PROJECT DOCUMENT ON EXERGY final

Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 56
CHAPTER-6
6.0 TABLES AND CALCULATIONS
Enthalpy (H): The sum total of internal energy (U) and (PV) is called Enthalpy.
Denoted by “H”
H= U+ PV
Where H = total enthalpy J,
U = total internal energy
P = absolute pressure
V = Specific Volume
Entropy: The disorderness of molecules of the system of molecules is called Entropy.
Denoted by “S”
Unit of Measure
Enthalpy H = h×m×10-3
=
𝑘𝐽
𝑘𝑔
x
𝑘𝑔
𝑠
=
𝑘𝐽
𝑆
×103
= kW x 103
= Mw
Entropy S = S x m x 10-3
=
𝑘𝐽
𝑘𝑔−𝐾
×
𝑘𝑔
𝑆
=
𝑘𝐽
𝑆−𝐾
=
𝐾𝑊
𝐾
×10-3
S = Mw / K
In this chapter, calculations related to various tables have been shown and the various directly
measured and derived parameters have been tabulated.
An interpolation has been done extensively in order to obtain properties which were not readily
available in the steam tables.

6.1 Boiler Inlet:
At boiler inlet
Pressure P = 157.8 bar, Temperature T = 574 K, mass flow rate, m= 175 kg/s
From steam tables
Specific volume
v = 0.001694
m3
𝑘𝑔
Pressure difference
dP = 157.8 – 157.8 = 0
Temperature difference
dT = 0k
At 574k, hf =13450.4
𝑘𝐽
𝑘𝑔
Enthalpy in Mw, H = 1345.4x175x10-3 =235.445 Mw
At 574 k, sf =3.271.
Entropy in Mw S = sf x mx10-3
= 3.271 x 175 x 10-3= 0.5724
𝑀𝑤
𝐾
6.2 BOILER OUTLET
PRESSURE P=146.05 bar, TEMPERATURE T = 613 K,
mass flow rate = 175
𝑘𝑔
𝑠
.
Form steam tables
Enthalpy h=2626.2
𝑘𝐽
𝑘𝑔
Entropy s = 5.3429
𝑘𝐽
𝑘𝑔−𝐾
Enthalpy in Mw H = hxmx10-3
H = 2626.2 x 175 x 10-3 = 459.58 Mw
Enthalpy in Mw S = sxmx10-3
= 5.3429 x 175 x 10-3 = 0.9349 Mw/K

6.3 SUPER HEATER INLET
PRESSURE P = 146.05 bar, TEMPERATURE T = 613 K,
mass flow rate mass flow rate = 177.78
𝑘𝑔
𝑠
From steam tables
Enthalpy h = 2626.2
𝑘𝐽
𝑘𝑔
Entropy s = 5.3429
𝑘𝐽
𝐾𝑔−𝐾
Enthalpy in Mw
H = h x m x 10-3
= 2626.2 x 177.78 x 10-3 = 466.88 Mw
Entropy in Mw
S = s x m x 10-3
= 5.3429 x 177.78 x 10-3 = 0.9497
𝑀𝑤
𝐾
6.4 SUPERHEATER OUTLET
PRESSURE P = 146.05 bar, TEMPERATURE T = 540 c, mass flow rate m= 177.18
𝑘𝑔
𝑠
From steam tables
At 140.05 bar Enthalpy
h = 3472.12+
3482.62−3472.12
10
x (150-146.05) [By interpolation] = 3476.26
𝑘𝐽
𝑘𝑔
Entropy values
Pressure 500 o
c 600 o
c 540 o
c
140 bar 6.394 6.714 6.5872
150 bar 6.349 6.676 6.5452
At 146.05 bar, interpolation values at 540o c
Entropy s = 6.5452 +
6.5872 −6.5452
10
x (150 – 146.05) = 6.56179
𝑘𝐽
𝑘𝑔−𝐾
= 3476.26x177.78x10-3 = 618.009 Mw
Entropy in Mw S = sxmx10-3
= 6.56179x177.78x10-3 = 1.1665
𝑀𝑤
𝐾

6.5 High Pressure Turbine [HPT] INLET
PRESSURE P = 146.05 bar, TEMPERATURE T = 540 o c, mass flow rate m= 177.18
𝑘𝑔
𝑠
For above values
Enthalpy h = 3476.26
𝑘𝐽
𝑘𝑔
Entropy s = 6.56179
𝑘𝐽
𝑘𝑔−𝐾
= 3476.26x177.78x10-3 = 618.009 Mw
= 6.56179x177.78x10-3= 1.1665
𝑀𝑤
𝐾
6.6 High Pressure Turbine [HPT] OUTLET
PRESSURE P=35.30 bar, TEMPERATURE T = 630 K =330 o c,
mass flow rate m= 161.11
𝑘𝑔
𝑠
From steam tables, enthalpy values
Pressure 300 o
c 350 o
c
34 bar 2982.2 3108.7
36 bar 2975.6 3108.7
At 34 bar, 330o c
Enthalpy h = 2982.2+
3108 .7−2982.2
50
x20 = 3032.8 kJ/kg
At 36 bar, 330 o c
h= 2975.6+
3104.2−2975.6
50
x20 = 3027.04 kJ/kg
At 35.3 bar, 330 o c
h = 3027.04+
3032.8−3027 .04
20
x (36-35.3) = 3029.056 kJ/kg
Entropy values
Pressure 300 o
c 350 o
c
34 bar 6.467 6.679
36 bar 6.432 6.647

At 34 bar, 330o c
s = 6.467+
6.679 −6.467
50
x 20 = 6.5518 kJ/kg- K
At 36 bar, 330 o c
s = 6.432+
6.647 −6.432
50
x 20 = 6.518
At 35.30 bar, 330 o c
s = 6.518+
6.5518−6.518
2
x(36 − 35.3) = 6.52983 kJ/kg
Enthalpy in Mw
H = h xm x10-3
=3029.056x161.11x10-3 = 488.011Mw
Entropy in Mw
S = sxmx10-3
=6.52983x161.11x10-3 = 1.05202 Mw/K
6.7 REHEATER INLET
PRESSURE P=35.30 bar, TEMPERATURE T = 630 K, =330 o c,
mass flow rate m= 161.11
𝑘𝑔
𝑠
For above values from steam tables from steam tables from steam tables
Enthalpy h = 3029.056 kJ/kg
Entropy s = 6.52983 kJ/kg
=3029.056x161.11x10-3 = 488.011Mw
Entropy
S = sxmx10-3
=6.52983x161.11x10-3 = 1.05202 Mw/K
6.8 REHEATER OUTLET
PRESSURE P=34.32 bar, TEMPERATURE T=813 K = 540 oc ,
mass flow rate m= 161.11 kg/s

