AREN 2110
FALL 2006
HOMEWORK ASSIGNMENTS 6, 7 and 8
SOLUTIONS
SOLUTIONS: HOMEWORK #6
Chapter 5 Problems
5-45 A number of brass balls are to be quenched in a water bath at a specified rate. The rate at which heat
needs to be removed from the water in order to keep its temperature constant is to be determined.
Assumptions 1 The thermal properties of the balls are constant. 2 The balls are at a uniform temperature
before and after quenching. 3 The changes in kinetic and potential energies are negligible.
Properties The density and specific heat of the brass balls are given to be ρ = 8522 kg/m3 and Cp = 0.385
kJ/kg.°C.
Analysis We take a single ball as the system. The energy balance for this closed
system can be expressed as
E − Eout
1in424
3
=
Net energy transfer
by heat, work, and mass
ΔEsystem
1
424
3
Brass balls, 120°C
Change in internal, kinetic,
potential, etc. energies
−Qout = ΔU ball = m (u2 − u1 )
Water bath, 5°C
Qout = mC (T1 − T2 )
The total amount of heat transfer from a ball is
m = ρV = ρ
Qout
πD 3
= (8522 kg / m 3 )
π (0.05 m) 3
= 0.558 kg
6
6
= mC(T1 − T2 ) = (0.558 kg)(0.385 kJ / kg. ° C)(120 − 74 )° C = 9.88 kJ / ball
Then the rate of heat transfer from the balls to the water becomes
Q&
= n& Q = (100 balls / min) × (9.88 kJ / ball) = 988 kJ / min
total
ball
ball
Therefore, heat must be removed from the water at a rate of 988 kJ/min in order to keep its temperature
constant at 50°C since energy input must be equal to energy output for a system whose energy level
remains constant. That is, Ein = Eout when ΔEsystem = 0 .
5-58C It is mostly converted to internal energy as shown by a rise in the fluid temperature.
5-59C The kinetic energy of a fluid increases at the expense of the internal energy as evidenced by a
decrease in the fluid temperature.
AREN 2110
FALL 2006
HOMEWORK ASSIGNMENTS 6, 7 and 8
SOLUTIONS
5-61 Air is accelerated in a nozzle from 30 m/s to 180 m/s. The mass flow rate, the exit temperature, and
the exit area of the nozzle are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas with
constant specific heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus heat
transfer is negligible. 5 There are no work interactions.
Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). The specific heat of air at the
anticipated average temperature of 450 K is Cp = 1.02 kJ/kg.°C (Table A-2).
&1 = m
&2 = m
& . Using the ideal gas relation, the
Analysis (a) There is only one inlet and one exit, and thus m
specific volume and the mass flow rate of air are determined to be
v1 =
RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K )(473 K )
=
= 0.4525 m 3 /kg
P1
300 kPa
m& =
1
1
A1 V1 =
(0.008m 2 )(30m/s) = 0.5304 kg/s
v1
0.4525m 3 /kg
(b) We take nozzle 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
E& − E& out
=
=0
ΔE& system Ê0 (steady)
1in
424
3
144
42444
3
Rate of net energy transfer
by heat, work, and mass
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
& ≅ Δpe ≅ 0)
m& (h1 + V12 / 2) = m& (h2 + V22 /2) (since Q& ≅ W
0 = h2 − h1 +
V22 − V12
V 2 − V12
⎯
⎯→ 0 = C p , ave (T2 − T1 ) + 2
2
2
Substituting,
0 = (1.02 kJ/kg ⋅ K )(T2 − 200 o C) +
It yields
(180 m/s) 2 − (30 m/s) 2
2
⎛ 1 kJ/kg
⎜
⎜ 1000 m 2 /s 2
⎝
⎞
⎟
⎟
⎠
T2 = 184.6°C
(c) The specific volume of air at the nozzle exit is
v2 =
m& =
RT2 (0.287 kPa ⋅ m 3 /kg ⋅ K )(184.6 + 273 K )
=
= 1.313 m 3 /kg
P2
100 kPa
1
1
A2 V2 ⎯
⎯→ 0.5304kg/s =
A2 (180m/s )
v2
1.313m 3 /kg
P1 = 300 kPa
T1 = 200°C
V1 = 30 m/s
A1 = 80 cm2
A2 = 0.00387 m2 = 38.7 cm2
5-77C Yes. Because energy (in the form of shaft work) is being added to the air.
AIR
P2 = 100 kPa
V2 = 180 m/s
AREN 2110
FALL 2006
HOMEWORK ASSIGNMENTS 6, 7 and 8
SOLUTIONS
5-79 Steam expands in a turbine. The change in kinetic energy, the power output, and the turbine inlet area
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 The device is adiabatic and thus heat transfer is negligible.
Properties From the steam tables (Tables A-4 through 6)
P1 = 10MPa ⎫⎪ v1 = 0.02975m 3 /kg
⎬
T1 = 450 o C ⎪⎭ h1 = 3240.9kJ/kg
P1 = 10 MPa
T1 = 450°C
V1 = 80 m/s
and
P2 = 10 kPa ⎫
⎬ h2 = h f + x 2 h fg = 191.83 + 0.92 × 2392.8 = 2393.2kJ/kg
x 2 = 0.92 ⎭
Analysis (a) The change in kinetic energy is determined from
V 2 − V12 (50m/s)2 − (80m/s) 2
Δke = 2
=
2
2
⎛ 1kJ/kg
⎜
⎜ 1000m 2 /s 2
⎝
=
Rate of net energy transfer
by heat, work, and mass
ΔE& system Ê0 (steady)
144
42444
3
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
& ≅ Δpe ≅ 0)
m& (h1 + V12 / 2) = W& out + m& (h2 + V22 /2) (since Q
⎛
V 2 − V12
W& out = − m& ⎜⎜ h2 − h1 + 2
2
⎝
⎞
⎟
⎟
⎠
Then the power output of the turbine is determined by substitution to be
W& out = −(12 kg/s)(2393.2 − 3240.9 − 1.95)kJ/kg = 10.2 MW
(c) The inlet area of the turbine is determined from the mass flow rate relation,
m& =
·
⎞
⎟ = −1.95kJ/kg
⎟
⎠
&1 = m
&2 = m
& . We take the
(b) There is only one inlet and one exit, and thus m
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
E& − E& out
1in
424
3
STEAM
·m = 12 kg/s
m& v
(12 kg/s )(0.02975 m 3 /kg )
1
A1 V1 ⎯
⎯→ A1 = 1 =
= 0.00446 m 2
v1
V1
80 m/s
W
P2 = 10 kPa
x2 = 0.92
V2 = 50 m/s
AREN 2110
FALL 2006
HOMEWORK ASSIGNMENTS 6, 7 and 8
SOLUTIONS
5-90 Helium is compressed by a compressor. For a mass flow rate of 90 kg/min, the power input 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 negligible. 3 Helium is an ideal gas with constant specific heats.
