, SOLUTIONS MANUAL
Wastewater Engineering:
Treatment and Resource Recovery Fifth
Edition
McGraw-Hill Book Company, Inc.
New York
, 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 18-1
Opportunities
v
,iii
, 1
INTRODUCTION TO
WASTEWATER TREATMENT
PROBLEM 1-1
Instructors Note: The first six problems are designed to illustrate the
application of the mass balance principle using examples from hydraulics with
which the students should be familiar.
Problem Statement - See text, page 53
Solution
1. Write a materials balance on the water in the tank
Accumulation = inflow – outflow + generation
dV dh
A Q –Q 0
in out
dt dt
2. Substitute given values for variable items and solve for h
dh t
A 0.2 m3 / s – 0.2 1 cos m3 / s
dt 43,200
A = 1000 m2
t
3. dh 2 x10 4 cos dt
43,200
Integrating the above expression yields:
(43,200) (2 x10 4 ) t
h ho sin
43,200
4. Determine h as a function of time for a 24 hour cycle
1-1
,Chapter 1 Introduction to Wastewater Treatment and Process Analysis
t, hr t, s h, m t, hr t, s h, m
0 0 5.00 14 50,400 3.62
2 7200 6.38 16 57,600 2.62
4 14,400 7.38 18 64,800 2.25
6 21,600 7.75 20 72,000 2.62
8 28,800 7.38 22 79,200 3.62
10 36,000 6.38 24 84,400 5.00
12 43,200 5.00
5. Plot the water depth versus time
PROBLEM 1-2
Instructors Note: The first six problems are designed to illustrate the
application of the mass balance principle using examples from hydraulics with
which the students should be familiar.
Problem Statement - See text, page 53
Solution
1. Write a materials balance on the water in the tank
Accumulation = inflow – outflow + generation
dV dh
A Q –Q 0
in out
dt dt
2. Substitute given values for variable items and solve for h
1-2
, Chapter 1 Introduction to Wastewater Treatment and Process Analysis
dh t
A 0.33 m3 / s – 0.2 1 cos m3 / s
dt 43,200
dh t
A 0.13 m3 / s 0.2 cos m3 / s
dt 43,200
A = 1600 m2
dh t 3
(1600) 0.13 m3 / s 0.2 cos m /s
dt 43,200
3. Integrating the above expression yields:
3
h ho ( 0.13 m / s) t ( 0.2)(43,200) t
1600 1600sin 43,200 m3 / s
Determine h as a function of time for a 24 hour cycle
t, hr t, s h, m t, hr t, s h, m
0 0 5.00 14 50,400 8.24
2 7200 6.44 16 57,600 8.19
4 14,400 7.66 18 64,800 8.55
6 21,600 8.47 20 72,000 9.36
8 28,800 8.83 22 79,200 10.58
10 36,000 8.78 24 84,400 12.02
12 43,200 8.51
4. Plot the water depth versus time
1-3
,Chapter 1 Introduction to Wastewater Treatment and Process Analysis
PROBLEM 1-3
Instructors Note: The first six problems are designed to illustrate the
application of the mass balance principle using examples from hydraulics with
which the students should be familiar.
Problem Statement - See text, page 53
Solution
1. Write a materials balance on the water in the tank
Accumulation = inflow – outflow + generation
dV dh
A Q –Q 0
in out
dt dt
2. Substitute given values for variable items and solve for h
dh t
A 0.3 1 cos m3 / s 0.3 m3 / s
dt 43,200
A = 1000 m2
t
dh 3 x10 4 cos dt
43,200
3. Integrating the above expression yields:
(43,200) (3x10 4) t
h ho sin
43,200
1. Determine h as a function of time for a 24 hour cycle
t, hr t, s h, m t, hr t, s h, m
0 0 5.00 14 50,400 2.94
2 7200 7.06 16 57,600 1.43
4 14,400 8.57 18 64,800 0.87
6 21,600 9.13 20 72,000 1.43
8 28,800 8.57 22 79,200 2.94
10 36,000 7.06 24 84,400 5.00
12 43,200 5.00
1-4
, Chapter 1 Introduction to Wastewater Treatment and Process Analysis
5. Plot the water depth versus time
PROBLEM 1-4
Instructors Note: The first six problems are designed to illustrate the
application of the mass balance principle using examples from hydraulics with
which the students should be familiar.
Problem Statement - See text, page 53
Solution
1. Write a materials balance on the water in the tank
Accumulation = inflow – outflow + generation
dV dh
A Q –Q 0
in out
dt dt
2. Substitute given values for variable items and solve for h
dh t
A = 0.35 1 +cos m3 /s - 0.35 m3 /s
dt 43,200
A = 2000 m2
t
dh = 1.75 x 10-4 cos dt
43,200
3. Integrating the above expression yields:
1-5
, Chapter 1 Introduction to Wastewater Treatment and Process Analysis
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1-6
Wastewater Engineering:
Treatment and Resource Recovery Fifth
Edition
McGraw-Hill Book Company, Inc.
