SOLUTIONS &
ST
TeSTbaNk
UV
Basic Environmental
IA
Technology:
Water Supply, Waste
AP
Management and Pollution
PR
Control 6th Edition
Authors: Jerry A. Nathanson, PE, Richard A. Schneider
RO
◊ aLL CHaPTeRS
VE
◊ INSTaNT PDF DOWNLOaD💯💯💯
◊ ORIGINaL FROM PUbLISHeR
D
MEDGEEK
??
Email:
Study material
Prepared by
Medconnoisseur
©2025
, Table of Contents
Chapter 1 1
ST
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
AP
Supplemental Problems 35
Multiple Choice and True/False 36
Answers to Multiple Choice and True/False 50
PR
Supplemental Problems 52
RO
VE
D ??
, 1
Basic Environmental Technology - Solutions Manual Sixth Edition
ST
CHAPTER 1 - BASIC CONCEPTS
Review Question Page References
(1) 1 (17) 15
UV
(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
IA
(8) 9 (24) 19
(9) 9, 10 (25) 19
(10) 9, 10 (26) 20
(11) 10 (27) 20
(12) 10 (28) 20
AP
(13) 11 (29) 13
(14) 12 (30) 14
(15) 12, 13 (31) 20, 21
(16) 12
PR
(There are no Practice Problems for Chapter 1)
RO
VE
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
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
PR
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
RO
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
VE
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.
D
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
TeSTbaNk
UV
Basic Environmental
IA
Technology:
Water Supply, Waste
AP
Management and Pollution
PR
Control 6th Edition
Authors: Jerry A. Nathanson, PE, Richard A. Schneider
RO
◊ aLL CHaPTeRS
VE
◊ INSTaNT PDF DOWNLOaD💯💯💯
◊ ORIGINaL FROM PUbLISHeR
D
MEDGEEK
??
Email:
Study material
Prepared by
Medconnoisseur
©2025
, Table of Contents
Chapter 1 1
ST
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
AP
Supplemental Problems 35
Multiple Choice and True/False 36
Answers to Multiple Choice and True/False 50
PR
Supplemental Problems 52
RO
VE
D ??
, 1
Basic Environmental Technology - Solutions Manual Sixth Edition
ST
CHAPTER 1 - BASIC CONCEPTS
Review Question Page References
(1) 1 (17) 15
UV
(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
IA
(8) 9 (24) 19
(9) 9, 10 (25) 19
(10) 9, 10 (26) 20
(11) 10 (27) 20
(12) 10 (28) 20
AP
(13) 11 (29) 13
(14) 12 (30) 14
(15) 12, 13 (31) 20, 21
(16) 12
PR
(There are no Practice Problems for Chapter 1)
RO
VE
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
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
PR
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
RO
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
VE
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.
D
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
