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Title307991349-Wastewater-Engineering-Treatment-5th-Edition-Solutions-Manual.pdf
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Table of Contents
                            aME5 SM Title Page (2).pdf
	SOLUTIONS MANUAL
		SOLUTIONS MANUAL
	Wastewater Engineering:
		Wastewater Engineering:
			Wastewater Engineering:
				Wastewater Engineering:
aME5 Contents   (2)
aME5 SM Preface (2)
	PREFACE
M&E5 SM Chap. 01 FINAL REVgt
	INTRODUCTION TO
		INTRODUCTION TO
			INTRODUCTION TO
				INTRODUCTION TO
				WASTEWATER TREATMENT
	PROBLEM1-1
	Solution
	PROBLEM1-2
	Solution
	PROBLEM1-3
	Solution
	PROBLEM1-4
	Solution
	PROBLEM1-5
	Solution
	PROBLEM1-6
	Solution
	PROBLEM1-7
	Solution: Graphical Approach
	Solution: Mathematical Approach
	PROBLEM1-8
	Solution
	PROBLEM1-9
	Solution
	PROBLEM1-10
	Solution
	PROBLEM1-11
	Solution
	PROBLEM1-12
	Solution
	PROBLEM1-13
	Solution
	PROBLEM1-14
	Solution
	PROBLEM1-15
	Solution
	PROBLEM1-16
	Solution
	PROBLEM1-17
	Solution
	PROBLEM1-18
	Solution
	PROBLEM1-19
	Solution
	PROBLEM1-20
	Solution
	PROBLEM1-21
	Solution
	PROBLEM1-22
	Solution
	PROBLEM1-23
	Solution
	PROBLEM1-24
	Solution: Part 1 (r = -kC
	Solution: Part 2 (r = - kC)
	Solution: Part 3 (r = -k)
	PROBLEM1-25
	Solution
	PROBLEM1-26
	Solution
	PROBLEM1-27
	Solution
M&E5 SM Chap. 02 FINAL REV
	PROBLEM2-1
	Solution
	PROBLEM2-2
	Solution
	PROBLEM2-3
	Solution
	PROBLEM2-4
	Solution
	PROBLEM2-5
	Solution
	PROBLEM2-6
	Solution
	PROBLEM2-7
	Solution
	PROBLEM2-8
	Solution
	PROBLEM2-9
	Solution
	PROBLEM2-10
	Solution
	PROBLEM2-11
	Solution
	PROBLEM2-12
	Solution
	1Rearrange Eq. (2-40) to solve for the hydrogen ion concentration
	PROBLEM2-13
	Solution
	1Compare the saturation concentrations of O
	PROBLEM2-14
	Solution
	PROBLEM2-15
	Solution
	PROBLEM2-16
	Solution
	PROBLEM2-17
	Solution
	PROBLEM2-18
	Solution
	PROBLEM2-19
	Solution
	PROBLEM2-20
	Solution
	PROBLEM2-21
	Solution
	PROBLEM2-22
	Solution
	PROBLEM2-23
	Solution
	PROBLEM2-24
	Solution
	PROBLEM2-25
	Solution
	PROBLEM2-26
	Solution
		Comment
	PROBLEM2-27
	Solution
	PROBLEM2-28
	Solution
	PROBLEM2-29
	Solution
	PROBLEM2-30
	Solution
	PROBLEM2-31
	Solution
	PROBLEM2-32
	Solution
	PROBLEM2-33
	Solution
	PROBLEM2-34
	Solution
	PROBLEM2-35
	Solution
	PROBLEM2-36
	Solution
	PROBLEM2-37
	Solution
	PROBLEM2-38
	Solution
	PROBLEM2-39
	Solution
	PROBLEM2-40
	Solution
M&E5 SM Chap. 03 FINAL REV
	PROBLEM 3-2
		PROBLEM 3-2
			PROBLEM 3-2
	Solution
		Solution
			Solution
				PROBLEM3-3
				PROBLEM3-4
				PROBLEM3-5
		Totals
			Flow weighted values
				PROBLEM3-6
	Solution
	PROBLEM3-7
	Solution
		Solution
			Solution
				PROBLEM3-8
				PROBLEM3-9
	PROBLEM3-10
	PROBLEMS3-11 to 3-12
		PROBLEMS3-11 to 3-12
			PROBLEMS3-11 to 3-12
				PROBLEM3-13
	Solution
		Solution
			Solution
				PROBLEM3-14
				PROBLEM3-15
	PROBLEM3-16
	Solution for City 1
M&E5 SM Chap. 04 FINAL REV
	PROBLEM4-1
	Instructional Guidelines
	PROBLEM4-2
	Instructional Guidelines
