The Presentation describes and calculates the LMW 210 MW Steam Turbine with actual parameters of steam, condensate and feed water

1. 210 MW LMW Turbine Cycle
Heat Rate Calculations
Heaters, Condensers Designs
Heat Balance, Features
Turbine Heat Rate
3/2/2021 1
Manohar Tatwawadi

2. 3/2/2021 Manohar Tatwawadi 2
210 MW LMW Turbine

3. 3/2/2021 Manohar Tatwawadi 3
HPT Glands
HPT
AS
39
To GC 1 Steam from
Deaerator/PRDS
AS55
OR
AS57
To LPH 4
thru ES12
To HPH 6
Sealing Chambers
Sealing Chambers
1 2 3 4 5
4 3 2 1
To GC 2

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IPT Glands
To GC 1
4 3 2 1
Steam from
Deaerator/PRDS
To GC 2
AS
40
AS55
OR
AS57
Sealing Chambers
Sealing Chambers
1 2 3
IPT
To LP4

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GLANDS
SHAFT

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LPT Glands
To GC 1
Steam from
Deaerator/PRDS
AS55
OR
AS57
1 2
2 1
Sealing
Chambers
Sealing Chambers
LPT

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Ejector
P- 4.5, T - 155
F - 1500, E-659.83
Steam from Deaerator
Condensate
to G.C.1
T - 47.27
F - 1500
T - 45.07
F - 480106
T - 43.16
F - 480106
Condensate
from CEP
Discharge
Drain to Flash
Chamber
Output Heat taken by condensate
Input heat given by steam
480106(45.07-43.16)*100
1500 (659.83-47.27)
= = 99.8 %

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GLAND STEAM CONDENSER GC1
T - 100
F - 1660
Drain to Cond
Flash Tank
T - 45.07
F - 480106
Condensate
from Ejector
P- 4.5, T - 155
F - 400, E-659.83
Steam from Deaerator
Condensate
to LPH 1
T - 47.18
F - 480106
Ejector
GLAND STEAM
COOLER (GC1)
Gland Steam
Condenser Effy.
480106(47.18-45.07)*100
1260(728.20-100)+400(659.83-100)
= = 99.7%
Steam Air
Mixture from
glands
P- 0.97, T - 280
F - 1260, E-728.20

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LOW PRESSURE HEATER NO 1
Condensate
to GC 2
T - 61.05
F - 480106
T - 47.18
F - 480106
Condensate
from GC 1
T - 65.93
F - 11989
Drain to flash
chamber in Cond.
(hd1)
P- 0.89, T - 85
F - 11989, E-622.05
Steam from
Extraction e1
from LPT
Effy of LPH No 1
480106(61.05-47.18)*100
11989(622.05-65.93)
= = 99.8%
LPH 1

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GLAND STEAM CONDENSER NO 2
Condensate
to LPH 2
T - 68.10
F - 480106
T - 61.5
F - 480106
Condensate
from LPH 1
T - 72.94
F - 4949
Drain to flash
chamber on Cond.
(gd2)
P- 0.50, T - 325
F - 4949, E-756.90
Steam from HP
& IP Glands g2
Effy of Gland
Steam
Condenser GC 2
0.8*480106(68.10-61.05)*100
4949(756.80-72.94)
= =80% (20% of
condensate is
bypassed)
80%
20%
Flow through GC2
GC 2

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LOW PRESSURE HEATER NO 2
Condensate
to LPH 3
T - 101.52
F - 555281
T - 68.10
F - 480106
Condensate
from GC 2
T - 105.78
F - 75175
Drain to
Condensate line
(hd2)
P- 1.369, T - 183
F - 25369, E-678.41
Steam from IPT
out e2
Effy of LPH 2
480106(100.86-68.10)*100
25369(678.41-105.78)+49806(130.26-105.78)
= = 99.8%
LPH 2
Drip Pump
Drain from
LPH3 (hd3)
T - 130.26
F - 49806
T-100.86, F-480106

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LOW PRESSURE HEATER NO 3
Condensate
to LPH 4
T - 125.27
F - 555281
T - 101.52
F - 555281
Condensate
from LPH2 &
Drip Pump
T - 130.26
F - 49806
Drain to LPH 2 (hd3)
P- 2.974, T - 264
F - 21033, E-715.68
Steam from IPT
Extr (e3)
Effy of LPH 3
555281(125.27-101.52)*100
21033(715.68-130.26)+28773(161.73-130.26)
= = 99.8%
LPH 3
Drain from
LPH4 (hd4)
T - 161.73
F - 28773