Pressure 500 oc 600 oc
34 bar 3451.7 3677.7
36 bar 3449.5 3676.1
At 34 bar, 540 o c
Enthalpy h = 3451.7 +
36677 .7−3451 .7
100
(600-540) = 3587.3 kJ/kg
At 36 bar, 540 o c
h = 3449.5 +
3679.6.1−3449.5
100
x 60 = 3585.46 kJ/kg
At 34.32bar, 540 o c
h = 3585.4+
3676.1−3585 .4
2.0
(36-34.32) = 3587.0056 kJ/kg
Entropy values
Pressure 540 o c 550 o c
34 bar 7.712 7.447
36 bar 7.144 7.420
At 34bar, 540 o c
Entropy s =7.712+
7.447−7.172
100
x 100 =7.337kJ/kg-K
At 36 bar, 540 o c
s =7.144+
7.420 −7.144
100
∗ 100 = 7.3096 kJ/kg - K
At 34.32 bar, 540 o c
s = 7.3096 +
7.337 −7.3096
2.0
(36-34.32) = 7.3326 kJ/kg – K
= 3587.0056-161.11x10-3 = 577.902 Mw
= 7.3326x16111x10-3
= 1.1813 Mw/K
6.9 Intermediate Pressure Turbine [IPT] INLET
PRESSURE P=34.32 bar, TEMPERATURE T=813 K = 540 oc,

mass flow rate m= 161.11 kg/s
For above values
h = 3587.0056 kJ/kg
s = 7.3326 kJ/kg - k
Enhalpy in MW
H = hxmx10-3
= 3587.0056-161.11x10-3= 577.902 Mw
Entropy in Mw
S = sxmx10-3
= 7.3326x16111x10-3 = 1.1813 Mw/K
6.10 Intermediate Pressure Turbine [IPT] TURBINE OUTLET
PRESSURE P = 6.87 bar, TEMPERATURE T 613 k, mass flow rate m=152.78 kg/s.
From the steam tables, enthalpy values
6bar 3062.3 3166.2
7bar 3059.8 3164.3
At 6bar, 340 o c
Enthalpy h = 3062.3+
3166 .2−3062.3
50
x10
= 3083.08 kJ/kg
At 7 bar, 340 o c
h = 3059.8+
3164.3−3059.8
50
x10
= 3080.7 kJ/kg
At 6.87 bar, 340 o c
h =3080.7+
3083.08−3080.7
1.0
∗ (7 − 6.87)
= 3081.0094 kJ/kg

Entropy values
6 bar 7.374 7.548
7 bar 7.300 7.475
At 6 bar, 340 o c
Entropy s = 7.374+
7.548−7.374
50
x10
= 7.48088 KJ/Kg – k
At 7 bar, 340 o c
Entropy s =7.300+
7.475−7.300
500
x 10
= 7.335 kJ/kg-K
At 6.87 bar, 340 o c
s = 7.335+
7.4088−7.335
1.0
x0.13= 7.344594 kJ/kg - K
Enthalpy in Mw
H = hxmx10-3
= 3081.0094x152.8x10-3 = 470.716 Mw
Entropy in Mw
S = sxmx10-3
= 7.3445x152.78x10-3 = 1.12209 Mw/K
6.11 Low Pressure Turbine [
LPT] INLET
PRESSURE P = 6.87 bar, TEMPERATURE T 613 K, m=152.78 kg/s
Entropy s = 7.344594 kJ/kg - K
Enthalpy in Mw
H = hxmx10-3
= 3081.0094x152.8x10-3 = 470.716 Mw
= 7.3445x152.78x10-3 = 1.12209 Mw/K

6.12 Low Pressure Turbine [LPT] OUTLET
PRESSURE P = 0.0873 bar, TEMPERATURE T 322 K = 49 o c, mass flow rate m=125 kg/s
0.085 bar 2579.2
0.09 2581.1
At 0.0873 bar,
Entropy s = 8.1988 kJ/kg-k
Enthalpy in Mw
H = hxmx10-3
=2580.226x125x10-3 = 2580.226 Mw
Entropy in Mw
S = sxmx10-3
= 8.1988x125x10-3 =1.024 Mw/K
6.13 CONDENSER INLET
P=0.0892 bar, T= 320 K = 47 o c, M=144.44kg/s
Entropy s = 8.192 kJ/kg-K
Enthalpy H = hxmx10-3
= 2579.504x144.44 x10-3 = 372.48 Mw
Entropy
S = s x m x10-3
=8.192x144.44x 10-3 = 1.1813 Mw/K
6.14 CONDENSER OUTLET
PRESSURE P= 0.0873 bar, T=316K = 43 o c, M=144.44kg/s
Enthalpy h = hf - Cpw ( Tsat- T)
= 181.184-4.187(43.28-43) = 179.9948 kJ/kg

Entropy s = sf - Cpw ln (
Tsat
𝑇
)
= 0.61556-4.187 ln (
43.28
43
) = 0.61179 kJ/kg K
Enthalpy in Mw
H = hxmx10-3
= 179.99x144x.44x10-3 = 25.998Mw
Entropy S = sxmx10-3
= 0.61556x144.44x10-3 = 0.0883Mw/K
6.15 Condensate extraction pump [CEP] INLET
PRESSURE P= 0.0873 bar, T=316K = 43 o c, M=144.44kg/s
Enthalpy h = hf - Cpw ( Tsa t- T)
= 181.184-4.187(43.28-43)
= 179.9948 kJ/kg
Entropy s = sf - Cpw ln (
Tsat
𝑇
)
= 0.61556-4.187 ln (
43.28
43
)
= 0.61179 kJ/kg K
= 179.99x144x.44x10-3= 25.998Mw
Entropy inMw
S = sxmx10-3
= 0.61556x144.44x10-3= 0.0883 Mw/K
6.16 Condensate extraction pump [CEP] OUTLET
PRESSURE P= 18.63 bar, Temperature T= 320 K = 47 o c, mass flow rate m=144.44kg/s
At 18.63bar,
Specific Volume v=0.00117148 m3/kg
H = 25.99844+ (vdPx102+Cpwdt) x 10-3 ×m
Pressure difference
dP = 18.63-0.0873 = 18.5427 bar

dT =320-316 K = 4 K
Enthalpy in Mw
H = 25.99844+ (vdPx102+Cpwdt)x10-3 × m
H = 25.99844+ (0.00117148x28.7312x102+4.187x4) ×144.4= 28.7312 Mw
Entropy in Mw
S = 0.0883 + (
mCpwdT∗10−3
Tprev
)
= 0.0883 + (
144 .44∗4.187 ∗4∗10−3
316
) = 0.096022 Mw/K
6.17 Ejector Inlet
PRESSURE P= 18.63 bar, Temperature T= 320 K = 47 o c, mass flow rate m=144.44 kg/s
At 18.63bar,
Enthalpy in Mw
H = 25.99844+ (vdPx102+Cpwdt)10-3 ×m
Pressure difference
dP = 18.63-0.0873= 18.5427.
dT =320-316 k= 4 K
Enthalpy in Mw H = 25.99844+ (vdPx102+Cpwdt) x10-3x m
H = 25.99844+ (0.00117148x28.7312x102+4.187x4)×144.4 = 28.7312 Mw
Entropy in Mw
S = 0.0883 + (
mCpwdT∗10−3
Tprev
)
S = 0.0883+ (
144 .44∗4.187∗4∗10−3
316
) = 0.096022 Mw/K
6.18 Ejector Outlet
PRESSURE P= 17.65 bar, Temperature T=325K=52o c, m=144.44kg/s
At 17.65 bar
Pressure difference
dP = 17.65-18.63 = -0.98 bar

dT = 325-320 = 5K
Enthalpy in Mw
H = 28.7312 + (vdPx102+CpwdT)10-3xm
= 28.7312 + (0.0011674x102+4.187x5)10-3x144.44 =31.7385 MW
Entropy in Mw
S = 0.096022 + (
mCpwdT∗10−3
Tprev
)
S = 0.096022 + (
144 .44∗4.187∗5∗10−3
320
) = 0.10547 Mw/K
6.19 Gland Steam Condenser [GSC] Inlet
PRESSURE P= 17.65 bar, T=325K=52 o c, m=144.44kg/s
At 17.65 bar
v=0.0011674 m3/kg
dP = 17.65-18.63 = -0.98 bar
dT = 325-320 = 5K
H = 28.7312 + (vdPx102+CpwdT)10-3x m
= 28.7312 + (0.0011674x102+4.187x5)10-3x144.44 =31.7385 Mw
S = 0.096022 + (
mCpwdT∗10−3
Tprev
)
S = 0.096022 + (
144 .44∗4.187∗5∗10−3
320
) = 0.10547 Mw/K
6.20 Gland Steam Condenser [GSC] outlet
PRESSURE P = 16.67bar, T=327 k = 54 c m=144.4kg/s
At 16.67 bar
Specific Volume
v= 0.00116232 m3/kg
Pressure difference
dP = 16.67-17.65 = -0.98 bar
dT = 327-325 o c = 2K
H = 31.7385+ (vdPx102+Cpwdt) x10-3 ×m
H = 31.7385+ (0.00116232x10.98x102+4.187x2)144.4x10-3 = 32.9315Mw.

