Solutions Manual to accompany

Water and Wastewater Technology Seventh Edition

Mark J. Hammer Mark J. Hammer Jr.

Upper Saddle River, New Jersey Columbus, Ohio

__________________________________________________________________________________ Copyright © 2012 by Pearson Education, Inc., Upper Saddle River, New Jersey 07458. Pearson Prentice Hall. All rights reserved. Printed in the United States of America. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department. Pearson Prentice Hall™ is a trademark of Pearson Education, Inc. Pearson® is a registered trademark of Pearson plc Prentice Hall® is a registered trademark of Pearson Education, Inc. Instructors of classes using Hammer and Hammer, Water and Wastewater Technology, 7e, may reproduce material from the instructor’s manual for classroom use.

10 9 8 7 6 5 4 3 2 1

ISBN-13: 978-0-13-511405-6 ISBN-10: 0-13-511405-5

Water Treatment

Water

Distribution

Sewage

Collection

Primarysettling

Gritchamber

Biologicaltreatment

Pumpingstation

Finalsettling

Secondary Treatment Preliminary and PrimaryTreatment

Return activated sludge

or recycle flow

Wastewater Treatment

Advanced Wastewater Treatment

SludgeRecirculation of NO , Q = Q3 S

AnoxicZone

NO N3 2 Effluent

Mechanically mixed anoxic chambers

Air mixedaerobic chambers

Aerobic Zone

NH NO3 3

P Cell Mass

FinalSettling

NO N3 2

N03

PrimaryEffluent

SettlingtankFlocculation

Rapidmix

FilterSecondary

effluent

CoagulantAuxiliary chemicals

Chlorine Chlorine

Tertiaryeffluent

Rapidmix

Chlorinationbasin

Mechanically mixed anaerobic chambers

Return Activated Sludge, Q = 0.3QR

N2

gasN2

gasN2

gas

AnaerobicZone

Organic P P

Recycled Water

Water Reclamation

Chlorine

Secondaryeffluent

UltravioletDisinfection

Breaktank

Cartridgefilters

ReverseOsmosis

Units

Packed-towerdecarbonator

High-pressurepumps

Sulfuric acidThreshold inhibitor

Cartridgefilters

MicroFilters

Compressed air

Injectionwell

Irrigation

SettlingtankFlocculation

Rapidmix

FilterRaw water

supply

CoagulantAuxiliary chemicals

Chlorine

Potablewater

Rapidmix Chlorination

basin

Lake

River

SurfacewaterDischarge

This solutions manual has been prepared by the authors for classroom use by instructors teaching from Water and Wastewater Technology, Seventh Edition. Textbook problems are an effective method of measuring student’s understanding and performance; therefore, the safekeeping of solutions is important to all instructors. Under no circumstance should any solution in this manual be reproduced and distributed or otherwise released in any form for student use. Obviously, solutions must be presented in a classroom by writing calculations on the classroom board or by projecting calculations using a projector. Your cooperation in maintaining the integrity of the homework solutions is appreciated. The inside book cover is intended as a reference and as a teaching aid to the overall treatment processes. Mark J. Hammer Mark J. Hammer Jr.

mhammer

Rectangle

3-9. Although impounded waters are oxygenated by diffusion from the atmosphere

and algal photosynthesis, the major periods of oxygenation are during spring and autumn circulations (turn-over).

3-10. Oligotrophic lakes are nutrient poor and biologically unproductive. Mesotrophic

lakes have an increased nutrient level to support some aquatic plants, greenish water from algae in the summer, and moderate populations of sport fish. Eutrophic lakes are nutrient rich with heavy weed growth along shores, blooms of algae, and tolerant fish.

3-11. Thermal stratification of an eutrophic results in reduced quality of the impounded

water in the hypolimnion. Since the water below the themocline is not oxygenated, decomposition of organic matter can delete the dissolved oxygen concentration.

3-12. The best way to prevent or retard the rate of eutrophication of a lake is to reduce

the nitrogen and phosphorus inputs. For an oligotrophic lake, either nutrient can promote plant growth, whereas for an eutrophic lake, phosphorus is the primary nutrient to control.

3-13. Pathogens are disease-producing organisms including viruses, bacteria, protozoa,

and helminthes (parasitic worms). 3-14. The fecal-oral route is the transmission of pathogens in the feces of an infected

person into the mouth of another person by person-to-person contact with contaminated fingers or through water and food contaminated by feces.

3-15. Latency is the period of time between excretion of a pathogen in feces and its

becoming infective to a new host. Persistence is measured by the length of time that a pathogen remains viable in the environment outside a human host. Infective does is the number of organisms that must be ingested to result in disease. Ascaris (roundworm) is III: latent, persistent, one egg produces one intestinal worm. Salmonella (bacteria) are II: non-latent, moderately persistent, medium to high infective dose.

8a

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37

6 WATER DISTRIBUTION SYSTEMS

6-1. The average municipal water use is 600 gpd per metered service including residential, commercial and industrial customers. Western regions use the most water with an average of 460 gpd per household service. In semiarid climates, lawn sprinkling and air conditioning are major factors in high water consumption.

6-2. Average residential water demand = 100 gpcd Maximum daily water demand = 1.8 • 100 = 180 gpcd Maximum hourly rate = 3 • 100 = 100 gpcd

6-3. The range of recommended water pressure is 65 to 75 psi. The minimum and maximum pressures for residential service are 40 psi and 100 psi.

6-4. Needed fire flow (NFF) is the rate of water flow required for fire fighting to confine a major fire to the buildings within a block or other group complex with minimal loss. The key considerations for NFF are: kind of construction, floor area of building, occupancy, and exposure and communication with other buildings.

6-5. Wood-frame construction has F equal to 1.5. The effective area Ai is the ground floor plus 50% of the second floor, equal to 2250 + 450 = 2700 sq ft. Ci = 18 · 15 (2700)0.5 = 1400 gpm Ci = 1500 gpm (rounded) The cabinet-making shop occupies over 25% of the total floor area of the building. Percentage = [900 / (2250 + 900)] / 100 = 28.6 From Table 6-1, Oi = 1.15 The exposure from Table 6-1, for building A at a distance of 11 ft and length-height of 240 ft · stories is 0.14. The exposure from Table 6-2 for building B at a distance of 12 ft and length-height of 240 ft · stories is 0.17. The building with the largest factor is B.

Xi = 0.17 No communication exists so Pi = 0 Using Equation 6-3, (X + P)i = (0.17 + 0) = 0.17 Using Equation 6-4, the needed fire flow is NFF = 1500 • 1.15 • 0.17 = 201 gpm or 250 gpm (rounded to the nearest 250 gpm)

38

6-6. Ai = 250 • 450 + 0.25(4 • 250 • 450) = 225,000 sq ft or Ai = 250 • 450 + 2(250 • 450) = 338,000 sq ft Ci = 18 • 0.6(225,000)0.5 = 5120 gpm Including Oi = 0.85, Ci • Oi = 0.85 • 5120 = 4350 gpm Duration = 3 hr Length-height = 450 • 5 = 2250 ft • stories From Table 6-3, Xi = 0.06 With no communication, Pi = 0 (X • P) i = 1.0 + 1.0(0.06 + 0) = 1.06 NNF = 5120 • 1.06 = 5430 gpm = 5500 gpm (rounded to the nearest 250 gpm) A sprinkler system is justified since NNF exceeds the practical limit 3500 gpm for fire flow. Flow requirement for sprinkler systems is generally in the range of 150-1600 gpm. Two hydrants within 120 ft of the building does reduce the NNF. “Credit up to 1000 gpm can be allowed for each hydrant within 300 ft.” Therefore, without a sprinkler system, the NNf could be lowered by 2000 gpm. NFF = 5500 - 2 • 1000 = 3500 gpm

6-7. For the restaurant: Ci = 18 • 0.8(115 • 105 + 0.5 • 115 • 105)0.5 = 2130 gpm (rounded 2250 gpm) Oi = 1.00 for C-3 Xi = 0 (blank wall) and Pi = 0 NFF = 1.0 • 2250 = 2250 gpm Duration = 2 hr Fire hydrants = 1000 gpm (200 ft) + 670 gpm (500 ft) = 1670 gpm Since the allowable hydrant capacity is less than NFF, the restaurant should be protected by a sprinkler system, such as, one that can be connected to the nearest hydrant by a fire hose. For the office: Ci = 18 • 0.8(75 • 55)0.5 = 930 gpm (rounded 900 gpm) Oi = 0.85 for C-2 Xi = 0.13 (Class 4, 48 ft, 230 ft-stories) Pi = 0.10 (Open, 48 ft, unprotected) (X + P)i = 1.0 + 1.0 (0.13 + 0.10) = 1.23 NFF = 900 • 0.85 • 1.23 = 940 gpm (rounded 900 gpm) Fire hydrants = 2 • 670 gpm (350 ft, 450 ft) = 1340 gpm (OK)

6-8. Supermarket F = 1.0, Ai = 65.6 • 98.4 = 41,900 sq ft Ci = 18 • 1.0(41,900)0.5 = 3680 gpm (rounded 3250 gpm)

6-9. From Table 6-4, 1000 gpm Lineal distance between hydrants is normally 600 ft with a maximum of 800 ft.

6-10. The maximum fire flow that most are likely to be able to reliably provide is 3500 gpm (220 l/s). For major structures, adequate fire suppression can be provided by automatic sprinklers, or for isolated properties private protection can be provided by on-site water storage and pumps.

6-11. The major components evaluated for reliability of a water system are: water supply capacity (pumping capacity in conjunction with storage), supply mains, treatment plant, and power source.

39

6-12. In rotary drilling the borehole is held open by using a viscous dense mud of bentonite clay as a drilling fluid. A well is developed by hydraulic cleaning and flushing to remove the residue of mud. Aquifer sands are prevented from entering a finished well by installation of a gravel pack.

6-13. The cylindrical intake in Figure 6-2 is placed parallel to the water flow with the cone end (debris deflector) pointed upstream. The wire is wound with narrow openings perpendicular to the water flow to attempt entry through the narrow slots.

6-14. For a max velocity of 10 ft/s, min velocity of 3 ft/s, and average velocity of 5 ft/s, pipe sizes range as follows:

� � � � �� · 449 ·

4

�

· 12

Velocity v (ft/s) q (gpm) di (in.)

Max 10 2500 10 Min 3 800 10

Average 5 1600 11

Low and high velocities suggest a pipe diameter of 10 inches. At average flow, a pipe diameter of 12 inches is preferred. Use 12 inch pipe to decrease the operating cost at average flow rates. The optimum pipe diameter:

� � 1.802 · ���� ������

�� �� �� ��������� �����

� 1.802 · 1600��� �����

120 ���� ����� �0.15 � 20

3.5 � 0.8 � �! "�#�$

� 14.5 inches

Round to 14 inches.

6-15. A gridiron pipe network permits water circulation that results in better flow patterns and pressure distribution. If a pipe break occurs, water can be supplied through other lines while the broken pipe is valved-off and repaired. The minimum diameter pipe for residential areas is 6 in. (150 mm). The minimum for commercial districts is 8 in. (200 mm) with intersecting lines of 12 in. (300 mm) or larger.

6-16. The major components of a house connection are the corporation stop installed in the main, service pipe, service box with curb stop valve, pipe into house, and valved water meter (Figure 6-6).

6-17. Compression joint and boltless restrained joint are used in distribution system piping. Although less common, the mechanical joint can also be used in distribution system piping. Flanged joints are used for interior piping in water plants and pump stations. PVC distribution piping is joined by a compression-type joint with a rubber gasket seal. Plastic pipe for service connections and household plumbing can be PVC, ABS, or PE. Reinforced concrete pipe for pressure conduits can be steel cylinder for internal pressures 40-260 psi, prestressed with steel cylinder for internal pressures 50-350 psi and noncylinder reinforced not prestressed for internal pressure less than 45 psi.

40

6-18. (a) & (b)

(c) At 6.0 mgd, 4 pumps are operating at a discharge head of 175 ft. (d) If a 5th pump is turned on, the operating point--which must be along that pumping curve--will initially move toward 6.0 mgd and 220 ft. However, looking at Figure 6-21 one might assume that inflow to the storage tank will increase, thus the pumping rate into the system will increase above 6 mgd and the pressure drop to less than 225; perhaps the new operating point will be 7 mgd and 205 ft. If one pump is turned off (4 to 3), the discharge head will decrease, pumping rate decrease, and reduction in pumping rate compensated for by flow from elevated storage. The operating head-discharge point will be on the pump curve for 3 pumps but not at the plotted system-head curve. A system-head curve depends on the response of a distribution system to the pumping applied to the system; in this case, a shift in pumping pressure influences flow into and out of elevated storage.

6-19.

100110120130140150160170180190200

200 300 400 500

Hea

d, ft

Discharge, gpm

Head versus Discharge

41

6-20. Pump operation is scheduled to meet the variation in diurnal water consumption. The required on-off sequencing of pumps for a given system is related to the degree of equalization provided by distribution storage and head-discharge characteristics (discharge pressure) of the pumps. The following schedules are probable, normal pump operations. (a) Winter average consumption (140 l/s) requires two of 1, 2 or 3 pumps during the majority of the day. Except during hours of low demand, one pump cannot maintain adequate discharge head (pressure). (b) Annual average consumption (180 l/s) requires two of 1, 2, or 3 pumps. (c) Summer average consumption (225 l/s) requires 1 and 2 for a low water level in elevated storage or during low diurnal demand and 1 and 4 or high water level or high demand. The system head-discharge curve allows operation of both the smaller and larger pumps. (d) Maximum daily consumption (270 l/s) requires 4 and 5 for higher discharge head, although 1 and 4 may be suitable for periods of low demand. An adequate system is able to maintain the maximum daily consumption rate plus fire flow, at minimum pressure, with one or two pumps out of service. Assume the quantity of stored water is adequate and the pump capacity for maximum daily with fire flow is 350 l/s. This discharge can only be provided with all three large pumps (4, 5 and 6) in operation. The discharge heads for pumps 1, 2, and 3 are too low for operation at the required system head at 350 l/s. Therefore, the pumping capacity is not adequate.

42

0

1

2

3

4

5

6

7

8

1 4 7 10 13 16 19 22 25

Time, hours

Cum

ula

tive

co

nsu

mp

tion

, mil

gal

Storage requiredfor 24-hr pumping

= 1.47 mil gal

Storage requiredfor 12-hr pumping

= 2.76 mil gal

6-21. Storage volumeTotal consumption

percent= =⋅500 000 1002 685 000

18 6,

, ,.

The rule-of-thumb range is 15 to 20 percent.

6-22. Time Consumption Time Consumption

12 PM 0 1 PM 2,790,000 1 AM 132,000 2 3,174,000

2 258,000 3 3,552,000 3 366,000 4 3,936,000 4 450,000 5 4,320,000 5 528,000 6 4,722,000 6 600,000 7 5,166,000 7 720,000 8 5,718,000 8 930,000 9 6,222,000 9 1,230,000 10 6,522,000 10 1,590,000 11 6,714,000 11 1,974,000 12 6,882,000 12 2,394,000

43

(a) Average = 6,882,000/1440 = 4779 gpm Storage capacity required from graph = 1.47 mil gal

Storage capacity

Total consumptionpercent= =⋅147 100

6 882214

..

.

gpm57256012

000,600000,722,4Average)b( =

•−=

On graph, 6 A.M. = Time 7 and 6 P.M. = Time 19 Storage capacity required from graph = 2.76 mil gal

percent1.40882.6

10076.2nconsumptioTotal

capacityStorage =•=

6-23. The principal functions of distribution and storage are: to permit continuous water treatment, uniform pumping rates of water into the distribution system, pressure stabilization, and reserve supply available for contingencies. Storage capacity considered to be available for fire fighting is only the normal minimum daily amount of water maintained in storage.

6-24. (a) Yes, based on calculated consumption per capita. Average daily = 120,000/900 = 130 gpcd Maximum daily = 280,000/900 = 310 gpcd (b) Maximum daily flow = 280,000/1440 = 200 gpm Required flow = 1500 + 200 = 1700 gpm The fire reserve in storage is estimated by subtracting the capacity needed for equalization (20%) from the total storage volume which equals 100,000 minus 56,000 = 44,000 gal. If discharged uniformly during the 2 hr duration for fire flow, the average flow available = 44,000/(2 • 60) = 370 gpm. Assume the demand of 1700 gpm must be met by storage plus pumping with the two most important pumps out of service. Based on this criterion, the flow is: 370 + 400 + 400 = 1170 gpm, which is less than 1700 gpm. With only the largest pump inoperative, this available flow increases to : 1170 + 600 = 1770 gpm, which is sufficient. Therefore, the system can apparently meet fire demand.

6-25. The design feature of a fire hydrant the prevents discharge if the barrel is broken is an upward seating valve with a break-away valve stem. In a cold climate, a hydrant is inspected after use to check for water remaining in the barrel that can freeze and break the barrel.

6-26. In a check valve, the cushion chamber is to prevent valve slam by controlling the closing speed and the counterweight assists in preventing valve slam.

6-27. Refer to Figure 6-21. Pressure head (energy) is lost as inlet water is forced through the restricted opening between the valve disk and the valve body. The valve opening automatically adjusts to maintain constant discharge pressure with variations of inlet pressure. For a given pressure setting, if the inlet pressure increases the resulting momentary increase in outlet pressure forces the diaphragm upward compressing the spring, lifting the stem, and closing the valve opening to increase the pressure loss. The action is reversed for a decrease in inlet pressure.

6-28. Refer to Figure 6-23. Turning the handwheel counterclockwise allows the pilot valve to increase the restriction of flow. This increases the pressure in the chamber above the piston increasing closure of the automatic valve, thus, reducing the outlet pressure.

6-29. Refer to Figures 6-22 and 6-24. When the system side pressure is lower, water drains out of the chamber above the pilot diaphragm. This permits the stem of the pilot to be forced upward when the piston is pushed up by high pressure water entering the valve from the tank. Water flows out of the tank through the open valve into the system. When the direction of flow reverses, the valve stays open allowing the tank to fill. When the water level reaches the maximum elevation, the water pressure to the chamber above the pilot diaphragm forces the stem downward. This allows

44

the higher pressure from the system side to force water into the chamber above the piston to lower the piston closing the valve.

6-30. Water level in an elevated storage tank can be monitored by analog sensors, pressure measuring diaphragms, long probes, or sonic signals. These measuring devices can energize solenoids that in turn control pre-programmed operation of pumps.

6-31. The backflow preventer usually installed on a lawn sprinkler is the atmospheric vacuum breaker. For larger landscape irrigation systems, either the pressure-vacuum breaker or double check valve assembly is used.

6-32. The water service to a mortuary contains either an air gap or reduced pressure principle backflow preventer.

6-33. The name of the reduced-pressure-principle backflow preventer is derived from the reduced pressure zone between the inlet and outlet check valves. During back siphonage, the water in this zone, and any water seeping into this intermediate zone, drains out through the relief valve. The reduced-pressure-principle backflow preventer is preferred to air-gap separation because the backflow preventer allows system pressure to pass through it to the piping in the building. Based on height of the building, repumping may not be required. If increase in pressure is required, in-line booster pumps may be installed on the discharge side of the preventer.

6-34. A pressure-vacuum breaker must be installed at least 12 in. above the highest outlet. For basement installation, a reduced-pressure-principle backflow preventer is required.

6-35. In Figure 6-34a, elevated storage is located near the load centers to supply water during peak demands and to support the hydraulic gradient to stabilize pressure. In Figure 6-34b, water is supplied directly from the wells and the hydraulic gradient is supported by the discharge pressure of well pumps, which operate based on demand for pressure.

45

7 WATER PROCESSING

7-1. Surface water contains a wider variety of contaminants from land drainage, industrial wastes, and human excreta--these are often limited to protect human health rather than just aesthetics. The quality of a surface water can be highly variable season to season and even day to day in the case of a river. The impurities commonly removed from groundwater are hardness (calcium and magnesium), iron, manganese, carbon dioxide and sometimes nitrate, arsenic, and radionuclides. The impurities commonly removed from surface water are turbidity (silt, clay, dead organic matter, algae, and other microorganisms), pathogenic bacteria and viruses, heavy metals and chemicals from industrial pollutants, pesticides and other contaminants from land drainage, taste and odor compounds, and color.

7-2. T logeCo/Ct

1.0 0.40 2.0 0.78 3.0 0.99 4.0 1.30

7-3. V = 48 • 52 • 12 • 7.48 = 224,000 gal t = (224,000 • 1440)/10,000,000 = 32 min > 30 min (OK) v = 48/32 = 1.5 ft/min = 1.5 ft/min (OK)

7-4. 30 mgd = 1.25 mil gal/hr = 20,800 gpm = 347 gal/sec V of rapid mix chambers < 30 • 347 = 10,400 gal V of flocculation basins = 30 • 20,800 = 624,000 gal V of settling tanks = 4 • 1.25 = 5.00 mil gal (Some plants are permitted by the State regulatory agency to reduce t for settling tanks to as low as 2 hr.)

7-5. t =1.0 24

8.0hr, V gpd / sq fto

⋅ = = =3 08 000 000

12 500640.

, ,,

7-6. ftsqday/gal

905daysec

400,86ft

gal48.7

ft

ftsecft

0014.0V 32

2

0 =•••=

hr0.23600014.0

10tmin =

•=

0

0.5

1

1.5

2

0 1 2 3 4 5

t, min

log e

C0/

Ct

k = 1.1/4 = 0.28 per min

46

7-7. ( )OKhr0.4hr04.4000,400

2448.71015302t ==•••••=

)OK(5.0min/ft25.01015144048.7

000,400vhorizontal <=

•••=

)OK(000,20ftday/gal

670060

000,400loadingWeir <==

7-8. t =2 27 5.0 3.8 24

6000= 4.1 h

⋅ ⋅ ⋅ ⋅

V =6000

2 27 5.0m m ho

3 2

⋅ ⋅ = ⋅22 /

weir loading = 6000/50 = 120 m3/m � d

7-9. Overflow rate = 0.000,24 • 86,400 = 20.7 m/d

( )m64.7

7.20

3800radiusTank 5.0 =

•π=

Diameter = 15.3 m and Area = 184 m2

m6.224184

38000.3Depth =

••=

7-10. A =15,000

16m Diameter =

4 938m2= ⋅�

��

��� =938 35

0 5

π

.

4 0938 24

. = ⋅ ⋅Depth15,000

Depth = 2.7 m

7-11.

0 40 80 120 160 200

t50

0

0.5

1.0

Time, min

Con

c ent

ratio

n, g

/l

1.5

t

t50 = 505/5.06 = 99.8 = 100 min

47

7-12.

1 2 3 4 5 6

t50

= 137 min +

0

50

100

Time, min

Con

c ent

ratio

n, g

/l

t50 = 85,200/624 = 137 min

7-13. Volume of entire tank = � • 10 • 202 = 12,600 cu ft = 94,000 gal Areas of cone-shaped skirt: A1 = �(12)2 = 452 sq ft. A2 = �(6.0)2 = 113 sq ft Volume of skirt = {8.0/3[452 + 113 + (452 • 113)0.5]} = 2109 sq ft = 15,800 gal Water surface area = �(20)2 - �(6.0)2 = 1140 sq ft 750,000 gpd = 31,300 gph = 521 gpm

)hr2to1,OK(hr0.3300,31000,94

)ktanentire(t ==

min)30,OK(min30521

800,15)onflocculatiandmixing(t ==

)ftsq/gpm75.1(ftsq

gpm46.0

1140521

rateUpflow <==

7-14. vs = 2.1 mm/s = 0.0068 ft/sec Vo = 1.75 gpm/sq ft = 1.75/(7.48 • 60) = 0.0039 ft/sec Depth = (vs - vo) t = (0.0068 - 0.0039)3600 = 10 ft

7-15. Direct filtration included rapid mixing but does not include sedimentation and often no flocculation (if provided, is usually for less than 30 min) prior to filtration. The flow scheme for conventional filtration includes rapid mixing, flocculation, and sedimentation prior to filtration (Figure 7-1).

7-16. A rate of flow controller maintains a constant filtration rate preventing excessive velocity through a clean bed. The discharge valve opens and the head loss through the bed increases with accumulation of coagulated floc.

7-17. Head available for filtration includes both the depth of water above the filter and the suction head between the filter and water level in the clear well. This total head, which is commonly 9 to 12 ft, is available for head loss during filtration.

7-18. A filter run is terminated when the head loss increases to a specified limit (approximately 8 ft.) or when the turbidity in the filtrate becomes excessive because of impurities passing through the filter.

48

7-19. A dual-media anthracite-sand filter provides better “in-depth” filtration than a single-medium sand filter. More water can be filtered between backwashes, therefore, design filter rates are higher and filter runs longer. The anthracite particles being lighter than the sand particles stratify above the sand layer after backwashing.

