SOLUTIONS &
TeSTbaNk
M
ED
Basic Environmental
PA
Technology:
PE
Water Supply, Waste
Management and Pollution
RS
Control 6th Edition
_S
Authors: Jerry A. Nathanson, PE, Richard A. Schneider
TU
◊ aLL CHaPTeRS
V
◊ INSTaNT PDF DOWNLOaD💯💯💯
IA
◊ ORIGINaL FROM PUbLISHeR
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MEDGEEK
Email:
Study material
Prepared by
Medconnoisseur
©2025
, Table of Contents
Chapter 1 1
Chapter 2 2
M
Chapter 3 5
Chapter 4 8
Chapter 5 10
Chapter 6 12
ED
Chapter 7 14
Chapter 8 18
Chapter 9 20
Chapter 10 23
PA
Chapter 11 26
Chapter 12 29
Chapter 13 29
Chapter 14 32
PE
Supplemental Problems 35
Multiple Choice and True/False 36
RS
Answers to Multiple Choice and True/False 50
Supplemental Problems 52
_S
TU
VI
A?
?
, 1
Basic Environmental Technology - Solutions Manual Sixth Edition
CHAPTER 1 - BASIC CONCEPTS
M
Review Question Page References
(1) 1 (17) 15
(2) 2, 3 (18) 15
ED
(3) 6 (19) 16
(4) 6 (20) 16, 17
(5) 6 (21) 17
(6) 7 (22) 17
PA
(7) 8 (23) 18
(8) 9 (24) 19
(9) 9, 10 (25) 19
(10) 9, 10 (26) 20
(11) 10 (27) 20
PE
(12) 10 (28) 20
(13) 11 (29) 13
(14) 12 (30) 14
(15) 12, 13 (31) 20, 21
RS
(16) 12
_S
(There are no Practice Problems for Chapter 1)
TU
VI
A?
?
, 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
M
(4) 25 (11) 33 (18) 44
(5) 27 (12) 35 (19) 45
(6) 28 (13) 36 (20) 45
ED
(7) 30 (14) 40-42 (21) 46
(22) www.iihr.uiowa.edu/research
Solutions to Practice Problems
PA
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
PE
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)
RS
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)
_S
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
TU
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
VI
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?
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
M
ED
Basic Environmental
PA
Technology:
PE
Water Supply, Waste
Management and Pollution
RS
Control 6th Edition
_S
Authors: Jerry A. Nathanson, PE, Richard A. Schneider
TU
◊ aLL CHaPTeRS
V
◊ INSTaNT PDF DOWNLOaD💯💯💯
IA
◊ ORIGINaL FROM PUbLISHeR
??
MEDGEEK
Email:
Study material
Prepared by
Medconnoisseur
©2025
, Table of Contents
Chapter 1 1
Chapter 2 2
M
Chapter 3 5
Chapter 4 8
Chapter 5 10
Chapter 6 12
ED
Chapter 7 14
Chapter 8 18
Chapter 9 20
Chapter 10 23
PA
Chapter 11 26
Chapter 12 29
Chapter 13 29
Chapter 14 32
PE
Supplemental Problems 35
Multiple Choice and True/False 36
RS
Answers to Multiple Choice and True/False 50
Supplemental Problems 52
_S
TU
VI
A?
?
, 1
Basic Environmental Technology - Solutions Manual Sixth Edition
CHAPTER 1 - BASIC CONCEPTS
M
Review Question Page References
(1) 1 (17) 15
(2) 2, 3 (18) 15
ED
(3) 6 (19) 16
(4) 6 (20) 16, 17
(5) 6 (21) 17
(6) 7 (22) 17
PA
(7) 8 (23) 18
(8) 9 (24) 19
(9) 9, 10 (25) 19
(10) 9, 10 (26) 20
(11) 10 (27) 20
PE
(12) 10 (28) 20
(13) 11 (29) 13
(14) 12 (30) 14
(15) 12, 13 (31) 20, 21
RS
(16) 12
_S
(There are no Practice Problems for Chapter 1)
TU
VI
A?
?
, 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
M
(4) 25 (11) 33 (18) 44
(5) 27 (12) 35 (19) 45
(6) 28 (13) 36 (20) 45
ED
(7) 30 (14) 40-42 (21) 46
(22) www.iihr.uiowa.edu/research
Solutions to Practice Problems
PA
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
PE
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)
RS
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)
_S
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
TU
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
VI
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?
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
