 Diagram for determining correction to be subtracted from the free discharge to obtain rate of submerged flow through 1- to 8-ft Parshall flumes (U.S. Department of the Interior, 1997).

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.309

FIGURE 5.16 Diagram for determining correction to be subtracted from discharge flow to obtain rate of submerged flow through 10- to 50-ft Parshall flumes (U.S. Department of the Interior, 1997).

Step 2. Determine discharge with 0.72 submergence for (b) From Fig. 5.14, enter the plot at Hb/Ha 0.72 and proceed horizontally to the line representing Ha 1.1. Drop downward and read a discharge of Qs 3.33 ft3/s EXAMPLE 2:

Hb 1.14 ft.

Determine the discharge of a 2-ft Parshall flume with gage readings Ha 1.55 ft and

Solution: Step 1. Compute Hb/Ha ratio Hb 1.14 ft 0.74 or 74% Ha 1.55 ft This ratio exceeds 0.7, therefore the flume is submerged. Step 2. Compute the free discharge of the 2-ft flume Q 4WH1.522W a

0.26

4(2) (1.55)1.5222 8 2.223 17.78 (ft3/s)

0.26

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS 1.310

CHAPTER 1.5

Step 3. Determine the correction discharge for a 1-ft flume from Fig. 5.15, enter the plot at a value of Ha 1.55 ft and proceed horizontally to the line representing the submergence curve of 74 percent. Then drop vertically downward and read a discharge correction for the 1-ft flume of Q cl 0.40 ft3/s Step 4. Compute discharge correction for the 2-ft flume, M 1.8 (Fig. 5.15) Q c2 1.8 0.40 ft3/s 0.72 ft3/s Step 5. Compute the actual discharge Q Q (at Step 2 at Step 4) 17.78 ft3/s 0.72 ft3/s 17.06 ft3/s

REFERENCES American Society of Civil Engineers (ASCE) and Water Environment Federation (WEF). 1992. Design and construction of urban stormwater management systems. New York: ASCE & WEF. Benefield, L. D., Judkins, J. F. Jr and Parr, A. D. 1984. Treatment plant hydraulics for environmental engineers. Englewood Cliffs, New Jersey: Prentice-Hall. Brater, E. F., King, H. W., Lindell, J. E. and Wei, C. Y. 1996. Handbook of hydraulics, 7th edn. New York: McGraw-Hill. Brisbin, S. G. 1957. Flow of concentrated raw sewage sludges in pipes. J. Sanitary Eng. Div., ASCE, 83: 201. Chadwick, A. J. and Morfett, J. C. 1986. Hydraulics in civil engineering. London: Allen & Unwin. Chow, V. T. 1959. Open-channel hydraulics. New York: McGraw-Hill. Cross, H. 1936. Analysis of flow in networks of conduits or conductors. University of Illinois Bulletin no. 286. Champaign-Urbana: University of Illinois. Ductile Iron Pipe Research Association. 1997. Ductile Iron Pipe News, fall/winter 1997. Fair, G. M., Geyer, J. C. and Okun, D. A. 1966. Water and wastewater engineering, Vol. 1: Water supply and wastewater removal. New York: John Wiley. Fetter, C. W. 1994. Applied hydrogeology, 3rd edn. Upper Saddle River, New Jersey: Prentice-Hall. Horvath, I. 1994. Hydraulics in water and waste-water treatment technology. Chichester: John Wiley. Hwang, N. H. C. 1981. Fundamentals of hydraulic engineering systems. Englewood Cliffs, New Jersey: PrenticeHall. Metcalf & Eddy, Inc. 1972. Wastewater engineering: Collection, treatment, disposal. New York: McGraw-Hill. Metcalf & Eddy, Inc. 1990. Wastewater engineering: Collection and pumping of wastewater. New York: McGraw-Hill. James M. Montgomery, Consulting Engineers, Inc. 1985. Water treatment principles and design. New York: John Wiley. Morris, H. M. and Wiggert, J. M. 1972. Applied hydraulics in engineering, 2nd edn. New York: John Wiley. Parshall, R. L. 1926. The improved Venturi flume. Trans. Amer. Soc. Civil Engineers, 89: 841–851. Parshall, R. L. 1932. Parshall flume of large size. Colorado Agricultural Experiment Station, Bulletin 386. Parshall, R. L. 1950. Measuring water in irrigation channels with Parshall flumes and small weirs. US Soil Conservation Service, Circular 843.

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

1.311

Perry, R. H. 1967. Engineering manual: A practical reference of data and methods in architectural, chemical, civil, electrical, mechanical, and nuclear engineering. New York: McGraw-Hill. Streeter, V. L. and Wylie, E. B. 1975. Fluid mechanics, 6th edn. New York: McGraw-Hill. US Department of the Interior, Bureau of Reclamation. 1997. Water measurement manual, 3rd edn. Denver, Colorado: US Department of the Interior. Williams, G. S. and Hazen, A. 1933. Hydraulic tables, 3rd edn. New York: John Wiley. Zipparro, V. J. and Hasen, H. 1993. Davis handbook of applied hydraulics. New York: McGraw-Hill.

FUNDAMENTAL AND TREATMENT PLANT HYDRAULICS

Source: HANDBOOK OF ENVIRONMENTAL ENGINEERING CALCULATIONS

CHAPTER 1.6

PUBLIC WATER SUPPLY Shun Dar Lin

SOURCES AND QUANTITY OF WATER 1.314

POPULATION ESTIMATES 1.316 2.1 Arithmetic Method 1.317 2.2 Constant Percentage Growth Rate Method 1.317 2.3 Declining Growth Method 1.318 2.4 Logistic Curve Method 1.318

WATER REQUIREMENTS 1.320 3.1 Fire Demand 1.320 3.2 Leakage Test 1.324

REGULATION FOR WATER QUALITY 1.324 4.1 Federal Drinking Water Standards 1.324 4.2 Atrazine 1.332

WATER TREATMENT PROCESSES 1.332

AERATION AND AIR STRIPPING 1.333 1.1 Gas Transfer Models 1.336 1.2 Diffused Aeration 1.344 1.3 Packed Towers 1.345 1.4 Nozzles 1.352

SOLUBILITY EQUILIBRIUM 1.355

COAGULATION 1.357 8.1 Jar Test 1.359 8.2 Mixing 1.361

FLOCCULATION 1.363

SEDIMENTATION 1.367 10.1 Overflow Rate 1.370 10.2 Inclined Settlers 1.371

FILTRATION 1.373 11.1 Filter Medium Size 1.374 11.2 Mixed Media 1.375 11.3 Hydraulics of Filter 1.375 11.4 Washwater Troughs 1.379 11.5 Filter efficiency 1.381

12 WATER SOFTENING 1.382 13 ION EXCHANGE 1.388 13.1 Leakage 1.393 13.2 Nitrate Removal 1.394 14 IRON AND MANGANESE REMOVAL 1.402 14.1 Oxidation 1.402 15 ACTIVATED CARBON ADSORPTION 1.406 15.1 Adsorption Isotherm Equations 1.406 16 MEMBRANE PROCESSES 1.409 16.1 Reverse Osmosis 1.409 16.2 Silt Density Index 1.413 17 RESIDUAL FROM WATER PLANT 1.414 17.1 Residual Production and Density 1.416 18 DISINFECTION 1.418 18.1 Chemistry of Chlorination 1.418 18.2 Ozonation 1.425 18.3 Disinfection Kinetics 1.426 18.4 CT Values 1.427 19 WATER FLUORIDATION 1.439 19.1 Fluoride Chemicals 1.439 19.2 Optimal Fluoride Concentrations 1.440 19.3 Fluoride Feed Rate (Dry) 1.442 19.4 Fluoride Feed Rate for Saturator 1.443 19.5 Fluoride Dosage 1.443 REFERENCES 1.445

1.313

PUBLIC WATER SUPPLY 1.314

CHAPTER 1.6

1 SOURCES AND QUANTITY OF WATER The sources of a water supply may include rainwater, surface waters, and groundwater. During the water cycle, rainwater recharges the surface waters and groundwater. River, stream, natural pond, impondment, reservoir, and lake waters are major surface water sources. Some communities use groundwater, such as galleries, wells, aquifers, and springs as their water supply sources. Water supplies may be drawn from a single source or from a number of different ones. The water from multiple sources could be mixed before distribution or separately distributed. Any new source of water has to be approved by the federal, state, and related authorities. The quantity of water needed varies with reason, geography, size and type of community, and culture. A water supply system may provide for domestic, industrial, and commercial usage; public services; fire demand; unaccounted losses; as well as farm uses. The design flow for a water supply system is discussed in the sections of water requirement and fire demand. The engineers should examine the availability of water sources and the quantities in the area. The prediction of the future population should be made for the design purpose. Based on long-term meteorological data, the amount of water stored in lakes and reservoirs can be estimated. Runoff refers to the precipitation that reaches a stream or river. Theoretically, every unit volume of water passing the observation station should be measured and the sum of all these units passing in a certain period of time would be the total runoff. However, it is not available due to cost. Observations should be carried out at reasonably close intervals so that the observed flow over a long period of time is the sum of flows for the individual shorter periods of observation and not the sum of the observed rates of flow multiplied by the total period. EXAMPLE 1: Records of observations during a month (30 days) period show that a flow rate 2.3 cfs (0.065 m3/s) for 8 days, 3.0 cfs (0.085 m3/s) for 10 days, 56.5 cfs (16.0 m3/s) for 1 days, 12.5 cfs (0.354 m3/s) for 2 days, 5.3 cfs (0.150 m3/s) for 6 days, and 2.65 cfs (0.075 m3/s) for 3 days. What is the mean flow rate?

Solution: Q

2.3 8 3.0 10 56.5 1 12.5 2 5.3 6 2.65 3 30

5.66 ft3/s 0.160 m3/s If the mean annual rainfall is 81 cm (32 in). A horizontal projected roof area is 300 m2 (3230 ft2). Make a rough estimate of how much water can be caught.

EXAMPLE 2:

Solution: Step 1. Calculate annual gross yield, Vy Vy 300 m2 0.81 m/yr 243 m3/yr 0.666 m3/d Step 2. Estimate net yield Vn 2 According to Fair et al. (1966), Vn 3 Vy

Vn 23 Vy 23 (243 m3/yr) 162 m3/yr 0.444 m3/d

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.315

Step 3. Estimate the water that can be stored and then be used, Vu Vu 0.5 Vn 0.5(162 m3/yr) 81 m3/yr 2860 ft3/yr

0.5(0.444 m3/d) 0.222 m3/d 7.83 ft3/d

EXAMPLE 3: Determine the rainfall-runoff relationship. Using a straight line by method of average. The given values are as follows: Year

Rainfall, in

1988 1989 1990 1991 1992 1993 1994 1995 1996 1997

15.9 19.4 23.9 21.0 24.8 26.8 25.4 24.3 27.3 22.7

Runoff, cfs/mile2 26.5 31.2 38.6 32.5 37.4 40.2 38.7 36.4 39.6 34.5

Solution: Step 1. Write a straight line equation as y mx b in which m is the slope of the line, and b is the intercept on the Y axis. Step 2. Write an equation for each year in which the independent variable is X (rainfall) and the dependent variable is Y (runoff). Group the equations in two groups, and alternate the years. Then total each group. 1988 1990 1992 1994 1996

26.5 15.9 m b 38.6 23.9 m b 37.4 24.8 m b 38.7 25.4 m b 39.6 27.3 m b 180.8 117.3 m 5b

1989 1991 1993 1995 1997

31.2 19.4 m b 32.5 21.0 m b 40.2 26.8 m b 36.4 24.3 m b 34.5 22.7 m b 174.8 114.2 m 5b

Step 3. Solve the total of each group of equations simultaneously for m and b 180.8 117.3m 5b (174.8 114.2m 5b) 6.0 3.1m m 1.94 Step 4. Substituting for m and solving for b. 180.8 117.3(1.94) 5b 5b 46.76 b 9.35

PUBLIC WATER SUPPLY 1.316

CHAPTER 1.6

Step 5. The equation of straight line of best fit is Y 1.94X 9.35 where X rainfall, in Y runoff, cfs/mile2 A watershed has a drainage area of 1000 ha (2470 acres). The annual rainfall is 927 mm (36.5 in). The expected evaporation loss is 292 mm (11.5 in) per year. The estimated loss to groundwater is 89 mm (3.5 in) annually. Estimate the amount of water that can be stored in a lake and how many people can be served, assuming 200 L/(c d) is needed.

EXAMPLE 4:

Solution: Step 1. Using a mass balance R (rainfall excess) P (precipitation) E (evaporation) G (loss to groundwater) R 927 mm 292 mm 89 mm 546 mm Step 2. Convert R from mm to m3 (volume) and L R 564 mm

10,000 m2 1m 1000 ha 1000 mm 1 ha

5.46 106 m3 5.46 106 m3 m3 L/1 m3 5.46 109 L Step 3. Compute the people that can be served Annual usage per capita 200 L/(c # d) 365 day 7.3 104 L/c No. of people served

5.46 109 L 7.3 104 L/c

74,800 c

2 POPULATION ESTIMATES Prior to the design of a water treatment plant, it is necessary to forecast the future population of the communities to be served. The plant should be sufficient generally for 25 to 30 years. It is difficult to estimate the population growth due to economic and social factors involved. However, a few methods have been used for forecasting population. They include the arithmetic method and uniform percentage growth rate method (Clark and Viessman, 1966; Steel and McGhee, 1979; Viessman and Hammer, 1993). The first three methods are short-term ( 10 years) forecasting.

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.317

2.1 Arithmetic Method This method of forecasting is based upon the hypothesis that the rate of increase is constant. It may be expressed as follows: dp ka dt

(6.1)

where p population t time, yr ka arithmetic growth rate constant Rearrange and integrate the above equation, p1 and p2 are the populations at time t1 and t2 respectively. p2

3 dp 3 kadt p1

We get p2 p1 ka(t2 t1) p2 p1 p ka t t t 2 1

(6.2a)

pt p0 ka t

(6.2b)

where pt population at future time p0 present population, usually use p2 (recently censused) 2.2 Constant Percentage Growth Rate Method The hypothesis of constant percentage or geometric growth rate assumes that the rate increase is proportional to population. It can be written as dp kp p dt

(6.3a)

Integrating this equation yields ln p2 ln p1 kp (t2 t1) kp

ln p2 ln p1 t2 t1

(6.3b)

The geometric estimate of population is given by ln p ln p2 kp (t t2)

(6.3c)

PUBLIC WATER SUPPLY 1.318

CHAPTER 1.6

2.3 Declining Growth Method This is a decreasing rate of increase on the basis that the growth rate is a function of its population deficit. Mathematically it is given as dp kd (ps p) dt

(6.4a)

where ps saturation population, assume value Integration of the above equation gives p2

2 dp 3 ps p kd 3 dt

ps p2 ln p kd (t2 t1) s p1 Rearranging ps p2 1 ln p kd t t s p1 2 1

(6.4b)

P P0 (Ps P0)(1 ekdt)

(6.4c)

The future population P is

where P0 population of the base year 2.4 Logistic Curve Method The logistic curve-fitting method is used for modeling population trends with an S-shape for large population center, or nations for long-term population predictions. The logistic curve form is P where

ps 1 eab t

(6.5a)

Ps saturation population a, b constants

They are: Ps

2p0 p1 p2 p21( p0 p2) p0 p2 p21

a ln

ps p0 p0

1 p0(ps p1) b n ln p1(ps p0)

(6.5b) (6.6) (6.7)

where n time interval between successive censuses Substitution of these values in Eq. (6.5a) gives the estimation of future population on P for any period t beyond the base year corresponding to P0.

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.319

EXAMPLE: A mid-size city recorded populations of 113,000 and 129,000 in the April 1980 and April 1990 census, respectively. Estimate the population in January 1999 by comparing (a) arithmetic method, (b) constant percentage method, and (c) declining growth method.

Solution: Step 1. Solve with the arithmetic method Let t1 and t2 for April 1980 and April 1990, respectively t t2 t1 10 yr p2 p1 129,000 113,000 1600 Ka t 10 2 t1

Using Eq. (6.2a)

Predict pt for January 1999 from t2, using Eq. (6.2b) t 8.75 yr pt p2 kat 129,000 1600 8.75 143,000 Step 2. Solve with constant percentage method, using Eq. (6.3b) ln p2 ln p1 ln 129,000 ln 113,000 t2 t1 10 0.013243

Then, using Eq. (6.3c)

ln P ln P2 kp (t t2) ln 129,000 0.013243 8.75 11.8834 p 144,800

Step 3. Solve with declining growth method Assuming ps 200,000 Using Eq. (6.4b)

ps p2 1 kd t ln p s p1 2 t1

200,000 129,000 1 ln 10 200,000 113,000

0.02032 From Eq. (6.4c) P P0 (Ps P0)(1 ekdt) 129,000 (200,000 129,000)(1 e0.02032 × 8.75) 129,000 71,000 × 0.163 140,600

PUBLIC WATER SUPPLY 1.320

CHAPTER 1.6

3 WATER REQUIREMENTS The uses of water include domestic, commercial and industrial, public services such as fire fighting and public buildings, and unaccounted pipeline system losses and leakage. The average usage in the United States for the above four categories are 220, 260, 30, and 90 liters per capita per day (L/(c d)), respectively (Tchobanoglous and Schroeder 1985). These correspond to 58, 69, 8, and 24 gal/(c d), respectively. Total municipal water use averages 600 L/(c d) or 160 gal/(c d) in the United States. The maximum daily water use ranges from about 120 to 400 percent of the average daily use with a mean of about 180 percent. Maximum hourly use is about 150 to 12,000 percent of the annual average daily flow; and 250 to 270 percent are typically used in design. 3.1 Fire Demand Fire demand of water is often the determining factor in the design of mains. Distribution is a shortterm, small quantity but with a large flow rate. According to uniform fire code, the minimum fire flow requirement for a one- and two-family dwelling shall be 1000 gallons per minute (gpm). For the water demand for fire fighting based on downtown business districts and high-value area for communities of 200,000 people or less, the National Board of Fire Underwriters (1974) recommended the following fire flow rate and population relationship: Q 3.86 2p (1 0.01 2p) (SI units)

(6.8a)

Q 1020 2p (1 0.01 2p) (US customary units)

(6.8b)

where Q discharge, m3/min or gal/min (gpm) P population in thousands The required flow rate for fire fighting must be available in addition to the coincident maximum daily flow rate. The duration during the required fire flow must be available for 4 to 10 h. National Board of Fire Underwriters recommends providing for a 10-h fire in towns exceeding 2500 in population. The Insurance Services Office Guide (International Fire Service Training Association, 1993) for determination of required fire flow recommends the formula F 18 C 2A (for US customary units)

(6.9a)

F 320 C 2A (for SI units)

(6.9b)

where F required fire flow, gpm or m3/d C coefficient related to the type of construction C value 1.5 1.0 0.9 0.8 0.6

Construction wood frame ordinary heavy timber type building noncombustible fire-resistant

Maximum flow, gpm (m3/d) 8000 (43,600) 8000 (43,600) 6000 (32,700) 6000 (32,700)

A total floor area, ft2 or m2 EXAMPLE 1: A four-story building of heavy timber type building of 715 m2 (7700 ft2) of ground area. Calculate the water fire requirement.

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.321

Solution: Using Eq. (6.9b): F 320 C 2A 320 0.9 24 715 15,400 (m3/d) 2800 gpm

A five-story building of ordinary construction of 7700 ft2 (715 m2) of ground area communicating with a three-story building of ordinary construction of 9500 ft2 (880 m2) ground area. Compute the required fire flow. EXAMPLE 2:

Solution: Using Eq. (6.9a) Total area

A 5 7700 3 9500 67,000 ft2 F 18 1.0 267,000 4660 (gpm) 4750 (gpm)

rounded to nearest 250 gpm

25,900 m3/d

Assuming a high-value residential area of 100 ha (247 acres) has a housing density of 10 houses/ha with 4 persons per household, determine the peak water demand, including fire, in this residential area.

EXAMPLE 3:

Solution: Step 1. Estimate population P P (4 capita/house) (10 houses/ha) (100 ha) 4000 capita Step 2. Estimate average daily flow Qa Qa residential public service unaccounted (220 30 90) 340 (L/(c d)) Step 3. Estimate maximum daily flow Qmd for the whole area Using the basis of Qmd is 180 percent of Qa Qmd (340 L/(c d))(1.8) (4000 c) 2,448,000 L/d 2400 m3/d Step 4. Estimate the fire demand using Eq. (5.8a) Qf 3.86 !p (1 0.01 !p) m3/ min

3.86 !4 (1 0.01 !4) 7.57 m3/min 7.57 m3/min 60 min/h 10 h/d 4540 m3/d

PUBLIC WATER SUPPLY 1.322

CHAPTER 1.6

Step 5. Estimate total water demand Q Q Qmd Qf 2400 m3/d 4540 m3/d 6940 m3/d Note: In this area, fire demand is a control factor. It is measuring to compare Q and peak daily demand. Step 6. Check with maximum hourly demand Qmh The Qmh is assumed to be 250% of average daily demand. Qmh 2400 m3/d 2.5 6000 m3/d Step 7. Compare Q versus Qmh Q 6940 m3/d Qmh 6000 m3/d Use Q 6940 m3/d for the main pipe to this residential area. EXAMPLE 4:

Estimate the municipal water demands for a city of 225,000 persons.

Solution: Step 1. Estimate the average daily demand Qavg Qavg 600 L/(c d) 225,000 c 135,000,000 L/d 1.35 105 m3/d Step 2. Estimate the maximum daily demand Qmd ( f 1.8) Qmd 1.35 105 m3/d 1.8 2.43 105 m3/d Step 3. Calculate the fire demand Qf, using Eq. (5.8a) Qf 3.86 !p (1 0.01 !p) m3/min

3.86 !225 (1 0.01 !225) m3/min 49.215 m3/min

For 10 h duration of daily rate Qf 49.215 m3/min 60 min/h 10 h/d 0.30 105 m3/d Step 4. Sum of Qmd and Qf (fire occurs coincident to peak flow) Qmd Qf (2.43 0.30) 105 m3/d 2.73 105 m3/d Step 5. Estimate the maximum hourly demand Qmh ( f 2.7) Qmh f Qavg 2.7 1.35 105 m3/d 3.645 105 m3/d 96.3 MGD

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.323

Step 6. Compare steps 4 and 5 Qmh 3.645 105 m3/d Qmd Qf 2.73 105 m3/d Use Qmh 3.65 105 m3/d to design the plant’s storage capacity. Fire flow tests. Fire flow tests involve discharging water at a known flow rate from one or more hydrants and examining the corresponding pressure drop from the main through another nearby hydrant. The discharge from a hydrant nozzle can be determined as follows (Hammer, 1975): Q 29.8 C d 2 !p

(6.10)

where Q hydrant discharge, gpm C coefficient, normally 0.9 d diameter of outlet, in p pitot gage reading, psi The computed discharge at a specified residual pressure can be computed as Qp Qf

Hp Hf

Qp Qf a

0.54

Hp Hf

(6.11a) b

0.54

(6.11b)

where Qp computed flow rate at the specified residual pressure, gpm Qf total discharge during fire flow test, gpm Hp pressure drop from beginning to special pressure, psi Hf pressure drop during fire flow test, psi EXAMPLE: All four hydrant numbers 2, 4, 6, and 8, as shown in the following figure, have the same nozzle size of 2.5 in and discharge at the same rate. The pitot tube pressure at each hydrant during the test is 25 psi. At this discharge the residual pressure at hydrant #5 dropped from the original 100 psi to 65 psi. Compute the flow rate at a residual pressure of 30 psi based on the test.

PUBLIC WATER SUPPLY 1.324

CHAPTER 1.6

Solution: Step 1. Determine total discharge during the test Using Eq. (6.10): Qf 4 29.8 C d 2 !p 4 29.8 0.9 (2.5)2 !25 3353 (gpm)

Step 2. Compute Qp at 30 psi by using Eq. (6.11b) Qp Q f a

Hp Hf

0.54

3353 a

100 30 0.54 b 100 65

4875 (gpm) 3.2 Leakage Test After a new main is installed, a leakage test is usually required. The testing is generally carried out for a pipe length not exceeding 300 m (1000 ft). The pipe is filled with water and pressurized (50 percent above normal operation pressure) for at least 30 min. The leakage limit recommended by the American Water Work Association is determined as: L (ND !p)/326

L (ND !p )/1850

for SI units

(6.12a)

for US customary units

(6.12b)

where L allowable leakage, L/(mm diameter km d) or gal/(in mile d) N number of joints in the test line D normal diameter of the pipe, mm or in P average test pressure Leakage allowed in a new main is generally specified in the design. It ranges from 5.6 to 3.2 L/mm diameter per km per day (60 to 250 gal/(in diameter/mile/d). Nevertheless, recently, some water companies are not allowing for any leakage in a new main due to the use of better sealers.

4 REGULATION FOR WATER QUALITY 4.1 Federal Drinking Water Standards Water Quality is a term used to describe the physical, chemical, and biological characteristics of water with respect to its suitability for a particular use. National Primary Drinking Water Regulations (NPDWRs or primary standards) are legally enforceable standards that apply to public water systems. The NPDWRs rotect drinking water quality by limiting the levels of specific contaminants that can adversely affect public health and are known or are anticipated to occur in public water systems. The NPDWRs, as shown in Table 6.1, divides these contaminants into inorganic chemicals, organic chemicals, radionuclides, and microorganisms (USEPA, 2000). All the primary standards are based on health effects to the customers; and they are mandatory standards. The USEPA (2000) also promulgated secondary drinking water regulations as shown in Table 6.2. The secondary drinking water regulations pertain to those contaminants such as taste, odor, color, and some chemicals that may adversely affect the aesthetic quality of drinking water. They are intended as guidelines for the states. The states may establish higher or lower values as appropriate to their particular circ*mstances, which are adequate to protect public health and welfare.