Properties The constant pressure specific heat of helium is Cp = 5.1926 kJ/kg·K (Table A-2a).
&1 = m
&2 = m
& . We take the compressor as the
Analysis There is only one inlet and one exit, and thus m
system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as
E& − E& out
1in
424
3
ΔE& system Ê0 (steady)
144
42444
3
=
Rate of net energy transfer
by heat, work, and mass
=0
P2 = 700 kPa
T2 = 430 K
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
& 1 = Q& out + mh
& 2 (since Δke ≅ Δpe ≅ 0)
W& in + mh
&
&
& p (T2 − T1 )
Win − Qout = m& (h2 − h1 ) = mC
He
·
m=90kg/mi
·
W
Thus,
W& in = Q& out + m& C p (T2 − T1 )
= (90/60kg/s)(20 kJ/kg) + (90/60kg/s)(5.1926kJ/kg ⋅ K)(430 − 310)K
= 965kW
P1 = 120 kPa
T1 = 310 K
5-104 A hot water stream is mixed with a cold water stream. For a specified mixture temperature, the mass
flow rate of cold water is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The mixing chamber is well-insulated so that heat loss
to the surroundings is negligible. 3 Changes in the kinetic and potential energies of fluid streams are
negligible. 4 Fluid properties are constant. 5 There are no work interactions.
Properties Noting that T < Tsat @ 250 kPa = 127.44°C, the water in all three streams exists as a compressed
liquid, which can be approximated as a saturated liquid at the given temperature. Thus,
h1 ≅ hf @ 80°C = 334.91 kJ/kg
h2 ≅ hf @ 20°C = 83.96 kJ/kg
h3 ≅ hf @ 42°C = 175.92 kJ/kg
Analysis We take the mixing chamber as the system, which is a control
volume. The mass and energy balances for this steady-flow system can be
expressed in the rate form as
Mass balance:
Ê0
m& in − m& out = ΔE& system
(steady)
=0
⎯
⎯→
m& 1 + m& 2 = m& 3
Energy balance:
E& − E& out
1in4243
=
Rate of net energy transfer
by heat, work, and mass
©0 (steady)
ΔE& system
1442443
=0
T1 = 80°C
· 1 = 0.5 kg/s
m
Rate of change in internal, kinetic,
potential, etc. energies
H2O
(P = 250 kPa)
T3 = 42°C
E& in = E& out
m& 1 h1 + m& 2 h2 = m& 3 h3 (since Q& = W& = Δke ≅ Δpe ≅ 0)
T2 = 20°C
m· 2
AREN 2110
FALL 2006
HOMEWORK ASSIGNMENTS 6, 7 and 8
SOLUTIONS
5-104 CONTINUED
& 2 gives
Combining the two relations and solving for m
m& 1h1 + m& 2 h2 = (m& 1 + m& 2 )h3
h1 − h3
&1
m
h3 − h2
Substituting, the mass flow rate of cold water stream is determined to be
&2 =
m
m& 2 =
(334.91 − 175.92)kJ/kg (0.5 kg/s ) = 0.864
(175.92 − 83.96)kJ/kg
kg/s
5-106 Feedwater is heated in a chamber by mixing it with superheated steam. If the mixture is saturated
liquid, the ratio of the mass flow rates of the feedwater and the superheated vapor 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. 3 There are no work interactions. 4 The device is adiabatic and thus heat
transfer is negligible.
Properties Noting that T < Tsat @ 800 kPa = 170.43°C, the cold water stream and the mixture exist as a
compressed liquid, which can be approximated as a saturated liquid at the given temperature. Thus,
h1 ≅ hf @ 50°C = 209.33 kJ/kg
h3 ≅ hf @ 800 kPa = 721.11 kJ/kg
and
P2 = 800kPa ⎫⎪
⎬ h2 = 2839.3kJ/kg
T2 = 200 o C ⎪⎭
Analysis We take the mixing chamber as the system, which is a control volume since mass crosses the
boundary. The mass and energy balances for this steady-flow system can be expressed in the rate form as
& in − m
& out = Δm
& system Ê0
m
Mass balance:
Energy balance:
E& − E& out
1in
424
3
=
Rate of net energy transfer
by heat, work, and mass
(steady)
=0
ΔE& system Ê0 (steady)
144
42444
3
→
& in = m
& out
m
=0
&1 + m
&2 = m
&3
→ m
T1 = 50°C
·1
m
Rate of change in internal, kinetic,
potential, etc. energies
H2O
(P = 800 kPa)
Sat. liquid
E& in = E& out
& 1h1 + m
& 2 h2 = m
& 3h3
m
Combining the two,
& 2 yields
Dividing by m
(since Q& ≅ W& ≅ Δke ≅ Δpe ≅ 0)
m& 1h1 + m& 2 h2 = (m& 1 + m& 2 )h3
y h1 + h2 = ( y + 1)h3
y=
Solving for y:
h3 − h2
h1 − h3
&1 / m
& 2 is the desired mass flow rate ratio. Substituting,
where y = m
y=
721.11 − 2839.3
= 4.14
209.33 − 721.11
T2 = 200°C
m· 2
AREN 2110
FALL 2006
HOMEWORK ASSIGNMENTS 6, 7 and 8
SOLUTIONS
SOLUTIONS: HOMEWORK #7
Chapter 6 Problems
6-2C Transferring 5 kWh of heat to an electric resistance wire and producing 5 kWh of electricity.