7-20. Refer to Table 7-1. (a) The kinds of underdrains that allow air scour are pipe laterals with nozzles and plastic nozzles. (b) The underdrains that allow concurrent air-and-water scour are precast concrete T-Pees, plastic dual-lateral block, and plastic nozzles.

7-21. Backwashing by air scouring and water backwashing is accomplished as follows: (1) the water is lowered in the filter box to near the surface of the media to restrict expansion of the bed during air scour, (2) air is forced up through the bed to loosen impurities from the media by particle-to particle scrubbing, (3) water backwash lifts the impurities out of the bed, and (4) after clean water appears the backwash is stopped and the media allowed to restratify.

7-22. After a powdered chemical is mixed with water, the solution is pumped to the point of application by a solution feeder.

7-23. (a) The lowest recommended dosage is 15 mg/l. (b) With distilled water, the lack of alkalinity prevented flocculation of the alum.

7-24. One equivalent of alum reacts with one equivalent of alkalinity, lime, and soda ash, Equations 7-6, 7-7, and 7-8. 10 100

5010. .meq / l alum

1.0 meq / alkmg / l alum

mg / l CaCO3

mg / l alum0.50 mg / l alkalinity

= =

10 100 10. .meq / l alum1.0 meq / CaO 28.0

mg / l alum0.28 mg / l CaO

= =

10 100 10. .meq / l alum

1.0 meq / Na2CO3 53.0mg / l alum

0.53 mg / l Na2CO3= =

7-25. Alkalinity reacted = 20 • 0.5 = 10 mg/l The pH decreases because of the carbon dioxide produced in the reaction.

7-26. (a) Alkalinity reacted = 40 • 0.5 = 20 mg/l.

(b) Sulfate added = 40 3 96

60019

⋅ ⋅ = mg / l

Bicarbonate destroyed = 40 2 3 61

60024

⋅ ⋅ =( )mg / l

Decrease in pH because of CO2 formation from HCO3.

7-27. meq/l of alum added = 30 / 100 = 0.30 meq/l meq of soda ash applied is the same = 0.30 meq/l The aluminum precipitates out of solution, but all of the sulfate ion remains in solution. The sodium from the soda ash remains in the water, while the carbonate is converted to carbon dioxide. Therefore, the changes are: 0.30 meq/l of SO4 = 0.30 • 48 = 14.4 mg/l of SO4 0.30 meq/l of Na = 0.30 • 23 = 6.9 mg/l of Na Production of CO2 reduces the pH

49

7-28. Using equivalent weights, Alum reacting with 10 mg/l of alkalinity = 10(100/50) = 20 mg/l Amount of alum remaining to react with lime = 40 – 20 = 20 mg/l Lime dosage required = 20(28/100) = 5.6 mg/l as CaO

7-29. (a) Alkalinity reacted = 12(50/100) = 12 • 0.50 = 6.0 mg/l CO2 released = 12[(6 • 44)/(600)] = 5.28 mg/l (b) Soda ash dosage = 12(53/100) = 12 • 0.53 = 6.36 mg/l CO2 released = 12[(3 • 44)/(600)] = 2.64 mg/l (c) Yes. Addition of lime slurry results in neither a reduction in alkalinity nor production of CO2. Lime slurry is much more difficult to prepare and feed than sodium carbonate solution, therefore, for small treatment plants, soda ash is often the chemical of choice and the carbon dioxide produced can be removed by air stripping in a counter-current packed tower.

7-30. Ferrous sulfate dosage = 30 mg/l = 30 • 8.34 = 250 lb/mil gal By Eq. 7-8, 2 • 278 lb + 2 • 74 lb � 2 • 107 lb FeSO4•7H2O + Ca(OH)2 � Fe(OH)3 Therefore, 556 = ___148___ = ___214___ 250 lb Y lb Ca(OH)2 Z lb Fe(OH)3 Y = 250 • (148/556) = 66.5 lb/mil gal (8.0 mg/l) Pounds of 70% CaO = 66.5 • (56/74) • (1/0.70) = 72 lb/mil gal (8.6 mg/l) Z = 250 • (214/556) = 96 lb/ mil gal (12 mg/l)

7-31. 2FeCl3 + 3Ca(OH)2 = 2Fe(OH)3 + 3CaCl2

324 222 214

(a) 40222324

56120 7

2⋅ ⋅ =. (

).

CaO)74.1(Ca(OH)

mg / l CaO

(b) 40214324

8 34 220⋅ ⋅ =. lb / mil gal

7-32. From Equation 7-11, CaSO4 is formed in an amount equivalent to the ferrous sulfate added which equals: 35/139 = 0.252 meg/l Therefore, Ca added = 0.252 • 20 = 5.0 mg/l SO4 added = 0.252 • 48 = 12.0 mg/l

7-33. Alkalinity increase = 20.0(50/53) = 18.9 mg/l in the form of bicarbonate ion.

7-34. Activated carbon is the most effective in taste and odor control. Heavy prechlorination forms trihalomethanes.

7-35. (a) Presedimentation is to remove silt, clay and other suspended particles that are settleable. (b) Chemical coagulation agglomerates colloidal impurities including pathogens, such as, Giardia cysts and Crytoporidium oocysts. (c) Activated carbon is for taste and odor control and for the reduction of soluble organic compounds. (d) Sedimentation removes settleable floc. (e) Filtration removes non-settleable floc to reduce turbidity with the goal of clarifying the water to ensure significant reduction of pathogenic protozoal cysts. (f) Postchlorination is for disinfection of remaining pathogens and to establish a chlorine residual in the distribution system.

50

7-36. Activated carbon can be added at any point before filtration, but usually between flocculation and sedimentation or sedimentation and filtration. Alum is applied in rapid mixing. Chlorine may be added anywhere during treatment but commonly after filtration and clear well storage. If trihalomethane formation is not a problem, prechlorination of the raw water can be practiced. Fluosilicic acid is generally added after filtration. Polyelectrolyte can be applied immediately after rapid mixing and/or at some point during flocculation before sedimentation.

7-37. SOCs are manufactured chemicals used in industry, agriculture, and household applications. Conventional water treatment by coagulation-sedimentation-filtration provides limited removal of SOCs and powdered activated carbon for taste and odor control is unsuccessful in absorbing SOCs. The limitations in using aeration to remove VOCs from contaminated well water are the low allowable maximum contaminant levels for drinking water, poor removal at low temperature, and ice formation in stripping towers.

7-38. Assume a commercial purity of 30 percent fluosilicic acid.

galmil/lb25l/mg0.330.079.03.00.1

Dosage ==•−=

7-39. Fluoride concentration = 82 0 30 0 79

2 3 8 3410

⋅ ⋅⋅ =. .

. .. mg / l

7-40. Gram of fluoideGram of solution

=+

=⋅ ⋅5 0 0 98 0 615 0 95

0 030. . .

..

Solution dosage = 0.80/0.030 = 26.7 g/m3

7-41. From Table 5-2, the optimum fluoride concentration = 0.9 mg/l

Fluosilicic acid usage = ( . . ) .

. .,

0 9 0 2 550 8 340 25 0 79

16 300−

⋅=⋅

lb / yr

7-42. Yes. Refer to Eqs. 7-12 and 7-13.

7-43.

7-44. Available chlorine = 8.34 • 0.15 = 1.25 lb/gal Volume required = (0.60 • 6.0 • 8.34)/1.25 = 24 gal

7-45. Dry hypochlorite = (400 • 1000 • 0.01)/0.70 = 5700 g

51

7-46. Quantity of water pumped per day = 400 • 60 • 18 = 432,000 gal Chlorine usage = 0.432 • 0.50 • 8.34 = 1.80 lb/day Duration of chlorine gas from 100-lb cylinder = 100/1.80 = 56 days Duration of liquid sodium hypochlorite = (0.10 • 100 • 8.34)/1.80 = 46 days Calcium hypochlorite powder added to 50 gal of water for a 15% available chlorine solution = (0.15 • 50 • 8.34)/0.70 = 89 lb Duration of solution = 89/1.80 = 35 days For a cold climate, chlorine gas is recommended provided design and operation minimizes hazards of handling, feeding, and storing chlorine. The well house should be designed with the safety features of a chlorination room. The risk of liquid chlorine in a cold climate is freezing of the solution. For a warm climate, liquid sodium hypochlorite is recommended provided it can be delivered be reliably by a supplier. Chlorine gas can be used in a warm climate if the design recognizes the risk of high temperature on chlorine cylinders. Use of calcium hypochlorite is not recommended since it is hazardous to handle and difficult to prepare in solution.

7-47. Chlorine dosage = 80 000

100 0000 80 0 80

,,

. .= =g / m3 mg / l

7-48. Regulating and compensating valves are needed to maintain a constant chlorine feed with varying outlet pressure, supply pressure, and cylinder temperature.

7-49. Refer to Figure 7-20. The four-function valve with a return line to the chemical container is for safety. The rate of chemical feed is controlled by speed of rotation of the disk in the head and stroke length of the shaft pressing on the diaphragm.

7-50. Pounds of hypochlorite = 40 8 34 0 01

0 704 8

⋅ ⋅ =. ..

. lb / 40gal

Feed rate = 50 1mg / l

10,000 mg / lvolume of 1.0% solution200 volumes of water

=

7-51. Commercial hypochlorite = 950 70

1000 0 7095

⋅⋅ =

.kg

7-52. Refer to Section 7-13, Disinfection By-Products The suspected health risk of THMs and HAA5 in drinking water is cancer. This risk has been demonstrated by carcinogenicity in laboratory animals. THMs and HAA5 in treated water are by-products of chlorination of humic substances in water. Remedial actions to reduce formation of THMs and HAA5 in water treatment are: (1) move the point of chlorine application to the finished water and reduce the applied dosage, (2) optimize coagulation to reduce the turbidity and organic content of the water prior to chlorination, (3) apply powdered activated carbon to adsorb humic substances, and (4) consider the application of alternative disinfectants.

7-53. Ozone is a strong oxidizing gas (O3) generated by passing dry air or oxygen between two electrodes with a high voltage potential. (Refer to Section 7-15, Ozone). Ozone is applied in an ozone contactor with fine bubbles rising through the downward flow of water (Figure 7-31). Ozone can be applied in water treatment for (1) inactivating microorganisms, (2) oxidizing iron, manganese, sulfide, nitrite, and selected trace organic compounds, (3) taste, odor, or color control, (4) oxidation of precursors that form trihalomethanes upon chlorination, and (5) destabilization of selected colloids.

52

Ozonation does not produce a disinfecting residual, therefore, chlorine is added to treated potable water before distribution to establish a disinfecting residual.

7-54. The process of chemical coagulation and granular-media filtration is essential to disinfection of surface waters to remove protozoal cysts, remove suspended and colloidal solids that protect bacteria and viruses from chlorine inactivation, and reduce the numbers of bacteria and viruses.

7-55. Giardia lamblia and Crytosporidium species cause diarrhea that can be life-threatening to persons with immunodeficiency syndrome (Section 3-7). Giardiasis and crytosporidiosis are transmitted by ingestion of protozoal cysts in polluted water. Other forms of transmission by the fecal-oral route are by contaminated food and person-to-person transmission. Waterborne sources are from feces of diseased humans and, in the case of Giardia, by beavers and, in the case of Crytosporidium, by feces of infected cattle and sheep.

7-56. The C·t product is the residual disinfectant concentration times the contact time (duration time) that the disinfection concentration is maintained in the water prior to use. It is expressed in the units of milligrams/liter·minute. The major factors that affect the C·t value used in design are the kind of disinfectant, temperature, pH, the kind and viability of the microorganisms to be inactivate, the presence of interfering substances, such as, suspended solids. Bacteria are most readily inactivated by free chlorine and protozoal cysts the most difficult to inactivate.

7-57. The chemical disinfectants in order of most effective to least effective, and the C·t values for 90% (1.0 log) inactivation of Giardia cysts at 10o C and pH 7 are as follows: Ozone 0.48 (mg/l)·min Chlorine dioxide 7.7 (mg/l)·min Free chlorine 37. (mg/l)·min Preformed chloramine 620. (mg/l)·min

7-58. The C·t from Table 7-3 is 134 (mg/l)·min at 10o C. At 20o C, the C·t = 67, which is 1/2 the 10o C value. At 5o C, the C·t = 179, which is 4/3 the 10o C value.

7-59. C = 2.0 mg/l t C/Co t C/Co T C/Co 0 0 25 0.21 55 0.76 5 0 30 0.35 65 0.82

10 0 35 0.44 75 0.92 15 0 40 0.57 85 0.91 20 0.05 45 0.64 95 0.92

53

7-60. The requirements specified by the EPA SWTR are: (a) At least 99.9% (3 log) removal and/or inactivation of Giardia and at least 99.99% (4 log) removal and/or inactivation of enteric viruses. (b) Turbidity less than or equal to 0.5 NTU in 95% of monthly measurements and not to exceed 5 NTU at anytime. (c) Chlorine residual not less than 0.2 mg/l for water entering the distribution system for more than 4 hr.

7-61. For conventional treatment, a 2.5-log removal of G. lamblia is achieved by filtration, leaving 0.5-log inactivation required by disinfection. The required C·t value for 1.0-log inactivation from Table 7-4 is 50 mg/l·min at pH 7 and 5o C . (Note that this value is based on a free residual of 1.0 mg/l. If a C·t value for inactivation were given for a lower chlorine residual, it would be lower. In other words, C·t values at a residual of 1.0 mg/l are conservative for postchlorination, since levels above about 0.5 mg/l in drinking water are intolerable for most customers.) C·t for 0.5-log inactivation = (1/2)50 = 25 (mg/l)·min From Figure 7-32c for a peak-hourly flowrate of 3.0 mgd, the t10 is 90 min. Therefore, the required free chlorine residual in the effluent from the baffled reservoir is

mg/l28.0min90

minmg/l)(25=C =

⋅

This chlorine residual also satisfies the requirement of a minimum residual of 0.2 mg/l for water entering the distribution system.

7-62. The chlorine disinfection after direct filtration treatment is 1.0-log Giardia inactivation and 3.0 log virus inactivation (Figure 7-33). From Table 7-4 at pH 7.5 and 15o C for 1.0-log Giardia, C·t = (25 + 36)/2 = 30.5 mg/l·min From Table 7-5 at pH 7.5 and 15o C for 3.0-log virus, C·t = 3 mg/l·min Time t in pipeline = 4000/(5 • 60) = 13.3 min C = 30.5 mg/l·min/(22 + 13.3 min) = 0.9 mg/l During this period of peak hourly flow, some customers living in the area where the water enters the distribution system are likely to complain of excess chlorine odor in the water.

7-63. (a) 0 3.7 4.7 5.7 6.2

Ca Mg Na K HCO3 SO4 Cl

0 4.0 5.2 6.2 3.7 Ca(HCO3)2 , 0.3 Mg(HCO3)2 , 0.7 MgSO4 , 0.5 Na2SO4 , 0.5 NaCl , 0.5 KCl Hardness = 4.7 • 50 = 235 mg/l (b)

Component meq/l Lime Soda Ash Ca(HCO3)2 3.7 3.7 0 Mg(HCO3)2 0.3 0.6 0

MgSO4 0.7 0.7 0.7 5.0 0.7

Lime dosage = 5.0 • 28 + 35 (excess) = 175 mg/l CaO

54

Soda Ash = 0.7 • 53 = 37 mg/l Na2CO3

0 0.6 0.8 2.5 3.0 Ca Mg Na K

CO3 HCO3 SO4 Cl 0 0.8 2.0 3.0

(d) Hardness to be removed = 4.7 - (120/50) = 2.3 meq/l Lime dosage to precipitate 2.3 meq/l of Ca(HCO3)2 = [2.3 + 0.6 (practical limit)] 28 = 81 mg/l CaO

7-64.

7-65. First stage recarbonation converts OH¯ to CO3= .

OH¯ from excess lime = 35/28 = 1.25 meq/l OH¯ from Mg(OH)2 = 10/50 = 0.20 meq/l CO2 required = (1.25 + 0.20)22.0 = 31.9 mg/l CO2 [22.0 is the equivalent weight of CO2] Second stage recarbonation converts 60 percent of CO3

= to HCO3¯ . CO2 required = (0.80 meq/l of CO3

= ) • (0.60) • (22.0) = 10.6 mg/l CO2 Total CO2 reacted in recarbonation = 31.9 + 10.6 = 42.5 mg/l Assuming 20% of CO2 gas applied escapes, the total CO2 applied = 42.5/0.80 = 53 mg/l

55

7-66. Calcium = 63/20.0 = 3.15 meq/l Magnesium = 15/12.2 = 1.23 Sodium = 20/23.0 = 0.87 Potassium = 10/39.1 = 0.26 Total cations = 5.51 meq/l Carbonate = 16/30.0 = 0.53 Bicarbonate = 189/61.0 = 3.10 Sulfate = 80/48.0 = 1.67 Chloride = 10/35.5 = 0.28 Total anions = 5.58 Component meq/l Lime CaCO3 0.53 0 Ca(HCO3)2 2.62 2.62 Lime = 2.62 • 28 = 73 mg/l Final hardness = (0.60 as Ca + 1.23 as Mg) 50 = 91 mg/l

7-67. Carbon dioxide = 15/22.0 = 0.7 meq/l Calcium = 60/20.0 = 3.0 Magnesium = 24/12.2 = 2.0 Sodium = 46/23.0 = 2.0 Alkalinity = 200/50 = 4.0 Sulfate = 96/48.0 = 2.0 Chloride = 35/35.5 = 1.0 0.7 0 3.0 5.0 7.0 0.7 0 4.0 6.0 7.0 Dosages for excess lime treatment of 2/3 flow Component meq/l Lime Soda Ash CO2 0.7 0.7 • 2/3 = 0.47 0 Ca(HCO3)2 3.0 3.0 • 2/3 = 2.00 0 Mg(HCO3)2 1.0 2.0 • 2/3 = 1.33 0 MgSO4 1.0 1.0 • 2/3 = 0.67 0.67 4.47 0.67 Chemical additions to first stage flow Lime dosage = 4.47 • 28 + 35 = 160 mg/l CaO Soda ash dosage = 0.67 • 53 = 36 mg/l 1.25 0 0.6 0.8 3.8 Excess lime 0 0.2 0.8 2.8 3.8 Bar graph of water after excess lime softening

CO2 CaClHCO3

Mg NaSO4

Ca Mg Na OH OH SO4 ClCO3

Ca

56

Blending of first stage softened water and by-passed raw water in second stage split treatment Components in by-passed water in meq/l CO2 = (1/3)0.7 = 0.23 HCO3 = (1/3)4.0 = 1.33 Ca = (1/3)3.0 = 1.0 Mg = (1/3)2.0 = 0.67 Na = (1/3)2.0 = 0.67 Components in excess-lime treated water Ca = (2/3)(0.6 + 1.25) = 1.23 Mg = (2/3)0.2 = 0.13 Na = (2/3)3.0 = 2.0 CO3 = (2/3)0.6 = 0.4 OH = (2/3)(1.25 + 0.20) = 0.97 Reactions when waters are blended CO2 is eliminated by OH, 0.97 - 0.23 = 0.74, which reduces both Ca and OH by 0.23 0.74 Ca(OH)2 reacts with Ca(HCO3)2 reducing HCO3 by 0.74 and precipitating 1.48 CaCO3 minus the practical limit of 0.6 meq/l Calculating concentrations in finished water after sedimentation and filtration Ca = 1.0 + 1.23 – 1.48 + 0.6 = 1.35 Mg = 0.67 + 0.13 = 0.80 Na = 0.67 + 2.0 = 2.67 Total cations = 4.9 HCO3 = 1.33 – 0.74 = 0.59 CO3 = 1.35 (same as Ca) SO4 = 2.0 (same as raw water) Cl = 1.0 (same as raw water) Total anions = 4.9 0 1.35 2.15 4.9 0 1.35 2.15 4.9 Estimated bar graph of finished water after split treatment

7-68. (a) 0 3.5 4.3 4.9 5.0

Ca Mg Na K HCO3 SO4 Cl

0 4.0 4.6 5.0

Component Meq/l Lime Soda Ash Ca(HCO3)2 3.5 3.5 0 Mg(HCO3)2 0.5 1.0 0 MgSO4 0.3 0.3 0.3 4.8 0.3

Lime dosage = 4.8 • 28 + 35 = 169 mg/l CaO Soda ash = 0.3 • 53 = 16 mg/l Na2CO3 0 0.6 0.8 1.7 1.8

Ca Mg Na K CO3 HCO3 SO4 Cl

0 0.8 1.4 1.8

HCO3 ClSO4CO3

Ca NaMg

57

(b) Lime dosage = 3.5 • 28 = 98 mg/l 0 0.6 1.4 2.0 2.1

Ca Mg Na K CO3 HCO3 SO4 Cl

0 0.6 1.1 1.7 2.1 Selective calcium is possible since magnesium hardness does not exceed 40 mg/l and the water contains sufficient alkalinity. (c) Lime dosage = 0.65 • 169 = 110 mg/l CaO Soda ash dosage = 0.65 • 16 = 10 mg/l Na2CO3 Excess hydroxide = 0.65(1.25 + 0.20) = 0.94 meq/l Ca(HCO3)2 = 0.35 • 3.5 = 1.23 meq/l Calcium = [(1.23 - 0.94) + 0.6]50 = 45 mg/l Magnesium = (0.35 • 0.8 + 0.65 • 0.2)50 = 21 mg/l Total hardness = 45 + 21 = 66 mg/l

7-69. 0 4.7 6.7 7.3

Ca Mg Na HCO3 SO4 Cl

0 5.2 6.6 7.3

Component meq/l Lime Soda Ash Ca(HCO3)2 4.7 4.7 0 Mg(HCO3)2 0.5 1.0 0 MgSO4 1.4 1.4 1.4 MgCl2 0.1 0.1 0.1 7.2 1.5

Lime = 7.2 • 28 + 35 = 240 mg/l CaO Soda ash = 1.5 • 53 = 80 mg/l Na2CO3 First-stage CO2 = 1.25 + 0.2 = 1.45 meq/l Second-stage CO2 = 0.75(0.6 + 0.2) = 0.60 meq/l CO2 reacted = (1.45 + 0.60)22 = 45 mg/l 0 0.6 0.8 2.9

Ca Mg Na CO3 HCO3 SO4 Cl

0 0.6 0.8 2.1 2.9

58

7-70. To calculate the quantity of by-passed flow consider:

XQ • 24 + (Q - XQ) • 10 = Q • 40, X = 1/3 Chemical doses needed: Lime = (2/3) • 240 = 160 mg/l CaO Soda ash = (2/3)80 = 53 mg/l Na2CO3 Estimating the final water hardness, Excess OH = (2/3) (1.25 + 0.20) = 0.97 meq/l Ca(HCO3)2 = (1/3) • 4.7 = 1.57 meq/l Calcium hardness = [(1.57 - 0.97) + 0.60]50 = 60 mg/l Magnesium hardness = (2/3) • 10 + (1/3) • 10 = 40 mg/l Total hardness = 60 + 40 = 100 mg/l

7-71. 0 3.0 4.2 5.0

Ca Mg Na HCO3 SO4 Cl

0 3.4 4.4 5.0 Hardness = 4.2 • 50 = 210 mg/l, which is a moderatley hard water for most consumers. Since the magnesium hardness is 60 mg/l, which is greater than the recommended maximum concentration of 40 mg/l, selective calcium removal is not recommended. Excess lime softening is possible, however, split treatment is recommended to reduce chemical consumption and eliminate the need for recarbonation. Although only 0.40 of Q requires first-stage treatment to reduce the magnesium to 40 mg/l, the estimated final hardness would be 130 mg/l. A better split flow is 0.60 Q given first-stage softening for an estimated final hardness of 77 mg/l.

7-72. Manganese not taken out by the aeration-filtration process pass into the distribution system and the manganese in the treated water used in the solution-feed chlorinator can precipitate in the injector mechanism.

7-73. Chlorine is oxidizing the iron and manganese forming precipitates that settle out in the distribution system. Periodically these are flushed out of the mains and storage tanks to cause “rusty” water that results in staining. Apparently prior to chlorination most of the iron and manganese remained in solution and colloidal suspension and passed through the system unnoticed.

7-74. The manganese zeolite process (Section 7-9) requires manganese-treated greensand to prevent passage of manganese through the filter from either inadequate oxidation of the manganese in the raw water or overdosing of potassium permanganate. In small water systems, the concentrations of iron and manganese in the raw water can change rapidly if different wells supply water of varying quality, which is often the case. Plain sand filtration is not satisfactory since the permanganate dosing cannot be easily paced to the varying demand. If overdosing of permanganate is the main problem, prechlorination is a competitive oxidizing reaction that will make the overdosing worse. A complete study is recommended including: iron and manganese testing of the raw, chemically treated and filtered water; and experimentation with a manganese-treated greensand filter medium.