	PROBLEM4-3
	Solution
	PROBLEM4-4
	PROBLEM4-5
	Solution
	PROBLEM4-6
	PROBLEM4-7
	Solution
	PROBLEM4-8
	Solution
	PROBLEM4-9
	Solution
	PROBLEM4-10
	Solution
	PROBLEM4-11
	Solution
	PROBLEM4-12
	Solution
	PROBLEM4-13
	Solution
	PROBLEM4-14
M&E5 SM Chap. 05.FINAL REV
	PROBLEM5-2
		PROBLEM5-2
			PROBLEM5-2
				PROBLEM5-3
				PROBLEM5-4
				PROBLEM5-5
				PROBLEM5-6
	PROBLEM 5-7
		PROBLEM 5-7
			PROBLEM 5-7
				PROBLEM 5-8
				PROBLEM 5-11
	Solution
	Solution
		Solution
			PROBLEM5-13
			PROBLEM5-14
	Solution
		Solution
			Solution
				Solution
	Solution
	Solution
		Solution
			Solution
				PROBLEM 5-19
				PROBLEM 5-20
				PROBLEM 5-21
	PROBLEM 5-22
		PROBLEM 5-22
			PROBLEM 5-22
				PROBLEM 5-23
	PROBLEM 5-24
	Solution
	PROBLEM 5-25
	PROBLEM 5-26
		References
			References
				PROBLEM 5-27
	Instructors Note:  There are many possible solutions to this problem; a range of typical values is presented below.  The student should be advised that other reference sources would have to be consulted, as some of the required information cannot be found in this text.
	PROBLEM 5-28
	Instructors Note:  The detailed solution is provided for mixed liquor.  Values calculated for settled activated sludge and primary sludge with activated sludge are summarized in the table.
		Solution
	PROBLEM 5-29
		Solution
			1.Determine K
			2.Plot dC/dt versus C
				The slope is equal to -KLa, so
	PROBLEM 5-30
		Solution
			Solution
				The maximum rate of oxygen transfer, then, is
	PROBLEM 5-31
		Solution
		Solution
	PROBLEM 5-33
		PROBLEM 5-33
			PROBLEM 5-33
				PROBLEM 5-33
M&E5 SM Chap. 06 FINAL REV
	PROBLEM6-1
	Solution
	PROBLEM6-2
	Solution
	PROBLEM6-3
	Solution – Part a
	Solution – Part b
	PROBLEM6-4
	Solution – Part a
	Solution – Part b
	Solution – Part c
	PROBLEM6-5
	PROBLEM6-6
	PROBLEM6-7
	PROBLEM6-8
	Solution
	PROBLEM6-9
	Solution
	PROBLEM6-10
	Solution
	PROBLEM6-11
	Solution
	PROBLEM6-12
	Solution
	PROBLEM6-13
	Solution
	PROBLEM6-14
	Solution
	PROBLEM6-15
	Solution
	PROBLEM6-16
	Solution
	PROBLEM6-17
	Solution
	PROBLEM6-18
	Solution
	PROBLEM6-19
	Solution
M&E5 SM Chap. 07 FINAL REV
	Instructors Note:  In many of the problems where constituent concentrations are used, the units mg/L and g/m3 are used interchangeably to facilitate computations without introducing additional conversion factors.
	PROBLEM 7-1
	Solution
	Solution
	PROBLEM 7-5
	Solution
	Solution
	Solution
		Solution
			Solution
				PROBLEM 7-10
	Solution
		Solution
			Solution
				PROBLEM 7-11
	Solution
		Organic compound
	PROBLEM 7-13
	Solution
	PROBLEM 7-14
	Solution for diameter = 1 µm
	PROBLEM 7-15
	PROBLEM 7-34
	Solution
	PROBLEM 7-35
	Solution
	PROBLEM 7-36
	Solution
	PROBLEM 7-37
	Solution (100 mg/L acetate)
	PROBLEM 7-38
		PROBLEM 7-38
			Solution
	PROBLEM 7-40
	Solution (Influent COD = 2000 mg/L)
	PROBLEM 7-42
	Solution
	PROBLEM 7-43
	Solution
	PROBLEM 7-44
	Solution
M&E5 SM Chap. 08 ALL FINAL
