978-0078027680 Chapter 6 Part 7

subject Type Homework Help
subject Pages 10
subject Words 5158
subject Authors John Cimbala, Robert Turner, Yunus Cengel

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6-119
6-156 Helium is compressed by a compressor. The power required 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
Analysis There is only one inlet and one exit, and thus
mmm 21
. We take the compressor 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
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
  
0)peke (since
21in
hmhmW
Compressor
400 kPa
200°C
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6-122
6-159 Problem 6-158 is reconsidered. The effect of the heat exchanger effectiveness on the money saved as the
effectiveness ranges from 20 percent to 90 percent is to be investigated, and the money saved is to be plotted against the
C = 4.18 [kJ/kg-C]
T_1 = 16 [C]
T_2 = 43 [C]
T_max = 39 [C]
T_min = T_1
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6-125
6-162 Problem 6-161 is reconsidered. The effects of turbine exit area and turbine exit pressure on the exit velocity
and power output of the turbine as the exit pressure varies from 10 kPa to 50 kPa (with the same quality), and the exit area
to varies from 1000 cm2 to 3000 cm2 is to be investigated. The exit velocity and the power output are to be plotted against
the exit pressure for the exit areas of 1000, 2000, and 3000 cm2.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
Vel[2]=m_dot*v[2]/(A[2]*convert(cm^2,m^2))
v[2]=volume(Fluid$, x=0.95, P=P[2]) "specific volume of steam at state 2"
T[2]=temperature(Fluid$, P=P[2], v=v[2]) "[C]" "not required, but good to know"
"[conservation of Energy for steady-flow:"
"Ein_dot - Eout_dot = DeltaE_dot" "For steady-flow, DeltaE_dot = 0"
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6-126
10 15 20 25 30 35 40 45 50
6000
8000
10000
12000
14000
16000
18000
P[2] [kPa]
Power [kW]
A2=3000 cm2
A2=2000 cm2
A2=1000 cm2
10 15 20 25 30 35 40 45 50
200
400
600
800
1000
1200
1400
1600
1800
2000
2200
P[2] [kPa]
Vel[2] [m/s]
A2=3000 cm2
A2=2000 cm2
A2=1000 cm2
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6-127
6-163 Air is preheated by the exhaust gases of a gas turbine in a regenerator. For a specified heat transfer rate, the exit
temperature of air and the mass flow rate of exhaust gases are 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. 3 There are no work interactions. 4 Heat loss from the regenerator to the surroundings is negligible and thus heat
transfer from the hot fluid is equal to the heat transfer to the cold fluid. 5 Exhaust gases can be treated as air. 6 Air is an
ideal gas with variable specific heats.
kJ/kg 02.607 K 600
44
hT
Analysis (a) We take the air side of the heat exchanger as the system,
which is a control volume since mass crosses the boundary. There is only
AIR
Exhaust
Gases
1
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6-128
6-164 Water is to be heated steadily from 20°C to 48°C by an electrical resistor inside an insulated pipe. The power rating
of the resistance heater and the average velocity of the water are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time at any point within the system and thus
 m E
CV CV
and  0 0
. 2 Water is an incompressible substance with constant specific heats. 3 The kinetic and potential
energy changes are negligible,
 ke pe 0
. 4 The pipe is insulated and thus the heat losses are negligible.
Properties The density and specific heat of water at room temperature are = 1000 kg/m3 and c = 4.18 kJ/kg·°C (Table A-
3).
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6-129
6-165 An insulated cylinder equipped with an external spring initially contains air. The tank is connected to a supply line,
and air is allowed to enter the cylinder until its volume doubles. The mass of the air that entered and the final temperature in
the cylinder are to be determined.
Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can
be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 The expansion process is
quasi-equilibrium. 3 Kinetic and potential energies are negligible. 4 The spring is a linear spring. 5 The device is insulated
and thus heat transfer is negligible. 6 Air is an ideal gas with constant specific heats.
Analysis We take the cylinder as the system, which is a control
Fspring
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6-132
6-168 Geothermal water flows through a flash chamber, a separator, and a turbine in a geothermal power plant. The
temperature of the steam after the flashing process and the power output from the turbine are to be determined for different
flash chamber exit pressures.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are
negligible. 3 The devices are insulated so that there are no heat losses to the surroundings. 4 Properties of steam are used
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6-133
6-169 A building is to be heated by a 30-kW electric resistance heater placed in a duct inside. The time it takes to raise the
interior temperature from 14C to 24C, and the average mass flow rate of air as it passes through the heater in the duct are
1.005 and c
v
= 0.718 kJ/kg·K (Table A-2).
Analysis (a) The total mass of air in the building is
 
 
m 400kPa 95
3
11
P
V
 
 
12avg,outinfan,ine,
TTmcQWWt
v
Solving for t gives
 
s 146
)kJ/s 450/60(kJ/s) 25.0()kJ/s 30(
C)1424)(CkJ/kg 0.718)(kg 461.3(
outinfan,ine,
12avg,
QWW
TTmc
t
v
(b) We now take the heating duct as the system, which is a control volume since mass crosses the boundary. There is only
one inlet and one exit, and thus
 
m m m
1 2
 
. The energy balance for this adiabatic steady-flow system can be expressed in
the rate form as
0
energies etc. potential,
kinetic, internal,in change of Rate
(steady ) 0
sy stem
mass and work,heat,by
nsferenergy tranet of Rate
outin
EEE
  
kg/s6.02
)C5)(CkJ/kg 1.005(
kJ/s 0.25)(30
infan,ine,
Tc
WW
m
p
6-170 6-174 Design and Essay Problems

250 W
450 kJ/min
T1

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