S = 0.0883+ (
mCpwdT∗10−3
Tprev
)
S = 0.10547+ (
144 .44∗4.187∗2∗10−3
325
) = 0.10919 Mw/K
6.21 Low Pressure Heater 1 [LPH] INLET
PRESSURE P=15.69bar, T=327K = 54 o c, m=144.44kg/s
For above values
Specific volume v =0.001157m3/kg
Pressure difference
dP = 15.69-16.67 =-0.98bar
dT = 327-327 = 0 K
Enthalpy in Mw
H = 32.9315 + (vdPx102+Cpwdt) x10-3 ×m
H=32.9315 + (0.001157x-0.98x102+4.187x0) 144.44x10-3 =32.91511Mw
Entropy in Mw
S = 01.0919 + (
mCpwdT∗10−3
Tprev
)
S = 0.10919 + (
144 .44∗4.187∗0∗10−3
327
) = 0.10919 Mw/K
6.22 Low Pressure Heater [LPH1] OUTLET
PRESSURE P=14.71 bar, Temperature T=346K, mass flow rate m=144.44kg/s
For above values from steam tables
Specific volume v =0.00115274 m3/kg
Pressure difference
dP =14.71-15.69 =-0.98bar
dT = 346-327 =19K
Enthalpy in Mw
H =32.91511+ (vdPx102+Cpwdt) x10-3 ×m
H= 32.91511+ (0.001152x-0.98x102+4.187x19) x144.44x10-3 = 44.3894Mw
Entropy in Mw
S = 0.10919+ (
mCpwdT∗10−3
Tprev
)

S = 0.10919+ (
144 .44∗.187∗19∗10−3
327
) =0.14432 Mw/K
6.23 Low Pressure Heater [LPH] 2 INLET
PRESSURE P=14.71 bar, Temperature T=346K, mass flow rate m=144.44kg/s
Specific volume v =0.00115274 m3/kg
Pressure difference
dP =14.71-15.69 =-0.98bar
dT = 346-327 =19K
Enthalpy in Mw
H= 32.91511+(0.001152x-0.98x102+4.187x19)x144.44x10-3
= 44.3894Mw
Entropy in Mw
S = 0.10919+ (
144 .44∗.187∗19∗10−3
327
)=0.14432 Mw/K
6.24 Low Pressure Heater [LPH] 2 OUT LET
PRESSURE P=13.73bar, Temperature T= 368 K, mass flow rate m=144.44 kg/s
Specific volume
V = 0.11476m m3/kg
Pressure difference
dP = 13.73-14.71 = -0.98 bar
dT = 368-346 = 22K
Enthalpy in Mw
H = 44.3894+ (vdPx102+Cpwdt) x10-3 ×m
H = 44.3894+ (0.0011476x102-0.98+4.187x22) x10-3x144.44 = 57.678Mw
Entropy in Mw
S = 0.14432+ (
mCpwdT∗10−3
Tprev
)
S = 0.14432 + (
144 .44∗4.187∗22∗10−3
346
) = 0.1827 Mw/K

6.25 Low Pressure Heater [LPH3] INLET
PRESSURE P = 14.71 bar, Temperature T = 346K, mass flow rate m = 144.44kg/s
Specific Volume
v=0.0011524 m3/kg
Pressure difference
dP = 14.71-15.69 = -0.98bar,
dT = 368-346 =22K
H = 44.3894+ (vdPx102+Cpwdt) x10-3 ×m
H = 44.3894 + (0.00114796x102x-0.98+4.187x22)10-3x144.44 = 57.678Mw
S = 0.14432+ (
mCpwdT∗10−3
Tprev
)
S = 0.14432 +
144 .44∗4.187∗22∗10−3
346
= 0.1827 Mw/K
6.26 Low Pressure Heater [LPH3] OUTLET
PRESSURE P = 6.37 bar, Temperature T = 398 K =125 o c, mass flow rate m =144.44kg/s
For above values
Specific Volume
v = 0.0011023m3/kg
Pressure difference
dP = 6.37-13.73 =-7.36bar
dT =398-368 = 20 K
H = 57.678+ (vdPx102+Cpwdt) x10-3 ×m
H = 57.678 + (0.0011023x-7.36x102+4.187x20)10-3x144.44 = 69.656Mw
S = 0.1827+ (
mCpwdT∗10−3
Tprev
)
S = 0.1827 + (
144 .44∗4.187∗20∗10 −3
368
) = 0.21556 Mw/K
6.27 DEAERATOR INLET
PRESSURE P = 6.37 bar, Temperature T = 398 K =125 o c, mass flow rate m =177.78 kg/s

Specific Volume
v = 0.0011023 m3/kg
Enthalpy
h =676.75 kJ/kg
Entropy
s =1.9458 kJ/kg – K
Enthalpy in Mw
H = hxmfx10-3
= 676.65x177.78x10-3 = 120.31Mw
Entropy in Mw
S = sxmx10-3
= 1.9458x177.78x10-3 = 0.3459 Mw/K
6.28 DEAERATOR OUTLET
PRESSURE P = 6.28bar, T = 433 K, m= 177.78 kg/s
Specific Volume
v = 0.0011032 m3/kg
Pressure difference
dP = 6.28-6.27 =-0.09bar
dT =433-396 = 37K
H =120.31+ (vdPx102+Cpwdt) x10-3×m
H = 120.31+ (0.0011032x-0.09-100+4.182x37)10-3x177.78 = 147.849 Mw
S = 0.3459+(
mCpwdT∗10−3
Tprev
)
S = 0.3459+ (
177.78∗4.187∗37∗10 −3
396
) =0.4154 Mw/K
6.29 BOOSTER PUMP INLET
PRESSURE P = 6.28bar, Temperature T = 433 K, m mass flow rate = 177.78 kg/s
Specific Volume v = 0.0011032 m3/kg
dP = 6.28-6.27 =-0.09bar

dT =433-396 = 37K
Enthalpy H = 120.31+ (0.0011032x-0.09-100+4.182x37)10-3x177.78= 147.849 Mw
Entropy S = 0.3459+ (
177.78∗4.187∗37 ∗10−3
396
)=0.4154 Mw/K
6.30 BOOSTER PUMP OUTLET
PRESSURE P = 14.71 bar, T = 440 K, m= 177.78 kg/s
Specific Volume
v =0.0011527 m3/kg
Pressure difference
dP = 14.1-6.28 = 8.34 bar
dT = 440-433 = 7K
H = 147.89+ (vdPx102+Cpwdt)x10-3×m
H = 147.89 + (0.0011527x8.34x102+4.187x7)177.78x10-3= 153.72 Mw
S = 0.415449+ (
mCpwdT∗10−3
Tprev
)
S = 0.415449+ (
177 .78∗4.187∗7∗10−3
433
) = 0.42748 Mw/K
6.31 Boiler Feed Pump [BFP] INLET
PRESSURE P = 14.71 bar, T = 440K, m= 177.78 kg/s
Specific Volume v =0.0011527 m3/kg
Pressure difference
dP = 14.1-6.28 = 8.34 bar
dT = 440-433 = 7 K
H = 147.89+ (vdPx102+Cpwdt) x10-3×m
H = 147.89 + (0.0011527x8.34x102+4.187x7)177.78x10-3 = 153.72Mw
S = 0.415449+(
mCpwdT∗10−3
Tprev
)
S = 0.415449+ (
177 .78∗4.187∗7∗10−3
433
) = 0.42748 Mw/K
6.32 Boiler Feed Pump [BFP] OUTLET