	Instructors Note:  In many of the problems where constituent concentrations are used, the units mg/L and g/m
	PROBLEM 8-3
	PROBLEM 8-8
	Solution
	PROBLEM 8-9
	Solution
	PROBLEM 8-10
	Solution – for wastewater #1
	PROBLEM 8-11
	Solution
	PROBLEM 8-25
M&E5 SM Chap. 09 FINAL
	Instructors Note:  In many of the problems where constituent concentrations are used, the units mg/L and g/m3 are used interchangeably to facilitate computations without introducing additional conversion factors.In the first print of the textbook, Eq. (9-15) should to be corrected to
	PROBLEM 9-1
	Solution
	PROBLEM 9-2
	Solution
	PROBLEM 9-3
	Solution (4 m packing depth)
	PROBLEM 9-4
	Solution (for test data collected at 12oC)
	PROBLEM 9-5
	Solution
	PROBLEM 9-6
	Solution (Use the NRC equation for rock trickling filters that can be found in the 4th edition of the Metcalf and Eddy Wastewater Engineering textbook or other references)
	PROBLEM 9-7
	Solution
	PROBLEM 9-8
	Solution
	PROBLEM 9-9
	Solution
	PROBLEM 9-10
	Solution
	PROBLEM 9-11
	Solution (Influent flowrate is 37,000 m3/d)
	PROBLEM 9-12
	Solution - Part A, 40 percent BOD removal in trickling filter
	Solution - Part B, 80 percent BOD removal in trickling filter
	PROBLEM 9-13
	Solution for Winter Condition (Temperature = 5°C and activated sludge SRT = 15 days)
	Solution for Summer Condition (Temperature = 26°C and activated sludge SRT = 5 days)
	PROBLEM 9-16
	Solution
	PROBLEM 9-17
	PROBLEM 9-18
	PROBLEM 9-19
	PROBLEM 9-20
	PROBLEM 9-21
	Solution (Effluent NO3-N concentration = 2.0 mg/L)
	PROBLEM 9-22
	Solution
M&E5 SM Chap. 10 FINAL
	ANAEROBIC SUSPENDED AND ATTACHED
		ANAEROBIC SUSPENDED AND ATTACHED
	PROBLEM10-3
	Solution (flowrate of 1000 m3/d)
	PROBLEM10-4
	Solution
	PROBLEM10-5
	Solution
	PROBLEM10-6
	PROBLEM10-7
	PROBLEM10-8
	PROBLEM10-9
	PROBLEM10-10
	PROBLEM10-11
		PROBLEM10-11
			Problem Analysis
	PROBLEM10-12
	PROBLEM10-13
M&E5 SM Chap. 11 FINAL REV
	REMOVAL OF RESIDUAL CONSTITUENTS
		REMOVAL OF RESIDUAL CONSTITUENTS
	Solution
	Part a
		Part a
			Modifying Stock Sand To Produce Desired Filter Sand
				Part c
	PROBLEM11-2
	Solution
	PROBLEM11-3
	Solution
	PROBLEM11-4
	Solution
	PROBLEM11-5
	Solution
	PROBLEM11-6
	Solution
	PROBLEM11-7
	Solution
	PROBLEM11-8
	Solution
	PROBLEM11-9
	Solution
	PROBLEM11-10
	PROBLEM11-11
	Solution
	PROBLEM11-12
	Problem Statement - See text, page 1281
	PROBLEM11-13
	PROBLEM11-14
	Solution
	PROBLEM11-15
	Solution
	PROBLEM11-16
	Solution
	PROBLEM11-17
	Solution
	PROBLEM11-18
	Solution
	PROBLEM11-19
	Solution
	PROBLEM11-20
	PROBLEM11-21
	Solution
	PROBLEM11-22
	Solution
	1.Review the three artciles related to disposal of nanofiltration, reverse osmosis, and elecetrodialysis.
	2.Prepare a table to discuss the type of process combinations that are being proposed accoring to the article review.
	Processes used to deal with brine from nanofiltration,  reverse osmosis and electrodialysis in no special order.
	PROBLEM11-23 - See text, page 1284
	Solution
	PROBLEM11-24
	Solution
	PROBLEM11-25
