$IfNot ParametricTable= P_ratio_gas. P ratio,gas = 14; Raport comprimare compresor aer - Pressure ratio for gas compressor (2) $EndIf

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1 P Equations Thermodynamics - An Engineering Approach (5th Ed) - Cengel, Boles - Mcgraw-Hill (2006) - pg. 598 Centrala cu cicluri combinate Se considera o centrala electrica cu ciclu combinat gaze-abur care are o putere neta de 450 MW. Raportul de comprimare al ciclului turbinei cu gaze este 14. Aerul intra în compresor la 300 K si în turbina la 1400 K. Gazele de ardere care ies din turbina cu gaze sunt folosite pentru a încalzi aburul la 8 MPa pâna la 400 C întrun schimbator de caldura. Gazele de ardere ies din schimbatorul de caldura la 460 K. Un preîncalzitor de apa de alimentare prin amestec (deschis) încorporat în ciclul cu abur functioneaza la o presiune de 0.6 MPa. Presiunea de condensare este de 20 kpa. Considerând toate procesele de comprimare si destindere ca fiind izoentropice, sa se determine (a) raportul dintre debitele masice de aer si abur, (b) fluxul necesar de caldura introdusa în camera de ardere, si (c) randamentul termic al ciclului combinat. Sa se studieze efectul raportului de comprimare al ciclului cu gaze daca acesta variaza de la 10 la 20 asupra raportului dintre debitele de gaze si abur si asupra randamentului termic al ciclului. Sa se reprezinte rezultatele în functie de raportul de comprimare al ciclului cu gaze, si sa se discute rezultatele. $UnitSystem K kpa Marimi de intrare: T 8 = 300 [K] ; P 8 = 14.7 [kpa] ; Gaze (aer la intrarea in compresor) - Gas compressor inlet (1) 1

2 $IfNot ParametricTable= P_ratio_gas P ratio,gas = 14; Raport comprimare compresor aer - Pressure ratio for gas compressor (2) $EndIf T 10 = 1400 [K] ; Intrare turbina cu gaze - Gas turbine inlet (3) T 12 = 460 [K] ; Gas exit temperature from Gas-to-steam heat exchanger (4) P 12 = P 8 ; Presiune iesire aer - Assumed air exit pressure (5) kw Ẇ net = 450 [MW] 1000 MW ; (6) $IfNot ParametricTable= eta η comp = 1.0; Eta gas,turb = 1.0; η pump = 1.0; Eta steam,turb = 1.0; (7) (8) $EndIf P 5 = 8000 [kpa] ; DELT AT 5,11 = 40 [K] Diferenta de temperatura la capatul cald al HEx (9) ; Abur la intrare in turbina - Steam turbine inlet (10) $IfNot ParametricTable= P[6] P 6 = 600 [kpa] ; Extraction pressure for steam open feedwater heater (11) $EndIf P 7 = 20 [kpa] ; Presiunea de condensare a aburului - Steam condenser pressure (12) Ciclul de forta cu gaze (aer) - GAS POWER CYCLE ANALYSIS Gas Compressor anaysis Starea [8]: h 8 = h (Air, T = T 8 ) ; s 8 = s (Air, T = T 8, P = P 8 ) ; (13) Starea [9s]: s s,9 = s 8 ; For the ideal case the entropies are constant across the compressor (14) P 9 = P ratio,gas P 8 ; (15) T s,9 = T (Air, s = s s,9, P = P 9 ) ; Ts9 is the isentropic value of T[9] at compressor exit (16) 2

