CH 12
CH 12
CH 12
By: S K Mondal
Chapter 12
Rankine Cycle
Q1
4
WP
WT
3
Q2
h 1 = W T + h2
WT = h1 h2
(iii)
or
h3 + W P = h 4
WP = h4 h3
p2
dh = vdp as v = constant
h = vp
or
h 4 h3 = v(p1 p2 ) = WP
(iv)
B.
WP = h4 h3 = v(p1 p2 ) kJ/kg
Wnet
(h1 h 2 ) (h 4 h3 )
W NP
= T
=
Q1
(h1 h 4 )
Q1
3600
kg
WT WP kWh
C.
Steam rate =
D.
3600 Q1 kJ
3600 kJ
=
WT WP kWh
kWh
By: S K Mondal
E.
Chapter 12
About Turbine Losses: If there is heat loss to the surroundings, h2 will decrease,
accompanied by a decrease in entropy. If the heat loss is large, the end state of steam
from the turbine may be 2.(figure in below).
It may so happen that the entropy increase due to frictional effects just balances the
entropy decrease due to heat loss, with the result that the initial and final entropies of
steam in the expansion process are equal, but the expansion is neither adiabatic nor
reversible.
F.
Isentropic Efficiency:
isen =
h1 h 2
h1 h 2s
Q1
Wp
WT
3
Q2
2s 2
S
G.
Q1 = h1 h 4 s = Tm (s1 s4 s )
Tm =
h1 h 4 s
s1 s4 s
1
5
T
Tm
4s
2s
3
S
By: S K Mondal
H.
Chapter 12
1 kg
12
11
10
m1 kg
4
(1m1 )kg
5
(1m1 m2 )kg
m2 kg
(1m1 m2 )kg
S
WT = (h1 h2) + (1 m1) (h2 h3) + (1 m1) (h4 h5) + (1 m1 m2) (h5 h6) kJ/kg
WP = (1 m1 m2) (h8 h7) + (1 m1) (h10 h9) + 1(h12 h11) kJ/kg
Q1 = (h1 h12) + (1 m1) (h4 h3) kJ/kg
Energy balance of heater 1 and 2
m1 h2 + (1 m1) h10 = 1 h11 For calculation of m1
And m2 h5 + (1 m1 m2) h8 = (1 m1 ) h9 ... For calculation of m2 .
I.
m kg
a
1
1 kg 6
4
3
2
S
m= 6
hb hc
By: S K Mondal
J.
Chapter 12
= n1 + n 2 n1 n 2
K.
(b) Wp in kJ/kg,
Type of Cycle
Condenser Pressure
-do-
-do-
Neglect Wp
-do-
-do-
-do-
0.1 bar
10 bar, 300c
-do-
-do-
-do-
-do-
-do-
-do-
-do-
-do-
Reheat to 600C
maximum intermediate
pressure to limit end
moisture to 15%
-do- but with 85% tur- bine
efficiencies
Isentropic pump process ends
on satura
Type of Cycle
Condenser Pressure
-do-
-do-
-do-
-do-
-do-
-do-
-do-
By: S K Mondal
Solution:
Chapter 12
p = 10 bar
T
1
4
p = 1 bar
3
2
S
s1 = 6.5865 kJ/kg-K
x = 0.8724
h 2 = 417.46 + 0.8724 2258 = 2387.3 kJ/kg
h3 = 417.46 kJ/kg
(a)
(b)
(c)
(d)
(e)
(f)
h4 = h3 + WP
WP = 1.043 103 [1000 100] kJ/kg = 0.94 kJ/kg
h4 = 418.4 kJ/kg
WT = h1 h2 = (2778.1 2387.3) kJ/kg = 390.8 kJ/kg
WP = 0.94 kJ/kg
Q1 = (h1 h4) = (2778.1 418.4) kJ/kg = 2359.7 kJ/kg
W
W NP
390.8 0.94
Cycle efficiency () = net = T
=
2359.7
Q1
Q1
= 16.52%
3600
3600
Steam rate =
kJ / kWh =
= 9.234 kg/kWh
Wnet
390.8 0.94
Moisture at the end of turbine process
= (1 x) = 0.1276 12.76%
Q.12.2
Solution:
p1 = 4.5 bar
T1 = 175C
From super heated STEAM TABLE.
