Solutions & Test Bank - Basic Environmental Technology Water Supply, Waste Management and Pollution Control 6th Edition by Jerry Nathanson, All Chapters 1-14

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



SOLUTIONS &
TESTBANK
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Basic Environmental
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Technology: Water Supply,
Waste Management, and
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Pollution Control, 6th edition
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Authors:
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Jerry A. Nathanson, Richard A. Schneider
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ISBN-13: 978-0-13-284048-4
ISBN-10: 0-13-284048-0

Table of Contents
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Chapter 1 1
Chapter 2 2
Chapter 3 5
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Chapter 4 8
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Chapter 5 10
Chapter 6 12
Chapter 7 14
Chapter 8 18
Chapter 9 20
Chapter 10 23
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Chapter 11 26
Chapter 12 29
Chapter 13 29
Chapter 14 32
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Supplemental Problems 35

Multiple Choice and True/False 36
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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
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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
text or in this manual at: or )
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CHAPTER 1 - BASIC CONCEPTS

Review Question Page References
(1) 1 (17) 15
(2) 2, 3 (18) 15
(3) 6 (19) 16
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(4) 6 (20) 16, 17
(5) 6 (21) 17
(6) 7 (22) 17
(7) 8 (23) 18
(8) 9 (24) 19
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(9) 9, 10 (25) 19
(10) 9, 10 (26) 20
(11) 10 (27) 20
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(12) 10 (28) 20
(13) 11 (29) 13
(14) 12 (30) 14
(15) 12, 13 (31) 20, 21
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(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
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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
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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)
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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)

4. h = 2.3 x P = 2.3 x 50 = 115 ft, in the water main
h = 115 - 40 = 75 ft
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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
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Pressure head (in tube) = 0.1 x 40 kPa = 4 m

6. Q= A x V (Eq. 2-4), therefore V = Q/A
A = πD2/4 = π (0.3)2/4 = 0.0707 m2
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100L/s x 1 m3/1000L=0.1 m3/s
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
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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)
In the constriction, V2 = (2 m/s) x (4) = 8 m/s