Energy and the Environment
3rd Edition
ST
UV
SOLUTIONS
IA
_A
MANUAL
PP
Robert A. Ristinen
RO
Jack J. Kraushaar
Jeffrey T. Brack
VE
Comprehensive Solutions Manual for Instructors and Students
D?
© Robert A. Ristinen, Jack J. Kraushaar & Jeffrey T. Brack
All rights reserved. Reproduction or distribution without permission is prohibited.
?
© DREAMSHUB
, Solutions Manual for Energy and the Environment (3rd Edition)
Robert A. Ristinen, Jack J. Kraushaar, & Jeffrey T. Brack
ISBN: 9781119179238
ST
UNIT 1: ENERGY FUNDAMENTALS AND FOSSIL FUELS
1. Energy Fundamentals, Energy Use in an Industrial Society
2. The Fossil Fuels
UV
UNIT 2: ENERGY CONVERSION AND THERMAL SYSTEMS
3. Heat Engines
UNIT 3: RENEWABLE ENERGY SOURCES
4. Renewable Energy Sources I: Solar Energy
IA
5. Renewable Energy Sources II: Alternatives
UNIT 4: NUCLEAR ENERGY AND CONSERVATION
_A
6. The Promise and Problems of Nuclear Energy
7. Energy Conservation
UNIT 5: TRANSPORTATION AND ENVIRONMENTAL IMPACTS
8. Transportation
PP
9. Air Pollution
10. Global Effects
RO
VE
D?
?
© DREAMSHUB
, Ristinen, Kraushaar, Brack Energy and the Environment Chapt 1 Solutions
Chapter 1: Energy Fundamentals
Odd-numbered problems
ST
•Problem 1.3 : Cart on a horizontal surface
Given: F = 10lb and d = 10f t
F ×d=W
UV
10lb × 10f t = 100f t · lbs
j
100f t · lbs × 1.36 = 136 joules
f t · lb
IA
•Problem 1.7 : Tons of coal per person
From Table 1.1: 98 QBtu/yr used in U.S.
U.S. population approximately 300 × 106 people.
_A
1 ton coal
98 × 1015 Btu × 7
= 3.6 × 109 tons
2.7 × 10 Btu
3.6 × 109 tons tons
Then: = 12 , approximately.
PP
6
300 × 10 people yr · person
•Problem 1.11 : Windmill heats water
f t3 lb
RO
40 gallons × 0.1337 × 62.4 3 = 334 lb
gal ft
Btu
334 lb × 1 × 50°F = 16, 700 Btu needed
lb · °F
1, 055 j
16, 700 Btu × = 17.6 × 106 j needed
Btu
VE
j
1400 W = 1400
sec
6
17.6 × 10 j
= 12, 600 seconds = 210 minutes = 3.5 hours
1400j/s
D?
?
1
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, Ristinen, Kraushaar, Brack Energy and the Environment Chapt 1 Solutions
Multiple choice problems
•Problem 1.1 : Product with exponentials
Answer: f)
(5 × 6 × 7) × 10(5+6+7) = 210 × 1018 = 2.1 × 1020
ST
•Problem 1.3: Mass on string
UV
Answer: c)
Initially: P.E. = mgh (see page 113)
m
= 5kg × 9.8 2 × 2m
s
2
m
IA
= 98kg 2
s
= 98Joules
_A
•Problem 1.5 : U.S. consumption vs. India
Answer: d)
50
From Fig. 1.3: = 12.5
PP
4
•Problem 1.7 : Average personal power
Answer: b)
RO
3000C = 3 × 106 cal
cal j 1 day 1 hour j
3 × 106 × 4.184 × × = 145 = 145 watts
day cal 24 hours 3600 sec sec
VE
•Problem 1.9 : Fossil fuel percentage of energy use
Answer: c)
D?
From Table 1.1: 18.5 + 27.3 + 36.1 = 81.9
?
