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PROBLEM 3.117
As shown in Fig. P3.117, 20 ft3 of air at T1 = 600
oR, 100 lbf/in.
2 undergoes a polytropic
expansion to a final pressure of 51.4 lbf/in.2 The process follows pV
1.2 = constant. The work is
W = 194.34 Btu. Assuming ideal gas behavior for the air, and neglecting kinetic and potential
energy effects, determine
(a) the mass of air, in lb, and the final temperature, in oR.
(b) the heat transfer, in Btu.
KNOWN: Air undergoes a polytropic process in a piston-cylinder assembly. The work is
known.
FIND: Determine the mass of air, the final temperature, and the heat transfer.
SCHEMATIC AND GIVEN DATA:
ENGINEERING MODEL: 1. The air is a closed system.
2. Volume change is the only work mode. 3. The process is
polytropic, with pV1.2
= constant and W = 194.34 Btu.
4. Kinetic and potential energy effects can be neglected.
ANALYSIS: (a) The mass is determined using the ideal gas equation of state.
m =
=
= 9.00 lb
To get the final temperature, we use the polytropic process, pV1.2
= constant, to evaluate V2 as
follows.
V2 =
=
(20 ft
3) = 34.83 ft
3
Now
Air
W = 194.34 Btu
Q T1 = 600
oR
p1 = 100 lbf/in.2
V1 = 20 ft3
P2 = 51.4 lbf/in.2
pV1.2
= constant
p
v
100
51.4 2
1 T1
T2
pv1.2 = constant
.
.
PROBLEM 3.117 (CONTINUED)
T2 =
=
= 537
oR
Alternative solution for T2
The work for the polytropic process can be evaluated using W =
. For the process pV
1.2 =
constant, and incorporating the ideal gas equation of state, we get
W =
=
Solving for T2 and inserting values
T2 =
+ T1 =
+ (600
oR) = 537
oR
(b) Applying the energy balance; ΔKE + ΔPE + ΔU = Q – W. With ΔU = m(u2 – u1), we get
Q = m(u2 – u1) + W
From Table A-22E: u(600oR) = 102.34 Btu/lb and u(537
oR) = 91.53 Btu/lb. Thus,
Q = (9.00 lb)(91.53 – 102.34)Btu/lb + (194.34 Btu) = 97.05 Btu (in)
1
Corrected December, 2011
4.59 Refrigerant 134a enters an air conditioner compressor at 4 bar, 20°C, and is
compressed at steady state to 12 bar, 80°C. The volumetric flow rate of the refrigerant
entering is 4 m3/min. The power input to the compressor is 60 kJ per kg of refrigerant
flowing. Neglecting kinetic and potential energy effects, determine the heat transfer rate,
in kW.
KNOWN: Refrigerant 134a with known inlet and exit conditions flows through a
compressor operating at steady state.
FIND: Determine the heat transfer rate, in kW.
SCHEMATIC AND GIVEN DATA:
ENGINEERING MODEL:
(1) The control volume shown in the accompanying schematic operates at steady state.
(2) Potential and kinetic energy changes from inlet to exit can be neglected.
ANALYSIS:
To determine the heat transfer rate, begin with the steady state mass and energy balances.
Simplify based on assumptions and solve for cvQ .
Refrigerant 134a
p1 = 4 bar
T1 = 20oC
A1V1 = 4 m3/min
2
1
p2 = 12 bar
T2 = 80oC
Compressorm
W
cv
= –60 kJ/kg
cvQ = ?
21
2
2
2
121cvcv
21
2
VV0 zzghhmWQ
mmm
2
(1)
To obtain m , in kg/s, fix state 1 by referencing Table A-11 at 4 bar. T1 > Tsat and
therefore a superheated condition exists at state 1. From Table A-12 at T1 and p1: v1 =
0.05397 m3/kg.
From Table A-12: h1 = 262.96 kJ/kg and h2 = 310.24 kJ/kg. To obtain cvQ , in kW,
substitute into Eq. (1).
1. The negative sign indicates that there is energy rejected from the system by heat
transfer.
s
kg1.235
60s
min 1
kg
m0.05397
min
m4
AV3
3
1
1 v
m
12
cv
12cvcv hhm
WmhhmWQ
1# kW 71.15
s
kJ1
kW 1
kg
kJ262.96310.24
kg
kJ60
s
kg1.235cv
Q
PROBLEM 6.45
Steam undergoes an adiabatic expansion in a piston-cylinder assembly from 100 bar, 360oC to 1
bar, 160oC. What is work in kJ per kg of steam for the process? Calculate the amount of entropy
produced, in kJ/K per kg of steam. What is the maximum theoretical work that could be obtained
from the given initial state to the same final pressure. Show both processes on a properly-labeled
sketch of the T-s diagram.