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.325

TABLE 6.1 National Primary Drinking Water Regulations MCLG1 (mg/L)2

MCL or TT1 (mg/L)2

Potential health effects from ingestion of water

Contaminant

Sources of contaminant in drinking water

Microorganisms 3

Cryptosporidium

Giardia lamblia

TT3

Heterotrophic plate count

n/a

TT3

Legionella

TT3

Total Coliforms (including fecal coliform and Escherichia Coli )5 Turbidity

5.0%4

n/a

TT3

Viruses (enteric)

TT3

Gastrointestinal illness (e.g., diarrhea, vomiting, cramps) Gastrointestinal illness (e.g., diarrhea, vomiting, cramps) HPC has no health effects; it is an analytic method used to measure the variety of bacteria that are common in water. Lower the concentration of bacteria in drinking water, better maintained is the water system. Legionnaire’s disease, a type of pneumonia Not a health threat in itself; it is used to indicate whether other potentially harmful bacteria may be present5

Human and fecal animal waste Human and animal fecal waste HPC measures a range of bacteria that are naturally present in the environment

Turbidity is a measure of the cloudiness of water. It is used to indicate water quality and filtration effectiveness (e.g., whether disease-causing organisms are present). Higher turbidity levels are often associated with higher levels of disease-causing microorganisms such as viruses, parasites, and some bacteria. These organisms can cause symptoms such as nausea, cramps, diarrhea, and associated headaches. Gastrointestinal illness (e.g., diarrhea, vomiting, cramps)

Found naturally in water; multiplies in heating systems Coliforms are naturally present in the environment, as well as feces, fecal coliforms and E. coli only come from human and animal fecal waste. Soil runoff

Human and animal fecal waste

Disinfection Byproducts Bromate Chlorite

0 0.8

0.010 1.0

Haloacetic acids (HAA5) Total Trihalomethanes (TTHMs)

n/a6

0.060

none7 n/a6

0.10 0.080

MRDL1 (mg/L)2

Increased risk of cancer Anemia; infants & young children: nervous system effects Increased risk of cancer Liver, kidney, or central nervous system problems; increased risk of cancer

MRDL1 (mg/L)2

Potential health effects from ingestion of water

Contaminant Chloramines (as Cl2) Chlorine (as Cl2) Chlorine dioxide (as ClO2)

Byproduct of drinking water disinfection Byproduct of drinking water disinfection Byproduct of drinking water disinfection Byproduct of drinking water disinfection Sources of contaminant in drinking water

Disinfectants MRDLG 4

MRDL 4.0

MRDLG 4.01 MRDL 4.01 MRDLG 0.81 MRDL 0.81

Eye/nose irritation; stomach discomfort, anemia Eye/nose irritation; stomach discomfort Anemia; infants & young children: nervous system effects

Water additive used to control microbes Water additive used to control microbes Water additive used to control microbes

(Continued)

PUBLIC WATER SUPPLY 1.326

CHAPTER 1.6

TABLE 6.1 National Primary Drinking Water Regulations (Continued) MCLG1 (mg/L)2

MCL or TT1 (mg/L)2

Contaminant

Potential health effects from ingestion of water

Sources of contaminant in drinking water

Inorganic Chemicals

Antimony

0.006

Increase in blood cholesterol; decrease in blood sugar

Arsenic

0.010 as of 01/23/06

Asbestos (fiber 10 m)

7 MFL

Barium

7 million fibers per liter 2

Skin damage or problems with circulatory systems, and may have increased risk of getting cancer Increased risk of developing benign intestinal polyps

Increase in blood pressure

Beryllium

0.004

Intestinal lesions

Cadmium

0.005

Kidney damage

Chromium (total) Copper

0.1

Allergic dermatitis

1.3

TT8; Action Level 1.3

Cyanide (as free cyanide) Fluoride

0.2

Short term exposure: gastrointestinal distress Long term exposure: liver or kidney damage People with Wilson’s Disease should consult their personal doctor if the amount of copper in their water exceeds the action level Nerve damage or thyroid problems

4.0

Lead

Mercury (inorganic)

0.002

Infants and children: delays in physical TT8; Action Level 0.015 or mental development; children could show slight deficits in attention span and learning abilities Adults: kidney problems; high blood pressure 0.002 Kidney damage

Bone disease (pain and tenderness of the bones); Children may get mottled teeth

Discharge from petroleum refineries; fire retardants; ceramics; electronics; solder Erosion of natural deposits; runoff from orchards, runoff from glass & electronicsproduction wastes Decay of asbestos cement in water mains; erosion of natural deposits

Discharge of drilling wastes; discharge from metal refineries; erosion of natural deposits Discharge from metal refineries and coal-burning factories; discharge from electrical, aerospace and defense industries Corrosion of galvanized pipes; erosion of natural deposits; discharge from metal refineries; runoff from waste batteries and paints Discharge from steel and pulp mills; erosion of natural deposits Corrosion of household plumbing systems; erosion of natural deposits

Discharge from steel/metal factories; discharge from plastic and fertilizer factories Water additive which promotes strong teeth; erosion of natural deposits; discharge from fertilizer and aluminum factories Corrosion of household plumbing systems; erosion of natural deposits

Erosion of natural deposits; discharge from refineries and factories; runoff from landfills and croplands

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.327

TABLE 6.1 National Primary Drinking Water Regulations (Continued) MCLG1 (mg/L)2

MCL or TT1 (mg/L)2

Contaminant

Potential health effects from ingestion of water

Sources of contaminant in drinking water

Inorganic Chemicals

Nitrate (measured as nitrogen)

Nitrite (measured as nitrogen)

Selenium

0.05

Thallium

0.0005

0.002

Infants below the age of 6 months who drink water containing nitrate in excess of the MCL could become seriously ill and, if untreated, may die. Symptoms include shortness of breath and blue-baby syndrome. Infants below the age of 6 months who drink water containing nitrite in excess of the MCL could become seriously ill and, if untreated, may die. Symptoms include shortness of breath and blue-baby syndrome. Hair or fingernail loss; numbness in fingers or toes; circulatory problems Hair loss; changes in blood; kidney, intestine, or liver problems

Runoff from fertilizer use; leaching from septic tanks, sewage; erosion of natural deposits

Discharge from petroleum refineries; erosion of natural deposits; discharge from mines Leaching from ore-processing sites; discharge from electronics, glass, and drug factories

Organic Chemicals Acrylamide

TT9

Alachlor

0.002

Atrazine

0.003

Benzene

0.005

Benzo(a)pyrene (PAHs) Carbofuran

0.0002

0.04

Carbon tetrachloride Chlordane

0.005

0.002

Chlorobenzene

0.1

Liver or nervous system problems; increased risk of cancer Liver or kidney problems

2,4-D

0.07

Kidney, liver, or adrenal gland problems

Dalapon

0.2

Minor kidney changes

0.0002

Reproductive difficulties; increased risk of cancer

0.6

Liver, kidney, or circulatory system problems

1,2-Dibromo-30 chloropropane (DBCP) o-Dichlorobenzene 0.6

Nervous system or blood problems; increased risk of cancer Eye, liver, kidney or spleen problems; anemia; increased risk of cancer Cardiovascular system or reproductive problems Anemia; decrease in blood platelets; increased risk of cancer Reproductive difficulties; increased risk of cancer Problems with blood, nervous system, or reproductive system Liver problems; increased risk of cancer

Added to water during sewage/ waste water treatment Runoff from herbicide used on row crops Runoff from herbicide used on row crops Discharge from factories; leaching from gas storage tanks and landfills Leaching from linings of water storage tanks and distribution lines Leaching of soil fumigant used on rice and alfalfa Discharge from chemical plants and other industrial activities Residue of banned termiticide Discharge from chemical and agricultural chemical factories Runoff from herbicide used on row crops Runoff from herbicide used on rights of way Runoff/leaching from soil fumigant used on soybeans, cotton, pineapples, and orchards Discharge from industrial chemical factories

(Continued)

PUBLIC WATER SUPPLY 1.328

CHAPTER 1.6

TABLE 6.1 National Primary Drinking Water Regulations (Continued) MCLG1 MCL or TT1 (mg/L)2 (mg/L)2 Contaminant

Potential health effects from ingestion of water

Sources of contaminant in drinking water

Organic Chemicals

p-Dichlorobenzene

0.075

1,2-Dichloroethane

0.005

Anemia; liver, kidney, or spleen damage; changes in blood Increased risk of cancer

1,1-Dichloroethylene

0.007

Liver problems

cis-1,2-Dichloroethylene

0.07

Liver problems

trans-1,2Dichloroethylene Dichloromethane

0.1

Liver problems

0.005

1,2-Dichloropropane

0.005

Liver problems; increased risk of cancer Increased risk of cancer

Di(2-ethylhexyl) adipate

0.4

Weight loss, liver problems, or possible reproductive difficulties. Reproductive difficulties; liver problems; increased risk of cancer Reproductive difficulties

Di(2-ethylhexyl) phthalat 0

0.006

Dinoseb

0.007

Dioxin (2,3,7,8-TCDD)

0.00000003 Reproductive difficulties; increased risk of cancer

Diquat Endothall Endrin Epichlorohydrin

0.02 0.1 0.002 0

0.02 0.1 0.002 TT9

Cataracts Stomach and intestinal problems Liver problems Increased cancer risk; and over a long period of time, stomach problems

Ethylbenzene

0.7

Liver or kidneys problems

Ethylene dibromide

0.00005

Glyphosate

0.7

Heptachlor Heptachlor epoxide Hexachlorobenzene

0 0

0.0004 0.0002

Problems with liver, stomach, reproductive system, or kidneys; increased risk of cancer Kidney problems; reproductive difficulties Liver damage; increased risk of cancer Liver damage; increased risk of cancer

0.001

Liver or kidney problems; reproductive difficulties; increased risk of cancer

Hexachlorocyclopentadiene Lindane

0.05

Kidney or stomach problems

0.0002

Liver or kidney problems

Discharge from industrial chemical factories Discharge from industrial chemical factories Discharge from industrial chemical factories Discharge from industrial chemical factories Discharge from industrial chemical factories Discharge from drug and chemical factories Discharge from industrial chemical factories Discharge from chemical factories Discharge from rubber and chemical factories Runoff from herbicide used on soybeans and vegetables Emissions from waste incineration and other combustion; discharge from chemical factories Runoff from herbicide use Runoff from herbicide use Residue of banned insecticide Discharge from industrial chemical factories; an impurity of some water treatment chemicals Discharge from petroleum refineries Discharge from petroleum refineries Runoff from herbicide use Residue of banned termiticide Breakdown of heptachlor Discharge from metal refineries and agricultural chemical factories Discharge from chemical factories Runoff/leaching from insec ticide used on cattle, lumber, gardens

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.329

TABLE 6.1 National Primary Drinking Water Regulations (Continued) MCLG1 (mg/L)2

MCL or TT1 (mg/L)2

Contaminant

Potential health effects from ingestion of water

Sources of contaminant in drinking water

Organic Chemicals

Methoxychlor

0.04

Reproductive difficulties

Oxamyl (Vydate)

0.2

Slight nervous system effects

Polychlorinated biphenyls (PCBs)

0.0005

Pentachlorophenol Picloram Simazine Styrene

0.001

0.5 0.004 0.1

Tetrachloroethylene Toluene

0.005

Toxaphene

0.003

Skin changes; thymus gland problems; immune deficiencies; reproductive or nervous system difficulties; increased risk of cancer Liver or kidney problems; increased cancer risk Liver problems Problems with blood Liver, kidney, or circulatory system problems Liver problems; increased risk of cancer Nervous system, kidney, or liver problems Kidney, liver, or thyroid problems; increased risk of cancer Liver problems Changes in adrenal glands

2,4,5-TP (Silvex) 0.05 1,2,4-Trichlorobenzene 0.07

0.05 0.07

1,1,1-Trichloroethane

0.20

0.2

1,1,2-Trichloroethane

0.003

0.005

Trichloroethylene

0.005

Vinyl chloride

0.002

Liver, nervous system, or circulatory problems Liver, kidney, or immune system problems Liver problems; increased risk of cancer Increased risk of cancer

Xylenes (total)

Nervous system damage

Runoff/leaching from insecticide used on fruits, vegetables, alfalfa, livestock Runoff/leaching from insecticide used on apples, potatoes, and tomatoes Runoff from landfills; discharge of waste chemicals

Discharge from wood preserving factories Herbicide runoff Herbicide runoff Discharge from rubber and plastic factories; leaching from landfills Discharge from factories and dry cleaners Discharge from petroleum factories Runoff/leaching from insecticide used on cotton and cattle Residue of banned herbicide Discharge from textile finishing factories Discharge from metal degreasing sites and other factories Discharge from industrial chemical factories Discharge from metal degreasing sites and other factories Leaching from PVC pipes; discharge from plastic factories Discharge from petroleum factories; discharge from chemical factories

Radionuclides Alpha particles

none7 0

Beta particles and photon emitters

none7 0

15 picocuries per liter (pCi/L) 4 millirems per year

Increased risk of cancer

Erosion of natural deposits of certain minerals that are radioactive and may emit a form of radiation known as alpha radiation Decay of natural and man-made deposits of certain minerals that are radioactive and may emit forms of radiation known as photons and beta radiation

(Continued)

PUBLIC WATER SUPPLY 1.330

CHAPTER 1.6

TABLE 6.1 National Primary Dringking Water Regulations (Concluded) MCLG1 (mg/L)2

MCL or TT1 (mg/L)2

Potential health effects from ingestion of water

Contaminant Radium 226 and Radium 228 (combined) Uranium

Sources of contaminant in drinking water

Radionuclides none7 0 0

5 pCi/L

Increased risk of cancer

Erosion of natural deposits

30 g/L as of 12/08/03

Increased risk of cancer, kidney toxicity

Erosion of natural deposits

Notes: 1 Definitions: Maximum contaminant level (MCL): The highest level of a contaminant that is allowed in drinking water. MCLs are set as close to MCLGs as feasible using the best available treatment technology and taking cost into consideration. MCLs are enforceable standards. Maximum contaminant level goal (MCLG): The level of a contaminant in drinking water below which there is no known or expected risk to health. MCLGs allow for a margin of safety and are non-enforceable public health goals. Maximum residual disinfectant level (MRDL): The highest level of a disinfectant allowed in drinking water. There is convincing evidence that addition of a disinfectant is necessary for control of microbial contaminants. Maximum residual disinfectant level goal (MRDLG): The level of a drinking water disinfectant below which there is no known or expected risk to health. MRDLGs do not reflect the benefits of the use of disinfectants to control microbial contaminants. Treatment technique: A required process intended to reduce the level of a contaminant in drinking water. 2 Units are in milligrams per liter (mg/L) unless otherwise noted. Milligrams per liter are equivalent to parts per million. 3 EPA’s surface water treatment rules require systems using surface water or ground water under the direct influence of surface water to (1) disinfect their water and (2) filter their water or meet criteria for avoiding filtration so that the following contaminants are controlled at the following levels: Cryptosporidium: (as of 1/1/02 for systems serving 10,000 and 1/14/05 for systems serving 10,000) 99 percent removal. Giardia lamblia: 99.9 percent removal/inactivation Viruses: 99.99% removal/inactivation Legionella: No limit, but EPA believes that if Giardia and viruses are removed/inactivated, Legionella will also be controlled. Turbidity: At no time can turbidity (cloudiness of water) go above 5 nephelolometric turbidity units (NTU); systems that filter must ensure that the turbidity go no higher than 1 NTU (0.5 NTU for conventional or direct filtration) in at least 95percent of the daily samples in any month. As of January 1, 2002, turbidity may never exceed 1 NTU, and must not exceed 0.3 NTU in 95 percent of daily samples in any month. • HPC: No more than 500 bacterial colonies per milliliter. • Long Term 1 enhanced surface water treatment (Effective Date: January 14, 2005); Surface water systems or (GWUDI) systems serving fewer than 10,000 people must comply with the applicable Long Term 1 Enhanced Surface Water Treatment Rule provisions (e.g. turbidity standards, individual filter monitoring, Cryptosporidium removal requirements, updated watershed control requirements for unfiltered systems). • Filter Backwash Recycling; The Filter Backwash Recycling Rule requires systems that recycle to return specific recycle flows through all processes of the system’s existing conventional or direct filtration system or at an alternate location approved by the state. 4 more than 5.0 percent samples total coliform-positive in a month. (For water systems that collect fewer than 40 routine samples per month, no more than one sample can be total coliform-positive per month.) Every sample that has total coliform must be analyzed for either fecal coliforms or E. coli if two consecutive TC-positive samples, and one is also positive for E. coli fecal coliforms, system has an acute MCL violation. 5 Fecal coliform and E. coli are bacteria whose presence indicates that the water may be contaminated with human or animal wastes. Disease-causing microbes (pathogens) in these wastes can cause diarrhea, cramps, nausea, headaches, or other symptoms. These pathogens may pose a special health risk for infants, young children, and people with severely compromised immune systems. 6 Although there is no collective MCLG for this contaminant group, there are individual MCLGs for some of the individual contaminants:

• • • • •

• •

Trihalomethanes: bromodichloromethane (zero); bromoform (zero); dibromochloromethane (0.06 mg/L). Chloroform is regulated with this group but has no MCLG. Haloacetic acids: dichloroacetic acid (zero); trichloroacetic acid (0.3 mg/L). Monochloroacetic acid, bromoacetic acid, and dibromoacetic acid are regulated with this group but have no MCLGs.

MCLGs were not established before the 1986 amendments to the Safe Drinking Water Act. Therefore, there is no MCLG for this contaminant. Lead and copper are regulated by a treatment technique that requires systems to control the corrosiveness of their water. If more than 10 percent of tap water samples exceed the action level, water systems must take additional steps. For copper, the action level is 1.3 mg/L and for lead is 0.015 mg/L. 9 Each water system must certify, in writing, to the state (using third-party or manufacturer’s certification) that when acrylamide and epichlorohydrin are used in drinking water systems, the combination (or product) of dose and monomer level does not exceed the levels specified, as follows: 8

• •

Acrylamide 0.05 percent dosed at 1 mg/L (or equivalent) Epichlorohydrin 0.01 percent dosed at 20 mg/L (or equivalent)

Source: www.epa.gov/OGWDW/regs.html, 2007 (modified)

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.331

TABLE 6.2 National Secondary Drinking Water Regulations Contaminants

Concentration

Aluminum Chloride Color Copper Corrosivity Fluoride Foaming Agents Iron Manganese Odor pH Silver Sulfate Total dissolved solids Zinc

0.05 to 0.2 mg/L 250 mg/L 15 (color units) 1.0 mg/L Non-corrosive 2.0 mg/L 0.5 mg/L 0.3 mg/L 0.05 mg/L 3 threshold odor number 6.5–8.5 0.10 mg/L 250 mg/L 500 mg/L 5 mg/L

Note: National Secondary Drinking Water Regulations (NSDWRs or secondary standards) are non-enforceable guidelines regulating contaminants that may cause cosmetic effects (such as skin or tooth discoloration) or aesthetic effects (such as taste, odor, or color) in drinking water. EPA recommends secondary standards to water systems but does not require systems to comply. However, states may choose to adopt them as enforceable standards. Source: www.epa.gov/OGWDW/regs.html, 2007

The design engineer should be thoroughly familiar with and integrate the various regulatory requirements into a water treatment design. It is necessary to obtain a permit from the state regulatory agency. Meetings with the representatives of the necessary regulatory agencies and related parties are recommended. The state reviews the design based on meeting minimum design requirements or standards, local conditions, and good engineering practice and experience. The most widely used standards are the Recommended Standards for Water Works promulgated by the Great Lakes–Upper Mississippi River Board of State Sanitary Engineers (GLUMRB) which are the so-called Ten State Standards. This Board includes representatives from the states of Illinois, Indiana, Iowa, Michigan, Minnesota, Missouri, New York, Ohio, Pennsylvania, and Wisconsin. Recently, the Province of Ontario (Canada) has joined the Board. Many other states have used identical or similar versions of the Ten State Standards. Maximum contaminant level goal (MCLG) for water quality is set by the US Environmental Protection Agency (EPA). For agents in drinking water not considered to have carcinogenic potential, i.e. no effect levels for chromic and/or lifetime period of exposure including a margin of safety, are commonly referred to as reference dosages (RfDs). Reference dosages are the exposure levels estimated to be without significant risk to humans when received daily over a lifetime. They are traditionally reported as mg per (kg d) or mg/(kg d). For MCLG purposes, however, no effect level in drinking water is measured by mg/L which has been in terms of the drinking water equivalent level (DWEL). The DWEL is calculated as (Cotruvo and Vogt, 1990) DWEL

(NOAEL)(70) (UF)(2)

where NOAEL no observed adverse effect level, mg/(kg d) (70) assumed weight of an adult, kg UF uncertainty factor (usually 10, 100, or 1000) (2) assumed quantity of water consumed by an adult, L/d

(6.13)

PUBLIC WATER SUPPLY 1.332

CHAPTER 1.6

When sufficient data is available on the relative contribution from other sources, i.e. from food and air, the MCLG can be determined as follows: MCLG RfD contribution from (food air) In fact, comprehensive data on the contributions from food and air are generally lacking. Therefore, in this case, MCLG is determined as MCLG DWEL. The drinking water contribution often used in the absence of specific data for food and air is 20 percent of RfD. This effectively provides an additional safety factor of 5. EXAMPLE: The MCLG for nitrate-nitrogen is 10 mg/L. Assume UF 100. What is the “no observed adverse effect level for the drinking water?’’

Solution: DWEL since

(NOAEL) 70 (UF) 2

DWEL MCLG 10 mg/L (NOAEL, mg/(kg # d)) (70 kg) 10 mg/L 100 2 L/d 2000 70 28.6 (mg/(kg d))

NOAEL

4.2 Atrazine Atrazine is a widely used organic herbicide due to its being inexpensive compared to other chemicals. Often, excessive applications are used. When soil and climatic conditions are favorable, atrazine may be transported to the drinking water sources by runoff or by leaching into ground water. Atrazine has been shown to affect offspring of rats and the hearts of dogs. The USEPA has set the drinking water standard for atrazine concentration to 0.003 mg/L (or 3 g/L) to protect the public against the risk of the adverse health effects. When atrazine or any other herbicide or pesticide is detected in surfacewater or groundwater, it is not meant as a human health risk. Treated (finished) water meets the EPA limits with little or no health risk and is considered safe with respect to atrazine. EXAMPLE: The lifetime health advisory level of 0.003 g/L for atrazine is set at 1000 times below the amount that causes no adverse health effects in laboratory animals. The no observed adverse effect level is 0.086 mg/(kg d). How is the 0.003 mg/L of atrazine in water derived?

Solution: As previous Example 1, Using Eq. (6.13) (NOAEL)(70) (UF)(2) (0.086 mg/(kg # d))(70 kg) (1000)(2 L/d)

DWEL

0.003 mg/L

5 WATER TREATMENT PROCESSES As the raw surface water comes to the treatment plant, physical screening is the first step to remove coarse material and debris. Thereafter, following the basic treatment process of clarification, it would

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.333

include coagulation, flocculation, and sedimentation prior to filtration, then disinfection (mostly by the use of chlorination). With a good quality source, the conventional treatment processes may be modified by removing the sedimentation process and to just have the coagulation and flocculation processes followed by filtration. This treatment process scheme is called direct filtration. Groundwater is generally better quality; however, it is typically associated with high hardness, iron, and manganese content. Aeration or air stripping is required to remove volatile compounds in groundwater. Lime softening is also necessary to remove the impurities and recarbonation is used to neutralize excess lime and to lower the pH value. Ion exchange processes can also be employed for the softening of water and the removal of other impurities, if low in iron, manganese, particulates, and organics. For a groundwater source high in iron and manganese, but with acceptable hardness, the aeration process and/or chemical oxidation can be followed by filtration to remove these compounds. The reverse osmosis process is used to remove chemical constituents and salts in water. It offers the promise of conversion of salt water to fresh water. Other membrane processes, such as microfiltration, ultrafiltration, nanofiltration, and electrodialysis, are also applied in the water industry.

6 AERATION AND AIR STRIPPING Aeration has been used to remove trace volatile organic compounds (VOCs) in water either from surfacewater or groundwater sources. It has also been employed to transfer a substance, such as oxygen, from air or a gas phase into water in a process called gas adsorption or oxidation (to oxidize iron and/or manganese). Aeration also provides the escape of dissolved gases, such as carbon dioxide and hydrogen sulfide. Air stripping has also been utilized effectively to remove ammonia gas from wastewater and to remove volatile tastes and other such substances in water. The solubility of gases which do not react chemically with the solvent and the partial pressure of the gases at a given temperature can be expressed as Henry’s law: p Hx

(6.14)

where x solubility of a gas in the solution phase H Henry’s constant p partial pressure of a gas over the solution In terms of the partial pressure, Dalton’s law states that the total pressure (Pt) of a mixture of gases is just the sum of the pressures that each gas (Pi) would exert if it were present alone: Pt P1 P2 P3 · · · Pj Since

(6.15)

PV nRT Pna

RT b V

RT (n n2 c nj) V 1 n1Pt P1 n1 n2 c nj Pt

(6.16)

Combining Henry’s law and Dalton’s law, we get Yi

Hi xi Pt

(6.17)

PUBLIC WATER SUPPLY 1.334

CHAPTER 1.6

where Yi mole fraction of ith gas in air xi mole fraction of ith gas in fluid (water) Hi Henry’s constant for ith gas Pt total pressure, atm The greater the Henry’s constant, the more easily a compound can be removed from a solution. Generally, increasing temperature would increase the partial pressure of a component in the gas phase. The Henry’s constant and temperature relationship is (James M. Montgomery Consulting Engineering, 1985; American Society of Civil Engineers and American Water Works Association, 1990): log H

H J RT

(6.18)

where H Henry’s constant H heat absorbed in evaporation of 1 mole of gas from solution at constant temperature and pressure, kcal/kmol R universal gas constant, 1.9872 kcal/kmol K 0.082057 atm L/mol K T temperature, Kelvin J empirical constant Values of H, H, and J of some gases are given in Table 6.3. At 20C the partial pressure (saturated) of chloroform CHCl3 is 18 mm of mercury in a storage tank. Determine the equilibrium concentration of chloroform in water assuming that gas and liquid phases are ideal.

EXAMPLE 1:

Solution: Step 1. Determine mole fraction (x) of CHCl3 From Table 6.3, at 20C and a total pressure of 1 atm, the Henry constant for chloroform is H 170 atm TABLE 6.3 Henry’s Law Constant and Temperature Correction Factors Gas Ammonia Benzene Bromoform Carbon dioxide Carbon tetrachloride Chlorine Chlorine dioxide Chloroform Hydrogen sulfide Methane Nitrogen Oxygen Ozone Sulfur dioxide Trichloroethylene Vinyl chloride

Henry’s constant at 20C, atm

h, 103 kcal/kmol

0.76 240 35 1.51 102 1.29 103 585 54 170 515 3.8 104 8.6 104 4.3 104 5.0 103 38 550 1.21 103

3.75 3.68 – 2.07 4.05 1.74 2.93 4.00 1.85 1.54 1.12 1.45 2.52 2.40 3.41 –

6.31 8.68 – 6.73 10.06 5.75 6.76 9.10 5.88 7.22 6.85 7.11 8.05 5.68 8.59 –

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.335

Partial pressure of CHCl3, p, is p

18 mm 1 atm 0.024 atm 760 mm

Using Henry’s law p Hx x p/H 0.024 atm/170 atm 1.41 104 Step 2. Convert mole fraction to mass concentration. In 1 L of water, there are 1000/18 mol/L 55.6 mol/L. Thus n x n nw where n number of mole for CHCl3 nw number of mole for water Then nx nw x n n

nw x 55.6 1.41 104 1x 1 1.41 104

7.84 103 mol/L Molecular weight of CHCl3 12 1 3 35.45 119.4 Concentration (C) of CHCl3 is C 119.4 g/mol 7.84 103 mol/L 0.94 g/L 940 mg/L Note: In practice, the treated water is open to the atmosphere. Therefore the partial pressure of chloroform in the air is negligible. Therefore the concentration of chloroform in water is near zero or very low. EXAMPLE 2: The finished water in an enclosed clear well has dissolved oxygen of 6.0 mg/L. Assume it is in equilibrium. Determine the concentration of oxygen in the air space of the clear well at 17C.

