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 combustion 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

(b) Find the temperature at the exhaust of the heat exchanger, assuming no heat loss to the surroundings

(c) Find the rate of entropy generated in the heat exchanger (regenerator), assuming no heat loss to the surroundings.

(d) Find the back work ratio

(e) Find the cycle efficiency

(f) Find the net power generated by the engine.

Answer 1

Finding every point temperature we can easily Solve all parts. I have shown below-
Sol: We have to find out - = 288 K. Pressure at the inlet of compressor, P = 100 kp a f temperature Ti = 15°C - Isentropic efficiency of compressor, ne =89% = 0.82 flow = 5 m3/secis Effectiveness of heat exchanger (regenerator)= volumetric rate v 787 = 0.78 efficiency of turbine nt Temperature at the iniet to turbine, T3=600°c =873°C I sentropic 70% =0.7 → we have to find out - a) p_vf T-8 diagram. Exhaust temperature of air in regeneratoy. Entropy generation in the regenerator Back work ratio , d) e) cycle efficiency Net power generation. b c ) Assumption :- Air as an ideal gas. Air properties are constant with temperature. Regenerator pressure loss is negligible. mass of fuel is neglected. . Processies :- 1-2:- compression 5-31- Isobaric heat addition. 3-47:- Expansion.

6-) Isobaric heat rejection . a) P-v diagram : Chamber T 3 5 3 Regenerator combustion * 27 Pac 4 2 6 Pac 6 4' 4 그 U Actual cycle :-1-2-5-3-4-6-1. 8 Idel :-1-2-5-3-4-6-1 Process 1-2:- I/ Tz T к - Cent) or, Tz = 288 (4) .33 1.33 P2 [ = Yp=4 (given)] 406-23 Ri Now, na cp ( Tz - T) ( Tz'-TI) Cp or 0.82= 406 123 288 Tal -288 or Ta'a 432.18 K Process 3-4; Tay K-1 Pu P3 Ti

.33 or, Tu = 873 (t); 1.33 PM P3 re امی 618.91k Now, Nr Cp. (T3-T6') Cp (T3 - Tu 0,70 = 873 -Tul 873-618.91 or, or, Tul = 69584 Effectiveness of regenerator is defined as the actual temperature rise to the maximum possible temperature rise in the the regenerator, E Ta' Tu-Ta' or, Ts 0.78 432118 695.14-432.18 X5 ie 11 or, Ts z 637:28 ot) - 364.28'c Ans (6) at the exhaust of the so, temperature regenerator of air 364:28 'c, generation in the regenerator c) Rate of entropy is given by in [pen Is R en Ps P 2 in Cp en 11 Ts Ta' [":ls =PL]

と Piv RTI Cpen TS Te' RIS 0.223X1:33 100 x 5 x 1-1 en 0.2.3 X 288 1 637.28 932.18 1:4 K-1 •33 =ll RJ ligy 2717 KWIK Aus (C). d) compressor work, we = nice (Ta'-T) ", Wy = mcp (T3-Tw') Turbine i. Back work ratio, row Wc Wp 432.18-288 Tz - Ti T3-T4 873 - 695.14 0.81 Ans (d). e ) cycle efficiency, n Wz-Wc Qs rice (T3 - Tu) wilo (Ta'-T nice (T3-T5) 673 – 695.14) -(432-18-288) - 973- 6 37:28 Z 14.288 Ans (e),

f) Net power generated, Whet = Wy - Wc nico (T3 - Tú) - mice ( Tz -T) (873 695014)-(432-18-288 = TDOXS 11 X 11 0.273x288 235.60 kW (£).