6-3C An electric resistance heater which consumes 5 kWh of electricity and supplies 6 kWh of heat to a
room.
6-10C Heat engines are cyclic devices that receive heat from a source, convert some of it to work, and
reject the rest to a sink.
6-12C No. Because 100% of the work can be converted to heat.
6-15C No. Such an engine violates the Kelvin-Planck statement of the second law of thermodynamics.
6-20 The power output and fuel consumption rate of a power plant are given.
The overall efficiency is to be determined.
Assumptions The plant operates steadily.
Properties The heating value of coal is given to be 30,000 kJ/kg.
60 t/h
Furnace
coal
HE
Analysis The rate of energy supply (in chemical form) to this power plant is
Q& H = m& coal u coal = (60,000 kg/h )(30,000 kJ/kg ) = 1.8 × 10 9 kJ/h = 500 MW
Then the overall efficiency of the plant becomes
η overall
W& net ,out 150 MW
=
=
= 0.300 = 30.0%
500 MW
Q& H
sink
6-40C The difference between the two devices is one of purpose. The purpose of a refrigerator is to
remove heat from a cold medium whereas the purpose of a heat pump is to supply heat to a warm medium.
6-47C No. The refrigerator captures energy from a cold medium and carries it to a warm medium. It does
not create it.
AREN 2110
FALL 2006
HOMEWORK ASSIGNMENTS 6, 7 and 8
SOLUTIONS
6-54 The COP and the power consumption of a refrigerator are given. The time it will take to cool 5
watermelons is to be determined.
Assumptions 1 The refrigerator operates steadily. 2 The heat gain of the refrigerator through its walls,
door, etc. is negligible. 3 The watermelons are the only items in the refrigerator to be cooled.
Properties The specific heat of watermelons is given to be C = 4.2 kJ/kg.°C.
Analysis The total amount of heat that needs to be removed from the watermelons is
(
)
Q L = (mCΔT )watermelons = 5 × (10kg ) 4.2kJ/kg⋅ o C (20 − 8)o C = 2520kJ
Kitchen air
The rate at which this refrigerator removes heat is
(
R
)
Q& L = (COPR ) W& net ,in = (2.5)(0.45kW ) = 1.125kW
COP = 2.5
cool space
That is, this refrigerator can remove 1.125 kJ of heat per second. Thus the time
required to remove 2520 kJ of heat is
Δt =
450 W
QL
2520kJ
=
= 2240s = 37.3min
&
Q L 1.125kJ/s
This answer is optimistic since the refrigerated space will gain some heat during this process from the
surrounding air, which will increase the work load. Thus, in reality, it will take longer to cool the
watermelons.
6-81 The sink temperature of a Carnot heat engine and the rates of heat supply and heat rejection are given.
The source temperature and the thermal efficiency of the engine are to be determined.
Assumptions The Carnot heat engine operates steadily.
⎛Q
Analysis (a) For reversible cyclic devices we have ⎜⎜ H
⎝ QL
Thus the temperature of the source TH must be
⎞
⎛T ⎞
⎟ =⎜ H ⎟
⎟
⎜
⎟
⎠rev ⎝ TL ⎠
⎛Q ⎞
⎛ 650 kJ ⎞
⎟⎟(290 K ) = 942.5 K
TH = ⎜⎜ H ⎟⎟ TL = ⎜⎜
⎝ 200 kJ ⎠
⎝ Q L ⎠ rev
(b) The thermal efficiency of a Carnot heat engine depends on the source
and the sink temperatures only, and is determined from
η th,C = 1 −
TL
290 K
=1−
= 0.69 or 69%
942.5 K
TH
source
650 kJ
HE
200 kJ
17°C
AREN 2110
FALL 2006
HOMEWORK ASSIGNMENTS 6, 7 and 8
SOLUTIONS
6-95 The refrigerated space and the environment temperatures for a refrigerator and the rate of heat
removal from the refrigerated space are given. The minimum power input required is to be determined.
Assumptions The refrigerator operates steadily.
Analysis The power input to a refrigerator will be a minimum when the refrigerator operates in a reversible
manner. The coefficient of performance of a reversible refrigerator depends on the temperature limits in the
cycle only, and is determined from
COPR ,rev =
1
1
=
= 8.03
(TH / TL ) − 1 (25 + 273K )/ (− 8 + 273K ) − 1
25°C
The power input to this refrigerator is determined from the definition of the
coefficient of performance of a refrigerator,
W& net , in, min =
R
Q& L
300 kJ / min
=
= 37.36 kJ / min = 0.623 kW
8.03
COPR, max
300 kJ/min
-8°C
6-103 A heat pump maintains a house at a specified temperature. The rate of heat loss of the house and the
power consumption of the heat pump are given. It is to be determined if this heat pump can do the job.
Assumptions The heat pump operates steadily.
Analysis The power input to a heat pump will be a minimum when the heat pump operates in a reversible
manner. The coefficient of performance of a reversible heat pump depends on the temperature limits in the
cycle only, and is determined from
COPHP ,rev =
1
1 − (T L / T H )
=
1
= 14.75
1 − (2 + 273K )/ (22 + 273K )
The required power input to this reversible heat pump is determined from the
definition of the coefficient of performance to be
W& net ,in,min =
Q& H
110,000kJ/h ⎛ 1h ⎞
⎜⎜
⎟⎟ = 2.07kW
=
14.75
COPHP
⎝ 3600s ⎠
This heat pump is powerful enough since 8 kW > 2.07 kW.