Solution: Step 1. Calculate Henry’s constant at 17C (290 K) From Table 6.3 and Eq. (6.18) log H

H J RT 1.45 103 kcal/kmol 7.11 (1.987 kcal/kmol # K)(290 K)

4.59 103.3 kPa 1 atm 4.02 106 kPa

H 38,905 atm

PUBLIC WATER SUPPLY 1.336

CHAPTER 1.6

Step 2. Compute the mole fraction of oxygen in the gas phase Oxygen (MW 32) as gas i in water Ci 6 mg/L 103 g/mg 103 L/m3 32 g/mol 0.1875 mol/m3 At 1 atm for water, the molar density of water is Cw 55.6 kmol/m3 xi

Ci 0.1875 mol/m3 Cw 55.6 103 mol/m3

3.37 106 Pt 1 atm From Eq. (6.17) yi

Hixi 38,905 atm 3.37 106 Pt 1 atm

0.131 Step 3. Calculate partial pressure Pi Pi yi Pt 0.131 1 0.131 atm Step 4. Calculate oxygen concentration in gas phase Cg ni /V Using the ideal gas formula Pi V ni RT ni Pi 0.131 atm V RT 0.08206 atm # L/mol # K (290 K) 5.50 10–3 mol/L 5.50 10–3 mol/L 32 g/mol 0.176 g/L 176 mg/L 6.1 Gas Transfer Models Three gas-liquid mass transfer models are generally used. They are two-film theory, penetration theory, and surface renewal theory. The first theory is discussed here. The other two can be found elsewhere (Schroeder, 1977). The two-film theory is the oldest and simplest. The concept of the two-film model is illustrated in Fig. 6.1. Flux is a term used as the mass transfer per time through a specified area. It is a function of the driving force for diffusion. The driving force in air is the difference between the bulk concentration and the interface concentration. The flux gas through the gas film must be the same as the flux through the liquid film. The flux relationship for each phase can be expressed as: F

dW kg(Pg Pi) kl(Ci Cl) dtA

(6.19)

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.337

where F flux W mass transfer A a given area gas-liquid transferal t time kg local interface transfer coefficient for gas Pg concentration of gas in the bulk of the air phase Pi concentration of gas in the interface kl interface transfer coefficient for liquid Ci local interface concentration at equilibrium Cl liquid-phase concentration in bulk liquid Since Pi and Ci are not measurable, volumetric mass-transfer coefficients are used which combine the mass-transfer coefficients with the interfacial area per unit volume of system. Applying Henry’s law and introducing P* and C* which correspond to the equilibrium concentration that would be associated with the bulk-gas partial pressure Pg and bulk-liquid concentration Cl, respectively, we can obtain F KG (Pg P*) KL (C* C1)

(6.20)

Pg Bulk gas

Bulk liquid Pi Ci Cl

Gas film

Liquid film

(a) Cl Bulk liquid

Bulk air Ci Pi Pg

Liquid film

Air film

(b) FIGURE 6.1

Schematic diagram of two-film theory for (a) gas absorption; (b) gas stripping.

PUBLIC WATER SUPPLY 1.338

CHAPTER 1.6

Mole fraction of gas in air

Pg s2 Pi s1 P* Cl

FIGURE 6.2

Ci Concentration

Concentration differences of gas in air and water phases.

The concentration differences of gas in air and water is plotted in Fig. 6.2. From this figure and Eq. (6.19), we obtain: Pg P* (Pg Pi) (Pi P*) Pg Pi sl (Ci Cl)

s1F F kg kl

From Eq. (6.20): F Pg P* KG Thus s1F F F KG kg kl s1 1 1 KG kg kl Similarly, we can obtain 1 1 1 KL s2kg kl In dilute conditions, Henry’s law holds, the equilibrium distribution curve would be a straight line and the slope is the Henry constant: s1 s2 H Finally the overall mass-transfer coefficient in the air phase can be determined with 1 1 H KG kl kl

(6.21)

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.339

For liquid-phase mass transfer 1 H 1 KL k1 Hkg

(6.22)

where 1/k1 and 1/kg are referred to as the liquid-film resistance and the gas-film resistance, respectively. For oxygen transfer, 1/kl is considerably larger than 1/kg, and the liquid film usually controls the oxygen transfer rate across the interface. The flux equation for the liquid phase using concentration of mg/L is F

dc V KL (Cs Ct) dt A

(6.23a)

mass transfer rate: dc A KL (Cs Ct) V dt dc KLa (C* Ct) dt

(6.23b) (6.23c)

where KL overall mass-transfer coefficient, cm/h A interfacial area of transfer, cm2 V volume containing the interfacial area, cm3 Ct concentration in bulk liquid at time t, mg/L C* equilibrium concentration with gas at time t as Pt HCs mg/L a the specific interfacial area per unit system volume, cm1 KLa overall volumetric mass-transfer coefficient in liquid, h1 or [g mol/(h cm3 atm)] the volumetric rate of mass transfer M is A N KLa (C* C) V KG a(P P*)

(6.24) (6.25)

where N rate at which solute gas is transferred between phases, g mol/h EXAMPLE 1: An aeration tank has the volume of 200 m3 (7060 ft3) with 4 m (13 ft) depth. The experiment is conducted at 20C for tap water. The oxygen transfer rate is 18.7 kg/h (41.2 lb/h). Determine the overall volumetric mass-transfer coefficient in the liquid phase.

Solution: Step 1. Determine the oxygen solubility at 20C and 1 atm Since oxygen in air is 21%, according to Dalton’s law of partial pressure, P is P 0.21 1 atm 0.21 atm From Table 6.3, we find H 4.3 104 atm The mole fraction x from Eq. (6.14) is x

0.21 P 4.88 106 H 4.3 104

In 1 L of solution, the number of moles of water nw is nw

1000 55.6 g mol/L 18

PUBLIC WATER SUPPLY 1.340

CHAPTER 1.6

Let n moles of gas in solution x

n n nw

since n is so small n nw x 55.6 g mol/L 4.88 106 2.71 104 g mol/L Step 2. Determine the mean hydrostatic pressure on the rising air bubbles. Pressure at the bottom is pb 1 atm 10.345 m of water head pb 1 atm 4 m

1 atm (1 0.386) atm 10.345 m

1.386 atm Pressure at the surface of aeration tank is ps 1 atm Mean pressure p is p

ps pb 1 atm 1.386 atm 1.193 atm 2 2

Step 3. Compute C* C C* 2.71 104 g mol/L atm 1.193 atm 3.233 104 g mol/L The percentage of oxygen in the air bubbles decreases, when the bubbles rise up through the solution, assuming that the air that leaves the water contains only 19 percent of oxygen. Thus 19 C* C 3.233 104 g # mol/L 21 2.925 104 g mol/L Step 4. Compute KL a N 18,700 g/h 32/mol 584 g mol/h Using Eq. (6.24)

KLa

N V(C* C) 584 g # mol/h 200 m 1000 L/m3 2.925 104 g # mol/L 3

9.98 h1 Equation (6.23c) is essentially the same as Fick’s first law: dC KLa(Cs C) dt

(6.23d)

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

where

1.341

dC rate of change in concentration of the gas in solution, mg/L s dt KLa overall mass transfer coeffcient, s1 Cs saturation concentration of gas in solution, mg/L C concentration of solute gas in solution, mg/L

The value of Cs can be calculated with Henry’s law. The term Cs C is a concentration gradient. Rearrange Eq. (6.23d) and integrating the differential form between time from 0 to t, and concentration of gas from C0 to Ct in mg/L, we obtain: Ct

Cs Ct KL at Cs C

(6.24a)

Cs Ct eKLat Cs C

(6.24b)

dC 3 C C KLa 3 dt s

when gases are removed or stripped from the solution, Eq. (6.24b) becomes: C0 Cs e(KLat) Ct Cs

(6.24c)

Similarly, Eq. (6.25) can be integrated to yield ln

pt p* KGaRTt p0 p*

(6.25a)

KLa values are usually determined in full-scale facilities or scaled up from pilot-scale facilities. Temperature and chemical constituents in wastewater affect oxygen transfer. The temperature effects on overall mass-transfer coefficient KLa are treated in the same manner as they were treated in the BOD rate coefficient (Section 8.3 of Chap. 1.2). It can be written as KLa(T ) KLa(20)uT20

(6.26)

where KLa(T) overall mass-transfer coefficient at temperature TC, s1 KLa(20) overall mass-transfer coefficient at temperature 20C, s1 Values of range from 1.015 to 1.040, with 1.024 commonly used. mass-transfer coefficient are influenced by total dissolved solids in the liquid. Therefore, a correction factor is applied for wastewater (Tchobanoglous and Schroeder, 1985): a

KLa (wastewater) KLa (tap water)

(6.27)

Values of range from 0.3 to 1.2. Typical values for diffused and mechanical aeration equipment are in the range of 0.4 to 0.8 and 0.6 to 1.2. The third correction factor () for oxygen solubility is due to particulate, salt, and surface-active substances in water (Doyle and Boyle, 1986): b

C*(wastewater) C*(tap water)

(6.28)

PUBLIC WATER SUPPLY 1.342

CHAPTER 1.6

Values of range from 0.7 to 0.98, with 0.95 commonly used for wastewater. Combining all three correction factors, we obtain (Tchobanoglous and Schroeder 1985): AOTR SOTR(a)(b)(uT20)a

Cs Cw b Cs20

(6.29)

where AOTR actual oxygen transfer rate under field operating conditions in a respiring system, kg O2/kW h SOTR standard oxygen transfer rate under test conditions at 20C and zero dissolved oxygen, kg O2/kW h , , defined previously Cs oxygen saturation concentration for tap water at field operating conditions, g/m3 Cw operating oxygen concentration in wastewater, g/m3 Cs20 oxygen saturation concentration for tap water at 20C, g/m3 EXAMPLE 2: Aeration tests are conducted with tap water and wastewater at 16C in the same container. The results of the tests are listed below. Assume the saturation DO concentrations (Cs) for tap water and wastewater are the same. Determine the values of KLa for water and wastewater and values at 20C. Assume 1.024

DO concentration, mg/L

Cs Ct Cs C0

Contact time, min

Tap water

Wastewater

Tap water

Wastewater

0 20 40 60 80 100 120

0.0 3.0 4.7 6.4 7.2 7.9 8.5

0.0 2.1 3.5 4.7 5.6 6.4 7.1

0 0.36 0.65 1.05 1.32 1.69 2.01

0 0.24 0.44 0.65 0.84 1.05 1.28

Solution: Step 1. Find DO saturation concentration at 16C From Table 2.2 Cs 9.82 mg/L Step 2. Calculate ln(Cs Ct)/(Cs C0), C0 0 The values of ln(Cs Ct)/(Cs C0) calculated are listed with the raw test data Step 3. Plot the calculated results in step 2 in Fig. 6.3. Step 4. Find the KL a at the test temperature of 16C, (T) For tap water

KLa

60 min 2.01 120 min 1h

1.00 h1

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.343

-ln (Cs - Ct)/(Cs - Co)

2.5 2

1.5

Tap water Wastewater

0.5 0 0

FIGURE 6.3

60 TIME, min

100

Functional plot of the data from aeration test.

For wastewater: 60 min 1.28 120 min 1h 0.64 h1

KLa

Step 5. Convert the values of KL a at 20C using 1.024 Using Eq. (6.26) For tap water KLa(T ) KLa(20C) 20T KLa(20) KLa(T) 20T

1.00 h1 (1.024)2016 1.10 h1 For wastewater: KLa(20) 0.64 h1(1.024)2016 0.70 h1 Step 6. Compute the value using Eq. (6.27) a

KLa for wastewater KLa for tap water 0.70 h1 1.10 h1

0.64

120

PUBLIC WATER SUPPLY 1.344

CHAPTER 1.6

6.2 Diffused Aeration Diffused aeration systems distribute the gas uniformly through the water or wastewater, in such processes as ozonation, absorption, activated sludge process, THM removal, and river or lake reaeration, etc. It is more costly using diffused aeration for VOC removal than the air stripping column. The two-resistance layer theory is also applied to diffused aeration. The model proposed by Mattee-Müller et al. (1981) is based on mass transfer flux derived from the assumption of diffused bubbles rising in a completely mixed container. The mass transfer rate for diffused aeration is F QGHuCe a1 exp where

KLaV b HuQL

(6.30)

F mass transfer rate QG gas (air) flow rate, m3/s or ft3/s QL flow rate of liquid (water), m3/s or ft3/s Hu unitless Henry’s constant, see section Design of Packed tower Ce effluent (exit) gas concentration, g/L KLa overall mass-transfer coefficient, per time V reaction volume (water), m3 or ft3

Assuming that the liquid volume in the reactor is completely mixed and the air rises as a steady state plug flow, the mass balance equation can be expressed as Ce 1 Ci 1 HuQG/QL[1 exp(KLaV/HuQG)]

(6.31a)

Ce 1 Ci 1 HuQG/QL[1 exp(u)]

(6.31b)

where Ci initial concentration, g/L u

KLaV HuQG

If 1, the transfer of a compound is with very low Henry’s constant such as ammonia. Air bubbles exiting from the top of the liquid surface is saturated with ammonia in the stripping process. Ammonia removal could be further enhanced by increasing the air flow. Until 4, the exponent term becomes essentially zero. When the exponent term is zero, the air and water have reached an equilibrium condition and the driving force has decreased to zero at some point within the reactor vessel. The vessel is not fully used. Thus the air-to-water ratio could be increased to gain more removal. On the other hand, if 1, the mass-transfer efficiency could be improved by increasing overall mass-transfer coefficient by either increasing the mixing intensity in the tank or by using a finer diffuser. In the case for oxygenation, 0.1, the improvements are required. A groundwater treatment plant has a capacity of 0.0438 m3/s (1 MGD) and is aerated with diffused air to remove trichloroethylene with 90 percent design efficiency. The detention time of the tank is 30 min. Evaluate the diffused aeration system with the following given information. EXAMPLE :

T 20C Ci 131 g/L (expected Ce 13.1 g/L) Hu 0.412 KLa 44 h1

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.345

Solution: Step 1. Compute volume of reactor V V 0.0438 m3/s 60 s/min 30 min 79 m3 Step 2. Compute QG, let QG 30 QL 30 V QG 30 79 m3 2370 m3 Step 3. Compute u

KLaV 44 79 3.56 HuQG 0.412 2370

Step 4. Compute effluent concentration Ce with Eq (6.31b). Ce

Ci 1 Hu QG/QL[1 exp(u)] 131 mg/L 1 0.412 30[1 exp(3.56)]

10.0 mg/L % removal

(131 10) 100 92.4 131

This exceeds the expected 90%. 6.3 Packed Towers Recently, the water treatment industry used packed towers for stripping highly volatile chemicals, such as hydrogen sulfide and VOCs from water and wastewater. It consists of a cylindrical shell containing a support plate for the packing material. Although many materials can be used as the packing material, plastic products with various shapes and design are most commonly used due to less weight and lower cost. Packing material can be individually dumped randomly into the cylinder tower or fixed packing. The packed tower or columns are used for mass transfer from the liquid to gas phase. Figure 6.4 illustrates a liquid-gas contacting system with a downward water velocity L containing c1 concentration of gas. The flows are counter current. The air velocity G passes upward through G p2 L

Partial z

c2 FIGURE 6.4

Pressure

p2 p p2

Schematic diagram of packed column.

c2 c c1 Concentration

PUBLIC WATER SUPPLY 1.346

CHAPTER 1.6

the packed material containing influent p1 and effluent p2. There are a variety of mixing patterns, each with a different rate of mass transfer. Removal of an undesirable gas in the liquid phase needs a system height of z and a selected gas flow rate to reduce the mole fraction of dissolved gas from c1 to c2. If there is no chemical reaction that takes place in the packed tower, the gas lost by water should be equal to the gas gained by air. If the gas concentration is very dilute, then Lc Gp

(6.32)

where L liquid velocity, m/s, m3/(m2 s), or mol/(m2 s) G gas velocity, same as above c change of gas concentration in water p change in gas fraction in air From Eq. (6.23d), for stripping (c and cs will be reversed): dc KLa (c cs) dt

(6.23e)

dc KLa(c cs)

(6.33)

then Multiplying each side by L, we obtain

Ldt dz

Ldc KLa(c cs)

(6.34)

where dz is the differential of height. The term (c cs) is the driving force (DF) of the reaction and is constantly changing with the depth of the column due to the change of c with time. Integrating the above equation yields c2

dc L 3 dz KLa 3 d(c c ) s c1

dc L KLa 3 3 d(DF)

(6.35)

c1 DF

The integral of DF is the same as the log mean (DFlm) of the influent (DFi) and effluent driving forces (DFe). Therefore z

L(ci ce) KLaDFlm

(6.36)

DFe DFi ln(DFe /DFi)

(6.37)

and DFlm where

z height of column, m L liquid velocity, m3/m2 h ci, ce gas concentration in water at influent and effluent, respectively, mg/L KLa overall mass-transfer coefficient for liquid, h1 DFi, DFe driving force at influent and effluent, respectively, mg/L DFlm log mean of DFi and DFe, mg/L

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

EXAMPLE 1:

1.347

Given T 20C 293 K L 80 m3 water/(m2 h) G 2400 m3 air/(m2 h) Ci 131 g/L trichloroethylene concentration in water at entrance Ce 13.1 g/L CCHCl3 concentration at exit KLa 44 h1

Determine the packed tower height to remove 90 percent of trichloroethylene by an air stripping tower. Solution: Step 1. Compute the molar fraction of CCHCl3 in air p2 Assuming no CCHCl3 present in the air the entrance, i.e. pi 0 MW of CCHCl3 131, 1 mole 131 g of CCHCl3 per liter. ci 131 g/L 131

g 1 mol/g 103 L 6 L 1 m3 131 10 g/g

1 103 mol/m3 ce 13.1 g/L 0.1 103 mol/m3 (with 90% removal) Applying Eq. (6.32) Lc Gp 80 m3/(m2 h)(1 0.1) 103 mol gas/m3 2400 m3 air/(m2 h)(pe 0) pe 3.0 105 mol gas/m3 air Step 2. Convert pe in terms of mol gas/mol air Let V volume of air per mole of air V

(1 mole)(0.08206 L atm/mol # K) (293 K) nRT P 1 atm

24.0 L 0.024 m3 From Step 1 pe 3.0 105 mol gas/m3 air 3.0 105

mol gas 0.024 m3 air 1 mol air m3 air

7.2 107 mol gas/mol air

PUBLIC WATER SUPPLY 1.348

CHAPTER 1.6

Step 3. Compute driving force, DF, for gas entrance and exit At gas influent (bottom) pi 0 ce 13.1 g/L 0.0131 mg/L cs 0 DFi ce cs 0.0131 mg/L At gas effluent (top) pe 7.2 107 mol gas/mol air ci 131 g/L 0.131 mg/L cs to be determined From Table 6.3, Henry’s constant H at 20C H 550 atm Convert atm to atm L/mg, Hd Hd

550 atm H 55,600 MW 55,600 131 mg/L

7.55 105 atm L/mg cs

pe pt 7.2 107 mol gas/mol air 1 atm Hd 7.55 105 atm # L/mg

0.0095 mg/L DFe ci cs 0.131 mg/L 0.0095 mg/L 0.1215 mg/L Step 4. Compute DFlm From Eq. (6.37) DFlm

DFe DFi 0.1215 mg/L 0.0131 mg/L ln(DFe/DFi) ln(0.1215/0.0131)

0.0487 mg/L Step 5. Compute the height of the tower z From Eq. (6.36) z

L(ci ce ) 80 m/h (0.131 0.0131) mg/L KLaDFlm 44 h1 0.0487 mg/L

4.4 m In a groundwater remediation study, volatile organic carbon removal through the vapor phase, the total hydrocarbons analyzer measured 1,1,1,-trichloroethane (Jones et at., 2000). The extraction pump and air stripper combined air flow averaged 0.40 m3/min (14 ft3/min). The concentration of 1,1,1,-trichloroethane averaged 25 parts per million by volume (ppmv) of air. Determine the amount of 1,1,1,-thrichloroethane removed daily. Temperature is 20C. EXAMPLE 2:

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.349

Solution: Step 1. Calculate 1,1,1,-trichloroethane concentration using the ideal gas law T 20C (20 273)K 293 K n

(1 atm)/1000 L/m3 PV RT 0.082(L # atm/mol # K) 293 K

41.6 mol/m3 This means 41.6 mol of total gases in 1 m3 of the air. The MW of 1,1,1,-trichloroethane (CH3CCl3) 133 One ppmv of 1,1,1,-trichloroethane 41.6 10–6 mol/m3 133 g/mol 1000 mg/g 5.53 mg/m3 25 ppmv of 1,1,1,-trichloroethane 5.53 mg/m3 25 138 mg/m3 Step 1. Calculate the daily removal Daily removal 138 mg/m3 0.40 m3/min 1440 min/d 79,500 mg/d 79.5 g/d Design of packed tower. Process design of packed towers or columns is based on two quantities: the height of the packed column, z, to achieve the designed removal of solute is the product of the height of a transfer unit (HTU) and the number of transfer unit (NTU). It can be expressed as (Treybal, 1968): z (HTU)(NTU)

(6.38)

The HTU refers the rate of mass transfer for the particular packing materials used. The NTU is a measure of the mass transfer driving force and is determined by the difference between actual and equilibrium phase concentrations. The height of a transfer unit is the constant portion of Eq. (6.35): HTU

L KLa

(6.39)

The number of transfer units is the integral portion of Eq. (6.35). For diluted solutions, Henry’s law holds. Substituting the integral expression for NTU with p1 0, the NTU is: NTU where

(c1/c2)(R 1) 1 R ln R R1

HuG L stripping factor, unitless when Hu is unitless c1, c2 mole fraction for gas entrance and exit, respectively R

(6.40) (6.41)

Convert Henry’s constant from terms of atm to unitless: Hu cH

atm(mol gas/mol air) 1 L of water 1 mol air b d a ba mol gas/mol water 0.082 T atm L of air 55.6 mol

H/4.56 T

(6.42)

PUBLIC WATER SUPPLY 1.350

CHAPTER 1.6

When T 20C 293 K Hu H/4.56 293 7.49 104 H, unitless

(6.42a)

G superficial molar air flow rate (k mol/s m ) 2

L superficial molar water flow rate (k mol/s m2) The NTU depends upon the designed gas removal efficiency, the air-water velocity ratio, and Henry’s constant. Treybal (1968) plotted the integral part (NTU) of Eq. (6.35) in Fig. 6.5. By knowing the desired removal efficiency, the stripping factor, and Henry’s constant, the NTU in a packed column can be determined for any given stripping factor of the air to water flow rate ratios. It can be seen from Fig. 6.5 that when the stripping factor, R, is greater than 3, little improvement for the NTU occurs. EXAMPLE 3:

Using the graph of Fig. 6.5 to solve Example 1. Given: T 20C 293 K L 80 m3 water/(m2 column cross section h) G 2400 m3 air/(m2 column cross section h) c1 131 g/L of CCHCl3 c2 13.1 g/L of CCHCl3 KLa 44 h1

Solution: Step 1. Compute HTU with Eq. (6.39) HTU

80 m/h L 1.82 m KLa 44 h1

Step 2. Compute Hu, the unitless Henry’s constant at 20C From Table 6.1 H 550 atm Using Eq. (6.42a) Hu 7.49 104 H 7.49 104 550 0.412 Step 3. Compute the stripping factor R using (Eq. 6.41) R

Hu G 0.412 2400 m/h L 80 m/h

12.36 Step 4. Find NTU from Fig. 6.5 since

c2 13.1 g/L c1 131 g/L 0.1

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

Using R 12.36 and c2/c1 0.1, from the graph in Fig. 6.5, we obtain NTU 2.42 Step 5. Compute the height of tower z by Eq. (6.38) z (HTU) (NTU) 1.82 m 2.42 4.4 m

FIGURE 6.5 Number of transfer units for absorbers or strippers with constant absorption or stripping factors (D. A. Cornwell, Air Stripping and aeration. In: AWWA, Water Quality and Treatment. Copyright 1990, McGraw-Hill, New York, reprinted with permission of McGraw-Hill).

1.351

PUBLIC WATER SUPPLY 1.352

CHAPTER 1.6

6.4 Nozzles There are numerous commercially available spray nozzles. Nozzles and tray aerators are air-water contact devices. They are used for iron and manganese oxidation, removal of carbon dioxide and hydrogen sulfide from water, and removal of taste and odor causing materials. Manufacturers may occasionally have mass transfer data such as KL and a values. For an open atmosphere spray fountain, the specific interfacial area a is (Calderbrook and Moo-Young, 1961): a

6 d

(6.43)

where d is the droplet diameter which ranges from 2 to 10,000 m. Under open atmosphere conditions, Cs remains constant because there is an infinite air-to-water ratio and the Lewis and Whitman equation can express the mass transfer for nozzle spray (Fair et al., 1968; Cornwell, 1990) Ce Ci (Cs Ci) [1 exp(KLat)] where Ci, Ce solute concentrations in bulk water and droplets, respectively, g/L t time of contact between water droplets and air twice the rise time, tr 2V sin f g or

tr V sin f/g

where

f angle of spray measured from horizontal V velocity of the droplet from the nozzle

where

V Cv 22gh Cv velocity coefficient, 0.40 to 0.95, obtainable from the manufacturer g gravitational acceleration, 9.81 m/s2

Combine Eq. (6.45b) and Eq. (6.46)

(6.44)

(6.45a) (6.45b)

(6.46)

gtr CV 22gh sin f

The driving head h is h gt r2 /(2Cv2 sin2 f)

(6.47)

Neglecting wind effect, the radius of the spray circle is r Vtr cos f V(V sin f/g) cos f V 2/g sin f cos f Cv2 2gh/g sin 2 f [from Eq. (6.46)] 2Cv2 h sin 2f

(6.48)

and the vertical rise of the spray is hr

1 2 gt 2 r

1 h (2Cv2 sin2f) 2

From Eq. (6.47)

Cv2 h sin2f

(6.49)

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.353

Determine the removal percentage of trichloroethylene from a nozzle under an operating pressure of 33 psi (228 kPa). The following data is given:

EXAMPLE 1:

d 0.05 cm Cv 0.50 f 30 KL 0.005 cm/s Ci 131 g/L Solution: Step 1. Determine the volumetric interfacial area a by Eq. (6.43): a

6 6 120 cm1 d 0.05 cm

Step 2. Compute the velocity of the droplet V by Eq. (6.46) The pressure head h 33 psi

70.3 cm of waterhead 1 psi

2320 cm Using Eq. (6.46) V Cv 22 gh 0.5 22 981 2320 1067 cm/s Step 3. Compute the time of contact by Eq. (6.45a) t

2V sin f 2 1067 cm/s sin 30 g 981 cm/s2 2 1067 0.5 s 981

1.09 s Step 4. Compute the mass transfer Ce Ci by Eq. (6.44) Ce Ci (Cs Ci)[1 exp(KLat)] Ce 131 (0 131)[1 exp(0.005 120 1.09)] Ce 131 131 0.48 68.1 ( g/L) 52% remained 48 percent removal Calculate the driving head, radius of the spray circle, and vertical rise of spray for (a) a vertical jet and (b) a jet at 45 angle. Given the expose time for water droplet is 2.2 s and Cv is 0.90.