	PROBLEM11-26
	Solution
	PROBLEM11-27
	Solution
	PROBLEM11-28
	PROBLEM11-29
	PROBLEM11-30
	Solution – Part A
	PROBLEM11-31
	PROBLEM11-32
	Solution
	PROBLEM11-33
	PROBLEM11-34
	PROBLEM11-35
	PROBLEM11-36
M&E5 SM Chap. 12 FINAL REV
	PROBLEM12-1
	Solution
	PROBLEM12-2
	Solution
	PROBLEM12-3
	Solution
	Part a
	Part b
	PROBLEM12-5
	Solution
	PROBLEM12-6
	Solution
	PROBLEM12-7
	Solution
	PROBLEM12-8
	Solution
	PROBLEM12-9
	Solution
	PROBLEM12-10
	Solution
	PROBLEM12-11
	Solution
	PROBLEM12-12
	Solution
	PROBLEM12-13
	Instructors Note: Assume the contact time in each reactor is 3 min and the t10/t ratio is 0.6.
	Solution
	PROBLEM12-14
	PROBLEM12-15
	PROBLEM12-16
	PROBLEM12-17
	PROBLEM12-21
	PROBLEM12-22
	PROBLEM12-23
	PROBLEM12-25
	Instructor Note:  Cleaning is a major issue in the application of low-pressure low-intensity and low-pressure high-intensity UV disinfection systems.  In low-pressure low-intensity lamp systems, the UV lamps are cleaned externally from the UV reactor whereas in low-pressure high-intensity cleaning of the lamps are cleaned in place with an integral wiper system.
	This problem may not be feasible unless the date is set back to 1995.  Little comparative work on the difference between low-pressure low-intensity and low-pressure high-intensity UV lamps has been published in the last 5-10 years.  Alternatively, the students could be asked to review the City of Ames and/or the URS Corporation et al. reports, given below, and comment briefly on how the UV disinfection option was assessed and/or faired against other disinfection technologies.
	PROBLEM12-26
	PROBLEM12-27
M&E5 SM Chap. 13 FINAL REV
	Solution
	Solution
		Solution
			Solution
				Solution
	Solution
		Solution
			Solution
				Solution
	Problem Analysis and/or Resolution
		Problem Analysis and/or Resolution
			Problem Analysis and/or Resolution
				Problem Analysis and/or Resolution
M&E5 SM Chap. 14 FINAL REV
	PROBLEM 14-1
		PROBLEM 14-1
			PROBLEM 14-1
				PROBLEM 14-1
	PROBLEM 14-4
	PROBLEM 14-5
	Solution
		Solution
			Solution
				Solution
	Instructors Note:  This problem is intended to have the student consider the various factors, economic as well as non-economic, in developing alternative solutions and recommendations for a practical application.  In making a recommendation in actual practice, a number of conditions will have to be considered that are site specific and include cost and environmental factors.  Some of those conditions are cited under the solution below.  The instructor may want to add other factors or constraints to the problem.  Because a community size of 200,000 people was specified, the treatment plant is of medium size and a reasonable level of operating skills can be assumed in the analysis.
	Solution – for Data Set 1
		Solution – for Data Set 1
			Solution – for Data Set 1
				Solution – for Data Set 1
	Solution – for Parameter series A
	Solution
		Solution
			Solution
				Solution
	Solution
		Solution
