This document summarizes an energy audit conducted at a thermal power plant. It provides background on factors that influence energy costs and efficiency at thermal plants. It then describes the objectives, areas, and parameters that were analyzed during the audit. These include analyzing the boiler, turbine, and auxiliary systems to calculate energy consumption and efficiency. The audit aims to identify areas of energy waste, quantify the waste, set benchmarks, and recommend measures to reduce waste and optimize energy usage. Key areas like boiler efficiency, turbine heat rate, and auxiliary power consumption were monitored. The document provides examples of calculations used to assess performance and efficiency of different plant components. Overall, the energy audit seeks to improve the plant's energy usage and lower energy costs and environmental impacts

1. Energy Audit in Thermal Power Plant
M. G. Morshad, ADGM / ELECT
Energy Manager / TPS II ( 7 X210MW)

2. Power Generation

3. Thermal Power Plant
Environmental
Pollution
Faster consumption of
natural energy source Energy Cost (Tariff)
Major issues involve with thermal power plant

4. Tariff components
Tariff ( Cost of one KWhr Energy ) = Capacity Charges (CC) + Energy Charges (EC)
Capacity Charges CERC Norms Remarks
1 Return on Equity (%) 15.5 Fixed for 5 years
2 Interest on Loan Capital (%) 10 Fixed for 5 years
3 Depreciation (%) 5.28 Fixed for 5 years
4 Interest on Working Capital (%) 10 Fixed for 5 years
5 O& M cost( Rs Lakh / MW) 22.74 > Loss , < Profit
Energy Charges CERC Norms Remarks
1 Availability Factor (%) 80 > Profit , < Loss
2 Station Gross Heat Rate ( Kcal / Kwhr) 2900 > Loss , < Profit
3 Fuel Oil Consumption (ml/KWHr) 2 > Loss , < Profit
4 Auxiliary Energy Consumption (%) 10 > Loss , < Profit
5 Cost of Lignite (Rs/Kg) 1.74 Varies in every year
6 GCV of lignite ( Kcal / Kg) { Internal norms) 2620 > Profit , < Loss
7 Cost of FO (Rs / L) 100 Varies in every year

5. Factors control the Energy charges
Factors control the capacity charges
1. Interest on Working Capital – Maintaining minimum inventory of spares
2. O& M cost – Minimum manpower and reliability & quality of maintenance

6. Plant EC CC Total (EC + CC)
TPS-1 4.65 1.21 5.86
TPS I Expn 3.01 1.28 4.29
TPS-II -1 3.27 0.86 4.13
TPS-II -2 3.27 1.07 4.34
TSII Expn 2.97 2.85 5.82
Barsingsar 1.53 2.43 3.96
NTPL 2.79 2.13 4.92
Monthly Power Bills (Tariff) in
Rs/Kwhr (June 2017)

7. Controlling of three major factors
Environmental
Pollution
EnergyCost
(Tariff)
Decreaseinfuel
consumption
Depletionof
energysource
IncreaseinEnergyConservationand
EnergyEfficiencymeasures

8. Fuel consumption and system efficiency
Boiler 80%
Turbine
40%
Generator
98.5%
Transmission &
Distribution
90%
Utilization 85%
 Power system = Generation + Transmission + Distribution + Utilization
 Efficiency of the power generation = 0.8 x 0.4 x 0.985 x0.9 x0.85 = 24%
 For utilizing 1 KWhr of energy ,Required fuel consumption -
a. Lignite consumption = (1/0.24)(1/3.25) = 1.28 Kg, [ 1Kg Lignite = 3.25 KWhr]
b. Coal consumption = (1/0.24)(1/4.65) = 0.89 Kg [ 1Kg coal = 4.65 KWhr ]

9. Major steps for improving energy efficiency &
conservation in Indian power sector
Energy
Conservation
Act 2001
Making availability
of energy efficient
technology in the
market
Creating manpower
for evaluating
energy efficiency
Creating
infrastructure ,
reporting authority
Energy
Efficiency
and
Energy
Conservation
Measures
PAT
( Perform,
Achieve
Trade)
Cycle

10. Target
not
achieved
Target
achieved
Net Heat
Rate Target
line set by
BEE
Number of Energy
saving certificate
to be issued in
proportion to
target
achievement
above target line
Indian Power
Exchange for
trading of energy
saving certificate
Number of Energy
saving certificate
to be procured in
proportion to
target
achievement
below target line
PAT Cycle mechanism
Sale
Buy
 PAT cycle period – 3 years (2012-15 : PAT Cycle I, 2016-19 : PAT Cycle II
 Mandatory Energy Audit at the end of each PAT Cycle
 Purpose to increase energy efficiency

11. Concept of energy audit
Production
( Kwhr)
Energy consumption (Thermal , Electrical)
X
Y
C
1. Determining the operating energy consumption line with fixed, variable &
specific energy consumption.
2. Suggesting measures for maintaining energy consumption within the
optimum range with cost benefit analysis
3. Implementing suggestions for maintaining energy consumptions within the
optimum range.