House
22°C
HP
110,000
kJ/h
8 kW
AREN 2110
FALL 2006
HOMEWORK ASSIGNMENTS 6, 7 and 8
SOLUTIONS
6-125 A Carnot heat engine drives a Carnot refrigerator that removes heat from a cold medium at a
specified rate. The rate of heat supply to the heat engine and the total rate of heat rejection to the
environment are to be determined.
Analysis (a) The coefficient of performance of the Carnot refrigerator is
1
1
COPR,C =
=
= 6.14
(TH / TL ) − 1 (300K )/ (258K ) − 1
Then power input to the refrigerator becomes
Q& L
400 kJ / min
W& net , in =
=
= 65.1 kJ / min
6.14
COPR, C
&
which is equal to the power output of the heat engine, W
net , out .
The thermal efficiency of the Carnot heat engine is determined from
η th, C = 1 −
750 K
-15°C
·
400 kJ/min
QH, HE
HE
R
·
QL, HE
·
QH, R
300 K
TL
300 K
= 1−
= 0.60
750 K
TH
Then the rate of heat input to this heat engine is determined from the definition of thermal efficiency to be
W& net , out 65.1 kJ / min
Q& H , HE =
=
= 108.5 kJ / min
0.60
η th, HE
(b) The total rate of heat rejection to the ambient air is the sum of the heat rejected by the heat engine
&
&
(Q
L , HE ) and the heat discarded by the refrigerator ( Q H , R ),
Q& L , HE = Q& H , HE − W& net ,out = 108.5 − 65.1 = 43.4kJ/min
Q& H , R = Q& L , R + W& net ,in = 400 + 65.1 = 465.1kJ/min
and
Q& Ambient = Q& L, HE + Q& H , R = 43.4 + 465.1 = 508.5 kJ / min
AREN 2110
FALL 2006
HOMEWORK ASSIGNMENTS 6, 7 and 8
SOLUTIONS
Chapter 7 problems
7-10C No. An isothermal process can be irreversible. Example: A system that involves paddle-wheel work
while losing an equivalent amount of heat.
7-17C Increase.
7-21C Yes. This will happen when the system is losing heat, and the decrease in entropy as a result of this
heat loss is equal to the increase in entropy as a result of irreversibilities.
7-26 Heat is transferred isothermally from a source to the working fluid of a Carnot engine. The entropy
change of the working fluid, the entropy change of the source, and the total entropy change during this
process are to be determined.
Analysis (a) This is a reversible isothermal process, and the entropy change during such a process is given
by
Q
ΔS =
T
Noting that heat transferred from the source is equal to the heat transferred to the working fluid, the entropy
changes of the fluid and of the source become
ΔS fluid =
(b)
Q fluid
ΔSsource =
T fluid
=
Qin, fluid
T fluid
=
900 kJ
= 1.337 kJ / K
673 K
Q
Qsource
900 kJ
= − out, source = −
= −1.337 kJ / K
Tsource
Tsource
673 K
(c) Thus the total entropy change of the process is
Source
400°C
900 kJ
S gen = ΔS total = ΔS fluid + ΔS source = 1.337 − 1.337 = 0
7-34 An insulated rigid tank contains a saturated liquid-vapor mixture of water at a specified pressure. An
electric heater inside is turned on and kept on until all the liquid vaporized. The entropy change of the
water during this process is to be determined.
Analysis From the steam tables (Tables A-4 through A-6)
P1 = 100kPa ⎫ v1 = v f + x1v fg = 0.001 + (0.25)(1.694 − 0.001) = 0.4243m 3 /kg
⎬
x1 = 0.25 ⎭ s1 = s f + x1 s fg = 1.3026 + (0.25)(6.0568) = 2.8168kJ/kg ⋅ K
v 2 = v1
⎫
⎬ s 2 = 6.8649kJ/kg ⋅ K
sat.vapor ⎭
Then the entropy change of the steam becomes
ΔS = m(s 2 − s1 ) = ( 4 kg )(6.8649 − 2.8168 ) kJ/kg ⋅ K = 16.19 kJ/K
H2O
4 kg
100 kPa
We
AREN 2110
FALL 2006
HOMEWORK ASSIGNMENTS 6, 7 and 8
SOLUTIONS
7-51 An aluminum block is brought into contact with an iron block in an insulated enclosure. The final
equilibrium temperature and the total entropy change for this process are to be determined.
Assumptions 1 Both the aluminum and the iron block are incompressible substances with constant specific
heats. 2 The system is stationary and thus the kinetic and potential energies are negligible. 3 The system is
well-insulated and thus there is no heat transfer.
Properties The specific heat of aluminum at the anticipated average temperature of 450 K is Cp = 0.973
kJ/kg.°C. The specific heat of iron at room temperature (the only value available in the tables) is Cp = 0.45
kJ/kg.°C (Table A-3).
Analysis We take the iron+aluminum blocks as the system, which is a closed system. The energy balance
for this system can be expressed as
E −E
1in424out
3
=
Net energy transfer
by heat, work, and mass
ΔE system
1
424
3
Change in internal, kinetic,
potential, etc. energies
0 = ΔU
Iron
20 kg
100°C
Aluminum
20 kg
200°C
or,
ΔU alum + ΔU iron = 0
[mC (T2 − T1 )] alum + [mC (T2 − T1 )] iron = 0
Substituting,
(20kg)(0.45kJ/kg ⋅ K )(T2 − 100 o C) + (20kg )(0.973kJ/kg ⋅ K )(T2 − 200 o C) = 0
T2 = 168.4 o C = 441.4K
The total entropy change for this process is determined from
⎛T ⎞
⎛ 441.4K ⎞
⎟⎟ = 1.515kJ/K
ΔS iron = mC ave ln⎜⎜ 2 ⎟⎟ = (20kg )(0.45kJ/kg ⋅ K )ln⎜⎜
⎝ 373K ⎠
⎝ T1 ⎠
⎛T ⎞
⎛ 441.4K ⎞
⎟⎟ = −1.346kJ/K
ΔS alum = mC ave ln⎜⎜ 2 ⎟⎟ = (20kg )(0.973kJ/kg ⋅ K )ln⎜⎜
⎝ 473K ⎠
⎝ T1 ⎠
Thus,
ΔStotal = ΔSiron + ΔSalum = 1515
.
