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 is 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

Answer 1

I have done the above question by two methods:

1- By assuming constant specific heat

2-By variable specific heat.

8 By consielewing Constant specific heat propouties : Qin 3 Wout given: Ti= 250k P = 25 kPa Press wre ratio (Mlp) = 10 T3= 1800 K: 2 4 more y lour. fou aje Standard Cycle Cp = 1:005 ks/kg T-S diagram of Aire plankers Brayton Cycle. Cvs 0.718 r= 1.4 Kalkg since, V-1 V-1 (MP) То Ta T ie Tz = 250x (10) *** T2=482-6744K alio, 3. T4 Pol .P4 Lep) 1800 T3 T4= Tap (10)** Tz= 932-3054 K Work in peer in compresor (Win) = Cp. ( T2 - To ) = 1.005 (482.6744 -- 250) Win = 233-85 712 Kulky Hear input in Combustion Chamber ( Pin)= Gp ( T3-T) = 1.005( 1800-482.6144) Pin = 1323.9122 K I) kolky

Work output by Turbine (Wout) = Cp (T3-Tu) = 1:005(1900 - 932.3054) Wout- 812.0330 K3 K3/kg Heat leeg'ecnol from Cycle (Clour) Cr(T-T) = 1005 (932:3054 -- 250) Yout 685.71692 kalky The thermal efficiency =) Whet Clin hout -hlin Qin 872.0330-233.83772 50.48205 1323-9122 Th1 = 48-2054. به Win Wout Back work ratio is defined defined as (8B/w) Defesa .26815 233.83772 872.0330 pew=-26815

By asuming variable specific heat properties: given: 3 TA T=250k P = 25kPa 73=1800k Pueywele Mario Pe Pi y الاحمر olid - 10 & Clout 승 T: 250 K from properties of air table; th-250-05 KJ/kg PM = 0.7329 for isentropic Compresion proces 1-2 we have; P82 IP2 = 10 Pri PH2 = 10X 0.7329 PH2= 7.329 at Prz = 47.329 by using in ten polation ; at, Py = 7:268 h=4821471 the 482-49+ 7.329-7.268 (492.74-482-49)P4= 1- 7.824 h = 492:747 (7.824-7-268 KI Similarly (490-480) 17.824-7-268) Px=7.268 T: 480 A: 1:24 1: 40 T2 = 481.091 K U o from properties of air table ; at T3 = 1800 K thz = 2003.3 PA3: 1310 KJ/kg

fone isentropic Expansion process 3-4 We have PHI Poru P3 P4 cie, Pry= Pu3 - 1310 10 = 131 131 - 123:4 133-3 - 123-4 (361) ar P44 - 131 by. in terpolation we have Be: 1234 T:1020 K hy: 1068.89 + ) (109195–1068-39) 1086-5158 kilka Kalkg Tq: (1040-1020) T4= 1035:353 K Pr= h=1068.89 kJ/er = 133.3 T-1040K na 1091.85 KITS Tha 1020 + 131-1234 133.3-123.4 OM - KJ/kg Work input in Compresor ha-hi = 4483-6145- 250.05* = 233-5645 KI/kg Win Min = 233.5645 kg Heat transfer in combustion Chamber = hz-hz 2003.3-483.6145 =1579.6855 Qin : 1519-6855 KIKA XJ/kg Turbine done work hg-h4 by compresor 2003:3-1086.5158 = 916-7842 KI Kolly TWout= 916.7842 Kulke Heat rejected from cycle ha-hi 1086.5158-250.05: 836-46578 Pout= 8364658 kJ/kg les /

Wout-Win Qin Clin The back cork ratio: Thermal Efficiency (ht) = Whet Th7:44.9574 916.7842 - 233:56457 1519.6855 044957 C7 - Win 233.5645 916.4842 Wout 8B/w = 125476