			Advantages
				Disadvantages
				Liquid biosolids transport
				Dewatered biosolids transport
	PROBLEM 14-13
		PROBLEM 14-13
			PROBLEM 14-13
				PROBLEM 14-13
	PROBLEM 14-14
M&E5 SM Chap. 15 FNAL REV
M&E5 SM Chap. 16 FINAL
	AIR EMISSIONS FROM WASTEWATERTREATMENT FACILITIES ANDTHEIR CONTROL
		AIR EMISSIONS FROM WASTEWATERTREATMENT FACILITIES ANDTHEIR CONTROL
			AIR EMISSIONS FROM WASTEWATERTREATMENT FACILITIES ANDTHEIR CONTROL
				AIR EMISSIONS FROM WASTEWATERTREATMENT FACILITIES ANDTHEIR CONTROL
	PROBLEM16-1
	Instructors Note: The total alkalinity required should be 14.68 instead of 10.87 mg/L as CaCO
	Solution
	PROBLEM16-2
	Solution
	PROBLEM16-3
	Solution
	PROBLEM16-4
	Solution
	PROBLEM16-5
	Solution
	PROBLEM16-6
	Solution
	PROBLEM16-7
	Instructors Note: The problem statement specified that average load by fuel oil is 35 percent.  For the convenience of using the data from AP42, however, it should be corrected to 5 percent (to be consistent with the basis of emission factors presented in AP42.  The emission factors in Table 16-15 have errors for SI units (US customary units are correct).  For the dual fuel engines, the emission factors for CO, NOx, and SO2 are 0.58, 1.16, and 0.0216S1+0.386S2, respectively.
	Because the emission factors presented in Table 16-15 are based on the fuel input, assumptions must be made for the average output and the efficiency of the engine.  In this solution, 38% efficiency for the reciprocating engine (electrical power output to fuel input).  For SO2, S1 is weight percent of sulfur in the fuel oil, and S2 is weight percent of sulfur in natural gas.  In this solution, 1.0 percent for fuel oil and 0.00077 percent for natural gas are assumed.  Instructor may provide these values, or students may be tasked to find out the typical values and state their assumptions.
	Alternatively, the original AP42 document may be used to find the emission factors based on the power output.
	Solution
	PROBLEM16-8
	Instructors Note: The reference to the example problem is to be corrected to Example 16-5.  The available digester gas is more than the natural gas use in the existing condition from Example 16-5.  The students may solve assuming only natural gas is replaced with the digester gas, or it can be assumed that the excess natural gas is used to generate electricity onsite, thereby reducing the purchase of electricity from the electrical grid.  Both scenarios are presented in this solution.
	Solution
M&E5 SM Chap. 17 FINAL
	ENERGY CONSIDERATIONS IN
		ENERGY CONSIDERATIONS IN
			ENERGY CONSIDERATIONS IN
				ENERGY CONSIDERATIONS IN
				WASTEWATER MANAGEMENT
M&E5 SM Chap. 18 FINAL REV
	WASTEWATER MANAGEMENT:
		WASTEWATER MANAGEMENT:
			WASTEWATER MANAGEMENT:
				WASTEWATER MANAGEMENT:
				FUTURE CHALLENGES AND OPPORTUNITIES
	PROBLEM18-1
	Solution
	PROBLEM18-2
	Solution
	PROBLEM18-3
	Solution
	PROBLEM18-4
	Solution
	PROBLEM18-5
	Solution
	PROBLEM18-6
	Solution
                        