PRESSURE P = 176.52 bar, Temperature T = 443K, mass flow rate m=177.78 kg/s
Specific Volume v = 0.001828 m3/kg
Pressure difference
dP = 176.52-14.71= 161.81 bar
dT = 443-440 = 3K
H =153.72+ (vdPx102+Cpwdt)x10-3×m
H = 153.72 + (0.001828x161.81x102×4.187x3)177.78x10-3 =161.2237Mw
S = 0.42748+ (
mCpwdT∗10−3
Tprev
)
S = 0.42748 + (
177 .78∗4.187∗10−3
440
) = 0.43255 Mw/K
6.33 HPH 5 INLET
PRESSURE P = 175.54 bar, Temperature T = 443 o c, mass flow rate m = 177.78 kg/s
Specific Volume
v =0.001836 m3/kg
Pressure difference
dP = 175.54-176.52 = -0.98 bar
dT = 443-443 = 0K
H =161.2237+ (vdPx102+Cpwdt) x10-3×m
H = 161.2237+ (0.001836x-0.98x102+4.187x0)177.78x 10-3 = 161.188 Mw
S = 0.43255+ (
mCpwdT∗10−3
Tprev
)
S = 0.43255 + (
4.187 ∗177 .78∗0∗10−3
443
) = 0.43255 Mw/K
6.34 HPH5 OUT LET
PRESSURE P = 173.58 bar, T Temperature =480K, mass flow rate m = 177.78kg/s
At above values from steam tables
Specific Volume
v = 0.001779 m3/kg

Pressure difference
dP = 173.58-175.54 =-1.96 bar
dT = 480-443 =37 K
H = 161.188+ (vdPx102+Cpwdt) x10-3 ×m
H = 161.188 + (0.001779x-1.96x102+4.187x37)177.78x10-3 = 188.674 Mw
S = 0.43255+ (
mCpwdT∗10−3
Tprev
)
S = 0.43255 +(
177.78∗4.187 ∗37∗10−3
443
) = 0.49472 Mw/K
6.35 HPH6 INLET
PRESSURE P = 171.62 bar, Temperature T =480K, mass flow rate m = 177.78kg/s
At above values from steam tables,
Specific Volume v = 0.001792m3/kg
Pressure difference
dP = 171.62-173.58 =-1.96 bar
dT = 480-480 = 0K
H = 188.66+ (vdPx102+Cpwdt) x10-3 ×m
H = 188.66 + (0.001792x-1.96x102+4.187x37)177.78x10-3 = 188.6624 Mw
S = 0.49472+ (
mCpwdT∗10−3
Tprev
)
S = 0.49472 +(
177.78∗4.187 ∗0∗10−3
480
) = 0.49472 Mw/K
6.36 HPH6 OUTLET
PRESSURE P = 171.62 bar, Temperature T=518K, mass flow rate m=177.78 kg/s
Specific volume v = 0.001792m3/kg
Pressure difference
dP = 171.62-171.62 =0 bar
dT = 518-480 = 38 K
H = 188.6624+ (vdPx102+Cpwdt) x10-3 ×m

H = 188.6624 + (0.001792 x 0 x102+40187x38)177.78x10-3 = 216.948 Mw
S = 0.4942+ (
mCpwdT∗10−3
Tprev
)
S = 0.4942+ (
177.78∗4.187∗38∗10 −3
480
) = 0.55364 Mw/K
6.37 FRS INLET
PRESSURE P = 171.62 bar, T Temperature = 518K, mass flow rate m=177.78 kg/s
Specific volume
v = 0.001792m3/kg
Pressure difference
dP = 171.62-171.62 =0 bar
dT = 518-480 = 38 K
H = 188.6624+ (vdPx102+Cpwdt)x10-3 ×m
H = 188.6624 + (0.001792 x 0 x102+40187x38)177.78x10-3 = 216.948 Mw
S = 0.4942+ (
mCpwdT∗10−3
Tprev
)
S = 0.4942+ (
177.78∗4.187∗38∗10 −3
480
) = 0.55364 Mw/K
6.38 FRS OUTLET
PRESSURE P = 163.77 bar, T=518K, m=175 kg/s
Specific Volume v = 0.001717m3/kg
Pressure difference
dP = 163.77-171.62 =-7.85 bar
dT = 518-518 = 0 K
H = 216.948+ (vdPx102+Cpwdt) x10-3 ×m
H = 216.948+ (0.001717x-7.87x102+4.187x0) x175x10-3 =216.712 Mw
S = 0.55364+ (
mCpwdT∗10−3
Tprev
)
S = 0.55364+ (
175 ∗4.187 ∗0∗10−3
518
) = 0.55364 Mw/K

6.39 ECONOMISER INLET
PRESSURE P = 163.77 bar, T, Temperature =518K, m mass flow rate =175 kg/s
Specific volume, v = 0.001717m3/kg
Pressure difference
dP = 163.77-171.62 =-7.85 bar
dT = 518-518 = 0 K
H = 216.948+ (vdP102+Cpwdt) x10-3×m
H = 216.948+ (0.001717x-7.87x102+4.187x0) x175x10-3 =216.712 Mw
S = 0.55364+ (
mCpwdT∗10−3
Tprev
)
S = 0.55364+ (
175 ∗4.187 ∗0∗10−3
518
) = 0.55364 Mw/K
6.40 ECONOMISER OUTLET
PRESSURE P = 157.89 bar, Temperature T=574K, mass flow rate m=175 kg/s
At above values from steam tables,
Specific volume v = 0.001694 m3/kg
Pressure difference dP = 157.89-163.77 = -5.88 bar
Pressure Temperature dT = 574-518= 56 K
H = 216.948+ (vdPx102+Cpwdt) x10-3 ×m
H = 216.948+ (0.001694x-5.88x102+4.187x56) x175x10-3 =257.57 Mw
S = 0.55364+ (
mCpwdT∗10−3
Tprev
)
S = 0.55364+ (
175 ∗4.187 ∗56 ∗10−3
518
) = 0.6328 Mw/K