EXAMPLE 2:

PUBLIC WATER SUPPLY 1.354

CHAPTER 1.6

Solution: Step 1. For question (a). Using Eq. (6.45b) 1 1 t 2.2 s 1.1 s 2 2 f 90 tr

sin f sin 90 1 sin 2 f sin 180 0 Using Eq. (6.47) h gt 2r /(2 Cv2 sin2 ) 9.81 m/s2 (1.1 s)2/(2 0.92 12) 7.33 m Using Eq. (6.48) r 2Cv2 h sin 2 f 0 Since sin 2f 0, and using Eq. (6.49) 1 1 hr g t2r 9.81 m/s2 (1.1 s)2 2 2 5.93 m Step 2. For question (b) tr 1.1 s f 45 sin f sin 45 1/ 22 0.707 sin 2 f sin 90 1 Using Eq. (6.47) h gt 2r /(2Cv2 sin f) 9.81(1.1)2/(2 0.92 0.7072) 14.66 (m) Using Eq. (6.48) r 2Cv2 h sin 2 f 2(0.9)2 (14.66 m) (1) 23.7 m and using Eq. (6.49) hr C v2 h sin f (0.9)2 14.66 m (0.707)2 5.93 m

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.355

7 SOLUBILITY EQUILIBRIUM Chemical precipitation is one of the most commonly employed methods for drinking water treatment. Coagulation with alum, ferric sulfate, or ferrous sulfate, and lime softening involve chemical precipitation. Most chemical reactions are reversible to some degree. A general chemical reaction which has reached equilibrium is commonly written by the law of mass action as aA bB 4 cC d D where

(6.50a)

A, B reactants C, D products a, b, c, d stoichiometric coefficients for A, B, C, D, respectively

The equilibrium constant Keq for the above reaction is defined as Keq

[C]c[D]d [A]a[B]b

(6.50b)

where Keq is a true constant, called the equilibrium constant, and the square brackets signify the molar concentration of the species within the brackets. For a given chemical reaction, the value of equilibrium constant will change with temperature and the ionic strength of the solution. For an equilibrium to exist between a solid substance and its solution, the solution must be saturated and in contact with undissolved solids. For example, at pH greater than 10, solid calcium carbonate in water reaches equilibrium with the calcium and carbonate ions in solution: consider a saturated solution of CaCO3 that is in contact with solid CaCO3. The chemical equation for the relevant equilibrium can be expressed as CaCO3 (s) 4 Ca2 (aq) CO32(aq)

(6.50c)

The equilibrium constant expression for the dissolution of CaCO3 can be written as Keq

[Ca2][CO2 3 ] [CaCO3]

(6.50d)

Concentration of a solid substance is treated as a constant called Ks in mass-action equilibrium, thus [CaCO3] is equal to Ks. Then Keq, Ks [Ca2][CO 2 ] Ksp 3

(6.50e)

The constant Ksp is called the solubility product constant. The rules for writing the solubility product expression are the same as those for the writing of any equilibrium constant expression. The solubility product is equal to the product of the concentrations of the ions involved in the equilibrium, each raised to the power of its coefficient in the equilibrium equation. For the dissolution of a slightly soluble compound when the (brackets) concentration is denoted in moles, the equilibrium constant is called the solubility product constant. The general solubility product expression can be derived from the general dissolution reaction Ax By (s) 4 xAy yBx

(6.50f)

Ksp [Ay]x [Bx]y

(6.50g)

and is expressed as

Solubility product constants for various solutions at or near room temperature are presented in Appendix C.

PUBLIC WATER SUPPLY 1.356

CHAPTER 1.6

If the product of the ionic molar concentration [Ay]x[Bx]y is less than the Ksp value, the solution is unsaturated, and no precipitation will occur. In contrast, if the product of the concentrations of ions in solution is greater than Ksp value, precipitation will occur under supersaturated condition. Calculate the concentration of OH– in a 0.20 mole (M) solution of NH3 at 25C. Keq 1.8 105

EXAMPLE 1:

Solution: Step 1. Write the equilibrium expression NH3 (aq) H2O (l) ↔ NH 4 (aq) OH(aq) Step 2. Find the equilibrium constant Keq

[NH 4 ][OH ] 1.8 105 [NH3]

Step 3. Tabulate the equilibrium concentrations involved in the equilibrium: Let x concentration (M) of OH NH3 (aq) H2O (1) ↔ NH 4 (aq) OH(aq) Initial: Equilibrium:

0.2 M (0.20 x)M

0M xM

Step 4. Inserting these quantities into step 2 to solve for x [NH 4 ][OH ] [NH3]

Keq

(x)(x) 1.8 105 (0.20 x)

x2 (0.20 x)(1.8 105)

Neglecting x term, then approximately x 2 0.2 1.8 105 x 1.9 103 M EXAMPLE 2:

Ca3(PO4)2.

Write the expression for the solubility product constant for (a) Al(OH) 3 and (b)

Solution: Step 1. Write the equation for the solubility equilibrium (a) Al(OH)3 (s) ↔ Al3 (aq) 3OH(aq) (b) Ca3(PO4)2 (s) ↔ 3Ca2 (aq) 2PO3 (aq) 4 Step 2. Write Ksp by using Eq. (6.50g) and from Appendix C (a) Ksp [Al3] [OH]3 2 1032 (b) Ksp [Ca2]3 [PO43]2 2.0 1029

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.357

The Ksp for CaCO3 is 8.7 109 (Appendix C). What is the solubility of CaCO3 in water in grams per liter?

EXAMPLE 3:

Solution: Step 1. Write the equilibrium equation CaCO3 (s) 4 Ca2(aq) CO32(aq) Step 2. Set molar concentrations For each mole of CaCO3 that dissolves, 1 mole of Ca2 and 1 mole of CO 32 enter the solution. Let x be the solubility of CaCO3 in moles per liter. The molar concentrations of Ca2 and CO 2 3 are [Ca2] x and [CO 32] x Step 3. Solve solubility x in moles per liter Ksp [Ca2][CO32] 8.7 109 (x)(x) 8.7 109 x 9.3 105 mol/L 1 mole of CaCO3 100 g/L of CaCO3 then

x 9.3 105(100 g/L) 0.0093 g/L 9.3 mg/L

8 COAGULATION Coagulation is a chemical process to remove turbidity and color producing material that is mostly colloidal particles (1 to 200 millimicrons, m ) such as algae, bacteria, organic and inorganic substances, and clay particles. Most colloidal solids in water and wastewater are negatively charged. The mechanisms of chemical coagulation involve the zeta potential derived from double-layer compression, neutralization by opposite charge, interparticle bridging, and precipitation. Destabilization of colloid particles is influenced by the Van der Waals force of attraction and Brownian movement. Detailed discussion of the theory of coagulation can be found elsewhere (American Society of Civil Engineers and American Water Works Association 1990). Coagulation and flocculation processes were discussed in detail by Amirtharajah and O’Melia (1990). Coagulation of water and wastewater generally add either aluminum, or iron salt, with and without polymers and coagulant aids. The process is complex and involves dissolution, hydrolysis, and polymerization; pH values play an important role in chemical coagulation depending on alkalinity. The chemical coagulation can be simplified as the following reaction equations: Aluminum sulfate (alum): Al2(SO4) 18H2O 3Ca (HCO3)2 S 2Al(OH)3 T 3CaSO4 6CO2 18H2O

(6.51a)

When water does not have sufficient total alkalinity to react with alum, lime or soda ash is usually also dosed to provide the required alkalinity. The coagulation equations can be written as below: Al2(SO4)3 18H2O 3Ca(OH)2 S 2Al(OH)3 T 3CaSO4 18H2O Al2(SO4)3 ⋅ 18H2O 3Na2 CO3 3H2O S 2Al(OH)3 T 3Na2SO4 3CO2 18H2O

(6.51b) (6.51c)

PUBLIC WATER SUPPLY 1.358

CHAPTER 1.6

Ferric chloride: 2FeCl3 3Ca(HCO3)2 S 2Fe(OH)3 T 3CaCl2 6CO2

(6.51d)

Fe(SO4)3 3Ca(HCO3)2 S 2Fe(OH)3 T 3CaSO4 6CO2

(6.51e)

Ferric sulfate:

Ferrous sulfate and lime: FeSO4 ⋅ 7H2O Ca(OH)2 S Fe(OH)2 CaSO4 7H2O

(6.51f)

followed by, in the presence of dissolved oxygen 4Fe(OH)2 O2 2H2O S 4Fe(OH)3 T

(6.51g)

3FeSO4 ⋅ 7H2O 1.5Cl2 S Fe2(SO4)3 FeCl3 2lH2O

(6.51h)

Chlorinated copperas:

followed by reacting with alkalinity as above Fe2(SO4)3 3Ca(HCO3)2 S 2Fe(OH)3 T 3CaSO4 6CO2

(6.51i)

2FeCl3 3Ca(HCO3)2 S 2Fe(OH)3 T 3CaCl2 6CO2

(6.51j)

and

Each of the above reaction has an optimum pH range. EXAMPLE 1: What is the amount of natural alkalinity required for coagulation of raw water with dosage of 15.0 mg/L of ferric chloride?

Solution: Step 1. Write the reaction equation (Eq. 6.51j) and calculate MW 2FeCl3 3Ca(HCO3)2 S 2Fe(OH)3 3CaCl2 6CO2 2(55.85 3 35.45)

3[40.08 2(1 12 48)]

324.4

486.2

The above equation suggests that 2 moles of ferric chloride react with 3 moles of Ca(HCO3)2. Step 2. Determine the alkalinity needed for X mg/L Ca(HCO3)2 486.2 1.50 324.4 mg/L FeCl2 X 1.50 mg/L FeCl2 1.50 15.0 mg/L 22.5 mg/L as Ca(HCO3)2 Assume Ca(HCO3)2 represents the total alkalinity of the natural water. However, alkalinity concentration is usually expressed in terms of mg/L as CaCO3. We need to convert X to a concentration of mg/L as CaCO3. MW of CaCO3 40.08 12 48 100.1 MW of Ca(HCO3)2 162.1

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.359

Then X 22.5 mg/L

100.1 13.9 mg/L as CaCO3 162.1

EXAMPLE 2: A water with low alkalinity of 12 mg/L as CaCO3 will be treated with the alumlime coagulation. Alum dosage is 55 mg/L. Determine the lime dosage needed to react with alum.

Solution: Step 1. Determine the amount of alum needed to react with the natural alkalinity From Example 1 Alkalinity 12 mg/L as CaCO3

162.1 as Ca(HCO3)2 100.1 as CaCO3

19.4 mg/L as Ca(HCO3)2 MW of Al2(SO4)3 18H2O 27 2 3(32 16 4) 18(2 16) 666 MW of Ca(HCO3)2 162.1 Equation (6.51a) suggests that 1 mole of alum reacts with 3 moles of Ca(HCO3)2. Therefore, the quantity of alum to react with natural alkalinity is Y: Y 19.4 mg/L as Ca(HCO3)2

666 as alum 3 162.1 as Ca(HCO3)2

26.6 mg/L as alum Step 2. Calculate the lime required. The amount of alum remaining to react with lime is (dosage – Y) 55 mg/L 26.6 mg/L 28.4 mg/L MW of Ca(OH)2 40.1 2(16 1) 74.1 MW of CaO 40.1 16 56.1 Equation (6.51b) indicates that 1 mole of alum reacts with 3 moles of Ca(OH)2 Ca(OH)2 required: 28.4 mg/L alum

3 74.1 as Ca(OH)2 666 alum

9.48 mg/L as Ca(OH)2 Let dosage of lime required be Z Z 9.48 mg/L as Ca(OH)2

56.1 as CaO 74.1 as Ca(OH)2

7.2 mg/L as CaO 8.1 Jar Test For the jar test, chemical (coagulant) is added to raw water sample for mixing in the laboratory to simulate treatment-plant mixing conditions. Jar tests may provide overall process effectiveness,

PUBLIC WATER SUPPLY 1.360

CHAPTER 1.6

particularly to mixing intensity and duration as it affects floc size and density. It can also be used for evaluating chemical feed sequence, feed intervals, and chemical dilution ratios. A basic description of the jar test procedure and calculations is presented elsewhere (APHA 1995). It is common to use six 2 L Gator jars with various dosages of chemical (alum, lime, etc); and one jar as the control (without coagulant). Appropriate coagulant dosages are added to the 2 L samples before the rapid mixing at 100 revolution per minute (rpm) for 2 min. Then the samples and the control were flocculated at 20 rpm for 20 or more minutes and allowed to settle. Water temperature, floc size, settling characteristics (velocity, etc.), color of supernatant, pH, etc. should be recorded. The following examples illustrate calculations involved in the jar tests; prepare stock solution, jar test solution mixture. EXAMPLE 1: Given that liquid alum is used as a coagulant. Specific gravity of alum is 1.33. One gallon of alum weighs 11.09 lb and contains 5.34 lb of dry alum. Determine: (a) the alum concentration, (b) mL of liquid alum required to prepare a 100 mL solution of 20,000 mg/L alum concentration, (c) the dosage concentration of 1 mL of stock solution in a 2000 mL Gator jar sample.

Solution: Step 1. Determine alum concentration in mg/mL Alum (mg/L)

(5.34 lb)(453,600 mg/lb) (1 gal)(3785 mL/gal)

640 mg/mL Step 2. Prepare 100 mL stock solution having a 20,000 mg/L alum concentration Let x mg of alum required to prepare 100 mL stock solution 20,000 mg x 100 mL 1000 mL x 2000 mg Step 3. Calculate mL (y) of liquid alum to give 2000 mg 2000 mg y mL 1 mL 640 mg y 3.125 mL Note: Or add 6.25 mL of liquid alum into the 200 mL stock solution, since 3.125 mL is difficult to accurately measure. Step 4. Find 1 mL of alum concentration (z) in 2000 mL sample (jar) (z mg/L) (2000 mL) (20,000 mg/L) (1 mL) z

20,000 mg/L 2000

10 mg/L Note: Actual final volume is 2001 mL; using 2000 mL is still reasonable. Assuming that 1 mL of potassium permanganate stock solution provides 1 mg/L dosage in a 2 L jar, what is the weight of potassium permanganate required to prepare 100 mL of stock solution? Assume 100 percent purity.

EXAMPLE 2:

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.361

Solution: Step 1. Find concentration (x mg/L) of stock solution needed (x mg/L) (1 mL) (1 mg/L) (2000 mL) x 2000 mg/L Step 2. Calculate weight (y mg) to prepare 100 mL (as 2000 mg/L) y mg 2000 mg 100 mL 1000 mL y 200 mg Add 200 mg of potassium permanganate into 100 mL of deionized water to give 100 mL stock solution of 2000 mg/L concentration. Liquid polymer is used in a 2 L Gator jar test. It weighs 8.35 pounds per gallon. One mL of polymer stock solution provides 1 mg/L dosage. Compute how many mL of polymer are required to prepare 100 mL of stock solution.

EXAMPLE 3:

Solution: Step 1. Calculate (x) mg of polymer in 1 mL x

(8.35 lb)(453,600 mg/lb) (1 gal)(3785 mL/gal)

1000 mg/mL Step 2. As in Example 2, find the weight (y mg) to prepare 100 mL of 2000 mg/L stock solution 2000 mg y mg 100 mL 1000 mL y 200 mg (of polymer in 100 mL distilled water) Step 3. Compute volume (z mL) of liquid polymer to provide 200 mg z

200 mg 1 mL 200 mg x 1000 mg

0.20 mL Note: Use 0.20 mL of polymer in 100 mL of distilled water. This stock solution has 2000 mg/L polymer concentration. One mL stock solution added to a 2 L jar gives 1 mg/L dosage. 8.2 Mixing Mixing is an important operation for the coagulation process. In practice, rapid mixing provides complete and uniform dispersion of a chemical added to the water. Then follows a slow mixing for flocculation (particle aggregation). The time required for rapid mixing is usually 10 to 20 s. However, recent studies indicate the optimum time of rapid mixing is a few minutes. Types of mixing include propeller, turbine, paddle, pneumatic, and hydraulic mixers. For a water treatment plant, mixing is used for coagulation and flocculation, and chlorine disinfection. Mixing is also used for biological treatment processes for wastewater. Rapid mixing for coagulant in raw water and activated sludge process in wastewater treatment are complete mixing. Flocculation basins after rapid mixing are designed based on an ideal plug flow using first-order kinetics. It is very

PUBLIC WATER SUPPLY 1.362

CHAPTER 1.6

difficult to achieve an ideal plug flow. In practice, baffles are installed to reduce short-circuiting. The time of contact or detention time in the basin can be determined by: For complete mixing: V 1 Ci C e b a K Q Ce

(6.52a)

Ci V 1 L y aln b K Q Ce

(6.52b)

t For plug flow: t

where t detention time of the basin, min V volume of basin, m3 or ft3 Q flow rate, m3/s or cfs K rate constant Ci influent reactant concentration, mg/L Ce effluent reactant concentration, mg/L L length of rectangular basin, m or ft horizontal velocity of flow, m/s or ft/s EXAMPLE 1: Alum dosage is 50 mg/L K 90 per day based on laboratory tests. Compute the detention times for complete mixing and plug flow reactor for 90 percent reduction.

Solution: Step 1. Find Ce Ce (1 0.9)Ci 0.1 Ci 0.1 50 mg/L 5 mg/L Step 2. Calculate t for complete mixing Using Eq. (6.52a) t

50 mg/L 5 mg/L 1 Ci Ce 1 b a a b K Ce 90/day 5 mg/L 1 day 1440 min 9 90 1 day

144 min Step 3. Calculate t for plug flow Using Eq. (6.52b) t

Ci 50 1440 min 1 aln b aln b K Ce 90 5

36.8 min Power requirements. Power required for turbulent mixing is traditionally based on the velocity gradient or G values proposed by Camp and Stein (1943). The mean velocity gradient G for mechanical mixing is G a

P 1/2 b

(6.53)

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.363

where G mean velocity gradient; velocity (ft/s)/distance (ft) is equal to per second P power dissipated, ft lb/s or N m/s (W)

absolute viscosity, lb s/ft2 or N s/m2 V volume of basin, ft3 or m3 The equation is used to calculate the mechanical power required to facilitate rapid mixing. If a chemical is injected through orifices with mixing times of approximately 1.0 s, the G value is in the range of 700 to 1000/s. In practice, G values of 3000 to 5000/s are preferable for rapid mixing (ASCE and AWWA, 1990). Eq. (6.53) can be expressed in terms of horsepower (hp) as 550 hp 1/2 b

(6.54a)

kW 10 1/2 b

(6.54b)

G a For SI units the velocity gradient per second is: G a

where kW energy input, kW V effective volume, m3

absolute viscosity, centipoise, cp 1 cp at 20C (see Table 5.1a, 1 cp 0.001 N ⋅ s/m2) The equation is the standard design guideline used to calculate the mechanical power required to facilitate rapid mixing. Camp (1968) claimed that rapid mixing at G values of 500/s to 1000/s for 1 to 2 min produced essentially complete flocculation and no further benefit for prolonged rapid mixing. For rapid mixing the product of Gt should be 30,000 to 60,000 with t (time) generally 60 to 120 s. EXAMPLE 2: A rapid mixing tank is 1 m 1 m 1.2 m. The power input is 746 W (1 hp). Find the G value at a temperature of 15C.

Solution: At 10C, 0.00113 N s/m2 (from Table 5.1a) V 1 m 1 m 1.2 m 1.2 m3 P 746 W 746 N m/s Using Eq. (6.53): G (P/ V)0.5 a

0.5 746 N m/s b 2 3 # 0.00113 N s/m 1.2 m

742 s1

9 FLOCCULATION After rapid mixing, the water is passed through the flocculation basin. It is intended to mix the water to permit agglomeration of turbidity settled particles (solid capture) into larger flocs which would have a mean velocity gradient ranging 20 to 70 s1 for a contact time of 20 to 30 min taking place in the flocculation basin. A basin is usually designed in four compartments (ASCE and AWWA 1990).

PUBLIC WATER SUPPLY 1.364

CHAPTER 1.6

The conduits between the rapid mixing tank and the flocculation basin should maintain G values of 100 to 150 s1 before entering the basin. For baffled basin, the G value is G a

QgH 0.5 62.4H 0.5 b a t b

(6.55)

where G mean velocity gradient, s1 Q flow rate, ft3/s specific weight of water, 62.4 lb/ft3 H head loss due to friction, ft

absolute viscosity, lb s/ft2 V volume of flocculator, ft3 t detention time, s For paddle flocculators, the useful power input of an impeller is directly related to the dragforce of the paddles (F). The drag force is the product of the coefficient of drag (Cd) and the impeller force (Fi). The drag force can be expressed as (Fair et al. 1968): F Cd Fi Fi r A

(6.56a)

y 2

(6.57)

then F 0.5 Cd A 2

(6.56b)

where F drag force, lb Cd dimensionless coefficient of drag mass density, lb s2/ft4 A area of the paddles, ft2 velocity difference between paddles and water, fps The velocity of paddle blades (p) can be determined by yp

2p r n 60

(6.58)

where p velocity of paddles, fps n number of revolutions per minute, rpm r distance from shaft to center line of the paddle, ft The useful power input is computed as the product of drag force and velocity difference as below: P F 0.5 Cd A 3

(6.59)

EXAMPLE 1: In a baffled basin with detention time of 25 min. Estimate head loss if G is 30 per second, 2.359 10–5 lb s/ft2 at T 60F (Table 5.1b).

Solution: Using Eq. (6.55) G a

62.4 H 1/2

t b

G2mt (30 s1)2(2.359 105 lb # s/ft2) (25 60 s) 62.4 62.4 lb/ft3 0.51 ft

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.365

A baffled flocculation basin is divided into 16 channels by 15 around-the-end baffles. The velocities at the channels and at the slots are 0.6 and 2.0 fps (0.18 and 0.6 m/s), respectively. The flow rate is 12.0 cfs (0.34 m3/s). Find (a) the total head loss neglecting channel friction; (b) the power dissipated; (c) the mean velocity gradient at 60F (15.6C); the basin size is 16 15 80 ft3; (d) the Gt value, if the detention (displacement) time is 20 min, and (e) loading rate in gpd/ft3.

EXAMPLE 2:

Solution: Step 1. Estimate loss of head H Loss of head in a channel Loss of head in a slot

y21 0.62 0.00559 (ft) 2g 2 32.2 22 0.0621 (ft) 64.4

H 16 0.00559 15 0.0621

(a):

1.02 (ft) Step 2. Compute power input P P QgH 12 ft3/s 62.37 lb/ft3 1.02 ft

(b):

763 ft lb/s Note: g 62.37 lb/ft at 60F (Table 5.1b) 3

Step 3. Compute G From Table 5.1b At 60F, 2.359 105 lb s/ft2 V 16 ft 15 ft 80 ft 19,200 ft3 Using Eq. (6.54a) (c):

G a

1/2 763 ft # lb/s P 1/2 b a b mV 2.359 105 lb # s/ft2 19,200 ft3

41.0 s1 Step 4. Compute Gt Gt 41 s1 20 min 60 s/min

(d):

49.200 Step 5. Compute loading rate Q 12 cfs 12 cfs 0.646 MGD/cfs 7.75 MGD 7.75 106 gpd (e):

Loading rate Q/V 7.76 106 gpd/19,200 ft3 404 gpd/ft3

A flocculator is 16 ft (4.88 m) deep, 40 ft (12.2 m) wide, and 80 ft (24.4 m) long. The flow of the water plant is 13 MGD (20 cfs, 0.566 m3/s). Rotating paddles are supported parallel to four horizontal shafts. The rotating speed is 2.0 rpm. The center line of the paddles is 5.5 ft (1.68 m) EXAMPLE 3:

PUBLIC WATER SUPPLY 1.366

CHAPTER 1.6

from the shaft (mid-depth of the basin). Each shaft equipped with six paddles. Each paddle blade is 10 in (25 cm) wide and 38 ft (11.6 m) long. Assume the mean velocity of the water is 28 percent of the velocity of the paddles and their drag coefficient is 1.9. Estimate: (a) the difference in velocity between the paddles and water (b) the useful power input (c) the energy consumption per million gallons (Mgal) (d) the detention time (e) the value of G and Gt at 60F (f) the loading rate of the flocculator Solution: Step 1. Find velocity differential Using Eq. (6.58) yp

2 3.14 5.5 ft 2 2p rn 60 60 s

1.15 fps p (1 0.28) 1.15 fps 0.72

(a)

0.83 fps Step 2. Find P A paddle area 4 shaft 6 pads/shaft 38 ft 10/12 ft/pad 760 ft2 g 62.4 lb/ft3 r g 1.938 lb # s2/ft4 32.2 ft/s2 Using Eq. (6.59): (b)

P 0.5Cd A 3 0.5 1.9 1.938 lb s2/ft4 760 ft2 (0.83 fps)3 800 ft lb/s

(800/550) hp

1.45 0.746 kW

1.45 (hp) 1.08 kW Step 3. Determine energy consumption E (c)

1.45 hp 24 h 13 Mgal/d day 2.68 hp h/Mgal 1.08 kW 24 h E 13 Mgal/d day 1.99 kW h/Mgal

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.367

Step 4. Determine detention time t Basin volume, V 16 ft 40 ft 80 ft 5.12 104 ft3 5.12 104 ft3 7.48 gal/ft3 3.83 105 gal 3.83 105 gal V 0.0295 day Q 13 106 gal/d 0.0295 day 1440 min/d

(d)

42.5 min Step 5. Compute G and Gt Using Eq. (6.53): (e)

G (P/ V)0.5 (800/2.359 105 5.12 104)0.5 25.7 (fps/ft, or s1) Gt (25.7 s1) (42.5 60 s) 65,530

Step 6. Compute the loading rate (f)

13 106 gpd Q V 5.12 104 ft3 254 gpd/ft3

Loading rate

10 SEDIMENTATION Sedimentation is one of the most basic processes of water treatment. Plain sedimentation, such as the use of a pre-sedimentation basin (grit chamber) and sedimentation tank (or basin) following coagulation—flocculation, is most commonly used in water treatment facilities. The grit chamber is generally installed upstream of a raw water pumping station to remove larger particles or objects. It is usually a rectangular horizontal-flow tank with a contracted inlet and the bottom should have at minimum a 1:100 longitudinal slope for basin draining and cleaning purposes. A trash screen (about 2-cm opening) is usually installed at the inlet of the grit chamber. Sedimentation is a solid–liquid separation by gravitational settling. There are four types of sedimentation: discrete particle settling (type 1); flocculant settling (type 2); hindered settling (type 3); and compression settling (type 4). Sedimentation theories for the four types are discussed in Chapter 1.7 and elsewhere (Gregory and Zabel, 1990). The terminal settling velocity of a single discrete particle is derived from the forces (gravitational force, buoyant force, and drag force) that act on the particle. The classical discrete particle settling theories have been based on spherical particles. The equation is expressed as u a

4g(rp r)d 3Cd r

where u settling velocity of particles, m/s or ft/s g gravitational acceleration, m/s2 or ft/s2 p density of particles, kg/m3 or lb/ft3 density of water, kg/m3 or lb/ft3 d diameter of particles, m or ft Cd coefficient of drag

1/2

(6.60)

PUBLIC WATER SUPPLY 1.368

CHAPTER 1.6

The values of drag coefficient depend on the density of water (), relative velocity (u), particle diameter (d), and viscosity of water ( ), which gives the Reynolds number R as: r ud R

(6.61)

The value of Cd decreases as the Reynolds number increases. For R less than 2 or 1, Cd is related to R by the linear expression as follows: Cd

24 R

(6.62)

At low values of R, substituting Eq. (6.61) and (6.62) into Eq. (6.60) gives u

g(rp r)d 2 18

(6.63)

This expression is known as the Stokes equation for laminar flow conditions. In the region of higher Reynolds numbers (2 R 500–1000), Cd becomes (Fair et al. 1968): Cd

3 24 0.34 R 2R

(6.64)

In the region of turbulent flow (5001000 R 200,000), the Cd remains approximately constant at 0.44. The velocity of settling particles results in Newton’s equation (ASCE and AWWA 1990): u 1.74 c

(rp r)gd 1/2 d r

(6.65)

When the Reynolds number is greater than 200,000, the drag force decreases substantially and Cd becomes 0.10. No settling occurs at this condition. Estimate the terminal settling velocity in water at a temperature of 15C of spherical silicon particles with specific gravity 2.40 and average diameter of (a) 0.05 mm and (b) 1.0 mm

EXAMPLE:

Solution: Step 1. Using the Stokes equation (Eq. (6.63)) for (a), d .05 mm From Table 5.1a, at T 15C 999 kg/m3, and 0.00113 N s/m2

(a)

d 0.05 mm 5 105 m g(rp r)d2 18

9.81 m/s2 (2400 999) kg/m3 (5 105 m)2 18 0.00113 N # s/m2 0.00169 m/s Step 2. Check with the Reynolds number ((Eq. (6.61)) rud 999 0.00169 5 105 R 0.00113 0.075 (a) The Stokes’ law applies, since R 2.