7-75. Aeration strips out dissolved gases and adds oxygen. Oxidation chemicals are free chlorine residual (HOCl, OCl¯ ) and potassium permanganate (KMnO4). Flocculent oxides (MnO2 and FeOx) are not large or heavy enough to settle by gravity.

59

7-76. In corrosion of iron, Eq. 7-24 represents corrosion and Eq. 7-26 represents scaling. Ductile-iron pipe is lined with a thin layer of cement mortar placed during manufacture to protect against corrosion.

7-77. Cathodic protection prevents internal corrosion of a steel water tank by hanging sacrificial anodes in the water that are energized by an electric potential (flow of electrons) between the anodes and the steel tank. The sacrificial anodes go into solution releasing ions of the anodic material and electrons. The reverse potential prevents release of electrons and Fe++ ions thus preventing corrosion Eq. 7-24.

7-78. Lead toxicity effects the red blood cells, nervous system, and kidneys with young children, infants, and fetuses being most vulnerable. First-flush samples are used to assess lead contamination since the primary sources are lead solder joining copper pipe, lead goosenecks in old service lines, and old brass fixtures. The recommended methods of controlling excessive lead are corrosion control and replacement of long lead service lines.

7-79. Groundwater disinfection is based on inactivation of viruses since they are more likely to be carried to a well by groundwater flow through the soil. Giardia cysts being much larger are removed by soil filtration. Natural disinfection is the natural attenuation of enteric viruses by inactivation with time and sorption on soil particles as water filters through subsurface soils. A variance exempts compliance with a treatment technique for justifiable reasons provided no risk to public health results. From Table 7-5, the C·t for a 3.0-log inactivation by free chlorine at 5o C is 6 (mg/l)·min.

7-80. The contact time in the pipeline = =⋅4200 8 12

2000 013422

2[ ( / ) ].

πmin

The required chlorine residual = 6/22 = 0.3 mg/l.

7-81. From Table 7-5, the C·t for 3.0-log inactivation by free chlorine at 10° C is 4 (mg/l)·min. C·t = 1.8 • 0.8 + 8.0 • 0.4 = 4.6 (mg/l)·min (OK)

7-82. 0 5.7 6.2

Na K HCO3 SO4 Cl

0 4.0 5.2 6.2 The disadvantages of this water for a municipal supply are corrosivity (absence of calcium), the high sodium content for persons on a low-sodium diet, and total dissolved solids of about 450 mg/l.

7-83. The fluoride ion (F¯ ) is not removed by the cation exchange resin used in home water softeners since only hardness ions are replaced by sodium ions.

7-84. Excessive nitrate can cause methemoglobinemia in infants.

7-85. The water is very high in sulfate ion that is preferentially removed by the anion resin. By replacing both the sulfate and nitrate ions with chloride, excessive regeneration of the resin is required and the finished water is high in chloride ion.

60

7-86. Review Section 7-26 for an explanation of flow paths. System A was more effective in cleaning the fibers and required significantly less filtered water for backwash.

7-87. The forced passage of water through a membrane is called reverse osmosis since the water flows in the reverse direction of the natural osmotic pressure.

7-88. Refer to Figure 7-38. Saline feed water enters one end of a module and flows through mesh spacers between membrane envelopes, Under high pressure, water is forced through the membranes and flows through the porous collectors to a central permeate tube from which it discharges as product water. The saline feed water that does not pass through the membrane leaves the module at the discharge end as reject brine.

7-89. Acid added to feed water entering RO modules lower the pH, often to pH 5.8, and converts bicarbonate ions to carbon dioxide. The presence of carbon dioxide makes the water corrosive preventing formation of scale such as calcium carbonate.

7-90.

(a) The quality of this water is poor for a municipal supply. The fluoride concentration of 1.4

mg/l is too high for a hot climate and will result in mottling of teeth. The nitrate concentration exceeds the maximum contaminant level and should not be used for infant feeding. The sodium content is too high for drinking by persons on a low sodium diet for hypertension. Esthetics problems are related to excessive dissolved solids and chloride that can effect the gastrointestinal tract of sensitive persons and the iron and manganese concentrations are high enough to cause red water and staining problems. (b) The following is the calculated bar graph after excess lime softening.

The fluoride ion may be reduced by excess lime softening by the precipitation of CaF2, but the degree of removal cannot be predicted. Nitrate is not removed. The sodium content is increased. The total dissolved solids still exceed 500 mg/l (estimated TDS is about 800 mg/l) and the chloride ion is unchanged. The advantages are reduction of calcium and magnesium and removal of iron and manganese. (c) The high dissolved solids concentration must be reduced and the most suitable process is reverse osmosis. Pretreatment is required to remove the iron, manganese, and any suspended solids that result from silt drawn through the well screen or corrosion of piping. The processes employed depend on the size of the treatment plan. In a small system, the iron and manganese can be removed by chemical oxidation, sedimentation, and filtration. (Section 7-9). Coagulation may be included in the process to flocculate suspended solids if further investigation indicates that the well water contains silt and metal oxides. Aeration may also be added to aid in oxidation and for cooling if the water temperature is likely to exceed 35° C as a result of above ground storage. RO membrane life is shortened if the water temperature exceeds 35° C to 40° C. Suitable pretreatment for a large plant is lime precipitation with the limited addition of soda ash to provide adequate floc formation followed by granular-media filtration. This processing also removes iron, manganese, and suspended solids. Recarbonation may be used before filtration to lower the pH if carbon dioxide is readily available from an adjoining power plant, otherwise, sulfuric acid can be applied to lower the pH. The following is a general flow scheme for RO treatment and post treatment.

61

Sulfuric acid lowers the pH to 5.8 to make the water corrosive and hexametaphosphate at about 10 mg/l is a corrosion inhibitor. The cartridge filters remove particles down to a size of 5 micron. With a feed pump pressure of about 350 psi, a recovery of product water with satisfactory quality (250 to 350 mg/l TDS) equals 75% of the feed water. The forced-air stripping tower removes the majority of the carbon dioxide to stabilize the water. The final adjustment to near pH 8 is done by addition of about 10 mg/l of soda ash. Fluoride compound is added to restore the concentration to the optimum level, and chlorine is added to establish an adequate protective residual in the distribution system. Sludge from lime treatment, filter backwash, and waste brine are pumped to an evaporation pond for disposal. (d) Ion removals are generally more than 90% for multi-valent ions and between 80 and 90% for mono-valent ions. With lime precipitation as pretreatment, the concentration of calcium ion may be too low for a stable water. One method of increasing the calcium content is to add 10 to 20% groundwater after direct filtration, with coagulant addition if necessary, to the RO product water after air stripping. With an approximate blend of 10% raw groundwater and 90% RO product water, an estimated finished bar graph with about 500 mg/l of TDS is shown below. 0 6.5

Ca Mg Na HCO3 SO4 Cl

0 6.5

7-91. Using Equation 7-37, sludge solids = 8.34 (40/4 + 12) = 180 lb/mil gal

V =sludge solids

solids fraction 8.34gal / mil gal⋅ ⋅= =180

0 001 8 3422 000

. .,

7-92. Using Equation 7-18, 2CaCO3/CaO = 200/56.1 CaCO3 formed = (126 • 0.78) (200/56.1) = 350 mg/l

7-93. Component meq/l CaCO3 Mg(OH)2 Ca(HCO3)2 4.7 9.4 -- Mg(HCO3)2 0.5 1.0 0.5

MgSO4 1.4 1.4 1.4 MgCl2 0.1 0.1 0.1

Excess lime 1.3 1.3 -- 13.6 2.0 Practical limits -0.6 -0.2

13.0 meq/l 1.8 meq/l

62

Residue = 13.0 • 50 + 1.8 • 29.2 = 703 mg/l Pounds of dry solids = 703 • 8.34 = 5900 lb/mil gal Assuming a sludge specific gravity of 1.0,

Using Equation 11-37, Sludge volume = 703 8 34010 8 34

⋅⋅

.. .

= 7000 gal/mil gal

7-94. (a) Waste sludge from excess lime softening Component meq/l CaCO3 Mg(OH)2

Ca(HCO3)2 3.5 7.0 -- Mg(HCO3)2 0.5 1.0 0.5 MgSO4 0.3 0.3 0.3 Excess lime 1.3 1.3 -- 9.6 0.8 Practical limits -0.6 -0.2

9.0 meq/l 0.6 meq/l

Residue = 9.0 • 50 + 0.6 • 29.2 = 467 mg/l = 3900 lb/mil gal Sludge volume = 3900/(0.8 • 8.34) = 5800 gal/mil gal (b) Waste sludge from selective calcium carbonate removal. The lime addition for 3.5 meq/l of Ca(HCO3)2 produces 7.0 meq/l of CaCO3 of which 6.4 meq/l (7.0 - 0.6) are precipitated, therefore, the residual produced is 6.4 • 50 = 320 mg/l = 2670 lb/mil gal. Sludge volume = 2670/(0.12 • 8.34) = 2700 gal/mil gal. (c) Waste sludge from split-treatment softening. CaCO3 precipitate + Mg(OH)2 precipitate = total residue (9.6 • 0.65 - 0.6)50 + (0.8 • 0.65 - 0.2)29.2 = 290 mg/l Alternate solution is: calcium reacted + lime added - practical limit = CaCO3 precipitate 0.65 • 3.5 + 110/28 - 0.6 = 5.6 meq/l magnesium reacted - practical limit = Mg(OH)2 precipitate 0.65 • 0.8 - 0.2 = 0.32 meq/l Total residue = 5.6 • 50 + 0.32 • 29.2 = 290 mg/l

7-95. The majority (50 to 90%) of the iron and manganese oxides are trapped in the filter bed.

7-96. The easiest of the three sludges to dewater is CaCO3 slurry.

7-97. The advantages of pressure filtration relative to centrifugation are (b) dewatering to a solid cake and (c) ease of dewatering hydroxide sludges.

63

8 OPERATIONS OF WATERWORKS

8-1. The major steps in inspection of a fire hydrant are: 1. Remove outlet-nozzle cap and check for water in barrel. 2. Detect for leaks from joints and nozzle caps after opening the main valve with caps in place. 3. Check drainage from barrel. 4. Check lubrication of packings. 5. Inspect nozzle caps for drainage, cleaning, and lubricating. Preventive maintenance of valves involves checking that the location is correct on maps, inspection of valve box and cover, and turn valve closed and compare the number of turns required against the number recorded on valve records. Backflow preventers are inspected to ensure proper operation. The primary purposes of leak detection are to locate and repair small defects before failures occur and to reduce water loss from the system.

8-2. Cross connection control is important to prevent contamination of potable water in the distribution pipe network. Mechanical backflow preventers are described in Section 6-9. Back siphonage is backflow resulting from reduced pressure in the potable system or excessive pressure from a source contaminated with chemicals or pathogens.

8-3. Sources of real water loss are associated with leakage in the water distribution system, leaks in storage tanks, or leaks in the service connections before the customer meters. Apparent losses include connections to the water system that are unauthorized and unmetered. Most apparent losses are related to inaccurate metering.

8-4. Using Eq. 8-1,

( ) ( ) gpm723122.00.929.8psi12atQ 0.52 =•= ( ) gpm12013723psi02atQ ==

Using Eq. 8-2,

gpmQR 230043502050

112050.0

=���

����

�

−−=

2300 gpm > 1500 gpm, Yes.

8-5. Using Eq. 8-2,

gpmQR 94027452045

80050.0

=���

����

�

−−=

8-6. From Figure 4-7 (Hazen-Williams nomograph) loss of head in a 12 in. pipe at a flow of 2500 gpm is 25 ft/1000 ft. Therefore, head loss in the 3400 ft main = 85 ft / 2.31 = 37 psi. Water pressure remaining = 63 – 37 = 27 psi In Section 6-2, ISO specifies adequate minimum pressure is 20 psi during fire flow.

64

8-7. A single station controller allows the operator to enter a setpoint value and control such functions as water level and chemical addition. Programmable logic controllers perform the same functions but are able to control more than one operation. SCADA (supervisory control and data acquisition) systems use a personal computer for data storage and display and the entire control system together with a computer network.

8-8. Poor water supply quality is identified by: deviations in quality parameters, vulnerability to contaminants from land use activities in the watershed, and unusual events, such as sewage spills, heavy runoff, and droughts.

8-9. Water barriers include watershed management, monitoring quality of water supply, testing of treated water, and protection against contamination in the distribution pipe network. (Refer to Multiple-barrier Concept in Section 5-1).

8-10. Before issuing a boil water notice, epidemiologic evidence should be observed suggesting that the outbreak is waterborne. (Refer to Water Survey in Section 8-3)

8-11. A sanitary survey is a physical review of the watershed, water supply, treatment plant, distribution system, and operation and maintenance. (Refer to Sanitary Survey in Section 8-3).

8-12. An annual report summarizes laboratory analyses, operational tests, maintenance, business, chemical dosages, and cost data. (Refer to Section 8-4)

8-13. Water conservation, oriented toward reducing consumer usage is based on public education, installation of water-efficient plumbing fixtures, (Refer to High Efficiency Plumbing Fixtures) and establishing water rates that encourage conservation (Section 8-6).

8-14. Logging-computer technology is described in the third paragraph under the heading “High Efficiency Plumbing Fixtures” in Section 8-5. The average decrease in indoor water use of single-family homes in the Tampa Study was 49.6 percent.

8-15. SCADA and electronic computer files and programs. Physical protection of water sources. Physical and electronic surveillance of the perimeter of the utility property. Inside and outside television systems with visible cameras. Background checks of prospective employees.

8-16. The costs used to determine water rates include amount of water consumed, rate of use or demand for use, debt service including interest and stipulated reserves, utility extensions and improvements, expense involved in customer accounts, and plant replacement for perpetuation of the system.

65

9 WASTEWATER FLOWS AND CHARACTERISTICS

9-1. Wastewater flow and loading calc

No. per day GPD/UnitSubtotal

GPD

BOD lb/day /Unit

Subtotal BOD

ResturantEmployees 30 30 900 gpd 0.10 3.0 lb/dayGuests 200 8 1,600 gpd 0.06 12.0 lb/dayHotelEmployees 5 50 250 gpd 0.10 0.5 lb/dayGuests 40 50 2,000 gpd 0.10 4.0 lb/dayResidentialHomes 15 80 1,200 gpd 0.17 2.6 lb/dayApartments 20 75 1,500 gpd 0.17 3.4 lb/day

7,450 gpd 25 lb/day

Using Equation 9-1, BOD = 25/(0.007,45•8.34) = 410 mg/l

Note on selecting per unit values:Engineers must be conservative and should select the largest valuesthat are reasonable. Selecting average versus maximum values arealways a judgement decision that should be based in part on otherconservative values selected. Selecting conservative flow and loadingvalues followed by conservative sizing criteria will exaggerate theresults.

In this solution, the typical home was selected, while the highestvalue for apartments was selected to represent the best judementgiven a lack of actual field data.

From Table 9-1 each meal served is 4 gpcd, serving lunch and dinnerresults in a total of 8 gpcd and 0.06 lb BOD/person/day

9-2. Suspended solids %05240

100)120240( =−

BOD = 35%200

100)130200( =−

Suspended solids ( )240 30 100

240− = 88%

BOD = ( )200 30 100

200− = 85%

9-3. Flow = 12 MGD SS = 12 • 240 • 8.34 = 24,000 lb/day BOD = 12 • 200 • 8.34 = 20,000 lb/day TN = 12 • 35 • 8.34 = 3,500 lb/day

66

TP = 12 • 7 • 8.34 = 700 lb/day

9-4. Pretreatment programs address three types of discharge: prohibited discharges, categorical discharges, and local limits. Prohibited discharges apply to all dischargers regardless of the locally issued NPDES permit requirements. Categorical standards are listed in the NPDES permit. Local limits address specific industries and capabilities of the treatment plant to limit discharges and protect receiving waters.

9-5. See Table 9-3, under Milk production: Flow = 0.05 MGD BOD = 0.05 • 1000 • 8.34 = 417 lb/day SS = 0.05 • 300 • 8.34 = 125 lb/day TN = 0.05 • 50 • 8.34 = 21 lb/day TP = 0.05 • 12 • 8.34 = 5 lb/day Comparison with sewer ordinance limits: BOD > 500 mg/l limit SS is ok N > 35 mg/l limit P is ok

9-6. Refer to the data for meat processing in Table 9-4 Flow equivalent: 1,200,000 gpd/120 gpcd = 10,000 people BOD: 1300 mg/l / 200 mg/l = 6.5 times = 65,000 people SS : 960 mg/l / 240 mg/l = 4 times = 40,000 people

9-7. Average of 700 and 1500 gpd/ac = 1100 gpd/ac Infiltration = 500 • 0.25 • 1100 = 137,500 gpd Wastewater flow = 500 • 5 • 100 = 250,000 gpd Ratio = 137,000 / 250,000 • 100 = 55% as compared with 10% also given in the text. This may not be inconsistent because local climate and ground water levels vary greatly It is important to use locally acceptable values for infiltration and inflow rather than text book values.

9-8.

Infiltration using acres3.4•1500 = 5100 gpd

Infiltration using pipe

gal/day /in.-mile Length (ft)

Pipe infiltration

(gpd)8-in. 500 1200 910

10-in. 500 800 76014-in. 500 430 570

2240

Number of lots = 3.4•3 = 10.2Using 4 people per house = 10.2•4•120 gpcd = 5,000 gpd

Dry weather flow = 5,000 gpdWet Weather flow = 5,000 + 5,100 = 10,100 gpd or

= 5,000 + 2,240 = 7,200 gpd

67

9-9. Average dry weather flow = 125,000 • 80 gpcd / 1000000 = 10 MGD Peaking Factor = 2.5 / (10^0.145) = 1.8 Peak flow is 18 MGD

9-10. Flow from population 35,000

100 gpcd3,500,000 gpd from population

Peak flow from population (Eq. 9-3) PF = 2.08

7,296,560 gpd from population

Max Infiltration based on acres78000 acres

1500 gal/day/acre117,000,000 gpd from infiltration

OR 10% of average flow350,000

Max Inflow based on population5 gpcd

175,000 gpd from infiltration

Dry-weather flow 3,500,000 gpd or 3.5 mgd or 2,431 gpm

Wet-weather flow 124,471,560 gpd or 124.5 mgd or 86,439 gpm

Pumping Station Using 5 pumps, 4 to meet the firm capacity:

Each pump = 86,400 / 4 = 21,610 gpm

9-11. Table 9-2, for normal domestic values BOD = 0.75 • 850 + 0.25 • 200 = 690 mg/l N = 0.75 • 30 + 0.25 • 35 = 31 mg/l P = 0.75 • 0 + 0.25 • 7 = 1.75 mg/l BOD/N/P = 690/31/1.75 = 100/4.5/0.25 No, the generally accepted BOD/N/P = 100/5/1. N may be ok, but wastewater is show on P

9-12. From the table, 1500/30/0 = 100/2/0. The waste is inadequate in both N and P. Adding Raw wastewater is 100/17/2 / 3 = 33/5.7/1. The total is 133/7.7/1 or 100/5.8/0.8. The wastewater remains short of P.

68

9-13. From Table 9-4

Flow BOD SSPotato chip 90,400 gpd 450 lb/day 510 lb/dayMilk 65,100 gpd 760 lb/day 170 lb/daySoft-drink 16,000 gpd 64 lb/day 64 lb/day

172,000 gpd 1,300 lb/day 700 lb/day

For a sewered population of 12,500Table 9-2: 120 gpcd 200 mg/l 240 mg/lPopulation 1,500,000 gpd

BOD concentration = 1300 lb/day / 1.5 mgd / 8.34 = 2,500 lb/daySS concentration = 700 lb/day / 1.5 mgd / 8.34 = 3,000 lb/day

Industrial 172,000 gpd 1,300 lb/day 700 lb/dayPopulation 1,500,000 gpd 2,500 lb/day 3,000 lb/dayTotal 1,672,000 gpd 3,800 lb/day 3,700 lb/dayConcentration 270 mg/l 270 mg/l

9-14. Flow (mgd) Volume (ml)

8 AM 17.5 1459 AM 21.8 180

10 AM 24.7 20411 AM 26.9 22312 AM 25.5 211

1 PM 23.3 1932 PM 20.4 1693 PM 17.5 1454 PM 16 1325 PM 16 1326 PM 16.7 1387 PM 17.5 1458 PM 16.7 1389 PM 16 132

10 PM 14.6 12111 PM 13.1 10812 PM 10.2 841 AM 7.3 602 AM 5.8 483 AM 5.5 464 AM 5.5 465 AM 5.8 486 AM 7.3 607 AM 11.6 96

3004 ml

Average 15.1 mgdHourly sample portion needed = 3000/(15.1•24) = 8.3 ml/mgd

69

9-15. 3-hr Flow

(mgd)Average

Flow (mgd) MultiplierComposite

(ml)6 AM 0.97

0.323 * 460 = 1509 AM 2.91

0.970 * 460 = 45012 AM 3.4

1.133 * 460 = 52015 AM 2.33

0.777 * 460 = 36018 AM 2.23

0.743 * 460 = 34021 PM 2.14

0.713 * 460 = 33024 PM 1.55

0.517 * 460 = 2403 PM 0.74

0.247 * 460 = 110Total 16.27 = 2,500 mlAverage 0.68 0.68

total volume of composite sample desiredMultiplier = ------------------------------------------------------------

average flow rate • number of portions

Given 8 samples per day:Multiplier = 2500/0.68/8 = 460 ml/mgd

9-16. The presence of inhibiting or toxic substances, resulting from industrial wastewaters, is often indicated by increasing BOD values with increasing dilution, lag periods at the beginning of properly seeded tests, and erratic test results. Wastewater flow, day of the week, weather and rainfall, and abnormal wastewater discharges caused, for example, by the breakdown of pretreatment facilities at a major industry are essential for future interpretation of recorded testing data. Additional discussion is provided in Chapter 3: Very few industrial wastewaters have sufficient biological populations to perform BOD testing without providing an acclimated seed. Industrial wastes frequently have high strengths that make it difficult, or even impossible, to pipette accurately the small quantity desired for a single test bottle. Wastewaters high in suspended solids may be difficult to mix with water; one alternative is to homogenize the sample in a blender to aid dispersion in the dilution water.

70

10 WASTEWATER COLLECTION SYSTEMS

10-1. Combined sewers convey surface runoff from rainfall, domestic wastewater, and industrial discharges in one pipe collection system. Storm sewers are those pipes in a dual network that carry only storm runoff and unpolluted water, such as foundation drainage and cooling water from air conditioning and refrigeration units. Separate sanitary sewers allow collection of domestic and industrial wastewaters for treatment prior to discharge. Combined wastewater flows often exceed treatment capacities during rain storms resulting in by-passing untreated wastewaters to surface watercourses.

10-2. See text. Source control measures include street sweeping and cleaning, controlling litter and erosion, spills, leaks, overflow control, limits on the use of chemicals, and monitoring for illegal dumping. Retention ponds are designed to attenuate storm flow, settle heavy solids and provide a degree of treatment for organics, pesticides, nutrients, oils, and grease.

10-3. Wastewater flow = 500 • 480 = 240,000 gpd • 3.5 = 840,000 gpd or 580 gpm peak flow. From Table 10-1, a 10” pipe is too small at 490 gpm and a 12” pipe has the capacity of 700 gpm and should be selected.

10-4. The minimum slope listed in Table 10-1 is based on a velocity of 2.0 ft/sec using Manning’s equation with an n = 0.013.

10-5. The average sanitary wastewater produced is between 80 and 120 gpcd, which includes reasonable infiltration, peak (instantaneous) discharges from residential areas are many times greater. Hydraulic design is based on a maximum flow up to 4 times greater than the average or by detailed calculation of infiltration and inflow and wastewater peak flow.

10-6. Development Access road Additonal homes

Homes 520 180 440Unit flow (gpd) 750 750 750 Total flow = Homes * Unit flowTotal flow (gpd) 390,000 135,000 330,000AccumulatedTotal flow (gpm) 270 360 590 From Table 10-1Sewer dia. 8 10 12Capacity (gpm) 310 490 700

10-7. From Table 10-1, the flow capacity of a 24” sewer is 2820 gpm or 4.06 MGD 4,060,000 gpd /3.5 PF /4.5 people/home /120 gpcd = 2,100 homes

10-8. Multiple pipes are used to maintain higher cleaning velocities through the siphon. Pipes are sized to maintain a velocity of 3 ft/s. Average flow = 8 MGDPF = 2.5 / (8^0.145) = 1.85Peak flow = 14.8 MGDMinimum flow = 4 MGDSize one pipe for minimum flow and one pipe for the net peak flow

cu ft/secArea at 3 ft/s

(sq ft)Calculated pipe

dia. (in.)Net Peak flow = 10.8 MGD 23 5.6 64Minimum flow = 4 MGD 6.2 2.1 39

Small pipe, use 40" pipeLarge pipe, use 64" pipe

71

10-9. The minimum horizontal separation between sewer and water main is 10 ft and the minimum vertical spacing is 18” with the water main above the sewer.