6.1.1 TABULATED VALUES OF ENTHALPY AND ENTROPY OF THE
COMPONENTS
S.NO Description Pressure
(bar)
Temperature
(K)
Mass flow
rate
(Kg/S)
Enthalpy (H)
MW
Entropy (S)
Mw/k
1. Boiler inlet 157.8 574 175 235.445 0.5724
2. Boiler outlet 146.05 613 175 459.58 0.9349
3. Super heater inlet 146.05 613 177.78 466.88 0.9497
4. Super heater outlet 146.05 813. 177.7 618.0098 1.1665
5. HPH inlet 146.05 813 177.7 618.0098 1.1665
6. HPH outlet 35.30 603 161.11 488.011 1.05202
7. Reheater inlet 35.30 603 161.11 488.011 1.05202
8. Reheater outlet 34.32 813 161.11 577.902 1.1813
9. IPT Inlet 34.32 813 161.11 577.902 1.1813
10. IPT Outlet 6.87 613 152.78 470.716 1.12209
11. LPT Inlet 6.87 613 152.78 470.716 1.12209
12. LPT Outlet 0.0873 322 125 322.53 1.024
13. Condenser Inlet 0.0873 322 125 322.53 1.024
14. Condenser Outlet 0.0873 316 144.44 25.998 0.0883
15. CEP Inlet 0.0873 316 144.44 25.998 0.0883
16. CEP Outlet 18.63 320 144.44 28.7312 0.096022
17. Ejector Inlet 18.63 320 144.44 28.7312 0.096022
18. Ejector Outlet 17.65 325 144.44 34.7385 0.10547
19. GSC Inlet 17.65 325 144.44 34.7385 0.10547
20. GSC Outlet 16.67 327 144.44 32.9315 0.10919
21. LPH 1 Inlet 15.69 327 144.44 32.9151 0.10919
22. LPH 1 Outlet 14.71 346 144.44 44.3894 0.14432
23. LPH 2 Inlet 14.71 346 144.44 44.3894 0.14432
24. LPH 2 Outlet 13.73 368 144.44 57.678 0.1827
25. LPH 3 Inlet 13.73 368 144.44 57.678 0.1827
26. LPH 3 Outlet 6.37 398 144.44 69.656 0.21556
27. Deaerator Inlet 6.37 398 17.78 120.31 0.3459
28. Deaerator Outlet 6.28 433 177.78 147.849 0.415449
29. Booster Pump Inlet 6.28 433 177.78 147.849 0.415449
30. BosterPump Outlet 14.71 440 177.78 153.72146 0.42748
31. BFP Inlet 14.71 440 177.78 153.721 0.42748
32. BFP Outlet 176..52 443 177.78 161.2237 0.43255
33. HPH 5 Inlet 175.54 443 177.78 161.18801 0.43255
34. HPH 5 Outlet 173.58 480 177.78 188.6674 0.49472
35. HPH 6 Inlet 177.62 480 177.78 188.6624 0.49472
36. HPH 6 Outlet 171.62 518 177.78 216.948 0.55364
37. FRS Inlet 171.62 518 175 216.948 0.55364

38. FRS Outlet 163.77 518 175 216.712 0.55364
39. Economiser Inlet 163.77 518 175 216.712 0.55364
40. Economiser Outlet 157.89 574 175 257.57 0.6328

6.2 Thermodynamic properties of steam at extractions
Extraction
The principle of regeneration can be practically utilized by extracting steam from turbine at
several locations and supply it to the regenerative heater. The most advantageous condensate
heating temperature is selected depending on the throttle conditions and this determines the
number of heaters to be used. Figure shows the layout of condensing steam power plant in which
a surface condenser is used to condense all the steam that is not extracted for feed water heating.
The turbine is double extracting and boiler is equipped with a super heater.
Extraction 1 at Low Pressure Turbine
Pressure P=0.216 bar, Temperature T=346 k=73oc, m=5.56 kg/s
h = hg =2612.42 kJ/kg
Enthalpy at extraction1
H = h×m×10-3
= 2612.42×5.56×10-3 =14.5250 MW
s = sg =7.8824 kJ/kg-K
Entropy at extraction1
S = s×m×10-3
= 7.8824×5.56×10-3 =0.0438 MW/K
Pressure P = 0.858 bar, Temperature T = 380 K, m= 6.94 kg/s
At above values,
h =18.515 kJ
s =7.411 kJ/kg-K
Enthalpy at extraction2 ,
H = h×m× 10-3
= 2668×6.94×10-3= 18.515 MW
Entropy at extraction2,
S = s×m×10-3
= 7.4118×6.94×10-3 =0.0514 MW/K

Pressure P = 2.37 bar, Temperature T =473K , m=6.94 kg/s
At above values,
h =2713.17 kJ/kg
H = h×m×10-3
= 2713.17×6.94×10-3= 18.8293 MW
s = 7.0702 kJ/kg-K
S = s×m×10-3
S = 7.072×6.94×10-3= 0.0490 MW/K
Extraction 4 at Intermediate Pressure Turbine
Pressure P = 6.87 bar, Temperature T=613 K, m = 8.33 kg/s
At above values,
h = 2761.58 kJ/kg
Enthalpy at exraction4
H = h×m×10-3
= 2761.58×8.33×10-3= 23.0039 MW
S = s×m×10-3
= 6.7115×8.33×10-3 = 0.0559 MW/K
Extraction 5 at High Pressure Turbine
Pressure P= 16.7 bar, Temperature T= 706k, m = 8.33 kg/s
At above values,
h = 2792.6 kJ/kg
H = h×m×10-3
=2793.28×8.33×10-3=23.26MW
s = 6.4026 kJ/kg-K
S = s×m×10-3 = 6.4026×8.33×10-3 = 0.0533 MW/K

Extraction 6 at High Pressure Turbine
Pressure P= 39.23 bar, Temperature T= 616 K = 343 C, m= 16.67 kg/s
For above values,
h =2800.685 kJ/kg
H = h×m×10-3
= 2800.685×16.67×10-3 =46.687 MW
s =6.0767 kJ/kg-K
S = s×m×10-3
= 6.0767×16.67×10-3 = 0.1013 MW/K
6.2.1 TABULATED VALUES OF THERMODYNAMIC EXTRACTIONS
Pressure
(bar)
Temperature
(K)
Mass
(Kg/S)
Enthalpy
(h)
KJ/Kg
Enthalpy
(H)
MW
Entropy
(s)
KJ/Kg-k
Entropy
(S)
Mw/k
Extraction
1(LPT)
0.216 346 5.56 2612.42 14.5250 7.8824 0.0438
Extraction
2(LPT)
0.858 380 6.94 2668 18.515 7.4118 0.0514
Extraction
3(LPT)
2.37 473 6.94 2713.17 18.8293 7.0702 0.0490
Extraction
4(IPT)
6.87 613 8.33 2761.58 23.0039 6.7115 0.0559
Extraction
5 (HPT)
16.70 706 8.33 2793.28 23.26 6.4026 0.0533
Extraction
6 (HPT)
39.23 616 16.67 2800.68 46.687 6.0767 0.10129

6.3 TABULATED VALUES OF TURBINES
6.3.1 HIGH PRESSURETURBINE (HPT)
HPT h (kJ/kg) v (m3
/kg) m (kg/s)
Inlet 3476.26 0.023959 177.78
Extraction 6 2880.68 0.0507564 161.11
Outlet 3029.056 0.071115 161.11
Work done by HPT = ( h in – h out ) × mass
= (3476.26-3029.056)×177.78 = 79.503 MW
6.3.2 INTERMEDIATE PRESSURETURBINE (IPT)
IPT h (kJ/kg) v (m3
/kg) m(kg/s)
Inlet 3587.0056 0.1097 161.11
Extraction 4 2793.28 0.118692 152.78
Outlet 3081.0094 16.68792 152.78
Work done by IPT = ( hin – hout ) × mass
Work done by the IPT = (3587.00-3081.00)×161.11
= 81.51 MW
6.3.3 LOW PRESSURETURBINE (LPT)
LPT h (kJ/kg) v (m3
/kg) m (kg/s)
Inlet 3081.0094 3865.93 144.44
Extraction 3 2713.17 0.75549 137.50
Extraction 2 2668 1.95474 130.56
Extraction 1 2612.42 7.11948 125
Outlet 2580.26 16.68792 125
Work done by LPT = ( hin – hout ) × mass
= (3081-2580.26)×144.4