12. Organizing Energy Audit in
Thermal Power Plant

13. Objective
To optimize the energy consumption of the plant by
1. Identifying the area of energy waste.
2. Quantifying energy waste.
3. Setting bench mark for energy waste.
4. Finding out suitable measures for the reduction
of energy waste

14. Area to be covered under energy audit
Main
Plant
Boiler System
Turbine system
Electrical system
1. Boiler
2. Economizer
3. RAPH
4. FD & ID Fan
5. ESP
1. HP,IP & LP Turbine
2. HP & LP Heaters
3. Condensers
4. CEP,BFP, CWP
5. Cooling tower
1. HT & LT Load distribution
2. Auxiliary Power Consumption
Balance of
Plant
1. AC Plant & ventilation system
2. Compressed air system
3. Plant lighting system
4. Ash disposal system
5. Lignite Handling system
Common
Auxiliary system

15. Energy audit activities
ENERGY AUDIT
Collecting / measuring operating
parameters
Calculating energy consumption
pattern for all energy intensive
system
Comparing with design
parameters
Setting Bench Mark
Recommending suitable measures
with cost benefits analysis
Submission of Report
Analyzing the reasons for
deviations in parameters

16. Parameters to be monitored for energy audit in
thermal power plant

17. Boiler
Efficiency
(%)
Turbine
Heat Rate
(Kcal /Kwhr)
Fuel
input
(Ton)
Gross
Generation
(Kwhr)
GCV
Fuel
(Kcal/Kg)
Gross Heat Rate
( Kcal/Kwhr)
1.(Fuel Input x GCV / Gross generation)
2. (Turbine Heat rate / Boiler Efficiency)
3. ( Sp Fuel consumption x GCV)
Auxiliary Power Consumption
(%)
(Total consumption / Total generation)
Net Heat Rate ( Target)
( Kcal/Kwhr)
[Gross Heat Rate / (1- % APC)]

18. BOILER
EFFICIENCY
AIR PRE HEATERS
PERFORMANCE
ECONOMIZER
PERFORMANCE
TURBINE
HEAT RATE
HP & LP HEATERS
PERFORMANCE
CONDENSER
PERFORMANCE
COOLING TOWER
PERFORMANCE
TURBINE CYLINDER
EFFICIENCY
TURBINE CYCLE
EFFICIENCY
SH & RH Spray, MS
pressure and Temp

19. Unit Auxiliary
Power
Consumption
SEC - ID & FD Fans )
SEC – BFP
SEC – CEP
SEC – Mill
SEC – CWP
Boiler & Turbine LT Load -
Energy Consumption
Excitation - Energy Consumption
ESP – SEC

20. Station
Auxiliary
Power
Consumption
SEC - Air compressor
SEC - AC & Ventilation
SEC - Fuel Handling System
SEC - DM water plant
SEC - Ash Disposal System
SEC - LHS
STRAY
ENERGY
LOSSES
Steam leaks
Peeling off thermal Insulation
Water Leaks

21. % Efficiency
[(Out Put /Input) x100]
% Performance
[(Out Put /Target Input) x100]
Specific Energy
Consumption (SEC)
[Energy consumption
/ Unit of product)
% Effectiveness
[(Out Put / Max. expected
output ) x100]
MEASURING UNITS

22. Variation Impact (kCal/kWh)
200 MW Unit 500 MW Unit
SH Spray 20 T/hr 0.30 0.81
RH Spray 20 T/hr 12.00 4.76
MS Pr. 10 kg 12.00 7.30
MS Temp. 10oC 6.00 6.20
RH Steam 10oC 6.00 5.55
Load 10 MW 12.00 4.36
Cond. Vacuum 5 mm Hg 8.00 6.70
FW Temp. 10oC 8.00 10
Variation of Gross heat rate (GHR) with
operating parameters

23. Reasons for higher net heat rate (NHR)
1. Variation in fuel quality ( GCV, Moisture, Hydrogen & Ash Content)
2. Error in fuel quantity measurement
3. Lower Boiler efficiency
4. Higher turbine heat rate.
5. Partial loading (Lower PLF ) as a result of –
a) Fuel shortage
b) Technical minima ( Low load schedule)
c) Poor condenser performance
d) Any other technical reasons.
6. Aging of the plant
7. Frequent shutdown and startup of units
8. Higher % auxiliary power consumption due to
a) Lower PLF
b) Fuel quality variation that leads to loading of ID & FD Fans
c) Idle running of equipment
d) Operating non energy efficient equipment

24. Energy Audit in Boiler system

25. Boiler Efficiency – Direct Method
Steam output : 660 T/Hr
Steam temp : 540 deg C
Steam Pressured : 150 Bar
Enthalpy : 815 Kcal /Kg
Fuel input : 208 T/Hr
NCV :2307 Kcal/Kg
FW temp : 252 Deg C
Enthalpy : 252 Kcal/Kg
Boiler Efficiency (NCV Basis)
= [MS Flow x ( MS enthalpy – FW enthalpy)] / (Fuel Flow x GCV)
= [660 x 1000 x (815-252)] / (208 x 1000 x 2307) = 77%
Boiler Efficiency (Fuel analysis basis )
= [92.5-{50*A +630*(M+9*H)}/GCV] = [92.5-{50*12 +630*(47+9*2)}/2650] =77.5%

