Water and Wastewater Technology 7e - Solutions

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

Rapid mix Flocculation

Raw water supply

River

Filter

Settling tank

Water Distribution

Rapid mix Chlorination basin Potable water

Coagulant Auxiliary chemicals

Chlorine

Water Treatment Surfacewater Discharge

Secondary Treatment

Final settling

Preliminary and Primary Treatment Primary settling

Biological treatment

Pumping station

Grit chamber

Sewage Collection

Return activated sludge or recycle flow

Wastewater Treatment Anaerobic Zone Primary Effluent

Organic P

Anoxic Zone

N2 gas P

NO3

Mechanically mixed anaerobic chambers

NH3 NO3 P Cell Mass

N2

Mechanically mixed anoxic chambers

N03

Lake

Final Settling

Aerobic Zone

N2 gas

NO3

N2 gas

N2

Effluent

Air mixed aerobic chambers Sludge

Recirculation of NO3, QS = Q Return Activated Sludge, QR = 0.3Q

Advanced Wastewater Treatment Rapid mix Flocculation

Secondary effluent

Filter

Settling tank

Irrigation Rapid Chlorination mix basin

Coagulant Auxiliary chemicals

Tertiary effluent Chlorine

Chlorine

Recycled Water Secondary effluent

Cartridge filters

Compressed air

MicroFilters

Break tank

Cartridge filters

Highpressure pumps

Chlorine Sulfuric acid Threshold inhibitor

Water Reclamation

Reverse Osmosis Units

Ultraviolet Disinfection

Packed-tower decarbonator

Injection well

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.

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

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)

37

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.

38

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:


Velocity Max Min Average

v (ft/s) 10 3 5




 4 · · 12  · 449

q (gpm) 2500 800 1600

di (in.) 10 10 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 66).

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.

39

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.

Head, ft

Head versus Discharge 200 190 180 170 160 150 140 130 120 110 100 200

300

400

Discharge, gpm

40

500

6-20.

Pump operation is scheduled to meet the variation in diurnal water consumption. The required onoff 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.

41

6-21.

⋅

Storage volume 500,000 100 = = 18.6 percent Total consumption 2,685,000

The rule-of-thumb range is 15 to 20 percent. 6-22. Time 12 PM 1 AM 2 3 4 5 6 7 8 9 10 11 12

Consumption 0 132,000 258,000 366,000 450,000 528,000 600,000 720,000 930,000 1,230,000 1,590,000 1,974,000 2,394,000

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

Consumption 2,790,000 3,174,000 3,552,000 3,936,000 4,320,000 4,722,000 5,166,000 5,718,000 6,222,000 6,522,000 6,714,000 6,882,000

8

Storage required for 12-hr pumping = 2.76 mil gal

6

5

Storage required for 24-hr pumping = 1.47 mil gal

4

3

2

1

42

25

22

Tim e, hours

19

16

13

10

7

4

0 1

Cumulative consumption, mil gal

7

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

⋅

Storage capacity 147 . 100 = = 214 . percent Total consumption 6.882

4,722,000 − 600,000 = 5725 gpm 12 • 60 On graph, 6 A.M. = Time 7 and 6 P.M. = Time 19 Storage capacity required from graph = 2.76 mil gal

( b) Average =

Storage capacity 2.76 •100 = = 40.1 percent Total consumptio n 6.882

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

43

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, inline 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.

44

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 2.0 3.0 4.0

0.40 0.78 0.99 1.30

2 loge C0/Ct

7

1.5 1 0.5

k = 1.1/4 = 0.28 per min

0 0

1

2

3

4

5

t, m in

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=

7-6.