= 72.328 MW
CHAPTER-7
7.0 EXERGY AND ENERGYANALYSIS ON THE COMPONENTS
7.1 EXERGY ANALYSIS:
7.1.1 Exergybalance of High pressure turbine:
The exergy balance for the high pressure turbine is given by :
10→HPT inlet
11→HPT ext
14→HPT outlet
Ψ = H-To.S
Work done by high pressure turbine,
W HPT = ˙ m10(Ψ10 − Ψ11) + (˙ m10 −˙ m11) (Ψ11 − Ψ14) – To× ˙Sgen
This gives:
T0 ×˙Sgen = ˙ m10(Ψ10 − Ψ11) + (˙ m10 −˙ m11) ( Ψ11 − Ψ14) –WHPT
˙ m10(Ψ10 − Ψ11) + (˙ m10 −˙ m11) ( Ψ11 − Ψ14) = Exergy input
And the entropy generation rate is:
˙Sgen = ˙ m10(s11 − s10) + (˙ m10 −˙ m11)(s14 − s11)
Irreversibility destroyed = exergy loss is:
˙I destroyed = To. ˙Sgen = To[˙ m10(s11 − s10) + (˙ m10 −˙ m11)(s14 − s11)]
The second law efficiency is:
ήII,HPT = 1 − (˙Idestroyed ÷ ˙m10(Ψ10 −Ψ11) + (˙ m10 −˙ m11)( Ψ11 −Ψ14))
= WHpT ÷ ˙ m10(Ψ10 −Ψ11) + (˙ m10 −˙ m11)(Ψ11 − Ψ14)
=1-(Exergy loss/Exergy input)

= WHPT/Exergy input
Data:
Exergy of high pressure turbine inlet
Ψ10 = 618.009 − (301×1.1665) = 266.89 kW
Exergy of high pressure turbine extraction
Ψ11 = 23.26 − (301×0.0533) = 7.2167 MW
Exergy of high pressure turbine outlet
Ψ14 = 488.011 − (301×1.0520) = 171.359 MW
To = 301K
By substituting in the above equations we get:
˙Sgen = 177.78 (6.56179−6.4026) + (177.78 − 8.33) (6.529 − 6.4026)
=28.3 + 21.418 = 49.718 kW/K
To ˙Sgen = 301×49.718 = 14965.26 kW =14.96 MW
Work done by high pressure turbine
W HPT = 177(266.89 −7.2167) + (177.78 − 8.33) (171.359 − 7.216) – 14965.26
= 73978.749 −14965.26 = 59013.489 kW
Second law efficiency of high pressure turbine,
ήII,HPT = 59013.489 ÷ 73978.74 = 0.7977×100 = 79.77 %
7.1.2 Exergybalance of Intermediate pressure turbine:
The exergy balance for the Intermediate pressure turbine is given by:
15→IPT inlet
19→IPT outlet
23→IPT ext (deaerator)
W IPT = ˙ m15 (Ψ15 −Ψ19) +˙ m23 ( Ψ19 −Ψ23) – To ×˙Sgen
This gives:

To× ˙Sgen = .m15 (Ψ15 −Ψ19) +˙ m23( Ψ19 −Ψ23) –WIPT
.m15 (Ψ 15 −Ψ 19) +˙ m23 ( Ψ 19 −Ψ 23) = Exergy input
and the entropy generation rate is:
˙Sgen = ˙ m23(s23 − s19) + ˙ m15(s19 − s15)
Irreversibility = exergy loss is:
˙I destroyed = To .˙Sgen
= To[˙m23(s23 − s19) + ˙ m15(s19 − s15)]
ήII,IPT = 1 – (˙Idestroyed ÷ ˙ m15 (Ψ15 −Ψ19) +˙ m 23( Ψ 19 –Ψ 23))
= W IPT ÷ ˙ m 15 (Ψ 15 –Ψ 19) +˙ m 23( Ψ 19 –Ψ 23)
= 1-(Exergy loss/Exergy input)
= W IPT /Exerg input
Data:
Exergy of intermediate pressure turbine inlet
Ψ 15=577.902−(301×1.1813)=222.33 MW
Exergy of intermediate pressure turbine outlet,
Ψ 19 =470.716−(301×1.122)=132.994 MW
Exergy of intermediate pressure turbine extraction,
Ψ 23 =23.2039−(301×0.0559)=6.378 MW
Entropy generated, ˙Sgen=8.33(7.34−6.71)+161.11(7.34−7.3326)
=7.189 kW/K
To ˙Sgen=301×7.189=2163.95 Kw =2.16 mW
Work done by intermediate pressureturbine,
W IPT =161.11(222.33−132.994)+8.33(132.994−6.378)−2163.95

=15448.278−2163.95=13284.328 kW
Second law efficiency of intermediate pressure turbine,
ήII,IPT=13284.328÷15448.278 =0.8599×100 =85.99%
7.1.3 Exergybalance of Low pressure turbine:
The exergy balance for the Low pressure turbine is given by :
1→LPTinlet
2→LPToutlet
3→LPText3
4→LPT ext2
5→LPText1
W LPT = .m 1 (Ψ1−Ψ3)+(.m1−.m3) ( Ψ3−Ψ4)+(.m1−.m3−.m4 )+(.m1−.m3−.m4−.m5)
( Ψ5−Ψ2) − To×˙Sgen
This gives:
To×˙Sgen = .m1(Ψ1−Ψ3)+(.m1−.m3) ( Ψ3−Ψ4)+(.m1−.m3−.m4)+(.m1−.m3−.m4−.m5)
( Ψ5−Ψ2) −W LPT
and the entropy generation rate is
˙Sgen = .m1(s2−s1)+.m3(s3−s2)+.m4(s4−s2)+.m5(s5−s2)
Irreversibility destroyed = exergy loss is:
˙ I destroyed = To ˙Sgen
= To(.m 1(s2−s1)+.m3(s3−s2)+.m4(s4−s2)+.m5(s5−s2))
ήII,LPT = 1 –( ˙Idestroyed ÷ .m1(Ψ1−Ψ3)+(.m1−.m3)(Ψ3−Ψ4)
+(.m1−.m3−.m4)+(.m1−.m3−.m4−.m5) ( Ψ5−Ψ2)

= W LPT ÷ .m1(Ψ1−Ψ3)+(.m1−.m3)(Ψ3−Ψ4)
+(.m1−.m3−.m4)+(.m1−.m3−.m4−.m5) ( Ψ5−Ψ2)
= W LPT /Exergy input
Data:
Mass of low pressure turbine inlet,
.m1=144.4 kg/s
Exergy of low pressure turbine inlet,
Ψ1→445.02−(301×1.0608)=125.71 MW
Mass of low pressure turbine outlet,
.m2=125 kg/s
Exergy of low pressure turbine outlet,
Ψ2→322.53−(301×1.024)=14.306 MW
Mass of low pressure turbine extraction3,
.m3=6.49 kg/s
Exergy of low pressure turbine extraction3,
Ψ3→18.829−(301×0.049)=4.08 MW
.m4=6.94 kg/s
Ψ4→18.515−(301×0.0513)=3.073 MW
.m5=5.56 kg/s
Ψ5→14.525−(301×0.0438)=1.3412 MW

Entropy generated,
˙Sgen=144.4(7.68−7.45)+6.94(7.68−7.421)+6.94(7.68−7.65)+5.56(7.88−7.68)
=31.4 kW/K
To ˙Sgen=301×31.4=9451.4 kW
Work done by low pressure turbine,
WLPT =144.4(125.7−14.3)+6.94(14.3−4.85)+6.94(14.3−4.12)+5.56(14.3−2.4)
=16288.526−9451.4=6837.126 kW
Efficiency of low pressure turbine,
ήII,LPT=6837.126÷16288.52 =0.4197×100 =41.97%
7.1.4 Exergybalance of condenser:
The exergy balance for the condenser is given by :
Ψ1→Condenserinlet
Ψ30→CEP inlet/condenser outlet
Ψw=˙ m30(Ψ30 – Ψ1) − ∑n
k=1 (1 – (To ÷Tk) )Qk − To× ˙Sgen
0= ˙ m30(Ψ30 – Ψ1) − ∑ n
k=1 (1 – (To ÷Tk) )Qk − To ×˙Sgen
This gives:
To ×˙Sgen = ˙ m30(Ψ30 –Ψ1) − ∑n
k=1 (1 – (To÷ Tk))Qk
˙ m30(Ψ30 –Ψ1) =Exergy input
Irreversibility destroyed= exergy loss is:
˙Idestroyed=To ˙Sgen = [{˙ m30(h30 – h1)} − To{˙ m30(s30 – s1)}] − ∑n
k=1 (1 –
(To÷Tk))Qk
ήII,Condenser = 1 – (˙Idestroyed ÷ ˙ m30(Ψ30 – Ψ1))

