Solutions to Chapter 5 Problems

5.1 (C) A reservoir must supply (or accept) continual thermal energy without a change in temperature.
The combustion in an engine occurs, the temperature increases markedly.

5.2 (D) The energy required to freeze the ice with a refrigeration process cannot be recovered by melting
the ice. It is not reversible.

5.3 (B) The Kelvin-Planck statement has to do with the cyclic conversion of heat to work, which a
rotating shaft implies. And, not all of the heat can be converted into work.

5.4 (C) The first law demands that, for a cycle, the energy in must equal the energy out. Recognizing
that kJ/s = kW,
output 100
QH = W + QL = 100 + 50 = 150 kW. ∴η = = = 0.667 or 66.7%
input 150

5.5 (A) Convert 5 hp to kJ/s: W = 5 hp = 5 hp ×746 W/hp = 3730 J/s. The COP is then

desired effect Q 24 000 / 3600

∴ COP = = L = = 1.79
purchased energy W 3.73

The 3600 converts hours to seconds.

5.6 (D) The ideal engine is a Carnot engine for which

QL TL TH 473
= . ∴ QH = Q L = 50 × = 80.7 kJ
QH TH TL 293

5.7 (C) The high-temperature reservoir would be at 100°C and the low-temperature reservoir would be
at 12°C. The maximum efficiency is that of a Carnot cycle, which is calculated to be

TL 285
η = 1− = 1− = 0.236
TH 373

For an actual engine to operates with an efficiency of 22% is possible but not very probable.

5.8 (D) The heat transfer, which occurs across an infinitesimal temperature difference, is given by Eq.
5.10 to be
V3 300
QH = mRTH ln = 0.2 × 0.287 × 313 × ln = 41.4 kJ
V2 30

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 5.9 (B) The desired effect is the heat transfer to the high-temperature reservoir. It is calculated to be

Desired effect 17
COP = = = 4.56
purchased energy 5 × 0.746

5.10 (A) Let us find the heat transfer from the low-temperature reservoir. It is

Q L = Q H − W
 = 17 − 5 × 0.746 = 13.27 kJ/s

Then, the ratio of heat transfers must equal to the ratio of temperatures for a Carnot cycle:

Q H TH Q L 13.27
= . ∴ TL = TH = 433 × = 338 K or 65°C
Q L TL 
QH 17

5.12 Convert 500 ft-lbf/s to watts:

ft-lbf 1 hp W
500 × × 746 = 678 W
s 550 ft-lbf/s hp

Only generator (i) is possible. The generator cannot produce more energy than what it accepts. That
even violates the first law of thermodynamics.

15 hp
5.13 b) The first law: QL = QH − W = 30 kW − = 9.89 kJ/s
0.746 hp/kW
output 15 / 0.746
The efficiency is η = = = 0.67 or 67.0%
input 30

5.14 The first law applied to the power plant requires

 + Q = 20 000 + 14 × 10 = 58 890 kW
7
Q H = W L
3600

output 20 000
The efficiency is η = = = 0.340 or 34%
input 58 890

kJ s
In one hour the heat supplied is QH = 58 890 × 3600 = 212 × 106 kJ
s hr

5.15 b) The definition of efficiency allows (refer to Appendix A)

output 200,000 / 60
η= = = 0.75. ∴Q H = 4440 Btu/s
input Q H

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 W  200 / 0.746
5.16 The efficiency relation is used: η = . Q H = = 957 kJ/s

QH 0.28

MJ 1 kg s
 = 0.957
The mass flux of gasoline is m × × 3600 = 68.9 kg/hr
s 50 MJ hr

5.18 Use Appendix A to convert the units.

 = 500, 000 − 70,000 × 2545 = 176,000 Btu/hr

b) The first law requires Q L = Q H − W
550
W  (70, 000 / 550) × 2545
The efficiency is η = = = 0.648 or 64.8%

QH 500, 000

ft 3 lbm Btu
5.19 The energy supplied is Q H = 1.0 × 45 3 × 21,000 = 945,000 Btu/hr
hr ft lbm


W  × 2545
W
b) The efficiency relation η =  . 0.25 = .  = 92.8 hp
∴W
QH 945, 000

miles 1 gal 1 ft 3 lbm Btu

5.20 The energy input is Q H = 60 × × × 45 3 × 21, 000 = 253, 000 Btu/hr
hr 30 miles 7.48 gal ft lbm

W  14 × 2545
b) The efficiency relation provides: η = = = 0.141 or 14.1%

QH 253, 000

kg MJ hr
5.22 The heat input is QH = 50 000 × 26 × 24 = 31.2 ×106 MJ each day
hr kg day
The work output is W = ηQH = 0.34 × (31.2 × 106 ) = 10.6 × 106 MJ each day

The heat dumped in the river is QL = QH − W = (31.2 − 10.6) × 106 = 20.6 ×106 MJ each day

5.23 Draw a box around the power plant, the “engine,” with Q B and W
 entering the engine and
P
 
W and Q leaving the engine.
T C

b) The 1st law is written as

 = 3000 × 10 − 120 × 10 + 270 = 16 940 kW or 22 700 hp

3 6
 = Q − Q + W
WT H L P
60 3600
The pump power is only 1.6% of the turbine power and could be ignored in the above.