10-10. The two methods of trenchless construction are horizontal boring and micro-tunneling. The horizontal boring machine remains in the drilling pipe while pushing segments of carrier pipe and removing material using a cutting wheel and auger. The micro-tunneling machine leads the pipe into the tunnel using a cutting wheel. Material is cut, mixed with a slurry material, and pumped from the tunnel.

10-11. Figure 10-5 shows three manhole types. The first is a standard manhole used for access for inspection and cleaning. They are placed at changes in sewer grade, pipe size or alignment, where lines intersect, at the end of lines, and at regular intervals not exceeding the length of cleaning tools. Drop manholes are necessary to lower the elevation of a sewer in a manhole. They are used to protect workers from the flowing wastewater when entering a manhole and to eliminate nuisance created by solids from splashing wastewater adhering to the walls of the manhole. Drop manholes are used when the vertical separation at the sewers are greater than 24”. Vortex sewers reduce the “water fall,” odors, and are typical when separation is greater than six ft.

10-12. Plumbing traps on house drains prevent sewer gasses from backing up into the house. The vertical vent pipe draws air in the plumbing trap allowing drainage to the sewer. Ventilation of lateral sewers is primarily through building vents of service connections.

10-13. The Palmer-Bowlus flume is preferred for measuring wastewater flow in a manhole because it can be installed for temporary use in a half section of a sewer pipe. A Parshall flume cannot be fitted into a sewer pipe, and weirs are difficult to install and collect stringy solids. Correct installation of a Palmer-Bowlus flume produces a smooth upstream flow and swift flow downstream to indicate free discharge (Section 10-3). Flow measurement is made using an ultrasonic sensor mounted above the measuring location, which is upstream a distance of one-half the diameter of the pipe for the flume (Figure 10-8). A submerged probe is mounted under the flowing water on the bottom of the channel at the measuring location. A bubbler is attached to a rigid tube with the outlet end submerged in the flow channel.

10-14. Sewer pipes may be corroded by electrochemical and chemical reactions with the surrounding soil and with chemicals in the wastewater. Biological activity in the wastewater produces hydrogen sulfide, which when released from solution, is converted to sulfuric acid at the crown of the pipe. The most common corrosion resistant material is VCP and PVC. Protective coatings include coatings of coal tar, vinyl, or epoxy.

10-15. The goal of stormwater runoff is to generate no more runoff after development than before. For green projects, at least 15% reduction in runoff must be met through pervious asphalt, sub surface piping, and ponding to reduce runoff.

10-16. Clay pipe is preferred for sanitary sewer because the material is corrosion resistant. Concrete pipe is used for storm sewers because of it’s relatively low cost, abrasion resistance, sizes, and high crushing strength.

10-17. Compression joints are shown in Figure 10-13 and consist of polyester casting with an “O” ring or bonded polyurethane. Minor change in pipe direction are made using curves, elbows, or by deflecting the joints.

10-18. To support the pipe, bedding of aggregate material like that in Figure 10-14 (d) and (f). In areas of poor soil, concrete may be used as shown in Figure 10-14 (e). Crushed rock is typically used with VCP.

10-19. For PVC pipe, the most appropriate bedding is Figure 10-14 (c) using sand to avoid puncturing the pipe.

72

10-20. Bedding types Figure 10-14 (a) and (b) are not appropriate for clayey-silt soils. For PVC and polyethylene-encased ductile-iron pipe, sand should be used to avoid puncturing the pipe or casing. For clay and concrete pipe, crushed rock should be used. To avoid settlement that may occur with native soils, the backfill aggregate may be extended up to the roadway.

10-21. When soil falls away, it is likely granular and the slope (given in Table 10-2) would need to be 1.5:1 or an angle of 34 degrees.

10-22. When soil falls away in chunks, it is likely cohesive. The degree to which the soil is cohesive, cannot be determined. From the information in Table 10-2, the safe slope is 1:1 or 45 degrees. The slope may be reduced to ¼:1 (53 degrees) if the compressive strength is > 1.5 ton/sq ft (>144 kPa).

10-23. For installation, see laser system, refer to explanation in Section 10-6 and Figure 10-16.

10-24. Trench compaction is tested using the ASTM Modified Proctor Test to determine the density and percent moisture content. The result is a graph showing moisture percent versus dry density. Trench compaction is used to calculate the percentage of the maximum density as compared with the engineer’s requirement. Equipment included sand cones, balloon meters, probes, and nuclear meters, see Figure 10-18.

10-25. Hole, 6" dia by 6" deep Dry Wt = 9.65 lb

Sample: 173 cu in or 0.1001157 cu ft = 96.4 Dry lb/cu ftOptimal density is 12% at 111 Dry lb/cu ft

As a percentage of optimal dry density, the field sample is 96.4/111 = 87%This compaction is < 90% and is inadequate.

Wet Wt = 10.8 lbField sample water weight = (10.8-9.65)/9.65 = 11.9 percent

The water content is close to optimal, therefore problems with compaction are not related to moisture content

The contractor needs to change compaction techniques and work effort to meet compaction requirements

10-26. Hole, 6" dia by 6" deep Dry Wt = 11.2 lb

Sample: 166 cu in or 0.0960648 cu ft = 116.6 Dry lb/cu ftOptimal density is 8% at 128 Dry lb/cu ft

As a percentage of optimal dry density, the field sample is 116.6/128 = 91%This compaction is < 95% and is inadequate.

Wet Wt = 11.6 lbField sample water weight = (11.8-11.2)/11.2 = 5.4 percent

The water content is not optimal and needs to be increased to improve compaction

73

10-27. Refer to Figure 10-19. Vibratory plates and ride-on rollers are used for granular materials, while sheeps foot, trench rollers, and roller attachments are used for cohesive soils.

10-28. If 13 gallons is lost in 30 min. is 26 gal/hr. The exfiltration rate is 26/10 or 2.6 gal/hr/100 ft. Working backwards in Table 10-3, for a 12 in. pipe, the 2.6 rate is a little less than 300 gal/in.dia/mile/day. If the limit is 200, the exfiltration rate is too great.

10-29. From Table 10-3, at 100 gal/in./mi/day and 8” pipe is 0.63 gal/hr/100 ft. The allowable exfiltration is 1.26 gallons. For air testing, see Table 10-4 to find a test time of 1.2 minutes and the total allowable time for 200 ft is 2.4 minutes.

10-30. From Table 10-3, for 8 in. pipe, the allowable leakage is 1.2 min/100 ft. For 250 ft, the allowable time is 1.2 • 2.5 • 60 = 180 sec. The time was also 180 sec., therefore the pipe is acceptable at the allowable leakage rate.

10-31. From Table 10-5, for a 20 ft-deep, 48” MH, the minimum test time is 50 sec. or 0.83 min. < 2.2 min. therefore the manhole is acceptable.

10-32. The common types of pumping stations are the submersible and self-priming as shown in Figure 10-21. In the case of the submersible, the pumps are in the wet well. To work on the pumps, they must be lifted to the surface. The self-priming pump is located at the ground and draws the wastewater up and out of the wet well.

74

11 WASTEWATER PROCESSING

11-1. Wastewater treatment can be viewed as a thickening process since each unit operation concentrates the remaining solids into a smaller volume of wastewater until the dewatered sludge is a cake. The volume of the raw wastewater of 120 gpcd with less than 0.1% solids is reduced to a filter cake of less than 0.5 pint per capita per day with 30% solids.

11-2. See Figure 11-1 Preliminary treatment consists of flow measurement, screening, pumping, and grit removal. Primary treatment consists of settling in a primary clarifier. Secondary treatment consists of biological treatment and settling in a secondary clarifier. Disinfection consists of the addition of chlorine, mixing, chlorine contact, in-plant water pumping, and discharge. Solids treatment consists of thickening, digestion, dewatering, and disposal. Screening is to remove large solids protecting pumps and other treatment units; a shredder can perform the same function in small plants. Pumping is required to lift the wastewater. Grit removal to remove sand and other heavy particulate matter that can cause abrasive wear on mechanical equipment and settle in tanks and pipes. Flow measuring to monitor wastewater flow. Typical arrangements of preliminary units are illustrated in Figure 11-4. Preliminary treatment does not impact flow, BOD, or suspended solids. Primary treatment is to remove settable solids such as paper, food waste, and the heavy fraction of suspended solids. Primary treatment is intended to remove what suspended solids will settle under quiescent conditions. A portion of the BOD is associated and removed with those solids reducing the load on secondary treatment. A small portion of plant flow is removed with the solids. Secondary treatment contains processes to biologically remove BOD and suspended solids. BOD reduction can be reduced in fixed film or suspended growth processes. BOD and suspended solids are reduced sufficiently for discharge. Solids settled in the Secondary Clarifier are removed with a small portion of the plant flow.

11-3. See Table 11-1 The pumping station is sized based on peak flow to establish firm capacity at 8.5 mgd and minimum flow using the smallest pump at 2 mgd. Primary treatment is sized using max. monthly flow of 4 mgd. Secondary treatment is sized based on BOD loading rather than flow. Disinfection is sized for the peak flow of 8.5 mgd. Solids treatment is based on max sludge flow and in some cases max. monthly solids loading. Numbers best representing capacity are stated above. Why? Specific numerical values of design parameters are frequently recommended in pulibshed standards

11-4. Screens protect pumps and prevent large solids from fouling subsequent units. Bar screens with openings from ½ to 2¼ in. remove large solids. Fine screens from 1 to 3 mm removes fibrous material and hair. Processes are similar in their removal of solids of a specific size. Processes are different in that bar screens have a rake to collect the solids and fine screens use a doctor blade and pressurized backwash to solids off of the screen.

11-5. Figure 11-8: Screenings treatment units prepare screenings for disposal by removing free water and reducing the amount of organic material in the screenings.

11-6. Figure 11-7: The bar screen at 3/4” spacing removes and average of 4.75 cu ft/mgd or 31 cu ft/day, the maximum value is over 7.6 cu ft/mgd or 49 cu ft/day. The fine screen removes about 11 cu ft/mgd or 72 cu ft to a max. value of about 16.5 or 17 cu ft/mgd or 111 cu ft/day. The total low estimate is 31 + 72 = 103 cu ft/day to a max. of 49 + 111 = 160 cu ft/day.

11-7. Grit removal units remove heavy particulate matter that cause abnormal abrasive wear on mechanical equipment, clog pipes, and accumulate in tanks. Early grit chambers consisted of long,

75

narrow channels in which the grit settled. Other grit removal devices include grit chambers, grit clarifiers, and forced-vortex grit units.

11-8. Grit processing is important because poor processing returns grit to the treatment plant. Augers become less efficient with age. Vortex grit removal and dewatering removes more grit and reduces grit in return flow.

11-9. Using equation 11-1, Grit = 670 • 1.2 (1.2/.93 – 1) = 233 lb/day

11-10. Figure 11-11 shows 6 pumps, 5 operate and 1 is standby. Each pump is 18/5 = 3.6 mgd or 2500 gpm. At 10 mgd, 2.8 pumps operate, actually 3 pumps operate at 92.6% speed.

11-11. For the ultimate capacity, 60 / 5 = 12 mgd or 8,300 gpm per pump. Existing peak flow will take 32 / 12 =2.7 pumps or 3 pumps at 89% speed. At the average flow of 15 mgd, 2 pumps will operate at a speed of 75% speed. The min. flow is 6.5 mgd and with 1 pump operating, the speed would be 54% speed which is greater than the minimum speed of 50%.

11-12. 150 homes at 850 gpd/home = 88.5 gpm, two pumps operating would be a firm capacity of 44.3 gpm each. The initial development of 60 homes at 850 gpd/home is 35.4 gpm. Therefore, one pump at 44.3 with a second pump as standby is adequate. The wet well is sized at ½Q = 44.3 gpm, 6 starts per hour = 10 min, 44.3 • 10 = 440 gallons

11-13. The easiest way to solve the problem is by spreadsheet. The static head is 1035 – 1020 = 15 ft. The dynamic head uses equation 4-8. From 500 gpm to 1600 gpm, the head losses are:

Flow (gpm)

Dyn + Static (ft)

Dynamic Hl (ft)

500 17.0 2.0600 17.8 2.8700 18.7 3.7800 19.8 4.8900 21.0 6.0

1000 22.3 7.31100 23.7 8.71200 25.2 10.21300 26.8 11.81400 28.6 13.61500 30.5 15.51600 32.4 17.4

The graph, with the pump data points, is constructed as follows:

15.0

17.0

19.0

21.0

23.0

25.0

27.0

29.0

31.0

33.0

35.0

400 600 800 1000 1200 1400 1600

Flow (gpm)

Hea

d (ft

)

Head curve

Pump curve

Operatingpoint

76

11-14. The static lift as the high water level is 20 ft and 30 ft at the low water level. Using a equation 4-8, the dynamic and total headloss is calculated as follows: Flow (gpm) z1 z2 Dym C=100 Flow (gpm) z1 z2 Dym C=140

0 30.0 20.0 0.0 0 20.0 30.0 0.0100 30.8 20.8 0.8 100 20.4 30.4 0.4200 32.7 22.7 2.7 200 21.5 31.5 1.5300 35.8 25.8 5.8 300 23.1 33.1 3.1400 40.0 30.0 10.0 400 25.3 35.3 5.3500 45.1 35.1 15.1 500 28.1 38.1 8.1600 51.2 41.2 21.2 600 31.3 41.3 11.3700 58.3 48.3 28.3 700 35.1 45.1 15.1800 66.3 56.3 36.3 800 39.4 49.4 19.4900 75.3 65.3 45.3 900 44.2 54.2 24.2

1000 85.1 75.1 55.1 1000 49.4 59.4 29.41100 95.8 85.8 65.8 1100 55.1 65.1 35.11200 107.4 97.4 77.4 1200 61.3 71.3 41.3

11-15.

0

200

400

600

800

1000

1200

1400

0 4 8 12 16 20 24 28 32 36 40 44 48

Influent Water Level (in.)

Pum

p C

apac

ity

(gpm

)

0

10

20

30

40

50

60

70

80

Pum

p E

ffici

ency

(Per

cent

)

| The table data is plotted above. At 60% of operating capacity is 0.6 • 48 = 29 in. for a capacity of about 400 gpm. Pump efficiency is over 70% from about 450 gpm and greater.

11-16. The purpose of primary clarification is to hold the wastewater quiescently to allow solids to settle out of suspension. Design criteria include: weir loading, over flow rate, detention time, and tank depth. Weir loading is the quantity of water flowing over the unit length of weir. The overflow rate (surface settling rate) is the flow divided by the area. It is a measure of the average upflow velocity of the clarifier. Detention time is the time the wastewater is held in the clarifier. Tank depth is a

0

20

40

60

80

100

120

0 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300

Pump curve

Dynamic lossesat C=100

Dynamic lossesat C=140

Operating points at intersection of pump and headloss curves

77

measure of the water depth at the side wall. Hydraulic criteria are set to achieve about 50 percent suspended solids removal and 35 percent BOD removal. Tank depth is used to provide additional sludge storage.

11-17. Three PCs have an area of 8,480 sq ft. The weir length total is 510 linear ft. The overflow rate at average flow is 542 gpd/sq ft and at peak flow is 1030 gpd/sq ft. The weir loading is 9,000 average and 17,000 gpd/linear ft peak. The average detention time is 3.0 hrs and the peak flow detention time is 1.6 hrs. The units are adequately sized: EPA 800-1200 avg 2000-3000 peak, GLURMB 1000 avg, 1500 peak, weir loadings are between 10,000 and 40,000 gpd/linear ft (EPA). Detention time is less than 2 hours. With one unit out of service, the area is 5,660 sq ft with overflow rates of 800 and 1,500 gpd/sq ft and remain within EPA design limits.

11-18. Q SS

1.2 230MGD mg/l

PRIMARY CLARIFIERNumber 1Diameter 65 Area 3,318 Depth 7 EPA 10-13, GLUMRB 7, EPA with secondary solids 13-16

Overflow Vo 362 gpd/sf EPA 800-1200 avg 2000-3000 peak, GLURMB 1000 avg, 1500 peakEPA 600-800 avg 12000-1500 peak with secondary solids

Det Time 3.5 hrs using 11-4 calcWeir loading 5,876 gpd/lf EPA 10k to 40k, GLUMRB 10k

without chemical additionremoval SS 70 % soilds removalin 2,302 lbsout 691 lbsunder 1,611 lbsWith chemical additionremoval SS 75 % soilds concentration removal SS 90 % soilds concentrationin 2,302 lbs in 2,302 lbsout 575 lbs out 230 lbsunder 1,726 lbs under 2,072 lbs

11-19. Avg Peak

Overflow Vo 1,000 2,500 gpd/sf EPA 800-1200 avg 2000-3000 peak, GLURMB 1000 avg, 1500 peakOverflow Vo 700 1,350 gpd/sf EPA 600-800 avg 1200-1500 peak with Secondary Solids

PRIMARY CLARIFIERWithout Secondary SolidsAverage 1 Peak 1Diameter 56 ft Diameter 53 ftDepth 11.5 ft EPA 10-13, GLUMRB 7, EPA with secondary solids 13-16

With Secondary SolidsArea 7,857 sq ft Area 4,074 sq ft

Diameter 100 ft Diameter 72 ftDepth 11.5 ft EPA 10-13, GLUMRB 7, EPA with secondary solids 13-16

11-20. Primary and secondary clarifiers are similar, the major difference is that the overflow rates of secondary clarifiers are less than those for primary clarifiers. The Secondary Clarifier for use with biological aeration (Figure 11-15) has uptake pipes along the collector arm for rapid return of activated sludge and an inboard weir channel. The primary clarifier (Figure 11-13) has a collector arm with scrapers to push the settled solids to a central hopper and a single outboard weir. The rapid return through uptake pipes increases solids

78

concentration and returns a “fresher” activated sludge, and the inboard weir channel reduces the approach velocity of the overflow to reduce carryover of solids.

11-21. Two clarifiers at 45 ft in diameter have an area of 3,180 sq ft. The overflow rate at average flow is 1,070 gpd/sq ft and at peak flow is 1,640 gpd/sq ft. The rates are within EPA 800-1200 avg 2000-3000 peak, close to the GLURMB 1000 avg, 1500 peak. If one unit is removed from service, the average overflow increases to 2,140 gpd/sq ft and the peak flow increases to 3,270 sq ft. The rates exceed EPA and GLURMB standards during peak flow.

11-22. Eight clarifiers have a total area of 14, 400 sq ft, a volume of 1.72 million gallons, and 880 linear feet of weir length. The average flow is 6 mgd and peak flow is 25 mgd. The overflow rate is 420 and 1740 gpd/sq ft The overflow is with the rates of EPA 800-1200 avg 2000-3000 peak, GLURMB 1000 avg, 1500 peak and is ok or close The detention time is 6.9 and 1.7 hours The recommended time of 2 to 3 hours is exceeded at average flow The weir loading rate is 6,800 and 28,400 gpd/linear ft EPA 10,000 to 40,000, GLUMRB 10,000 and is ok

11-23. Design flow = 10.5 mgd, BOD = 205 mg/l, TSS = 200 mg/l, Note, there was a typo and the flow should have been 1.05 mgd with 900 gpm recirculation. 2 Trickling Filters: Volume = 2 • 60^2 • π /4 • 7 /1000 = 40 (1000)cu ft BOD load = 10.5 mgd • 205 mg/l • 8.34 / 28 = 450 lb/1000 cu ft Too high and outside of the range of 30-90 lb/1000 cu ft, could cause odor problems Note, this should be a 1.05 mgd plant where the BOD loading rate is 45 lb/1000 cu ft 9000 gpm = 13 mgd recirculation flow + 10.5 = 23.5 mgd / 2 = 11.75 mgd each filter Area = 60^2 • π /4 = 2800 sq ft / 43560sq ft/acre = 0.065 acres Hydraulic loading = 2.9 gpm/sq ft or 180 mgd/acre; loading are well above those in Table 11-3 Again, the flow is 10 times too great and if corrected would result in 0.29 gpm/sq ft and 18 mgd/acre, which would have been acceptable.

11-24. Design flow = 2.8 mgd, BOD = 215 mg/l, TSS = 240 mg/l, PC removal increases SS removal from 50% to 75 or 90%. If 50% SS removal represents 35% BOD removal because 70% of BOD = SS, therefore BOD removal is as follows: SS Removal BOD Removal 75% 52.5% 90% 63% 2 Trickling Filters: Volume = 2 • 60^2 • π /4 • 5 /1000 = 28 (1000)cu ft At 75% SS Removal BOD load on TF = 2.8 mgd • 215 mg/l (1-0.525) • 8.34 / 28 = 85 lb/1000cu ft and is within of the range of 30-90 lb/1000 cu ft, odor problems could be eliminated, but loading is greater than adding a third trickling filter BOD load = 2.8 mgd • 215 mg/l (1-0.63) • 8.34 / 28 = 52 lb/1000cu ft and is within the range of 30-90 lb/1000 cu ft, odor problems should be eliminated based on loading calcs.

79

11-25. Primary Clarifier Effluent

Q SS BOD sBOD Temp0.8 280 450 250 18

MGD mg/l mg/l mg/l °C

BIOLOGICAL FILTERk 20 0.0035T 20As 42D 20Q 556Area 480Qp 1.16R 1n 0.5Sp 250Total to soluable ratio 0.54

At T=20

-1.932At T=18

-1.804

Soluable effluent BOD

Total effluent BOD

0.078 20 mg/l 36 mg/l

0.090 22 mg/l 42 mg/l

nps

T RQDAk )]1(/[2020 +− −ϑ

nps

T RQDAk )]1(/[2020 +− −ϑ

804.1

804.1

1)11( −

−

⋅−+=

e

eSS

p

e

932.1

932.1

1)11( −

−

⋅−+=

e

eSS

p

e

11-26.

So 120 mg/lSe 2 mg/lQ 0.3 mgdY 0.8 lb VSS/lb BODQc 20 daysX 5000 mg/lkd 0.04 1/day

V = 0.06 mil gal or ������������ cu ftNo safety factor for membranes: ������������ cu ft Aeration volume

Px = 131 lb/day (note, multiply by 8.34 lb/day/mgd for english units)

Aeration Volume = 8,000 cu ft = 0.060 mgDetention time = V/Q*24 = 4.8 hours

FM = So/dt/X 0.120 soluable BOD loading0.221 total BOD loading

80

11-27. Design flow = 7.8 mgd, BOD = 240 mg/l Oxidation Ditches: Volume = 3 • 2 mil gal = 6 mil gal Volume = 6 • 1000000 / 7.48 / 1000 = 802 (1000s) cu ft Detention time = 6 / 7.8 • 24 = 18.5 hr BOD load = 7.8 mgd • 205 mg/l (1-0.25) • 8.34 / 281 = 48 lb/1000cu ft F/M ratio = 7.8 • 240 / 6 / 1800 = 0.17 lb BOD/lb MLSS From Figure 11-30, sludge is in the Good Settleability range

11-28. Q SS BOD12 212 190

MGD mg/l mg/l

PRIMARY CLARIFIERSS BOD

Effluent 10,608 14,261 lbs

ACTIVATED SLUDGENumber 10

L W DDimensions 40 100 15.0 Volume 600,000 cf 4.488 mil galMLSS 2,350 mg/lEffluent SS 20 mg/l

Liquid Detention Time 9.0 hrsBOD loading 24 lb/1000 cu ft/dayF/M 0.16 Waste sludge = 5330 lb/daySludge Age 12 days

According to Figure 11-30, at an F/M of 0.16, the sludge should settle well.