− 1346
.
= 0.169 kJ / K
AREN 2110
FALL 2006
HOMEWORK ASSIGNMENTS 6, 7 and 8
SOLUTIONS
7-62 An insulated tank contains CO2 gas at a specified pressure and volume. A paddle-wheel in the tank
stirs the gas, and the pressure and temperature of CO2 rises. The entropy change of CO2 during this process
is to be determined using constant specific heats.
Assumptions At specified conditions, CO2 can be treated as an ideal gas with constant specific heats at
room temperature.
Properties The specific heat of CO2 is Cv = 0.657 kJ/kg.K (Table A-2).
Analysis Using the ideal gas relation, the entropy change is determined to be
P2V P1V
T
P
120kPa
=
⎯
⎯→ 2 = 2 =
= 1.2
T2
T1
T1
P1 100kPa
Thus,
⎛
T
V
ΔS = m(s 2 − s1 ) = m⎜ C v , ave ln 2 + R ln 2
⎜
T1
V1
⎝
= (2.7 kg )(0.657 kJ/kg ⋅ K )ln (1.2 )
= 0.323 kJ/K
©0
⎞
⎟ = mC v , ave ln T2
⎟
T1
⎠
CO2
1.5 m3
100 kPa
1.2 kg
AREN 2110
FALL 2006
HOMEWORK ASSIGNMENTS 6, 7 and 8
SOLUTIONS
7-110 Steam is condensed by cooling water in the condenser of a power plant. The rate of condensation of
steam and the rate of entropy generation are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to
the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the
cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties
are constant.
Properties The enthalpy and entropy of vaporization of water at 50°C are hfg =2382.7 kJ/kg and sfg= 7.3725
kJ/kg.K (Table A-4). The specific heat of water at room temperature is Cp = 4.18 kJ/kg.°C (Table A-3).
Analysis (a) We take the cold water tubes as the system, which is a control
volume. The energy balance for this steady-flow system can be expressed in
the rate form as
E& − E& out
1in424
3
=
Rate of net energy transfer
by heat, work, and mass
ΔE& system Ê0 (steady)
144
42444
3
Steam
50°C
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
& 1 = mh
& 2 (since Δke ≅ Δpe ≅ 0)
Q& in + mh
&
& p (T2 − T1 )
Qin = mC
18°C
Water
Then the heat transfer rate to the cooling water in the condenser becomes
Q& = [m& C p (Tout − Tin )]cooling water
50°C
= (101 kg/s)(4.18 kJ/kg.°C)(27°C − 18°C) = 3800 kJ/s
The rate of condensation of steam is determined to be
Q&
3800kJ/s
Q& = (m& h fg ) steam ⎯
⎯→ m& steam =
=
= 1.595 kg/s
h fg 2382.7 kJ/kg
(b) The rate of entropy generation within the condenser during this process can be determined by applying
the rate form of the entropy balance on the entire condenser. Noting that the condenser is well-insulated
and thus heat transfer is negligible, the entropy balance for this steady-flow system can be expressed as
S& in − S& out
1424
3
+
Rate of net entropy transfer
by heat and mass
S& gen
{
Rate of entropy
generation
= ΔS& system ©0 (steady)
1442443
Rate of change
of entropy
m& 1 s1 + m& 3 s 3 − m& 2 s 2 − m& 4 s 4 + S& gen = 0 (since Q = 0)
m& water s1 + m& steam s 3 − m& water s 2 − m& steam s 4 + S& gen = 0
S& gen = m& water ( s 2 − s1 ) + m& steam ( s 4 − s 3 )
Noting that water is an incompressible substance and steam changes from saturated vapor to saturated
liquid, the rate of entropy generation is determined to be
T
T
S& gen = m& water C p ln 2 + m& steam ( s f − s g ) = m& water C p ln 2 − m& steam s fg
T1
T1
= (101 kg/s)(4.18 kJ/kg.K)ln
27 + 273
− (1.595 kg/s)(7.3725 kJ/kg.K) = 1.100 kW/K
18 + 273
AREN 2110
FALL 2006
HOMEWORK ASSIGNMENTS 6, 7 and 8
SOLUTIONS
7-132 Liquid water is heated in a chamber by mixing it with superheated steam. For a specified mixing
temperature, the mass flow rate of the 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 Kinetic and potential
energy changes are negligible. 3 There are no work interactions.