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5.24 The heat transfers from the reservoirs are Q H = 15 kW and Q L = 10 kW . The first law provides the
compressor power:
 = Q − Q = 15 − 10 = 5 kW or 6.70 hp
WC H L

   5
5.25 b) The rejected heat transfer is Q cond = Qevap + Wcomp = 10 + = 16.7 kJ/s
0.746
desired effect 15
The performance is COP = = = 2.24
energy input 5 / 0.746

5.26 The performance parameter is used as follows:

desired effect Q cond

COP = =5= . ∴ Q cond = 33.5 kJ/s
energy input 5 / 0.746
The first law provides
5
Q evap = Q cond − W

comp = 33.5 − = 26.8 kJ/s
0.746

  
5.28 The 1st law is used as follows: Q cond = Qevap + Wcomp = 1500 × 60 + 7 × 2545 = 108,000 Btu/hr

desired effect Q evap 1500 × 60

The COP is COP = = = = 5.05
energy input W  7 × 2545
comp

5.30 b) The first law is written as

Q evap
Q cond = Q evap + W

comp . 100 = 
Q evap + = (1 + 0.167) Q evap . ∴Q evap = 85.69 kJ/s
6
The compressor power is found using the coefficient of performance:

 Q evap 85.69
W comp = = = 14.28 kW or 19.1 hp
COP 6

TL 500
5.34 The maximum possible efficiency is η = 1 − = 1− = 0.0741 or 7.41% .
TH 540

TL 500
5.36 The maximum possible efficiency is η = 1 − = 1− = 0.689 or 68.9% .
TH 1610

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5.38 The work produced is W = QH − QL = 1000 − 400 = 600 MJ/s.

W 600
The efficiency is η = = = 0.6 or 60%
QH 1000

TL T
The low-temperature reservoir follows: η = 1 − . 0.6 = 1 − L . ∴TL = 509 K or 236°C
TH 1273

TL 313 W W
5.40 η = 1 − = 1− = 0.390. η= = = 0.390. ∴ W = 156 kW or 209 hp
TH 513 QH 400

5.42 The maximum amount of energy that can be extracted from the hot water is
450 lbm Btu
Q = mC
 p (T1 − T2 ) = ×1.0 × (200 − 55) °F = 1088 Btu/s
60 s lbm-°F
TL 515
Using the maximum efficiency, η = 1 − = 1− = 0.220 , the maximum power that could be
TH 660
produced would be

 = ηQ = 0.22 ×1088 × 3600 = 339 hp

W
2545

P1v1 200 × 0.5

5.43 Using P1 and v1 the low temperature is T1 = = = 348 K.
R 0.287
W 200 T 348
b) Q = = = 333 kJ and 0.6 = 1 − L = 1 − . ∴TH = 870 K or 597°C
η 0.6 TH TH

TL 293
5.44 The maximum possible efficiency would be ηmax = 1 − = 1− = 0.073.
TH 373

The maximum heat that could be extracted from the hot water is

150 kg kJ
Q = mC
 p (T1 − T2 ) = × 4.18 × (100 − 20) °C = 836 kJ/s
60 s kg ⋅ °C

W 46 / 0.746
The proposed efficiency would be η = = = 0.074. Impossible.
QH 836

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  = ηQ = 0.6Q . Q = Q + W
5.45 b) W  = 50 + 0.6Q . ∴ Q = 125 kJ/s and W
 = 0.6 × 125 = 75 kW
H H H L H H

TL 293
ηmax = 1 − . 0.6 = 1 − . ∴TH = 732 K or 460°C
TH TH

TL TL 1 1
5.46 η = 1 − = 0.75. ∴ = 0.25. COPR = = = 0.333
TH TH TH 1
−1 −1
TL 0.25

T2 293
5.48 η1 = 1 − = 1− . T22 = 293 × 673. ∴T2 = 444 K or 171°C
673 T2

TH 296 Q 700
5.50 b) COP = = = 16.4 = H =
TH − TL 296 − 278 W W

 = 42.7 kW or 57.2 hp Q = Q − W
∴W  = 700 − 42.7 = 657 kJ/s
L H

TH 528 Q 72 000 / 3600  = 2.95 kJ/s

5.52 COP = = = 6.77 = H = . ∴W or 3.96 hp
TH − TL 528 − 450 W W

TH 542 Q 800, 000  = 44, 200 Btu/hr or 17.4 hp

5.54 COP = = = 18.1 = H = . ∴W
TH − TL 542 − 512 W W

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