Question

In a gas turbine engine, the compressor takes in air at a temperature of 15°C, pressure of 100 kPa, and a volumetric flow rate of 5 m3/s and compresses it to four times the initial pressure with an isentropic efficiency of 82%. The air then passes through a heat exchanger heated by the turbine exhaust before reaching the combustion chamber. In the heat exchanger 78% of the available heat is given to the air. The maximum temperature after constant pressure combustio is 600°C, and the efficiency of the turbine is 70%. Neglecting all losses except those mentioned, and assuming the working fluid throughout the cycle to have the average properties (k=1.33, R=0.273 kJ/kg.K). Neglect the effects of the mass of fuel: (a) Plot the cycle on a p-V and a T-s diagrams (2 mark) (b) Find the temperature at the exhaust of the heat exchanger, assuming no heat loss to the surroundings (3 marks) (c) Find the rate of entropy generated in the heat exchanger (regenerator), assuming no heat loss to the surroundings. (3 marks) (d) Find the back work ratio (2 marks) (e) Find the cycle efficiency (2 marks)

Answer 1

Drawing the.T-s diagram and then finding Every point temperature we can solve this question as below-
Sol - The data provided in the question are as below - Inlet Pressure to compressure -100 kpa 15'c tempo = 289 k. PA T 5 m3/sec. volumetric frow rate of air, V = Isentropic efficiency of compressor (Mis) = 0.82 Etectiveness of regenerator, E = 0.78 Inlet temperature to turbine, I = 600c = 873 K. Isentropic efficiency of turbine his) ) = 0.70, T we have to find out a) .P-V and T-s diagram of the cycle Exhaust temperature ら of the heat exchanger. heat exchanger. c) Entropy generated in the Back work ratio d ) e) Cycle efficiency $) Net Power generated by the engine. Assumption - Air as an ideal gas } 6 Aiy Properties are constant With temperature. Pressure to sses in the Average properties of air : K= 1:33, 6301273 kg/1974 regenerator neglected. mass of fuel is neglected heat loss in the regenerator. NO

Process 1-2':- compression . A-3:- constant Pressure heat addition. 3-41:- Expansion . bol:- constant pressure heat rejection Ideal cycle 1-2-2-3-4-6-1 0 Actual cycle 1-2'- Q-3-4-6-j. P- v dia a ) diagram: Is diag ram : P С. T A 3 2 2 3 *Regenerator - 873 K th=695.140k To=637.290k/ a Th=618.915k1 TI!=432-186K T2=406.2331 6 2 K 4 1 b 41 T= 288k 8 helf of all sailent First of all, we have to find out temperature cuith the given data · points Process 1-2- Tz P2 (é k-1 k TI PI :33 1 33 = 4 4 (Gi ( Giren) or, Tz 288x (4) 406. 233 ki

Is entropic efficiency of compressor, Qis) C Ideal work! Actual work = Tz - Ti to'rti 406-233 Ta' 288 Oy, 0.82= -288 or, Tz' E 432.186 R. Process 3-4:- Tu To Pur K-1 K 23 Pu .33 1.33 [ U or, Tu = 873 ( ( T3 P P2 4 4 618.915 K. Now, Isentropic efficiency of turbine, of turbine, (Mis) T Actual work Ideal work = T3 - Ty' 13 - 14 or, o. 70 873 - Ty' 873-618.915 or, t'a 695.140 ki AS, in the is to heat air, then heat exchanger 78% of the available given effectiveness of heat exchanger (regenerator) will be Ta - Ta E = the Ty-Ta

0.78 - = or, Ta 432186 695.140 – 432.186 oy, (TQ 637 290 K Ans (6) There fore , temperature at the exhaust exhaust of the heat 6 37.2gok exchanger From Ideal gas equation / mass flow rate of air, in Pivi U RT 100x 5 O.273x288 6.3594 297 seci generated in the regene Therefore, rate of entropy ratoy To in [Cp en R en Pa Ta Pa mice en Ta Ta' [ Pa = P₂ 637.290 6.3594x lolo en Cp Rik K-1 0.273 x 1.33 .432-186 1.33-1 19 kJ/kgk = 2.7167 KW/K [un Unit : Ans KJ (C). sec legk kw d) compressor. work, we = nCp (T2-To) )

56.3594 x lil x (432 • 186 –288) 1008.63 KW. Turbine work, WT = mcp (T3 - Tu') = 6:3594 x 1+1 X (873-6950140) 1244.991 KW, = 1 : Back work ratio, row Wc 1008.63 Wy [244:191 - = 11 0.811 Ans @ ). e) cycle efficiency, n. Whet Qs Wp - We = Qs 1244.191- 100 8.63 il 6.3594x1'1X (873-637-290 290) Qs in C, (T3-Ta] =114.286 %. le).

f Net power generated by the engine, whet =Wy -We (12.441191 - 1008563 Il ) KW ㄹ 235.561 kw Ans. 6).