26. Boiler Efficiency - Indirect
Fuel Analysis
1. GCV,
2. % Ash,
3. %Moisture,
4. % carbon
5. % Oxygen
6. % Hydrogen
7. % Nitrogen
8. % sulpher
Flue gas
BOILER
Air for fuel
combustion
R
A
P
H
% O2, % Co2 , % Co
Environmental condition
1. Wind speed
2. Humidity
3. Ambient temp
Physical condition
1. Surface area
2. Surface temperature
1. % Bottom ash
2. GCV bottom Ash

27. Data collection
Parameters Symbol Data
GCV – Lignite (Kcal / Kg) CVL 2560
GCV – Bottom ash (Kcal / Kg) CVBA 1100.28
GCV – Fly ash (Kcal / Kg) CVFA 532
% Bottom ash BAP 20
% Fly ash FAP 80
% Ash Content A 12.98
% Moisture content M 47.00
% Carbon content C 24.18
% Hydrogen content H 2.2
% Nitrogen content N 0.21
% Oxygen content O 12.58
% Sulphur content S 0.85
Parameters Symbol Data
Flue gas temp (Deg. C) GT 160
% CO2 Before RAPH ACo2 14.00
% CO Before RAPH CO 0.00
% O2 Before RAPH O2 6.8
% O2 after RAPH O2A 9.2
% CO2 after RAPH CO2A 11.2
Parameters Symbol Data
Ambient temp. (Deg.C) T 36
Air humidity (Kg / Kg dry air) HU 0.025
Boiler surface temp (Deg C) ST 80
Wind velocity (m / Sec) V 2.1
Boiler surface area (m2) M2 3442
1. Fuel data 2. Flue gas data
3. Physical & Environmental data

28. Performance Calculation
Parameters Formula Value
Theoretical air
required
TA =
[(11.6* C) + {34.8 (H – O/ 8)} + (4.35 *S)] / 100
3.06 (Kg / Kg of Lignite)
Excess air
supplied
EA = [O2 / (21 – O2)] X 100 47.88 %
Mass of air
supplied
MA = (1 + EA / 100) X TA 4.52 ( Kg / Kg of Lignite)
Mass of CO2 in
gas
A = (C/100) X (44 / 12) 0.88( Kg / Kg of Lignite)
Mass of SO2 in
fuel
B = (S /100)x(64/32) 0.017 ( Kg / Kg of Lignite)
Mass of N2 in air
supplied
C = (MA X 0.77)
3.48 ( Kg / Kg of Lignite)
Mass of O2 in gas
D = (MA – TA) X 0.23
0.33 ( Kg / Kg of Lignite)
Mass of dry flue
gas
MDFG = (A + B + C + D)
4.70 ( Kg / Kg of Lignite)

29. Loss calculation & Efficiency
Parameters Formula Actual
Heat loss in dry flue
gas (L1)
[MDFG X 0.23 X (TG – T) / CVL] X 100 5.088 %
Heat loss due to
formation of water
(L2)
[9 X (H / 100) X {584 + 0.45 (TG – T) / CVL] X 100 4.969 %
Heat loss due to
moisture in fuel (L3)
[ (M / 100) X { 584 + 0.45 ( TG – T ) / CVL] X 100 11.795 %
Heat loss due to
moisture in air (L4)
[MA X HU X 0.45 (TG – T) / CVL] X 100 0.238 %
Heat loss due to partial
conversion of C to CO
(L5)
{CO X (C / 100) X 5744 X 100} / {(CO + ACo2) X CVL} 0
Heat loss due to
radiation and
convection (L6)
[0.548 {((273+ GT) / 55.55} 4 - ((273+ T) / 55.55} 4 ] + 1.957 ( GT –T) 1.25
√(196.85 V + 69.8) / 69.8] x M2 x 0.86 X 100 ] / ( CVL X LQ)
0.38%
Heat loss due to un
burnt in fly ash (L7)
[(FAP / 100) X (A / 100) X (CVFA) X 100] / CVL 2.157 %
Heat loss due to un
burnt in bottom ash
(L8)
[(BAP / 100) X (A / 100) X (CVBA) X 100] / CVL 1.116%
Boiler efficiency 100 – (L1 + L2 + L3 + L4 + L5 + L6 + L7 + L8) 74.26%

30. Uncontrollable Loss = 18 %
Moisture in fuel = 12%
Moisture in Air = 0.23%
Water formation = 5%
Radiation = 1%
Controllable Loss = 9 %
Dry flue gas = 5%
Un burnt fly ash = 2.5%
Un burnt bottom ash = 1.5%
Boiler Efficiency =74%
Losses =(100 – 74) = 26%