New York
, 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 18-1
Opportunities
v
,iii
, 1
INTRODUCTION TO
WASTEWATER TREATMENT
PROBLEM 1-1
Instructors Note: The first six problems are designed to illustrate the
application of the mass balance principle using examples from hydraulics with
which the students should be familiar.
Problem Statement - See text, page 53
Solution
1. Write a materials balance on the water in the tank
Accumulation = inflow – outflow + generation
dV dh
A Q –Q 0
in out
dt dt
2. Substitute given values for variable items and solve for h
dh t
A 0.2 m3 / s – 0.2 1 cos m3 / s
dt 43,200
A = 1000 m2
t
3. dh 2 x10 4 cos dt
43,200
Integrating the above expression yields:
(43,200) (2 x10 4 ) t
h ho sin
43,200
4. Determine h as a function of time for a 24 hour cycle
1-1
,Chapter 1 Introduction to Wastewater Treatment and Process Analysis
t, hr t, s h, m t, hr t, s h, m
0 0 5.00 14 50,400 3.62
2 7200 6.38 16 57,600 2.62
4 14,400 7.38 18 64,800 2.25
6 21,600 7.75 20 72,000 2.62
8 28,800 7.38 22 79,200 3.62
10 36,000 6.38 24 84,400 5.00
12 43,200 5.00
5. Plot the water depth versus time
PROBLEM 1-2
Instructors Note: The first six problems are designed to illustrate the
application of the mass balance principle using examples from hydraulics with
which the students should be familiar.
Problem Statement - See text, page 53
Solution
1. Write a materials balance on the water in the tank
Accumulation = inflow – outflow + generation
dV dh
A Q –Q 0
in out
dt dt
2. Substitute given values for variable items and solve for h
1-2
, Chapter 1 Introduction to Wastewater Treatment and Process Analysis
dh t
A 0.33 m3 / s – 0.2 1 cos m3 / s
dt 43,200
dh t
A 0.13 m3 / s 0.2 cos m3 / s
dt 43,200
A = 1600 m2
dh t 3
(1600) 0.13 m3 / s 0.2 cos m /s
dt 43,200
3. Integrating the above expression yields:
3
h ho ( 0.13 m / s) t ( 0.2)(43,200) t
1600 1600sin 43,200 m3 / s
Determine h as a function of time for a 24 hour cycle
t, hr t, s h, m t, hr t, s h, m
0 0 5.00 14 50,400 8.24
2 7200 6.44 16 57,600 8.19
4 14,400 7.66 18 64,800 8.55
6 21,600 8.47 20 72,000 9.36
8 28,800 8.83 22 79,200 10.58
10 36,000 8.78 24 84,400 12.02
12 43,200 8.51
4. Plot the water depth versus time
1-3
,Chapter 1 Introduction to Wastewater Treatment and Process Analysis
PROBLEM 1-3
Instructors Note: The first six problems are designed to illustrate the
application of the mass balance principle using examples from hydraulics with
which the students should be familiar.
Problem Statement - See text, page 53
Solution
1. Write a materials balance on the water in the tank
Accumulation = inflow – outflow + generation
dV dh
A Q –Q 0
in out
dt dt
2. Substitute given values for variable items and solve for h
dh t
A 0.3 1 cos m3 / s 0.3 m3 / s
dt 43,200
A = 1000 m2
t
dh 3 x10 4 cos dt
43,200
3. Integrating the above expression yields:
(43,200) (3x10 4) t
h ho sin
43,200
1. Determine h as a function of time for a 24 hour cycle
t, hr t, s h, m t, hr t, s h, m
0 0 5.00 14 50,400 2.94
2 7200 7.06 16 57,600 1.43
4 14,400 8.57 18 64,800 0.87
6 21,600 9.13 20 72,000 1.43
8 28,800 8.57 22 79,200 2.94
10 36,000 7.06 24 84,400 5.00
12 43,200 5.00
1-4
, Chapter 1 Introduction to Wastewater Treatment and Process Analysis
5. Plot the water depth versus time
PROBLEM 1-4
Instructors Note: The first six problems are designed to illustrate the
application of the mass balance principle using examples from hydraulics with
which the students should be familiar.
Problem Statement - See text, page 53
Solution
1. Write a materials balance on the water in the tank
Accumulation = inflow – outflow + generation
dV dh
A Q –Q 0
in out
dt dt
2. Substitute given values for variable items and solve for h
dh t
A = 0.35 1 +cos m3 /s - 0.35 m3 /s
dt 43,200
A = 2000 m2
t
dh = 1.75 x 10-4 cos dt
43,200
3. Integrating the above expression yields:
1-5
, Chapter 1 Introduction to Wastewater Treatment and Process Analysis
THOSE WERE PREVIEW PAGES
TO DOWNLOAD THE FULL PDF
CLICK ON THE L.I.N.K
ON THE NEXT PAGE
1-6
