Solution Manual for Basic Environmental Technology 6th Edition by Jerry A. Nathanson – Water Supply, Waste Management & Pollution Control, Newest edition 2025.

Instructor’s Manual and Testbank
To accompany



Basic Environmental Technology:
Water Supply, Waste Management and Pollution Control
6th Edition



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
Chapter 2 2
Chapter 3 5
Chapter 4 8
Chapter 5 10
Chapter 6 12
Chapter 7 14
Chapter 8 18
Chapter 9 20
Chapter 10 23
Chapter 11 26
Chapter 12 29
Chapter 13 29
Chapter 14 32

Supplemental Problems 35

Multiple Choice and True/False 36
Answers to Multiple Choice and True/False 50

Supplemental Problems 52
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Basic Environmental Technology - Solutions Manual Sixth Edition

This manual provides instructors with (a) text page references where answers to the end-of-chapter
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
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
text or in this manual at: or )


CHAPTER 1 - BASIC CONCEPTS

Review Question Page References
(1) 1 (17) 15
(2) 2, 3 (18) 15
(3) 6 (19) 16
(4) 6 (20) 16, 17
(5) 6 (21) 17
(6) 7 (22) 17
(7) 8 (23) 18
(8) 9 (24) 19
(9) 9, 10 (25) 19
(10) 9, 10 (26) 20
(11) 10 (27) 20
(12) 10 (28) 20
(13) 11 (29) 13
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(14) 12 (30) 14
(15) 12, 13 (31) 20, 21
(16) 12
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(There are no Practice Problems for Chapter 1)
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CHAPTER 2 - HYDRAULICS

Review Question Page References
(1) 24 (8) 30 (15) 42
(2) 24 (9) 31 (16) 44
(3) 25 (10) 32 (17) 44
(4) 25 (11) 33 (18) 44
(5) 27 (12) 35 (19) 45
(6) 28 (13) 36 (20) 45
(7) 30 (14) 40-42 (21) 46
(22) www.iihr.uiowa.edu/research

Solutions to Practice Problems

1. P = 0.43 x h (Equation 2-2b)
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)

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)

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)

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
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6. Q= A x V (Eq. 2-4), therefore V = Q/A
A = πD2/4 = π (0.3)2/4 = 0.0707 m2
100L/s x 1 m3/1000L=0.1 m3/s
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V = 0.1 m3/s 0.707m2 = 1.4 m/s

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)
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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)
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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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