Properties Noting that T < Tsat @ 200 kPa = 120.23°C, the cold water and the exit mixture streams exist as a
compressed liquid, which can be approximated as a saturated liquid at the given temperature. From tables,
600 kJ/min
P1 = 200kPa ⎫⎪ h1 ≅ h f @ 20o C = 83.96kJ/kg
⎬s ≅ s
o
=
⋅
0.2966
kJ/kg
K
T1 = 20 C ⎪⎭ 1
f @ 20o C
P2 = 200kPa ⎫⎪ h2 = 3071.8kJ/kg
⎬
T2 = 300 o C ⎪⎭ s 2 = 7.8926kJ/kg ⋅ K
1
20°C
2.5 kg/s
MIXING
CHAMBER
60°C
P3 = 200kPa ⎪⎫ h3 ≅ h f @60o C = 251.13kJ/kg
200 kPa
300°C
⎬s ≅ s
2
o
=
0.8312
kJ/kg
⋅
K
o
T3 = 60 C ⎪⎭ 3
f @60 C
Analysis (a) We take the mixing chamber as the system, which is a control volume. The mass and energy
balances for this steady-flow system can be expressed in the rate form as
Mass balance:
Ê0
m& in − m& out = ΔE& system
(steady)
=0
⎯
⎯→
m& 1 + m& 2 = m& 3
Energy balance:
E& − E& out
1in4243
Rate of net energy transfer
by heat, work, and mass
©0 (steady)
ΔE& system
1442443
=
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
m& 1 h1 + m& 2 h2 = Q& out + m& 3 h3
Combining the two relations gives
Q& out = m& 1 h1 + m& 2 h2 − (m& 1 + m& 2 )h3 = m& 1 (h1 − h3 ) + m& 2 (h2 − h3 )
& 2 and substituting, the mass flow rate of the superheated steam is determined to be
Solving for m
Q& − m& 1 (h1 − h3 ) (600/60kJ/s) − (2.5kg/s )(83.96 − 251.13)kJ/kg
=
= 0.152kg/s
m& 2 = out
(3071.8 − 251.13)kJ/kg
h2 − h3
Also,
m& 3 = m& 1 + m& 2 = 2.5 + 0.152 = 2.652 kg / s
(b) The rate of total entropy generation during this process is determined by applying the entropy balance
on an extended system that includes the mixing chamber and its immediate surroundings so that the
boundary temperature of the extended system is 25°C at all times. It gives
S& in − S& out
+ S& gen
= ΔS& system ©0 = 0
1424
3
{
14243
Rate of net entropy transfer
by heat and mass
m& 1 s1 + m& 2 s 2 − m& 3 s 3 −
Rate of entropy
generation
Rate of change
of entropy
Q& out
+ S& gen = 0
Tb, surr
Substituting, the rate of entropy generation during this process is determined to be
Q&
S& gen = m& 3 s 3 − m& 2 s 2 − m& 1 s1 + out
Tb, surr
= (2.652kg/s )(0.8312 kJ/kg ⋅ K ) − (0.152kg/s )(7.8926 kJ/kg ⋅ K )
(600 / 60 kJ/s)
− (2.5kg/s )(0.2966 kJ/kg ⋅ K ) +
298 K
= 0.297kW/K
3
AREN 2110
FALL 2006
HOMEWORK ASSIGNMENTS 6, 7 and 8
SOLUTIONS
7-143 Air is compressed in a two-stage ideal compressor with intercooling. For a specified mass flow rate
of air, the power input to the compressor is to be determined, and it is to be compared to the power input to
a single-stage compressor. √
Assumptions 1 The compressor operates steadily. 2 Kinetic and potential energies are negligible. 3 The
compression process is reversible adiabatic, and thus isentropic. 4 Air is an ideal gas with constant specific
heats at room temperature.
Properties The gas constant of air is R =0.287 kPa.m3/kg.K (Table A-1). The specific heat ratio of air is k =
1.4 (Table A-2)
Analysis The intermediate pressure between the two stages is
100 kPa
27°C
(100kPa )(900kPa ) = 300kPa
Px = P1 P2 =
900 kPa
The compressor work across each stage is the same, thus total
compressor work is twice the compression work for a single stage:
(
)
(
)
kRT1
(Px P1 )(k −1) / k − 1
k −1
0.4/1.4
⎞
1.4)(0.287kJ/kg ⋅ K )(300K ) ⎛⎜ ⎛ 300kPa ⎞
(
⎟
⎜
⎟
=2
−
1
⎜ ⎜ 100kPa ⎟
⎟
1.4 − 1
⎝
⎠
⎝
⎠
= 222.2kJ/kg
wcomp,in = (2 ) wcomp,in, I = 2
and
W
Stage I
Stage II
27°C
Heat
W& in = m& wcomp ,in = (0.02 kg/s )(222.2 kJ/kg ) = 4.44kW
The work input to a single-stage compressor operating between the same pressure limits would be
wcomp,in =
and
0.4/1.4
⎛⎛
⎞
⎞
kRT1
(P2 P1 )(k −1) / k − 1 = (1.4)(0.287kJ/kg ⋅ K )(300K ) ⎜⎜ ⎜⎜ 900kPa ⎟⎟
− 1⎟ = 263.2kJ/kg
⎟
k −1
1.4 − 1
100kPa ⎠
⎝⎝
⎠
(
)
W& in = m& wcomp ,in = (0.02 kg/s )(263.2 kJ/kg ) = 5.26 kW
Discussion Note that the power consumption of the compressor decreases significantly by using 2-stage
compression with intercooling.
AREN 2110
FALL 2006
HOMEWORK ASSIGNMENTS 6, 7 and 8
SOLUTIONS
SOLUTIONS: HOMEWORK #8
Chapter 8 Problems
8-95 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The
thermal efficiency, the pressure at the turbine inlet, and the net work output are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) The thermal efficiency is determined from
η th, C = 1 −
TL
60 + 273 K
= 46.5%
= 1−
TH
350 + 273 K
(b) Note that s2 = s3 = sf + x3sfg
1
2
4
3
350°C
= 0.8312 + 0.891 × 7.0784 = 7.138 kJ/kg·K
Thus ,
T2 = 350 °C
⎫
⎬ P2 = 1.40 MPa
s 2 = 7.138 kJ/kg ⋅ K ⎭
60°C
(c) The net work can be determined by calculating the enclosed area on the T-s diagram,
s 4 = s f + x 4 s fg = 0.8312 + (0.1)(7.0784 ) = 1.539 kJ/kg ⋅ K
Thus,
wnet = Area = (TH − TL )(s3 − s 4 ) = (350 − 60)(7.138 − 1.539 ) = 1624 kJ/kg
Drawing from solution (1-2-3-4) is incorrect. Yellow-shaded area is correct. Water is
compressed liquid at the boiler inlet, not a mixture, since s1 < sg at 350 oC.
Numerical answers are okay.
8-97C Heat rejected decreases; everything else increases.
