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CANKAYA UNIVERSITY
FACULTY OF ENGINEERING AND ARCHITECTURE
MECHANICAL ENGINEERING DEPARTMENT
ME 212 THERMODYNAMICS II
CHAPTER 8
EXAMPLES SOLUTIONS
Example 1:
Consider a steam power plant that operates on a simple ideal Rankine cycle. Steam enters the
turbine at 7 MPa and 450 oC and is condensed in the condenser at a pressure of 10 kPa by
running cooling water from a lake through the tubes of the condenser at a rate of 1600 kg/s.
Rankine cycle has a net power output of 40 MW. Show the cycle on T-s diagram with respect
to saturation lines, and determine (a) the thermal efficiency of this cycle, (b) the back work
ratio, (c) the mass flow rate of the steam, in kg/s, and (d) the temperature rise of the cooling
water, in oC.
Solution:
Assumptions : (1) Steady state conditions exits.
(2) Kinetic and Potential energy changes are negligible.
Analysis:
State 1
For P1=7 MPa and T1=450 °C , Table A-4 gives
kg
kJ57.3286h1 and
K.kg
kJ6359.6s1
State 2
K.kg
kJ6359.6ss 12
From Table A-3
K.kg
kJ1502.8s kPa10@g
K.kg
kJ6493.0s kPa10@f
7981.06493.01502.8
6493.06359.6
10@
10@2
2
kPafg
kPaf
s
ssx
So,
s
T
3 2
4 7MPa=70 bar
10kPa=0.1 bar
450°C
1
kPa10@fg2kPa10@f2 hxhh
From Table A-3, kg
kJ83.191h kPa10@f and
kg
kJ8.2392h kPa10@fg
Thus,
kg
kJ52.2101)8.2392(7981.083.191h 2
State 3: From table A-3
kg
kJ83.191hh kPa10@f3 and
kg
m100102.1
33
kPa10@f3
State 4:
)PP(hh
m
W34334
p
)PP(hh 34334
kg
kJ89.198
m
kN)107000)(
kg
m100102.1(
kg
kJ83.191h
2
33
4
Thus;
kg
kJ05.1185)52.2101()57.3286(hh
m
W21
t
kg
kJ06.7)83.191()89.198(hh
m
W34
p
kg
kJ68.3087)89.198()57.3286(hh
m
Q41
in
kg
kJ69.1909)83.191()52.2101(hh
m
Q32
out
kg
kJ99.117769.190968.3087
m
W
m
W
m
Q
m
Q
m
W ptoutinnet
(a) The thermal efficiency of the cycle is
)2.38(%382.068.3087
69.19091
m
Q
m
Q
1
in
out
(b) The back-work ratio is determined from
)60.0(%0060.005.1185
06.7
m
W
m
W
bwr
t
p
(c) The mass flow rate of the steam is
s
kg96.33
kg
kJ99.1177
s
kJ40000
m
W
Wm
net
net
(d) The rate of heat rejection to the cooling water and its temperature rise are
s
kJ07.853,64)
kg
kJ69.1909)(
s
kg96.33(
m
QmQ out
out
C70.9
)K.kg
kJ18.4)(
s
kg1600(
s
kJ07.853,64
)cm(
QT o
watercoolingp
out
watercooling
Example 2:
A Rankine steam power plant uses water as the working fluid. Steam enters the turbine at 7
MPa and 450 oC and is condensed in the condenser at a pressure of 10 kPa by running cooling
water from a lake through the tubes of the condenser at a rate of 1600 kg/s. Cycle has a net
power output of 40 MW. Isentropic efficiency of the turbine is 85 percent, and the isentropic
efficiency of the pump is 90 percent. Assuming no pressure losses in the condenser and boiler,
show the cycle on T-s diagram with respect to saturation lines, and determine (a) the thermal
efficiency of this cycle, (b) the back work ratio, (c) the mass flow rate of the steam, in kg/s,
and (d) the temperature rise of the cooling water, in oC.
Solution:
Assumptions: (1) Steady state conditions exist.
(2) Kinetic and potential energy changes are negligible.
