SOLUTIONS &
TeSTbaNk
Basic Environmental
?M
Technology:
Water Supply, Waste
ED
Management and Pollution
GE
Control 6th Edition
EK
Authors: Jerry A. Nathanson, PE, Richard A. Schneider
??
◊ aLL CHaPTeRS
◊ 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
Chapter 4 8
Chapter 5 10
Chapter 6 12
Chapter 7 14
Chapter 8 18
Chapter 9 20
Chapter 10 23
?M
Chapter 11 26
Chapter 12 29
Chapter 13 29
Chapter 14 32
Supplemental Problems 35
ED
Multiple Choice and True/False 36
Answers to Multiple Choice and True/False 50
GE
Supplemental Problems 52
EK
??
, 1
Basic Environmental Technology - Solutions Manual Sixth Edition
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
?M
(8) 9 (24) 19
(9) 9, 10 (25) 19
(10) 9, 10 (26) 20
(11) 10 (27) 20
(12) 10 (28) 20
ED
(13) 11 (29) 13
(14) 12 (30) 14
(15) 12, 13 (31) 20, 21
(16) 12
GE
(There are no Practice Problems for Chapter 1)
EK
??
, 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
(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
?M
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)
ED
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)
GE
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
EK
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
??
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)
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
Basic Environmental
?M
Technology:
Water Supply, Waste
ED
Management and Pollution
GE
Control 6th Edition
EK
Authors: Jerry A. Nathanson, PE, Richard A. Schneider
??
◊ aLL CHaPTeRS
◊ 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
Chapter 4 8
Chapter 5 10
Chapter 6 12
Chapter 7 14
Chapter 8 18
Chapter 9 20
Chapter 10 23
?M
Chapter 11 26
Chapter 12 29
Chapter 13 29
Chapter 14 32
Supplemental Problems 35
ED
Multiple Choice and True/False 36
Answers to Multiple Choice and True/False 50
GE
Supplemental Problems 52
EK
??
, 1
Basic Environmental Technology - Solutions Manual Sixth Edition
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
?M
(8) 9 (24) 19
(9) 9, 10 (25) 19
(10) 9, 10 (26) 20
(11) 10 (27) 20
(12) 10 (28) 20
ED
(13) 11 (29) 13
(14) 12 (30) 14
(15) 12, 13 (31) 20, 21
(16) 12
GE
(There are no Practice Problems for Chapter 1)
EK
??
, 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
(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
?M
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)
ED
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)
GE
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
EK
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
??
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)
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