11-29. BEFORE CHEMICAL ADDITION BEFORE CHEMICAL ADDITION

Q SS BOD Q SS BOD18 212 190 18 212 190

MGD mg/l mg/l MGD mg/l mg/l

PRIMARY CLARIFIER PRIMARY CLARIFIER50% SS removal 75% SS removal25% BOD Removal 38% BOD Removal

SS BOD SS BODin 31,825 28,523 lbs in 31,825 28,523 lbsEffluent 15,913 21,392 lbs Effluent 7,956 17,684 lbsunder 15,913 7,131 lbs under 23,869 10,839 lbs

ACTIVATED SLUDGE ACTIVATED SLUDGENumber 10 Number 10

L W D L W DDimensions 40 100 15.0 Dimensions 40 100 15.0 Volume 600,000 cf 4.5 MG Volume 600,000 cf 4.5 MGMLSS 2,350 mg/l MLSS 2,350 mg/lEffluent SS 20 mg/l Effluent SS 20 mg/l

Liquid Detention Time 6.0 hrs Liquid Detention Time 6.0 hrsBOD loading 36 lb/1000 cu ft/day BOD loading 29 lb/1000 cu ft/dayF/M 0.24 F/M 0.20

According to Figure 11-26, at an F/M of 0.24, According to Figure 11-26, at an F/M of 0.20, the sludge should settle fairly well. the sludge should settle well.

81

As flow increase to 18 mgd, BOD loading increases to 36 lb/1000 cu ft, which is still acceptable, but close to the upper limit (Table 11-4). Adding chemicals improves the suspended solids removal. The existing clarifier removes 50% of SS with 25% of the BOD, therefore the solids associated BOD is about 50%. Increasing the SS removal to 75% increases the BOD removal by 25%/2 = 13% for a total BOD removal of 38%.

11-30. Primary Clarifier has no secondary solids return, use 1000 gpd/sq ft from Table 11-2 Conventional Step Aeration Basin, use 30 lb/1000 cu ft from Table 11-4 Secondary Clarifier use 800 gpd/sq ft Primary Clarifier = 2,750,000 gpd / 1000 gpd/sq ft / 2 = 1375 sq ft each Primary Clarifiers: 2 at 47 ft diameter Secondary Clarifier = 2,750,000 gpd / 800 gpd/sq ft /2 = 1720 sq ft Secondary Clarifiers: 2 at 47 ft dia PC removal 35% BOD Aeration Basin Volume = 2.75 mgd (195 mg/l (1-0.35) 8.34 / 30 lb /1000 cu ft = 9,000 cu ft Dt = 97,000 (7.487) / 106 / 2.75 (24) = 6.3 hr F/M = 2.75 (195) (1-0.35) / (2000 (6.3) / 24) = 0.66

11-31. Rock trickling filters are significantly influenced by temperature. During winter, trickling filter efficiency drops 5% or more. The performance of biological towers is adjusted by 1.035(T-20) and drops 3% with a 1 deg C change in temperature. For activated sludge, biological activity doubles or halves for every 10 to 15 deg C temperature change. Submerged aerators are better than surface aerators. Clarifier effectiveness also fluctuates with changes in temperature.

11-32. This question has no answer. Discussion of high purity oxygen was removed from the 7th Edition because of its inability to be filtered and perform nitrification-denitrification. While some facilities like Tampa, Florida have been modified with nitrification and denitrification using methanol, future use of high purity oxygen may be avoided in favor of activated sludge biological nutrient removal or membrane treatment.

11-33. Both systems are impacted by BOD loading rates which can be controlled by chemical addition to the primary clarifier. But, more to the question: Fixed film systems are only controlled by increasing or decreasing the recirculation ratio. The adjustment changes the hydraulic loading rate and BOD loading rate. Increased flushing may thing the slime layer. Aeration systems are controlled by increasing or decreasing the wasting rate to adjust the MLSS which changes the F/M ratio and sludge age. In addition, the dissolved oxygen level in the aeration basin and the recirculation ratio can be adjusted. Recycle rates are based on returning sufficient organisms rather than hydraulic considerations as in bio-filters.

82

11-34. From the text, the minimum aeration period is 24 hours and the clarifier overflow rate is 600 gpd/sq ft. Trailer Park 140 unitsOccupancy 2 per unitWastewater 80 gpcdFlow 22,400 gpd

Q SS BOD22,400 190 170

MGD mg/l mg/l

ACTIVATED SLUDGE

Detention time = 24 hours Table 11-4, 20 to 30Volume of aeration basin = 22,400 / 24 * 24 / 7.48 = 2990 cu ftBOD loading = 11 lb/1000 cu ft Table 11-4, loading of 10 to 20F/M = 170 * 0.0224 / 1000 / 0.0224 = 0.170 Table 11-4, F/M 0.05 to 0.2Vo = 600 gpd/sf Area = 37 sf Dia = 7 ft

11-35. Flow 11.5 MGDBODi 200 mg/lBODe 10 mg/lNH4 35 mg/l

Oxygen demand = 11.5(200+4.6*35-10)8.34 = 33,664 lbs/dayTemp 16 °C

Brush Aerators

Number 2 ditches 4 aerators each�F 0.5 � 0.9DO residual 4 mg/l

From Table 11-5, SOTR O2 lb/Hp-hr 2.75

Cs 10 mg/l, see Table 2-5Cinf = CsCt 4 mg/l

OTR 0.61 lb/hp-hrHp 104 for each aerator

���

����

� −⋅⋅⋅⋅=

∞

−

CCC

FSOTROTR tst βθα 20

11-36. Flow 21.5 MGDBODi 180 mg/lBODe 10 mg/lNH4 31 mg/l

Oxygen demand = 30,483 lbs/day without nitrificationOxygen demand = 56,052 lbs/day with nitrification

Temp 14.8 °C winterTemp 18.5 °C summer

Fine Bubble Aerators

Depth 15 feet�F 0.75 text for equation 11-17� 0.925 text for equation 11-18DO residual 2 mg/lSOTR 5.85 lb/hp-hr Average of values in Table 11-5Cs 10.2 mg/l, see Table 2-5 14.8 °C

83

Depth adjustedCs 9.1 mg/l, see Table 2-5 20 °C 10.7 mg/lCs 10.1 mg/l, see Table 2-5 14.8 °C 11.9 mg/lCs 9.4 mg/l, see Table 2-5 18.5 °C 11.1 mg/l

Cde/C = 1 + 0.01205(diffuser depth)(1+5.6·10-7(site elevation))Depth adjustment Cde/C 1.2

OTR at 14.8 3.1 lb/hp-hr HP at 14.8 412 hpOTR at 18.5 3.2 lb/hp-hr HP at 18.5 729 hp

QAir = 2,308 cfm winter

QAir = 4,082 cfm summer

14400174.02

⋅⋅=

OTEQ

Q OAir

( )( )FC

CCSOTROTR T

de

LdePT αθβ 20

,1,20

,, −��

�

�

��

�

� −⋅=

11-37. Flow 6.5 MGDBODi 143 mg/l (220 * (1-0.35))BODe 30 mg/l

Oxygen demand = 6,126 lbs/day without nitrificationTemp 16 °C winter

Fine Bubble Aerators

Depth 15 feet� 0.65 Example 11-10F 0.75 Example 11-10� 0.95 Example 11-10DO residual 2 mg/l Example 11-10SOTR 5.85 lb/hp-hr Average of values in Table 11-5

Depth adjusted

Cs at STP 9.1 mg/l, see Table 2-5 20 °C 10.7 mg/lCs 9.9 mg/l, see Table 2-5 16 °C 11.7 mg/l

Cde/C = 1 + 0.01205(diffuser depth)(1+5.6·10-7(site elevation))Depth adjustment Cde/C 1.2 mg/l for a site elevation = 0

OTR at 14.8 2.0 lb/hp-hr HP = 130

QO2 = 255 lb/hr or 6,126 lb/day

QAir = 244 cu ft/min

14400174.02

⋅⋅=

OTEQ

Q OAir

( )( )FC

CCSOTROTR T

de

LdePT αθβ 20

,1,20

,, −���

����

� −⋅=

11-38. The following calculations are for the first test period. Using Equation 11-20,

θC = 12 1730

0 30 1730 412 0 30 9 4⋅

⋅ + −. ( . . ) . = 23 d

10 043

θc

-1d= .

Using Equations 11-24 and 11-25,

rsu = (126 - 5.2)/(12/41.2) = 414 mg/l•d U =414

1730d -1= 0 24.

1 1

0 244 2

152

019U

d1

Smg / l)

e

-1= = = =.

..

. (

84

Calculated values for all test runs are as follows:

θc (d)

rsu

(mg/l•d) U

(d-1) 1/θc (d-1)

1/U (d)

1/Se [(mg/l)-1]

23 414 0.24 0.043 4.2 0.19 19 407 0.27 0.053 3.7 0.14

9.2 396 0.41 0.11 2.4 0.095 8.3 392 0.46 0.12 2.2 0.087

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.2 0.25 0.3 0.35 0.4 0.45 0.5

U, day

1/Q

c, d

ay-1

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

2 2.25 2.5 2.75 3 3.25 3.5 3.75 4 4.25 4.5

1/Se, 1/mg/l

1/U

, 1/d

ay

Y = 0.35 mg VSS/mg/BOD kd = 0.04d-1 k = 1/0.25 = 4.0 d-1 KS = 22.5 • 4.0 = 90 mg/l BOD

11-39.

So 200 mg/lSe 7 mg/lQ 3 mgdY 0.6 lb VSS/lb BODQc 8 daysX 2500 mg/lkd 0.06 1/day

V = 0.75 mil gal or 100,000 cu ftUsing a safety factor of 2x = 200,000 cu ft

Px = 235 g/day

11-40. The biological activity in an unmixed facultative stabilization pond is shown in Figure 11-38. Because of the lack of mixing, solids settle and decompose anaerobically releaseing carbon dioxide and hydrogen sulfide. The surface is aerobic, but depend on wind to provide aeration. Algae grow at the surface due to the lack of mixing and sunlight. Aerated lagoons are similar to activated sludge, but typically without Secondary Clarifiers and sludge recycle or mixed liquor control. Biological activity is aerobic, but if aeration equipment is

85

inadequate for the lagoon size, deposition of solids can result in anaerobic decomposition and foul odors.

11-41. BOD applied = 0.65 • 165 • 8.34 = 894 lb/day, Acres = 894 / 25 = 36 acres. Water loss = (36) 0.7 in./week / 12 in./ft / 7 • 43560 = 13,000 cu ft/day or 0.1 mgd.

11-42. BOD loading = 0.2 • 230 • 8.34 = 384 lb/day/acre. Water loss = 0.05 in./day/12 • 16 • 43560 • 7.48 = 21,700 gpd. Volume of pond = (5-2) • 16 • 43560 = 2.1 mil cu ft OR 15.6 mil gal. Storage time = 15.6 / (0.2 - .0217) = 87 days.

11-43. Detention time = 175,000 • 7.48 / 1000000 / 0.2 = 6.5 days which is in the range of 3 to 8 days. The influent BOD is strong, a 50 mg/l effluent represents a 89% reduction. Using the highest k of 0.68/day, the Effluent BOD/Influent BOD = 1 / (1 + 0.68 • 6.5) = 0.18 • 450 = 81 mg/l > 50 mg/l. Nominal reduction at 20 deg C is 85% and is less than the reduction needed. A larger aerated cell seems warranted. Check aerator capacity: From Table 11-5, the maximum aeration transfer is 3.6 lb/hp-hr or 180 lb/hr or 4320 lb/day using 2-25hp aerators. From Table 2-5, Cs at 10 deg C = 11.3 mg/l, and at 30

deg C = 7.6 mg/l. At 10 deg C ( ) daylbR /24409.002.12.9

23.118.04320 2010 =−⋅= −

At 30 deg C, R = 2100 lb/day versus a BOD load of 0.175 • 450 • 8.34 = 660 lb/day, therefore aerators are adequate for the temperature extremes. Check the lowest SOTR of 2.0 for floating aerators: At 10 deg C, R = 1,360 lb/day Ok At 30 deg C, R = 1,170 lb/day Ok

11-44. The BOD demand is 0.375 • 220 • 8.34 = 688 lb/day. Using an � = 0.9, � = 0.8, at DO residual of

2.0 mg/l and T = 10 deg C, ( ) 56.09.002.12.9

23.118.0 2010 =−⋅= −RoR

Using the low oxygen transfer of 2 lb/hp-hr, 688 / 0.56 / 10 / 24 / 2 = 2.6 or 3 aerators Using the high oxygen transfer of 3.6 lb/hp-hr, 688 / 0.56 / 10 / 24 / 3.6 = 1.4 or 2 aerators

11-45. The primary reason for disinfecting wastewater discharges is to protect public health. Hypochlorous acid + ammonia → chloramines HOCl + NH3 → H2O + NH2Cl (and NHCl2) Chlorination inactivates the majority of viruses in a wastewater effluent but not all of them. Virus removal requires filtration of the wastewater prior to chlorination and a longer contact time at a higher residual than is necessary to achieve a fecal coliform count of 200/100 ml. (Refer to Section 13-10). Chloramines are extremely toxic to aquatic life. The commonly recommended limit for chlorine residual in a receiving stream after effluent mixing is 0.02 mg/l. Sulfur dioxide (gas form) or sodium metabisulfite (liquid form) are used to destroy any chlorine residual.

11-46. Long, narrow tanks maintain a plug flow condition that limits mixing, short-circuiting, and allows the chlorine to contact organisms over the minimum 30 min time.

11-47. According to the text, the minimum detention time is 20 to 30 min. For 20 min, the total volume would be 14.6 • 20/60 = 4.87 mil gal • 1000000 /7.481 = 651,000 cu ft. Using a cross section of 10 ft x 10 ft, the length would be 6,510 feet. For 30 min, the total volume would be 14.6 • 30/60 = 7.3 mil gal • 1000000 /7.481 = 976,000 cu ft. Using a cross section of 10 ft x 10 ft, the length would be 98,000 cu ft. Chlorine use = 14.6 • 5 • 8.34 = 610 lb/day Number of days for 1 ton = 2000 / 610 = 3.4 days

86

11-48. UV light strikes the DNA of organisms creating dimmers (links of cytosine and thymine) that inhibit replication. Viral destruction requires pretreatment with hydrogen peroxide prior to UV or post treatment with chlorine.

11-49. Unrestricted use of recycled water requires a disinfection level <2.2 MPN. From Table 11-7, for California and Hawaii, 48 lamps/mgd * 6 mgd = 288 lamps. For other states, 32,4 lamps/mgd * 6 = 195 lamps are required.

11-50. Individual household treatment typically consists of a septic tank and absorption field. Solids settle in the septic tank where anaerobic biological activity reduces the volume of solids. An absorption field allows liquids to be disposed of underground, thus eliminating odors. Tanks must be pumped out periodically to keep solids from plugging the drain field.

11-51. Primary Sludge = 220 (0.48) 5.2 mgd • 8.34 = 4,580 lb/day Sludge flow = 4580/0.055/8.34 = about 10,000 gpd Dt = V / Q • 24, therefore V = Dt / 24 • Q and F = primary effluent concentration • Q / M = MLSS • Dt/24 • Q and the Qs cancel, so F/M = 220 mg/l (1-0.32) / (3000 mg/l • (8/24)) = 0.15, using Figure 11-50, K = 0.38, sludge solids = 0.38 • 5.2 • 200 (1-0.32) 8.34 = 2,240 lb/day /0.005/8.34 = 53,700 gpd + 10,000 PC sludge flow, for a total of 63,700.

11-52. 1. PC Sludge = 0.24 lb/capita-day (0.5) = 0.12 lb/capita-day Secondary Solids = 0.2 lb/capita-day (1-0.35)(0.31) = 0.04 lb/capita-day Total Sludge = 0.16 lb/capita-day Total Sludge Flow = 0.16/0.05/8.34 = 0.38 gpcd 2. PC Sludge Lb = 0.24 lb/capita-day (0.5) = 0.12 lb/capita-day Q = 0.24 lb/capita-day (0.5) / 0.05 / 8.34 = 0.29 gal/capita-day F/M = 0.2 lb/capita-day (1-0.30) / 8.34 / 0.0001 / (1500 • 8/24) = 0.33 K = 0.48 Secondary Solids = 0.2 lb/capita-day (1-0.30)(0.48) = 0.067 lb/capita-day Secondary Sludge Q = 0.067 / 0.04 / 8.34 = 0.20 gal/capita-day Total sludge = 0.12 + 0.067 = 0.187 lb/capita-day Total Sludge Flow = 0.29 + 0.20 = 0.49 lb/capita-day 3. No Primary Clarifier F/M = 0.2 lb / 8.34 / 0.0001 / (1200 mg/l • 24/24) = 0.20 K = 0.42 Waste Sludge = 2 (0.42) 0.2 lb/capita-day = 0.17 lb/capita-day Total Sludge Flow = 0.17/ 0.01 / 8.34 = 2.0 gal/capita-day

87

11-53.

WAS Digested Primary0.5 D ft 0.5 D ft 0.5 D ft

1000 L ft 1000 L ft 1000 L ft

4 % solids 4 % solids 4 % solids0.62 Sy 0.12 Sy 0.27 Sy0.17 n 0.016 n 0.0096 n

600.00 Q gpm 600.00 Q gpm 600.00 Q gpm

6.81 Velocity ft/s 6.81 Velocity ft/s 6.81 Velocity ft/s

16.7 Vlc ft/s 5.0 Vlc ft/s 7.0 Vlc ft/s22.7 Vuc ft/s 6.3 Vuc ft/s 8.7 Vuc ft/s

0.1746 H/L ft/ft 0.0266 H/L ft/ft 0.0488 H/L ft/ft174.6 Headloss ft water 26.6 Headloss ft water 48.8 Headloss ft water75.57 Headloss psi 11.53 Headloss psi 21.12 Headloss psi

Q (gpm) Headloss (ft) Q (gpm) Headloss (ft) Q (gpm) Headloss (ft)0 103 0 20 0 45

50 109 50 20 50 45100 115 100 21 100 45150 121 150 22 150 46200 127 200 22 200 46250 133 250 23 250 46300 139 300 23 300 47350 145 350 24 350 47400 151 400 24 400 47450 157 450 25 450 48500 163 500 26 500 48550 169 550 26 550 48600 175 600 27 600 49

WATERQ (gpm) Headloss (ft)

0 050 0

100 2150 4200 6250 9300 13350 17400 22450 28500 34550 40600 47

HEADLOSS VS. FLOW

0

20

40

60

80

100

120

140

160

180

200

0 100 200 300 400 500 600

FLOW (gpm)

Hea

dlos

s (f

t. of

wat

er)

Primary SludgeDigested Sludge

Waste Activated Sludge

Water

88

11-54. 4.5% Primary Sludge, Sy = 0.3, n = 0.013

4.64)12/8(4.643.02)^12/8(2^013.094103)013.0(1000

⋅⋅⋅+⋅+

=lcV

9.426.80016.010313 ++=lcV = 7.3 ft/sec

Actual velocity: Q = 500 gpm in a 8 inch pipe, velocity is about 3.2 ft/s < lower critical, therefore, the flow is laminar

ftftLH /039.00015.0037.0)12/8(4.64)2.3(013.0

)12/8)(4.64(3)3.0(16

/ 2 =+=+=

Head loss = 850 ft • 0.039 = 33 ft or 14.3 psi

No. K1 K?

Check valves 1 1500 1500 1.5 1.590 deg elbows 24 800 19200 0.25 6Tees 8 800 6400 0.8 6.4Plug valves 14 1000 14000 0.25 3.5

41100 17.4

Laminar Re = ( ) =

+⋅⋅

ySdvd

)12/(16312/3 2

ηρ ( ) =

+⋅⋅

)3.0)(12/8(16)013.0(34.642.312/83 2

1320 / (0.039 + 3.2) = 408

K = K1/Re + K(1 + 1/d) = 41,100/408 + 17.4(1 + 1/8) = 120 Hf = K • v2 / 2 / g = 120 3.22 /64.4 = 19 ft or 8.3 psi Total head = 14.3 + 8.3 = 22.6 psi 0.5% WAS is a Newtonian fluid that follows the headloss of water. Using the Hazen-Williams equation, headloss is calculated as follows: H = 0.002083 • 420 • (100/100)1.85 • (500 gpm)1.85 / 84.8655 = 3.5 ft or 1.5 psi 5% TWAS, Sy = 1.7, n = 0.2

4.64)12/8(4.647.12)^12/8(2^2.094103)2.0(1000

⋅⋅⋅+⋅+

=lcV

9.426.4876.3103200 ++=lcV = 22 ft/sec

Actual velocity: Q = 100 gpm in a 8 inch pipe, velocity is about 0.64 ft/s < lower critical, therefore, the flow is laminar

ftftLH /214.00045.021.0)12/8(4.64)64.0(2.0

)12/8)(4.64(3)7.1(16

/ 2 =+=+=

Head loss = 1200 ft • 0.214 = 257 ft or 112 psi

89

No. K1 K?

Check valves 1 1500 1500 1.5 1.590 deg elbows 18 800 14400 0.25 4.5Tees 3 800 2400 0.8 2.4Plug valves 14 1000 14000 0.25 3.5

32300 11.9

Laminar Re = ( ) =

+⋅⋅

ySdvd

)12/(16312/3 2

ηρ ( ) =

+⋅⋅

)3.0)(12/8(16)013.0(34.642.312/83 2

1320 / (0.039 + 3.2) = 408

K = K1/Re + K(1 + 1/d) = 41,100/408 + 17.4(1 + 1/8) = 120 Hf = K • v2 / 2 / g = 120 3.22 /64.4 = 19 ft or 8.3 psi Total head = 112 + 8 = 120 psi 3.5% Digested Sludge, Sy = 0.1, n = 0.012

4.64)12/8(4.641.02)^12/8(2^012.094103)012.0(1000

⋅⋅⋅+⋅+

=lcV

9.4286.20135.010312 ++=lcV = 4.3 ft/sec

Actual velocity: Q = 500 gpm in a 8 inch pipe, velocity is about 2.23 ft/s < lower critical, therefore, the flow is laminar

ftftLH /013.00009.0012.0)12/8(4.64)23.2(012.0

)12/8)(4.64(3)1.0(16

/ 2 =+=+=

Head loss = 520 ft • 0.013 = 6.8 ft or 3 psi just for comparison, H = 0.002083 • 520 • (100/100)1.85 • (350 gpm)1.85 / 84.8655 = 2.2 ft or 1 psi

No. K1 K?

Check valves 1 1500 1500 1.5 1.590 deg elbows 14 800 11200 0.25 3.5Tees 3 800 2400 0.8 2.4Plug valves 6 1000 6000 0.25 1.5

21100 8.9

Laminar Re = ( ) =

+⋅⋅

ySdvd

)12/(16312/3 2

ηρ ( ) =

+⋅⋅

)1.0)(12/8(16)012.0(34.6423.212/83 2

640 / (0.036 + 1.07) = 579

K = K1/Re + K(1 + 1/d) = 21,100/579 + 8.9(1 + 1/8) = 46.3 Hf = K • v2 / 2 / g = 46.3 2.232 /64.4 = 3.6 ft or 1.6 psi Total head = 2.2 + 1.6 = 3.8 psi

11-55. PC sludge = 20 (215)(0.6) 8.34 = 21,500 lb/day. Flow at 2% = 21,500 / 0.02 / 8.34 = 129,000 gpd. Area = 21,900 / 2 / 9 = 1190 sq ft or about 39 ft dia.

11-56. F/M = 185 / 2500 = 0.074, Figure 11-50, K = 0.32, Sludge produced = 2 • 0.32 • 28 • 185 • 8.34 = 27,600 lb/day at a concentration of 2.8%, the flow rate is = 442,000 gpd or 306 gpm for 24 hours. For 8 hr/day operation, flow would be 920 gpm. At 250 gpm per unit, 3.7, say 4 units would be required. Solids in the cake = 0.95 • 27,600 = 26,200 lb/day at 4.5%, the flow rate would be 26,200 / 0.045 / 8.34 = 69,000 gpd.

90

Solids in the filtrate = 0.05 • 27,600 = 1,380 lb/day, at a flow rate of 442,000 – 69,000 = 373,000 gpd

11-57. Class B biosolids treatment must consistently reduce coliform to 2 million MPN/g, but solids are not considered pathogen-free. Requirements are met by monitoring and processes that significantly reduce pathogens in conjunction with site restrictions for land application. Treatment processes include aerobic digestion, anaerobic digestion, air drying, composting, and lime stabilization. Disposal alternatives are shown in Figure 11-67. All biosolids are not suitable for land application, but may be disposed of in a landfill or monofill. Class B biosolids may be land applied to non-edible crops such as sod farms and for animal grazing with public access restrictions. Several restrictions application to edible crops apply.

11-58. The aerobic digester volume is 50,000 • 7.481 = 374,000 gal. The detention time is 0.374 / 0.032 = 11.7 days. The VSS loading rate is 0.032 • 1500 • 8.34 • 0.63 / 50,000 = 0.05 lb VSS/cu ft/day, which is lower than the maximum. The degree days are 280 • 15 = 4200 degree-days, which is more than adequate for digestion. Class B biosolids requirements are met, because 60 days of detention are required at 15 deg C and 900 degree days.