Analysis:
State 1:
With P1 = 7 MPa and T1 = 450 0C, Table A-4 gives
h1 = 3286.57 kJ/kg and s1 = 6.6359 kJ/kg
State 2:
s2s = s1 = 6.6359 kJ/kg
From Table A-3
sg@10kPa = 8.1502 kJ/kg.K
sf@10kPa = 0.6493 kJ/kg.K
X2s =
So,
h2s = hf@10kPa + X2s.hfg@10kPa
From Table A-3
hf@10kPa = 191.83 kJ/kg
hfg@10kPa = 2392.8 kJ/kg
Thus,
h2s = 191.83 + 0.7981(2392.8) = 2101.52 kJ/kg
Isentropic turbine efficiency is
Ƞt = =
h2 = h1 - Ƞt (h1-h2s)
= 3286.57-0.85(3286.57-2101.52)
= 2279.28 kJ/kg
State 3:
h3 = hf@10kPa = 191.83 kJ/kg
v3 = vf@10kPa = 1.0102 x 10-3
m3/kg
State 4:
s = h4s – h3 = v3(P4s - P3)
h4s = h3 + v3(P4s – P3)
= 191.83 kJ/kg + (1.0102 x 10-3
m3/kg) (7000 - 10) kN/m
2
= 198.89 kJ/kg
Isentropic pump efficiency is
Ƞp = =
h4 = + h3 = + 191.83
h4 = 199.67 kJ/kg
Hence,
= h1 – h2 = 3286.57 – 2279.28 = 1007.29 kJ/kg
= h4 – h3 = 199.67 – 191.83 = 7.84 kJ/kg
= h1 – h4 = 3286.57 – 199.67 = 3086.9 kJ/kg
= h2 – h3 = 2279.28 – 191.83 = 2087.45 kJ/kg
3086.7 - 2087.45 = 999.25 kJ/kg
(a) The thermal efficiency of the cycle is
Ƞ = 1 - = 1 – = 0.324 or % 32.4
(b) The back-work ratio is determined from
bwr = = = 0.0078
(c) The mass flow rate of the steam is
= 40.03 kg/s
(d) The rate of heat rejection to the cooling water and its temperature rise are
= = (40.03 kg/s) (2087.45 kJ/kg) = 83,560.62 kJ/s
= = = 12.49 0C
Example 3:
In a Rankine cycle, saturated liquid water at 10 kPa is compressed in a pump isentropically to
8 MPa. It is then heated, first in a boiler and then by superheating at a constant pressure of 8
MPa, to a temperature of 600 oC. After an adiabatic reversible expansion to 3 MPa, the steam
is reheated to 600 oC, and a second adiabatic reversible expansion to 10 kPa occurs.This is
essentially a reheat cycle. (a) What is the total work (kJ/kg) generated. (b) What is the
efficiency of the cycle (%)? (c) Sketch the cycle on a T-s diagram.
Solution:
(c)
State 1;
P1=7MPa
T1=700°C Table A-4 gives
kg
kjh 0.36421 and
Kkg
kjs
.0206.71
State 2:
P2=3MPa
Kkg
kjss
.0206.712
Kkg
kjs MPag
.1869.63@ , From Table A-3,
MPagss 3@2
600°C 1
2
3
4
6
5
8MPa
3MPa
10kPa= 0.10 bar
s
T
P2=3MPa
Kkg
kjs
.0206.72 ,
kg
kjh 8.32992 , From Table A-4
State 3;
P1=3MPa
T1=600°C Table A-4 gives
kg
kjh 3.36823 and
Kkg
kjs
.5085.71
State 4;
Kkg
kjs kPag
.1502.810@ From Table A-3
Kkg
kjs kPaf
.6493.010@
Kkg
kjss
.50853.734
kPagkPaf sss 10@410@
9145.06493.01502.8
6493.05085.7
10@10@
10@4
4
kPafkPag
kPaf
ss
ssx
kPafgkPaf hxhh 10@410@4
kg
kjh 1.2380)8.2392)(9145.0(83.1914 (Table A-3)
State 5;
kg
kjhh kPaf 83.19110@5 ,
kg
mkPaf
33
10@5 100102.1
State 6;
)( 56556 PPhh
m
Wp
)( 56556 PPhh
kg
kj
m
kN
kg
m
kg
kjh 9.199)108000)(100102.1(83.191
2
33
6
(a) The total work generated
m
W
m
W
m
W ttgeneratedtotal
21
kg
kjhh
m
W t2.342)8.3299()0.3642(21
1
kg
kjhh
m
W t2.1302)1.2380()3.3682(43
2
kg
kj
m
W generatedtotal
4.16442.13022.342
(b)
kg
kjhhhh
m
Qin 6.3824)8.32993.3682()9.1990.3642()()( 2361
kg
kjhh
m
W p07.8)83.191()9.199(56
428.06.3824
07.82.13022.342
21
m
Q
m
W
m
W
m
W
m
Q
m
W
in
ptt
in
net
or (%42.8)
Example 4:
Consider the steam power plant with an open feedwater heater operating under the conditions
shown in the following diagram. Saturated liquid exits the open feedwater heater at 1 MPa,
and saturated liquid exits the condenser. Each pump operates isentropically. Net power output
of the cycle is 100 MW.