Document Text Contents
Page 2

v

CONTENTS
1. Wastewater Engineering: An Overview 1-1

2. Constituents in Wastewater 2-1

3. Wastewater Flowrates and Constituent Loadings 3-1

4. Process Selection and Design Considerations 4-1

5. Physical Processes 5-1

6. Chemical Processes 6-1

7. Fundamentals of Biological Treatment 7-1

8. Suspended Growth Biological Treatment Processes 8-1

9. Attached Growth and Combined Biological Treatment
Processes

9-1

10. Anaerobic Suspended and Attached Growth Biological
Treatment Processes

10-1

11. Separation Processes for Removal of Residual Constituents 11-1

12. Disinfection Processes 12-1

13. Processing and Treatment of Sludges 13-1

14. Ultimate and Reuse of Biosolids 14-1

15. Treatment of Return Flows and Nutrient Recovery 15-1

16. Treatment Plant Emissions and Their Control 16-1

17. Energy Considerations in Wastewater Management 17-1

18. Wastewater Management: Future Challenges and
Opportunities

18-1

Page 360

Chapter 8 Suspended Growth Biological Treatment Processes


8-104

F

T

V
V

=
3

T

792.2m
V

= 0.167

VT =
3792.2m

0.167
= 4374 m3

4. Determine the SRT using Eqs. (8-20), (8-21), and (7-57).

a. From Eqs. (8-20) and (8-21)

PX,TSS =
H o d H H o

H H

QY (S S) (f )(b )QY (S S)SRT
[1 b (SRT)]0.85 [1 b (SRT)]0.85



n X
o o

n

QY (NO )
Q(nbVSS) Q(TSS VSS )

[1 b (SRT)]0.85


Substituting PX,TSS in Eq. (7-57).

2
H o d H H o

TSS
H H

QY (S S)SRT (f )(b )QY (S S)(SRT)
(X )(V)

[1 b (SRT)]0.85 [1 b (SRT)]0.85


n x

n

QY (NO )SRT
[1 (b )SRT]0.85

+ o oQ(nbVSS)SRT Q(TSS VSS )SRT

b. Define values for solution to above equation:

Influent bCOD = 1.6 (BOD) = 1.6 (250) = 400 mg/L

Assume So – S influent bCOD = 1.6(BOD) = 1.6(250 g/m3)

= 400 g/m3

Volume/tank = 4374 m3

Flow/tank =
3(5000m /d)

2
= 2500 m3/d

c. Develop coefficients from Tables 8-14 at 12°C:

Y = 0.45 g/g

Kn = 0.50(1.0)12-20 = 0.50 g/m3


bH = 0.12 (1.04)12-20 = 0.088 g/g • d

Yn = 0.20 g/g (including ammonia- and nitrite- oxidizers)

bn = 0.17 (1.029)12-20 = 0.135 g/g • d

max,AOB = 0.90 (1.072)12-20 = 0.516 g/g • d

Ko= 0.50 g/m3

Page 361

Chapter 8 Suspended Growth Biological Treatment Processes


8-105

d. Insert values and coefficients in equation developed in 4a and compute

SRT.

(4000 g/m3) (4374 m3) =
3 3(0.45g/g)(2500m /d)(400g/m )(SRT)

[1 (0.088g/g•d)SRT]0.85


3 3 2(0.15g/g)(0.088g/g•d)(2500m /d)(0.45g/g)(400g/m )(SRT)
[1 (0.088g/g•d)SRT]0.85



3 3
3 3(0.20g/g)(2500m /d)(36g/m )(SRT) (2500m /d)(120g/m )(SRT)

[1 (0.135g/g•d)SRT]0.85


3 3 (2500m /d)[(220 210)g/m ](SRT)

17,496,000 g =
2529,411(SRT) 6988(SRT) 21,176(SRT)

300,000(SRT) 25,000(SRT)
1 .088(SRT) 1 0.088(SRT) 1 0.135(SRT)



SRT = 33.5 d

5. Determine PX,bio using appropriate components of Eq. (8-15)

H o d H H o n x
X,bio

H H n

Q(Y )(S S) f (b )Q(Y )(S S)SRT QY (NO )
P

1 b (SRT) 1 b (SRT) 1 b (SRT)


