Equations P Se va nota cu y fractia din debitul masic care intra in turbina care e extrasa din turbina pentru preincalzitorul inchis

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P10-110 Equations Thermodynamics - An Engineering Approach (5th Ed) - Cengel, Boles - Mcgraw-Hill (2006) - pg.

1 P Equations Thermodynamics - An Engineering Approach (5th Ed) - Cengel, Boles - Mcgraw-Hill (2006) - pg. 602 Ciclul Rankine cu regenerare cu preincalzitoare deschise multiple - Regeneration using Multiple Open Feedwater Heaters Sa se studieze influenta numarului de trepte de regenerare asupra performantei unui ciclu Rankine ideal cu regenerare. Aburul intra in turbina cu15 MPa si 600 C, si in condensator cu 5 kpa. In fiecare caz se va mentine aceeasi diferenta de temperatura intre oricare doua trepte de regenerare. Sa se determine randamentul termic al ciclului si sa se traseze grafic in functie de nr de trepte de regenerare. Se va nota cu y fractia din debitul masic care intra in turbina care e extrasa din turbina pentru preincalzitorul inchis $UnitSystem C kpa procedure Reheat(NoF wh, T 5, P 5, P cond, η turb, η pump : q in, w net ) (1) T cond = T (steam, P = P cond, x = 0) [C] ; (2) T boiler = T (steam, P = P 5, x = 0) [C] ; (3) P 7 = P cond [kpa] ; s 5 = s (steam, T = T 5, P = P 5 ) ; (4) 1

2 h 5 = h (steam, T = T 5, P = P 5 ) ; h 1 = h (steam, P = P 7, x = 0) ; (5) P 4 1 = P 5 NOTICE THIS IS P4[i] WITH i= 1 (6) T cond,boiler = T boiler T cond [C] (7) If NoF W H = 0 then (8) the following are h7, h2, w net, and q in for zero feedwater heaters, NoFWH= 0 h7 = h (steam, s = s 5, P = P 7 ) (9) h2 = h 1 + v (steam, P = P 7, x = 0) P5 P 7 η pump (10) w net = η turb (h 5 h7) (h2 h 1 ) (11) q in = h 5 h2 else i = 0 i = i + 1 (12) (13) (14) (15) (16) The following maintains the same temperature difference between any two regeneration stages. T F W H,i = (NoF W H + 1 i) T cond,boiler NoF W H T cond [C] (17) P extract,i = P (steam, T = T F W H,i, x = 0) [kpa] (18) P 3 i = P extract,i ; P 6 i = P extract,i (19) If i > 1 then P 4 i = P 6 i 1 until i = NoF W H P 4 NoF W H+1 = P 6 NoF W H (20) (21) (22) h4 NoF W H+1 = h 1 + v (steam, P = P 7, x = 0) P 4 NoF W H+1 P 7 η pump (23) i = 0 (24) (25) 2

3 i = i + 1 (26) Boiler condensate pump or the Pumps 2 between feedwater heaters analysis h3 i = h (steam, P = P 3 i, x = 0) ; v3 i = v (ST EAM, P = P 3 i, x = 0) (27) w pump2,s = v3 i (P 4 i P 3 i ) SSSF isentropic pump work assuming constant specific volume (28) w pump2,i = w pump2,s /η pump Randamentul pompei - Definition of pump efficiency (29) h4 i = w pump2,i + h3 i Steady-flow conservation of energy (30) s4 i = s (ST EAM, P = P 4 i, h = h4 i ) ; T 4 i = T (ST EAM, P = P 4 i, h = h4 i ) ; (31) until i = NoF W H i = 0 i = i + 1 (32) (33) (34) (35) Preincalzitor deschis - Open Feedwater Heater analysis: s5 i = s 5 ; ss6 i = s5 i (36) hs6 i = h (ST EAM, s = ss6 i, P = P 6 i ) ; T s6 i = T (ST EAM, s = ss6 i, P = P 6 i ) ; (37) h6 i = h 5 η turb (h 5 hs6 i ) Randamentul turbinei - Definition of turbine efficiency for high pressure sta If i = 1 then y 1 = h3 1 h4 2 h6 1 h4 2 Steady-flow conservation of energy for the FWH (39) If i > 1 then js = i 1 j = 0 sumyj = 0 j = j + 1 sumyj = sumyj + y j ; until j = js (40) (41) (42) (43) (44) (45) (46) (47) y i = (1 sumyj) h3 i h4 i+1 h6 i h4 i+1 (48) 3

4 endif (49) T 3 i = T (steam, P = P 3 i, x = 0) Condensate leaves heater as sat. liquid at P[3] (50) s3 i = s (ST EAM, P = P 3 i, x = 0) (51) Turbine analysis T 6 i = T (ST EAM, P = P 6 i, h = h6 i ) ; s6 i = s (ST EAM, P = P 6 i, h = h6 i ) (52) yh6 i = y i h6 i until i = NoF W H ss 7 = s6 i (53) (54) (55) hs 7 = h (ST EAM, s = ss 7, P = P 7 ) ; T s 7 = T (ST EAM, s = ss 7, P = P 7 ) (56) h 7 = h6 i η turb (h6 i hs 7 ) Definition of turbine efficiency for low pressure stages (57) T 7 = T (ST EAM, P = P 7, h = h 7 ) ; s 7 = s (ST EAM, P = P 7, h = h 7 ) (58) sumyi = 0; sumyh6i = 0; wp2i = W pump2,1 (59) i = 0 i = i + 1 sumyi = sumyi + y i sumyh6i = sumyh6i + yh6 i (60) (61) (62) (63) (64) If NoF W H > 1 then wp2i = wp2i + (1 sumyi) W pump2,i (65) until i = NoF W H (66) Pompa condensator - Condenser Pump Pump 1 Anslysis: P 2 = P 6 NoF W H P 1 = P cond ; X 1 = 0; h 1 = h (ST EAM, P = P 1, x = 0) v1 = v (ST EAM, P = P 1, x = 0) s 1 = s (ST EAM, P = P 1, x = 0) T 1 = T (ST EAM, P = P 1, x = 0) (67) (68) (69) (70) (71) (72) 4