11-59. The flow rate = 5250 / 0.012 / 8.34 = 52,500 gpd. For Class B, the degree days is 900 / 15.5 = 58 days. At 58 days of detention, the digester volume = 52,200 • 58 = 3.1 mil gal or /7.481 = 414,000 cu ft. The degree dates are 3100000 / 52500 • 15.5 = 915 degree days.

11-60. For a flow rate of 180 gpm for 5 min/hr = 900 gal/hr or 21,600 gpd. Total digester volume = 3 • 40^2 � / 4 •20 = 75,400 cu ft or • 7.48 = 564,000 gal. The detention time is 564,000 / 21,600 = 26 days. VSS loading is 0.021 • 5000 • 8.34 • 0.65 / 75400 = 0.0075 lb/cu ft. Final solids = 5% (1-0.65 • 0.42) = 3.6%

11-61.

So 100 mg/lSe 0 mg/lQ 0.55 mgdY 0.8 lb VSS/lb BODQc 28 daysX 7000 mg/lkd 0.04 1/day

V = 0.08 mil gal or ����������� cu ftUsing no safety factor AB vol = ����������� cu ft

Px = 173 lb/day Volatile Solids

Volatile solids concentration = 0.8

Px = 216.3679 lb/day

Using a solids concentration of about 2.0 % solids, Flow rate = 216/0.02/8.34 = 1295 gpd

91

For 40 days, the volume is = 51,799 gal 6,924 cu ftFor 60 days, the volume is = 77,698 gal 10,387 cu ft

Using an average VSS loading rate of 0.015 lb VS/cu ft/day. Vol = 11,540 cu ft

Use 60 days and a volume of 10,400 cu ftEach digesters is = 13 ft square x 15 ft deep

11-62. Primary sludge + secondary sludge = 1000 + 400 = 1,400 lb/day. Using 20 lb/sq ft/year, the Area required is 1400 • 365 / 20 = 25,600 sq ft. Using beds that are 20 ft x 100 ft or 200 sq ft each, about 128 drying beds are necessary.

11-63. Sludge composting is a process of heat pasteurization using sludge and a bulking agent to allow an air flow through the compost pile. Piles are turned to allow all areas to the pile to reach temperatures of at least 104 deg F, avoid drying, and prevent overheating. Pathogen reduction is accomplished by a combination of heat and time.

11-64. Hydrated lime or quicklime is mixed with the sludge. The reaction with the water raises the pH and heats the sludge. At a pH greater than 11, all biological activity stops, but helminth ova remain resistant to treatment.

11-65. Class A treatment requires a 5 log removal of coliform and is essentially pathogen free. Class A biosolids requirements limit Salmonella sp. to less than 3 MPN/4 g total dry weight, enteric viruses to less than 1 PFU/4 g, and viable helminth ova to less than 1 ova/4 g. One of the vector attraction requirements must also be met. Evidence of compliance with Class A requirements may be established with a fecal coliform value of less than 1000 MPN/g or Salmonella sp. value of less than 3 MPN/4 g of total dry-weight solids. Class A biosolids have no restrictions on use and may be bagged and sold to the public.

11-66. Aerobic digestion occurs in an open tank where sludge is aerated to maintain aerobic conditions. Pathogens are reduced by holding the sludge. The best the process can do is Class B sludge with a minimum of 800 degree-days. The ATAD process is also aerobic using aeration, but because the process is accomplished in an enclosed insulated tank, the heat of the biological activity is retained. Temperatures obtained in the tank are adequate to accelerate pathogen destruction, thus creating a Class A biosolid.

11-67. Pasteurization may be accomplished in a separately heated tank ahead of anaerobic digestion as in Figure 11-63. As described in the text, lime and supplemental heat following anaerobic digestion to heat the sludge and achieve pasteurization

11-68. Hydraulic loading rate = 20 gpm / 0.5 m = 40 gpm/m Solids loading = 40 • 0.00144 mgd/gpm • 3500 • 8.34 / 24 = 70 lb/m/hr Polymer dose = 2000 lb/ton • 0.2% = 4 lb/ton Solids feed rate = 35 lb/hr Solids in recycle stream = 17 • 0.00144 mgd/gpm • 2000 • 8.34 + 19 • 0.00144 mgd/gpm • 500 • 8.34 = 522 lb/day or 22 lb/hr. Capture = (35 – 22) / 35 • 100 = 37%

11-69. Hydraulic loading rate = 150 / 2 = 75 gpm/m

Solids loading rate = m2

8.34 4000 mgd/gpm gpm/694 150 ⋅⋅ = 3600 lb/day/m or 150 lb/m/hr)

92

Polymer dosage = 6.4 gpm/694 mgd/gpm • 500 mg/l • 8.34 = 38.8 lg/day Polymer load = 38.4 / 7200 • 2000 = 11 lb/ton Wash solids = 60 / 694 • 1720 • 8.34 = 1240 lb Solids in Cake = 7200 – 1200 = 6000 lb, Solids capture = 6000 / 7200 = 83%

11-70. lb of solids applied = 250 (0.00144 mgd/gpm) 2600 • 8.34 = 7800 lb/day Removal efficiency = 1 - (2600 – 1720) / 2600 = 66% lbs of active polymer = 7800 lb/day / 2000 lb/ton • 22 lb/ton = 85.8 lb/day lbs to polymer solution = 85.8 / 0.45 = 190 lb/day at 9 lb/gal, flow rate = 190 / 9 = 21 gpd or 0.86 gal/hr

11-71. lb of dry weight sludge = 8000 • 0.26 = 2080 lb/day Nitrogen applied = 2080 / 2000 • 75 = 78 lb/day • 365 = 28,500 lb/year For Hay, the nitrogen demand is 400 lb/acre less 75 residual = 325 lb/acre Acres required for application = 28,500 / 325 = 88 acres

11-72. Design flow = 3.2 mgd, Peak flow = 9.3 mgd, Min. flow = 0.68 mgd, BOD = 200 mg/l, soluble BOD = 140 mg/l, TSS = 180 mg/l, WW temperature = 15°C Effluent BOD = 2 mg/l, suspended solids = 0 mg/l, <2.2 MPN/100 Influent Pumping Station With 1 pump in standby, firm capacity = 3 * 1600 = 4800 gpm or 6.9 mgd firm capacity. The PS capacity is not OK because the firm capacity is < the peak flow of 9.3 mgd. If fact, using all 4 pumps the pumping rate would be 6400 gpm or 9.2 mgd < 9.3 mgd peak flow. Pumps are variable speed, so wet well volume is assumed to be adequate. Bar screen with 3 mm openings Screening removal; 3 mm = ¼” ; from Figure 11-7, Screenings quantity = 8 cu ft/mgd * 3.2 = 26 cu ft/day Grit Removal Grit = 670 * 3.2 ((9.3 / 3.2) – 1) = 4100 lb/day Assumptions from Example 11-9: 80% volatile SS, Y=0.60 mgVSS/mg; kd=0.06 per day. From Table 11-4 conservative design; BOD loading = 50 lb BOD/day per 1000 cu ft; MLSS = 7000 mg/l; F/M = 0.1, sludge age = 15 days

93

Aeration Basin

So 140 mg/lSe 2 mg/lQ 3.2 mgdY 0.8 lb VSS/lb BODQc 15 daysX 7000 mg/lkd 0.06 1/day

V = 0.40 mil gal or 53,000 cu ftUsing a BOD loading of 50 lb BOD/1000 cu ft = 3.2 * 200 * 8.34 / 50 * 1000

= 107,000 cu ftBOD loading results in a greater volume, Vol = 107,000 cu ft or 0.80 mil gal

Detention time = V/Q*24 = 6.0 hr

BOD loading = 50 lb/1000 cu ft

F/M ratio = Q*BOD/MLSS/V = 0.11 lb BOD/lb VSS

UV DisinfectionFor CA and Hawaii, 48 lamps/mgd = 48 * 9.3 = 450 lamps; for other states, use 300 lamps

Sludge Solids

Px = 1550 lb/day Volatile Solids

Volatile solids concentration = 0.8

Px = 1940 lb/day

Using a solids concentration of about 2.5 % solids, Flow rate = 1940/0.025/8.34 = 9305 gpd

Aerobic DigestionFor 40 days, the volume is = 372,200 gal 49,800 cu ftFor 60 days, the volume is = 558,300 gal 74,600 cu ft

Using an average VSS loading rate of 0.015 lb VS/cu ft/day. Vol = 103,333 cu ft

Use 60 days and a volume of 74,600 cu ft Loading rate = 0.021 lb VSS/cu ft Land Application of Biosolids Amount of nitrogen = 35 mg/l * 9300/1000000 * 8.34 * 365 = 991 lb/year Using a nitrogen demand of 250 lb/acre, acres = 991 / 250 = 4 acres based on nitrogen loading. Yes.

94

11-73. Q = 0.12 mgd, Peak flow = 0.32 mgd, BOD = 360 mg/l, TSS = 280 mg/l Primary Clarifier: Area = 12^2 • π /4 = 113 sq ft • 5 • 7.481 / 106 = 0.004 mg Weir length = 12 • π = 38 ft Overflow at Peak Q = 320,000 / 113 = 2830 gpd/sq ft Too high, outside of the range 1200-1500 gpd/sq ft and above even the peak of 2500 gpd/sq ft Overflow at Avg Q = 120,000 / 113 = 1060 gpd/sq ft Too high, above the range 600-800 gpd/sq ft Detention time at avg Q = 0.004 / 0.12 • 24 = 0.8 Low, range 2-3 hr. Weir loading = 120,000 / 38 = 3,200 gpd/ft OK, weir loading 10,000-20,000 gpd/ft Biological Tower: Primary Clarifier Effluent

Q SS BOD sBOD Temp0.8 280 450 250 18

MGD mg/l mg/l mg/l °C

BIOLOGICAL FILTERk 20 0.0035T 20As 42D 20Q 556Area 480Qp 1.16R 1n 0.5Sp 250Total to soluable ratio 0.54

At T=20

-1.932At T=18

-1.804

Soluable effluent BOD

Total effluent BOD

0.078 20 mg/l 36 mg/l

0.090 22 mg/l 42 mg/l

nps

T RQDAk )]1(/[2020 +− −ϑ

nps

T RQDAk )]1(/[2020 +− −ϑ

804.1

804.1

1)11( −

−

⋅−+=

e

eSS

p

e

932.1

932.1

1)11( −

−

⋅−+=

e

eSS

p

e

Secondary Clarifier: Area = 4 • 130^2 • π /4 = 53,100 sq ft • 13 • 7.481 / 106 = 5.16 mg Weir length = 130 • π = 408 ft • 4 = 1632 ft Overflow at Peak Q = 22,500,000 / 53100 = 424 gpd/sq ft Low, could be as high as 1600 gpd/sqft Overflow at Avg Q = 10,500,000 / 53100 = 200 gpd/sq ft Low, could be as high as 800 gpd/sq ft Detention time at avg Q = 5.16 / 10.5 • 24 = 11.8 hr High, range 2-3 hr. may cause problems with anaerobic conditions and floating solids. Consider removing clarifiers from service Weir loading = 22,500,000 / 1632 = 13,800 gpd/ft OK, weir loading 10,000-20,000 gpd/ft Chlorine Contact: Detention time = 3 • 200 • 10 • 10 • 7.481 / 106 /22.5 • 24 • 60 = 27 min OK, range 20-30 min. Sludge: F/M = 10.5 • 290 / 3000 / 10.5 = 0.90 K = 0.34 WAS = 2 • 0.34 • 10.5 mgd • 290 mg/l • 8.34 = 17,300 lb/day Sludge flow = 17,300 lb/day / 0.015 /8.34 = 138,000 gal/day Sludge pumped continuously = 138,000 / 24 hrs/day / 60 min/hr = 96 gpm Aerobic Digestion: Digester volume 8 • 48^2 • 15 = 276,000 cu ft or 2.1 mil gal Detention time = 2,100,000 gal / 138,000 gpd = 15 days Digester loading assuming 75% VSS/TSS = 17,300 lb/day • 0.75 / 276,000 = 0.047 lb/cu ft

95

Dewatering: Needed dewatering volume = 96 gpm • 24 hours/day / 8 hours of dewatering = 288 gpm Use one machine Yes, plant will meet EPA requirements for effluent BOD < 30 mg/l based on aeration basin loading and performance. Poor final settling as a result of the long detention time may cause floating solids in the plant effluent and violate solids limits.

11-74. Design flow = 2.4 mgd, Peak flow = 4.2 mgd, Min. flow = 1.3 mgd, BOD = 260 mg/l, TSS = 210 mg/l, ammonia = 25 mg/l, temperature = 17.5°C Influent Pumping Station With 1 pump in standby, firm capacity = 4 • 3,000 gpm / 694 gpm/mgd = 17.3 mgd, is more than adequate to cover the peak flow. In fact one pump will cover the peak flow. One pump at minimum speed of 50% pumps 2.16 mgd and will not reach the minimum flow of 1.3 mgd without shutting off. At the minimum flow of 3,000 gpm and at 6 starts/hour, the minimum wet well volume is 3000 gpm • 10 min. = 30,000 gal or 4000 cu ft. The actual volume of 15 x 20 x 12 = 3600 cu ft and is just short of being adequate for on/off operation. Bar Screen: ¾ in openings, Figure 11-7, screenings quantity = 5 cu ft/mgd • 2.4 = 12 cu ft/day Oxidation Ditch: Volume = 4 • 80,000 = 320,000 cu ft • 7.481 / 103 = 2.4 mg BOD load = 2.4 mgd • 260 mg/l • 8.34 / 320 = 16.2 lb/1000cu ft This is right in the range of 10 to 20 lb/day/1000 cu ft (Table 11-4) Detention time = 2.4 mg / 2.4 mgd • 24 = 24 hours OK, within range of 20 to 30 for extended aeration Secondary Clarifier: Area = 2 • 60^2 • π /4 = 5,650 sq ft • 12 • 7.481 / 106 = 0.507 mg Weir length = 60 • π = 188 ft • 2 = 377 ft Overflow at Peak Q = 4,200,000 / 5,650 = 740 gpd/sq ft OK, < 1600 gpd/sq ft Overflow at Avg Q = 2,400,000 / 5,650 = 420 gpd/sq ft OK, < 800 gpd/sq ft Detention time at avg Q = 0.507 / 2.4 • 24 = 5.1 hr Higher than range of 2-3 hours Weir loading = 2,400,000 / 377 = 6,400 gpd/ft Higher than range of 10,000-20,000 gpd/ft Recirculation at 1 pump per ditch = 300 /694 = 0.43 mgd or about 20% of the influent flow which is a little short of the 50% typical. Chlorine Contact: Detention time = 4 • 20 • 10 • 10 • 7.481 / 106 / 2.4 • 24 • 60 = 36 min Ok, greater than range of 20-30 min. Sludge: Primary Solids = none F/M = 2.4 • 260 / 2000 / 2.4 = 0.13 Table K = 0.36 WAS = 2 • 0.36 • 2.4 mgd • 260 mg/l • 8.34 = 3,700 lb/day Aerobic Digestion: Digester volume 50 • 50 • 15 = 37,500 cu ft or 0.28 mil gal • 4 = 1.12 mg ASSUME 0.5% SOLIDS Sludge flow = 3700 / 500 / 8.34 = 0.89 mgd Detention time = 1.12 / 0.89 • 24 = 30 hours which is completely inadequate

96

11-75. Design flow = 10.3 mgd, Peak flow = 19 mgd, Min. flow = 5 mgd, BOD = 210 mg/l, TSS = 240 mg/l, ammonia = 22 mg/l, temperature = 17.5°C Influent Pumping Station With 1 pump in standby, firm capacity = 4 • 4,000 gpm / 694 gpm/mgd = 23 mgd, is adequate to cover the peak flow. One pump at minimum speed of 50% = 2000 gpm or 2.9 mgd and will reach the minimum flow of 5 mgd without shutting off. The wet well volume is 15 • 25 • 10 = 3750 cu ft or 0.028 mg, which is ok because pumps do not stop. Bar Screen: 1/2 in openings, Figure 11-7, screenings quantity = 6 cu ft/mgd • 10.3 = 62cu ft/day Grit Units: Horizontal area = 2 • 40 • 12 = 960 sq ft Horizontal velocity = 10,300,000 / 7.481 / 24 / 60 / 60 cfs / 960 = 0.017 ft/s < 1.0 design value even at peak flow, horizontal velocity = 29 cfs / 960 = 0.03 ft/s Primary Clarifier: Area = 2 • 65^2 • π /4 = 6600 sq ft • 7.5 • 7.481 / 106 = 0.37 mg Weir length = 65 • π = 204 ft • 2 = 410 ft Overflow at Peak Q = 19,000,000 / 6600 = 2900 gpd/sq ft Too small for the recommended range of 1200-1500 gpd/sq ft, even greater than the max of 2500 gpd/sq ft. Overflow at Avg Q = 10,500,000 / 6600 = 1600 gpd/sq ft Too small for the recommended range 600-800 gpd/sq ft Detention time at avg Q = 0.37 / 10.5 • 24 = 0.85 hr Too small Weir loading = 10,500,000 / 410 = 25,600 gpd/ft OK, in the range 10,000-40,000 gpd/ft, but overflow rate is the controlling design parameter. Aeration Basin: Volume = 8 • 80 • 40 • 14 = 358,000 cu ft • 7.481 / 103 = 2.7 mg BOD load = 10.3 mgd • 210 mg/l • 8.34 / 358 = 50 lb/1000 cu ft This is at the high end of the range of 30 to 50 lb/day/1000 cu ft for step aeration (Table 11-4) Detention time = 2.7 mg / 10.3 mgd • 24 = 6.3 hours OK, within range of 4.0 to 7.0 for step aeration F/M = 10.3 • 210 / 2.7 / 400 = 2.0. According to Figure 11-30, sludge will not settle well, and MLSS should be more in the range of 6,000, although 4,000 mg/l would be ok. Secondary Clarifier: Area = 2 • 90^2 • π /4 = 12,700 sq ft • 14 • 7.481 / 106 = 1.32 mg Weir length = 90 • π = 280 ft • 2 = 560 ft Overflow at Peak Q = 19,000,000 / 12,700 = 1500 gpd/sq ft OK, < 1600 gpd/sq ft Overflow at Avg Q = 10,300,000 / 12,700 = 810 gpd/sq ft OK, about equal to 800 gpd/sq ft Detention time at avg Q = 1.32 / 10.3 • 24 = 3 hr Ok, in the range of 2-3 hours Weir loading = 10,300,000 / 560 = 18,400 gpd/ft Ok, within the range of 10,000-20,000 gpd/ft Recirculation using 5 pumps = 5 • 3000 /694 = 21.6 mgd or about 200% of the influent flow which is more than adequate. Aeration System: Diffuser depth = 18 ft

97

Oxygen demand = 10,867 lbs/day without nitrificationOxygen demand = 19,560 lbs/day with nitrificationTemp 17.5 °C winter

Fine Bubble Aerators

Depth 18 feet�F 0.55 average value from text� 0.9 value from textDO residual 2 mg/lSOTR 5.85 lb/hp-hr Average of values in Table 11-5 4.0–7.7

depth adjustedCs STP 9.1 mg/l, see Table 2-4 20 °C 11.1Cs 9.6 mg/l, see Table 2-5 17.5 °C 11.7

Cde/C = 1 + 0.01205(diffuser depth)(1+5.6·10-7(site elevation))Diffuser depth Cde/C 1.2 mg/l 17.5 °C

OTR 2.3 lb/hp-hr HP for BOD 199 hpHP w/NH4 359 hp

QAir = 1,118 cfm winter2,012 cfm summer

14400174.02

⋅⋅=

OTEQ

Q OAir

( )( )FC

CCSOTROTR T

de

LdePT αθβ 20

,1,20

,, −���

����

� −⋅=

Sludge: Primary Solids = 25.0 mgd • 220 mg/l • 8.34 • 0.5 = 22,900 lb/day F/M = 25 • 210 (1-.35) / 2000 / 6.8 = 0.24 K = 0.44 WAS = 0.44 • 25 mgd • 210 (1-.035) mg/l • 8.34 = 12,500 lb/day Thickening PC Sludge flow = 22,900 lb/day / 0.05 / 8.34 = 54,900 gal/day FC Sludge flow = 12,500 lb/day / 0.005 / 8.34 = 300,000 gal/day or 208 gpm Operate one gravity belt unit at 208 gpm for 24 hours/day Thickened WAS = 12,500 • 0.95 capture / 0.05 /8.34 = 28,500 gpd Total Solids = 22,900 + 12,500 • 0.95 = 34,800 lb/day Total Solids flow = 54,900 + 28,500 = 83,400 gpd Anaerobic Digestion: Digester volume = 15 days • 83,400 gpd = 1.25 mil gal Digester volume based on loading assuming 75% VSS/TSS = 34,800 lb/day • 0.75 / 0.10 = 261,000 cu ft or 2.0 mil gal. Design flow = 2.4 mgd, Peak flow = 4.2 mgd, Min. flow = 1.3 mgd, BOD = 260 mg/l, TSS = 210 mg/l, ammonia = 25 mg/l, temperature = 17.5°C Influent Pumping Station With 1 pump in standby, firm capacity = 4 • 3,000 gpm / 694 gpm/mgd = 17.3 mgd, is more than adequate to cover the peak flow. In fact one pump will cover the peak flow. One pump at minimum speed of 50% pumps 2.16 mgd and will not reach the minimum flow of 1.3 mgd without shutting off. At the minimum flow of 3,000 gpm and at 6 starts/hour, the minimum wet well volume is 3000 gpm • 10 min. = 30,000 gal or 4000 cu ft. The actual volume of 15 x 20 x 12 = 3600 cu ft and is just short of being adequate for on/off operation. Oxidation Ditch: Volume = 4 • 80,000 = 320,000 cu ft • 7.481 / 103 = 2.4 mg BOD load = 23 mgd • 185 mg/l • 8.34 / 744 = 48 lb/1000cu ft Higher than conventional activated sludge, but within 40-60 lb/1000 cu ft range for step aeration Detention time = 5.6 mg / 23 mgd • 24 = 5.8 hours OK, within range of 4 to 7.0 for step aeration

98

Secondary Clarifier: Area = 3 • 130^2 • π /4 = 39,800 sq ft • 16 • 7.481 / 106 = 4.8 mg Weir length = 130 • π = 408 ft • 3 = 1225 ft Overflow at Peak Q = 55,000,000 / 39,800 = 1380 gpd/sq ft OK, < 1600 gpd/sq ft Overflow at Avg Q = 23,000,000 / 39,800 = 580 gpd/sq ft OK, < 800 gpd/sq ft Detention time at avg Q = 4.8 / 23 • 24 = 5.0 hr Higher than range of 2-3 hours Weir loading = 55,000,000 / 1225 = 45,000 gpd/ft Higher than range of 10,000-20,000 gpd/ft Chlorine Contact: Detention time = 320 • 10 • 10 • 7.481 / 106 /23.0 • 24 • 60 = 15 min Too Low, less than range of 20-30 min. Sludge: Primary Solids = 23.0 mgd • 210 mg/l • 8.34 • 0.5 = 20,100 lb/day F/M = 23 • 185 / 2100 / 5.6 = 0.36 K = 0.49 WAS = 0.49 • 23 mgd • 185 (1-.035) mg/l • 8.34 = 11,300 lb/day PC Sludge flow = 20,100 lb/day / 0.03 / 8.34 = 80,300 gal/day PC Sludge pumped = 80300 • 60 min/hr / 10 min pumped per hour / 1440 min/day = 335 gpm FC Sludge flow = 11,300 lb/day / 0.005 / 8.34 = 271,000 gal/day or 188 gpm Operate one gravity belt unit at 188 gpm for 24 hours/day or two units at 282 gpm for 8 hr/day Thickened WAS = 11,300 • 0.95 capture / 0.06 /8.34 = 21,500 gpd Total Solids = 20,100 + 11,300 • 0.95 = 30,800 lb/day Total Solids flow = 80,300 + 21,500 = 101,800 gpd Anaerobic Digestion: Digester volume 3 • 90^2 • π /4 • 20 = 382,000 cu ft or 2.9 mil gal Detention time = 2,900,000 gal / 101,800 gpd = 28 days Digester loading assuming 75% VSS/TSS = 31,400 lb/day • 0.75 / 382,000 = 0.062 lb/cu ft Dewatering: Needed dewatering volume = 101,800 gpd / 1440 min/day • 24 hours/day / 8 hours of dewatering = 212 gpm Use one machine No, plant will not meet EPA requirements for disinfection, will meet effluent BOD and suspended solids requirements depending on the performance of the Secondary Clarifiers. Sludge flow headloss calculations 3% Primary Sludge, Sy = 0.16, n = 0.00135

4.645.04.6416.02^5.02^00135.094103)00135.0(1000

Vlc ⋅⋅⋅+⋅+=

2.326.20002.010335.1

Vlc++= = 5.2 ft/sec

Actual velocity: Q = 355 gpm in a 6 inch pipe, velocity is about 4 ft/s < lower critical, therefore, the flow is laminar

ft/ft027.00003.0027.05.0)5.0(4.64)4(00135.0

5.0)4.64(3)16.0(16

L/H =+=+=

For a pipe length of 350 ft, H = 9.5 ft or 4.1 psi 0.5% WAS is a Newtonian fluid that follows the headloss of water. Using the Hazen-Williams equation, headloss is calculated as follows: H = 0.002083 • 800 • (188 gpm)^1.85 / 6^4.8655 = 4.4 ft or 2 psi

99

6% TWAS, Sy = 1.7, n = 0.2

4.645.04.647.12^5.02^2.094103)2.0(1000

Vlc ⋅⋅⋅+⋅+=

2.324.2776.3103200

Vlc++= = 24 ft/sec

Actual velocity: Q = 188 gpm in a 6 inch pipe, velocity is about 0.6 ft/s < lower critical, therefore, the flow is laminar

ft/ft29.0008.028.05.0)5.0(4.64

)6.0(2.05.0)4.64(3

)7.1(16L/H =+=+=

For a pipe length of 360 ft, H = 104 ft or 45 psi Digested sludge is composed of: Total Solids = 30,800 lb/day, 50% of the volatile suspended solids are destroyed, where volatile solids are 75% of the total solids, the solids remaining after digestion are = 30,800 (0.75 • 0.5 + 0.25) = 19,300 lb/day in a Total Solids flow = 101,800 gpd. The solids concentration is 19,300 lb/day / 101,800 gpd / 8.34 = 2.3% solids 2.3% Digested, Sy = 0.05, n = 0.007

4.645.04.6405.02^5.02^007.094103)007.0(1000

Vlc ⋅⋅⋅+⋅+=

2.32805.005.01037

Vlc++= = 3.1 ft/sec

Actual Velocities are less than the lower critical velocity: Recirculation velocity at 120 gpm = 1.36 ft/s Transfer to belt press = 212 gpm = 2.4 ft/s

Recirculation: ft/ft0086.00006.0008.05.0)5.0(4.64

)4.1(007.05.0)4.64(3

)05.0(16L/H =+=+=

For digester recirculation at a pipe length of 120 ft, H = 1 ft

Transfer: ft/ft009.0001.0008.05.0)5.0(4.64)4.2(007.0

5.0)4.64(3)05.0(16

L/H =+=+=

For transfer to the belt filter press at a pipe length of 400 ft, H = 3.6 ft or 1.1 psi

100

12 OPERATION OF WASTEWATER SYSTEMS

12-1. The Capacity assurance, Management, Operation, and Maintenance (CMOM) program is a comprehensive checklist of self-monitoring activities designed to improve system performance and specifically to reduce sanitary sewer overflows. The states in EPA Region IV developed a capacity assurance, management, operation, and maintenance program to help utilities identify needs for collection and treatment.