V0 = 0.0014

⋅

1.0 24 8,000,000 = 3.0 hr, Vo = = 640 gpd / sq ft 8.0 12,500

ft ft 2 gal sec gal / day • 2 • 7.48 3 • 86,400 = 905 sec ft day sq ft ft 10 t min = = 2.0 hr 0.014 • 3600

45

7-7.

t=

2 • 30 •15 •10 • 7.48 • 24 = 4.04 hr = 4.0 hr (OK ) 400,000

v horizontal =

400,000 = 0.25 ft / min < 0.5 (OK ) 7.48 •1440 •15 •10

Weir loading =

7-8.

t=

400,000 gal / day = 6700 < 20,000 (OK ) 60 ft

⋅ ⋅ ⋅ ⋅

2 27 5.0 3.8 24 = 4.1 h 6000

Vo =

6000 = 22 m3 / m2 ⋅ h 2 27 5.0

⋅ ⋅

weir loading = 6000/50 = 120 m3/m

d

7-9.

Overflow rate = 0.000,24 • 86,400 = 20.7 m/d 3800 Tank radius = = 7.64 m (π • 20.7 )0.5 Diameter = 15.3 m and Area = 184 m2 3.0 • 3800 Depth = = 2.6 m 184 • 24

7-10.

A=

15,000 = 938 m2 16

4.0 = 7-11.

Diameter =

⋅ ⋅

938 24 Depth 15,000

4 ⋅ 938

π

0.5

= 35 m

Depth = 2.7 m

Concentration, g/l

1.5

1.0

0.5 t

t

50

0 0

40

80

120 Time, min

t50 = 505/5.06 = 99.8 = 100 min

46

160

200

7-12.

Concentration, g/l

100

50 t

50

= 137 min +

0 1

2

3

4

5

6

Time, min

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

94,000 = 3.0 hr (OK , 1 to 2 hr ) 31,300 15,800 t (mixing and flocculation ) = = 30 min (OK, 30 min) 521 521 gpm Upflow rate = = 0.46 ( 1500 gpm, Yes. 8-5.

Using Eq. 8-2,

QR = 800 8-6.

45 − 20 45 − 27

0.50

= 940 gpm

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.

63

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 86).

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.

64

9

WASTEWATER FLOWS AND CHARACTERISTICS 9-1. Wastewater flow and loading calc

Resturant Employees Guests Hotel Employees Guests Residential Homes Apartments

No. per day GPD/Unit

Subtotal GPD

BOD lb/day /Unit

30 200

30 8

900 gpd 1,600 gpd

0.10 0.06

3.0 lb/day 12.0 lb/day

5 40

50 50

250 gpd 2,000 gpd

0.10 0.10

0.5 lb/day 4.0 lb/day

15 20

80 75

1,200 gpd 1,500 gpd 7,450 gpd

0.17 0.17

2.6 lb/day 3.4 lb/day 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 values that are reasonable. Selecting average versus maximum values are always a judgement decision that should be based in part on other conservative values selected. Selecting conservative flow and loading values followed by conservative sizing criteria will exaggerate the results. In this solution, the typical home was selected, while the highest value for apartments was selected to represent the best judement given a lack of actual field data. From Table 9-1 each meal served is 4 gpcd, serving lunch and dinner results in a total of 8 gpcd and 0.06 lb BOD/person/day 9-2.

(240 − 120)100 = 50% 240 (200 − 130)100 = 35% BOD = 200 Suspended solids

Suspended solids

BOD =

9-3.

Subtotal BOD

( 240 − 30)100 = 88% 240

( 200 − 30)100 = 85% 200

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

65

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 acres 3.4•1500 = 5100 gpd Infiltration using pipe

8-in. 10-in. 14-in.

gal/day /in.-mile 500 500 500

Pipe infiltration (gpd) Length (ft) 1200 910 800 760 430 570 2240

Number of lots = 3.4•3 = 10.2 Using 4 people per house = 10.2•4•120 gpcd = 5,000 gpd Dry weather flow = 5,000 gpd Wet Weather flow = 5,000 + 5,100 = 10,100 gpd or = 5,000 + 2,240 = 7,200 gpd 66

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 gpcd 3,500,000 gpd from population

Peak flow from population (Eq. 9-3) PF = 2.08 7,296,560 gpd from population Max Infiltration based on acres 78000 acres 1500 gal/day/acre 117,000,000 gpd from infiltration OR

10% of average flow 350,000

Max Inflow based on population 5 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.

