Question

In a gas turbine power plant, the gas enters the five-stage compressor at 300 K. The overall power ratio is 20. After the compression, the gas enters a regenerator with an effectiveness of 80%. For preheating, it is further heated in a combustion chamber until it reaches a temperature of 1600 K. The heated gas expands to the first turbine stage and is reheated to 1550 K. The temperatures after the reheating processes are 1500 K, 1450 K, 1400 K, and 1350 K for the remaining stages of the turbine, respectively. The gas is cooled by a pre-cooler before it enters the compressor again. a. Draw the schematic diagram of the gas turbine power plant b. Draw the T-S diagram c. Determine the thermal efficiency d. Resolve the problem by varying the number of stages of the compressor and turbine as shown below. Compare the efficiency of the five-stage compression and expansion. No of stages Overall Pressure ratio 20 300 20 Compressor Turbine inlet temperature/s (K) Inlet Temp (K) 300 1600 1600 and 1550 300 1600, 1550, and 1500 1600, 1550, 1450, and 1400 300 1600, 1550, 1450, 1400, and | 1350 20 Uw 300 20

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Answer 1

Given data! Inlet to compressor, The Book. Pressure ratio op=20. Turbine Inlet temp = 16ook. Effectiveness of Regenerator =0.8 To find Thermal efficiency, oh. soch Care I: Single stage conhesson & Jurtins essor. at m y At combustion Chamber, compressor a to TI

Process (1-2) compresor: It is an Isentropic process so Here we take Aiu as workery fluid serice Nothing is Guilh pori = 1.4. T2 = Cop) Metin => T2 = 300 (20286. Th=706667 Here since Regenerator is used so Exit tend. of the Gurbin will be equal to Inlet tend. of Regenerator (Ts) but Now Heat and requered is (1600-TI') rather than (1600-72). but do Regenerator effectiveness is 80%. so E= TG-TS 30.8=16 T3-TS T3-70 Process (3-2) Turbens: Isenbopic Expansio 7- (0 Tiga 2009 => [Ty z 679.24K.

In this case. Zurbin outlet tend. is less than the confressor outllet Tenk so. Regenerator will not be effective so Heat added in combustor = Cp (73-72) Qin =1005 (1000-706.6). =897.86 KJ/kg] Wook Hp to the Confressor = CP (72-71) =C6 ( 706.6-300). Tom. = 408.633 KIJKs. Work off from the Lubina cp (73.-74.] wy. =P ( 1600-679:24) W12925.36 kJ/K. aun Thermal efficiency= Wy-we Qin = 925.36-408.633 897.86 T 9th. = 37.85%

for Tur stage compressor 4 Turbine Regenerabi - on a combustion chamber. | Ti TL. 1 11401. Intercooler Reneath 3 6 2. -.-.• - . - - Ti6 - 2 Given datu: Tu=1550k. Rest all dater is same Since overall Pess. ratio is 20 so Intermediate Press. off for confress orl. P2 = JPxP3 =520 4.477 rpi=&p2= 4.47.

for compressor 1-2 I 2 = (20.286 Tha 300 (4.47 0.20 J Y2 460.34K. ti in cooled to Initil tenp. 3ook in entrcooler. - 72 = 300(4.747)0.286 72 = 460.37K. / for Turbine (3-4') = 14.470.286. (Pawberis = JZOX =6.47. u' - loa 634 K For Turbine 4-50 Ty = (4.4%)0*286. => Ts: 1010.05k Regenerator & Effectiveneis = 1600-1010-05 - 1600 - To. T6=862.56k] Win work Ilp to compressor 1. = CP (72'-7. =1005 (460-37-300). Wha= 161.17 KJ/14. Win 2=161.17K/kg Work opp for furbline 1, W7, = Cp C1600-1042.674). = 560.15 KS/1G. wook off for Zurbine 2, WT2 = CP C1550 - 1010.05). =542.64 kJ/kg.

Wer stat Heat HP in Combustion chamber CQ (T3-Te Qin =1005 (1600*862:56) Rina 7418242 KJ/kg.) Heat supplied in the Repeates = cp (Tu - tu) = 1.005 (1550-1042-634 QR = 509.9 KJ /kg) Thermal efficiency = (WT, W12) - (Wetwers Qintax) = (560.11+542.64) -(161.1774 741.1272) +(509.9) Thu = 62.36 Due to Reheating the Intercooling ore all efficiency oft P. the Power Plant As we i more no of Reheates & Intercooles more efficiency will t. 161-7