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Instructor’s Manual and Testbank
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To accompany
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Basic Environmental Technology:
Water Supply, Waste Management and Pollution Control
6th Edition
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Jerry A. Nathanson, PE
Richard A. Schneider
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Upper Saddle River, New Jersey
Columbus, Ohio
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, Table of Contents
Chapter 1 1
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Chapter 2 2
Chapter 3 5
Chapter 4 8
Chapter 5 10
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Chapter 6 12
Chapter 7 14
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Chapter 8 18
Chapter 9 20
Chapter 10 23
Chapter 11 26
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Chapter 12 29
Chapter 13 29
Chapter 14 32
Supplemental Problems 35
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Multiple Choice and True/False 36
Answers to Multiple Choice and True/False 50
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Supplemental Problems 52
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, 1
Basic Environmental Technology - Solutions Manual Sixth Edition
This manual provides instructors with (a) text page references where answers to the end-of-chapter
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Review Questions can be found and worked-out solutions to each of the Practice Problems.
Additional materials including supplemental problems and projects.
Generally, answers to end-of-chapter Practice Problems are rounded-off to reflect the precision of
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the data and/or the accuracy of the assumed factors in the problems. These answers are also listed
in Appendix G of the text for students to use in checking their work. (The authors have made every
attempt to keep errors to a minimum. They can be notified of any mistakes that may be found in the
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text or in this manual at: or )
CHAPTER 1 - BASIC CONCEPTS
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Review Question Page References
(1) 1 (17) 15
(2) 2, 3 (18) 15
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(3) 6 (19) 16
(4) 6 (20) 16, 17
(5) 6 (21) 17
(6) 7 (22) 17
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(7) 8 (23) 18
(8) 9 (24) 19
(9) 9, 10 (25) 19
(10) 9, 10 (26) 20
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(11) 10 (27) 20
(12) 10 (28) 20
(13) 11 (29) 13
(14) 12 (30) 14
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(15) 12, 13 (31) 20, 21
(16) 12
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(There are no Practice Problems for Chapter 1)
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, 2
CHAPTER 2 - HYDRAULICS
Review Question Page References
(1) 24 (8) 30 (15) 42
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(2) 24 (9) 31 (16) 44
(3) 25 (10) 32 (17) 44
(4) 25 (11) 33 (18) 44
(5) 27 (12) 35 (19) 45
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(6) 28 (13) 36 (20) 45
(7) 30 (14) 40-42 (21) 46
(22) www.iihr.uiowa.edu/research
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Solutions to Practice Problems
1. P = 0.43 x h (Equation 2-2b)
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P = 0.43 x 50 ft = 22 psi at the bottom of the reservoir
P = 0.43 x (50 -30) = 0.43 x 20 ft = 8.6 psi above the bottom
2. h = 0.1 x P = 0.1 x 50 = 5 m (Equation 2-3a)
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3. Depth of water above the valve: h = (78 m -50 m) + 2 m = 30 m
P = 9.8 x h = 9.8 x 30 = 294 kPa ≈ 290 kPa (Equation 2-2a)
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4. h = 2.3 x P = 2.3 x 50 = 115 ft, in the water main
h = 115 - 40 = 75 ft
P = 0.43 x 75 = 32 psi, 40 ft above the main (Equation 2-2b)
47
5. Gage pressure P = 30 + 9.8 x 1 = 39.8 kPa ≈ 40 kPa
Pressure head (in tube) = 0.1 x 40 kPa = 4 m
6. Q= A x V (Eq. 2-4), therefore V = Q/A
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A = πD2/4 = π (0.3)2/4 = 0.0707 m2
100L/s x 1 m3/1000L=0.1 m3/s
V = 0.1 m3/s 0.707m2 = 1.4 m/s
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7. Q = (500 gal/min) x (1 min/60 sec) x (1 ft3/7.5 gal) = 1.11 cfs
A = Q/V (from Eq. 2-4)
A = 1.11 ft3/sec /1.4 ft/sec = 0.794 ft2
A = πD2/4, therefore D = √4A/π = √(4)(0.794)/π = 1 ft = 12 in.
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8. Q=A1 x V1 = A2 x V2 (Eq.2-5)
Since A = πD2/4, we can write
D12 xV1 = D22 xV2 and V2 =V1 x (D12 /D22)
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In the constriction, V2 = (2 m/s) x (4) = 8 m/s
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d
Instructor’s Manual and Testbank
yg
To accompany
en
Basic Environmental Technology:
Water Supply, Waste Management and Pollution Control
6th Edition
iu
s2
Jerry A. Nathanson, PE
Richard A. Schneider
47
??