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

Step 3. Using the Stokes’ law for (b), d 1 mm 0.001 m 9.81 (2400 999) (0.001)2 18 0.00113 0.676 (m/s)

Step 4. Check the Reynold number Assume the irregularities of the particles 0.85 fr u d 0.85 999 0.188 0.001

0.00113 508

Since R 2, the Stokes’ law does not apply. Use Eq. (6.60) to calculate u. Step 5. Using Eqs. (6.64) and (6.60) 3 3 24 24 0.34 0.34 R 508 2R 2508 0.52 4g(rp r)d u2 3Cdr

4 9.81 (2400 999) 0.001 3 0.52 999 u 0.188 (m/s)

Step 6. Re-check R fr u d 0.85 999 0.188 0.001

0.00113 141

Step 7. Repeat Step 5 with new R 3 24 0.34 141 2141 0.76 4 9.81 1401 0.001 u2 3 0.76 999 u 0.155 (m/s)

Step 8. Re-check R 0.85 999 0.155 0.001 0.00113 116

R Step 9. Repeat Step 7

3 24 0.34 116 2116 0.83 4 9.81 1401 0.001 u2 3 0.83 999 u 0.149 (m/s)

(b) The estimated velocity is around 0.15 m/s based on steps 7 and 9.

1.369

PUBLIC WATER SUPPLY 1.370

CHAPTER 1.6

10.1 Overflow Rate For sizing the sedimentation basin, the traditional criteria used are based on the overflow rate, detention time, weir loading rate, and horizontal velocity. The theoretical detention time is computed from the volume of the basin divided by average daily flow (plug flow theory): t where

24V Q

(6.66)

t detention time, h 24 24 h per day V volume of basin, m3 or million gallon (Mgal) Q average daily flow, m3/d or Mgal/d (MGD)

The overflow rate is a standard design parameter which can be determined from discrete particle settling analysis. The overflow rate or surface loading rate is calculated by dividing the average daily flow by the total area of the sedimentation basin as follows: u where

Q Q A lw

(6.67)

u overflow rate, m3/(m2 d) or gpd/ft2 Q average daily flow, m3/d or gpd A total surface area of basin, m2 or ft2 l and w length and width of basin, respectively, m or ft

For alum coagulation u is usually in the range of 40 to 60 m3/(m2 d) (or m/d) (980 to 1470 gpd/ft2) for turbidity and color removal. For lime softening, the overflow rate ranges 50 to 110 m/d (1230 to 2700 gpm/ft2). The overflow rate in wastewater treatment is lower, ranging from 10 to 60 m/d (245 to 1470 gpm/ft2). All particles having a settling velocity greater than the overflow rate will settle and be removed. It should be noted that rapid particle density changes due to temperature, solid concentration, or salinity can induce density current which can cause severe short-circuiting in horizontal tanks (Hudson, 1972). EXAMPLE: A water treatment plant has four clarifiers treating 4.0 MGD (0.175 m3/s) of water. Each clarifier is 16 ft (4.88 m) wide, 80 ft (24.4 m) long, and 15 ft (4.57 m) deep. Determine: (a) the detention time, (b) overflow rate, (c) horizontal velocity, and (d) weir loading rate assuming the weir length is 2.5 times the basin width.

Solution: Step 1. Compute detention time t for each clarifier Q

1 day 1,000,000 gal 4 MGD 1 ft3 4 day 7.48 gal 24 h

5570 ft3/h 92.83 ft3/min (a)

V 16 ft 80 ft 15 ft Q 5570 ft 3/h

3.447 h

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.371

Step 2. Compute overflow rate u, using Eq. (6.67) u

(b)

1,000,000 gpd Q lw 16 ft 80 ft

781 gpd/ft2 31.83 m3/m2 d Step 3. Compute horizontal velocity Q 92.83 ft3/min wd 16 ft 15 ft

(c)

0.387 ft/min 11.8 cm/min Step 4. Compute weir loading rate uw 1,000,000 gpd Q 2.5w 2.5 16 ft

(d)

25,000 gpd/ft 310 m2/m d 10.2 Inclined Settlers Inclined (tube and plate) settlers are sedimentation units that have been used for more than two decades. A large number of smaller diameter (20 to 50 mm) tubes are nested together to act as a single unit and inclined with various angles (7C to 60). The typical separation distance between inclined plates for unhindered settling is 2 in (5 cm) with inclines of 3 to 6 ft (1 to 2 m) height. The solids or flocs settle by gravitational force. It is not necessary to use tubes and can take various forms or plates also. The materials are lightweight, generally PVC or ABC plastic (1 m 3 m in size). Tube settlers have proved as effective units. However, there is a tendency of clogging. Inclined settling systems can be designed as cocurrent, countercurrent, and cross-flow. Comprehensive theoretical analyses of various flow geometries have been discussed by Yao (1976). The flow velocity of the settler module () and the surface loading rate for the inclined settler (u) (Fig. 6.6) are calculated as (James M. Montgomery Consulting Engineering, 1985): y

Q A sin u

(6.68)

Qw A(H cos u w cos2 u)

(6.69)

v H u w FIGURE 6.6

Schematic diagram of settling action.

PUBLIC WATER SUPPLY 1.372

CHAPTER 1.6

where velocity of the water in settlers, m/s or ft/s Q flow rate, m3/s or MGD A surface area of basin, m2 or ft2 inclined angle of the settlers u settling velocity, m/s or ft/s w width of settler, m or ft H vertical height, m or ft Two flocculators treat 1.0 m3/s (22.8 MGD) and remove flocs larger than 0.02 mm. The settling velocity of the 0.02 mm flocs is measured in the laboratory as 0.22 mm/s (0.67 in/min) at 15C. Tube settlers of 50.8 mm (2 in) square honeycombs are inclined at a 50 angle, and its vertical height is 1.22 m (4 ft). Determine the basin area required for the settler module and the size of each flocculator at 15C.

EXAMPLE:

Solution: Step 1. Determine the area needed for the settler modules Q (1 m3/s)/2 0.5 m3/s 30 m3/min w 50.8 mm 0.0508 m H 1.22 m 50 Using Eq. (6.69) u

Qw A(H cos u w cos2 u)

0.5(0.0508) A(1.22 0.643 0.0508 0.6432)

0.315 A

Step 2. Determine A In practice, the actual conditions in the settlers are not as good as under controlled laboratory ideal conditions. A safety factor of 0.6 may be applied to determine the designed settling velocity. Thus u 0.6 0.00022 m/s

0.0315 A

A 238.6 m2 (use 240 m2) Step 3. Find surface loading rate Q/A Q/A (0.5 24 60 60 m3/d)/240 m2 180 m3/(m2 d) 3.07 gpm/ft2 (Note: 1m3/(m2 d) 0.017 gpm/ft2) Step 4. Compute flow velocity in the settlers, using Eq. (6.68) Q/A sin 180/0.766 235 (m/d) 0.163 m/min 0.0027 m/s

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.373

Step 5. Determine size of the basin Two identical settling basins are designed. Generally, the water depth of the basin is 4 m (13.1 ft). The width of the basin is chosen as 8.0 m (26.2 ft). The calculated length of the basin covered by the settler is l 240 m2/8 m 30 m 98 ft In practice, one-fourth of the basin length is left as a reserved volume for future expansion. The total length of the basin should be 30 m

4 40 m (131 ft) 3

Step 6. Check horizontal velocity Q/A (30 m3/min)/(4 m 8 m) 0.938 m/min 3 ft/min Step 7. Check Reynolds number (R) in the settler module Hydraulic radius R

0.05082 0.0127 m 4 0.0508

(0.0027 m/s)(0.0127 m) yR R 0.000001131 m2/s 30 2000, thus it is in a lamella flow

11 FILTRATION The conventional filtration process is probably the most important single unit operation of all the water treatment processes. It is an operation process to separate suspended matter from water by flowing it through porous filter medium or media. The filter media may be silica sand, anthracite coal, diatomaceous earth, garnet, ilmenite, or finely woven fabric. In early times, a slow sand filter was used. It is still proved to be efficient. It is very effective for removing flocs of bacteria, algae, other microorganisms, and protozoa, such as Giardia, and Cryptosporidium. Rapid filtration has been very popular for several decades. Filtration usually follows the processes coagulation—flocculation—sedimentation. However, for some water treatment, direct filtration is used due to the high quality of raw water processes. Dual-media filters (sand and anthracite, activated carbon, or granite) give more benefits than single-media filters and became more popular; even triple-media filters have been used. In Russia, up-flow filters are used. All filters need to clean out the medium by backwash after a certain period (most are based on head loss) of filtration. The filters are also classified by allowing loading rate. Loading rate is the flow rate of water applied to the unit area of the filter. It is the same value as the flow velocity approaching the filter surface and can be determined by Q/A

(6.67a)

where loading rate, m3/(m2 d) or gpm/ft2 Q flow rate, m3/d or ft3/d or gpm A surface area of filter, m2 or ft2 On the basis of loading rate, the filters are classified as slow sand filters, rapid sand filters, and high-rate sand filters. With each type of filter medium or media, there are typical design criteria for the range of loading rate, effective size, uniform coefficient, minimum depth requirements, and

PUBLIC WATER SUPPLY 1.374

CHAPTER 1.6

backwash rate. The typical loading rate for rapid sand filters is 120 m3/(m2 d), (83 L/(m2 min) or 2 gpm/ft2). For high-rate filters, the loading rate may be four to five times this rate. EXAMPLE: A city is to install rapid sand filters downstream of the clarifiers. The design loading rate is selected to be 160 m3/(m2 d), (2.7 gpm/ft2). The design capacity of the water works is 0.35 m3/s (8 MGD). The maximum surface per filter is limited to 50 m2. Design the number and size of filters and calculate the normal filtration rate.

Solution: Step 1. Determine the total surface area required Q 0.35 m3/s (86,400 s/d) A y 160 m3/m2 # d 189 m2 Step 2. Determine the number (n) of filters n

189 m2 3.78 50 m

Select four filters. The surface area (a) for each filter is a 189 m2/4 47.25 m2 We can use 7 m 7 m or 6 m 8 m, or 5.9 m 8 m (exact) Step 3. If a 7 m 7 m filter is installed, the normal filtration rate is y

Q 0.35 m3/s 86,400 s/d A 4 7m 7m

154.3 m3/(m2 d) 11.1 Filter Medium Size Before a filter medium is selected, a grain size distribution analysis should be performed. The sieve size and percentage passing by weight relationships are plotted on logarithmic-probability paper. A straight line can be drawn. Determine the geometric mean size ( g) and geometric standard deviation size ( g). The most common parameters used in the United States to characterize the filter medium are effective size (ES) and uniformity coefficient (UC) of medium size distribution. The ES is that the grain size for which 10 percent of the grain (d10) are smaller by weight. The UC is the ratio of the 60-percentile (d60) to the 10-percentile. They can be written as (Fair et al., 1968; Cleasby, 1990): ES d10 g/ g1.282

(6.70)

1.535 g

(6.71)

UC d60/d10

The 90-percentile, d90, is the size for which 90 percent of the grains are smaller by weight. It is interrelated to d10 as (Cleasby 1990) d90 d10(101.67 log UC) The d90 size is used for computing the required filter backwash rate for a filter medium.

(6.72)

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.375

EXAMPLE: A sieve analysis curve of a typical filter sand gives d10 0.54 mm and d60 0.74 mm. What are its uniformity coefficient and d90?

Solution: Step 1. UC d60/d10 0.74 mm/0.54 mm 1.37 Step 2. Find d90 using Eq. (6.72) d90 d10(101.67 log UC) 0.54 mm(101.67 log 1.37) 0.54 mm(100.228) 0.91 mm 11.2 Mixed Media Mixed media are popular for filtration units. For the improvement process performance, activated carbon or anthracite is added on the top of the sand bed. The approximate specific gravity (s) of ilmenite sand, silica sand, anthracite, and water are 4.2, 2.6, 1.5, and 1.0, respectively. For equal settling velocities, the particle sizes for media of different specific gravity can be computed by d1 s2 s 2/3 as sb d2 1 where

(6.73)

d1, d2 diameter of particles 1 and 2, respectively s1, s2, s specific gravity of particles 1, 2, and water, respectively

Estimate the particle sizes of ilmenite (specific gravity 4.2) and anthracite (specific gravity 1.5) which have same settling velocity of silica sand 0.60 mm in diameter (specific gravity 2.6)

EXAMPLE:

Solution: Step 1. Find the diameter of anthracite by Eq. (6.73) d (0.6 mm)a

2.6 1 2/3 b 1.5 1

1.30 mm Step 2. Determine diameter of ilmenite sand d (0.6 mm)a

2.6 1 2/3 b 4.2 1

0.38 mm 11.3 Hydraulics of Filter Head loss for fixed bed flow. The conventional fixed-bed filters use a granular medium of 0.5 to 1.0 mm size with a loading rate or filtration velocity of 4.9 to 12.2 m/h (2 to 5 gpm/ft2). When the

PUBLIC WATER SUPPLY 1.376

CHAPTER 1.6

clean water flows through a clean granular (sand) filter, the loss of head (pressure drop) can be estimated by the Kozeny equation (Fair et al., 1968): k (1 )2 h A a by L V gr3 where

(6.74)

h head loss in filter depth L, m or ft k dimensionless Kozeny constant, 5 for sieve openings, 6 for size of separation g acceleration of gravity, 9.81 m/s or 32.2 ft/s

absolute viscosity of water, N s/m2 or lb s/ft2 density of water, kg/m3 or lb/ft3 porosity, dimensionless A/V grain surface area per unit volume of grain specific surface S (or shape factor 6.0 to 7.7) 6/d for spheres 6/ deq for irregular grains

grain sphericity or shape factor deq grain diameter of spheres of equal volume filtration (superficial) velocity, m/s or fps

The Kozeny (or Carmen–Kozeny) equation is derived from the fundamental Darcy—Waeisback equation for head loss in circular pipes (Eq. (5.17)). The Rose equation is also used to determine the head loss resulting from the water passing through the filter medium. The Rose equation for estimating the head loss through filter medium was developed experimentally by Rose in 1949 (Rose, 1951). It is applicable to rapid sand filters with a uniform near spherical or spherical medium. The Rose equation is h

1.067 Cd L y2 w g d 4

(6.75)

where h head loss, m or ft shape factor (Ottawa sand 0.95, round sand 0.82, angular sand 0.73, pulverized coal 0.73) Cd coefficient of drag (Eq. (6.64)) Other variables are defined previously in Eq. (6.74) Applying the medium diameter to the area to volume ratio for hom*ogeneous mixed beds the equation is h

1.067 Cd Ly2 w g 4

x ad g

(6.76)

For a stratified filter bed, the equation is h

Cd x 1.067 Ly2 4 a d wg g

(6.77)

where x percent of particles within adjacent sizes dg geometric mean diameter of adjacent sizes A dual medium filter is composed of 0.3 m (1 ft) anthracite (mean size of 2.0 mm) that is placed over a 0.6 m (2 ft) layer of sand (mean size 0.7 mm) with a filtration rate of 9.78 m/h EXAMPLE 1:

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.377

(4.0 gpm/ft2). Assume the grain sphericity is 0.75 and a porosity for both is 0.40. Estimate the head loss of the filter at 15C. Solution: Step 1. Determine head loss through anthracite layer Using the Kozeny equation (Eq. (6.74)) k (1 )2 A 2 h a by L V g r 3 where

k6 g 9.81 m/s2

/ v 1.131 106 m2 s (Table 5.1a) at 15C 0.40 A/V 6/0.75d 8/d 8/0.002 9.78 m/h 0.00272 m/s L 0.3 m

then h6

2 (1 0.4)2 1.131 106 8 b a (0.00272)(0.3) 9.81 0.002 0.43

0.0508 (m) Step 2. Compute the head loss passing through the sand Most input data are the same as in Step 1, except k5 d 0.0007 m L 0.6 m then h5

0.62 8 2 1.131 106 a b (0.00272)(0.6) 9.81 0.43 d

0.3387 106/d 2 0.3387 1 06/(0.0007)2 0.6918 (m) Step 3. Compute total head loss h 0.0508 m 0.6918 m 0.743 m Using the same data given in Example 1, except the average size of sand is not given. From sieve analysis, d10 (10 percentile of size diameter) 0.53 mm, d30 0.67 mm, d50 0.73 mm, d70 0.80 mm, and d90 0.86 mm, estimate the head of a 0.6 m sand filter at 15C. EXAMPLE 2:

PUBLIC WATER SUPPLY 1.378

CHAPTER 1.6

Solution: Step 1. Calculate head loss for each size of sand in the same manner as Step 2 of Example 1 h10 0.3387 106/d 2 0.3387 106/(0.00053)2 1.206 (m) h30 0.3387 106/(0.00067)2 0.755 (m) Similarly

h50 0.635 m h70 0.529 m h90 0.458 m

Step 2. Taking the average of the head losses given above h (1.206 0.755 0.635 0.529 0.458)/5 0.717 (m) Note: This way of estimation gives slightly higher values than Example 1. Head loss for a fluidized bed. When a filter is subject to back washing, the upward flow of water travels through the granular bed at a sufficient velocity to suspend the filter medium in the water. This is fluidization. During normal filter operation, the uniform particles of sand occupy the depth L. During backwashing the bed expands to a depth of Le. When the critical velocity (c) is reached, the pressure drops (p hg gc) and is equal to the buoyant force of the grain. It can be expressed as h (s ) (1 e)Le or h Le (1 e) (s )/

(6.78)

The porosity of the expanded bed can be determined by using (Fair et al., 1963) e (u/s)0.22

(6.79)

where u upflow (face) velocity of the water s terminal settling velocity of the particles A uniform bed of particles will expand when 4.5

u s e

(6.80)

The depth relationship of the unexpanded and expanded bed is L e La

1 b 1 e

(6.81)

The minimum fluidizing velocity (umf) is the superficial fluid velocity needed to start fluidization. The minimum fluidization velocity is important in determining the required minimum backwashing flow rate. Wen and Yu (1966) proposed the Umf equation excluding shape factor and porosity of fluidization: Umf

33.7

(1135.69 0.0408 Gn)0.5 r deq r deq

(6.82)

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.379

where Gn Galileo number 3

deq (s ) g/ 2

(6.83)

Other variables used are expressed in Eq. (6.74). In practice, the grain diameter of spheres of equal volume deq is not available. Thus the d90 sieve size is used instead of deg. A safety factor of 1.3 is used to ensure adequate movement of the grains (Cleasby and Fan, 1981). EXAMPLE: Estimate the minimum fluidization velocity and backwash rate for the sand filter at 15C. The d90 size of sand is 0.88 mm. The density of sand is 2.65 g/cm3.

Solution: Step 1. Compute the Galileo number From Table 5.1a, at 15C r 0.999 g/cm3

0.00113 N s/m2 0.00113 kg/m s 0.0113 g/cm s (See Example 2 in Section 1.4 in Chapter 1.5, page 1.247)

/r 0.0113 cm2/s g 981 cm/s2 d 0.088 cm rs 2.65 g/cm3 Using Eq. (6.83) Gn (0.088)3 (0.999) (2.65 0.999) (981)/(0.0113)2 8635 Step 2. Compute Umf by Eq. (6.82) 0.0113 33.7 0.0113 (1135.69 0.0408 8635)0.5 0.999 0.088 0.999 0.088 0.627 (cm/s)

Umf

Step 3. Compute backwash rate Apply a safety factor of 1.3 to Umf as backwash rate Backwash rate 1.3 0.627 cm/s 0.815 cm/s 0.00815 m/s 0.00815 m/s 86,400 s/d 704.16 m/d or (m3/m2 d) 704 m/d (0.01705 gpm/ft2)/1 (m/d) 12.0 gpm/ft2 11.4 Washwater Troughs In United States, in practice washwater troughs are installed at even spaced intervals (5 to 7 ft apart) above the gravity filters. The washwater troughs are employed to collect spent washwater. The total

PUBLIC WATER SUPPLY 1.380

CHAPTER 1.6

rate of discharge in a rectangular trough with free flow can be calculated by (Fair et al. 1968, ASCE and AWWA 1990) Q C w h1.5

(6.84a)

where Q flow rate, cfs C constant (2.49) w trough width, ft h maximum water depth in trough, ft For rectangular horizontal troughs of such a short length friction losses are negligible. Thus the theoretical value of the constant C is 2.49. The value of C may be as low as 1.72 (ASCE and AWWA, 1990). In European practice, backwash water is generally discharged to the side and no troughs are installed above the filter. For SI units, the relationship is Q 0.808 w h1.5

(6.84b)

where Q flow rate, m3/(m2 s) w trough width, m h water depth in trough, m EXAMPLE 1: Troughs are 20 ft (6.1 m) long, 18 in (0.46 m) wide, and 8 ft (2.44 m) to the center with a horizontal flat bottom. The backwash rate is 24 in/min (0.61 m/min). Estimate (1) the water depth of the troughs with free flow into the gullet, and (2) the distance between the top of the troughs and the 30 in sand bed. Assuming 40 percent expansion and 6 in of freeboard in the troughs and 8 in of thickness.

Solution: Step 1. Estimate the maximum water depth (h) in trough 24 (in/min) 2 ft/60 s 1/30 fps A 20 ft 8 ft 160 ft2 Q VA 160/30 cfs 5.33 cfs Using Eq. (6.84a) Q 2.49 w h1.5, w 1.5 ft h (Q/2.49 w)2/3 [5.33/(2.49 1.5)]2/3 1.27 (ft) Say h 16 in 1.33 ft Step 2. Determine the distance (y) between the sand bed surface and the top troughs Sand bed 30 in 2.5 ft Height of expansion 2.5 ft 0.40 Freeboard 6 in 0.5 ft Thickness 8 in 0.67 ft (the bottom of the trough) y 2.5 ft 0.4 1.33 ft 0.5 ft 0.67 ft 3.5 ft

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.381

A filter unit has surface area of 16 ft (4.88 m) wide and 30 ft (9.144 m) long. After filtering 2.88 Mgal (10,900 m3) for 50 h, the filter is backwashed at a rate of 16 gpm/ft2 (0.65 m/min) for 15 min. Find: (a) the average filtration rate, (b) the quantity of washwater, (c) percent of washwater to treated water, and the flow rate to each of the four troughs.

EXAMPLE 2:

Solution: Step 1. Determine flow rate Q Q

2,880,000 gal 50 h 60 min/h 16 ft 30 ft

2.0 gpm/ft2

2 gal 0.0037854 m3 1440 min 10.764 ft2 min ft2 1 gal 1 day 1 m2

117.3 m/d 120 m/d

say

Step 2. Determine quantity of wash water q q (16 gal/min ft2) 15 min 16 ft 30 ft 115,200 gal Step 3. Determine percentage (%) of wash water to treated water % 115,200 gal 100%/2,880,000 gal 4.0% Step 4. Compute flow rate (r) in each trough r 115,200 gal/(15 min 4 troughs) 1920 gpm 11.5 Filter efficiency The filter efficiency is defined as the effective filter rate divided by the operation filtration rate as follows (AWWA and ASCE, 1998): E where

Re UFRV UBWV Ro UFRV

(6.84c)

E filter efficiency, % Re effective filtration rate, gpm/ft2 or m3/m2/h (m/h) R0 operating filtration rate, gpm/ft2 or m3/m2/h (m/h) UFRV unit filter run volume, gal/ft2 or L/m2 UBWV unit backwash volume, gal/ft2 or L/m2

A rapid sand filter operating at 3.7 gpm/ft2 (9.0 m/h) for 46 hours. After this filter run, 295 gal/ft2 (12,000 L/m2) of backwash water is used. Find the filter efficiency.