5 w pump1,s = v1 (P 2 P 1 ) SSSF isentropic pump work assuming constant specific volume (73) w pump1 = w pump1,s /η pump Definition of pump efficiency (74) h 2 = w pump1 + h 1 Steady-flow conservation of energy (75) s 2 = s (ST EAM, P = P 2, h = h 2 ) ; T 2 = T (ST EAM, P = P 2, h = h 2 ) (76) Generator de abur - Boiler analysis q in = h 5 h4 1 SSSF conservation of energy for the Boiler (77) w turb = h 5 sumyh6i (1 sumyi) h 7 SSSF conservation of energy for turbine (78) Condensator - Condenser analysis q out = (1 sumyi) (h 7 h 1 ) SSSF First Law for the Condenser (79) Ciclu - Cycle Statistics w net = w turb ((1 sumyi) w pump1 + wp2i) (80) endif (81) end (82) Marimi de intrare - Input Data: $IfNot ParametricTable= NoFWH NoF W H = 2; $EndIf P 5 = [kpa] ; T 5 = 600 [C] ; P cond = 5 [kpa] ; (83) (84) (85) (86) η turb = 1.0 Randament turbina - Turbine isentropic efficiency (87) η pump = 1.0 Randament pompa - Pump isentropic efficiency (88) P 1 = P cond [kpa] (89) Rezolvare: P 4 = P 5 ; (90) Condenser exit pump or Pump 1 analysis call Reheat(NoF wh, T 5, P 5, P cond, η turb, η pump : q in, w net ); (91) η th = w net /q in ; (92) Data$ = Date$; (93) Solution Variables in Main program 5

6 Data$ = η pump = 1 η th = η turb = 1 NoF W H = 2 P cond = 5 [kpa] q in = 2534 [kj/kg] w net = 1242 [kj/kg] Variables in Procedure Reheat NoF wh = 2 T [5] = 600 [C] P [5] = [kpa] P cond = 5 [kpa] η turb = 1 η pump = 1 q in = 2534 [kj/kg] w net = 1242 [kj/kg] T cond = [C] T boiler = [C] P [7] = 5 [kpa] s[5] = [kj/kg-c] h[5] = 3581 [kj/kg] h[1] = [kj/kg] P 4[1] = [kpa] T cond,boiler = [C] h7 = notdefined [kj/kg] h2 = notdefined [kj/kg] i = 2 w pump2,s = [kj/kg] y[1] = h3[1] = 1033 [kj/kg] h4[2] = [kj/kg] h6[1] = 3099 [kj/kg] js = 1 j = 1 sumyj = ss[7] = [kj/kg-c] hs[7] = 2035 [kj/kg] T s[7] = [C] h[7] = 2035 [kj/kg] T [7] = [C] s[7] = [kj/kg-c] sumyi = sumyh6i = [kj/kg] wp2i = [kj/kg] W pump2,1 = [kj/kg] P [2] = 322 [kpa] P [1] = 5 [kpa] X[1] = 0 v1 = [m 3 /kg] s[1] = [kj/kg-c] T [1] = [C] w pump1,s = [kj/kg] w pump1 = [kj/kg] h[2] = 138 [kj/kg] s[2] = [kj/kg-c] T [2] = [C] h4[1] = 1047 [kj/kg] w turb = 1271 [kj/kg] q out = 1281 [kj/kg] T F W H,1 = [C] P EXT RACT,1 = 3292 [kpa] P 3[1] = 3292 [kpa] P 6[1] = 3292 [kpa] T F W H,2 = 136 [C] P EXT RACT,2 = 322 [kpa] P 3[2] = 322 [kpa] P 6[2] = 322 [kpa] P 4[2] = 3292 [kpa] P 4[3] = 322 [kpa] H4[3] = 138 [kj/kg] V 3[1] = [m3/kg] S4[1] = [kj/kg-c] T 4[1] = [C] H3[2] = [kj/kg] V 3[2] = [m 3 /kg] W P UMP 2,2 = [kj/kg] S4[2] = [kj/kg-c] T 4[2] = [C] S5[1] = [kj/kg-c] SS6[1] = [kj/kg-c] HS6[1] = 3099 [kj/kg] T S6[1] = [C] T 3[1] = [C] S3[1] = [kj/kg-c] T 6[1] = [C] S6[1] = [kj/kg-c] Y H6[1] = 562 [kj/kg] S5[2] = [kj/kg-c] SS6[2] = [kj/kg-c] HS6[2] = 2609 [kj/kg] T S6[2] = 136 [C] H6[2] = 2609 [kj/kg] Y [2] = T 3[2] = 136 [C] S3[2] = [kj/kg-c] T 6[2] = 136 [C] S6[2] = [kj/kg-c] Y H6[2] = 375 [kj/kg] Arrays Row P i T i [kpa] [C]

7 NoFWH Run NoF W H q in w net η th [kj/kg] [kj/kg] T-s: Steam 7

8 NoFWH 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.