SOLUTIONS &
TeSTbaNk
ST
U
Basic Environmental
VI
Technology:
A
Water Supply, Waste
AP
Management and Pollution
Control 6th Edition
PR
Authors: Jerry A. Nathanson, PE, Richard A. Schneider
OV
◊ aLL CHaPTeRS
ED
◊ INSTaNT PDF DOWNLOaD💯💯💯
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MEDGEEK
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Study material
Prepared by
Medconnoisseur
©2025
, Table of Contents
Chapter 1 1
Chapter 2 2
Chapter 3 5
ST
Chapter 4 8
Chapter 5 10
Chapter 6 12
Chapter 7 14
U
Chapter 8 18
Chapter 9 20
Chapter 10 23
VI
Chapter 11 26
Chapter 12 29
Chapter 13 29
A
Chapter 14 32
Supplemental Problems 35
AP
Multiple Choice and True/False 36
Answers to Multiple Choice and True/False 50
Supplemental Problems 52
PR
OV
ED
??
, 1
Basic Environmental Technology - Solutions Manual Sixth Edition
CHAPTER 1 - BASIC CONCEPTS
ST
Review Question Page References
(1) 1 (17) 15
(2) 2, 3 (18) 15
(3) 6 (19) 16
U
(4) 6 (20) 16, 17
(5) 6 (21) 17
(6) 7 (22) 17
VI
(7) 8 (23) 18
(8) 9 (24) 19
(9) 9, 10 (25) 19
(10) 9, 10 (26) 20
A
(11) 10 (27) 20
(12) 10 (28) 20
(13) 11 (29) 13
AP
(14) 12 (30) 14
(15) 12, 13 (31) 20, 21
(16) 12
PR
(There are no Practice Problems for Chapter 1)
OV
ED
??
, 2
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
ST
(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
U
(22) www.iihr.uiowa.edu/research
Solutions to Practice Problems
VI
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)
AP
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
PR
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
OV
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
V = 0.1 m3/s 0.707m2 = 1.4 m/s
ED
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.
??
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
TeSTbaNk
ST
U
Basic Environmental
VI
Technology:
A
Water Supply, Waste
AP
Management and Pollution
Control 6th Edition
PR
Authors: Jerry A. Nathanson, PE, Richard A. Schneider
OV
◊ aLL CHaPTeRS
ED
◊ INSTaNT PDF DOWNLOaD💯💯💯
◊ ORIGINaL FROM PUbLISHeR
? ?
MEDGEEK
Email:
Study material
Prepared by
Medconnoisseur
©2025
, Table of Contents
Chapter 1 1
Chapter 2 2
Chapter 3 5
ST
Chapter 4 8
Chapter 5 10
Chapter 6 12
Chapter 7 14
U
Chapter 8 18
Chapter 9 20
Chapter 10 23
VI
Chapter 11 26
Chapter 12 29
Chapter 13 29
A
Chapter 14 32
Supplemental Problems 35
AP
Multiple Choice and True/False 36
Answers to Multiple Choice and True/False 50
Supplemental Problems 52
PR
OV
ED
??
, 1
Basic Environmental Technology - Solutions Manual Sixth Edition
CHAPTER 1 - BASIC CONCEPTS
ST
Review Question Page References
(1) 1 (17) 15
(2) 2, 3 (18) 15
(3) 6 (19) 16
U
(4) 6 (20) 16, 17
(5) 6 (21) 17
(6) 7 (22) 17
VI
(7) 8 (23) 18
(8) 9 (24) 19
(9) 9, 10 (25) 19
(10) 9, 10 (26) 20
A
(11) 10 (27) 20
(12) 10 (28) 20
(13) 11 (29) 13
AP
(14) 12 (30) 14
(15) 12, 13 (31) 20, 21
(16) 12
PR
(There are no Practice Problems for Chapter 1)
OV
ED
??
, 2
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
ST
(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
U
(22) www.iihr.uiowa.edu/research
Solutions to Practice Problems
VI
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)
AP
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
PR
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
OV
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
V = 0.1 m3/s 0.707m2 = 1.4 m/s
ED
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.
??
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