By: S K Mondal
Chapter 12
p1
1
T
p2
2
S
At 4 bar
150C
h = 2752.8
s = 6.9299
200C
h = 2860.5
s = 7.1706
at 5 bar
152C
h = 2748.7
s = 6.8213
200C
h = 2855.4
s = 7.0592
at 4 bar 175C
at 5 bar, 175C
1
h = 2752.8 + (2860.5 2752.8)
2
175 152
h = 2748.7 +
(2855.4 2748.7)
200 152
= 2806.7 kJ/kg
= 2800 kJ/kg
1
23
s = 6.9299 + (7.1706 6.9299) s = 6.8213 + (7.0592 6.8213)
2
48
= 7.0503 kJ/kg K
= 6.9353
hfg = 2392.8
sg = 8.1502
sg = 8.0085
hfg = 2373.1
sg = 8.1502 +
15 10
By: S K Mondal
Chapter 12
15 13.342
h f = 191.83 +
(225.94 191.83) = 203.14 kJ/kg
15 10
15 13.342
h fg = 2392.8 +
(2373.1 2392.8) = 2386.3 kJ/kg
15 10
h2s = hf + x hfg = 203.14 + 0.85033 2386.3 = 2232.3 kJ/kg
isentropic =
h1 h 2
h1 h 2s
m. WT = 12.5 103
or
m =
12.5 103
= 29.18 kg/s
428.36
Q.12.3
A simple steam power cycle uses solar energy for the heat input. Water
in the cycle enters the pump as a saturated liquid at 40C, and is pumped
to 2 bar. It then evaporates in the boiler at this pressure, and enters the
turbine as saturated vapour. At the turbine exhaust the conditions are
40C and 10% moisture. The flow rate is 150 kg/h. Determine (a) the
turbine isentropic efficiency, (b) the net work output (c) the cycle
efficiency, and (d) the area of solar collector needed if the collectors pick
up 0.58 kW/ m 2 .
(Ans. (c) 2.78%, (d) 18.2 m 2 )
Solution:
T1 = 120.23C = 393.23 K
h1 = 2706.7 kJ/kg
s1 = 7.1271 kJ/kg K
2 bar
1
T
4
2s 2
S
hfg = 2406.7
By: S K Mondal
Chapter 12
sf = 0.5725
sg = 8.2570
x = 0.853
Q1 m
Required area A =
collection picup
2539 150
= 182.4 m2
=
0.58 3600
Q.12.4
Solution :
p3 = 0.1 bar
T = 45.8C
h f = 191.8 kJ/kg
h fg = 2392.8 kJ/kg
By: S K Mondal
Chapter 12
p1
1
p2
3
2
6
5
p3
4
S
h4 = hf + x hfg = 191.8 + 0.95 2392.8 = 2465 kJ/kg
sf = 0.649: Sfg = 7.501
15 bar 550C
s = 7.7045
p 10
7.775 = 7.8955 +
(7.7045 7.8955)
15 10
0.1205 = (p 10) (0.0382)
= 1.505 kJ/kg
h6 = h5 + WP = 193.3 kJ/kg
WT = (h1 h2) + (h3 h4) = 1768.6 kJ/kg
WP = 1.50 kJ/kg
Wnet = 1767.5 kJ/kg
Q = (h1 h6) + (h3 h2) = 4040.3 kJ/kg
W
1767.5
cycle = net =
100 % = 43.75%
4040.3
Q
Steam rate =
Q.12.5
3600
3600
kg / kWh = 2.0368 kg/kWh
=
1767.5
WT WP
By: S K Mondal
Solution:
Chapter 12
ga
Na
7
8
6
1
1 kg
m kg
(1m) kg
5
4
(1m)kg
S
From super heated S.T. at 55 bar 650C
at 50 bar 600C,
700C
By calculation at 650C
h = 3666.5
h = 3900.1
h = 3783.3
s = 7.2589
s = 7.5122
s = 7.3856
At 60 bar 600
h = 3658.4
s = 7.1677
C = 700C
h = 3894.2
s = 7.4234
by calculation
h = 3776.3
s = 7.2956
sf = 0.573
sfg = 7.685
x = 0.8806
h6 = 604.7 kJ/kg
By: S K Mondal
Chapter 12
80 103
kg / s = 76.638 kg/s
1049.8
Solution:
Q.12.7
Solution:
Q.12.8
Solution:
Q.12.9
By: S K Mondal
Chapter 12
515.5
72.23
363.0
0.1478
0.5167
0.2
277.3
38.35
336.55
0.0967
0.6385
80.9 x 106
0.0333
77.4 x 106
1.163
Solution:
Try please.
