Chem E 260
Solution #6
6-82E The sink temperature of a Carnot heat engine, the rate of heat rejection, and
the thermal efficiency are given. The power output of the engine and the source
temperature are to be determined.
Assumptions The Carnot heat engine operates steadily.
Analysis (a) The rate of heat input to this heat engine is determined from the definition of
thermal efficiency,
 th  1 
Q L
800 Btu/min

 0.75  1 

 Q H  3200 Btu/min

QH
Q H
Then the power output of this heat engine can be determined from
W net,out   th Q H  0.75 3200 Btu/min   2400 Btu/min  56.6 hp
(b) For reversible cyclic devices we have
 Q H

 Q
 L

T
  H


 rev  TL





 3200 Btu/min
 TL  
 800 Btu/min


 rev
HE
800 Btu/min
Thus the temperature of the source TH must be
 Q
TH   H
 Q
 L
TH
60°F

520 R   2080 R

6-84E The claim of an inventor about the operation of a heat engine is to be evaluated.
Assumptions The heat engine operates steadily.
Analysis If this engine were completely reversible, the thermal efficiency would be
 th,max  1 
TL
550 R
 1
 0.45
TH
1000 R
When the first law is applied to the engine above,
 2544.5 Btu/h 
  15,000 Btu/h  27 ,720 Btu/h
Q H  W net  Q L  (5 hp)
1 hp


15,000
The actual thermal efficiency of the proposed heat engine is then Btu/h
 th
W
 2544.5 Btu/h 
5 hp

  0.459
 net 

27,720 Btu/h 
1 hp
QH

1000
Q R
H
HE
5
hp
550
R
Since the thermal efficiency of the proposed heat engine is greater than that of a
completely reversible heat engine which uses the same isothermal energy reservoirs, the
inventor's claim is invalid.
6-97 The rate of cooling provided by a reversible refrigerator with specified reservoir
temperatures is to be determined.
Assumptions The refrigerator operates steadily.
300 K
Analysis The COP of this reversible refrigerator is
COP R, max 
TL
250 K

5
TH  TL 300 K  250 K
Rearranging the definition of the refrigerator coefficient of
performance gives
Q H
R
10
kW
Q L
Q L  COP R, max W net,in  (5)(10 kW)  50 kW
250 K
6-109E A heat pump maintains a house at a specified temperature. The rate of heat loss
of the house is given. The minimum power inputs required for different source
temperatures are to be determined.
Assumptions The heat pump operates steadily.
Analysis (a) The power input to a heat pump will be a minimum when the heat pump
operates in a reversible manner. If the outdoor air at 25°F is used as the heat source, the
COP of the heat pump and the required power input are determined to be
COP HP,max  COP HP,rev 
1
1  TL / TH 
1

 10.15
1  25  460 R / 78  460 R 
House
78F
55,000
Btu/h
and
Wnet,in,min 

Q H
55,000 Btu/h 
1 hp

  2.13 hp

COP HP,max
10.15
 2545 Btu/h 
(b) If the well-water at 50°F is used as the heat source,
the COP of the heat pump and the required power input
are determined to be
COPHP,max  COPHP,rev 
HP
25F or
50F
1
1

 19.2
1  TL / TH  1  50  460 R / 78  460 R 
and
Wnet,in,min 

Q H
55,000 Btu/h 
1 hp

  1.13 hp

COP HP,max
19.2
2545
Btu/h


7-25 Heat is transferred directly from an energy-source reservoir to an energy-sink. The
entropy change of the two reservoirs is to be calculated and it is to be determined if the
increase of entropy principle is satisfied.
Assumptions The reservoirs operate steadily.
Analysis The entropy change of the source and sink is given by
S 
Q H Q L  100 kJ 100 kJ



 0.0833 kJ/K
TH
TL
1200 K 600 K
Since the entropy of everything involved in this process has
increased, this transfer of heat is possible.
7-31 R-134a enters an evaporator as a saturated liquid-vapor at a specified pressure. Heat
is transferred to the refrigerant from the cooled space, and the liquid is vaporized. The
entropy change of the refrigerant, the entropy change of the cooled space, and the total
entropy change for this process are to be determined.
Assumptions 1 Both the refrigerant and the cooled space involve no internal
irreversibilities such as friction. 2 Any temperature change occurs within the wall of the
tube, and thus both the refrigerant and the cooled space remain isothermal during this
process. Thus it is an isothermal, internally reversible process.
Analysis Noting that both the refrigerant and the cooled space undergo reversible
isothermal processes, the entropy change for them can be determined from
S 
Q
T
(a) The pressure of the refrigerant is maintained constant. Therefore, the temperature of
the refrigerant also remains constant at the saturation value,
(Table A-12)
T  Tsat@160 kPa  15.6C  257.4 K
Then,
Srefrigeran t 
Qrefrigeran t,in
Trefrigeran t

180 kJ
 0.699 kJ/K
257.4 K
R-134a
160
kPa
180 kJ
(b) Similarly,
S space  
Qspace,out
Tspace

180 kJ
 0.672 kJ/K
268 K
5C
(c) The total entropy change of the process is
Sgen  Stotal  Srefrigeran t  Sspace  0.699  0.672  0.027 kJ/K
7-37 Water is compressed in a compressor during which the entropy remains constant.
The final temperature and enthalpy are to be determined.
Analysis The initial state is superheated vapor and the entropy is
T1  160 C  h1  2800.7 kJ/kg
(from EES)

P1  35 kPa  s1  8.1531 kJ/kg  K
Note that the properties can also be determined from Table
A-6 by interpolation but the values will not be as accurate
as those by EES. The final state is superheated vapor and
the properties are (Table A-6)
T
2
1
s
P2  300 kPa
 T2  440.5 C

s 2  s1  8.1531 kJ/kg  K  h2  3361.0 kJ/kg
7-58 Steam is expanded in an isentropic turbine. The work produced is to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 The
process is isentropic (i.e., reversible-adiabatic).
Analysis There is only one inlet and one exit, and thus m1  m 2  m . We take the turbine as
the system, which is a control volume since mass crosses the boundary. The energy
balance for this steady-flow system can be expressed in the rate form as
2 MPa
E  E

E system0 (steady)
0
inout



360°C
Rate of net energy transfer
Rate of change in internal,kinetic,
by heat, work, and mass
potential,etc. energies
E in  E out
Turbin
e
m h1  m h2  W out
W  m (h  h )
out
1
2
T
The inlet state properties are
100 kPa
P1  2 MPa  h1  3159 .9 kJ/kg
(Table A - 6)

T1  360 C  s1  6.9938 kJ/kg  K
2
MPa
For this isentropic process, the final state properties are (Table A-5) 100 kPa
s2  s f
6.9938  1.3028
P2  100 kPa

 0.9397
 x2 
s
6.0562

fg
s 2  s1  6.9938 kJ/kg  K 
h2  h f  x 2 h fg  417 .51  (0.9397 )( 2257 .5)  2538 .9 kJ/kg
Substituting,
wout  h1  h2  (3159 .9  2538 .9) kJ/kg  621.0 kJ/kg
1
2
s