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