Using the conditions shown on the diagram above and values obtained from the steam tables,
determine
a. the enthalpy values at all eight stations,
b. the mass fraction, y, of steam bled from the turbine set at station (2),
c. the thermal efficiency, , of this power plant, and
d. the mass flow rate of steam entering the first turbine stage, in kg/h.
e. Show the cycle on T-s diagram with respect to saturation lines.
Solution:
Assumptions: (1) Steady-state
(2) Each pump operates isentropically.
(3) The turbines, pumps, and feedwater heater operate adiabatically.
(4) Kinetic and potential energy changes are negligible.
(5) Saturated liquid exits the open feedwater heater, and saturated liquid exits the condenser.
Analysis:
a) State 1: T1 = 520 0Cand P1 = 10 MPa, Table A-4 gives h1 = 3425.1 kJ/kg
State 2: With P2 = 1 MPa and T2 = 200 0C, Table A-4 gives h2 = 2827.9 kJ/kg
State 3: With 1 MPa and 520 0C, Table A-4 gives h3 = 3522.1 kJ/kg
State 4: With 6 kPa and 50 0C, Table A-4 gives h4 = 2593.5 kJ/kg
State 5: With 6 kPa, Table A-3 gives h5 = hf@6kPa = 151.53 kJ/kg
v5 = vf@6kPa = 1.0064 x 10-3
m3/kg
State 6: h6 = h5 + v5 (P6 – P5)
= 151.53 kJ/kg + (1.0064 x 10-3
m3/kg) (1000 – 6) kN/m
2
= 152.53 kJ/kg
State 7: With 1 MPa, Table A-3 gives
h7 = hf@10bar = 762.81 kJ/kg
v7 = vf@10bar = 1.1273 x 10-3
m3/kg
State 8: h8 = h7 + v7 (P8 – P7) = 762.81 kJ/kg + (1.1273 x 10-3
m3/kg) (10000 – 1000) kN/m
2
= 772.96 kJ/kg
(b) Energy rate balance for open feedwater heater gives
yh2 + (1-y)h6 = h7
(c) On the basis of a unit of mass passing through the high pressure turbine, the total turbine
work output is
(h1 – h2) + (1-y) (h3 – h4)
= (3425.1 – 2827.9) + (1 – 0.2281) (3522.1 – 2593.5)
= 1314.0 kJ/kg
The total pump work per unit mass passing through the high pressure turbine is
(h8 – h7) + (1 – y) (h6 – h5)
= (772.96 – 762.81) + (1 – 0.2281) (152.53 – 151.53)
= 10.92 kJ/kg
The heat added in the steam generator per unit of mass passing through the high pressure
turbine is
(h1 – h8) + (1 – y) (h3 – h2)
= (3425.1 – 772.96) + (1 – 0.2281) (3522 – 2827.9)
= 3187.9 kJ/kg
The thermal efficiency is then
ƞ = or % 40.9
(d) The mass flow rate of the steam entering the high pressure turbine, ṁ1, can determined
using the given value for the net output, 100 MW. Since Ẇcycle = Ẇt - Ẇp
and
1314.0 kJ/kg and 10.32 kJ/kg
it follows that
ṁ1 = = = 76.74 kg/s
= 2.76 x 105 kg/h
(e)
Example 5:
Consider a water-ammonia binary vapor cycle consisting. In the steam cycle,
superheated vapor enters the turbine at 7 MPa, 450 oC, and saturated liquid exits the
condenser at 55 oC. The heat rejected from the steam cycle is provided to the ammonia
cycle, producing saturated vapor at 45 oC, which enters the ammonia turbine. Saturated
liquid leaves the ammonia condenser at 1 MPa. For a net power output of 24 MW from
the binary cycle, determine (a) the mass flow rates for the steam and ammonia cycles,
respectively, in kg/s, (b) the power output of the steam and ammonia turbines,
respectively, in MW. (c) the rate of heat input to the ammonia cycle, in MW, (d) the
rate of heat addition to the binary cycle, in MW, and (e) the thermal efficiency of the
binary vapor cycle. (f) Show the cycle on T-s diagram with respect to saturation lines.