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LOW PRESSURE HEATER NO 4
Condensate
to Deairator
T - 156.52
F - 555281
T - 125.27
F - 555281
Condensate
from LPH3
T - 161.73
F - 28773
Drain to LPH 3 (hd4)
P- 6.911, T - 364
F - 23838, E-762.69
Steam from IPT
Extr (e4)
Effy of LPH 4
555281(156.52-125.27)*100
23838(762.69-161.73)+4395(782.31-161.73)
= = 99.7%
LPH 4
Leak off
steam HP
& IP (g3)
P-6.9, T - 161.73
F - 4935, E-782.31

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DEAERATOR - FEED TANK
Deaerator
pegging steam
From PRDS or
Extr.6(ed)
Drain from HPH
5/6 (hd5 or hd6)
Spindle leak off
from ESV/IV
Steam supply to
glands (d1)
Steam supply to
Ejectors (main
& GC1) d2+d4
Condensate
from LPH4
P-7.5, T - 479
F - 2300, E-820.28
P-1.03, T - 141
F - 2930, E-659.83
P-4.5, T - 155
F - 1900, E-659.83
T - 181.51
F - 93256
E - 802.51
F - 5984
T - 156.52
F - 555281
Feed water
to HPH5
T - 173.12
F - 652000
(555281(166.71-156.52)+4830(820.28-659.83))*100
2300(820.28-166.71)+93256(181.51-166.71)+5984(802.92-166.71)
Deaerator Effy = = 96.8%
T - 166.71
F - 652000
BFP
FEED TANK

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Deaerator Working
Note:-
1. Below 150 MW, Deaerator heating steam is tapped from HPT
at HPT exhaust.(Extraction 6)
2. Above 150 MW, Heating steam supplied from Extraction No 5
from IPT.
3. Above 150 MW, Aux Steam Supply to Main Ejector, Gland
Steam Ejector and sealing steam of Turbine Glands is changed
over from Aux PRDS to Deaerator.
4 Gland leakage Steam from ESV & IV, Drains of Heater No 5
and 6 are also fed to the deaerator.
5. Pegging steam supply is initially from PRDS and changed
over to Extraction 6 at about 100 MW Load.

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Boiler Feed Pump
The total quantity of feed water supplied from feed water tank
to boiler feed water pump is equal to the main steam flow to
the HP Turbine
Heat Added by Boiler feed pump due to mechanical churning
of water is also taken into consideration and equals to:-
T=166.71
T=173.12
Flow= 652000
= 652000(173.12-166.71) = 4179320 Kcal.
BFP

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Feed Water
to HPH 6
T - 189.69
F - 652000
T - 173.12
F - 652000
Feed water
from BFP
Discharge
T - 181.51
F - 93256
Drain to Deaeator (hd5)
P- 12.96, T - 444
F - 16640, E-802.92
Steam from IPT
Extr (e5)
Effy of HPH 5
652000(189.69-173.12)*100
16640(802.92-181.51)+76616(198.05-181.51)
= = 93.7%
HPH 5
Drain from
HPH6 (hd6)
T - 198
F - 76616, E-198
HIGH PRESSURE HEATER NO 5

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Feed Water
to HPH 7
T – 228.80
F - 652000
T – 189.69
F - 652000
Feed water
from HPH 5
T - 198.01
F - 76616
Drain to HPH 5 (hd6)
P- 28.07, T - 327
F - 40391, E-732.99
Steam from
HPT Exh. (e6)
Effy of HPH 6
652000(228.80-189.69)*100
40391(732.99-198.05)+31844(238.65-198.05)+4381(806.50-198.05)
= = 99.7%
HPH 6
Drain from
HPH7 (hd7)
T - 238
F - 31844, E-238.65
HIGH PRESSURE HEATER NO 6
Spindle
Leakage from
HPT (g5)
P- 28.07, T - 464
F - 4381, E-806.50

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HIGH PRESSURE HEATER NO 7
Feed Water to
Economiser T – 253.94
F - 652000
T – 228.80
F - 652000
Feed water
from HPH 6
T – 238.65
F - 31844
Drain to HPH 6 (hd7)
P- 42.18, T - 381
F - 31844, E-755.97
Ext. Steam
from HPT (e7)
Effy of HPH 7
652000(253.94-228.80)*100
31844(755.97-238.65)
= = 99.5%
HPH 7