Data:
Exergy of condenserinlet,
Ψ1 =25.998−(301×0.0883)=−0.5803 MW
Exergy of condenseroutlet,
Ψ30 =372.48−(301×1.183)=16.397 MW
To ˙Sgen= 4.05×1000 kW
Ψw=144.4(16.397+0.5803)−(6.5051×1000)−4.05×1000=0
Second law efficiency,
ήII,Condenser= 1− (4.05×103 ÷2.45×103 ) = 0.6530×100 =65.30%
7.1.5 Exergybalance of super heater:
m.
g(Ψ gi−Ψ go)+m.
s(Ψ si−Ψ so)−E.
Qsh= I.
SH
m.
g(Ψ gi−Ψ go)+m.
s(Ψ si−Ψ so)−(1−To÷Tk)Qk=I.
SH
m.
g(Ψ gi−Ψ go) = Exergy input
Irreversibility of superheater = exergy loss
I.
SH=243.05(958.95−68.47)+177.78(266.89−18.02)−(111.18×1000)
=216430.92+15265.96−(111.18×1000) =120516.88 kW =120.516 MW
To.s.
gen=m.
s(hsi−hso)−To(m.
s(ssi−sso))+∑(1−(To÷Tk))Qk
=177.78(2626.2−3476.2)−301(177.78(5.34−6.56))+111180
=−151123.66−301(−216.89)+111180=25340.71 kW =25.34 MW
Exergy input= 243.05.(958.95-68.47)=243.05.890.47=216430.92 kW
ή II=1−(Exergy loss ÷ Exergy input)

= 1−(120516.881 ÷ 216430.92) =0.443×100 =44.3%
7.1.6 Exergy analysis of boiler:
m.
g(Ψgi−Ψgo)+m.
b(Ψbi−Ψbo)−E.
b= I.
destroyed
m.
g(Ψgi−Ψgo)+m.
b(Ψbi−Ψbo)−(1−(To ÷ Tk))Qk= I.
destroyed
m.
g(Ψgi−Ψgo) = Exergy input =243.05*890.47=216430.92 kW
I.
destroyed = 243.05(958.95−68.47)+175(178.17−63.717)−163770
=216430.92+20029.27−163770
=72690.19 kW =72.69 MW= exergy loss
To
.s.
gen = mw (hbi−hbo) −To(mw(sbi−sbo))+∑(1−(To ÷ Tk))Qk
=175(1345.4−301 (175(−5.3+3.2))+(16.377×1000)
=−224140+110617.5+163770
=50247.5 kW =50.24 MW
ή II = 1−(Exergy loss ÷ Exergyinput)
= 1−(72690.19 ÷ 216430.92)
=0.6641×100 =66.41%
7.2 ENERGY ANALYSIS
7.2.1 Energybalance of High PressureTurbine:
The energy balance for the high pressure turbine is given by :
10→HPT inlet
11→HPT ext
14→HPT outlet
WHPT = ˙ m 10(h10 – h11) + (˙ m10 −˙ m11)(h11 – h14) − Energy loss
This gives :

Energy loss = ˙ m10(h10 – h11) + (˙ m10 −˙ m11)(h11 – h14) –WHPT
˙ m10(h10 – h11) + (˙ m10 −˙ m11)(h11 – h14) = Energy input
Energy input= 177.78(3476.26−2983.41)+(177.78−161.11)(2983.41−2793.28)
=90788.33 kW
WHPT = 90788.33− Energy loss
79.505×1000 = 90788.33 –Energy loss
Energy loss = 11283.33 kW
The first law efficiency of high pressure turbine is:
ηI,HPT = 1 – (Energy loss ÷ ( ˙ m10(h10 − h11) + (˙ m10 −˙ m11)(h11 − h14))
= = WHPT ÷ (˙ m10(h10 − h11) + (˙ m10 −˙ m11)(h11 − h14))
=1-(Energy loss/Energy input)
=WHPT/Energy input
= 1−(11283.33 ÷ 90788.33)
= 0.875×100 =87.5%
7.2.2 Energybalance of Intermediate Pressure Turbine:
The energy balance for the intermediate pressure turbine is given by :
15→IPT inlet
19→IPT outlet
23→IPT ext (deaerator)
WIPT = m.
15(h15−h19)+m.
23(h19−h23) − Energy loss
This gives :
Energy loss = m.
15(h15−h19)+m.
23(h19−h23) –WIPT
Energy input = m.
15(h15−h19)+m.
23(h19−h23) =84182.42 kW
WIPT = 161.11(3587.00−3081) + 8.3(3081−2761.58) – Energy loss

(81.51×1000) = 84182.42−Energyloss
The first law efficiency of intermediate pressure turbine is :
ηI,IPT = WIPT ÷ m.
15(h15−h19)+m.
23(h19−h23)
= WIPT /Energy input
= (81.51×103) ÷ 84182.42 =0.968×100 =96.8%
7.2.3 Energybalance of Low Pressure Turbine:
The energy balance for the low pressure turbine is given by:
1→LPT inlet
2→LPT outlet
3→LPT ext3
4→LPText2
5→LPT ext1
WLPT = m.
1 (h1−h2) +m.
3 (h2−h3) +m.
4 (h2−h4)+m.
5(h2−h5)− Energy loss
This gives:
Energy loss = m.
1(h1−h2)+m.
3(h2−h3)+m.
4(h2−h4)+m.
5(h2−h5)−WLPT
Energy input = m.
1(h1−h2)+m.
3(h2−h3)+m.
4(h2−h4)+m.
5(h2−h5) =74014.62 kW
WLPT =144.4(3081−2580.26)+6.94(2580.26−2713.17)+6.94(−2580.26+2668)+5.56(2612−2580.26)
−Energy loss
72.32×10
3
= 74014.62 − Energy loss
The first law efficiency is :
ηI,LPT = 1− ( Energy loss ÷ m.
1(h1−h2)+m.
3(h2−h3)+m.
4(h2−h4)+m.
5(h2−h5) )

= 1-(Energy loss/ Energy input )
=1− (1694.62 ÷ 74014.62)=0.977×100 =97.7%
7.2.4 Energybalance of condenser:
The energy balance for the condenseris given by :
2→condenserinlet
3→CEP inlet /condenser outlet
0 = ˙m30(h30 – h1) – Qk − Energy loss
This gives :
Energy loss = ˙m30(h30 – h1) − Qk
= 144.4(2579.5−179.99)−(307.74×103)
= 38845.22 kW
Energy input = ˙m30(h30 – h1) =144.4(2579.5-179.99)= 346585.22 kW
The first law efficiency is :
ηI,Condenser = 1 – (Energy loss ÷ ˙m30(h30 − h1))
= 1-(Energy loss/Energy input)
=1−(38845.22 ÷ 346585.22)=0.8879×100 =88.79%
7.2.5 Energybalance of super heater:
The energy balance for the super heater are :
Wsup = Qk − m.
g (hgi−hgo) − m.
s (hsi−hso) – Energy loss
O = Qk − m.
s (hsi−hso) – Energy loss
Energy loss = Qk − m.
s (hsi−hso)
Energy loss=(148.908×103)−243.05(1111.969−509.789) −(177.78(2626.2−3476.26))
=148908−292189.849−(−151123.66)
=148908−292189.849+151123.66