3 η comp = w gas,comp,isen /w gas,comp compressor adiabatic efficiency, w comp > w comp,isen (17) h 8 +w gas,comp,isen = h s,9 h s,9 = h (Air, T = T s,9 ) ; SSSF conservation of energy for the isentropic compressor, assuming: adiabatic, ke=pe=0 (19) Starea [9]: h 8 +w gas,comp = h 9 ; SSSF conservation of energy for the actual compressor, assuming: adiabatic, ke=pe=0(20) T 9 = T (Air, h = h 9 ) ; s 9 = s (Air, T = T 9, P = P 9 ) ; (21) Gas Cycle External heat exchanger analysis Starea [10]: P 10 = P 9 ; Assume process 9-10 is SSSF constant pressure (22) h 10 = h (Air, T = T 10 ) ; s 10 = s (Air, T = T 10, P = P 10 ) ; (23) h 9 +q in = h 10 ; Q in = ṁ gas q in ; SSSF conservation of energy for the external heat exchanger, assuming W=0, ke=pe=0(24) (25) Gas Turbine analysis Starea [11s]: P 11 = P 10 P ratio,gas ; s s,11 = s 10 ; For the ideal case the entropies are constant across the turbine (26) h s,11 = h (Air, T = T s,11 ) ; T s,11 = T (Air, s = s s,11, P = P 11 ) ; Ts11 is the isentropic value of T[11] at ga Starea [11]: h 10 = w gas,turb,isen +h s,11 ; SSSF conservation of energy for the isentropic gas turbine, assuming: adiabatic, ke= η gas,turb = w gas,turb /w gas,turb,isen ; gas turbine adiabatic efficiency, w gas,turb,isen > w gas,turb (29) h 10 = w gas,turb +h 11 ; SSSF conservation of energy for the actual gas turbine, assuming: adiabatic, ke=pe=0(30) T 11 = T (Air, h = h 11 ) ; s 11 = s (Air, T = T 11, P = P 11 ) ; (31) Gas-to-Steam Heat Exchanger Starea [12]: h 12 = h (Air, T = T 12 ) ; s 12 = s (Air, T = T 12, P = P 12 ) ; (32) 3

4 SSSF conservation of energy for the gas-to-steam heat exchanger, assuming: adiabatic, W=0, ke=pe=0 ṁ gas h 11 + ṁ steam h 4 = ṁ gas h 12 + ṁ steam h 5 ; ec 1/2 ec -> debitele (33) Ciclul de forta cu abur - STEAM CYCLE ANALYSIS Starea [1]: Steam Condenser exit pump or Pump 1 analysis P 1 = P 7 ; (34) h 1 = h (ST EAM, P = P 1, x = 0) ; v1 = v (ST EAM, P = P 1, x = 0) ; (35) s 1 = s (ST EAM, P = P 1, x = 0) ; T 1 = T (ST EAM, P = P 1, x = 0) ; (36) Starea [2]: P 2 = P 6 ; (37) w pump1,s = v1 (P 2 P 1 ) ; SSSF isentropic pump work assuming constant specific volume (38) w pump1 = w pump1,s /η pump ; Definition of pump efficiency (39) h 1 + w pump1 = h 2 ; Steady-flow conservation of energy (40) s 2 = s (ST EAM, P = P 2, h = h 2 ) ; T 2 = T (ST EAM, P = P 2, h = h 2 ) ; (41) Starea [3]: P 3 = P 6 ; Condensate leaves heater as sat. liquid at P[3] (42) h 3 = h (steam, P = P 3, x = 0) ; T 3 = T (steam, P = P 3, x = 0) ; (43) s 3 = s (ST EAM, P = P 3, x = 0) ; v3 = v (ST EAM, P = P 3, x = 0) ; (44) Boiler condensate pump or Pump 2 analysis P 4 = P 5 ; (45) w pump2,s = v3 (P 4 P 3 ) ; SSSF isentropic pump work assuming constant specific volume (46) w pump2 = w pump2,s /η pump ; Definition of pump efficiency (47) h 3 + w pump2 = h 4 ; Steady-flow conservation of energy (48) s 4 = s (ST EAM, P = P 4, h = h 4 ) ; T 4 = T (ST EAM, P = P 4, h = h 4 ) ; (49) 4

