Basic Environmental Technology: Water Supply,
ST
Waste Management, and Pollution Control – 6th
Edition
UV
TEST BANK
IA
_A
PP
Jerry Nathanson
RO
Comprehensive Test Bank for Instructors and
VE
Students
D?
© Jerry Nathanson
All rights reserved. Reproduction or distribution without permission is prohibited
??
©STUDYSTREAM
, Table of Contents
ST
Chapter 1 1
Chapter 2 2
Chapter 3 5
Chapter 4 8
Chapter 5 10
UV
Chapter 6 12
Chapter 7 14
Chapter 8 18
Chapter 9 20
Chapter 10 23
IA
Chapter 11 26
Chapter 12 29
Chapter 13 29
Chapter 14 32
_A
Supplemental Problems 35
Multiple Choice and True/False 36
Answers to Multiple Choice and True/False 50
PP
Supplemental Problems 52
RO
VE
D?
??
, 1
Basic Environmental Technology - Solutions Manual Sixth Edition
This manual provides instructors with (a) text page references where answers to the end-of-chapter
ST
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
UV
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 )
IA
CHAPTER 1 - BASIC CONCEPTS
Review Question Page References
_A
(1) 1 (17) 15
(2) 2, 3 (18) 15
(3) 6 (19) 16
(4) 6 (20) 16, 17
(5) 6 (21) 17
PP
(6) 7 (22) 17
(7) 8 (23) 18
(8) 9 (24) 19
(9) 9, 10 (25) 19
(10) 9, 10 (26) 20
RO
(11) 10 (27) 20
(12) 10 (28) 20
(13) 11 (29) 13
(14) 12 (30) 14
(15) 12, 13 (31) 20, 21
VE
(16) 12
(There are no Practice Problems for Chapter 1)
D?
??
, 2
CHAPTER 2 - HYDRAULICS
Review Question Page References
(1) 24 (8) 30 (15) 42
ST
(2) 24 (9) 31 (16) 44
(3) 25 (10) 32 (17) 44
(4) 25 (11) 33 (18) 44
(5) 27 (12) 35 (19) 45
UV
(6) 28 (13) 36 (20) 45
(7) 30 (14) 40-42 (21) 46
(22) www.iihr.uiowa.edu/research
Solutions to Practice Problems
IA
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
_A
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)
PP
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)
RO
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
A = πD2/4 = π (0.3)2/4 = 0.0707 m2
VE
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
A = Q/V (from Eq. 2-4)
D?
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.
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
ST
Waste Management, and Pollution Control – 6th
Edition
UV
TEST BANK
IA
_A
PP
Jerry Nathanson
RO
Comprehensive Test Bank for Instructors and
VE
Students
D?
© Jerry Nathanson
All rights reserved. Reproduction or distribution without permission is prohibited
??
©STUDYSTREAM
, Table of Contents
ST
Chapter 1 1
Chapter 2 2
Chapter 3 5
Chapter 4 8
Chapter 5 10
UV
Chapter 6 12
Chapter 7 14
Chapter 8 18
Chapter 9 20
Chapter 10 23
IA
Chapter 11 26
Chapter 12 29
Chapter 13 29
Chapter 14 32
_A
Supplemental Problems 35
Multiple Choice and True/False 36
Answers to Multiple Choice and True/False 50
PP
Supplemental Problems 52
RO
VE
D?
??
, 1
Basic Environmental Technology - Solutions Manual Sixth Edition
This manual provides instructors with (a) text page references where answers to the end-of-chapter
ST
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
UV
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 )
IA
CHAPTER 1 - BASIC CONCEPTS
Review Question Page References
_A
(1) 1 (17) 15
(2) 2, 3 (18) 15
(3) 6 (19) 16
(4) 6 (20) 16, 17
(5) 6 (21) 17
PP
(6) 7 (22) 17
(7) 8 (23) 18
(8) 9 (24) 19
(9) 9, 10 (25) 19
(10) 9, 10 (26) 20
RO
(11) 10 (27) 20
(12) 10 (28) 20
(13) 11 (29) 13
(14) 12 (30) 14
(15) 12, 13 (31) 20, 21
VE
(16) 12
(There are no Practice Problems for Chapter 1)
D?
??
, 2
CHAPTER 2 - HYDRAULICS
Review Question Page References
(1) 24 (8) 30 (15) 42
ST
(2) 24 (9) 31 (16) 44
(3) 25 (10) 32 (17) 44
(4) 25 (11) 33 (18) 44
(5) 27 (12) 35 (19) 45
UV
(6) 28 (13) 36 (20) 45
(7) 30 (14) 40-42 (21) 46
(22) www.iihr.uiowa.edu/research
Solutions to Practice Problems
IA
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
_A
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)
PP
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)
RO
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
A = πD2/4 = π (0.3)2/4 = 0.0707 m2
VE
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
A = Q/V (from Eq. 2-4)
D?
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.
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