2
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3rd Edition
ST
UV
SOLUTIONS
IA
_A
MANUAL
PP
Robert A. Ristinen
RO
Jack J. Kraushaar
Jeffrey T. Brack
VE
Comprehensive Solutions Manual for Instructors and Students
D?
© Robert A. Ristinen, Jack J. Kraushaar & Jeffrey T. Brack
All rights reserved. Reproduction or distribution without permission is prohibited.
?
© DREAMSHUB
, Solutions Manual for Energy and the Environment (3rd Edition)
Robert A. Ristinen, Jack J. Kraushaar, & Jeffrey T. Brack
ISBN: 9781119179238
ST
UNIT 1: ENERGY FUNDAMENTALS AND FOSSIL FUELS
1. Energy Fundamentals, Energy Use in an Industrial Society
2. The Fossil Fuels
UV
UNIT 2: ENERGY CONVERSION AND THERMAL SYSTEMS
3. Heat Engines
UNIT 3: RENEWABLE ENERGY SOURCES
4. Renewable Energy Sources I: Solar Energy
IA
5. Renewable Energy Sources II: Alternatives
UNIT 4: NUCLEAR ENERGY AND CONSERVATION
_A
6. The Promise and Problems of Nuclear Energy
7. Energy Conservation
UNIT 5: TRANSPORTATION AND ENVIRONMENTAL IMPACTS
8. Transportation
PP
9. Air Pollution
10. Global Effects
RO
VE
D?
?
© DREAMSHUB
, Ristinen, Kraushaar, Brack Energy and the Environment Chapt 1 Solutions
Chapter 1: Energy Fundamentals
Odd-numbered problems
ST
•Problem 1.3 : Cart on a horizontal surface
Given: F = 10lb and d = 10f t
F ×d=W
UV
10lb × 10f t = 100f t · lbs
j
100f t · lbs × 1.36 = 136 joules
f t · lb
IA
•Problem 1.7 : Tons of coal per person
From Table 1.1: 98 QBtu/yr used in U.S.
U.S. population approximately 300 × 106 people.
_A
1 ton coal
98 × 1015 Btu × 7
= 3.6 × 109 tons
2.7 × 10 Btu
3.6 × 109 tons tons
Then: = 12 , approximately.
PP
6
300 × 10 people yr · person
•Problem 1.11 : Windmill heats water
f t3 lb
RO
40 gallons × 0.1337 × 62.4 3 = 334 lb
gal ft
Btu
334 lb × 1 × 50°F = 16, 700 Btu needed
lb · °F
1, 055 j
16, 700 Btu × = 17.6 × 106 j needed
Btu
VE
j
1400 W = 1400
sec
6
17.6 × 10 j
= 12, 600 seconds = 210 minutes = 3.5 hours
1400j/s
D?
?
1
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Distribution of this document is illegal extra per year?
, Ristinen, Kraushaar, Brack Energy and the Environment Chapt 1 Solutions
Multiple choice problems
•Problem 1.1 : Product with exponentials
Answer: f)
(5 × 6 × 7) × 10(5+6+7) = 210 × 1018 = 2.1 × 1020
ST
•Problem 1.3: Mass on string
UV
Answer: c)
Initially: P.E. = mgh (see page 113)
m
= 5kg × 9.8 2 × 2m
s
2
m
IA
= 98kg 2
s
= 98Joules
_A
•Problem 1.5 : U.S. consumption vs. India
Answer: d)
50
From Fig. 1.3: = 12.5
PP
4
•Problem 1.7 : Average personal power
Answer: b)
RO
3000C = 3 × 106 cal
cal j 1 day 1 hour j
3 × 106 × 4.184 × × = 145 = 145 watts
day cal 24 hours 3600 sec sec
VE
•Problem 1.9 : Fossil fuel percentage of energy use
Answer: c)
D?
From Table 1.1: 18.5 + 27.3 + 36.1 = 81.9
?
2
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