EXAMPLE:

Solution: Step 1. Calculate operating filtration rate, Ro Ro 3.7 gal/min/ft2 60 min/h 46 h 10,212 gal/ft2

PUBLIC WATER SUPPLY 1.382

CHAPTER 1.6

Step 2. Calculate effective filtration rate, Re Re (10,212 295) gal/ft2 9917 gal/ft2 Step 3. Calculate filter efficiency, E, using Eq. (5.84c) E 9917/10,212 0.97 97%

12 WATER SOFTENING Hardness in water is mainly caused by the ions of calcium and magnesium. It may also be caused by the presence of metallic cations of iron, sodium, manganese, and strontium. These cations are present with anions such as HCO3 , SO2 , Cl, NO3 , and SiO2 . The carbonates and bicarbonates of 4 4 calcium, magnesium, and sodium cause carbonate hardness or temporary hardness since it can be removed and settled by boiling of water. Noncarbonate hardness is caused by the chloride and sulfate slots of divalent cations. The total hardness is the sum of carbonate and noncarbonate hardness. The classification of hardness in water supply is shown in Table 6.4. Although hard water has no health effects, using hard water would increase the amount of soap needed and would produce scale on bath fixtures, cooking utensils, also cause scale and corrosion in hot-water heaters, boilers, and pipelines. Moderate hard water with 60 to 120 mg/L as CaCO3 is generally publicly acceptable (Clark et al., 1977). The hardness in water can be removed by precipitation with lime, Ca(OH)2 and soda ash, Na2CO3 and by an ion exchange process. Ion exchange is discussed in another section. The reactions of limesoda ash precipitation for hardness removal in water and recarbonation are shown in the following equations: Removal of free carbon dioxide with lime c

CO2 d Ca(OH)2 S CaCO3 T H 2O H 2CO3

(6.85)

Removal of carbonate hardness with lime Ca

2HCO 3 Ca(OH)2 → 2CaCO3 ↓ 2H2O

2HCO 3

(6.86)

2Ca(OH)2 → 2CaCO3 ↓ Mg(OH)2 ↓ 2H2O

TABLE 6.4 Classification of Hard Water mg/L as CaCO3 Hardness classification Soft Moderate soft Slightly hard Moderate hard Hard Very hard

U.S. 0–60

61–120 121–180 180

International 0–50 51–100 101–150 151–200 201–300 300

(6.87)

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.383

Removal of noncarbonate hardness with soda ash and lime Ca2 c

2Cl 2Cl d Na2 CO3 S CaCO3 T 2Na c 2 d 2 SO4 SO4

Mg 2 c

2Cl 2Cl d Ca(OH)2 S Mg(OH)2 T Ca2 c 2 d 2 SO 4 SO 4

(6.88) (6.89)

Recarbonation for pH control (pH 8.5) CO 3 CO2 H2O S 2HCO3 2

(6.90)

Recarbonation for removal of excess lime and pH control (pH 9.5) Ca(OH)2 CO2 S CaCO3 T H2O

(6.91)

Mg(OH)2 CO2 S MgCO3 H2O

(6.92)

Based on the above equations, the stoichiometric requirement for lime and soda ash expressed in equivalents per unit volume are as follows (Tchobanoglous and Schroeder 1985):

Lime required (eq/m3) CO2 HCO 3 Mg excess Soda ash required (eq/m3) Ca2 Mg2 alkalinity

(6.93) (6.94)

Approximately 1 eq/m of lime in excess of the stoichiometric requirement must be added to bring the pH to above 11 to ensure Mg(OH)2 complete precipitation. After the removal of precipitates, recarbonation is needed to bring pH down to a range of 9.2 to 9.7. The treatment processes for water softening may be different depending on the degree of hardness and the types and amount of chemical added. They may be single stage lime, excess lime, single stage lime-soda ash, and excess lime-soda ash processes. Coldwell—Lawrence diagrams are based on equilibrium principles for solving water softening. The use of diagrams is an alternative to the stoichiometric method. It solves simultaneous equilibria equations and estimates the chemical dosages of lime-soda ash softening. The interested reader is referred to the American Water Works Association (1978) publication: Corrosion Control by Deposition of CaCO3 Films, and to Benefield and Morgan (1990). EXAMPLE 1: Water has the following composition: calcium 82 mg/L, magnesium 33 mg/L, sodium 14 mg/L, bicarbonate 280 mg/L, sulfate 82 mg/L, and chloride 36 mg/L. Determine carbonate hardness, noncarbonate hardness, and total hardness, all in terms of mg/L of CaCO3.

Solution: Step 1. Convert all concentration to mg/L of CaCO3 using the following formula: The species concentration in milliequivalents per liter (meq/L) is computed by the following equation meq/L

mg/L 100 50 for CaCO3 2 equivalent weight

The species concentration expressed as mg/L of CaCO3 is calculated from mg/L as CaCO3 meq/L of species 50

(mg/L of species) 50 equivalent weight of species

PUBLIC WATER SUPPLY 1.384

CHAPTER 1.6

Step 2. Construct a table for ions in mg/L as CaCO3 Concentration Molecular weight

Equivalent weight

mg/L

meq/L

mg/L as CaCO3

Ca Mg2+ Na+

40 24.3 23

20.0 12.2 23.0

82 33 14

HCO3– Cl– SO2– 4

61 35.5 96.1

61 35.5 48

280 36 82

4.0 2.7 0.6 Total: 7.3 4.6 1.0 1.7 Total: 7.3

200 135 30 365 230 50 85 365

Ion species 2+

Step 3. Construct an equivalent bar diagram for the cationic and anionic species of the water. The diagram shows the relative proportions of the chemical species important to the water softening process. Cations are placed above anions on the graph. The calcium equivalent should be placed first on the cationic scale and be followed by magnesium and other divalent species and then by the monovalent species sodium equivalent. The bicarbonate equivalent should be placed first on the anionic scale and immediately be followed by the chloride equivalent and then by the sulfate equivalent.

Step 4. Compute the hardness distribution Total hardness 200 135 335 (mg/L as CaCO3) Alkalinity (bicarbonate) 230 mg/L as CaCO3 Carbonate hardness alkalinity 230 mg/L as CaCO3 Noncarbonate hardness 365 230 165 (mg/L as CaCO3) or Na Cl SO2– 4 (single-stage lime softening): Raw water has the following composition: alkalinity 248 mg/L as CaCO3, pH 7.0, 1 0.77 (at T 10C), calcium 88 mg/L, magnesium 4 mg/L. Determine the necessary amount of lime to soften the water, if the final hardness desired is 40 mg/L as CaCO3. Also estimate the hardness of the treated water. EXAMPLE 2:

Solution: Step 1. Calculate bicarbonate concentration Alkalinity [HCO3] [CO2 ] [OH] [H] 3 At pH 7.0, assuming all alkalinity is in the carbonate form HCO–3 248

61 eq wt of HCO3 mg 1g 1 mole L 1000 mg 61 g 50 eq wt of alkalinity

4.96 103 mol/L

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.385

Step 2. Compute total carbonate species concentration CT (Snoeyink and Jenkins, 1980) CT [HCO3 ]/1 4.96 103/0.77 6.44 103 mol/L Step 3. Estimate the carbonate acid concentration ] [CO2 ] CT [H2CO3] [HCO 3 3 Rearranging, and [CO32] 0 [H2CO3] (6.44 4.96) 103 mol/L 1.48 103 mol/L 1.48 103 1000 62 mg/mol mol/L 92 mg/L or

148 mg/L as CaCO3 (92 50/32)

Step 4. Construct a bar diagram for the raw water converted concentration of Ca and Mg as CaCO3 Ca 88 100/40 220 mg/L as CaCO3 Mg 4 100/24.3 16 mg/L as CaCO3

Step 5. Find hardness distribution Calcium carbonate hardness 220 mg/L Magnesium carbonate hardness 16 mg/L Total hardness 220 16 236 (mg/L) Step 6. Estimate the lime dose needed Since the final hardness desired is 40 mg/L as CaCO3, magnesium hardness removal would not be required. The amount of lime needed (x) would be equal to carbonic acid concentration plus calcium carbonate hardness: x 148 220 368 mg/L as CaCO3 74 as Ca(OH)2 100 as CaCO3

368 mg/L as CaCO3

272 mg/L as Ca(OH)2 56 368 mg/L as CaO 100 206 mg/L as CaO

PUBLIC WATER SUPPLY 1.386

CHAPTER 1.6

The purity of lime is 70 percent. The total amount of lime needed is 206/0.70 294 mg/L as CaO Step 7. Estimate the hardness of treated water The magnesium hardness level of 16 mg/L as CaCO3 remains in the softened water theoretically. The limit of calcium achievable is 30 to 50 mg/L of CaCO3 and would remain in the water unless using 5 to 10 percent in excess of lime. Hardness levels less than 50 mg/L as CaCO3 are seldom achieved in plant operation. EXAMPLE 3: (excess lime softening): Similar to Example 2, except for concentrations of calcium and magnesium: pH 7.0, 1 0.77 (at T 10C). Alkalinity, calcium, and magnesium concentrations are 248, 158, and 56 mg/L as CaCO3, respectively. Determine the amount of lime needed to soften the water.

Solution: Step 1. Estimate H2CO3 as Example 2 H2CO3 148 mg/L as CaCO3 Step 2. Construct a bar diagram for the raw water

Step 3. Find hardness distribution Calcium carbonate hardness 158 mg/L Magnesium carbonate hardness 66 mg/L Total carbonate hardness 158 66 224 (mg/L) Comment: Sufficient lime (in excess) must be dosed to convert all bicarbonate alkalinity to carbonate alkalinity and to precipitate magnesium as magnesium hydroxide (needs 1 eq/L excess lime). Step 4. Estimate lime required (x) is the sum for carbonic acid, total alkalinity, magnesium hardness, and excess lime (say 60 mg/L as CaCO3) to raise pH 11.0. x (148 248 66 60) mg/L as CaCO3 522 mg/L as CaCO3 522 74/100 mg/L as Ca(OH)2 386 mg/L as Ca(OH)2 or

522 0.56 mg/L as CaO 292 mg/L as CaO

EXAMPLE 4: (single stage lime-soda ash softening): A raw water has the following analysis: pH 7.01, T 10C, alkalinity 248 mg/L, Ca 288 mg/L, Mg 12 mg/L, all as CaCO3. Determine amounts of lime and soda ash required to soften the water.

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.387

Solution: Step 1. As in previous example, construct a bar diagram

Step 2. Find the hardness distribution Total hardness 288 12 300 (mg/L as CaCO3) Calcium carbonate hardness, CCH 248 mg/L CaCO3 Magnesium carbonate hardness 0 Calcium noncarbonate hardness, CNH 288 248 40 (mg/L as CaCO3) Magnesium noncarbonate hardness, MNH 12 mg/L Step 3. Determine lime and soda ash requirements for the straight lime-soda ash process. Refer to Eqs. (6.85) and (6.86) Lime dosage H2CO3 CCH (148 248) mg/L as CaCO3 396 mg/L as CaCO3 396 0.74 mg/L as Ca(OH)2

293 mg/L as Ca(OH)2 Refer to Eq. (6.88) and soda ash dosage CNH 40 mg/L as CaCO3 (40 106/100) mg/L as Na2CO3 42.4 mg/L as Na2CO3 EXAMPLE 5: (excess lime-soda ash process): Raw water has the following analysis: pH 7.0, T 10C, alkalinity, calcium, and magnesium are 248, 288, and 77 mg/L as CaCO3, respectively. Estimate the lime-and soda-ash dosage required to soften the water.

Solution: Step 1. Construct a bar diagram for the raw water. Some characteristics are the same as the previous example.

PUBLIC WATER SUPPLY 1.388

CHAPTER 1.6

Step 2. Define hardness distribution Total hardness 288 77 365 (mg/L as CaCO3) Calcium carbonate hardness, CCH 248 mg/L as CaCO3 Magnesium carbonate hardness, MCH 0 Calcium noncarbonate hardness, CNH 288 mg/L 248 mg/L 40 mg/L Magnesium noncarbonate hardness, MNH 77 mg/L Step 3. Estimate the lime- and soda-ash requirements by the excess lime-soda ash process. Lime dosage carb. acid CCH 2(MCH) MNH excess lime (148 248 2(0) 77 60) mg/L as CaCo3 533 mg/L as CaCO3 or

533 74/100 [mg/L as Ca(OH)2] 394 mg/L as Ca(OH)2

and soda ash dosage for excess lime-soda ash (Y ) is Y CNH MNH (40 77) mg/L as CaCo3 117 mg/L as CaCO3 or

87 mg/L as Ca(OH)2

13 ION EXCHANGE Ion exchange is a reversible process. Ions of a given species are displaced from an insoluble solid substance (exchange medium) by ions of another species dissolved in water. In practice, water is passed through the exchange medium until the exchange capacity is exhausted and then it is regenerated. The process can be used to remove color, hardness (calcium and magnesium), iron and manganese, nitrate and other inorganics, heavy metals, and organics. Exchange media which exchange cations are called cationic or acid exchangers, while materials which exchange anions are called anionic or base exchangers. Common cation exchangers used in water softening are zeolite, greensand, and polystyrene resins. However, most ion exchange media are currently in use as synthetic materials. Synthetic ion exchange resins include four general types used in water treatment. They are strong- and weak-acid cation exchangers and strong- and weak-base anion exchangers. Examples of exchange reactions as shown below (Schroeder, 1977): Strong acidic 2R—SO3H Ca2 ↔ (R—SO3)2Ca 2H

(6.95)

2R—SO3Na Ca2 ↔ (R—SO3)2Ca 2Na

(6.96)

2R—COOH Ca2 ↔ (R—COO)2Ca 2H

(6.97)

Weak acidic

2R—COONa Ca

↔ (R—COO)2Ca 2Na

(6.98)

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.389

Strong basic 2R—X3NOH SO42 4 (R—X3N)2SO4 2OH 2R—X3NCl

SO42

4 (R—X3N)2SO4 2Cl

(6.99) (6.100)

Weak basic 2R—NH3OH SO42 4 (R—NH3)2SO4 2OH

(6.101)

2R—NH3Cl SO 42 4 (R—NH3)2SO4 2Cl

(6.102)

where in each reaction, R is a hydrocarbon polymer and X is a specific group, such as CH2. The exchange reaction for natural zeolites (Z) can be written as Ca2 Na 2Z μ Mg 2 Fe 2

Ca2 Mg 2 ∂ Z 2Na Fe 2

(6.103)

In the cation-exchange water softening process, the hardness-causing elements of calcium and magnesium are removed and replaced with sodium by a strong-acid cation resin. Ion-exchange reactions for softening may be expressed as

Na2R

(HCO3)2 Ca f SO4 Mg Cl2

2 NaHCO3 Ca S Mg f R μ Na2SO4 2 Na Cl

(6.104)

where R represents the exchange resin. They indicate that when a hard water containing calcium and magnesium is passed through an ion exchanger, these metals are taken up by the resin, which simultaneously gives up sodium in exchange. Sodium is dissolved in water. The normal rate is 6 to 8 gpm/ft2 (350 to 470 m/d) of medium. After a period of operation, the exchanging capacity would be exhausted. The unit is stopped from operation and regenerated by backwashing with sodium chloride solution and rinsed. The void volume for backwash is usually 35 to 45 percent of the total bed volume. The exchange capacity of typical resins are in the range of 2 to 10 eq/kg. Zeolite cation exchangers have the exchange capacity of 0.05 to 0.1 eq/kg (Tchobanoglous and Schroeder 1985). During regeneration, the reaction can be expressed as: Ca Ca f R 2 NaCl S Na2R f Cl 2 Mg Mg

(6.105)

The spherical diameter in commercially available ion exchange resins is of the order of 0.04 to 1.0 mm. The most common size ranges used in large treatment plant are 20 to 50 mesh (0.85 to 0.3 mm) and 50 to 100 mesh (0.3 to 0.15 mm) (James M. Montgomery Consulting Engineering 1985). Details on the particle size and size range, effective size, and uniform coefficient are generally provided by the manufacturers. The affinity of exchanges is related to charge and size. The higher the valence, the greater affinity and the smaller the effective size, the greater the affinity. For a given sense of similar ions, there is a general order of affinity for the exchanger. For synthetic resin exchangers, relative affinities of common ions increase as shown in Table 6.5. The design of ion exchange units is based upon ion exchange equilibria. The generalized reaction equation for the exchange of ions A and on a cation exchange resin can be expressed as nRA Bn ↔ Rn BnnA where

R an anionic group attached to exchange resin A, Bn ions in solution

(6.106)

PUBLIC WATER SUPPLY 1.390

CHAPTER 1.6

TABLE 6.5 Selectivity Scale for Cations on Eight Percent Cross-Linked StrongAcid Resin and for Anions on Strong-Base Resins Cation

Li H Na UO2 NH4 K Rb Cs Mg2 Zn2 Co2 Cu2 Cd2 Ni2 Be2 Mn2 Pb2 Ca2 Sr2 Ag Pb2 Ba2 Ra2

Selectivity 1.0 1.3 2.0 2.5 2.6 2.9 3.2 3.3 3.3 3.5 3.7 3.8 3.9 3.9 4.0 4.1 5.0 5.2 6.5 8.5 9.9 11.5 13.0

Anion

Selectivity

HPO2 4 CO32

0.01 0.03 0.06 0.1 0.15 0.2 0.4 0.65 1.0 1.0 1.3 1.3 1.6 3 4 8 9.1 17 100

OH (type I) F SO2 4 CH3COO HCO3 OH(type II) BrO 3 Cl CN NO2 HSO4 Br NO3 I SO2 4 SeO2 4 CrO2 4

Sources: James M. Montgomery Consulting Engineering (1985), Clifford (1990).

The equilibrium expression for this reaction is KASB where

n n [R qBCA n B ][A ] n qACB [R A ] [Bn]

(6.107)

KA S B KAB selectivity coefficient, a function of ionic strength and is not a true constant. [R A], [R Bn] mole fraction of A and B exchange resin, respectively, overbars represent the resin phase, or expressed as [ A], and [ B] [A], [Bn] concentration of A and B in solution, respectively, mol/L qA, qB concentration of A and B on resin site, respectively, eq/L CA, CB concentration of A and B in solution, respectively, mg/L

The selectivity constant depends upon the valence, nature, and concentration of the ion in solution. It is generally determined in laboratory for specific conditions measured. For monovalent/monovalent ion exchange process such as HR Na 4 NaR H The equilibrium expression is (by Eq. (6.107)) KHRSNaR

qNaCH [ NaR][H] qHCNa [HR][Na]

and for divalent/divalent ion exchange processes such as 2 (NaR) Ca2 4 Ca2 R2 2Na

(6.108)

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.391

then using Eq. (6.107) qCaC 2Na [Ca2R][Na]2 2 2 2 [ Na R] [Ca ] q NaCCa

K NaRSCaR

(6.109)

Anderson (1975) rearranged Eq. (6.107) using a monovalent/monovalent exchange reaction with concentration units to equivalent fraction as follows: 1. In the solution phase: Let C total ionic concentration of the solution, eq/L. The equivalent ionic fraction of ions A and B in solution will be [A]/C or [A] CXA X A

(6.110)

X B

(6.111)

[B ]/C

[B ]

CX B

then [A] [B] 1 [A] 1 [B]

(6.112)

2. In the resin phase: let C total exchange capacity of the resin per unit volume, eq/L. Then we get XA [RA]/C

[RA] CX A

(6.113)

equivalent fraction of the A ion in the resin and XB [RB]/C

[RB] CX B

(6.114)

also XA XB 1 XA 1 XB

(6.115)

Substitute Eqs. (6.110), (6.111), (6.113), and (6.114) into Eq. (6.107) which yields: K BA

[CXB][CXA] [CXA][CXB]

or KAB XB XA

[CXB][CXA] [CXA][CXB]

K BA

XB XA

(6.116)

Substituting for Eqs. (6.112) and (6.115) in Eq. (6.116) gives XB 1 XB

KBA

XB 1 XB

(6.117)

If the valence is n, Eq. (6.117) will become n1 XBn B C b K a A C (1 XBn)n (1 XBn)n

XBn

(6.118)

PUBLIC WATER SUPPLY 1.392

CHAPTER 1.6

EXAMPLE 1:

Determine the meq/L of Ca2 if Ca2 concentration is 88 mg/L in water.

Solution: Step 1. Determine equivalent weight (EW) EW molecular weight/electrical charge 40/2 20 grams per equivalent weight (or mg/meq) Step 2. Compute meq/L meq/L (mg/L)/EW (88 mg/L)/(20 mg/meq) 4.4 A strong-base anion exchange resin is used to remove nitrate ions from well water which contain high chloride concentration. Normally bicarbonate and sulfate is present in water (assume they are negligible). The total resin capacity is 1.5 eq/L. Find the maximum volume of water that can be treated per liter of resin. The water has the following composition in meq/L:

EXAMPLE 2:

Ca2 1.4 Mg2 0.8 Na 2.6 Total cations 4.8

Cl 3.0 SO 2 0.0 4 NO 3 1.8 Total anions 4.8

Solution: Step 1. Determine the equivalent fraction of nitrate in solution XNO3 1.8/4.8 0.38 Step 2. Determine selective coefficient for sodium over chloride from Table 6.5. NO3 4/1 4 KCl

Step 3. Compute the theoretical resin available for nitrate ion by Eq. (6.117) XNO3 1 XNO3

0.38 2.45 1 0.38

XNO3 0.71 It means that 71 percent of resin sites will be used. Step 4. Compute the maximum useful capacity Y Y 1.5 eq/L 0.71 1.065 eq/L 1065 meq/L Step 5. Compute the volume of water (V) that can be treated per cycle 1065 meq/L of resin 1.8 meq/L of water 592 L of water/L of resin 1 gal 28.32 L L 592 L 3.785 L 1 ft3 4429 gal of water/ft3 of resin

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.393

A strong-acid cation exchanger is employed to remove calcium hardness from water. Its wet-volume capacity is 2.0 eq/L in the sodium form. If calcium concentrations in the influent and effluent are 44 mg/L (2.2 meq/L) and 0.44 mg/L, respectively, find the equivalent weight (meq/L) of the component in the water if given the following:

EXAMPLE 3:

Cations

Anions

Ca Mg2+ Na+ Total cations

HCO3– –

2.2 1.0 3.0 6.2

Cl SO 2– 4 Total anions

2.9 3.1 0.2 6.2

Solution: Step 1. Determine the equivalent fraction of Ca2 XCa2 2.2/6.2 0.35 Step 2. Find the selectivity coefficient K for calcium over sodium from Table 6.5 Ca K Na 5.2/2.0 2.6

Step 3. Compute the theoretical resin composition with respect to the calcium ion C/C 6.2/6.2 1.0 Using Eq. (6.118), n 2 XCa2 (1 XCa2)2

2.6

0.35 2.15 (1 0.35)2

XCa2 0.51 This means that a maximum of 51 percent of the resin sites can be used with calcium ions from the given water. At this point, the water and resin are at equilibrium with each other. Step 4. Compute the limiting useful capacity of the resin (Y ) Y 2.0 eq/L 0.51 1.02 eq/L (or 1020 meq/L) Step 5. Compute the maximum volume (V ) of water that can be treated per cycle: 1020 meq/L of resin (for Ca) 2.2 meq/L of calcium in water 464 L of water/L of resin

3472 gal of water/ft3 of resin 13.1 Leakage Leakage is defined as the appearance of a low concentration of the undesirable ions in the column effluent during the beginning of the exhaustion. It comes generally from the residual ions in the bottom resins due to incomplete regeneration. Leakage will occur for softening process. A water softening column is usually not fully regenerated due to inefficient use of salt to completely regenerate the resin on the sodium form. The leakage depends upon the ionic composition of the bed bottom and the composition of the influent water.

PUBLIC WATER SUPPLY 1.394

CHAPTER 1.6

A detailed example of step-by-step design method for a fix bed ion exchange column for water softening is given by Benefield et al. (1982). EXAMPLE: The bottom of the water softener is 77 percent in the calcium form after regeneration. The strong-acid cation resin has a total capacity of 2.0 eq/L and selective coefficient for calcium over sodium is 2.6. Determine the initial calcium leakage the water composition has as follows (in meq/L):

Ca2 0.4

Cl 0.4

Mg2 0.2

SO 42 0.4

Na 1.2

HCO3 1.0

Total cations 1.8

Total anions 1.8

Solution: Step 1. Determine the equivalent fraction of calcium ion from the given: XCa2 0.77 C 2.0 eq/L C 2 meq/L 0.002 eq/L Ca K Na 2.6

Using Eq. (6.118) XCa 0.77 2 bc 2.6a d 0.002 (1 XCa)2 (1 0.77)2 XCa (1 XCa)2

0.0056

Step 2. Calculate the initial calcium leakage y y 0.0056 c 0.0056 1.8 meq/L 0.010 meq/L 20 mg as Ca meq meq L 0.20 mg/L as Ca 100 as CaCO3 0.20 mg/L as Ca 40 as Ca 0.50 mg/L as CaCO3 0.010

13.2 Nitrate Removal Nitrate, NO3, is a nitrogen—oxygen ion that occurs frequently in nature as the result of interaction between nitrogen in the atmosphere and living things on earth. It is a portion of the nitrogen cycle. When plant and animal proteins are broken down, ammonia and nitrogen gas are released. Ammonia is subsequently oxidized to nitrite (NO2 ) and nitrate by bacterial action (Fig. 2.1). High concentration of nitrate in drinking water may cause methemoglobinemia, especially for infants. Nitrates interfere with the body’s ability to take oxygen from the air and distribute it to body cells. The United States EPA’s standard for nitrate in drinking water is 10 mg/L. In Illinois, the water

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.395

purveyor must furnish special bottle water (low nitrate) to infants if the nitrate level of the finished water exceeds the standard. The sources of nitrates to water are due to human activities. Agricultural activities, such as fertilizer application and animal feedlots, are major contributors of nitrate and ammonia to be washed off (runoff) and percolate to soil with precipitation. The nitrates that are polluted can subsequently flow into surface waters (streams and lakes) and into groundwater. Municipal sewage effluents improperly treat domestic wastewaters, and draining of septic tanks may cause nitrate contamination in the raw water source of a water works. Identification of pollution sources and protective alternatives should be taken. If the nitrate concentration in the water supply is excessive, there are two approaches for solving the problem, i.e. treatment for nitrate removal and nontreatment alternatives. Nontreatment alternatives may include (1) use new water source, (2) blend with low nitrate waters, (3) connect to other supplier(s), and (4) organize in a regional system. Nitrate is not removed by the conventional water treatment processes, such as coagulation— flocculation, sedimentation, filtration, activated carbon adsorption, and disinfection (chlorination and ozonation). Some treatment technologies, such as ion exchange, reverse osmosis, electrodialysis, microbial denitrification, and chemical reduction, can remove nitrates from drinking water. An ion exchange process design is given as the following examples. Two types (fixed bed and continuous) of ion exchangers are used. The fixed bed exchange is used mostly for home and industry uses and is controlled by a flow totalizer with an automatic regeneration cycle at approximately 75 to 80 percent of the theoretical bed capacity. Continuous ion exchangers are employed by larger installations, such as water works, that provide continuous product water and require minimum bed volume. A portion of the resin bed is withdrawn and regenerated outside of the main exchange vessel. In the design of an ion exchange system for nitrate removal, raw water quality analyses and pilot testing are generally required. The type of resin and resin capacity, bed dimensions, and regenerant requirement must be determined. Basic data, such as the design flow rate through the exchanger, influent water quality, total anions, and suggested operating conditions for the resin selected (from the manufacturer) must be known. Analysis of water quality may include nitrate, sulfate, chloride, bicarbonate, calcium carbonate, iron, total suspended solids, and total organic carbon. For Duolite A-104, for example, the suggested design parameters are listed below (Diamond Shamrock Co., 1978): Parameters Minimum bed depth Backwash flow rate Regenerant dosage Regenerant concentration Regenerant temperature Regenerant flow rate Rinse flow rate Rinse volume Service flow rate Operating temperature pH limitation

Recommended values 30 in 2–3 gpm/ft2 15–18 lb sodium chloride (NaCl) per ft3 resin 10–12% NaCl by weight Up to 120F or 49C 0.5 gpm/ft3 2 gpm/ft3 50–70 gal/ft3 Up to 5 gpm/ft3 Salt form, up to 180F or 85C None

In design processes, resin capacity, bed dimensions, and regenerant requirements must be computed. Resin capacity determines the quantity of resin needed in the ion exchanger and is computed from a pilot study. Data is provided by the manufacturer. For example, the operating capacity of Duloite A-104 resin for nitrate removal is quite dependent upon the nitrate, sulfate, and total anion concentrations to calculate the corrected resin capacity. The raw or uncorrected resin capacity is determined by using the manufacturer’s graph (Fig. 6.7). This capacity must be adjusted downward to reflect the presence of sulfate in the water (Fig. 6.8), because sulfate anions will be exchanged before nitrate (USEPA 1983).