KNOWN: Steam undergoes an adiabatic expansion in a piston-cylinder assembly. Data are
given at the initial and final states.
FIND: Determine the work and entropy produced, each per unit mass of steam. Find the
maximum theoretical work that could be obtained from the given initial state to the same final
pressure. Show both processes on a T-s diagram.
SCHEMATIC AND GIVEN DATA:
ANALYSIS: The work is determined by reducing the energy balance, as follows.
ΔKE +ΔPE + ΔU = Q – W → W/m = (u1 – u2)
From Table A-3: p1 = 100 bar and T1 = 360oC; u1 = 2729.1 kJ/kg and s1 = 6.0060 kJ/kg∙K
From Table A-3: p2 = 1 bar and T2 = 160oC; u2 = 2597.8 kJ/kg and s2 = 7.6597 kJ/kg∙K
W/m = (2729.1 - 2597.8) = 131.3 kJ/kg (out, as expected)
Now, the entropy production can be evaluated using the entropy balance, as follows.
ΔS =
+ σ → σ/m = (s2 – s1) = (7.6597 – 6.0060) kJ/kg∙K = 1.6537 kJ/kg∙K
To find the maximum theoretical work, we note that since (s2 – s1) = σ/m ≥ 0; s2 ≥ s1
Steam
p1 = 100 bar
T1 = 360oC
p2 1 bar
T2 = 160oC
ENGINEERING MODEL: (1) The
steam is a closed system. (2) Q = 0.
(3) ΔKE +ΔPE = 0.
PROBLEM 6.45 (CONTINUED)
Graphically
From Table A-3: p2 = 1 bar, s2s = 6.0060 kJ/kg∙K gives
x2s =
=
= 0.7765
and
u2s = uf@1 bar + x2s(ug@1 bar – uf@1 bar) = 417.36 + (0.7765)(2506.1 – 417.36) = 2039.3 kJ/kg
Finally, the maximum theoretical work is
(W/m)max = (u1 – u2s) = (2729.1 – 2039.3) = 689.8 kJ/kg
T
s
100 bar
360oC
1 bar
160oC . .
.
(2)
(2s)
(1) Note that since s2 ≥ s1, state 2s is the limiting
case when s2s = s1 (reversible expansion with
no entropy production) . It is not possible to
access states to the left of state 2s adiabatically
from state 1.
Expansion to state 2s give the biggest
difference in u, and hence the maximum
theoretical work.
Not accessible
adiabatically
PROBLEM 6.140
m3/s and
1
8.43 Compare the results of Problem 8.40 with those for the same cycle whose processes of the
working fluid are not internally reversible in the turbines and pumps. Assume that both turbine
stages and both pumps have an isentropic efficiency of 83%.
KNOWN: A regenerative vapor power cycle with one open feedwater heater operates with
steam as the working fluid. Operational data are provided.
FIND: Determine (a) the cycle thermal efficiency, (b) the mass flow rate into the first turbine
stage, in kg/s, and (c) the rate of entropy production in the open feedwater heater, in kW/K.
Compare results with those of Problem 8.40.
SCHEMATIC AND GIVEN DATA:
tW
3
Condenser
outQ
Pump
2
p2W
4
7
1Steam
Generator
inQ
p4 = p3 = 6 kPa
x4 = 0 (saturated liquid)
p1 = 12 MPa
T1 = 560oC
p3 = 6 kPa
p7 = p1 = 12 MPa
MW 330cycle W
Open
Feedwater
Heater
Pump
1
p1W
56
2
(1)
(1)
(1–y)
(1–y)
(1–y)
(y)
p6 = p5 = p2 = 1 MPa
x6 = 0 (saturated liquid)
Turbine
p2 = 1 MPa
ht = 83%
hp2 = 83% hp1 = 83%
s
T
65s
2s
1
p = 6 kPa
p = 12 MPa
3s4
p = 1 MPa
7s
2
3
7
5
2
ENGINEERING MODEL:
1. Each component of the cycle is analyzed as a control volume at steady state. The control
volumes are shown on the accompanying sketch by dashed lines.
2. All processes of the working fluid are internally reversible except for processes in the
turbines and pumps and mixing in the open feedwater heater.
3. The turbines, pumps, and open feedwater heater operate adiabatically.
4. Kinetic and potential energy effects are negligible.
5. Saturated liquid exits the open feedwater heater, and saturated liquid exits the condenser.
ANALYSIS: First fix each principal state.
