Heat balance diagram

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Presentation on Heat & Mass Balance Diagram

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Heat balance diagram

  1. 1. Index PLANT BLOCK DIAGRAM PLANT RANKINE CYCLE HIP HEAT & MASS BALANCE LP HEAT & MASS BALANCE HEAT RATE Important Formulae
  2. 2. Plant BLOCK DIAGRAM HRH P= LPT Inlet 41, T = 566, G= P= 9.45, 1660.39, G= H= 1363.63, Main 3594.6 H= Steam P = 3161.5 HPT LPT LPT IPT 242, T = Gr. 566, G = 1984.36 LPT Outlet, H = 3396 CRH P= G= 1125.25, Condense 45, T = H = 2411.4 rLos BOILE Vc. Pr = R 318.7, Gs = .1 bar Hotwel 1660.39 H= DRTR, P = 0.916,Feed T=176.1 CEPWater 2999.1T= 280°C, G = 2044 TDBF CP U P MDBFP G S C HP- HP- HP- LP-5 LP-6 LP-7 LP-8 1 2 3
  3. 3. RH Outlet , T =566 ° C, P = 40.4, G = 1660.39, H = 3594.6 HPT Inlet , T =566 ° C, P Plant Rankine Cycle = 242, G= 1984.36, H = 3396 Expansion in HP Turbine, Stages = 9, Total Power Output = 186 MWT= Expansion in IPTemperature Turbine, Stage = 6, HP HTR-1 Outlet, Total Power Output = T=280,H=1229.6 = Feed Water 206 MW Temp. HP HTR-2 Outlet, T=255.5,H=1114 IP Outlet, P = 9.45, G=1363.63, HP HTR-3 Outlet, 3161.5 T=213.6,H=925.8 RH Inlet , T =318.7 ° C, P =DRTR Outlet, T=175.6,H= 45, G = 1660.39, H = 2999.1746.2LP HTR 8 Outlet,T=138.9,H= 585.2LP HTR 7 Outlet,T=121.9,H= 512.9LP HTR 6 Outlet, T=104, H= Expansion in LP437.1 Turbine, Double Flow,LP HTR 5 Outlet, T=79.2, Each Side Stages = 6,H= 332.8 Total Power Output = 268 MWLP HTR 5 Inlet,T=46.5, H= 194.6 LP Outlet, G=1125.25, H = 2411.4CEP Inlet, P = Condenser Vaccum Pr. = 0.1 Bar, make0.1, T = 46.1, H = up water = 60 G, Emergrncy Drain = .75192.9 G, Extractin Drains = 239.7 G T in ° C, P = Bar, S = Entropy Flow = G, Enthalpy = H
  4. 4. Heat Rate The heat rate is usually expressed as the amount of heat energy needed to produce a given amount of electrical energy. It can also be expressed as rate of Heat input required to produce unit shaft output. Guarantees Heat Rate = 1898.8 Kcal / KWHr or 8068.786 Kj / KW.h ( Conditions :- 660 MW, 76mm Hg TMCR, 0% Make up) Heat Rate for Actual Condition ( 660 MW, 76 mm Hg TMCR, 3 % Make up) = 8011.7 KJ/kW.h Heat Rate = H/E H = heat supplied = Heat in HP Turbine + Heat in IP Turbine H = M*∆h ( Kj / Hr) M = Mass Flow ( Tons/ Hr) or 1000Kg/Hr ∆h = Enthalpy Change ( Kj / Kg) E = Energy output ( KW) So HR = M*∆h/E Kj / KwHr
  5. 5. Vidarbha Heat Balance Diagram FOR HIP TURBINE P=242, H = 3396, T = 566, G = 1984.36 .63G A GV .15G B P=40.44, H = 3594.6, T = 566, G = 1660.39 To LPP=9.45, TurbinH=3161.5, e HPG=1363.63 IP Turbine TurbineG= 13.39 G= 29.88, P= 183.48, LOS P=9. P=21 H=3337.3 P=44. Boiler S 46 .6 93 .10 P=64 T=351.1, .6 .11 G H=3161.5, G 16.49G G= 81.79 L M N T=468, T=318.7,H=2999.1,G=1660.39 R P H=3395.2, 126.81 G= 90.03 T=280, G X A H H=1229.6,To BFPT 15.37G G= 2044 L T=318.7, T=366.7, H=2999.1, H=3081.7, G= 155.78 G= 121.65 T=182, T=213.6, T=255.5,T=138.9, H=790.7 H=925.8 H=1114H= 585.2 DRTR HTR-3 HTR-2 HTR-1FromLPHT=175.6, T=261, T=187.5,H=746.2 T=219.2, H=1140 H=796.5G=2044 H=939.8 G=121.65 G=383.94 BFP G=293.91
  6. 6. HEAT BALANCE DIAGRAM FOR LP TURBINE From IP Turbine H P=9.45, Fl H=3161.5, M G= as 1363.63 R h LPT LPT ta S n k .26 .26 H= 2411.4, G = 0.75G G G 1125.25 T=189.5, H=2846.9 T S T=83.2, , G= S T H=2596.8 T=132.5, 48.15 Condenser , G= H=2846.9 T=244, 79.93 , G= H=2952.1 Make up Water = 48.15 , G= 60G 46.75 .89 T=138.9,T=46.1, G H=585.2,H=192.9 T=46.8, G H=196 To S LPH-8 LPH-7 LPH-6 LPH-5 DRTR T=46.5, H=194.6 C CEP G=1562.7 4 T=52.4, T=52.4, T=52.4, T=52.4, H=219.2 H=219.2 H=219.2 H=219.2 G=239.79 G=239.79 G=239.79 G=239.79
  7. 7. Heat Rate The heat rate is usually expressed as the amount of heat energy needed to produce a given amount of electrical energy. It can also be expressed as rate of Heat input required to produce unit shaft output. Guarantees Heat Rate = 1898.8 Kcal / KWHr or 8068.786 Kj / KW.h ( Conditions :- 660 MW, 76mm Hg TMCR, 0% Make up) Heat Rate for Actual Condition ( 660 MW, 76 mm Hg TMCR, 3 % Make up) = 8011.7 KJ/kW.h Heat Rate = H/E H = heat supplied = Heat in HP Turbine + Heat in IP Turbine H = M*∆h ( Kj / Hr) M = Mass Flow ( Tons/ Hr) or 1000Kg/Hr ∆h = Enthalpy Change ( Kj / Kg) E = Energy output ( KW) So HR = M*∆h/E Kj / KwHr
  8. 8.  Boiler Efficiency ◦ η = Output / Input ◦ η = Steam Output * Enthalpy / Coal Input ◦ Η = M ₁* H / GCV * M₂ ◦ Were M ₁ = Flow ◦ Where M₂ = Coal Flow rate GCV for the design coal is 3300 Kcal/Kg Specific Heat Consumption ◦ It is defined as rate of steam flow ( kg/ s) required to produce unit shaft output. ◦ S.R. = Total Steam Flow / Total Unit Generated Cost of Power ◦ Cost of Power = HR*Fuel Price / GCV
  9. 9. Thank You

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