8-103C Yes, because the saturation temperature of steam at 10 kPa is 45.81°C, which is much higher than
the temperature of the cooling water.
s
AREN 2110
FALL 2006
HOMEWORK ASSIGNMENTS 6, 7 and 8
SOLUTIONS
8-105 A steam power plant that operates on a simple ideal Rankine cycle is considered. The quality of the
steam at the turbine exit, the thermal efficiency of the cycle, and the mass flow rate of the steam are to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
h1 = h f @ 10 kPa = 191.83 kJ/kg
v1 = v f @ 10 kPa = 0.00101 m 3 /kg
w p ,in = v1 (P2 − P1 )
(
3
)
⎛ 1 kJ ⎞
⎟
= 0.00101 m 3 /kg (10,000 − 10 kPa )⎜⎜
1 kPa ⋅ m 3 ⎟⎠
⎝
= 10.09kJ/kg
h2 = h1 + w p ,in = 191.83 + 10.09 = 201.92 kJ/kg
10 MPa
qin
2
10 kPa
1
P3 = 10 MPa ⎫ h3 = 3373.7 kJ/kg
⎬
T3 = 500 °C ⎭ s 3 = 6.5966 kJ/kg ⋅ K
s4 − s f
P4 = 10 kPa ⎫
6.5966 − 0.6493
=
= 0.793
⎬ x4 =
s 4 = s3
s fg
7.5009
⎭
h4 = h f + x 4 h fg = 191.83 + (0.793)(2392.8 ) = 2089.3 kJ/kg
(b)
qin = h3 − h2 = 3373.7 − 201.92 = 3171.78 kJ / kg
qout = h4 − h1 = 2089.3 − 191.83 = 1897.47 kJ / kg
wnet = qin − qout = 3171.78 − 1897.47 = 1274.31 kJ / kg
and
η th =
(c)
m& =
w net 1274.31kJ/kg
=
= 40.2%
q in
3171.78kJ/kg
210,000 kJ/s
W& net
=
= 165 kg/s
wnet 1274.31kJ/kg
qout
4
s
AREN 2110
FALL 2006
HOMEWORK ASSIGNMENTS 6, 7 and 8
SOLUTIONS
8-111 A steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits.
The thermal efficiency of the cycle, the mass flow rate of the steam, and the temperature rise of the cooling
water are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
h1 = h f @ 10 kPa = 191.83 kJ/kg
v1 = v f @ 10 kPa = 0.00101 m 3 /kg
w p ,in = v1 (P2 − P1 )
(
3
)
⎛ 1 kJ ⎞
⎟
= 0.00101 m 3 /kg (7,000 − 10 kPa )⎜⎜
3 ⎟
⎝ 1 kPa ⋅ m ⎠
= 7.06 kJ/kg
7 MPa
qin
2
10 kPa
1
h2 = h1 + w p ,in = 191.83 + 7.06 = 198.89 kJ/kg
qout
s
P3 = 7 MPa ⎫ h3 = 3410.5 kJ/kg
⎬
T3 = 500°C ⎭ s 3 = 6.7975 kJ/kg ⋅ K
s4 − s f
P4 = 10 kPa ⎫
6.7975 − 0.6493
=
= 0.820
⎬ x4 =
s 4 = s3
7.5009
s fg
⎭
h4 = h f + x 4 h fg = 191.83 + (0.820)(2392.8) = 2153.93 kJ/kg
Thus,
qin = h3 − h2 = 3410.5 − 198.89 = 3211.61 kJ / kg
. kJ / kg
qout = h4 − h1 = 2153.93 − 191.83 = 196210
. = 1249.51 kJ / kg
wnet = qin − qout = 3211.61 − 196210
and
η th =
(b)
m& =
4
w net 1249.51kJ/kg
=
= 38.9%
q in
3211.61kJ/kg
45,000 kJ/s
W& net
=
= 36.0kg/s
w net 1249.51kJ/kg
(c) The rate of heat rejection to the cooling water and its temperature rise are
Q& out = m& qout = (36.0 kg/s)(1962.1 kJ/kg ) = 70,636 kJ/s
Q& out
70,636 kJ/s
ΔTcooling water =
=
= 8.45°C
(m& C )coolingwater (2000 kg/s) 4.18 kJ/kg⋅o C
(
)
AREN 2110
FALL 2006
HOMEWORK ASSIGNMENTS 6, 7 and 8
SOLUTIONS
8-121 A steam power plant that operates on an ideal reheat Rankine cycle between the specified pressure
limits is considered. The pressure at which reheating takes place, the total rate of heat input in the boiler,
and the thermal efficiency of the cycle are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
h1 = hsat@ 10 kPa = 191.83 kJ/kg
v1 = v sat@ 10 kPa = 0.00101 m 3 /kg
w p ,in = v1 (P2 − P1 )
(
3
9 MPa
)
⎛ 1 kJ ⎞
⎟
= 0.00101 m 3 /kg (9 ,000 − 10 kPa )⎜
⎜ 1 kPa ⋅ m 3 ⎟
⎠
⎝
= 9.08 kJ/kg
h2 = h1 + w p ,in = 191.83 + 9.08 = 200.91 kJ/kg
4
2
10 kPa
1
P3 = 9 MPa ⎫ h3 = 3386.1 kJ/kg
⎬
T3 = 500°C ⎭ s 3 = 6.6576 kJ/kg ⋅ K
P6 = 10 kPa ⎫ h6 = h f + x 6 h fg = 191.83 + (0.90 )(2392.8) = 2345.4 kJ/kg
⎬
s 6 = s5
⎭ s 6 = s f + x 6 s fg = 0.6493 + (0.90 )(7.5009 ) = 7.4001 kJ/kg ⋅ K
T5 = 500°C ⎫ P5 = 2.146 MPa (the reheat pressure )
⎬
s5 = s 6
⎭ h5 = 3466.0 kJ/kg
P4 = 2.146 MPa ⎫
⎬ h4 = 2979.5 kJ/kg
s 4 = s3
⎭
(b) The rate of heat supply is
Q& in = m& [(h3 − h2 ) + (h5 − h4 )]
= (25 kJ/s )(3386.1 − 200.91 + 3466 − 2979.5)kJ/kg = 91,792 kJ/s
(c) The thermal efficiency is determined from
Q& out = m& (h6 − h1 ) = (25 kJ/s)(2345.4 − 191.83)kJ/kg = 53,839 kJ/s
Thus,
ηth = 1 −
5
Q& out
53,839 kJ/s
= 1−
= 41.3%
91,792 kJ/s
Q& in
6
s
AREN 2110
FALL 2006
HOMEWORK ASSIGNMENTS 6, 7 and 8
SOLUTIONS
8-123 A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered.