67

9-13. From Table 9-4 Potato chip Milk Soft-drink

Flow 90,400 gpd 65,100 gpd 16,000 gpd 172,000 gpd

BOD 450 lb/day 760 lb/day 64 lb/day 1,300 lb/day

SS 510 lb/day 170 lb/day 64 lb/day 700 lb/day

200 mg/l

240 mg/l

For a sewered population of 12,500 Table 9-2: 120 gpcd Population 1,500,000 gpd

BOD concentration = 1300 lb/day / 1.5 mgd / 8.34 = 2,500 lb/day SS concentration = 700 lb/day / 1.5 mgd / 8.34 = 3,000 lb/day Industrial Population Total Concentration

172,000 gpd 1,500,000 gpd 1,672,000 gpd

1,300 lb/day 2,500 lb/day 3,800 lb/day 270 mg/l

700 lb/day 3,000 lb/day 3,700 lb/day 270 mg/l

9-14. 8 AM 9 AM 10 AM 11 AM 12 AM 1 PM 2 PM 3 PM 4 PM 5 PM 6 PM 7 PM 8 PM 9 PM 10 PM 11 PM 12 PM 1 AM 2 AM 3 AM 4 AM 5 AM 6 AM 7 AM

Flow (mgd) Volume (ml) 17.5 145 21.8 180 24.7 204 26.9 223 25.5 211 23.3 193 20.4 169 17.5 145 16 132 16 132 16.7 138 17.5 145 16.7 138 16 132 14.6 121 13.1 108 10.2 84 7.3 60 5.8 48 5.5 46 5.5 46 5.8 48 7.3 60 11.6 96 3004 ml

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

68

9-15.

6 AM 9 AM 12 AM 15 AM 18 AM 21 PM 24 PM 3 PM Total Average

3-hr Flow Average Composite (mgd) Flow (mgd) Multiplier (ml) 0.97 0.323 * 460 = 150 2.91 0.970 * 460 = 450 3.4 1.133 * 460 = 520 2.33 0.777 * 460 = 360 2.23 0.743 * 460 = 340 2.14 0.713 * 460 = 330 1.55 0.517 * 460 = 240 0.74 0.247 * 460 = 110 16.27 = 2,500 ml 0.68 0.68

total volume of composite sample desired Multiplier = -----------------------------------------------------------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.

69

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 Homes 520 Unit flow (gpd) 750 Total flow = Homes * Unit flow Total flow (gpd) 390,000 Accumulated Total flow (gpm) 270 From Table 10-1 Sewer dia. 8 Capacity (gpm) 310

Access road 180 750

Additonal homes 440 750

135,000

330,000

360

590

10 490

12 700

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 MGD PF = 2.5 / (8^0.145) = 1.85 Peak flow = 14.8 MGD Minimum flow = 4 MGD Size one pipe for minimum flow and one pipe for the net peak flow Area at 3 ft/s Calculated pipe cu ft/sec (sq ft) dia. (in.) Net Peak flow = 10.8 MGD 23 5.6 64 Minimum flow = 4 MGD 6.2 2.1 39 Small pipe, use 40" pipe Large pipe, use 64" pipe

70

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.

71

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 Sample:

Dry Wt = 9.65 lb or 0.1001157 cu ft = Optimal density is 12% at

173 cu in

96.4 Dry lb/cu ft 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 lb Field 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 Sample:

Dry Wt = 11.2 lb or 0.0960648 cu ft = Optimal density is 8% at

166 cu in

116.6 Dry lb/cu ft 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 lb Field 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

72

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.