Q.12.10
Solution:
Q.12.11
sg = 0.48kJ/kg K
And
Solution:
For a supercritical steam cycle, the specific enthalpy and entropy at the
turbine inlet may be computed by extrapolation from the steam tables.
Try please.
By: S K Mondal
Q.12.12
Solution:
Q.12.13
Solution:
Q.12.14
Solution:
Q.12.15
Solution:
Chapter 12
3
65C
At 2 bar
2 bar
Q0
S
Page 205 of 265
By: S K Mondal
Chapter 12
Wnet + Q0
377.1 + 2189.4
=
100%
2972.2
Q1
= 86.35%
x = 0.85137
By: S K Mondal
Solution:
Chapter 12
1
8
T
30 bar
1 kg
3
m kg 2
(1m)kg
(1m)kg
0.7 bar
4
x = 0.8186
h1 2802.3 kJ/kg
h2 2380 kJ/kg
h3 2721.5 kJ/kg
h4 2113 kJ/kg
h5 = 121.5 kJ/kg
h6 = 124.5 kJ/kg
h7 = 551.4 kJ/kg
h8 = 554.3 kJ/kg
By: S K Mondal
Chapter 12
931
100% = 35.68% with turbine exhaust quality 0.8186
2609.5
If No separation is taking place, Then is quality of exhaust is x
x = 0.715
Then 6.1837 = 0.423 + x 8.052
=
=
100% = 35%
2677.8
Q.12.17
Solution:
1 80 bar 500C
3
400C
8
m kg
(1 m) kg
0.07 bar
5
(1 m) kg
h2 = 2759.5 kJ/kg
h3 = 3270.3 + 0.6(3268.7 3270.3) = 3269.3 kJ/kg
s3 = 7.708 + 0.6 (7.635 7.708)
= 7.6643 kJ/kg K
At 0.07 bar
hf = 163.4,
hfg = 2409.1
By: S K Mondal
hf = 0.559,
Chapter 12
sfg = 7.717
If quality is x then
7.6642 = 0.559 + x 7.717 x = 0.9207
h4 = 163.4 + 0.9207 2409.1 = 2381.5 kJ/kg
h7 = 686.8 kJ/kg h8
h5 = 163.4 kJ/kg h6,
From Heat balance of heater
m h2 + (1 m) h6 = h7
m = 0.2016 kg/kg of steam at H.P
(1 m) = 0.7984
WT = h1 h2 + (1 m) (h3 h4) = 1347.6kJ/kg
WP neglected
Q = (h1 h8) + (1 m) (h3 h2) = 3118.5 kJ/kg at H.P
(a) Reheat pr. 6.6 bar
80 103
kg/s = 59.36 kg/s
1347.6
W
1347.6
=
100% = 43.21%
3118.5
Q
By: S K Mondal
Solution:
Chapter 12
Separator mixes with the condensate from the process heater and the
combined flow enters the hotwell H at 50C. Traps are provided at the
exist from P and S. A pump extracts the condensate from condenser C
and this enters the hot well at 38C. Neglecting the feed pump work and
radiation loss, estimate the temperature of water leaving the hotwell
which is at atmospheric pressure. Also calculate, as percentage of heat
transferred in the boiler, (a) the heat transferred in the process heater,
and (b) the work done in the turbines.
Try please.
Q.12.19
In a combined power and process plant the boiler generates 21,000 kg/h
of steam at a pressure of 17 bar, and temperature 230 C . A part of the
steam goes to a process heater which consumes 132.56 kW, the steam
leaving the process heater 0.957 dry at 17 bar being throttled to 3.5 bar.
The remaining steam flows through a H.P. turbine which exhausts at a
pressure of 3.5 bar. The exhaust steam mixes with the process steam
before entering the L.P. turbine which develops 1337.5 kW. At the
exhaust the pressure is 0.3 bar, and the steam is 0.912 dry. Draw a line
diagram of the plant and determine (a) the steam quality at the exhaust
from the H.P. turbine, (b) the power developed by the H.P. turbine, and
(c) the isentropic efficiency of the H.P. turbine.