The coefficient of performance, the amount of heat absorbed from the refrigerated space, and the net work
input are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) Noting that TH = 30°C = 303 K and TL = Tsat @ 120 kPa = -22.36°C = 250.6 K, the COP of this
Carnot refrigerator is determined from
COPR,C =
1
1
=
= 4.78
TH / TL − 1 (303 K ) / (250.6 K ) − 1
(b) From the refrigerant tables (Table A-11),
4
h3 = hg @ 30°C = 263.50 kJ / kg
h4 = h f @ 30°C = 91.49 kJ / kg
QH
3
30°C
Thus,
120 kPa
q H = h3 − h4 = 263.50 − 91.49 = 172.01 kJ/kg
1 Q
L
2
s
and
⎛ 250.6 K ⎞
q H TH
T
⎟⎟(172.01 kJ/kg ) = 142.3 kJ/kg
=
⎯
⎯→ q L = L q H = ⎜⎜
q L TL
TH
⎝ 303 K ⎠
(c) The net work input is determined from
wnet = q H − q L = 172 .01 − 142.3 = 29.71 kJ/kg
8-132 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is
considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the
rate of heat rejection to the environment, and the COP are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the
refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser
as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13),
P1 = 120 kPa ⎫ h1 = h g @ 120 kPa = 233.86 kJ/kg
⎬s = s
sat.vapor
g @ 120 kPa = 0.9354 kJ/kg ⋅ K
⎭ 1
P2 = 0.7 MPa ⎫
⎬ h2 = 270.22 kJ/kg (T2 = 34.6°C )
s 2 = s1
⎭
P3 = 0.7 MPa ⎫
⎬ h3 = h f
sat.liquid
⎭
@ 0.7 MPa
·
QH
3
2
0.7 MPa
·
Win
= 86.78 kJ/kg
h4 ≅ h3 = 86.78 kJ/kg (throttling )
Then the rate of heat removal from the refrigerated space and the power
input to the compressor are determined from
Q& L = m& (h1 − h4 ) = (0.05 kg/s)(233.86 − 86.78) kJ/kg = 7.35 kW
and
W& in = m& (h2 − h1 ) = (0.05 kg/s)(270.22 − 233.86 ) kJ/kg = 1.82 kW
0.12 MPa
4s
4
·
QL
1
s
AREN 2110
FALL 2006
HOMEWORK ASSIGNMENTS 6, 7 and 8
SOLUTIONS
8-132 CONTINUED
(b) The rate of heat rejection to the environment is determined from
Q& H = Q& L + W& in = 7.35 + 1.82 = 9.17 kW
(c) The COP of the refrigerator is determined from its definition,
COPR =
Q& L
7.35 kW
=
= 4.04
W& in 1.82 kW
8-146 A heat pump with refrigerant-134a as the working fluid heats a house by using underground water as
the heat source. The power input to the heat pump, the rate of heat absorption from the water, and the
increase in electric power input if an electric resistance heater is used instead of a heat pump are to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the refrigerant tables (Tables A-12 and A-13),
P1 = 280 kPa ⎫
⎬h1 = 247.64 kJ/kg
T1 = 0°C
⎭
P2 = 1.0 MPa ⎫
h = 291.36 kJ/kg
T2 = 60°C ⎬⎭ 2
P3 = 1.0 MPa ⎫
h ≅ hf
T3 = 30°C ⎬⎭ 3
@ 30°C
= 91.49 kJ/kg
30°C
3
The mass flow rate of the refrigerant is
Q&
Q& H
60,000/3,600 kJ/s
m& R = H =
=
= 0.0834 kg/s
qH
h2 − h3 (291.36 − 91.49 ) kJ/kg
Then the power input to the compressor becomes
W&in = m& (h2 − h1 ) = (0.0834 kg/s)(291.36 − 247.64) kJ/kg = 3.65 kW
(b) The rate of hat absorption from the water is
Q& L = m& (h1 − h4 ) = (0.0834 kg/s)(247.64 − 91.49) kJ/kg = 13.02 kW
W&e = Q& H = 60,000 / 3600kJ/s = 16.67kW
Thus,
W&increase = W&e − W&in = 16.67 − 3.65 = 13.02kW
60°C
·
QH
h4 ≅ h3 = 91.49 kJ/kg (throttling )
(c) The electrical power required without the heat pump is
2
House
·
Win
1 MPa
0.28 MPa
4
·
QL
1
0°C
Water, 8°C
s
AREN 2110
FALL 2006
HOMEWORK ASSIGNMENTS 6, 7 and 8
SOLUTIONS
8-192 A heat pump water heater has a COP of 2.2 and consumes 2 kW when running. It is to be
determined if this heat pump can be used to meet the cooling needs of a room by absorbing heat from it.
Assumptions The COP of the heat pump remains constant whether heat is absorbed from the outdoor air or
room air.
Analysis The COP of the heat pump is given to be 2.2. Then the COP of the air-conditioning system
becomes
COPair-cond = COPheat pump − 1 = 2.2 − 1 = 1.2
Then the rate of cooling (heat absorption from the air) becomes
Q& cooling = COPair-cond W& in = (1.2)(2 kW) = 2.4 kW = 8640 kJ / h
since 1 kW = 3600 kJ/h. We conclude that this heat pump can meet the cooling needs of the room since its
cooling rate is greater than the rate of heat gain of the room.