31. y = 0.8394x + 34.494
R² = 0.9742
0
50
100
150
200
250
0 50 100 150 200 250
TPH
MW
Fuel consumption with Load variation
Unit load (MW) 210 180 150 125 110
Fuel Consumption (TPH) 208 185 170 135 125
GCV – Primary Fuel (Kcal/Kg) 2650 2650 2650 2650 2650
Fuel consumption pattern
Fuel consumption ( TPH) = 0.8349 x MW + 34.494

32. Parameters Unit Formula Design Unit 1 Unit 2
Fuel flow - Mill A TPH A 42 42 42
Fuel flow - Mill B TPH B 42 38 40
Fuel flow - Mill C TPH C 42 42 31
Fuel flow - Mill D TPH D 42 41 42
Fuel flow - Mill E TPH E 42 0 40
Fuel flow - Mill F TPH F 0 35 0
Total fuel flow TPH TF =∑A to F 210 198 195
GCV of fuel Kcal/Kg GCV 2650 2650 2650
Heat energy input 106
Kcal Hin = (TF x GCV)/1000 556.5 524.7 516.75
Generation MW G 210 190 180
Gross Heat rate Kcal / Kwhr (Hin X 1000)/G 2650 2761.58 2870.83
Specific fuel
consumption
Kg/ Kwhr TF / G 1 1.04 1.08
Gross heat Rate and specific fuel consumption

33. Economizer performance
Flue gas
Feed water flow
Inlet Water temp
243 Deg c (IW)
Outlet Water temp
336 Deg C (OW)
Inlet Flue gas temp
471 Deg C (IFG)
Outlet Flue gas temp
309 Deg c (OFG)
1. Heat loss in flue gas ( HL) = (IFG -FG) = (471-309) = 162 Deg C
2. Heat gain in feed water ( HG) = (OW – IW) = (336-243) = 93 Deg C
3. % Water side effectiveness = (HG) / (IFG – IW) = {93/(471-243)}x100 = 40%
4. % Gas side effectiveness = (HL)/(IFG –IW) = {162/(471-243)}x100 = 71 %
 Low % effectiveness in water side indicates scale formation in the tubes
 Low % effectiveness in gas side indicates formation of soot’s
Flue gas

34. BOILER
A
P
H
A
P
H FD
FAN
FD
FAN
ESP
ID
FAN
ESP
ID
FAN
Air and flue gas
circuit
Performance of APH
Boiler efficiency decreases by 1% for each 20 Deg C increase in exit flue gas temp

35. A
P
H
FG temp
before APH
(FG T1)
FG Temp
after APH
(FG T2)
Air Temp
before APH
(AT1)
Air Temp
after APH
(AT2
)
Temp Loss
(FG1 - FG2)
Temp Gain
(AT2-AT1)
Max Temp Head
(FGT1-AT1)BOILER
Gas side effectiveness
FG temp loss FGTL=(FGT1-FGT2) oC 161.00 116.60
Max temp head MTH = (FGT1-AT1) oC 276.60 261.30
Gas side %
effectiveness
(FGTL/MTH)x100 % 58.21% 44.62%
Low gas side effectiveness
indicates low heat transfer from
gas side to air side due to –
1. Dirty / aging of basket
2. Air leak
3. Air flow bypassing APH ( X Ratio)

36. A
P
HBOILER
Air Path
Air Leak
Before APH
% O2
% CO2
% CO
After APH
% O2
% CO2
% CO
FGT1 FGT2
AT1AT2
Air Leak
O2 Before APH O2 in % 4.20 6.30
O2 After APH O2 out % 5.50 9.20
Air leak (O2 out - O2 in ) / (21-O2 Out) % 8.39% 24.58%
CO2 Before APH CO2 in % 15.02 15.00
CO2 After APH CO2 out % 15.52 16.30
Air leak (CO2 out - CO2 in ) x 0.9 / (21-CO2 Out) % 8.21% 24.89%

37. A
P
H
FG temp
before APH
(FG T1)
FG Temp
after APH
(FG T2)
Air Temp
before APH
(AT1)
Air Temp
after APH
(AT2
)
Temp Loss
(FG1 - FG2)
Temp Gain
(AT2-AT1)
BOILER
FG temp loss FGTL=(FGT1-FGT2) oC 161.00 177.00
Air Temp gain ATG=(AT2-AT1) oC 239.00 220.00
X Ratio FGTL/ATG 0.67 0.80
X - Ratio
High x ratio indicates high air flow bypassing APH

38. Higher % Effectiveness in
Gas side
Higher Exit flue gas temp
Higher air
leak rate
High X RatioHigher
Differential
pressure
Indicates plugging of basket due to
soot deposition / corrosion /aging
Indicate a problem with the seals
(axial, radial, or Circumferential
Performance monitoring