3 3

X,bio
(2500m /d)(0.45g/g)(400g/m )

P
[1 (0.088g/g•d)(33.5d)]



3 3(0.15g/g)(0.088g/g•d)(2500m /d)(0.45g/g)(400g/m )(33.5d)


[1 0.088g/g•d(33.5d)]


3 3(2500m /d)(0.20g/g)(36g/m )


[1 0.135g/g•d(33.5d)]


PX,bio = (113,981 + 50,402 + 3259) g/d = 167,642 g/d

6. Determine NOx using Eq. (8-24):

NOx = TKN – Ne – 0.12 PX,bio/Q

Considering a long SRT, assume Ne = 0.5 mg/L NH4-N

NOx = 45.0 g/m3 – 0.5 g/m3 - 3
0.12(167,642g/d)

(2500m /d)

Page 720

Chapter 18 Wastewater Management: Future Challenges and Opportunities

18-2

PROBLEM 18-3
Problem Statement - See text, page 1899
Solution

1. Investigate whether the collection system has any storage capacity. If
excess capacity is available in gravity flow collection system, determine if
excess capacity can be utilized by installing control structures (e.g. dams
and weirs). If the collection system contains pump stations, it may be
possible to sequence the pumps so that the storage capacity in the
collection system can be utilized.

2. Investigate whether any unused tankage is available at the wastewater
treatment plant, and whether it could be utilized for peak flow equalization.

3. Investigate whether all of the secondary clarifiers are being utilized. Any
unused clarifiers should be brought online during storm events.

4. Investigate whether some of the primary sedimentation tanks could be
converted to secondary clarifiers during storm events.

5. Consider installation of tunnel storage (see Table 18-7) within the collection
system.

6. Consider installation of online or offline flow equalization on or near
wastewater treatment plant.

7. If a conventional plug-flow activated sludge process is used, consider
converting it to step feed prior to and during storm events.

PROBLEM 18-4
Problem Statement - See text, page 1899
Solution
1. Some advantages and disadvantages, in no special order, are:

Advantages Disadvantages

Improved treatment can be
provided for wastewater and
stormwater.

The cost of treating of
stormwater with complete
secondary treatment can be
reduced.

A higher level of treatment could
be provided for the smaller
wastewater flowrate.



Enormously expensive.
Disruptive to the public for extended period of

time that would be needed for the
construction of the separate stormwater
collections system.

Because of reduced flowrates, solids
deposition may occur in the wastewater
collection system, which would reqire periodic
flushing.

Right of way may be difficult to obtain.
More personnel would be required to manage

the separate wastewater and stormwater
treatment plants.

Securing appropriate rights of way
easements may be difficult and costly

Page 721

Chapter 18 Wastewater Management: Future Challenges and Opportunities

18-3

PROBLEM 18-5
Problem Statement - See text, page 1899
Solution
1. Some advantages and disadvantages, in no special order, are:

Advantages Disadvantages

A single treatment facility could
be operated.

Untreated stormwater discharges
could potentially be eliminated.

Depending of the rainfall pattern, excess
plant capacity would not be utilized during
extended periods of the year.

Disruptive to the public for extended period of
time that would be needed to construction a
combined collection system.

Serious operational problems can occur in a
plant subject to peak storm events if flow
equalization is not available.



PROBLEM 18-6
Problem Statement - See text, page 1899
Solution
1. Some benefits and drawbacks, in no special order, are:

Benefits Drawbacks

Separate collection systems (Problem 18-4)

Smaller treatment facility for the
treatment of wastewater system.


Two different agencies or departments may
be responsible for operation and
management functions.

Because more staff are required, a larger
supervisory staff would be required.

Enormously expensive to construct separate
collection system. Beyond the financial,
means of most communities.

Combined collection system (Problem 18-5)

Centralized agency responsible
for management of both
wastewater and stormwater

Potentially improved treatment of
stormwater.

New rate structure must be developed to deal
with combined flows

Enormously expensive to construct combined
collection system to handle both wastewater
and stormwater. Beyond the financial, means
of most communities

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