73

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, 74

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 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 Dyn + Static (gpm) (ft) 500 17.0 600 17.8 700 18.7 800 19.8 900 21.0 1000 22.3 1100 23.7 1200 25.2 1300 26.8 1400 28.6 1500 30.5 1600 32.4

Dynamic Hl (ft) 2.0 2.8 3.7 4.8 6.0 7.3 8.7 10.2 11.8 13.6 15.5 17.4

The graph, with the pump data points, is constructed as follows: 35.0 33.0 31.0 29.0 Head (ft)

11-13.

Pump curve

27.0 25.0 23.0 21.0 19.0

Operating point

Head curve

17.0 15.0 400

600

800

1000 Flow (gpm)

75

1200

1400

1600

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

z1 30.0 30.8 32.7 35.8 40.0 45.1 51.2 58.3 66.3 75.3 85.1 95.8 107.4

z2 20.0 20.8 22.7 25.8 30.0 35.1 41.2 48.3 56.3 65.3 75.1 85.8 97.4

Dym C=100 0.0 0.8 2.7 5.8 10.0 15.1 21.2 28.3 36.3 45.3 55.1 65.8 77.4

Flow (gpm) 0 100 200 300 400 500 600 700 800 900 1000 1100 1200

z1 20.0 20.4 21.5 23.1 25.3 28.1 31.3 35.1 39.4 44.2 49.4 55.1 61.3

z2 30.0 30.4 31.5 33.1 35.3 38.1 41.3 45.1 49.4 54.2 59.4 65.1 71.3

Dym C=140 0.0 0.4 1.5 3.1 5.3 8.1 11.3 15.1 19.4 24.2 29.4 35.1 41.3

120 100 Pump curve

80 60

Dynamic losses at C=100

40

Dynamic losses at C=140

20

Operating points at intersection of pump and headloss curves

0 0

100

200

300

400

500

600

700

800 900 1000 1100 1200 1300

1400

80

1200

70

1000

60 50

800

40

600

30

400

20

200

10

0

Pump Efficiency (Percent)

Pump Capacity (gpm)

11-15.

0 0 4 8 12 16 20 24 28 32 36 40 44 48 Influent Water Level (in.)

| 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 76

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 1.2 MGD

SS 230 mg/l

PRIMARY CLARIFIER Number 1 Diameter 65 Depth 7

Area

3,318 EPA 10-13, GLUMRB 7, EPA with secondary solids 13-16

Overflow Vo

362 gpd/sf

Det Time Weir loading

3.5 hrs 5,876 gpd/lf

without chemical addition removal SS in 2,302 out 691 under 1,611 With chemical addition removal SS in 2,302 out 575 under 1,726

EPA 800-1200 avg 2000-3000 peak, GLURMB 1000 avg, 1500 peak EPA 600-800 avg 12000-1500 peak with secondary solids using 11-4 calc EPA 10k to 40k, GLUMRB 10k

70 % soilds removal lbs lbs lbs 75 % soilds concentration

lbs lbs lbs

removal in out under

SS 2,302 lbs 230 lbs 2,072 lbs

90 % soilds concentration

11-19. Overflow Vo Overflow Vo

11-20.

Avg 1,000 700

Peak 2,500 1,350

gpd/sf gpd/sf

EPA 800-1200 avg 2000-3000 peak, GLURMB 1000 avg, 1500 peak EPA 600-800 avg 1200-1500 peak with Secondary Solids

PRIMARY CLARIFIER Without Secondary Solids Average 1 Diameter 56 ft Depth 11.5 ft

Peak 1 Diameter 53 ft EPA 10-13, GLUMRB 7, EPA with secondary solids 13-16

With Secondary Solids Area 7,857 sq ft Diameter 100 ft Depth 11.5 ft

Area 4,074 sq ft Diameter 72 ft EPA 10-13, GLUMRB 7, EPA with secondary solids 13-16

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 77

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 20003000 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 75% 90%

BOD Removal 52.5% 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.