39. SH
RH
ECO
A
P
H
FD
Fan
Wind
Box
ESP
ID
Fan
FG
Air
Performance of the draft system
Air leak at APH, Wind Box , PF ducts causes extra power loss in FD Fan
Air ingression in boiler, APH and poor efficiency of ESP causes extra power
consumption in ID Fans
 Excess power consumption in ID & FD fan increases APC

40. A
P
H
AirHeader
Actual air for
combustion
(measured)
Fan discharge
flow
Air Leak at
APH –A
Air Leak at
APH - B
Available air for
combustion
(Calculated)
Air Leak at wind
box, PF duct
Air Leak at wind
box, PF duct
FD-A
FD-B
FD Fan Performance

41. Measured discharge flow TPH DF1 525 450
Expected discharge flow TPH 3.5674 xMW +146.6 521.18 447.88
Air Temp Deg C T 36 36
Air Density Kg/m3 AD = (273X1.293)/(273+T) 1.14 1.14
Calculated discharge flow (m3/sec) DF2 = (DF1 x 1000) /(AD x 3600) 127.66 109.42
Measured discharge pressure mmWC P Dis 300 300
Measured suction pressure mmWC P suc 1.03 1.03
Measured total Head mmWC TH = (P Dis - P Suc) 298.97 298.97
Operating current Amps A 55.00 52.00
Operating Voltage KV KV 6.60 6.60
Operating PF PF 0.85 0.85
Motor input power KW Min = 1.732xKVXAXPF 534.41 505.26
Motor efficiency % E 94.00 94.00
Motor out put /
Fan Input power
KW F in = 1.732XAXKVXPFXE/100 502.34 474.94
Fan efficiency % FE = (DF2 x TH) /(102 X Fin) 74.49% 67.53%
Motor design capacity KW MC 1200 1200
Motor loading % (F in/MC) 42% 40%
Specific Power Consumption KW/Ton SPC = Min / DF1 1.02 1.12
Diff in SPC KW/Ton DSPC = SPC (Actual - Design) *** 0.10
Excess Power Consumption KW PL = DSPC x DF1 *** 47.19
FD Fan – Motor loading ,Efficiency and specific
power consumption

42. ID Fan Performance
ID
B
A
P
H
ESPFG discharge
flow (Gas +Ash +
Moisture)
Air
ingression
at APH –A
BOILER
Air
ingression
at APH –B
Removed
ash from
FG
Removed
ash from
FG
FG discharge flow (Gas +
Air +Ash + Moisture)
ID
A
Chimney

43. FG flow to ID Fans TPH Q8 = Q2+Q4+Q6+Q7 806.76 830.88
FG temp at ID inlet Deg C FGT 170 180
Gas Density Kg/m3 GD = (273X1.293)/(273+FGT) 0.80 0.78
FG flow to ID Fans m3/s Q9 = (Q8 x1000)/(GD x 3600) 281.25 296.19
Fan suction pressure mmWC Suc -450 -450
Fan discharge pressure mmWC Dis 4.5 4.5
Head mmWC H = Dis -suc 454.5 454.5
Motor current Amps A 160.00 180.00
Operating Voltage KV KV 6.60 6.60
Operating PF PF 0.89 0.89
Motor input power KW Min = 1.732xKVXAXPF 1627.80 1831.28
Motor efficiency % E 94.00 94.00
Motor out put /
Fan Input power
KW Fin = Min X E/100 1530.13 1721.40
Fan efficiency FE = (Q9 x H )/(102 x F in) 81.90% 76.67%
Motor design capacity KW MC 1800.00 1800.00
Motor loading % (F in/MC) 85.01 95.63
Specific Power consumption KW/Ton SPC = F in / (Q8) 1.90 2.07
Difference in SPC KW/Ton DSPC (Actual - Design) 0.00 0.18
Excess power consumption KW EPC = Q8 x DSP 0.00 145.53
ID Fan – Motor loading ,Efficiency and specific
power consumption

44. Energy Audit in Turbine system

45. Turbine Heat rate
HPT IPT LPT
210
MW
MS : 660 T/Hr, 535 0C,150 Ksc
,815 Kcal / Kg
CRH: 658 T/Hr, 341 0C,38 Ksc
,735 Kcal / Kg
HRH: 658 T/Hr, 535 0C,34 Ksc
,843 Kcal / Kg
FW: 658 T/Hr, 251Kcal / Kg
MS flow (MS enthalpy – FW enthalpy) + HRH Flow (HRH enthalpy – CRH enthalpy)
= 2112 Kcal / Kwhr
(Unit load in KW)

46. THR = 0.0153xMW2 - 6.3063xMW + 2656.5
1950
2000
2050
2100
2150
2200
2250
2300
2350
0 50 100 150 200 250
Unit Load Measured Turbine Heat rate
MW Kcal/Kwhr
60 2320.8
75 2280.3
100 2188.5
125 2108.7
150 2038.7
180 2018.4
210 2008.7
Turbine HBD Curve

