Question

Consider the Brayton Cycle for gas-turbine heat engines as shown below. Air is compressed from P,=100kPa, T,=20°C to P2=1400 kPa. The inlet temperature to the turbine is Tz=1100°C. The compressor and turbine are both adiabatic and both have isentropic efficiencies of 80%. in Heat exchanger Wool Compressor Turbine Heat exchanger 9 out

This is from a practice test and I want to be able to check my answers so any help you can give will be greatly appreciated!

Answer 1

Solu Heat exchanger furt T th Intercooler 3 T Os WT I Pin = Ps = Heat supplied Pourt = o p = » Wc = workdone in Rejected 4 Te We OR compressar WT = work done in Turbine fig = * T-S dignam 3 Given

Given To = host 20°c = 2936 T3 = 1100'c = dooooo Pi = 100 kPa 1373K P2 = 1400 kPa nisentropic 80% Cp = 1.005 kJ/kg. K Molecular weight = 29 g/mol K=1.4 Calculation > Gryland o'r 1-4-1 KA Tz a General relation. Ta PIK Ti PL 1400 100 293 T2 = 622.77K b specific compressor work W compresor Cp (T2-T1) kJ кя 1.005 ( 622.77 – 293) w comp =1331.426 kJ kg © The temp. at turbine (Ty) 91= Bussure ration - P2= 2 = 140o = 14 Pi 24 100 - 1.4-1 I3 = (ap) 1373 - (19) 5) Tu Tu Tu = 645.9596

directly (4) pecific turbine 'work done * Wurbine Cp (T3 - Ty) =1005( 1373-645-4596) = 730.675 KI continued ) The net work ( K3/ks) What = Wy. We (33) 426 X 756-675 242 164.6 kJ/kg ce (F73 - F)? = 1.005 ( 11373 - 1 293 ) 22 kJ 399.462 kg Continue What) maym SA maximum workdone ideal Condition on Ogupplied) and (9 out of Orejected) lin Ср ( T3-T2) = 1.005 ( 1373- 622.77) 753.98115 KJ qina

Yout col Thu. -To) = 1.005 ( 645,95 - 293) Yout = 354.714 kJ kg g) Thermal efficierney "thermal (Whet )actura 399.5% 3 kJ KS 753-9811 = 0.4238 42-38% mass flow rate (Kg/sc) to produc 3 Mo asp Power ma Whit k] κω kg/see kg '3003 KW ma kg see 319.56 9.3879 = ma mass flow rate

G The rate of Entropy generation in turbine. Yo 00 qs - Yout T 399.266 645.95 Sgene = 0.618 (9) 0.8 x Wc = actor 331-426X0.8 265.1408KT Kg d (w Turbing) actul 1150 = 0.8% WT = 0.8x730.675 584.54 KI Kg e What), total nx (Whet) m marh 20.8 X 39.462 319.56 kJ ks
In the end b,d,e are further solved for efficiency given and all other parts are solve for these efficiency