Question

Problem 1 (15 pts) A gas turbine cycle operates with a compressor pressure ratio of 12 and a mass flow rate of 5.0 kg/s. Air enters the compressor at 1 bar, 290 K. The maximum cycle temperature is 1600 K. For the compressor, the isentropic efficiency is 85%, and for the turbine the isentropic efficiency is 90%. Using an air-standard analysis with air as ideal gas with constant specific heats, calculate: a) the volumetric flow rate of air entering the compressor, in my/sec. b) the net power developed, in kW. c) the back work ratio of the cycle. d) the thermal efficiency of the cycle. e) the thermal efficiency, if the cycle includes a regenerator which has a regenerator effectiveness of 85%.

Answer 1

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600k - - - - - - 1 Pmox - L Pminc I 90k F E Solution Given that rp=12 ( pressure Ratio) m=stols , T, = 290k, P,= 1 bar, T3=1600.k. e nisem propria compe= 0.85, Mienpapie Sturbine = 0.90 2 T-S diagram ~ 1-2 isenpropic compression - T3=1600k 1-2' Actuel compression : 2-3 → P=c, heat addition 3-4 → ſsen hobic Expansion 3-4's Actual Expansion . . P=6, heet Rejection. - For Process 1-2 (isenkopic) For Air (p=1.005 ks To = 290(12) = 589.84 k Cr=0.718 kJ - ided work/ isenpropic work kgok lisen. comp. actol work . Go (12-1) r=14 0.85= Go (12'-11). . 0.85 = 589.84 290 → To'= 642-75 k To'- 290 - For prouss (3-4) isenpropic P= Py P2 = P3 * E- 97 (19)99 Leg-k R= 0.287 I lig ke 4 s a 9 = (12)*

> Tue 1600 0.4 = 786-65 k g (12) * Misen turbine = altul work - Storcat ptta-To') ided work - foto Ts) ep (T3 - Tu) 0.9= TEE 1600 - Tu! - 1600-786.65 Tu'= 867.98k Assumptias taken into account → I 0 working fluid as air only in the entire cycle. - ® > air os an ideal gos. 6 Dotriction Faltin ang tentonant Change in PE and 1. Change in KE are neglected. ③ → Cp, cv and & are constant for air. a or P, Vi=mrti V,= mrti 5x2878290 ē Ixlos Vilvop-flow rale) =4.16151 b) Net Power (P) = m (W turbine - Wcomp.) p=5 [cp (F3- Tul) - Cp (12'-1)] p-581005 [ 2600 - 867-98)-C642.75-290) 7 -> p = 1905.83 kW c): Back work ratio = the work . Compressor work turbine work the work

. row = Go tre'-T) So tTs-Tu!) = 642-75-290 1600 - 867-98 Now = 0.4818 d) nin = net work heat supplied im Co ( To-T2') 1905.83 th= 5xl wsx( 1660-642.75) nth= 0.3962 OR 39.62%