47. y = 2.8879x + 49.166
R² = 0.9997
0
50
100
150
200
250
0.00 100.00 200.00 300.00 400.00 500.00 600.00 700.00
SteamConsumption(TPH)
MW
Steam consumption with load variation
Generation MW G 210 180 150 130 110
Expected THR Kcal/Kwhr
ETHR = 0.0153 X (G)2 –
6.3063 X (G) +2656.5
2006.91 2017.09 2054.81 2095.25 2147.94
Heat energy
input
Kcal/Hr HE = ETHR x G 421.45 363.08 308.22 272.38 236.27
Steam enthalpy Kcal/Kg h 814.70 814.70 814.70 814.70 814.70
Required Steam
flow
TPH RQ = (HE /h) 517.31 445.66 378.32 334.33 290.01
Actual steam
flow
TPH AQ = 1.27 X RQ 656.98 565.98 480.47 424.61 368.32
Steam consumption

48. Turbine cycle efficiency
Formula Actual Design
Turbine Heat rate (Kcal / KWhr) HR 2112 2053
Generator load (KW) L 210000 210000
Turbine Output (KW) To = L / 0.9856 213068 213068
Turbine Input (KW) Ti = HR * L * (4.187/3600) 515838 501428
Turbine cycle loss (KW) LT = (Ti – To) 302770 288360
Turbine cycle efficiency (%) (To/Ti) X100 41.3 42.5
Recoverable loss in Turbine Ti (Actual – design) 14.410 MW
Turbine Generator
210000 KW(210000 / 0.9856) KW
= 213068 KW
2112x210000x(4.187/3600)
= 515838 KW
Condenser
(515838 – 213068)KW
= 302770 KW

49. Turbine cylinder efficiency
Inlet
Temperature
Inlet Pressure
Out let
Temperature
Outlet Pressure
Enthalpy (H)
Entropy (S)
H 1
H 2
H 3
Deg C, Ksc
Deg C, Ksc
Ksc , S
S
H1 = Enthalpy & entropy (s) of the inlet steam corresponding to pressure and temp.
H2 = Enthalpy of the out let steam corresponding to pressure and temp
H3 =Enthalpy of the outlet steam corresponding to inlet entropy (s) & outlet pressure
MOLLIER CHART –

50. Function Enthalpy (h) Entropy (s) HP Cylinder Efficiency
H1 ƒ (T = 537, P = 151) 815 S=0.0015493
[(H1 – H2) / (H1 – H3)] *100
=76.53%
H2 ƒ (T = 347, P = 36) 740 -----
H3 ƒ (P = 36, S) 717 -----
Function Enthalpy (h) Entropy (s) IP Cylinder Efficiency
H1 ƒ (T = 538, P = 34) 845 S=0.0017412
[(H1 – H2) / (H1 – H3)] *100
=90.59%
H2 ƒ (T = 318, P = 7) 739 -----
H3 ƒ (P=7,S) 728 -----
Function Enthalpy (h) Entropy (s) LP Cylinder Efficiency
H1 ƒ (T = 318, P = 7) 739 S=0.0017612
[(H1 – H2) / (H1 – H3)] *100
= 90.37%
H2 ƒ (P = 0.0856 Q = 0.92) 570 ----
H3 ƒ (P = 0.0856, S) 552 ----

51. Extracted steam Pressure
15.5Ksc , Flow 34 T/Hr , Temp
(Th1) 433 OC Saturation temp
(St) 201.7 oC
Out let Drain
Temp (Th2)
174.7 oC
Inlet condensate Temp (Tc1)
167 oC
Out let condensate Temp
(Tc2) 198 0C
TTD (Deg C) = (St – Tc2),
3.7 Deg C
DCA (Deg C) = (Th2 - Tc1)
7.7 Deg C
Performance of LP & HP Heaters

52. Significance of TTD (Terminal Temperature Difference)
1. It indicates feed water heater’s performance relative to heat transfer.
2. TTD (Deg C) = St – Tc2
3. An increase in TTD indicates a reduction in heat transfer, while a decrease indicates
an improvement.
4. Typical ranges for TTD on a high-pressure heater with and without a de superheating
zone are 3° C to 5°C and 0° C, respectively.
5. The TTD for low-pressure heaters is typically around 5° C.
6. increase of TTD in LPH by 0.56°Ccauses increases in heat rate approximately by
0.016%
7. increase of TTD in HPH by 0.56°C causes increases in heat rate approximately by
0.013%
Significance of DCA (Drain Cooler Approach)
1. It indicates heater levels based on the temperature difference between the drain
cooler outlet and the feed water inlet
2. DCA ( Deg C) = (Th2 - Tc1)
3. An increasing DCA temperature difference indicates the level is decreasing and a
decreasing DCA temperature indicates a rise in level.
4. If there is a 0.56°C Increase in DCA, the corresponding increase in heat rate is
0.005%. The impact can be less at part load.
5. Typical value for DCA 10 Deg C