PUBLIC WATER SUPPLY 1.396

CHAPTER 1.6

FIGURE 6.7

FIGURE 6.8

Relationship of NO3/TA and unadjusted resin capacity for A-104 resin (USEPA, 1983).

Correction factor for A-104 resin (USEPA, 1983).

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

FIGURE 6.9

1.397

Bed expansion curve for A-104 resin (modified from USEPA, 1983).

Once the adjusted resin capacity is determined for the water to be treated, the required bed volume of the ion exchange resin can be calculated. The bed volume is the amount of nitrate that must be removed in each cycle divided by the adjusted resin capacity. Using this bed volume and minimum depth requirement, the surface area of resin can be calculated. A standard size containment vessel can be selected with the closest surface area. Using this standard size, the depth of the resin can be recalculated.The final height of the vessel should be added to a bed expansion factor during backwashing with a design temperature selected (Fig. 6.9). Once the bed volume and dimensions are computed, the regeneration system can be designed. The regeneration system must determine: (1) the salt required per generation cycle, (2) the volume of brine produced per cycle, (3) the total volume of the brine storage tank, and (4) the time needed for the regeneration process. The design of an ion exchanger for nitrate removal is illustrated in the following three examples. EXAMPLE 1: A small public water system has the maximum daily flow rate of 100,000 gallons per day (gpd), maximum weekly flow of 500,000 gallons per week, and maximum nitrate concentration of 16 mg/L. The plant treats 100,000 gallons of water and operates only 7 h per day and 5 days per week (assume sufficient storage capacity for weekend demand). Finished water after ion exchange process with 0.6 mg/L of nitrate-nitrogen (NO3-N) will be blended with untreated water at 16 mg/L of NO3-N to produce a finished water of 8 mg/L or less of NO3-N. The NO3-N standard is 10 mg/L. Determine the flow rate of ion exchanger in gpm and the blending rate.

Solution: Step 1. Compute the quantity of water that passes the ion exchanger, q1 Let untreated flow q2 and C, C1, and C2 NO3-N concentrations in finished, treated (ion exchanger), and untreated waters, respectively.

PUBLIC WATER SUPPLY 1.398

CHAPTER 1.6

Then C 8 mg/L C1 0.6 mg/L C2 16 mg/L by mass balance C1q1 C2q2 100,000 C 0.6q1 16q2 100,000 8

(a)

q2 100,000 q1

(b)

since

Substituting (b) into (a) 0.6q1 16 (100,000 q1) 800,000 15.4q1 800,000 q1 51,950 (gpd) Step 2. Compute flow rate in gpm to the exchanger (only operates 7 hours per day) 1 day gal 24 h d 7h 1440 min 124 gpm

q1 51,950

Step 3. Compute blending ratio, br br

51,950 gpd 100,000 gpd

0.52 The given conditions are the same as in Example 1. The anion concentration of the influent of the ion exchanger are: NO3-N 16 mg/L, SO4 48 mg/L, Cl 28 mg/L, and HCO3 72 mg/L. Determine the total quantity of nitrate to be removed daily by the ion exchanger and the ratios of anions.

EXAMPLE 2:

Solution: Step 1. Compute total anions and ratios

Anions NO3-N SO4 Cl HCO3 Total

Concentration, mg/L

Milliequivalent weight, mg/meq

Concentration, meq/L

16 48 28 72

14.0 48.0 35.5 61.0

1.14 1.00 0.79 1.18 4.11

In the last column above for NO3-N % 1.14 100/4.11 27.7

Percentage 27.7 24.4 19.2 28.7

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.399

Step 2. Determine nitrate to be removed daily. The total amount of nitrates to be removed depends on the concentrations of influent and effluent, Ci and Ce and can be determined by nitrates removed (meq/d) [(Ci Ce) meq/L] q1(L/d) Ci 1.14 meq/L

Since

Ce 0.6 mg/L 14 mg/meq 0.043 meq/L Nitrates removed (1.14 0.043) meq/L 51,950 gpd 3.785 L/gal 215,700 meq/d If the ion exchanger is going to operate with only one cycle per day, the quantity of nitrate to be removed per cycle is 215,700 meq. Design a 100,000 gpd ion exchanger for nitrate removal using the data given in Examples 1 and 2. Find resin capacity, resin bed dimension, and regenerent requirements (salt used, brine production, the total volume of the brine storage tank, and regeneration cycle operating time) for the ion exchanger.

EXAMPLE 3:

Solution: Step 1. Determine the uncorrected (unadjusted) volume of resin needed (a) In Example 2, the nitrate to total anion ratio is 27.7%. Using 27.7% in Fig. 6.7, we obtain the uncorrected resin capacity of 0.52 meq/(NO3 mL) resin. (b) In Example 2, the sulfate to total anion ratio is 24.4%. Using 25% in Fig. 6.8, we obtain the correction factor of resin capacity due to sulfate is 0.70. (c) The adjusted resin capacity is down-warded to 0.52 meq/mL 0.70 0.364 meq/mL (d) Convert meq/mL to meq/ft3 0.364 meq/mL 3785 mL/gal 7.48 gal/ft3 10,305 meq/ft3 Step 2. Determine the bed volume, BV In Example 2, nitrate to be removed in each cycle is 215,700 meq BV 215,700 meq 10,305 meq/ft3 20.93 ft3 Step 3. Check the service flow rate, SFR In Example 1, the ion exchange unit flow rate 124 gpm SFR 124 gpm/BV 124 gpm/20.93 ft3 5.9 gpm/ft3 This value exceeds the manufacturer’s maximum SFR of 5 gpm/ft3. In order to reduce the SFR, we can modify it either by (a) increasing the exchanger operating time per cycle, or (b) increasing the bed volume.

PUBLIC WATER SUPPLY 1.400

CHAPTER 1.6

(a) Calculate adjusted service time, AST 3

5.9 gpm/ft 7h cycle 5.0 gpm/ft 3 8.3 h/cycle

AST

Calculate the adjusted flow rate for the new cycle In Example 1, q1 51,950 gpd or 51,950 gal/cycle then, flow rate q1/AST 51,950 gal/cycle/(8.3 h/cycle 60 min/h) 104 gpm Alternatively (b) Increase the bed volume and retain the initial unit flow rate Adjusted BV 20.93 ft3 (5.9 gpm/ft3/5.0 gpm/ft3) 24.7 ft3 25 ft3

say

In comparison between alternatives (a) and (b), it is more desirable to increase BV using 7 h/d operation. Thus BV 25 ft3 is chosen. Step 4. Determine bed dimensions (a) Select the vessel size Based on the suggested design specification, the minimum bed depth (z) is 30 in or 2.5 ft. The area, A, would be A BV/z 25 ft3/2.5 ft 10.0 ft2 For a circular vessel, find the diameter A D2/4 D 24A/p 24 10.0 ft2/3.14 3.57 ft The closest premanufactured size is 3.5 ft in diameter with a corresponding cross-sectional area of 9.62 ft2. The resin height, h, would then be re-calculated as h 25 ft3/9.62 ft2 2.6 ft (b) Adjusting for expansion during backwash Since allowable backwash flow rate is 2 to 3 gpm/ft 2 . If the backwash rate is selected as 2.5 gpm/ft2 with the backwash water temperature of 50F, the percent bed expansion would be 58 percent as read from the curve in Fig. 6.9. Thus, the design height (H) of the vessel should be at least H 2.6 ft (1 0.58) 4.11 ft Note: We use 3.5 ft diameter and 4.11 ft (at least) height vessel.

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.401

Step 5. Determine regenerant requirements (a) Choice salt (NaCl) required per regeneration cycle since regenerant dosage is 15 to 18 pounds of salt per cubic foot of resin. Assume 16 lb/ft3 is selected. The amount of salt required is Salt 16 lb/ft3 25 ft3 400 lb We need 400 pounds of salt per cycle. (b) Calculate volume of brine used per cycle On the basis of the specification, the regenerant concentration is 10 to 12 percent NaCl by weight. Assume it is 10 percent, then wt. of salt 100 total wt. of brine total wt. of brine 400 lb 100/10

Salt concentration, %

4000 lb (c) Calculate total volume of brine Weight of water (4000 400) lb/cycle 3600 lb 3600 lb wt. of water Volume of water density of water 62.4 lb/ft 3 57.69 ft3 The specific weight of salt is 2.165 Volume of salt 400 lb/(62.4 lb/ft3 2.165) 2.96 ft3 Total volume of the brine 57.69 ft3 2.96 ft3 60.65 ft3 Total volume in gallons 60.65 ft3 7.48 gal/ft3 454 gal Note: The total volume of brine generated is 454 gal per cycle. The total volume of the brine storage tank should contain sufficient capacity for 3 to 4 generations. Assume 3 cycles of brine produced should be stored, then the total brine tank volume (V) would be V 454 gal/cycle 3 cycle 1362 gal (d) Calculate time required for regeneration cycle The regeneration flow rate (Q) is 0.5 gpm/ft3 resin specified by the manufacturer. Then Q 0.5 gpm/ft3 25 ft3 12.5 gpm The regeneration time (t) is the volume of the brine per cycle divided by the flow rate Q, thus t 454 gal/12.5 gal/min 36.3 min

PUBLIC WATER SUPPLY 1.402

CHAPTER 1.6

14 IRON AND MANGANESE REMOVAL Iron (Fe) and manganese (Mn) are abundant elements in the earth’s crust. They are mostly in the oxidized state (ferric, Fe3, and Mn4) and are insoluble in natural waters. However, under reducing conditions (i.e., where dissolved oxygen is lacking and carbon dioxide content is high), appreciable amounts of iron and manganese may occur in surface water, groundwater, and in water from the anaerobic hypolimnion of stratified lakes and reservoirs. The reduced forms are soluble divalent ferrous (Fe2) and manganous (Mn2) ions that are chemically bound with organic matter. Iron and manganese get into natural water from dissolution of rocks and soil, from acid mine drainage, and from corrosion of metals. Typical iron concentrations in ground water are 1.0 to 10 mg/L, and typical concentrations in oxygenated surface waters are 0.05 to 0.2 mg/L. Manganese exists less frequently than iron and in smaller amounts. Typical manganese values in natural water range from 0.1 to 1.0 mg/L (James M. Montgomery Consulting Engineers, 1985). Voelker (1984) reported that iron and manganese levels in groundwaters in the American Bottoms area of southwestern Illinois ranged from 0.01 to 82.0 mg/L and 0.01 to 4.70 mg/L, respectively, with mean concentrations of 8.4 mg/L and 0.56 mg/L, respectively. Generally, iron and manganese in water are not a health risk. However, in public water supplies they may discolor water and cause tastes and odors. Elevated iron and manganese levels also can cause stains in plumbing, laundry, and cooking utensils. Iron and manganese may also cause problems in water distribution systems because metal depositions may result in pipe encrustation and may promote the growth of iron bacteria which may in turn cause tastes and odors. Iron and manganese may also cause difficulties in household ion exchange units by clogging and coating the exchange medium. To eliminate the problems caused by iron and manganese, the U.S. Environmental Protection Agency (1987) has established secondary drinking water standards for iron at 0.3 mg/L and for manganese at 0.05 mg/L. The Illinois Pollution Control Board (IPCB, 1990) has set effluent standards of 2.0 mg/L for total iron and 1.0 mg/L for total manganese. It is considered that iron and manganese are essential elements for plants and animals. The United Nations Food and Agriculture Organization recommended maximum levels for iron and manganese in irrigation water are 5.0 and 0.2 mg/L, respectively. The techniques for removing iron and manganese from water are based on the oxidation of relatively soluble Fe(II) and Mn(II) to the insoluble Fe(III) and Mn(III, IV) and the oxidation of any organic-complex compounds. This is followed by filtration to remove the Fe(III) and processes for iron and manganese removal are discussed elsewhere (Rehfeldt et al., 1992). Four major techniques for iron and manganese removal from water are: (1) oxidation–precipitation– filtration, (2) manganese zeolite process, (3) lime softening–settling–filtration, and (4) ion exchange. Aeration–filtration, chlorination–filtration, and manganese zeolite process are commonly used for public water supply.

14.1 Oxidation Oxidation of soluble (reduced) iron and manganese can be achieved by aeration, chlorine, chlorine dioxide, ozone, potassium permanganate (KMnO4) and hydrogen peroxide (H2O2). The chemical reactions can be expressed as follows: For aeration (Jobin and Ghosh, 1972; Dean, 1979) 1 1 Fe 2 4O2 2OH 2H 2O S Fe(OH)3

(6.119)

Mn2 O2 S MnO2 2e

(6.120)

In theory, 0.14 mg/L of oxygen is needed to oxidize 1 mg/L of iron and 0.29 mg/L of oxygen for each mg/L of manganese.

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.403

For chlorination (White, 1972): 2Fe (HCO3)2 Cl2 Ca(HCO3)2 → 2Fe(OH)3 CaCl2 6CO2

(6.121)

MnSO4 Cl2 4NaOH → MnO2 2NaCl Na2SO4 2H2O

(6.122)

The stoichiometric amounts of chlorine needed to oxidize each mg/L of iron and manganese are 0.62 and 1.3 mg/L, respectively. For potassium permanganate oxidation (Ficek 1980): 3Fe2 MnO4 4H → MnO2 2Fe3 2H2O

(6.123)

2MnO4 2H2O → 5MnO2 4H

(6.124)

3Mn

In theory, 0.94 mg/L of KMnO4 is required for each mg/L of soluble iron and 1.92 mg/L for one mg/L of soluble manganese. In practice, a 1 to 4 percent solution of KMnO4 is fed in at the low-lift pump station or at the rapid-mix point. It should be totally consumed prior to filtration. However, the required KMnO4 dosage is generally less than the theoretical values (Humphrey and Eikleberry 1962, Wong 1984). Secondary oxidation reactions occur as Fe2 MnO → Fe3 Mn2O3 Mn

MnO2 2H2O → Mn2O3 X(H2O)

(6.125) (6.126)

and 2Mn2 MnO2 → Mn3O4 X(H2O) MnO Mn2O3 X(H2O)

(6.127a) (6.127b)

For hydrogen peroxide oxidation (Kreuz, 1962): 1. Direct oxidation 3Fe2 H2O2 2H → 2Fe3 2H2O

(6.128)

2. Decomposition to oxygen 1

H2O2 → 2 O2 H2O

(6.129)

Followed by oxidation 1

3Fe2 2 O2 H2O → 2Fe3 2OH

(6.130)

In either case, 2 moles of ferrous iron are oxidized per mole of hydrogen peroxide, or 0.61 mg/L of 50 percent H2O2 oxidizes 1 mg/L of ferrous iron. For oxidation of manganese by hydrogen peroxide, the following reaction applies (H. M. Castrantas, FMC Corporation, Princeton, NJ, personal communication). 3Mn2 H2O2 2H 2e → 2Mn3 e 2H2O

(6.131)

To oxidize 1 mg/L of soluble manganese, 1.24 mg/L of 50 percent H2O2 is required. Studies on ground water by Rehfeldt et al. (1992) reported that oxidant dosage required to meet the Illinois recommended total iron standard follow the order of Na, O, Cl as Cl2 KMnO4 H2O2. The amount of residues generated by oxidants at the critical dosages follow the order of KMnO4 NaOCl H2O2, with KMnO4 producing the largest, strongest, and most dense flocs.

PUBLIC WATER SUPPLY 1.404

CHAPTER 1.6

For ozonation: It is one of the strongest oxidants used in the water industry for disinfection purposes. The oxidation potential of common oxidants relative to chlorine is as follows (Peroxidation Systems, 1990): Fluoride 2.32 Hydroxyl radical 2.06 Ozone 1.52 Hydrogen peroxide 1.31 Potassium permanganate 1.24 Chlorine 1.00 Ozone can be very effective for iron and manganese removal. Because of its relatively high capital costs and operation and maintenance costs, the ozonation process is rarely employed for the primary purpose of oxidizing iron and manganese. Since ozone is effective in oxidizing trace toxic organic matter in water, pre-ozonation instead of pre-chlorination is becoming popular. In addition, many water utilities are using ozone for disinfection purposes. Ozonation can be used for two purposes: disinfection and metal removal. Manganese zeolite process. In the manganese zeolite process, iron and manganese are oxidized to the insoluble form and filtered out, all in one unit, by a combination of sorption and oxidation. The filter medium can be manganese greensand, which is a purple-black granular material processed from glauconitic greens, and/or a synthetic formulated product. Both of these compositions are sodium compounds treated with a manganous solution to exchange manganese for sodium and then oxidized by KMnO4 to produce an active manganese dioxide. The greensand grains in the filter become coated with the oxidation products. The oxidized form of greensand then adsorbs soluble iron and manganese, which are subsequently oxidized with KMnO4. One advantage is that the greensand will adsorb the excess KMnO4 and any discoloration of the water. Regenerative-Batch Process. The regenerative-batch process uses manganese treated greensand as both the oxidant source and the filter medium. The manganese zeolite is made from KMnO4-treated greensand zeolite. The chemical reactions can be expressed as follows (Humphrey and Eikleberry, 1962; Wilmarth, 1968): Exchange: NaZ Mn2 → MnZ 2Na

(6.132)

MnZ KMnO4 → Z MnO2 K

(6.133)

Fe3 Fe2 # # S Z MnO3 Mn3 Z MnO2 Mn2 Mn4

(6.134)

Z MnO3 KMnO4 → Z MnO2

(6.135)

Generation:

Degeneration:

Regeneration:

where NaZ is greensand zeolite and Z MnO2 is manganese zeolite. As the water passes through the mineral bed, the soluble iron and manganese are oxidized (degeneration). Regeneration is required after the manganese zeolite is exhausted. One of the serious problems with the regenerative-batch process is the possibility of soluble manganese leakage. In addition, excess amounts of KMnO4 are wasted, and the process is not economical for water high in metal content. Manganese zeolite has an exchange capacity of 0.09 lb of iron or manganese per cubic foot of material, and the flow rate to the exchanger is usually 3.0 gpm/ft3 Regeneration needs approximately 0.18 lb KMnO4 per cubic foot of zeolite.

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.405

Continuous Process. For the continuous process, 1 to 4 percent KMnO4 solution is continuously fed ahead of a filter containing anthracite be (6 to 9 in. thick), manganese-treated greensand (24 to 30 in), and gravel. The system takes full advantage of the higher oxidation potential of KMnO4 as compared to manganese dioxide. In addition, the greensand can act as a buffer. The KMnO4 oxidizes iron, manganese, and hydrogen sulfide to the insoluble state before the water reaches the manganese zeolite bed. Greensand grain has a smaller effective size than silica sand used in filters and can result in comparatively higher head loss. Therefore, a layer of anthracite is placed above the greensand to prolong filter runs by filtering out the precipitate. The upper layer of anthracite operates basically as a filter medium. When iron and manganese deposits build up, the system is backwashed like an ordinary sand filter. The manganese zeolite not only serves as a filter medium but also as a buffer to oxidize any residual soluble iron and manganese and to remove any excess unreacted KMnO4. Thus a KMnO4 demand test should be performed. The continuous system is recommended for waters where iron predominates, and the intermittent regeneration system is recommended for groundwater where manganese predominates (Inversand Co., 1987). Theoretically, how many mg/L each of ferrous iron and soluble manganese can be oxidized by 1 mg/L of potassium permanganate?

EXAMPLE 1:

Solution: Step 1. For Fe2, using Eq. (6.123) 3Fe2 4H KMnO4 S MnO2 2Fe3 K 2H2O MW 3 55.85

39.1 54.94 4 16

167.55

158.14

X mg/L

1 mg/L

By proportion 167.55 X 1 158.14 X 1.06 mg/L Step 2. For Mn2, using Eq. (6.124) 3Mn2 2H2O 2KMnO4 S 5MnO2 4H 2K M1W1 3 54.94 164.82 y

2 158.14 316.28 1 mg/L

then y

164.82 0.52 mg/L 316.28

EXAMPLE 2: A groundwater contains 3.6 mg/L of soluble iron and 0.78 mg/L of manganese. Find the dosage of potassium permanganate required to oxidize the soluble iron and manganese.

Solution: From Example 1 the theoretical potassium permanganate dosage are 1.0 mg/L per 1.06 mg/L of ferrous iron and 1.0 mg/L per 0.52 mg/L of manganese. Thus 1.0 0.78 mg/L 1.0 3.6 mg/L 1.06 0.52 4.9 mg/L

KMnO4 dosage

PUBLIC WATER SUPPLY 1.406

CHAPTER 1.6

15 ACTIVATED CARBON ADSORPTION 15.1 Adsorption Isotherm Equations The Freundlich isotherm equation is an empirical equation which gives an accurate description of adsorption of organic adsorption in water. The equation under constant temperature equilibrium is (USEPA, 1976; Brown and LeMay, 1981; Lide, 1996) qe KC1/n e

(6.136)

log qe log K (1/n) log Ce

(6.136a)

where qe quantity of absorbate per unit of absorbent, mg/g Ce equilibrium concentration of adsorbate in solution, mg/L K Freundlich absorption coefficient, (mg/g) (L/mg)1/n n empirical coefficient The constant K is related to the capacity of the absorbent for the absorbate. 1/n is a function of the strength of adsorption. The molecule that accumulates, or adsorbs, at the surface is called an adsorbate; and the solids on which adsorption occurs is called adsorbent. Snoeyink (1990) compiled the values of K and 1/n for various organic compounds from the literature which is listed in Appendix D. From the adsorption isotherm, it can be seen that, for fixed values of K and Ce, the smaller the value of 1/n, the stronger the adsorption capacity. When 1/n becomes very small, the adsorption tends to be independent to Ce. For fixed values of Ce and 1/n, the larger the K value, the greater the adsorption capacity qe. Another adsorption isotherm developed by Langmuir assumed that the adsorption surface is saturated when a monolayer has been absorbed. The Langmuir adsorption model is a b Ce 1 b Ce

(6.137)

Ce Ce 1 qe ab a

(6.138)

qe or where a empirical coefficient b saturation coefficient, m3/g

Other terms are the same as defined in the Freundlich model. The coefficient a and b can be obtained by plotting Ce/qe versus Ce on arithmetic paper from the results of a batch adsorption test with Eq. (6.138). Adsorption isotherm can be used to roughly estimate the granular activated carbon (GAC) loading rate and its GAC bed life. The bed life Z can be computed as Z where

(qe)0 r (C0 C1)

(6.139)

Z bed life, L H2O/L GAC (qe)0 mass absorbed when Ce C0, mg/g of GAC apparent density of GAC, g/L C0 influent concentration, mg/L C1 average effluent concentration for entire column run, mg/L

C1 would be zero for a strongly absorbed compound that has a sharp breakthrough curve, and is the concentration of the nonadsorbable compound presented.

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.407

The rate at which GAC is used or the carbon usage rate (CUR) is calculated as follows: CUR (g/L)

C0 – C1 (qe)0

(6.140)

Contact Time. The contact time of GAC and water may be the most important parameter for the design of a GAC absorber. It is commonly used as the empty-bed contact time (EBCT) which is related to the flow rate, depth of the bed, and area of the GAC column. The EBCT is calculated by taking the volume (V) occupied by the GAC divided by the flow rate (Q): V Q

(6.141)

The depth of GAC bed H Q/A Loading rate

(6.142)

EBCT or EBCT

where EBCT empty-bed contact time, s V HA GAC volume, ft3 or m3 Q flow rate, cfs or m3/s Q/A surface loading rate, cfs/ft2 or m3/(m2 s) The critical depth of a bed (Hcr) is the depth which creates the immediate appearance of an effluent concentration equal to the breakthrough concentration, Cb. The Cb for a GAC column is designated as the maximum acceptable effluent concentration or the minimum datable concentration. The GAC should be replaced or regenerated when the effluent quality reaches Cb. Under these conditions, the minimum EBCT (EBCTmin) will be: EBCTmin

Hcr Q/A

(6.143)

The actual contact time is the product of the EBCT and the interparticle porosity (about 0.4 to 0.5). Powdered activated carbon (PAC) has been used for taste and odor removal. It is reported that PAC is not as effective as GAC for organic compounds removal. The PAC minimum dose requirement is D

Ci Ce qe

(6.144)

where D dosage, g/L Ci influent concentration, mg/L Ce effluent concentration, mg/L qe absorbent capacity KC1/n , mg/g e EXAMPLE 1: In a dual media filtration unit, the surface loading rate is 3 gpm/ft2. The regulatory requirement of EBCTmin is 5.5 min. What is the critical depth of GAC?

Solution: Q 3.74 gpm 3.74

gal 1 ft3 min 7.48 gal

0.5 ft3/min Q/A 0.5 ft3/min/1 ft2 0.5 ft/min

PUBLIC WATER SUPPLY 1.408

CHAPTER 1.6

From Eq. (6.143), the minimum EBCT is EBCTmin

Hcr Q/A

and rearranging Hcr EBCTmin (Q/A) 5.5 min 0.5 ft/min 2.75 ft EXAMPLE 2: A granular activated carbon absorber is designed to reduce 12 g/L of chlorobenzene to 2 g/L. The following conditions are given: K 100 (mg/g) (L/mg)1/n, 1/n 0.35 (Appendix D), and GAC 480 g/L. Determine the GAC bed life and carbon usage rate.