State 1: p1 = 12 MPa (120 bar), T1 = 560oC → h1 = 3506.2 kJ/kg, s1 = 6.6840 kJ/kg∙K
State 2s: p2s = p2 = 1 MPa (10 bar), s2s = s1 = 6.6840 kJ/kg∙K → h2s = 2823.3 kJ/kg
State 2: p2 = 1 MPa (10 bar), h2 = 2939.4 kJ/kg (see below) → s2 = 6.9174 kJ/kg∙K
kg
kJ)3.28232.3506)(83.0(
kg
kJ2.3506)( 21t12
21
21t
s
s
hhhhhh
hhhh = 2939.4 kJ/kg
State 3s: p3s = p3 = 6 kPa (0.06 bar), s3s = s2 = 6.9174 kJ/kg∙K → x3s = 0.8191, h3s = 2130.4
kJ/kg
State 3: p3 = 6 kPa (0.06 bar), h3 = 2267.9 kJ/kg (see below) → x3 = 0.8760, s3 = 7.3620
kJ/kg∙K
kg
kJ)4.21304.2939)(83.0(
kg
kJ4.2939)( 32t23
32
32t
s
s
hhhhhh
hhhh = 2267.9 kJ/kg
State 4: p4 = 6 kPa (0.06 bar), saturated liquid → h4 = 151.53 kJ/kg, v4 = 0.0010064 m3/kg, s4 =
0.5210 kJ/kg∙K
State 5: p5 = p2 = 1 MPa (10 bar), h5 = 152.74 kJ/kg (see below) → s5 ≈ 0.5249 kJ/kg∙K
(assuming the saturated liquid state corresponding to h5 = hf in Table A-2 and interpolating
for s5 = sf)
p1
45445
45
454p1
)()(
hh
pphh
hh
pp
vv
mN 1000
kJ 1
kPa 1m
N1000
83.0
kPa )61000)(kg
m0010064.0(
kg
kJ53.151
2
3
5
h = 152.74 kJ/kg
State 6: p6 = 1 MPa (10 bar), saturated liquid → h6 = 762.81 kJ/kg, s6 = 2.1387 kJ/kg∙K,
3
v6 = 0.0011273 m3/kg
State 7: p7 = p1 = 12 MPa (120 bar), h7 = 777.75 kJ/kg (see below)
p2
67667
67
676p2
)()(
hh
pphh
hh
pp
vv
mN 1000
kJ 1
kPa 1m
N1000
83.0
kPa )100012000)(kg
m0011273.0(
kg
kJ81.762
2
3
7
h = 777.75 kJ/kg
(a) Applying energy and mass balances to the control volume enclosing the open feedwater
heater, the fraction of flow, y, extracted at location 2 is
kJ/kg )74.1524.2939(
kJ/kg )74.15281.762(
52
56
hh
hhy = 0.2189
For the control volume surrounding the turbine stages
))(1()( 32211
t hhyhhm
W
kg
kJ)9.22674.2939)(2189.01(
kg
kJ)4.29392.3506(
1
t m
W
= 1091.3 kJ/kg
For the pumps
))(1()( 45671
phhyhh
m
W
kg
kJ)53.15174.152)(2189.01(
kg
kJ)81.76275.777(
1
p
m
W
= 15.89 kJ/kg
For the working fluid passing through the steam generator
kg
kJ)75.7772.3506(71
1
in hhm
Q
= 2728.5 kJ/kg
Thus, the thermal efficiency is
4
kJ/kg 5.2728
kJ/kg )89.153.1091(
/
//
1in
1p1t
mQ
mWmW
h = 0.394 (39.4%)
(b) The net power developed is
)//( 1p1t1cycle mWmWmW
Thus,
)//( 1p1t
cycle1
mWmW
Wm
MW 1
s
kJ1000
kg
kJ)89.153.1091(
MW 3301
m = 306.9 kg/s
(c) The rate of entropy production in the open feedwater heater is determined using the steady-
state form of the entropy rate balance:
cv0
e
e
ei
i
i
j j
jsmsm
T
Q
Since the feedwater heater is adiabatic, the heat transfer term drops. Thus,
552266cv smsmsmsmsm i
i
ie
e
e
])1([ 5261cv syyssm
kJ/s 1
kW 1
Kkg
kJ)]5249.0)(2189.01()9174.6)(2189.0(1387.2[
s
kg9.306cv
= 65.82 kW/K
Compared to the ideal cycle in problem 8.40, the presence of internal irreversibilities in the
turbine stages and the pumps results in lower cycle thermal efficiency, higher required mass
flow rate of steam entering the first-stage turbine, and greater rate of entropy production in
the open feedwater heater.