(Ans. (a) 0.96, (b) 1125 kW, (c) 77%)
Solution:
m = 21000 kg/h =
kg/s
6
1
HPT
4
2
3
5
LPT
6
BFP
From Steam Table at 17 bar 230C
15 bar 200C
250C
h = 2796.8
2923.3
s = 6.455
6.709
at 230C
30
h = 2796.8 +
(2623.3 2796.8) = 2872.7 kJ/kg
50
30
(6.709 6.455) = 6.6074 kJ/kg
s = 6.455 +
50
20 bar
212.4C
250C
Page 210 of 265
Con
By: S K Mondal
Chapter 12
h = 2797.2
h = 2902.5
s = 6.3366
s = 6.545
at 230C
17.6
h = 2797.2 +
(2902.5 2797.2) = 2846.4 kJ/kg
37.6
17.6
(6.545 6.3366) = 6.434 kJ/kg
s = 6.3366 +
37.6
at 17 bar 230C
2
h1 = 2872.7 + (2846.5 2872.7) = 2862.2 kJ/kg
5
2
s1 = 6.6074 + (6.434 6.6074) = 6.5381 kJ/kg
5
h2 = 871.8 + 0.957 1921.5 = 2710.7 kJ/kg h3
h4 = ?
WT = m(h5 h 6 )
WT
1337.5 3600
+ h6 =
+ 2419.8 = 2649.1 kJ/kg
21000
h5 =
(c)
WHPT = m2 (h1 h 4 )
17486.5
(2862.2 2636.7) kJ/kg = 1095 kW
=
3600
At 3.5 bar, sf 1.7273, sfg = 5.2119 quality is isentropic
Solution:
In a cogeneration plant, the power load is 5.6 MW and the heating load is
1.163 MW. Steam is generated at 40 bar and 500C and is expanded
isentropically through a turbine to a condenser at 0.06 bar. The heating
load is supplied by extracting steam from the turbine at 2 bar which
condensed in the process heater to saturated liquid at 2 bar and then
pumped back to the boiler. Compute (a) the steam generation capacity of
the boiler in tonnes/h, (b) the heat input to the boiler in MW, and (c) the
heat rejected to the condenser in MW.
(Ans. (a) 19.07 t/h, (b) 71.57 MW, and (c) 9.607 MW)
From steam table at 40 bar 500C
Page 211 of 265
By: S K Mondal
Chapter 12
1
1 kg
7
6
4
Q0
m kg
(1 m) kg
(1 m )k g
3
at 2 bar
so
so
h7 = h6 + WP = 508.7kJ/kg
For heating load Qo = h2 h6 = (2691.8 504.7) kJ/kg
= 2187.1 kJ/kg
For WT = (h1 h2) + (1 m) (h2 h3)
= 753.5 + (1 m) 508
= 1261.5 508 m
Wnet = WT WP4 5 (1 m) m WP6 7
= (1257.5 508 m) kJ/kg
If mass flow rate at 1 of steam is w kg/s then
w (1257.5 508m) = 5600
wm 2187.1 = 1163
From (i) & (ii) w = 4.668 kg/s = 16.805 Ton/h
...(i)
...(ii)
By: S K Mondal
Q.12.21
Solution:
Chapter 12
(b)
(c)
if condition of steam is x1
6.6625 = 1.7273 + x1 x 5.2119
x1 = 0.9469
h2 = 584.3 + 0.9469 2147.4 = 2617.7 kJ/kg
At 0.07 bar
sf = 0.559
sfg = 7.717
x 2 = 0.7909
1 35 bar 330C
6
4
3.5 bar
m kg
2
0.07 bar
(1 m)
Q0
(1 m) kg
By: S K Mondal
Chapter 12
w = 1.3311 kg/s
;
m = 0.51724 kg/kg of steam at H.P
Solution :
HPT = 83%
2
3
1
= 78%
4
T
6
2s
4s
5
S
At 2 bar
hf = 504.7 kJ/kg, hfg = 2201.6 kJ/kg
3
4
By: S K Mondal
Chapter 12
x1 = 0.99342
h2s = 504.7 + 0.99342 2201.6 kJ/kg = 2691.8 kJ/kg
h h2
isen. = 1
h1 h 2s
h2 h1 in (h1 h2s)