Solution:
Assumptions: (1) Each component is analyzed as a control volume at steady state
(2) All process of the working fluids is internally reversible, except in the interconnecting heat
exchanges.
(3) There are no stray heat transfers from the turbines or the heat exchanger.
(4) Kinetic and potential energy effects can be neglected.
Analysis: First, fix each of principle states. For steam, use tables A-2 and A-4.
State 1:
P1 = 70 bar, T1 = 450 0C → h1 = 3286.57 kJ/kg
s1 = 6.6359 kJ/kg.K
State 2:
T2 = T3 = 55 0C, s2 = s1 = 6.6359 kJ/kg
sg@55C = 7.9913 kJ/kg.K
sf@55C = 0.7679 kJ/kg.K
X2 =
hf@55C = 230.23 kJ/kg
hfg@55C = 2370.7 kJ/kg
h2 = hf@55C + X2.hfg@55C
h2 = 2309.23 + 0.8124 (2370.7) = 2156.19 kJ/kg
State 3:
T3 = 55 0C, saturated liquid.
h3 = 230.23 kJ/kg
v3 = 1.0146 x 10-3
m3/kg
P3 = 0.1576 bar = 15.76 kPa
State 4:
= h4 – h3 = v3 (P4 – P3)
h4 = h3 + v3 (P4 – P3)
= 230.23 + (1.0146 x 10-3
m3/kg) (7000 – 15.76) kN/m
2
= 237.32 kJ/kg
Now, for Ammonia, use Tables A-13 and A-14
State a:
Ta = 45 0C, saturated vapor
ha = 1470.96 kJ/kg
sa = 4.8125 kJ/kg.K
Pa = 17.819 bar = 1781.9 kPa
State b:
Pb = Pc = 10 bar
sb = sa = 4.8125 kJ/kg.K
sf@10bar = 1.1191 kJ/kg.K
sg@10bar = 5.0294 kJ/kg.K
Xb =
hb = hf@10bar + Xb.hfg@10bar
hf@10bar = 297.76 kJ/kg
hfg@10bar = 1165.42 kJ/kg
hb = 297.76 + 0.9445 (1165.22) = 1398.50 kJ/kg
State c:
Pc = 10 bar, saturated liquid.
hc = 297.76 kJ/kg
vc = 1.6584 x 10-3
m3/kg
State d:
hd = hc + vc (Pd – Pc)
= 297.76 + (1.6584 x 10-3
m3/kg) (1781.9 - 1000) kN/m
2
= 299.06 kJ/kg
(a) The mass flow rates of the steam and ammonia can be obtained using mass and energy
rate balances for the inter-connecting heat exchanges.
0 =
or
For the steam cycle
And, for the ammonia cycle
Noting that and combining
Solving
=
(b) The power output of the steam and ammonia turbines are
(c) The rate of heat input to the ammonia cycle is
= 37.27 MW
(d) The rate of heat addition to the binary cycle is
(e) The thermal efficiency of the binary cycle is
ƞ = or % 40.7
(f)
6) Water is the working fluid in a Rankine cycle. Superheated vapor enters the
turbine at 10 MPa, 480 0C, and the condenser pressure is 6 kPa. The turbine and
pump have isentropic efficiencies of 80 and 70%, respectively. Determine the
rate of exergy input, in kJ per kg of steam flowing, to the working fluid passing
through the steam generator. Perform calculations to account for all outputs,
losses, and destructions of this exergy. Let 0
0 0T 15 C and p 0.1MPa T0 5
158C, p0 5 0.1 MPa