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H.P.Turbine
MS Inlet P-130, T=535,
F=652000, E= 819.94
C Rh P 28.07 / T 327
F=566369 E=732.99
E6 P 28.07/ T 327
F- 40391 E-732.99
E7 P 42.18 / T 381
F-31844 E-755.97
• Total 12 Stages
• Extraction E7
after 6th Stage

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I P Turbine
H Rh P 24.69 / T 535
F- 566369 E- 845.83
P-1.369 / T- 183
F- 470857 E-678.41
E5 P 12.96/ T 444
F- 16640 E-802.92
E4 P 6.911/ T 364
F- 23838 E-762.69
E3 P 2.974/ T 264
F- 21033 E-756.79
E2 P 1.369/ T 183
F- 25369 E-678.41
• Total 11 Stages
• E5 after 3rd Stage
• E4 after 6th Stage
• E3 after 9th Stage
• E2 at 11th stage

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L P Turbine
Steam from LPT
P-0.0889, F-2*230004,
T= 430C, E- 589.18
P 1.369 / T 183
F-470857 E-678.41
E1 P 0.8895/ T 45
F-11989 E-622.05
• Total 4+4 = 8 Stages
• E1 after 3rd stage both sides

23. 3/2/2021 Manohar Tatwawadi 23
Condenser
Cooling
Water out
to Cooling
Towers
Cooling
Water from
CW Pumps
T – 30
F – 270000 m3
CEP
CEP Disc.
To
Ejector
P- 22, T – 43
F–480106, E-43.1
Drains to cond
Flash Tank.
gd1- 1660
gd2- 4949
hd1- 11989
ejd – 1500
Total = 20098
Steam from LPT
P-0.0889, F-2*230004,
T= 430C, E- 589.18
Steam to circ. water Ratio
=270000*1000/460008
=586Kg/kg steam.
T. T. D.. =110C Effy = 80%
Make up stopped
T – 410C
F – 270000 m3

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Condensers
Two Condensers connected in Parallel
Total condensate from Hotwell is extracted by Condensate
Extraction Pump and passes through the regenerative feed
cycle before entering the Economiser in the Boiler.
Heat Lost in Condensers
= (Enthalpy of Steam at LPT Exhaust-Enthalpy of
Hotwell water) * Qty of steam at LPT Exhaust
= 460008 ( 589.18 – 43.16) = 251173568 Kcal
OR 1185.32kcal/kwh

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UNIT HEAT RATE
860 KCAL
20 KCAL
1185 KCAL
2065 KCAL
331 KCAL
2396 KCAL
TURBINE HEAT RATE = 2065 KCAL/KWH
UNIT HEAT RATE = 2396 KCAL/KWH

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Heat given to Turbine Cylinders
HP Turbine Flow Enthalpy Heat kcal
Input Steam 652000 819.94 534600880
Output Steam 566369 732.99 415142813
Heat given to HP Turbine 119458067
IP Turbine Flow Enthalpy Heat kcal
Input Steam 566369 845.83 479051891
Output Steam 470857 678.41 319434097
Heat given to IP Turbine 159617794
LP Turbine Flow Enthalpy Heat kcal
Input Steam 470857 678.41 319434097
Output Steam 460008 589.18 271027514
Heat given to LP Turbine 48406583
Total Heat given to HP,IP & LP Cylinders 327482444 Kcal

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Turbine Cycle Heat Rate
Heat Rate =
Total Heat Supplied to Turbine in Kcal
Total Geneation in Kwh
Q1(I1-IWE) + Q(I3-I4)
W
Q1= Live steam flow 652000kg/hr
I1 – Enthalpy of Live Steam 819.94 Kcal/Kg
IWE – Enthalpy of Feed water at HP7 outlet 253.94 Kcal/kg
QR – Reheat Steam Flow 566369 Kg/hr
I3 – Enthalpy of Reheat Steam 845.83 Kcal/kg
I4 – Enthalpy of Cold Reheat Steam 732.99 Kcal/kg
W – Total power Developed at gen out in Kwh 211902kwh.