= 3445.3 0.83 (3445.3 2691.8) = 2819.9 kJ/kg
s2 = 7.31 kJ/kg K
From molier diagram
Adiabatic mixing
h5 = 504.7 + 0.87 2201.6 = 2420 kJ/kg
h2 5500 + h5 2700 = h3 (5500 + 2700)
h3 = 2688.3 kJ/kg from molier dia at 2 bar 2688.3 kJ/kg
quality of steam x3
Then 504.7 + x 2 2201.6 = 2688.3 x 3 = 0.9912
s3 = 1.5301 + 0.9918 5.5967 = 7.081 kJ/kg K
at 0.1 bar
sf = 0.649 + sfg = 7.501
x4 7.501 + 0.649 = 7.081
x4 = 0.8575
WT = 1745.6 kW
5500
WP =
0.001010 (4000 10) = 6.16 kW
3600
Q.12.23
Wnet = 1739.44 kW
h5 = 191.8 kJ/kg, h6 = h5 + WP = 195.8 kJ/kg
Heat input =
5500
(h1 h 6 ) = 4964.5 kW
3600
1739.44
100% = 35.04%
4964.5
By: S K Mondal
Solution :
Chapter 12
at 80 bar 500C
h1 = 3398.3 kJ/kg
s1 = 6.724 kJ/kg K
80 bar
1
1 kg
0.06 bar
4
22
a
m
kg
b
T
d
x 2 = 0.79434
h4 = 159.5 kJ/kg
For R-12
at 30C saturated vapour
ha = 199.6 kJ/kg, p = 7.45 bar sg = 0.6854 kJ/kg K
at 40C
sf = 0,
xb 0.7274 = 0.6854
hC = 0
x b = 0.94226
0.66 + 0.77
3
WP = Vc (pa pc ) =
10 (745 64.77) x = 0.4868 kJ/kg
2
hb = hC + WP = 0.4868 kJ/kg
By: S K Mondal
Q.12.24
Solution :
Chapter 12
h3 = 3230.9 kJ/kg
s 3 = 6.921 kJ/kg
From Molier diagram h2s = 3130 kJ/kg
h3 = 3218.8 kJ/kg
s3 = 6.875 kJ/kg
(From Molier diagram)
h4 = 2785 kJ/kg
h5s = 2140 kJ/kg
h5 = h3 (h5 h5s) = 3218.8 0.80 (3218.8 2140) kJ/kg
1
70 bar 500C
1kg
1m
2 kg
T
7
3 kg 6
3 3 30 bar 900C
2s
5 bar
m
(3 m)
(3 m) 0.06 bar
5s
S
From S.L in H.P. h4 = 2920 kJ/kg
From Heat balance & heater
m h4 + (3 m)h7 = 3hg
h7 = h6 + WP
Page 217 of 265
By: S K Mondal
Chapter 12
Q.12.25
Solution:
cycle =
= 642.6 kJ/kg
2493.6
100% = 31.4%
7940.3
An ideal steam power plant operates between 70 bar, 550C and 0.075
bar. It has seven feed water heaters. Find the optimum pressure and
temperature at which each of the heaters operate.
Try please.
Q.12.26
Solution:
x = 0.8533
1 p
1
3
T
6
5
p2
2
p3
4
S
Page 218 of 265
By: S K Mondal
Chapter 12
p3 = 0.0738 bar
(a)
h3 = 3247.6 kJ/kg
S2 = 6.3366 kJ/kg K
(b)
at 550C and 6.3366 kJ/kg k
From Steam Table
Pr = 200 bar
h1 = 3393.5 kJ/kg
h5 = 167.6 kJ/kg
WP = 0.001 (20000 7.38) kJ/kg
h6 = h5 + W = 187.6 kJ/kg
= 20 kJ/kg
(c)
Solution:
1602.7
100% = 43.83 %
3656.3
By: S K Mondal
Chapter 12
1
50 C
2
6
0.1 bar
h6 = 206.94 kJ/kg
=
42.7%
3665.86
Q.12.28
Solution:
By: S K Mondal
Chapter 12
1 kg
T
5
(1 m)kg
4
m kg
6
2s
Q0
(1 m)kg
2
(1m)kg
3s
S
This is also wet so
s2 = 1.6716 + x2 5.319.3 [2648.7 = 561.4 + x2 2163.2]
= 6.8042 kJ/kg K
If at 3s condition of steam is x3 then 40 kPa = 0.4 bar
x 3 = .8697
6.8042 = 1.0261 + x 3 6.6440
h5 = 318.7 kJ/kg
h6 = 562.1 kJ/kg, WP6 7 = 0.001 (1000 300)
= 0 0.7 kJ/kg
h7 = 562.1 kJ/kg