??
??
?
Upper Saddle River, New Jersey
Columbus, Ohio
?
, Table of Contents
Chapter 1 1
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Chapter 2 2
Chapter 3 5
Chapter 4 8
Chapter 5 10
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Chapter 6 12
Chapter 7 14
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Chapter 8 18
Chapter 9 20
Chapter 10 23
Chapter 11 26
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Chapter 12 29
Chapter 13 29
Chapter 14 32
Supplemental Problems 35
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Multiple Choice and True/False 36
Answers to Multiple Choice and True/False 50
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Supplemental Problems 52
47
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??
??
, 1
Basic Environmental Technology - Solutions Manual Sixth Edition
This manual provides instructors with (a) text page references where answers to the end-of-chapter
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Review Questions can be found and worked-out solutions to each of the Practice Problems.
Additional materials including supplemental problems and projects.
Generally, answers to end-of-chapter Practice Problems are rounded-off to reflect the precision of
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the data and/or the accuracy of the assumed factors in the problems. These answers are also listed
in Appendix G of the text for students to use in checking their work. (The authors have made every
attempt to keep errors to a minimum. They can be notified of any mistakes that may be found in the
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text or in this manual at: or )
CHAPTER 1 - BASIC CONCEPTS
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Review Question Page References
(1) 1 (17) 15
(2) 2, 3 (18) 15
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(3) 6 (19) 16
(4) 6 (20) 16, 17
(5) 6 (21) 17
(6) 7 (22) 17
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(7) 8 (23) 18
(8) 9 (24) 19
(9) 9, 10 (25) 19
(10) 9, 10 (26) 20
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(11) 10 (27) 20
(12) 10 (28) 20
(13) 11 (29) 13
(14) 12 (30) 14
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(15) 12, 13 (31) 20, 21
(16) 12
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(There are no Practice Problems for Chapter 1)
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, 2
CHAPTER 2 - HYDRAULICS
Review Question Page References
(1) 24 (8) 30 (15) 42
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(2) 24 (9) 31 (16) 44
(3) 25 (10) 32 (17) 44
(4) 25 (11) 33 (18) 44
(5) 27 (12) 35 (19) 45
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(6) 28 (13) 36 (20) 45
(7) 30 (14) 40-42 (21) 46
(22) www.iihr.uiowa.edu/research
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Solutions to Practice Problems
1. P = 0.43 x h (Equation 2-2b)
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P = 0.43 x 50 ft = 22 psi at the bottom of the reservoir
P = 0.43 x (50 -30) = 0.43 x 20 ft = 8.6 psi above the bottom
2. h = 0.1 x P = 0.1 x 50 = 5 m (Equation 2-3a)
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3. Depth of water above the valve: h = (78 m -50 m) + 2 m = 30 m
P = 9.8 x h = 9.8 x 30 = 294 kPa ≈ 290 kPa (Equation 2-2a)
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4. h = 2.3 x P = 2.3 x 50 = 115 ft, in the water main
h = 115 - 40 = 75 ft
P = 0.43 x 75 = 32 psi, 40 ft above the main (Equation 2-2b)
47
5. Gage pressure P = 30 + 9.8 x 1 = 39.8 kPa ≈ 40 kPa
Pressure head (in tube) = 0.1 x 40 kPa = 4 m
6. Q= A x V (Eq. 2-4), therefore V = Q/A
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A = πD2/4 = π (0.3)2/4 = 0.0707 m2
100L/s x 1 m3/1000L=0.1 m3/s
V = 0.1 m3/s 0.707m2 = 1.4 m/s
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7. Q = (500 gal/min) x (1 min/60 sec) x (1 ft3/7.5 gal) = 1.11 cfs
A = Q/V (from Eq. 2-4)
A = 1.11 ft3/sec /1.4 ft/sec = 0.794 ft2
A = πD2/4, therefore D = √4A/π = √(4)(0.794)/π = 1 ft = 12 in.
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8. Q=A1 x V1 = A2 x V2 (Eq.2-5)
Since A = πD2/4, we can write
D12 xV1 = D22 xV2 and V2 =V1 x (D12 /D22)
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In the constriction, V2 = (2 m/s) x (4) = 8 m/s
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