53. Parameters LPH-1 LPH -2 LPH- 3 HP - 5 HP-6
Saturation temperature – St ( Deg .C) 74.8 91.26 125.6 201.7 243.9
Condensat ourlet temp. - Tc2 (Deg ) 57.5 87.5 117.5 198.06 241
Actual TTD (St – TC2) 17.3 3.76 8.1 3.64 2.9
Designed TTD 3 3 3 3 3
Condensate temp rise (Tc2 – Tc1) 10.7 30 30 30.9 42.9
% Effectiveness [(TC2 – TC1) /(Th1- Tc1)] X 100 75.35 38.8 19.45 11.62 26.99
Sp Heat feed water (Kcal / Kg) 1 1 1 1 1
Heat Load [ Cond Flow *Sp heat*(Tc2 – Tc1)]/ 3600 KW 1531.2 4375.0 4375.0 5613.3 7793.5
Delta T1 = (Th1 – Tc2) 3.5 47.3 124.2 234.9 116
Delta T2 = (Th2 – Tc1) 13.6 30.7 27.4 7.6 7.34
(Delta T1 – Delta T2) -10.1 16.6 96.8 227.3 108.6
In [ ( Delta T1 / Delta T2)] -1.35 0.43 1.51 3.43 2.76
LMTD (Delta T1 – Delta T2) / In [ ( Delta T1 / Delta T2)] 7.48 38.60 64.10 66.26 39.34
Heating surface area (m2) 317 500 426 703 810
Heat Transfer Coefficient KW / ( Area X LMTD) 0.64 0.23 0.16 0.12 0.24
• Positive or higher TTD than design value indicates poor performance of the heater, which reduces
the cycle efficiency.
• Negative or lower TTD than the design value indicates better performance of the heater, which
increases the cycle efficiency
Performance calculation

54. Exhaust hood / Saturation Temp (Tsat)
CW Outlet Temp (T2)
CW inlet Temp (T1)
Hot well / Condensate Temp (Tc)
Temperature
rise TR = (T2-T1)
TTD (Approach)
= (Tsat – T2)
Range
= (Tsat – T1)
Sub cooling temp
TS = ( Tc - Tsat)
Heat
Transfer
capacity
Condenser
Temperature Profile

55. Reasons for low
condenser vacuum
CW Temperature rise,
TR = (T2-T1)
TTD = (Tsat – T2)
Sub cooling temp,
TS = (Tc - Tsat)
1. Low cooling water
flow
2. High inlet cooling
water temperature
Higher deviation with
respect to design
value (+)
Minor deviation with
respect to design
value
Minor deviation with
respect to design
value
Tube fouling
Minor deviation with
respect to design
value
Higher deviation with
respect to design
value (+ )
Minor deviation with
respect to design
value
Incondensable gases
Minor deviation with
respect to design
value
Minor deviation with
respect to design
value
Higher deviation with
respect to design
value (+)
Over heat duty
Higher deviation with
respect to design
value (+)
Higher deviation with
respect to design
value (+)
Minor deviation with
respect to design
value
Condenser Parameters monitoring

56. Parameters calculation
CW i/L Temp
31 Deg C
CW o/L Temp
37 Deg C
Saturation temp of exhaust
steam 44 Deg c
1. TTD = Sat temp – CW O/L Temp= 44-37 = 6 Deg C
2. CW temp gain = O/L – I/L = 37 – 31= 6 Deg C
3. Condenser cooling range = Sat temp – CWI/L temp = 44 – 31= 13 Deg C
4 LMTD = Log (Range/TTD) = Log 13/6 =7.42
5 Cleanness = Actual (LMTD / design LMTD)

57. Parameters Formula Actual Design
TTD (Deg.C) TTD = (SAT – CWTo) 4.64 3.47
Condenser cooling range (Deg C) R = (SAT – CWTi) 11.34 13.01
Cooling range / TTD R / TTD 2.44 3.75
LN ( Cooling range / TTD) LN = In (R/ TTD) 0.89 1.32
LMTD (Deg.C) LMTD = (CWTR / LN) 7.53 7.22
Condenser cleanliness factor (%) (Actual LN / ) *100 67.42% 85%
•The vacuum is optimum when condenser vacuum (Value of mmHg) is higher than the design
vacuum corresponding to the same exhaust hoot temperature.
•Vacuum loss occurs when condenser vacuum (Value of mmHg) is lower than the design vacuum
corresponding to the same exhaust hoot temperature.
•Vacuum loss of 1mmHG leads to increase of turbine heat rate by 1.6 Kcal / Kwhr
Performance calculation

58. Cooling Tower
Inlet Hot water temperature ( A)
Out let cold water temperature (B)
Ambient wet bulb temperature (C)
Range = (A – B)
Approach = (B – C)
Effectiveness = Range / (Range + Approach) = [(A –B) /(A-C)] x 100