12-2. The capacity of a treatment plant is determined by sizing treatment processes to meet equipment capacity and ultimately, NPDES requirements for flow and load limitations. Measurements of evaluation of treatment units include flow, BOD, SS concentrations of the influent, effluent, and for some processes, sludge or recycle streams. Evaluation is based on hydraulic, BOD, and SS loading rates.

12-3. First calculation to total is to determine in influent mass = 1.2 • 220 • 8.34 = 2202 lb/day. The separated water effluent is 2202 – 1010 = 1192 lb/day. Calculating the separated water flow = 1.2 – 1.181 = 0.019 mgd. Calculating the concentration of the sludge = 1010 / 0.019 / 8.34 = 6360 mg/l or about 6.4%. The final table is:

V S MLiquid influent 1,200,000 gpd 220 mg/l 2200 lb/daySeparated water 1,181,000 gpd 121 mg/l 1190 lb/dayThickened sludge 19,000 gpd 6360 mg/l 1010 lb/day

12-4. First calculation to total is to convert the gpm to gpd, 500 gpm = 720,000 gpd then determine the influent mass = 0.72 • 250 • 8.34 = 1500 lb/day. The mass of the thickened sludge is 1500 • 0.9 = 1351 lb and the mass of the recycled flow is 1500 • 0.1 = 150 lb. The flow of thickened sludge = 1351 / 4000 / 8.34 • 1000000 = 40,500 gpd. Recycled flow is 720,000 – 40,500 = 679,500 gpd. The recycled flow concentration is 150 / 0.6795 / 8.34 = 26 mg/l. The final table is:

V S MLiquid influent 720,000 gpd 250 mg/l 1500 lb/dayThickened sludge 40,500 gpd 4000 mg/l 1350 lb/dayRecycled flow 679,500 gpd 26 mg/l 150 lb/day

12-5. From equation 14-7, efficiency = Q·H/3960/hp = 2618·(76-13)·2.31/3960/(118/0.746) = 61% Loss of efficiency = 82% - 61% = 21% and pump should be refurbished.

12-6. Because flow monitoring suggests a direct connection, TV inspection should be used to determine if visual evidence of pipe defects can be seen. Inspection may not show the cause of water intrusion. Smoke testing is a good way to locate holes and the connection to the surface. Before testing, residents of buildings along a sewer line should be notified to reduce the chance of surprise. If smoke enters a building, the occupants should be evacuated and the interior ventilated while locating the point of entry.

12-7. Four kinds of wastes are strictly prohibited from disposal: 1) wastes that create a fire of explosion hazard, 2) substances that impair hydraulic capacity, 3) contaminants that create a hazard to people and/or to treatment processes, and 4) refractory toxic wastes that pass through the treatment system.

101

12-8. Sewer rates should be fair and in proportion of the use and benefits received. Residential charges should be based on a uniform rate or in proportion to winter water meter readings. Rates for industrial dischargers should include flow and wastewater strength surcharges.

12-9. Maintaining sewer capacity is necessary to avoid sewer overflows. Maintaining capacity requires regular flushing and cleaning. Sewer blockages may be caused by sand, greasy materials, sticks, stones, and tree roots, see figure 12-6.

12-10. Using a modification of equation 12-5: wastewater volume cost = 1200 • $0.35 = $420.00 BOD surcharge = 1200/1000000 • (4200 – 250) • $102 = $483.48 SS surcharge = 1200/1000000 • (800 – 300) • $180 = $108.00 Total daily charge is $1011.48

12-11. The following steps are used to develop an asset management program:

• Identify the target level of service

• Condition and capacity assessment

• O&M planning - Standardize procedures

• Identify solutions and activities to reduce long-term costs (repair, rehabilitation, or replacement)

• Development of structured programs to protect assets

12-12. Stream flow = 15 cfs or 9.7 mgd. The wastewater flow of 2 mgd represents a dilution factor of 4.8:1. Dilution credit may not be given and dilution ranges between zero and 4.8. Without dilution, effluent limits must be less than 1.48 mg/l N and 0.36 mg/l P. With dilution of 4.8, effluent limits increase to 7.1 mg/l N and 1.7 mg/l P.

102

13 ADVANCED WASTEWATER TREATMENT

13-1. Conventional secondary treatment standards are 30 mg/l BOD and 30 mg/l suspended solids monthly average and, in many locations, 200/100 ml coliform. Many of the facilities that use chlorine must also dechlorinate (refer to Chapter 11, Effluent Quality). Pollutants not treated during conventional secondary processes include coliform (incomplete removal even if disinfection processes are included), viruses, ammonia, phosphorus, and other soluble non-biodegradable chemicals. Toxins and metals are removed to varying degrees if co-precipitated with biological solids. The limits were technology-based standards for biological processes. Ammonia in its ionic form is toxic to fish. Phosphorus is a growth nutrient for algae and leads to algae blooms.

13-2. Pathogenic organisms include Giardia, Cryptosporidium, and Bacteria. Based on particle size, pathogenic organisms can be partially removed using granular filtration and microfiltration. All removed by ultrafiltration, nanofiltration, and reverse osmosis. Viruses are partially removed by ultrafiltration. Viruses are removed by nanofiltration and reverse osmosis. Although granular filtration will not remove bacteria or viruses, with coagulation, filtration will remove sufficient solids to allow effective disinfection of bacteria, viruses, and other pathogens.

13-3. Protection of public health is related to pathogens in wastewater. The concern is about body-contact recreation and drinking water supplies. Conventional treatment removes 99 to 99.9 percent of bacteria, but an unknown number of viruses and protozoa. Conventional treatment does not render effluent pathogen free. See section 13-3, disinfection of biologically treated wastewater requires coagulation and granular-media filtration followed by chlorination with extended contact time. Although chlorine is effective, pathogens can be protected by suspended and colloidal solids. The Pomona Virus Study determined that filtration with chemical addition followed by long-term chlorine contact is adequate to protect public health.

13-4. Gravity sedimentation along cannot remove small particles allowing bacteria and viruses to survive even with disinfection by chlorination. Variations in gravity filtration include pressure filters, traveling bridge, and upflow filters. Treatment results of filtration in combination with disinfection are similar. All filtration systems with chemical addition can physically remove protozoal cycts, and helminth eggs and inactive the remaining bacteria and viruses by extended chlorine contact.

13-5. Disinfection of biologically treated wastewater requires coagulation and granular-media filtration followed by chlorination with extended contact time. Although chlorination is effective in killing bacteria and inactivating enteric viruses, these pathogens can be protected in suspended and colloidal solids if the wastewater has not been filtered and if gravity settling does not remove sufficient solids. Gravity settling along does not provide sufficient solids removal for chlorination to be effective.

13-6. 1 mgd is equal to 694 gpm. PSI is converted to ft of water by multiplying by 2.31 ft of water per psi. HP is calculated using the equation given, because not efficiency is given, 100% is assumed (bad assumption because 80% is typical, but no value was given). kWh is 0.749 times HP, cost is kWh times 24 hours/day times 365 days per year times the cost of $0.15/kWh.

103

mgd gpm psi ft water HP kWh Cost/yr

1 694 10 1.8 1.31 $ 1,726 1 694 150 345 60.5 45.32 $59,544 1 694 5 12 2.0 1.51 $ 1,985 1 694 15 35 6.1 4.53 $ 5,954 1 694 80 184 32.3 24.17 $31,757 1 694 150 345 60.5 45.32 $59,544

13-7. Size the filters using the peak flow to calculate ¾ of the area. Design is based on peak flow. 9.6 mgd • 694 gpm/mgd = 6667 gpm / 3 = 2,222 gpm applied to each of 3 filters at 5 gpm/sq.ft., 2222/5 = 444 sq.ft. Each filter is 44.4 or say 21 ft by 21 ft.

13-8. To qualify as an effluent toxic substance, it must be hazardous to aquatic life or human health. Toxicity to human health may be related to either carcinogenicity or chronic disease through long-term consumption. Gross toxicity is evidenced by upset of the biological processes in wastewater treatment. The conventional approach to toxicity testing is to test raw and treated wastewater samples for specific toxic substances. This is costly because of the large number of individual substances that must be tested for and the test ignores the combined and synergistic effects of multiple contaminants. Another approach is to evaluate toxicity by bio-monitoring of effluent. Living organisms are placed in the wastewater sample and monitored. If the organisms die, additional testing must be conducted to determine the cause. Bioassays are conducted in a laboratory to determine specific toxic substances. Water fleas and fat-head minnows are used as indicator species.

13-9. Based on Figure 13-1, viruses are removed by reverse osmosis and nanofiltration. In Figure 13-2 all treatment processes are capable of virus removal. The key is solids removal through coagulation and filtration in order to provide effective disinfection. Coagulation is important in removing solids that may inhibit disinfection. Figure 13-10 shows rapid mixing, filtration, and chlorination along with chemical addition necessary for virus removal. Effluent water quality is < 2 NTU to allow effective disinfection of pathogens and viruses resulting in a coliform concentration less than 2.2/100 ml.

13-10. The treatment processes corresponding to each discharge are as follows: SS = 3; MPN 2.2 Chemical addition, flocculation, filtration, disinfection > 90 min. SS = 10; MPN 23 Chemical addition, flocculation, filtration, disinfection > 20 min. SS = 21; MPN 100 Filtration (no chemical addition), disinfection > 20 min. SS = 30; MPN 200 Secondary treatment, no filtration, disinfection > 20 min.

104

13-11. Phosphorus removal in a typical secondary treatment plant is shown in Figure 13-5. About 30% of the phosphorus is removed in biological sludge. For an influent of 7 mg/l, the typical effluent is about 5 mg/l. Phosphorus released into streams may not have a large impact in flowing and turbid waters because of the lack of algae. In pools and lakes, phosphorus becomes a growth nutrient for algae. Phosphorus may be removed by precipitation using metal salts or by enhanced biological processes.

13-12. Alum dose depends on the amount of phosphorus to be removed as listed in the following table: The table also shows the Alum does based on a 7 mg/l P concentration in the influent.

FeCl dose is reported to be about 15:1 and would require 105 mg/l.

13-13. Example 13-2 was based on a dose of 90 mg/l. The results of feeding 100 mg/l should be as good or slightly better. The plant could expect to have a chemical precipitation of aluminum phosphate and aluminum hydroxide. Concentrations would be slightly higher than those in the example. The suspended solids removal of 75% and BOD removal of 55% would also be the same or slightly higher. If the removals are proportional to the increase in chemical feed, solids and BOD removals would increase by 100/90 or about 10% better. An estimate of total solids production would be 253 mg/l * 100/90 or about 280 mg/l.

13-14. Alum does is 12 mgd • 90 mg/l • 8.34 = 8,900 lb/day (• 365 day/yr / 2000 lbs/ton) or about 1624 tons/year (• $210) for an annual cost of about $341,000/year.

13-15. Toxicity testing consists of testing waters for specific chemical constituents or biomonitoring. Specific chemical testing includes metals, organics, and other constituents as identified by the EPA. It measures the concentration of items tested. Biomonitoring measures the death of indicator organisms (typically fathead minnows and water fleas). The measure does not suggest a toxic cause, but indicates that additional testing must be conducted to determine toxic agent. Where industrial wastewater is not responsible, the most likely cause of toxicity is ammonia, particularly during the summer months.

13-16. The figure reference is incorrect; the problem should reference Figure 13-12. The effluent P(total) is 1.3 mg/l • 12 mgd • 8.34 = 130 lb/day or 47,500 lb/year. The alum dose is 90 mg/l • 12 mgd • 8.34 = 8,900 lb/day

Alum Dose to Phosphorus

Expected P Removal

Alum Dose (mg/l) based on 7 mg/l influent

13:1 75% 9116:1 85% 11222:1 95% 154

105

At 5 lb/gallon, the flow rate of alum is 8,900 lb/day / 5 lb/gallon = 1780 gpd or about 1.24 gpm Size pumps for at least 1.24 gpm

13-17. The organic portion consists of the primary sludge (180 mg/l) and waste activated sludge (49 mg/l) for a total organic portion of 229 mg/l • 6 mgd • 8.34 = 11,500 lb/day The inorganic portion consists of the AlPO4 precipitate (12 mg/l) and the aluminum hydroxide (15 mg/l) for a total of 27 mg/l • 6 mgd • 8.34 = 565 lbs/day Following anaerobic digestion: Solids = 50% reduction of VSS + non-VSS + inorganic 0.5 • 0.75 • 11,500 + 0.25 • 11,500 + 565 = 7,750 lb/day Cake dry weight solids = 0.95 • 7,750 = 7,360 lb/day or 3.7 tons/day Cake wet weight solids = 7,360/0.26 = 28,300 lb/day or 14 tons/day

13-18. Phosphorus reduction by biological growth in activated sludge is about 30%, see Figure 13-11. Chemical addition for phosphorus removal allows about 82% phosphorus removal without filtration, see Figure 13-12. With filtration, the 2% phosphorus contained in the biological cells is removed. For an additional 18 mg/l suspended solids, 0.4 mg/l phosphorus is removed. In addition, an unknown amount of inorganic P may be attached with the solids and may also be removed.

13-19. Determine the alum dose by first calculating the removal efficiency: Alum dose is based on P removal, P removed = (7 – 0.75) / 7 • 100 = 89% Alum dose for 95% removal is 22:1 or 22 • 7 = 154 mg/l Alum dose for 85% removal is 16:1 or 16 • 7 = 112 mg/l 89% is about ½ of the way between the two, split the difference and use 130 mg/l Use a flow of 1 mgd Using 5 lb/gallon, the flow rate would be = 130 mg/l • 1 mgd • 8.34 / 5 lb/gallon = 217 gal/day, size pump for 9 gal/hr. A storage tank sized for 9 days of storage would be 217 gal/day • 9 days or about 2,000 gal.

13-20. The increase in PC SS is proportional to the increased loading 210 mg SS with a P content of 0.009 lbP/lbSS 1.89 P in SS 3.15 P (precipitation) remains the same 5.04 P mg in the PC sludge, 265 lb/day The total P removal increases slightly with the increase in PC SS BOD removal would also be proportional 114 mg BOD in the settled wastewater 57.6 mg based on effluent 4 mg based on a 20 mg/l effluent 10 mg captured in SS 74 mg 1.4 mg P or 75 lb/day, Total P removal = 6.5 mg or 340 lb/day

106

107

13-21. The common forms of phosphorus in wastewater are orthophosphates (PO4), polyphosphates, and organically bound phosphates. Biological phosphorus is contained in human feces, food and organic material that may be discharged to the sewer. Like all chemicals, phosphates may be wasted from industrial manufacturing. Common forms of nitrogen are organic, ammonia, nitrate, and nitrite. Because of the lack of oxygen in the sewer, the most common forms are ammonia and organic nitrogen. Nitrogen results from human excreta, organic material in food waste, and industrial wastes. Refer to Table 13-2 nitrogen is about 35 mg/l, phosphorus is about 7 mg/l, lbs of nitrogen = 35 mg/l • 120/1000000 mgd • 8.34 = 0.035 lb/person/day lbs of phosphorus = 7 mg/l • 120/1000000 mgd • 8.34 = 0.007 lb/person/day

13-22. The question should have asked under which conditions does nitrification, denitrification, and biological phosphorus removal take place. Aerobic refers to a water containing dissolved oxygen. Fish, bacteria, and other aquatic organisms use dissolved oxygen to support biological activity. Anoxic refers to a water without dissolved oxygen, but containing chemically bound oxygen such as sulfate (SO3) and nitrate (NO3). Facultative bacterial can use the oxygen converting SO3 to H2S gas and NO3 to N2 gas. Anaerobic refers to a water without dissolved or chemically bound oxygen. Anaerobic bacteria do not use oxygen for biological activity. Nitrification requires an aerobic environment to convert ammonia (NH3) to nitrate (NO3). Denitrification requires an anoxic environment to convert nitrate (NO3) to nitrogen gas (N2). Biological phosphorus removal requires an anaerobic environment to release organic P and begin the process of luxurious phosphorus uptake.

13-23. Biological nitrification destroys alkalinity by using the carbonate in the conversion of ammonia to nitrate. About 7.2 lb of alkalinity is destroyed per pound of ammonia-nitrogen oxidized to nitrate.

13-24. Starting with the blower capacity, there are 3 electric and 2 engine drive blowers. Assuming one is provided for standby, use 4 blowers at 18,300 cfm for total capacity of 73,200 cfm. Using 0.0154 lbs/cfm, the oxygen delivery capacity of the blowers is 1,130 lb/minute Using an oxygen transfer efficiency of 6%, the dissolved oxygen rate is 67.8 lb/min or 97,600 lbs/day 4.6 lbs of oxygen is required per lb of ammonia removed, therefore, the blowers have a capacity of 21,200 lb/day of ammonia removal. Note: Using the information from Las Vegas, the ammonia removed is 18 mg/l – 0.8 mg/l or 17.2 mg/l at a flow rate of 46 mgd. The amount of oxygen required is 17.2 mg/l • 46 mgd • 8.34 = 6,600 lb/day of ammonia destroyed. The actual capacity of about 3 times this amount accounts for peak day demands.

13-25. Using equation 13-12, assume 2 mg/l of dissolved oxygen (the minimum), methanol dose = 0.9 • 2 mg/l (DO) + 2.5 • 32 mg/l (nitrate) = 82 mg/l • 4.3 mgd • 8.34 = 2,940 lb/day Methanol volume = 2940 lb/day / (8.34 • 0.82) / 0.95 = 450 gallons/day Cost = 450 gal/day • 30 day/mo • $1.3 = $17,500/month.

13-26. Using 22 mg/l of ammonia (per Table 13-2), the lbs of influent ammonia are 22 mg/l • 2.2 mgd • 8.34 = 404 lbs/day assuming that all of the ammonia is converted into nitrate, there would be 404 lb/day of nitrate and assume 30% is converted to nitrogen gas. Using 1 lb of BOD per 1 lb of

108

nitrate, BOD consumed is 404 lb/day • 0.3 = 121 lbs/day. Check if sufficient soluble BOD exists in the influent, influent soluble BOD = 200 mg/l • 0.4 • 2.2 mgd • 8.34 = 1,470 lbs/day. The BOD consumed by denitrification is 121 lbs/day.

13-27. Alum

Applied (mg/l)

Influent Phosphorous

(mg/l)

Al/P Weight Ratio

Phosphorous Removal (%)

BOD Removal

(%)

SS Removal

(%) 0 10.3 0 23 96 93

58 10.9 0.55 50 96 92 117 10.6 1.14 77 97 90 175 10.4 1.74 90 94 93 325 9.3 3.61 95 95 86

Data from text under Equation 13-2.

1.34 75 1.65 85 2.27 95

(b) Although in general the addition of alum increases the settleability of biological floc, this argument is not valid in this case since the BOD and SS removal efficiencies are independent of alum dosage from 0 to 175 mg/l. The factors that contributed to the high removal efficiencies in this laboratory study are: (1) the excellent hydraulic efficiency of the clarifier to separate the floc from the supernatant, (2) the gentle aeration of the mixed liquor and rapid return of settled sludge thus keeping it continuously aerobic, (3) constant rate of wastewater feed eliminating any variations in hydraulic and organic loading rates, (4) and high temperature of 22 to 24° C for the loadings that were an aeration period of 7.2 hr, approximate BOD loadings of 33 lb/1000 cu ft/day and 0.3 lb BOD/day/lb MLSS, and a sludge age of about 14 days. (c) The concentration of alkalinity decreases with increasing alum dosage since the excess alum reacts with the bicarbonate ion to reduce the alkalinity by producing carbon dioxide (Equation 13-2).

-3.0

-2.0

-1.0

0.0

1.0

2.0

3.0

4.0

1.0 2.0 3.0 4.0 5.0

Al/P

Phos

phor

ous

rem

oval

, per

cent

109

13-28. Based on the data, the following plots may be constructed:

13-29. Anoxic zones range from 0.5 to 3.0 hr detention time, aerobic zones range from 6 to 24 hours, and anaerobic zones range from 10 to 20% of the aeration tank volume. Use detention times shown for the Las Vegas plant:

Tank Volume Tank Detention Time Mil Gal Cubic Feet

Anaerobic 3 hours 14.4 1,925,134 Anoxic 6.75 hours 32.4 4,331,551 Aerobic 6 hours 28.8 3,850,267

Methanol and acetic acid addition may be required.

Distance Ammonia Alkalinity ln(NH4d/NH4o)/MLVSS/(1.024^(18.5-20))/(1800/133670)/D0 16.11 2436 16.46 185 (0.00013)

12 12 190 0.00120 18 8 190 0.00189 24 5.28 188 0.00225 30 4.48 189 0.00206 36 3.96 188 0.00188 42 4.55 191 0.00145 48 2.93 181 0.00171 54 2.64 178 0.00162 60 2.62 178 0.00146 66 2.59 178 0.00134 72 2.44 178 0.00126 78 2.31 179 0.00120 84 2.44 178 0.00108 90 2.44 180 0.00101 96 1.69 178 0.00113

102 2.01 175 0.00098 108 1.44 172 0.00108 114 0.87 172 0.00123 120 1.05 171 0.00110 126 1.25 172 0.00098 132 0.6 177 0.00120 138 0.8 173 0.00105 144 0.53 173 0.00114 150 0.42 174 0.00117 156 0.4 172 0.00114 162 0.59 169 0.00098 168 0.55 170 0.00097

Alkalinity Concentration vs. Distance from Inlet

0

50

100

150

200

250

300

0 12 24 36 48 60 72 84 96 108 120 132 144 156 168

Distance

Con

cent

ratio

n (m

g/l)

Ammonia vs. Distance

0

5

10

15

20

0 12 24 36 48 60 72 84 96 108 120 132 144 156 168

Distance

Con

cent

ratio

n (m

g/l)

k versus distance

-

0.00050

0.00100

0.00150

0.00200

0.00250

0 12 24 36 48 60 72 84 96 108 120 132 144 156 168

Distance

ln(N

d/N

o)/M

LVS

S/te

mp/

t

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13-30. The ammonia load applied to the basin is 25.5 mg/l • 10 mgd • 8.34 = 2130 lb/day for a loading rate of 10 lb/100 cu, the basin volume would be 2130 / 10 or 21,300 cu ft, thus the detention time is 3.8 hours. The detention time is just short of the 4 hours minimum recommended.