78

11-25. Primary Clarifier Effluent Q 0.8 MGD

SS 280 mg/l

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

BOD 450 mg/l

sBOD 250 mg/l

0.54

At T=20

− k 20 ϑ

T − 20

A s D /[ Q

p

(1 + R )] n

-1.932

− k 20 ϑ

T − 20

A s D /[ Q

p

(1 + R )] n

-1.804

At T=18

Se e−1.932 = S p (1+1) −1⋅ e−1.932

Soluable effluent Total effluent BOD BOD 20 mg/l 36 mg/l

0.078

Se e − 1.804 = Sp (1 + 1) − 1 ⋅ e −1.804

0.090

22 mg/l

42 mg/l

11-26.

So Se Q Y Qc X kd

120 2 0.3 0.8 20 5000 0.04 V=

mg/l mg/l mgd lb VSS/lb BOD days mg/l 1/day

0.06 mil gal or

No safety factor for membranes:

Px =

131 lb/day

Aeration Volume = Detention time = V/Q*24 = FM = So/dt/X

cu ft cu ft Aeration volume

(note, multiply by 8.34 lb/day/mgd for english units) 8,000 cu ft = 4.8 hours

0.120 soluable BOD loading 0.221 total BOD loading

79

0.060 mg

Temp 18

°C

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 12 MGD

SS 212 mg/l

PRIMARY CLARIFIER SS Effluent 10,608 ACTIVATED SLUDGE Number 10 L Dimensions 40 Volume 600,000 MLSS 2,350 Effluent SS 20

BOD 190 mg/l BOD 14,261 lbs

W

D 100

15.0 4.488 mil gal

cf mg/l mg/l

Liquid Detention Time 9.0 hrs BOD loading 24 lb/1000 cu ft/day F/M 0.16 Waste sludge = 5330 lb/day Sludge 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 Q SS BOD 18 212 190 MGD mg/l mg/l

BEFORE CHEMICAL ADDITION Q SS BOD 18 212 190 MGD mg/l mg/l

PRIMARY CLARIFIER 50% SS removal 25% BOD Removal SS in 31,825 Effluent 15,913 under 15,913

PRIMARY CLARIFIER 75% SS removal 38% BOD Removal SS in 31,825 Effluent 7,956 under 23,869

ACTIVATED SLUDGE Number 10 L Dimensions 40 Volume 600,000 MLSS 2,350 Effluent SS 20

BOD 28,523 lbs 21,392 lbs 7,131 lbs

W 100 cf mg/l mg/l

D 15.0 4.5 MG

ACTIVATED SLUDGE Number 10 L Dimensions 40 Volume 600,000 MLSS 2,350 Effluent SS 20

BOD 28,523 lbs 17,684 lbs 10,839 lbs

W

D 100

cf mg/l mg/l

15.0 4.5 MG

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

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

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

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

80

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.

81

11-34.

From the text, the minimum aeration period is 24 hours and the clarifier overflow rate is 600 gpd/sq ft. Trailer Park Occupancy Wastewater Flow

140 2 80 22,400

Q 22,400 MGD

units per unit gpcd gpd

SS 190 mg/l

BOD 170 mg/l

ACTIVATED SLUDGE Detention time = 24 hours Volume of aeration basin = 22,400 / 24 * 24 / 7.48 = 2990 BOD loading = 11 lb/1000 cu ft F/M = 170 * 0.0224 / 1000 / 0.0224 = 0.170 Vo = 600 gpd/sf Area = 37 sf Dia =

Table 11-4, 20 to 30 cu ft Table 11-4, loading of 10 to 20 Table 11-4, F/M 0.05 to 0.2 7 ft