59. Parameters Formula Actual Design
Inlet cooling water temp. (Deg .C ) A 36.5 42.73
Outlet cooling water temp. (Deg C) B 29.75 33.19
Air Wet bulb temp (Deg.C) C 24 28
Air Dry bulb temp (Deg C) D 30.5 35.8
Steam flow to the condenser (Kg / Hr) E 442000 441800
Total cooling water flow (Kg / Hr) F 30,000,000 30,000,000
Condenser vacuum (mmHg) G 700 684
Air velocity (m/S) H 2.1 1.171
Ave dia of cooling tower (m) I 56.06 56.06
Air density (Kg/m3) J = (273 * 1.293) / (273+ D) 1.16 1.14
Total cooling air flow (Kg/Hr) K= 3.143* (I/2)2 *H*J*3600 21655607 11867379
Water / Air ratio L = (F/ K) 1.38 2.52
Range ( Deg c) M = (A - B) 6.75 9.54
Approach ( Deg C) N = (B - C) 5.75 5.19
% Effectiveness O = [M / (M +N)] *100 54 64.76
Enthalpy – exhaust steam (Kcal /Kg) P 570 579
Heat load (Kcal / Hr) Q = P * E 251940000 255802200
Evaporation loss (m3 /Hr) R = 0.00085*1.8*(F/1000)*M 340.8 437.9
% Evaporation loss [R / (F/1000)] *100 1.03 1.46
Performance calculation

60.  WATER / AIR ratio increases with the decreases of air flow
 Low RANGE indicates –Insufficient water flow
 Low APPROACH indicates –Higher CT size
 Low % EFFECTIVENESS indicates –poor performance
 High EVAPORATION LOSS indicates –good performance
Performance Monitoring

61. Energy Audit in Electrical system

62. Auxiliary Power Consumption
(Station consumption + Unit Consumption )
% Auxiliary consumption = x 100
(Total Generation )
Export in
MU/day
Import in
MU / Day
Unit Auxiliary
Consumption
in MU / Day

63. Stage I Stage II
% UAC 8.38% 8.08%
%SAC 1.43% 1.17%
0.00%
1.00%
2.00%
3.00%
4.00%
5.00%
6.00%
7.00%
8.00%
9.00%
Typical % UAC & % SAC
% APC 9.81% 9.25%

64. 7.40
7.60
7.80
8.00
8.20
8.40
8.60
8.80
9.00
0.00 50.00 100.00 150.00 200.00 250.00
MW %UAC
154.74 8.83
178.06 8.53
185.21 8.40
192.06 8.23
199.18 7.96
206.44 7.70
214.76 7.63
For obtaining % UAC less than 8% - Unit has to operated with load higher than 192
MW
Break even point for unit Auxiliary Power Consumption

65. Auxiliary loads

79. Stray energy losses

80. Air leak in compressed air system
Time
Pressure
Cut in
6.8 Ksc
Cut out
7.5 Ksc
Unloading time 10.5 minutes
with motor load 54 KW
Loading time
1.5 minutes
with motor
load 188 KW
1.Compressor capacity ( C ) = 35 m3/ Minute
2.Electrical capacity ( KW) = 188 KW
3.Specific power consumption / min = (188/35 ) = 5.37 KW/m3
4. Air leak = [TL /( TL + TUL)] x C = (1.5/12) x35 = 4.37 m3/minute
5.Energy loss = 5.37 x 4.37 = 23.46 KW / Minute

81. Energy loss due to steam leak
Hole area A= π(d/2)2
Volume of the leaked steam
V = (A X Velocity x 3600) m3/Hr
Steam parameters : Pressure,
Temperature & Enthalpy
Mass of the leaked steam
M = (V / Specific volume of the steam) Kg/Hr
Energy contain in the leaked steam
E = (M x Enthalpy ) Kcal / Hr
Fuel loss due to steam leaking
L = (E/ GCV of fuel ) Kg / Hr
Steam
Velocity
1 Super heated Steam: 50-70m/sec
2. Saturated Steam : 30-40m/sec
3. Wet / Exhaust steam : 20-30m/Sec
Energy Loss due to steam leak
1. Pressure : 8 Ksc
2. Temp :170 Deg C
3. Enthalpy : 660.81 Kcal / Kg
4. Specific volume : 0.244 m3/Kg
5. Steam velocity 40 m/sec
6. Diameter of the hole : 3mm = 0.003m
7. GCV of the fuel = 2610 Kcal / Kg
8. Area of the hole = 7.1x10-6 m2
9. Volume = 10.224 m3/Hr
10. Mass= 41.9 Kg / Hr
11. Energy = 27987 Kcal / Hr
12. Fuel Loss = 10.6 Kg / Hr

82. Energy loss due to improper insulation
-2000
0
2000
4000
6000
8000
10000
12000
14000
16000
0 100 200 300 400 500
Kcal/m2/Hr
Difference in temperature between ambient & surface
Surface Temp
300 Deg C
Ambient Temp
50 Deg C
Non insulated
area 0.5m2
1. Diff in temp : 250 Deg C
2. Corresponding heat loss :7200 Kcal /m2/Hr
3. Non insulated area : 0.5 m2
4. Total heat loss : 0.5 x 7200 = 3600 Kcal / Hr
5. Fuel loss : (3600/2600) = 1.3 Kg / Hr

83. Thank You