Question

4. A piston/cylinder contains pure oxygen at ambient conditions 20°C, 100 kPa. The piston is moved to a volume that is seven times smaller than the initial volume in a polytropic process with exponent n = 1.25. Find the final pressure and temperature, the specific work, and the specific heat transfer. Use table A.8 to find the internal energy at each state. Plot the process in a P – v diagram.

Answer 1

P2 =1138.6 kPa

T2 = 476.6 K

Specific work = -190.8 kJ/kg

Specific heat transfer = -66.2 kJ/kg

u1 = 190.61 kJ/kg

u2 = 315 25 kJ/kg

Gn t = 20c P, = 100k le V V2 = P n=1.25 3 X2 = 7 Polytropic process, Pr = P q ve ? M"- (3) Pg 1.85 Pa pr diagram 100% 11.386 = 11386 kPa 1.25-1 1.25 n 1138-6 Ta Ti = 1.6266 Pi 100 Ta 1.6266 x J = 1.6266 * (273+20) Tg - 476. 5 C Specific Wa SPdv R = 8.314 kkmok mook Pq Ve Piri In R B = R (T2-T) Molecular weight 1_n of a (476.6-291) = 0.2598 kJ/kg-k =0.8698 kJ /kg-K 8-314 32 1-1.95 = -190.8 kJ/mey Speutic heat transfer qz All tw

Те 850 k и= Та 20 L и. L45. For 293k 162-4) 3/8- kiling 145- - 14 2:41 ( аяз қо) а, 2 10 г. 4) + Зоо •360 И, - 110-6) kola For T,= 76.3 к T=450k u = 296.52 kg lung Т: Бъо к 1. -47 64 и, = 29452 + (33-2-0 . 296.52) (4%.6- 45) 33). 72 km/s 500-60 и, = 316-25 ) Ди Ug U, = 3/5. 85 - 140.61 Ди 7 18+, 6344 КО) Ди + W - 18h. (3 64 - Ј90. 8 — 60. e kJ ke kola

For oxygen Note Can some book this method aloo) Ди a CTR-T) G = 0.662 x lng-k = 0.655 ( 476.6–293) ) 20.8088 kg Autw 120.8088 190.8 69,9124 T ) tag Am P2 = 1138.6 kPa = 203.6°C Tg = 476.6k Specific mork 110.8 kg 8 - 66:20 kJ/kg PV-diagram Specific heat tramber Piv, 25. Po va P, VE u, 1.90.61 koling ug - 315.85 kJing