Question

For all of the processes, you may assume the gases have the properties of pure air and are thermally and calorically perfect. For all processes except the exhaust nozzle, you may neglect kinetic and potential energy changes. Therefore, the static and stagnation properties are equal.

Answer 1

Solution : Griven, - 30ctm maz 50kg/s PL 1 atm Ti 300k P2 ne = 0.9 mf = 1 = mf = 80 so 80 50 = 0.625 kg/s 80 CV = 50,000 ma mf= KJ/kg quel t process 1-2 :- compression process 2-3 :- constant pressure combustion 34 PI Sa Cheat added] 2 P=Const 21 T 230 r = P2 s (e) and (a) Ts? isentropic compression, process 1-21 77 0.4/1.4 [y for air = 14 (9) q Ti 0.4/1.4 - 792.78K T2 = 300x (30) now no T2-T Tor-T, 0.9= 792.78-00 T2-300 T2 = 847.533k

from prouss 2-33 Sq = (ma+mg) Cp Tz - map Tz ma.cv = [(50 +0.625) T3 -50*947.522].4 [pforair = 1,005 KJ / kg k] 0.625 X 50, OUD 1.005 50.625 Tz - 42376.65 suna T3 = 1451.28K w Pr. C. Pz = B = P = const (6) Sz-k? now entropy change for process 2-3 Sz - f = ma [ Cpen Tz - ReyP Sg-2 = Map en to Ta Sz-z = 50x100s en 1451.28 847.533 Sz-P2 = 27.028 kJ/4/ Ang