13-31. This is a trick question: Figure 13-16 shows that ammonia removal results in an effluent total nitrogen of 28.5 mg/l, of which 1.2 mg/l is organically associated. If filters remove 20 mg/l SS to 5 mg/l SS for a 75% reduction, then the 1.2 becomes 1.2 * 0.25 = 0.3 mg/l for a total nitrogen effluent of 27.6 mg/l which is significantly greater than 10 mg/l. If methanol is added per Figure 13-7, the effluent total nitrogen can be reduced to 10 mg/l or less. If the treatment process is converted to nitrification-denitrification and subsequent treatment is provided with methanol addition, a final total nitrogen of 4-8 mg/l may be reached.

13-32. Without primary clarification there is no primary sludge, BOD is not removed, and the increase in BOD results in a doubling of waste activated sludge. To hold the same F/M, the concentration of MLSS must increase. 200 mg/l of BOD and 240 mg/l of SS enter the aeration basin along with 38.5 mg/l of total N K = 0.5 using the same 0.4 F/M, see Figure 11-45 Ws = 100 mg of sludge solids increased efficiency captures of 4 and 10 mg remain the same for a total of 114 mg A nitrogen content of the solids is 0.062 or N = 7.068 with an additional 3 mg of N in the effluent

13-33. The first step is anaerobic treatment designed to release organic phosphorus and change the behavior of the bacteria enhancing subsequent phosphorus uptake. An internal recycle brings fresh organisms back to anaerobic basin 1. During the anaerobic step, nitrate is converted to nitrogen gas, but this is a secondary consequence of the nitrate contained in the return activated sludge. The second step is an anoxic treatment is designed to remove the nitrate contained in the recirculation flow. The amount of nitrate removed is based on the recycle rate. The third step is aerobic treatment where BOD is removed, ammonia is converted to nitrate, and during the process, excess phosphorus is absorbed by the organisms. Final settling removes the floc for return and waste.

13-34. The main purposes for each process are listed after the name. Lime precipitation: precipitation of heavy metals, suspended organics and phosphates, and disinfection. Air stripping: ammonia and volatile organic compounds Recarbonation: stabilization by pH reduction with carbon dioxide Carbon adsorption: nonpolar organic compounds and some heavy metals Reverse osmosis: dissolved salts and trace organic compounds. The primary problem in operation of the reverse osmosis process at WF 21 is loss of productivity. The corrective maintenance is periodic cleaning with detergents and enzymes.

13-35. Microfiltration is a pretreatment for the reverse osmosis process and replaces the processes of flocculation, lime addition, settling, recarbonation, and filtration. The function is to remove small particles (down to the size of bacteria) that reduce the run-time of reverse osmosis membranes.

13-36. An effluent of 60 mg/l cannot be achieved with plain primary sedimentation and full secondary treatment typically produces an effluent of 30 mg/l or better. An effluent of 60 mg/l can be achieved with a coagulant aid to the primary clarifier of alum, iron, and /or polymer (advanced primary treatment). Typical removal rates are about 80%. Using the influent wastewater value of 240 mg/l, a primary clarifier with chemical addition could achieve an

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effluent of 48 mg/l. In large plants, split treatment using primary effluent and secondary effluent could create a combined discharge of 60 mg/l. If 1/3 of the primary effluent (120 mg/l) were blended with 2/3 of secondary effluent (30 mg/l) the combined effluent would be 1/3 • 120 + 2/3 • 30 = 60 mg/l. The most common toxicity is ammonia and requires treatment using nitrification. Other sources of toxicity must be eliminated prior to discharge through the Agency’s sewer use permit and pre-treatment program.

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14 WATER REUSE

14-1. From a disposal perspective, wastewater is something to be discharged and must meet water quality requirements for discharge or application requirements for reuse. The decision to reuse is based on the cost difference between treatment for a surface water discharge and treatment and distribution costs for recycling. As part of water resource planning, recycled water is valued in contrast to the cost or value of potable water. Recycled water is viewed as an alternative source of water for specific uses such as irrigation, groundwater replenishment, and non-potable uses.

14-2. The general uses for wastewater reuse are agricultural irrigation, recreational impoundments, industrial uses, urban reuse, environmental enhancement, seawater intrusion barriers, and other reuses The largest California use was agricultural irrigation (49%) followed by landscape irrigation (20%) and groundwater recharge (12%). In contrast, Florida reuses recreational impoundments (43%) followed by agricultural irrigation (19%) and groundwater recharge (17%). Differences exist in the amount used for recreational impoundments. California uses a greater percentage on landscape irrigation. Both are similar in their focus on agricultural irrigation and groundwater recharge.

14-3. Water quality requirements for recycled water are set on a state-by-state basis. Refer to Table 14-2. Unrestricted reuse applications include irrigation of food crops, parks, playgrounds, residential landscaping and other locations accessible to the public. Numeric water quality objectives include: turbidity < 2 NTU and no detectable fecal coliform. Treatment processes include secondary treatment followed by filtration using coagulation followed by disinfection. Examples of secondary treatment include activated sludge and pure oxygen, but not trickling filter. Examples of filtration include gravity filters, pressure filters, traveling bridge automatic, membrane filters. Disinfection includes chlorination and ultraviolet light.

14-4. Restricted versus unrestricted reuse refers to differences in the opportunity for human contact (Classification A versus Classification B). Where adequate buffer areas are established and public access is restricted, filtration and disinfection requirements prior to reuse are less than unrestricted reuse that includes potential for human contact with the treated water or crops eaten raw. The general requirements are listed in Table 14-2. A comparison of restricted versus unrestricted water quality requirements indicate that BOD and suspended solids removals are comparable for restricted reuse applications, but are more lax for restricted irrigation use. Coliform quantity varies greatly with the most restrictive applied to unrestricted reuse. Filtration and long-term disinfection is required to create <2.2 MPN/100 ml for unrestricted access by the public, where <200 MPN/100 ml can obtained without filtration for restricted irrigation reuse for limited access by the public, and <1000 MPN/100 ml may obtained without disinfection for some restricted reuse applications without access by the public.

14-5. Refer to Table 14-3, a preliminary design report must include information on land use, soils, hydrogeology, land management, and evaluation of long-term impacts, pretreatment, and reclaimed water characteristics. For groundwater recharge, the report must include travel time from the discharge point, impacts of water mounding, and assessment on other impacts. For industrial applications, the report must address any additional disinfection requirements and disposal of industrial wastewaters. The (industrial) pretreatment program is important because many industrial pollutants are not removed in conventional or advanced wastewater treatment. Source control is critical to keeping many toxic and carcinogenic substances out of the recycled water.

14-6. California Law, Title 22, Chapter 3, Water Recycling Criteria, refers to four levels of recycled water treatment: disinfected tertiary (equal to grade A), disinfected secondary – 2.2 total coliform /100ml (grade B with greater coliform removal), disinfected secondary – 23 total coliform/100ml (grade B with greater coliform removal), and undisinfected secondary (equal to grade C).

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Disinfected tertiary treatment refers to an oxidized wastewater (secondary treatment) that has been coagulated and passed through natural undisturbed soils or a bed of filter media at a rate that does not exceed 5 gpm/sq ft or does not exceed 2 gpm/sq ft in a traveling bridge automatic backwash filter. Filter discharge must not exceed a 2 NTU 24-hr average and must not exceed 5 NTU for more than 15 min and never exceed 10 NTU. Wastewater passed through a microfiltration, ultrafiltration, nanofiltration, or reverse osmosis membrane must not exceed 0.2 NTU for more than 15 min and never exceed 0.5 NTU. The chlorine disinfection process following filtration must provide a CT (the product of total chlorine residual and modal contact time measured at the same point) value of not less than 450 mg-min/l at all times with a modal contact time of at least 90 minutes, based on peak dry weather design flow or a disinfection process that when combined with filtration has been demonstrated to inactivate and/or remove 99.999 percent of plaque-forming units of F-specific bacteriophage MS2, or polio virus in the wastewater. The 7-day median (4 of 7 samples) of total coliform bacteria measured in the disinfected effluent must not exceed an MPN of 2.2/100ml and the number of total coliform bacteria must not exceed an MPN of 23/100ml in more than one sample in any 30 day period. No sample shall exceed an MPN of 240/100ml total coliform bacteria. The modal contact time is that amount of time elapsed between the time that a tracer dye is injected into the influent and the time that the highest concentration of the tracer is observed in the effluent from the chamber.

14-7. Check filtration rate, given 6 filters, one out of service is 5 at 30 ft in diameter for a total of 707 sq ft and a total area of 3534 sq ft. At a maximum flow rate of 5 gpm/sq ft, the maximum flow rate is 17,670 gpm or 25.4 mgd. A violation of the filtration rate occurred when the effluent flow hit 31.5 mgd. Check chlorine contact time, given a basin of 5.1 million cu ft and a 90 minute detention time, the volume is 38 million gallons and has a flow capacity of 610 mgd at 90 minutes. No flows approached 610 mgd. Check chlorine contact time, give a CT of 450, at 8 mg/l the minimum detention time would be 56 minutes which is less than the 90 minute minimum. Therefore, the CT time would be met by the minimum detention time of 90 minutes.

14-8. Restricted versus unrestricted reuse refers to differences in the opportunity for human contact. Where adequate buffer areas are established and public access is restricted, disinfection requirements prior to reuse are less than unrestricted reuse that includes potential for human contact with the treated water or crops eaten raw. The general requirements are listed in Table 14-1. A comparison of restricted versus unrestricted water quality requirements indicate that BOD and suspended solids removals are comparable, but the degree of disinfection increases from 200/100ml to 23/100ml for agricultural reuse and from 23/100ml to 2.2/100ml for urban reuse. Agricultural irrigation is divided into restricted and unrestricted reuse. Restricted reuse may only require primary or secondary treatment with limited disinfection. Unrestricted agricultural irrigation includes food eaten raw and requires secondary treatment with coagulation, filtration, and disinfection to limit the potential spread of disease causing organisms. Similar to problem #4, water-quality requirements for restricted non-agricultural irrigation are divided between cemeteries, freeways, golf courses, nursery stock and sod farms with access by the general public, but that access is controlled and irrigation can be limited to times when the public is not present. Restricted agricultural irrigation applications include orchards, vineyards, feed, fodder, and fiber crops that undergo commercial pathogen reduction treatment and were access by the public is almost non-existent. Treatment may be limited to a series of stabilization ponds or undisinfected secondary treatment. Water quality requirements are coliform < 1000 MPN/100ml, BOD and suspended solids limited to 40 to 60 mg/l. In contrast, unrestricted urban irrigation requires the greatest level of treatment with coliform < 2.2 MPN/100 ml and turbidity <2 NTU (or about 5 mg/l suspended solids).

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14-9. Using simple loading rate calculation: Convert mgd to ac ft/yr, 2.3 mgd • 106 / 7.48 • 365 / 43560 = 25,800 ac ft/yr. Using 4 in/wk • 52 wk/yr / 12 in/ft = 17.3 ft/yr. Acres = 25,500 ac ft/yr / 17.3 ft/yr = 1,500 acres. Based on limiting crop uptake to 320 lb nitrogen/year/acre, lb nitrogen applied = 2.3 mgd • 24 mg/l • 8.34 = 460 lb/day • 365 day/yr = 167,900 lb/year / 320 lb/yr/acre = 525 acres Accounting for percolation, 2.3 mg/l • 10 mg/l • 8.34 • 365 = 70,000 lb/year could percolate into the ground. Acres = (168,000 – 70,000) / 320 = 306 acres. The maximum loading rate is the controlling limitation is the hydraulic loading rate resulting in 1,500 acres, which is 3 to 5 times the area based on nitrogen limitations.

14-10. Using equation 14-1, water load + precipitation = evapotranspiration + percolation + runoff runoff = 0 wastewater applied = 2.3 mgd • 36.8 ac-in./mg = 59 ac-in/day • 30 days/mo. = 2,540 ac-in./mo. precipitation = 2.3 in./mo./acre evapotranspiration = 5.9 in./ac/mo. percolation = 14 in./ac/mo. 2540 ac-in./mo. + 2.3(acres) = (5.9 + 14)(acres) Acres = 2540/(5.9 + 14 – 2.3) = 144 acres required for water balance Nitrogen applied = 24 mg/l • 2.3 mgd • 8.34 • 30 days = 13,800 lb/mo. Percolation limited to 10 mg/l nitrogen • 2.3 mgd • 8.34 • 365 = 70,000 lbs/yr of nitrogen Acres based on nitrogen loading = (13800 lb/mo • 12 mo – 70000 lbs/yr) / 400 lb/yr/acre = 239 acres If 1 mgd is reduced to a nitrogen < 10 mg/l, the nitrogen to be applied is (1 mgd • 10 mg/l + (2.3 – 1) mgd • 24 mg/l) • 8.34 = 427 lb/day • 365 days/yr = 156,000 lb/year. Number of lbs that can flow to groundwater is 70,000 lb/year. Acres can be reduced to (156,000 – 70,000)/400 = 215 acres. The limiting consideration is nitrogen loading resulting in 215 acres If the state allowed a 20% reduction if nitrogen as the result of percolation through the soil. The acres could be further reduced to (156,000 * (1 - .2) – 70000)/400 = 137 acres, which is about the same as the calculation based on water balance.

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14-11. Construct table converting all units to in./ac/mo.

Total Acres = (Total WW flow) / (Evaporation + Irrigation Req’d – Precipitation) Acres = (31,719) / (78 + 106 – 24) = 198 acres

Using 198 acres, the maximum depth is 7.6 ft

NETwwf wwf precip evap irrig reqd irrig reqd irrig reqd

Month mgd ac in/mo in/mo in/mo in/ac/wk in/mo in/moOctoberNovember 1.64 1841 2 2 0 0.0 0.0December 2.25 2525 3 3 0 0.0 0.0January 3.05 3423 4 2 0 0.0 -2.0February 4.5 5051 6 1.8 0 0.0 -4.2March 3.15 3536 4 2.3 0 0.0 -1.7April 2.46 2761 2 6 2.5 10.8 14.8May 2.51 2817 1 8 3 13.0 20.0June 1.87 2099 0 10 4 17.3 27.3July 1.76 1975 0 12 4 17.3 29.3August 1.75 1964 0 12 4 17.3 29.3September 1.7 1908 0 12 4 17.3 29.3October 1.62 1818 2 7 3 13.0 18.0

NET NETirrg reqd feet

Month ac in/mo ac in/mo total of waterOctober 0 0 0.0November - 1,841 1,841 0.8December - 2,525 4,366 1.8January (396) 3,819 8,186 3.4February (832) 5,883 14,068 5.9March (337) 3,872 17,941 7.6April 2,938 (177) 17,764 7.5May 3,961 (1,144) 16,619 7.0June 5,414 (3,315) 13,304 5.6July 5,810 (3,835) 9,470 4.0August 5,810 (3,846) 5,624 2.4September 5,810 (3,902) 1,722 0.7October 3,565 (1,747) (25) 0.0

to / (from) storage

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14-12. States differ as to whether golf course irrigation is considered restricted or unrestricted reuse. Unrestricted reuse requires the higher degree of treatment, but is less restrictive regarding hours of irrigation and access to the public. To be considered restricted use, public access would have to be limited by fencing and buffer area, grass should be allowed to dry and excess water soaked into the ground before allowing public access. Buffer areas would have to be appropriate for the spray system used and public use of adjacent property.

14-13. Recycled water is the product of domestic wastewater treatment. Gray water from residential sources includes close washing, bathtub, or sink water collected separately from toilet, dish water, and kitchen sink water. Gray water from industrial and food processing results from processing facilities which are separated from sanitary facilities within the industry. The difference is that gray water is not contaminated with human pathogens and may be applied to land without the disinfection requirements required by domestic wastewater.

14-14. Water quality requirements for groundwater injection of recycled water depend on the water quality and beneficial use of the groundwater. Where groundwater is has TDS > 3000 mg/l, treatment requirements must meet drinking water requirements while for TDS < 3000 mg/l, treatment requires must meet drinking water standards and include limits on TOC, and TOX. In contrast, rapid infiltration is more lax in nitrate concentrations and suspended solids concentrations. Depending on hydrogeologic and ground water use (potable water supply), reuse water must meet drinking water standards and nitrogen limits. The assumption is that additional biological and chemical treatment will take place in the vadose zone and recycled water for rapid infiltration does not have to meet the same water quality requirements as direct groundwater injection.

14-15. The NRC report states that the best available water source should be used for drinking water. In some parts of the United States where quality water sources are becoming scarce, the Committee reported that planned indirect potable reuse may be viable, but only after careful, thorough contaminant monitoring, health and safety testing, and treatment reliability issues are addressed. Issues include chemical constituents, microbial contaminants, and evaluation of potential health effects. Planned indirect reuse into a potable water supply should be considered only after implementation of water conservation, a review of other water sources, and evaluation of non-potable reuse to replace existing potable water use.

14-16. This is kind of a trick problem, first construct a table:

mgd ac ft/mo ac ft/mo ac ft/mo ac ft/momonth Supply supply demand startage accum

11 3.89 363 0 363 36312 4.31 402 0 402 7651 6.38 596 0 596 13612 7.71 720 0 720 20813 6.03 563 17.28 546 26264 5.32 497 35.9 461 30875 4.95 462 71.81 390 34776 4.13 386 107.71 278 37557 4.19 391 143.61 248 40038 4.09 382 148.4 233 42369 3.77 352 143.6 208 4445

10 3.59 335 107.71 227 4672

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The total of the supply is significantly greater than the demand. Excess supply goes to storage but must be discharged. The condition can be best visualized in a graph, The graph indicates that supply can meet demand in every month without storage. Excess water must be discharged.

14-17. Backflow protection must be used to prevent contamination of a potable water supply. Where the public water system is used to supplement the recycled supply, an air gap must be used to prevent contamination of the water supply. An air gap system requires a tank with a float operated valve where the water supply discharges above the high water level and free-falls into the tank. Pumps and a hydropneumatic tank are required to repressurize water to its point of use. Where recycled water is piped separately and there is not interconnection with the potable water system, a reduced pressure backflow preventer may be used on the water supply. Under some circumstances such as the use of recycled water for landscape irrigation as part of a dual plumbed use area, a double check assembly may be used with the approval of the utility agency and Health Services.

14-18. Tallahassee found that surface discharge of wastewater could not be assimilated by the river or downstream lake. The solution was to develop a year-round crop irrigation system that generated a cash crop. Treatment was improved to include denitrification reducing the nitrogen content and allowing an increase in hydraulic loading. Zero discharge is achieved by providing storage and growing year-round grasses.

14-19. El Dorado Hills uses recycled water for front and backyard urban irrigation. Responsibilities for the homeowner include all liability and responsibility for operation, maintenance, and all phases of construction (design, review, and inspection) for modification to the irrigation system. Piping is wrapped with purple tape or colored purple and above ground components are tagged or colored purple to designate recycled water use. Irrigation times for spray systems are limited from 9:00 pm to 6:00 am to limit the opportunity for public contact, while drip irrigation systems may be operated at any time. Dual plumbed parcels cannot be converted from recycled water to potable water for landscape irrigation. Potable water may only be used indoors, through hose bibs outside the house for swimming pools and spas.

14-20. The most common industrial reuse applications are cooling water makeup. The important issues are related to the industrial water quality needs rather than the public disease risk. Because of the food processing industry in Hawaii, even cooling water must be treated with coagulation, filtration, and chlorination with additional disinfectant for Legionella and Klebsiella.

14-21. Reuse treatment requirements for surface percolation are less restrictive than groundwater injection because of the treatment by organisms in the vadose zone. Limits are placed where the depth to groundwater is shallow and the unsaturated soil profile is limited.

Supply and Demand

0

100

200

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400

500

600

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1 2 3 4 5 6 7 8 9 10 11 12Month

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w (

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Direct injection into the groundwater requires a high degree of treatment because there will be no treatment once underground. Treatment includes removal of pathogens, heavy metals, and inorganic salts. Refer to Figure 13-9

14-22. Note, the problem should refer to Figure 1-1. Attached is the graphic that should have been included in the text and was the original reference to this problem. Surface discharge for disposal by dilution takes place upstream of other users. Where stream flows are low in the summer, wastewater may constitute a significant portion of the stream flow resulting in little dilution prior to water-based recreation or surface water supply for the next community. Discharge to a lake tends to accumulate nutrients causing algal blooms. The importance of wastewater and advanced wastewater treatment is to meet water quality requirements for surface discharge. The importance of water quality requirements is to secure the beneficial uses of lakes, rivers, and groundwaters. Recycled water reduces or eliminates the amount of water required for disposal by using water for irrigation and other reuses. Reuse increases the water quality of surface water by reducing the amount of contaminants released. Water quality requirements for reuse depend on the potential for public contact. The purpose of the water quality goals are to protect potable water supplies and safeguard public health.

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15 SUSTAINABILITY AND CARBON FOOTPRINT

15-1. Carbon footprint is the total amount of green house gas emissions caused directly or indirectly by individuals, product production, and waste treatment. Key contributors include: Scope 1 direct fugitive emissions, Scope 2 purchased electricity, and Scope 3 chemicals, materials, construction, and transport. Of these, purchased electricity has the greatest impact for water and wastewater treatment.

15-2.

Supply Treatment Delivery Energy Intensity kWh/MG

Well 14,000 100 5,000 19,100 Desalination 16,000 0 5,000 21,000 The deep well is very close to the same energy intensity as desalination. Additional information is needed to make a final decision, but energy intensity being about equal, favors desalination which is more sustainable.

15-3. The most important step in reducing the carbon footprint of a wastewater treatment plant is to make the greatest use of anaerobic digester gas. Conventional digestion and engines represent a 35% reduction in a typical plant’s carbon footprint. Further optimization and digestion enhancements can result in a 72% reduction. Resolving overdesign and other energy management initiatives provides an additional 11%. Wastewater facilities can be carbon neutral by generating additional digester gas using FOG or food waste or by generating green energy using solar or wind. Water treatment does not generate carbon free energy source like wastewater digester gas. Energy conservation, chemical optimization, waste management, and renewable energy are readily available to reduce the carbon footprint of water facilities activities. One of the key opportunities is to use recycled water to off set a portion of new water supplies and in so doing, conserve water and save the energy associated with water supply. But the distribution carbon footprint still requires a renewable energy source to be carbon neutral.

15-4. From Figure 15-2: Water supply, treatment, and distribution have an energy intensity of

100+600+130 = 830 kW/mil gal. Wastewater collection, treatment, recycled water treatment, and distribution have an energy intensity of 100+2500+1000+600 = 4200 kW/mil gal, but should be credited for the water supply that was not used so the net is 4200 – 800 = 3400 kW/mil gal. One might further argue that wastewater treatment is required for discharge and the net energy intensity of recycled water treatment and distribution is 3400 – 2500 or 900 kW/mil gal. These values are significantly less than the 8000 kW/mil gal added by the homeowner.

15-5. Water supply and distribution would have an intensity of (800+600+130) * 1500 = 2.3 mil kW

Given that wastewater treatment is required, the recycled water intensity would be (1000+600)*1500 = 2.4 mil kW, which is about the same as the new water supply. Conservation would favor recycled water. In addition, to be fair, the recycled water distribution should be the same as the water supply distribution of 130 kW/mil gal, therefore recycled water would be (100+130)*1500 = 1.7 mil kW.

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15-6.

Refer to Example 15-2, the sludge flow is the same at 413,000 lb/day. BTU to heat raw sludge = 413,000 (98-42) / 24 = 960,000 BTU/hr. BTU to maintain temperature = 964,000 / 0.8 * 0.2 = 240,000 BTU/hr for a total head demand of 1.20 mil BTU/hr. At 80% boiler efficiency the boiler demand is 1.2 / 0.8 = 1.5 mil BTU/hr. Given the 2.24 mil BTU/hr from digester gas production, an additional 0.7 mil BTU/hr is available for other head demands.

15-7. Pretreatment includes removal of hydrogen sulfide and siloxane. Engines must be designed for low NOx emissions.