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 9-1
Processes
10. Anaerobic Suspended and Attached Growth Biological 10-1
Treatment Processes
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
, 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 • 3
A 0.2 m3
/ s – 0.2 1 cos
43,200
m /s
dt
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
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 9-1
Processes
10. Anaerobic Suspended and Attached Growth Biological 10-1
Treatment Processes
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
, 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 • 3
A 0.2 m3
/ s – 0.2 1 cos
43,200
m /s
dt
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