5 Steam Turbine analysis T 5 = T 11 T 5,11 ; (50) h 5 = h (steam, T = T 5, P = P 5 ) ; s 5 = s (ST EAM, P = P 5, T = T 5 ) ; (51) Starea [6s]: s s,6 = s 5 ; (52) h s,6 = h (ST EAM, s = s s,6, P = P 6 ) ; T s,6 = T (ST EAM, s = s s,6, P = P 6 ) ; (53) Starea [6]: h 6 = h 5 η steam,turb (h 5 h s,6 ) Definition of steam turbine efficiency (54) T 6 = T (ST EAM, P = P 6, h = h 6 ) ; s 6 = s (ST EAM, P = P 6, h = h 6 ) ; (55) y h 6 + (1 y) h 2 = 1 h 3 ; Steady-flow conservation of energy (56) w steam,pumps = (1 y) w pump1 + w pump2 ; Total steam pump work input/ mass steam (57) Starea [7s]: s s,7 = s 5 ; (58) h s,7 = h (ST EAM, s = s s,7, P = P 7 ) ; T s,7 = T (ST EAM, s = s s,7, P = P 7 ) ; (59) Starea [7]: h 7 = h 5 η steam,turb (h 5 h s,7 ) ; Randamentul turbinei - Definition of steam turbine efficiency (60) T 7 = T (ST EAM, P = P 7, h = h 7 ) ; s 7 = s (ST EAM, P = P 7, h = h 7 ) ; (61) SSSF conservation of energy for the steam turbine: adiabatic, neglect ke and pe h 5 = w steam,turb + y h 6 + (1 y) h 7 ; (62) Steam Condenser analysis (1 y) h 7 = q out + (1 y) h 1 ; SSSF conservation of energy for the Condenser per unit mass (63) Q out = ṁ steam q out ; (64) Cycle Statistics MassRatio gastosteam = ṁ gas /ṁ steam ; Ẇ net = ṁ gas (w gas,turb w gas,comp )+ṁ steam (w steam,turb w steam,pumps ) ; (65) definitia puterii nete a ciclului - net η th = Ẇnet/ Q in 100 [%] ; Cycle thermal efficiency, in percent (67) Bwr = ṁgas w gas,comp + ṁ steam w steam,pumps ṁ gas w gas,turb + ṁ steam w steam,turb ; Back work ratio (68) Ẇ net,steam = ṁ steam (w steam,turb w steam,pumps ) ; (69) Ẇ net,gas = ṁ gas (w gas,turb w gas,comp ) ; (70) NetW orkratio gastosteam = Ẇnet,gas/Ẇnet,steam; (71) Data$ = Date$; (72) 5

6 Solution Bwr = Data$ = T 5,11 = 40 [K] η comp = 1 η gas,turb = 1 η pump = 1 η steam,turb = 1 η th = 62.4 [%] MassRatio gastosteam = ṁ gas = [kg g as/s] ṁ steam = [kg s team/s] NetW orkratio gastosteam = P ratio,gas = 14 Q in = [kw] Q out = [kw] q in = 877 [kj/kg g as] q out = 1523 [kj/kg s team] v1 = [m 3 /kg] v3 = [m 3 /kg] Ẇ net = [kw] Ẇ net,gas = [kw] Ẇ net,steam = [kw] w gas,comp = [kj/kg g as] w gas,comp,isen = [kj/kg g as] w gas,turb = [kj/kg g as] w gas,turb,isen = [kj/kg g as] w pump1 = [kj/kg s team] w pump1,s = [kj/kg s team] w pump2 = [kj/kg s team] w pump2,s = [kj/kg s team] w steam,pumps = [kj/kg s team] w steam,turb = [kj/kg s team] y = Arrays Row P i T s,i T i h s,i h i s s,i s i [kpa] [K] [K] [kj/kg] [kj/kg] [kj/kg-k] [kj/kg-k]

7 T-s Plot all(pratiog as) 7

8 etath(p [6]) etath vs eta 8

Equations P $UnitSystem K kpa. F luid$ = Air (1) Input data for fluid. $If Fluid$= Air. C P = [kj/kg K] ; k = 1.

Equations P $UnitSystem K kpa. F luid$ = Air (1) Input data for fluid. $If Fluid$= Air. C P = [kj/kg K] ; k = 1. P09-169 Equations Thermodynamics - An Engineering Approach (5th Ed) - Cengel, Boles - Mcgraw-Hill (2006) - pg. 549 Ciclul Brayton cu regenerare (aer, heliu) cu comprimare si destindere in trepte, c p =ct.