8-102
8-144E Large brass plates are heated in an oven at a specified rate. The rate of heat transfer to the plates in the oven and the
rate of entropy generation associated with this heat transfer process are to be determined.
Assumptions 1 The thermal properties of the plates are constant. 2 The changes in kinetic and potential energies are
negligible.
Properties The density and specific heat of the brass are given to be = 532.5 lbm/ft3 and cp = 0.091 Btu/lbm.F.
)()( 1212platein
energies etc. potential,
kinetic, internal,in Change
system
mass and work,heat,by
nsferenergy traNet
outin
TTmcuumUQ
EEE
The mass of each plate and the amount of heat transfer to each plate is
system
in
gensystemgen
in
entropyin
Change
system
generation
Entropy
gen
mass andheat by
ansferentropy trNet
outin
S
T
Q
SSS
T
Q
SSSS
bb
Plates
75F
8-104
8-146 The inner and outer surfaces of a brick wall are maintained at specified temperatures. The rate of entropy generation
within the wall is to be determined.
Assumptions Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified
values.
Analysis We take the wall to be the system, which is a closed
8-106
8-148E Steam is decelerated in a diffuser from a velocity of 900 ft/s to 100 ft/s. The mass flow rate of steam and the rate of
entropy generation are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes are negligible. 3
There are no work interactions.
Properties The properties of steam at the inlet and the exit of the diffuser are (Tables A-4E through A-6E)
/lbmft 16.316
RBtu/lbm 1.7141
vaporsat.
F240
RBtu/lbm 1.7406
Btu/lbm 1162.3
F240
psia 20
3
2
2
2
2
1
1
1
1
v
s
T
s
h
T
P
Analysis (a) The mass flow rate of the steam can be
determined from its definition to be
lbm/s 6.129 ft/s 100ft 1
/lbmft 16.316
11 2
3
22
2
VAm
v
(b) We take diffuser 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
outin
energies etc. potential,
kinetic, internal,in change of Rate
(steady) 0
sy stem
mass and work,heat,by
nsferenergy tranet of Rate
outin 0
EE
EEE
2
0)peW (since /2)+()2/(
2
1
2
2
12out
2
22out
2
11
VV
hhmQ
VhmQVhm
Steam
P1 =20 psia
T1 = 240F
V1 = 900 ft/s
T2 = 240F
Sat. vapor
V2 = 100 ft/s
A2 = 1 ft2
Q
·
8-108
8-150 Steam expands in a turbine from a specified state to another specified state. The rate of entropy generation during this
process is to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are
negligible.
the boundary. The energy balance for this steady-flow system can be
expressed in the rate form as
kinetic, internal,in change of Rate
(steady) 0
sy stem
nsferenergy tranet of Rate
outin 0
EEE
8.2 MW
8-112
kJ/K 0.1237
K 273)(230
kJ 222.6
KkJ/kg 1.5279kg 169.76
KkJ/kg 1.5346kg 84.88KkJ/kg 1.5279kg 84.88
)(
source
outsource,
1122sourcetankgen T
Q
smsmsmSSsmS eeee
8-113
8-154E An unknown mass of iron is dropped into water in an insulated tank while being stirred by a 200-W paddle wheel.
Thermal equilibrium is established after 10 min. The mass of the iron block and the entropy generated during this process
are to be determined.
Assumptions 1 Both the water and the iron block are incompressible substances with constant specific heats at room
temperature. 2 The system is stationary and thus the kinetic and potential energy changes are zero. 3 The system is well-
insulated and thus there is no heat transfer.
Analysis (a) We take the entire contents of the tank, water + iron block, as the system. This is a closed system since no mass
crosses the system boundary during the process. The energy balance on the system can be expressed as
UW
EEE
inpw,
energies etc. potential,
kinetic, internal,in Change
system
mass and work,heat,by
nsferenergy traNet
outin
WATER
70F
8-114
Review Problems
8-155E The source and sink temperatures and the thermal efficiency of a heat engine are given. The entropy change of the
Btu 0.65Btu) 1)(35.01()1( HL QQ
when the thermal efficiency is 35%. The entropy change of everything involved
in this process is then
TL
total
LH
SSS
The total entropy change is then
total
LH
SSS
HE
QH
QL
Wnet
TH
8-116
8-158E Air is compressed reversibly and isothermally in a piston-cylinder device. The amount of heat transfer is to be
determined.
Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values. 2 The
kinetic and potential energy changes are negligible,
0peke
. 3 Constant specific heats at room temperature can be
used for nitrogen.
Properties For air, R = 0.3704 psia∙ft3/lbmR (Table A-1E).
Analysis We take the air as the system. This is a closed system since no mass crosses the boundaries of the system. The
energy balance for this system can be expressed as
energies etc. potential,
kinetic, internal,in Change
system
mass and work,heat,by
nsferenergy traNet
outin
EEE
P
(psia)
8-117
8-159 Water vapor condensed in a piston-cylinder device in an isobaric and reversible process. It is to be determined if the
process described is possible.
Analysis We take the water as the system. This is a closed system since no mass crosses the boundaries of the system. The
energy balance for this system can be expressed as
)(
12outb,out
energies etc. p otential,
kinetic, internal,in Change
system
mass and work,heat,by
nsferenergy traNet
outin
uumUWQ
EEE
H2O
Q
8-119
8-161 A horizontal cylinder is separated into two compartments by a piston, one side containing nitrogen and the other side
containing helium. Heat is added to the nitrogen side. The final temperature of the helium, the final volume of the nitrogen,
the heat transferred to the nitrogen, and the entropy generation during this process are to be determined.
Assumptions 1 Kinetic and potential energy changes are negligible. 2 Nitrogen and helium are ideal gases with constant
K 321.7
1
(b) The initial and final volumes of the helium are
3
3
1
1
He,1 m 6406.0
kPa 95
K) 273K)(20/kgmkPa kg)(2.0769 (0.1
P
mRT
V
3
3
2
2
He,2 m 5568.0
kPa 120
K) K)(321.7/kgmkPa kg)(2.0769 (0.1
P
mRT
V
Then, the final volume of nitrogen becomes
3
K) 273K)(20/kgmkPa (0.2968
3
1
K 1.525
K)/kgmkPa kg)(0.2968 (0.2185
)m 8kPa)(0.283 (120
3
3
22
N2,2
mR
P
T
V