11-35. Flow BODi BODe NH4

11.5 200 10 35

MGD mg/l mg/l mg/l

Oxygen demand = 11.5(200+4.6*35-10)8.34 = Temp 16 °C

33,664

lbs/day

Brush Aerators Number

2 ditches 0.5

F

4 aerators each

0.9 4 mg/l

DO residual

From Table 11-5, SOTR O2 lb/Hp-hr Cs Cinf = Cs Ct

10 mg/l, see Table 2-5 4 mg/l

β ⋅ C s − Ct

OTR = SOTR ⋅ α F ⋅ θ t − 20 ⋅ OTR Hp

2.75

C∞

0.61 lb/hp-hr 104 for each aerator

11-36. Flow BODi BODe NH4

21.5 180 10 31

Oxygen demand = Oxygen demand = Temp Temp

MGD mg/l mg/l mg/l 30,483 56,052

14.8 °C 18.5 °C

lbs/day lbs/day

without nitrification with nitrification

winter summer

Fine Bubble Aerators Depth

F DO residual SOTR Cs

15 feet 0.75 text for equation 11-17 0.925 2 5.85 10.2

text for equation 11-18 mg/l lb/hp-hr Average of values in Table 11-5 mg/l, see Table 2-5 14.8 °C

82

Cs Cs Cs

9.1 mg/l, see Table 2-5 10.1 mg/l, see Table 2-5 9.4 mg/l, see Table 2-5

Depth adjusted 10.7 mg/l 11.9 mg/l 11.1 mg/l

20 °C 14.8 °C 18.5 °C -7

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

OTR = SOTR

1.2

β ⋅ CT , P ,de − C L C 20 ,1,de

OTR at 14.8 OTR at 18.5

3.1 lb/hp-hr 3.2 lb/hp-hr

(θ

T − 20

)(αF )

HP at 14.8 HP at 18.5

QO 2 OTE ⋅ 0.0174 ⋅ 1440 QAir = 2,308 cfm

winter

QAir =

summer

412 hp 729 hp

Q Air =

4,082 cfm

11-37. Flow BODi BODe

6.5 MGD 143 mg/l 30 mg/l

Oxygen demand = Temp

16 °C

(220 * (1-0.35))

6,126 lbs/day winter

without nitrification

Fine Bubble Aerators Depth

15 feet 0.65 Example 11-10

F

0.75 Example 11-10

DO residual SOTR

0.95 Example 11-10 2 mg/l Example 11-10 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 9.9 mg/l, see Table 2-5

Cs

20 °C 16 °C

10.7 mg/l 11.7 mg/l

-7

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

OTR = SOTR

β ⋅ CT ,P ,de − C L C 20,1,de

OTR at 14.8 Q Air =

11-38.

1.2 mg/l

for a site elevation = 0

(θ )(αF ) T −20

2.0 lb/hp-hr

HP =

130

QO 2 OTE ⋅ 0 .0174 ⋅ 1440

QO2 =

255 lb/hr or

QAir =

244 cu ft/min

6,126 lb/day

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

⋅

12 1730 = 23 d 0.30 1730 + (41.2 − 0.30)9.4

⋅

1 = 0.043 d -1 θc

Using Equations 11-24 and 11-25, rsu = (126 - 5.2)/(12/41.2) = 414 mg/l•d

1 1 = = 4.2 d U 0.24

U=

414 = 0.24 d -1 1730

1 1 = = 019 . ( mg / l) -1 S e 5.2

83

Calculated values for all test runs are as follows: θc (d) 23 19 9.2 8.3

rsu (mg/l•d) 414 407 396 392

U (d-1) 0.24 0.27 0.41 0.46

1/U (d) 4.2 3.7 2.4 2.2

1/θc (d-1) 0.043 0.053 0.11 0.12

1/Se [(mg/l)-1] 0.19 0.14 0.095 0.087

0.14 0.12

1/Qc, day -1

0.1 0.08 0.06 0.04 0.02 0 0.2

0.25

0.3

0.35

0.4

0.45

0.5

U, day

0.2 0.18 0.16 1/U, 1/day

0.14 0.12 0.1 0.08 0.06 0.04 0.02 0 2

2.25

2.5

2.75

3

3.25

3.5

3.75

4

4.25

4.5

1/Se, 1/mg/l

kd = 0.04d-1 KS = 22.5 • 4.0 = 90 mg/l BOD

Y = 0.35 mg VSS/mg/BOD k = 1/0.25 = 4.0 d-1

11-39.