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Heat Rate Calculations
Turbine Cycle Heat Rate
Heat Rate =
652000(819.94-253.94)+566369(845.83-732.99)
211902
Heat Rate = 2043.11 Kcal/Kwh at 210 MW
2068 Kcal/kwh at 176 MW
2070 Kcal/Kwh at 150 MW
2156 Kcal/kwh at 100 MW
Heat Rate =
652000(566)+566369(112.84) 369032000+63909077.96
211902 211902
Heat Rate =
432941077.96
211902
= 2043.11 Kcal/Kwh
=

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Extractions Quantity & Heat
Tapped
Symbol Details Pressure Temp Flow kg/hr Enthalpy Total Heat
e1 LPH 1 0.8895 45 11989 622.05 7457757.45
e2 LPH2 1.369 183 25369 678.41 17210583.3
e3 LPH3 2.974 264 21033 715.68 15052897.4
e4 LPH4 6.911 364 23838 762.69 18181004.2
ed Deaerator 12.96 449 5984 802.92 4804673.28
e5 HPH5 12.96 444 16640 802.92 13360588.8
e6 HPH6 28.07 327 40391 732.99 29606199.1
e7 HPH7 42.18 381 31844 755.97 24073108.7
Total 177088 129746812
Quantity of Extraction flow = 177088 kg/hr
Percentage of Extraction =177088/650000 = 27.16%
Percent heat tapped = 129746812/327482444 = 39.61%

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Heat regained by Feed water
Upto LPH 2 480106 (100.86-43.16) = 27702116
After LPH 3 555281 (156.52-100.86)= 30906940
After Deaerator 555281 (166.71-156.52)= 5658313
In BFP 652000 (173.12-166.71)= 4179320
After HPH 7 652000 (253.94-173.12)= 52694640
Total Heat gained by Feed water =121141330
Heat Lost in Feedwater regenerative heating cycle
= Heat in Extractions – Heat gained by FW
=129746812-121141330 = 8605482 kcal
Efficiency of regenerative feed heating system
= 121141330*100 / 129746812 = 93.36%

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Material/Energy Balance
Statements
Material Balance is the statement on the
conversion of Mass for a chosen time interval
Energy Balance is the statement of Energy during
the Process for a chosen time interval

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Basic Principle
Material input = Material output + Material lost
Energy output = Energy Input – Energy Lost
PROCESS
WORKPLACE
PROCESS INPUT PROCESS OUTPUT
WORKPLACE
INPUT
WORKPLACE
OUTPUT
LOSSES

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PLANT WORKPLACES
Coal Handling Plants > River water Pumps
Boilers > Oil Unloading Plants
Turbines > High Voltage Boards
Generators > Low Voltage Boards
Ash Handling Plants > Ash Slurry Pumps
Oil Handling Plants > ETC
Water Treatment Plants
Plant Control Rooms

34. 3/2/2021 Manohar Tatwawadi 34
Plant Heat Rate
PLANT
WORKPLACES /
SECTIONS /
PROCESSES
F
U
E
L
ELECTRICITY
PLANT HEAT RATE =
FUEL QTY X HEAT VALUE
UNITS GENERATED
PLANT HEAT RATE = KCAL / KWH
LOSSES

35. 3/2/2021 Manohar Tatwawadi 35
Questions / Answers
1. The designed THR for a BHEL 210 MW LMW Turbine is
a) 2063 b) 2500 c) 1900 Kcal/Kwh
2. Steam at 470C comes out from the LPT and enters the
condenser. True / False
3. Heat rejected in the condenser is more than the mechanical
output of the turbine. True / False
4. The quantity of Steam trapped for Extraction is
a) 20.26% b) 10.15% c) 27.16% of the total steam.
5. Total amount of the heat tapped for the Feed Heating in
Heaters is a) 30.25% b) 25.36% c) 39.16% of the
total heat supplied by the boiler.

36. 3/2/2021 Manohar Tatwawadi 36
Questions / Answers
6. Steam in the condenser flows through the tubes and
cooling water flows over the tubes.. True / False
7. Condenser tube leakage can be detected by
a) online O2 Aanlyser b) online sodium analyser.
9. What is the effect of increase in heat load of the condenser ?
11. How the heat Transfer in the condenser will be affected
with reference to water chemistry.
8. The Economiser Inlet Feed water Temp determines the
efficiency of the feed heating system. True / False
10. What is the effect of air ingress in the condenser ?

37. 3/2/2021 Manohar Tatwawadi 37
Questions / Answers
12. Feed heaters improve the boiler Efficiency………….
True / False
13. Bypassing of HP Heaters will increase the Generator MW
…….Discuss..
14. The decrease in the Feedwater / Condensate at the outlet of
the heater may be due to
a) Feed/Condensate Bypass valve of the heater passing.
b) Extraction Steam Valve not full open.
c) Both of these
d) None of these
15. Gland Steam cooler No 1 is under vacuum…
True / False.

38. 3/2/2021 Manohar Tatwawadi 38
Thanks
Questions
Please

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