Question

8. An ideal air-standard Brayton cycle operates at steady state with compressor inlet conditions of 250 K and 25 kPa. The compressor pressure ratio is 10. The turbine inlet temperature 1800 K. For the cycle: (a) the heat addition and work done in each process, in kJ/kg, (b) the thermal efficiency (c) the back work ratio (35 points)

Answer 1

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solution giren data Compressor inlet temperature (T) = 250k Turbine iniet temperature CT3) = 180ok Compressure inlet pressure (Pi) = 25kpa Compressor pressure ratio (r) s 10 Asstemp Hon!! - al process is taken as reversible process. air is treated to be ideal gas. for air (ideal gas) Cp = 1.005 kJ/kg K , C = 0.78 kJ / tgk and Y=1.4 pooless taken as steady flow process. * Kinetic Energy and potential Energy changes to be are D negligible. Analysis! (T-S diagram) TCH) 个 P-C= 10x P, 15 Qs 2 > Wp 2 Tu 4 Wc To Pi=C2 25kpa. - 5,=52 S3 = 54 st

Now; Compsession ratio (2) = P2 pi 10 = P2 25 (kpa) P2 = 250 ckpa) sprocess (1-2) isentropic Compression process for isentropic process ( 12 에 To P2 Ti PI 1.4) 1.4 T2 = 250 (10) 1 T DE 482.67K

* applying steady How Energy equation in process 132 so that we get compressor work hit a = htWc @ 20 for adiabatic process ha fet) for ideal gas dha Cp dr hy h2 = We , = Wc Cp CT1-T2) = 1.005 (250-482.67) We = - 233.83 kol ng Where the sign indicates Work done is on the Compressor So, We = 233.83 kalkg

- process (2-3) Constant pressure heat addition (Qs) here for process (2-3) ( P2 =P3] and Work done in Combustion chamber is zero. so applying SFEE hat e hy tyto as 2 hz-h Cp CT3-T2) 71.005 (1600-482,6+] 1323. 91 kilkg → Process (3-4) ventropic Expansion, for isentropic Exponsion. 어 To - (14) Ty 1.4) 1800 J4 (io) 14 T4=932.30k → applying SFEE ; We get Turbine work hat +07 = hq tWq

Wr= hzrhy =lp (T3-T4) = 1.005 (1800 - 932. 30) The Wp 872.03 ki KJ/kg St by applying law of Thermodynamics for a cycle Ea Ew cycle cycle as-ap Wy - We OR= as - Wr the 1323.91 - 872,03 + 233,83 Op = 68504 kJ/kg Play heat addition as z 1323.91 kalkg heat rejection OR = 685.# kj lkg Work done on compresset WE = 2.33.83 Kylling W7 = 872.03 kilkg ans a Work done by turbine

(6) the thermal Efficiency Whet nih a supply Wz-Wc 11 als 872,03 -233.83 1323. 91 7th 0.4820 the U8.2013 sans-b (C) the back Work ratio (BWR) BWR = Compressor Work (WC) turbine work (WA) 233,83 0.2681 872.03 ane e BWR = 0.268