Solution: Step 1. Compute equilibrium adsorption capacity, (qe)0 (qe)0 KC1/n 100 (mg/g) (L/mg)1/n (0.002 mg/L)0.35 11.3 mg/g Step 2. Compute bed life, Z Assume: C 0 mg/L (qe)0 r Z (C0 C1) 11.3 mg/g 480 g/L of GAS (0.012 0.002) mg/L of water 542,400 L of water/L of GAC Step 3. Compute carbon usage rate, CUR. (C0 C1) (qe)0 (0.012 0.002) mg/L 11.3 mg/g 0.00088 g GAC/L water

CUR

The maximum contaminant level (MCL) for trichloroethylene is 0.005 mg/L. K 28 (mg/g) (L/mg)1/n, and 1/n 0.62. Compute the PAC minimum dosage needed to reduce trichloroethylene concentration from 33 g/L to MCL. EXAMPLE 3:

Solution: Step 1. Compute qe assuming PAC will equilibrate with 0.005 mg/L (Ce) trichloroethylene concentration. qe KC1/n e 28 (mg/g) (L/mg)1/n (0.005 mg/L)0.62 1.05 mg/g

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.409

Step 2. Compute required minimum dose, D D

(C0 Ce) qe 0.033 mg/L 0.005 mg/L 1.05 mg/g

0.027 g/L or

27 mg/L

16 MEMBRANE PROCESSES The use of semipermeable membranes having pore size as small as 3 Å (1 angstrom 108 cm) can remove dissolved impurities in water or wastewater. Desalting of sea water is an example. The process allowing water to pass through membranes is called “osmosis” or hyperfiltration. On the other hand, ions and solutes that pass through a membrane is a process called “dialysis” which is used in the medical field. The driving forces can be by pressure, chemical concentration, temperature, and by electrical charge. Various types and pore size of membranes incorporate the various driving forces. The membrane processes may include reverse osmosis (RO), electrodialysis (ED) or electrodialysis reversal (EDR), ultrafiltration (UF), and nanofiltration (NF). In the United States, RO and EDR have been used for water utilities ranging from 0.5 to 12 MGD (1.89 to 45.4 metric tons/d); most are located on the east and west coasts (Ionic, 1993). The electrodialysis process is when ions from a less concentrated solution pass through a membrane to a more concentrated solution under an electric current. Impurities are electrically removed from the water. Electrodialysis reversal is an ED process. A direct current transfers ions through the membranes by reversing polarity two to four times per hour. The system provides constant, automatic self-cleaning that enables improvements in treatment efficiency and less downtime for periodic cleaning. Ultrafiltration is to remove colloids and high molecular-weight material by pressure. The UF process retains nonionic material and generally passes most ionic matter depending on the molecular weight cutoff of the membrane. Nanofiltration is a process that removes all extreme fine matter and was an emerging technology in the 1980s. “Nano” means one thousand millionth (109), thus one nanometer is 1 nm 109 m. Water treated by nanofiltration can be used by special industries, such as electronics. Nowadays, the NF process has been used at full scale for treating filter backwash waters in New York. In Europe, hundreds of NF installations have been used for drinking water treatment. 16.1 Reverse Osmosis In the reverse osmosis process, the feedwater is forced by hydrostatic pressure through membranes while impurities remain behind. The purified water (permeable or product water) emerges at near atmospheric pressure. The waste (as brine) at its original pressure is collected in a concentrated mineral stream. The pressure difference is called the osmotic pressure which is a function of the solute concentration, characteristics, and temperature. The operating pressure ranges from 800 to 1000 psi (54.4 to 68 atm) for desalinating seawater and from 300 to 400 psi (20.4 to 27.2 atm) for brackish water desaltation (James M. Montgomery Consulting Engineering, 1985). The osmotic pressure theoretically varies in the same manner as the pressure of an ideal gas (James M. Montgomery Consulting Engineering, 1985; Conlon, 1990): nRT p y

(6.145a)

PUBLIC WATER SUPPLY 1.410

CHAPTER 1.6

or (Applegate 1984): p 1.12 T m where

(6.145b)

osmotic pressure, psi n number of moles of solute R universal gas constant T absolute temperature, C 273 K molar volume of water m sum of molarities of all ionic and nonionic constituent in solution

At equilibrium, the pressure difference between the two sides of the RO membranes equals the osmotic pressure difference. In low solute concentration, the osmotic pressure ( ) of a solution is given by the following equation (USEPA 1996):

Cs RT

(6.145c)

where osmotic pressure, psi Cs concentration of solutes in solution, mol/cm3 or mol/ft3 R ideal gas constant, ft lb/mol K T absolute temperature, K C 273 When dilute and concentrated solutions are separated by a membrane, the liquid tends to flow through the membrane from the dilute to the concentrated side until equilibrium is reached on both sides of the membrane. The liquid and salt passage (flux of water) through the membrane is a function of the pressure gradient (Allied Signal 1970): Fw W(P )

(6.146)

and for salt and solute flow through a membrane: Fs S(C1 C2) where

(6.147)

Fw, Fs liquid and salt fluxes across the membrane, respectively, g/(cm2 s) or lb/(ft2 s); cm3/(cm2 s) or gal/(ft2 d) W, S flux rate coefficients, empirical, depend on membrane characteristics, solute type, salt type, and temperature, s/cm or s/ft P drop in total water pressure across membrane, atm or psi osmotic pressure gradient, atm or psi P net driving force for liquid pass through membrane C1, C2 salt concentration on both sides of membrane, g/cm3 or lb/ft3 C1 C2 concentration gradient, g/cm3 or lb/ft3

The liquid or solvent flow depends upon the pressure gradient. The salt or solute flow depends upon the concentration gradient. Also, both are influenced by the membrane types and characteristics. The terms water flux and salt flux are the quantities of water and salt that can pass through the membrane per unit area per unit time. The percent of feedwater recovered is given in terms of water recovery: Recovery

Qp Qf

100%

(6.148)

where Qp product water flow, m3/d or gpm Qf feedwater flow, m3/d or gpm Water recovery for most RO plants are designed to be 75 to 80 percent for brackish water and 20 to 25 percent for seawater (James M. Montgomery Consulting Engineering 1985).

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.411

The percent of salt rejection is: Rejection 100 salt passage a1

product concentration b 100% feedwater concentration

(6.149)

Salt rejection is commonly measured by the total dissolved solids (TDS). Measurement of conductivity is sometime used in lieu of TDS. In the RO system, salt rejection can be achieved at 90 percent or more in normal brackish waters. Solute rejection varies with feedwater concentration, membrane types, water recovery rate, chemical valance of the ions in the solute, and other factors. EXAMPLE 1: A city’s water demand is 26.5 metric ton/d (7 MGD). What is the source water (feedwater) flow required for a brackish water RO, if the plant recovery rate is 78 percent?

Solution: Recovery

Qp Qf

100

Qf (26.5 ton/d) 100/78

rearranging:

34.0 ton/d 9 MGD

For a brackish water RO treatment plant, the feedwater applied is 53.0 ton/d (14 MGD) to the membrane and the product water yields 42.4 ton/day (11.2 MGD). What is the percentage of brine rate?

EXAMPLE 2:

Solution: Amount of brine produced feed rate product rate Qf Qp (53.0 42.4) ton/d 10.6 ton/d 10.6 ton/d 100% 53.0 ton/d 20%

% of brine rate

At a brackish water RO treatment plant, the total dissolved solids concentrations for the pretreated feedwater and the product water are 2860 and 89 mg/L, respectively. Determine the percentages of salt rejection and salt passage.

EXAMPLE 3:

Solution: Step 1. Compute salt rejection Salt rejection a1 a1 96.7%

product concentration b 100% feedwater concentration 89 b 100% 2680

PUBLIC WATER SUPPLY 1.412

CHAPTER 1.6

Step 2. Compute salt passage Salt passage 100 salt rejection (100 96.7)% 3.3% A pretreated feedwater to a brackish water RO process contains 2600 mg/L of TDS. The flow is 0.25 m3/s (5.7 MGD). The designed TDS concentration of the product water is no more than 450 mg/L. The net pressure is 40 atm. The membrane manufacturer provides that the membrane has a water flux rate coefficient of 1.8 106 s/m and a solute mass transfer rate of 1.2 106 m/s. Determine the membrane area required. EXAMPLE 4:

Solution: Step 1. Compute flux of water Given: P 40 atm 40 atm 101.325 kPa/atm 4053 kPa 4053 kg/m s2 W 1.8 106 s/m Using Eq. (6.146) Fw W (P ) 1.8 106 s/m (4053 kg/m s2) 7.3 103 kg/m2 s Step 2. Estimate membrane area Q Fw A rearranging: A Q/Fw (0.25 m3/s 1000 kg/m3)/(7.3 103 kg/m2 s) 34,250 m2 Step 3. Determine permeate TDS (Cp) with the above area Cp 2600 mg/L 2600 g/m3 2.6 kg/m3 Fs S(Cf Cp) QCp Fs A S(Cf Cp)A SCf A 1.2 106m/s 2.6 kg/m3 34,250 m2 Q SA 0.25 m3/s 1.2 106 m/s 34,250 m2 3 0.367 kg/m

367 g/m3 367 mg/L The TDS concentration of the product water is lower than the desired limit. For economic purposes, missing some feedwater into the product water can reduce membrane area. Step 4. Estimate blended flows Using Cp 0.367 kg/m3 as an estimated TDS concentration of the product water, the TDS in the blended water is 0.45 kg/m3.

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.413

Mass balance equation would be (Qf Qp)(0.45 kg/m3) Qf(2.6 kg/m3) Qp(0.367 kg/m3) 2.15Qf 0.083Qp 0

(a)

and Qf Qp 0.25 m3/s Qp 0.25 Qf

(b)

Substituting (b) into (a) 2.15Qf 0.083 (0.25 Qf) 0 2.233 Qf 0.02075 Qf 0.009 (m3/s) Qp 0.25 0.009 0.241 (m3/s)

then

The water supply will use a blend of 0.24 m3/s of RO product water and 0.01 m3/s of feedwater (4%). Step 5. Compute the required membrane area A

Qp fw 0.24m3/s 1000 kg/m3 7.3 103 kg/m2 # s

32,880 m2 16.2 Silt Density Index The silt density index (SDI) test involves water filtering in the direct-flow mode with no cross-flow at a constant pressure of 207 kPa (30 psig). It commonly uses disc membranes rated at 0.45 m for 15 min test with a 500 mL water sample. The SDI is calculated as (Chellam et al. 1997): ti 100 SDI a 1 t b 15 f

(6.150)

where SDI silt density index ti time initially needed to filter 500 mL of sample tf time needed to filter 500 mL at the end of the 15 min test period The SDI has been widely used as a rough estimation of the potential for colloidal fouling in nanofiltration. Generally, the feedwaters for nanofiltration (pore size 0.2 m) should have SDI of 5 or less. EXAMPLE: The time initially required to filter 500 mL of a dual-media filter effluent is 14.5 s. The time required to filter 500 mL of the same water sample at the end of the 15 min test period is 48 s. Calculate the SDI.

Solution: ti 100 100 14.5 s b SDI a1 t b a1 48 s 15 15 f 4.65

PUBLIC WATER SUPPLY 1.414

CHAPTER 1.6

17 RESIDUAL FROM WATER PLANT A water treatment plant not only produces drinking water, but is also a solids generator. The residual (solids or wastes) comes principally from clarifier basins and filter backwashes. These residuals contain solids which are derived from suspended and dissolved solids in the raw water with the addition of chemicals and chemical reactions. Depending on the treatment process employed, waste from water treatment plants can be classified as alum, iron, or polymer sludge from coagulation and sedimentation; lime sludge or brine waste from water softening; backwash wastewater and spent granular activated carbon from filtration; and wastes from iron and manganese removal process, ion exchange process, diatomaceous earth filters, microstrainers, and membranes. The residual characteristics and management of water plant sludge and environmental impacts are discussed in detail elsewhere (Lin and Green, 1987). Prior to the 1970s, residuals from a water treatment plant were disposed of in a convenient place, mostly in surface water. Under the 1972 Water Pollution Control Act, the discharge of a water plant residuals requires a National Pollution Discharge Elimination System (NPDES) permit. Residuals directly discharged into surface water is prohibited by law in most states. Proper residual (solids) handling, disposal, and/or recovery is necessary for each water work nowadays. Engineers must make their best estimate of the quantity and quality of residuals generated from the treatment units. A conventional water treatment plant treats an average flow of 0.22 m3/s (5.0 MGD). The total suspended solids (TSS) concentration in raw (river) water averages 88 mg/L. The TSS removal through sedimentation and filtration processes is 97 percent. Alum is used for coagulation/ sedimentation purpose. The average dosage of alum is 26 mg/L. Assume the aluminum ion is completely converted to aluminum hydroxide. (a) Compute the average production of alum sludge. (MW: Al 27, S 32, O 16, and H 1). (b) Estimate the sludge volume generated daily, assuming the specific gravity of 4% sludge is 1.02.

EXAMPLE 1:

Solution: Step 1. Determine quantity (q1) settled for TSS TSS 88 mg/L 88 g/m3 q1 flow TSS % removal 0.22 m3/s 88 g/m3 0.97 18.8 g/s Step 2. Determine Al(OH)3 generated, q2 The reaction formula for alum and natural alkalinity in water: (Eq. (6.51a)) Al2(SO4)3 18H2O 3Ca(HCO3)2 → 2Al(OH)3 3CaSO4 6CO2 18H2O 27 2 3(32 16 4) 18(18)

2(27 17 3)

666

156

The above reaction formula suggests that each mole of dissolved Al3 dosed will generate 2 moles of Al(OH)3 q2 alum dosage flow c 26 g/m3 0.22 m3/s 1.34 g/s

156 g Al(OH)3 d 666 g alum

156 666

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.415

Step 3. Compute total residual generated q q q1 q2 18.8 g/s 1.34 g/s 20.14 g/s 1 kg g 86,400 s 20.14 s 1000 g 1 day

1740 kg/d 1740

kg 2.194 lb day 1 kg

3818 lb/day Step 4. Estimate the sludge volume generated daily for (b) Volume 1740 kg/d (1000 kg/m3 1.02 0.04) 42.6 m3/d or

Volume 3818 lb/d (62.4 lb/ft3 1.02 0.04) 1500 ft3/d

EXAMPLE 2: A river raw water is coagulated with a dosage of 24 mg/L of ferrous sulfate and an equivalent dosage of 24 mg/L of ferrous sulfate and an equivalent dosage of lime. Determine the sludge generated per m3 of water treated.

Solution: From Eqs. (6.51f ) and (6.51g), 1 mole of ferrous sulfate reacts with 1 mole of hydrated lime and this produces 1 mole of Fe(OH)3 precipitate. MW of FeSO4 7H2O 55.85 32 4 16 7 18 277.85 MW of Fe(OH)3 55.85 3 17 106.85 by proportion sludge 106.85 0.3846 F.S. dosage 277.85 Sludge F.S. dosage 0.3846 24 mg/L 0.3846 9.23 mg/L mg 1g 1000 L 9.23 L 1000 mg 1 m3 3 9.23 g/m or 9.23

g 0.0022 lb 1 m3 1g m3 2.642 104 Mgal

77.0 lb/Mgal Note: Mgal million gallons

PUBLIC WATER SUPPLY 1.416

CHAPTER 1.6

A 2-MGD (0.088 m3/s) water system generates 220 lb/Mgal (26.4 mg/L) solids in dry weight basis. The residual solids concentration by weight is 0.48 percent. Estimate the rate of residual solids production and pumping rate if the settled residual is withdrawn every 5 minutes per hour.

EXAMPLE 3:

Solution: Step 1. Calculated residual production (220 lb/Mgal)(2 MGD) 10,990 gpd 7.63 gpm (8.34 lb/gal)(0.0048) Step 2. Calculate pump capacity required 7.63 gpm 60/5 92 gpm 17.1 Residual Production and Density The quantity of residuals (called sludge) generated from clarifiers are related to coagulants used, raw water quality, and process design. The residual production for whole water plant as the weight per unit volume treated (g/m3 or lb/Mgal) and its solids concentration should be determined. These values with the average flow rate will be used for the estimation of daily residual production. Then the residual production rate can be employed for the design of residual (sludge) treatment units, commonly intermittent sand drying bed. The density of residuals (pounds per gallon) can be estimated using the following equation: w

8.34 (US customary units) cs/rs c/r

(6.151a)

8.31 (SI units) cs/rs c/r

(6.151b)

where w density of residuals, lb/gal or kg/L cs weight fractional percent of solids c weight fractional percent of water s specific gravity of solid specific gravity of water (approximately 1.0) Calculate the sludge density (lb/gal) for a 6 percent sludge composed of solids with a specific gravity of 2.48.

EXAMPLE 1:

Solution: Step 1. Find specific gravity of 6% residual (sludge) r Weight fraction of sludge, solids, and water are 1.0, 0.06, and 0.94, respectively. Using mass balance, we obtain 0.94 0.94 0.06 0.06 1 rr rs r 2.48 1.0 r 1.037 Step 2. Determine the density of the residue Density r 8.34 lb/gal 1.037 8.34 lb/gal 8.65 lb/gal or

1037 g/L

Note: This solution is a kind of proof of Eq. (6.151).

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.417

EXAMPLE 2: A water plant treats 4.0 MGD (0.175 m3/s) and generates 240 lb of residuals per million gallons (Mgal) treated (28.8 g/m3). The solid concentration of the residuals is 0.5%. Estimate: (a) the daily production of residuals, (b) the pump capacity, if the pump operates for 12 min each hour, and (c) the density of the residuals, assuming the specific density of the residual is 2.0 with 4 percent solids.

Solution: Step 1. Determine residual production rate q q

(4 MGD) (240 lb/Mgal) (0.005) (8.34 lb/gal)

23,020 gpd

(a)

16.0 gpm Step 2. Find pump capacity qp qp 16 gpm (b)

60 min 12 min

80 gpm

Step 3. Estimate residual density w, using Eq. (6.151a) w (c)

8.34 0.04/2.0 0.96/1.0

8.51 lb/gal

EXAMPLE 3: In a 1 MGD (0.0438 m3/s) water works, 1500 lb of dry solids are generated per million gallons of water treated. The sludge is concentrated to 3 percent and then applied to an intermittent sand drying bed at a 18 in depth, with 24 beds used per year. Determine the surface area needed for the sand drying bed.

Solution: Step 1. Calculate yearly 3% sludge volume, 1500 lb/Mgal 1 Mgal/d/0.03 50,000 lb/d 50,000 lb/d 365 d/(62.4 lb/ft3) 292,500 ft3 8284 m3 Step 2. Calculate area A, (D depth, n number of applications/year) ADn or

A /D n 292,500 ft3/(1.5 ft 24) 8125 ft2 755 m2

PUBLIC WATER SUPPLY 1.418

CHAPTER 1.6

18 DISINFECTION Disinfection is a process to destroy disease-causing organisms, or pathogens. Disinfection of water can be done by boiling the water, ultraviolet radiation, and chemical inactivation of the pathogen. In the water treatment processes, pathogens and other organisms can be partly physically eliminated through coagulation, flocculation, sedimentation, and filtration, in addition to the natural die-off. After filtration, to ensure pathogen-free water, the chemical addition of chlorine (so called chlorination), rightly or wrongly, is most widely used for disinfection of drinking water. For the public health standpoint, chlorination of drinking water provides more benefits than its shortcoming due disinfection by-products (DBPs). The use of ozone and ultraviolet for disinfection of water and wastewater is increasing in the United States. Chlorination serves not only for disinfection, but as an oxidant for other substances (iron, manganese, cyanide, etc.) and for taste and odor control in water and wastewater. Other chemical disinfectants include chlorine dioxide, ozone, bromine, and iodine. The last two chemicals are generally used for personal application, not for the public water supply. 18.1 Chemistry of Chlorination Free available chlorine. Effective chlorine disinfection depends upon its chemical form in water. The influencing factors are temperature, pH, and organic content in the water. When chlorine gas is dissolved in water, it rapidly hydrolyzes to hydrochloric acid (HCl) and hypochlorous acid (HOCl)

Cl2 H2O ↔ H Cl HOCl

(6.152)

The equilibrium constant is KH

[H ][Cl–][HOCl] [Cl 2(aq)]

(6.153)

4.48 10–4 at 25C (White 1972) The dissolution of gaseous chlorine, Cl2(g), to form dissolved molecular chlorine, Cl2(aq) follows Henry’s law and can be expressed as (Downs and Adams, 1973) Cl2(g)

Cl2(aq) [Cl2(aq)] PCl2 H (mol/L # atm)

(6.154)

where [Cl2(aq)] molar concentration of Cl2 PCl partial pressure of chlorine in atmosphere 2

The distribution of free chlorine between HOCl and OCl is presented in Fig. 6.10. The disinfection capabilities of HOCl is generally higher than that of OCl (Water, 1978). H Henry’s law constant, mol/L atm 4.805 106 expa

2818.48 b T

(6.155)

Hypochlorous acid is a weak acid and subject to further dissociation to hypochlorite ions (OCl) and hydrogen ions: HOCl ↔ OCl H

(6.156)

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.419

1.0 Cl2

OCl−

HOCl

Fraction α

−2

pH FIGURE 6.10

Distribution of chlorine species (Snoeyink and Jenkins, 1980).

and its acid dissociation constant Ka is [OCl][H ] [HOCl] 3.7 108 at 25C

(6.157)

2.61 108 at 20C The values of Ka for hypochlorous acid is a function of temperature in kelvins as follows (Morris 1966): In Ka 23.184 0.058T 6908/T

(6.158)

EXAMPLE 1: The dissolved chlorine in the gas chlorinator is 3900 mg/L at pH 4.0. Determine the equilibrium vapor pressure of chlorine solution at 25C (use KH 4.5 10–4).

Solution: Step 1. Since pH 4.0 5, from Fig. 6.10, the dissociation of HOCl to OCl is not occurring. The available chlorine is [Cl2] [HOCl]: [Cl2] [HOCl] 3900 mg/L

3.9 g/L 70.9 g/mol

0.055 mol/L [HOCl] 0.055 [Cl2]

Step 2. Compute [H ] pH 4.0 log

1 [H]

[H] 104 Step 3. Apply KH at 25C KH 4.5 104

[H][Cl][HOCl] [Cl2]

Substitute [H] from Step 2 4.5 108

[Cl][HOCl] [Cl2]

PUBLIC WATER SUPPLY 1.420

CHAPTER 1.6

From Eq. (6.152), it can be expected that, for each mole of HOCl produced, one mole of Cl will also be produced. Assuming there is no chloride in the feedwater, then [Cl] [HOCl] 0.055 [Cl2] Step 4. Solve the above two equations 4.5 108

(0.055 [Cl2])2 [Cl2]

0.000212 [Cl2]1/2 0.055 [Cl2] [Cl2] 21,400 [Cl2]1/2 0.055 0 0.000212 20.0002122 0.22 2 0.2345

[ Cl2]1/2

[Cl2] 0.055 (mol/L) Step 5. Compute H using Eq. (6.155) H 4.805 106 expa

2818.48 b 25 273

0.062 mol/(L atm) Note: [HOCl] is negible Step 6. Compute the partial pressure of chlorine gas: PCl [Cl2]/H (0.055 mol/L)/(0.062 mol/(L atm)) 2

0.89 atm 0.89 atm 101,325 Pa/atm

90 Pa EXAMPLE 2: A water treatment plant uses 110 lb/d (49.0 kg/d) of chlorine to treat 10.0 MGD (37,850 m3/d or 0.438 m3/s) of water. The residual chlorine after 30 minutes contact time is 0.55 mg/L. Determine the chlorine dosage and chlorine demand of the water.

Solution: Step 1. Calculate chlorine demand Since 1 mg/L 8.34 lb/mil gal 1 mg/L 110 lb/d 10.0 Mgal/d 8.34 lb/Mgal 1.32 mg/L

Chlorine dosage or solve by SI units: Since 1 mg/L 1 g/m3

Chlorine dosage (49.9 kg/d) (1000 g/kg) (37,850 m3/d) 1.32 g/m3 1.32 mg/L

PUBLIC WATER SUPPLY PUBLIC WATER SUPPLY

1.421

Step 2. Calculate chlorine demand Chlorine demand chlorine dosage chlorine residual 1.32 mg/L 0.55 mg/L 0.77 mg/L Calcium hypochlorite (commercial high-test calcium hypochlorite, HTH) containing 70 percent available chlorine is used for disinfection of a new main. (a) Calculate the weight (kilograms) of dry hypochlorite powder needed to prepare a 2.0 percent hypochlorite solution in a 190 L (50 gal) container. (b) The volume of the new main is 60,600 L (16,000 gal). How much of the 2 percent hypochlorite solution will be used with a feed rate of 55 mg/L of chlorine?

EXAMPLE 3:

Solution: Step 1. Compute kg of dry calcium hypochlorite powder needed 190 L 1.0 kg/L 0.02 0.70 5.42 kg

Weight

Step 2. Compute volume of 2 percent hypochlorite solution needed 2% solution 20,000 mg/L 20 g/L Let X 2 percent hypochlorite solution needed Then (20,000 mg/L) (X L) (55 mg/L) (60,600 L) X 167 L Combined available chlorine. Chlorine reacts with certain dissolved constituents in water, such as ammonia and amino nitrogen compounds to produce chloramines. These are referred to as combined chlorine. In the presence of ammonium ions, free chlorine reacts in a stepwise manner to form three species: monochloramine, NH2Cl; dichloramine, NHCl2; and trichloramine or nitrogen trichloride, NCl3. NH 4 HOCl ↔ NH2Cl H2O H or

(6.159)

NH3(aq) HOCl ↔ NH2Cl H2O

(6.160)

NH2Cl HOCl ↔ NHCl2 H2O

(6.161)

NHCl2 HOCl ↔ NCl3 H2O

(6.162)

At high pH, the reaction for forming dichloramine from monochloramine is not favored. At low pH, other reactions occur as follows: NH2Cl H ↔ NH3Cl NH3Cl NH2Cl ↔ NHCl2 NH 4

(6.163) (6.164)

Free residual chlorine is the sum of [HOCl] and [OCl]. In practice, free residual chlorine in water is 0.5 to 1.0 mg/L. The term “total available chlorine” is the sum of “free chlorine” and “combined chlorine”. Chloramines are also disinfectants, but they are slower to act than that of hypochlorite. Each chloramine molecule reacts with two chlorine atoms. Therefore, each mole of mono-, di-, and trichloramine contains 71, 142, and 213 g of available chlorine, respectively. The molecular weight of these three chloramines are respectively 51.5, 85.9, and 120.4. Therefore, the chloramines contain 1.38, 1.65, and 1.85 g of available chlorine per gram of chloramine, respectively. The pH of the water is the most important factor on the formation of chloramine species. In general, monochloramine is formed at pH above 7. The optimum pH for producing monochloramine is approximately 8.4.

PUBLIC WATER SUPPLY 1.422

CHAPTER 1.6

Handbook of Environmental Engineering Calculations - PDF Free Download (2024)

References