So Se Q Y Qc X kd

200 7 3 0.6 8 2500 0.06

mg/l mg/l mgd lb VSS/lb BOD days mg/l 1/day

V= 0.75 mil gal or Using a safety factor of 2x =

Px =

11-40.

100,000 200,000

cu ft cu ft

235 g/day

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 84

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 0.8 ⋅ 11.3 − 2 1.0210−20 (0.9 ) = 2440 lb / day deg C = 7.6 mg/l. At 10 deg C R = 4320 9.2 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 0.8 ⋅11.3 − 2 2.0 mg/l and T = 10 deg C, R = Ro 1.0210−20 (0.9) = 0.56 9 .2 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 85

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 20 min. SS = 21; MPN 100 Filtration (no chemical addition), disinfection > 20 min. SS = 30; MPN 200 Secondary treatment, no filtration, disinfection > 20 min.

103

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. Alum Dose to Phosphorus 13:1 16:1 22:1

Expected P Removal 75% 85% 95%

Alum Dose (mg/l) based on 7 mg/l influent 91 112 154

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 104

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 105

106

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 107

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) 0 58 117 175 325

Influent Phosphorous (mg/l) 10.3 10.9 10.6 10.4 9.3

Al/P Weight Ratio

Phosphorous Removal (%)

0 0.55 1.14 1.74 3.61

23 50 77 90 95

BOD Removal (%) 96 96 97 94 95

SS Removal (%) 93 92 90 93 86

Data from text under Equation 13-2. 1.34 1.65 2.27

75 85 95

Phosphorous removal, percent

4.0 3.0 2.0 1.0 0.0 1.0

2.0

3.0

4.0

5.0

-1.0 -2.0 -3.0

Al/P

(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 132). 108

13-28.

Based on the data, the following plots may be constructed:

Concentration (mg/l)

Concentration (mg/l)

Distance Ammonia Alkalinity ln(NH4d/NH4o)/MLVSS/(1.024^(18.5-20))/(1800/133670)/D 0 16.11 243 6 16.46 185 (0.00013) Alkalinity Concentration vs. Distance from Inlet 12 12 190 0.00120 300 18 8 190 0.00189 24 5.28 188 0.00225 250 30 4.48 189 0.00206 200 36 3.96 188 0.00188 150 42 4.55 191 0.00145 48 2.93 181 0.00171 100 54 2.64 178 0.00162 50 60 2.62 178 0.00146 0 66 2.59 178 0.00134 0 12 24 36 48 60 72 84 96 108 120 132 144 156 168 72 2.44 178 0.00126 78 2.31 179 0.00120 Distance 84 2.44 178 0.00108 90 2.44 180 0.00101 96 1.69 178 0.00113 102 2.01 175 0.00098 Ammonia vs. Distance 108 1.44 172 0.00108 20 114 0.87 172 0.00123 120 1.05 171 0.00110 15 126 1.25 172 0.00098 132 0.6 177 0.00120 10 138 0.8 173 0.00105 144 0.53 173 0.00114 5 150 0.42 174 0.00117 156 0.4 172 0.00114 0 162 0.59 169 0.00098 0 12 24 36 48 60 72 84 96 108 120 132 144 156 168 168 0.55 170 0.00097 Distance

ln(Nd/No)/MLVSS/temp/t

k versus distance 0.00250 0.00200 0.00150 0.00100 0.00050 0

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

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

Anaerobic Anoxic Aerobic

Detention Time 3 hours 6.75 hours 6 hours

Methanol and acetic acid addition may be required.

109

Tank Volume Mil Gal Cubic Feet 14.4 32.4 28.8

1,925,134 4,331,551 3,850,267

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 110

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 pretreatment program.

111

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