=300031.66−292189.8 =7841.86 kW=7.841 MW
Energy input = 148908 kW
First law efficiency of condenser
ηI= 1− (Energy loss ÷ Energy input )
= 1− (7841.86 ÷ 148908)
= 0.947×100 =94.7%
7.2.6 Energy balance of boiler:
The energy balance for the combustion/ boiler is given by :
0 = Qk −˙mw(h10 – h9) −˙ms(h15 – h14)] − Energy loss
where mw is the mass flow rate of water, ms is the mass flow rate of steam
combustion which gives:
Energy loss = Qk −˙mw(h10 – h9) −˙ms(h15 – h14)
Energy loss = (203.04×10
3
)−177.78(3476.26−2626.2)−161.11(587.7−3029.056)
=241014.017−(203.04×10
3
) =37974.017 kW
Energy input = 203.04×1000 kW
The first law efficiency of boiler is defined as
ηI,Boiler =( Energy output ÷ Energy input)
= 1 – (Energy loss ÷ Energy input)
= 1−(37974.017÷(203.04×103))
=0.842×100 =84.2%

7.3 TABLES OF THE EXERGY, ENERGY EFFICIENCIES AND
LOSSES
7.3.1 First law and Second law Efficiency table:
Component
name
First law
efficiency(ηI)
Second law efficiency(ήII)
HPT 87.5 % 79.77%
IPT 96.8% 85.99%
LPT 97.4% 41.97%
Superheater 94.79% 44.3%
Condenser 88.79% 65.3%
Boiler 84.85% 66.41%
7.3.2 Energyand Exergylosses table:
Component name Energy loss(mW) Exergy loss(mW)
HPT 11.283 14.96
IPT 2.16 2.67
LPT 1.69 9.45
SUPERHEATER 7.84 25.34
CONDENSER 38.84 4.050
BOILER 37.97 72.6

CHAPTER-8
8.0 COMPARISON GRAPHS BETWEEN EXERGY AND ENERGY
8.1 EXERGY DESTRUCTION GRAPH
X-axis –components, Y-axis- Exergy loss in %
8.2 TURBINE EXERGYEFFICIENCYAND DESTRUCTIONGRAPH
X-axis- Turbines, Y-axis-Exergy loss in %
72.6
25.34
14.96
2.16
9.45
4.05
0
10
20
30
40
50
60
70
80
Exergy destruction
Series 1
Series 2
0
10
20
30
40
50
60
70
80
90
100
HP turbine IP turbine LP TURBINE
Exergy
efficiency
Turbine Exergy Efficiency &
distruction
Turbine Exergy

8.3 EXERGY Vs ENERGYEFFICIENCYGRAPH
X-axis- Exergyefficiency of components, Y-axis- Energyefficiency of
components
8.4 COMPARISON CHARTS
1
X-axis-Energy loss of components, Y-axis-Exergy loss of components
0
10
20
30
40
50
60
70
80
90
100
Exergy
Exergy
Exergy v/s energy efiiciency
Series 1
Series 2
Comparison charts

CHAPTER-9
9.0 CONCLUSION
Now-a-days there are few methods to measure the performance of a
power plant. Some researchers use the conservation of mass and conservation of energy
principles (first law of thermodynamics), however the evaluation is not actually complete. The
exergy analysis based on the second law of thermodynamics should be included in order to do a
complete analysis, which can also access accurately the utilization of energy. This method
provides the information which are useful for engineers or managers to know about the power
plant performance. The information obtained on the result of the analysis will form a basis for
the energy manager or operation engineer to make decisions on how he should operate the plant
in order to save cost and energy usage.
This project has presented the results of an exergy analysis performed
on 210 MW power plant. The analysis was applied on the unit with running load of 210 MW.
Exergy destruction on the plant components are also presented and energy losses are discussed.
The results of the exergy indicate that boiler produces highest exergy destruction of 72 MW.
Comparing the 3 turbine stages, the results of the analysis indicate that HPT produces highest
exergy destruction than IPT and LPT.
The exergy destroyed in the turbines, super-heaters are small compare
to exergy destroyed in the boiler. It is apparent from the analysis,72% of the total exergy
destruction occurs in the boiler. This large exergy loss is mainly due to the combustion reaction
and to the large temperature difference during heat transfer between the combustion gas and
steam. The factors that contribute to high amount of irreversibilities are tubes fouling, defective
burners, fuel quality, inefficient soot blowers, valves steam traps and air heaters fouling.
Inspections of this equipment need to be carried out during the boiler outage. Other factors like
heat loss, incomplete combustion and exhaust losses. This study pinpoints that boiler requires
necessary modification to reduce its exergy destruction, there by performance can be improved.
The exergy losses in the turbines are due to the frictional effects and
pressure drops across the turbine blades as well as the pressure and heat losses to the

surroundings. The HPT, IPT, LPT constitutes a combined 28% of the total exergy destruction
which indicates a need for reducing its irreversibilities. Other factors that may contribute to the
irreversibilities are most likely due to the throttling losses at the turbine governor valves, silica
deposited at the nozzles and blades. Amongst the three turbines, HPT produces the highest
exergy destruction. Overhauling of the turbine may be needed to check the real causes for
improving the plant performance. All this information complemented by the engineers intuition
and judgement, can assist in the improvement of efficiency and the reduction in generation cost.
9.1 RECOMMENDATIONS FOR FURTHER STUDIES
 An exergetic- economic analysis of the plant and of different potential options available
for the plant improvement.
 The measuring devices are necessary to measure the different values and to proceed
further analysis.

CHAPTER-10
10.0 BIBILOGRAPHY
 Cengel Y.A. Boles M.A.,”Thermodynamics :An Engineering Approach ,”2nd
edition,Mc.Graw Hill,1994.
 Kotas T.J., “ The exergy method of Thermal plant Analysis, ” Krieger Publishing
Co.,1995.
 S.C.Kaushik, V.Siva Reddy, S.K.Tyagi,”Renewable and Sustainable Energy Reviews”15
(2011)”
 T.Ganapathy, N.Alayamurthy, R.P.Gowkhar and K.Murugesan “ Journal of engg science
and Tech Review 2 (2009)”
 Mali Sanjay.D, Dr.Mehta NS, “International journal of advanced engg research and
studies E-ISSN2249-8974”
 I.Satyanarayana,A.V.S.S.K.S.Gupta and Dr.k.Govinda Rajulu, “International Journal of
Engineering (IJE)”
 A.Hepbasli, “Renewable and sustainable energy reviews 12(2008).”
 Vundala Siva Reddy, Shubash Chandra Kaushik, Sudhir kumar Tyagi, Naraya Lal
Panwar, “Smart grid and renewable energy,2010.”
 Ravi Prakash Kurkiya, Sharad Chaudhary, “International journal of scientific and energy
research volume 3,2012.”
 Vosough Amir, “2nd international conference on Mecanical,2012.”
 P.K.Nag, “Engineering Thermodynamics, 4th edition, Mc.Graw Hill, 1995.”
 Sam Cooper, Energy and exergy analysis. People.bath.ac.uk/en8c.
 A.GALOVIC, M.ZIVIC, M.kokanovic, “ Analysis of exergy destruction of condenser-
1987”
 A.Rashad, A.El Maihy, “13th International conference on Aerospaces and Aviation
Technology.”
 www.bhel.com (bhel maintenance manuals)
 www.suzlon.com and www.scribd.com

 www.plantmaintenance.com and www.apgenco.gov.in/

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