Question

A piston-cylinder assembly initially contains 0.8 kg of air at 100 kPa and 300 K. It is then compressed in a polytropic process PV3 = C to half the original volume. Assuming the ideal gas model for air and specific heat ratio is constant, k=1.4, determine (a) the final temperature, (b) work and heat transfer, each in kJ. R= 0.287 kJ/kg K. W, 82

Answer 1

M = 0.8 kg pi = 100 kpa Ti = 300 k v_2 = vi/2 pivi = mRT_i 100 times 10^3 times vi = 0.8 times 287 times 300 vi = 0.69 m^3 v_2 = 0.344 m^3 to find temperature(T_2) Tivi^n - 1 = T_2v_2^n - 1 300 times (vi)^1.3 - 1 = T_2 times (vi/02)^1.3 - 1 T_2 = 369.34 k work - donme = [p2v2 - pivi/n - 1] = mR[T_2 - Ti]/n - 1 = 0.8 times 287 times [369.34 - 300]/1.3 - 1 w = -53.070 KJ (work done is negative because of compression) now for the using 1st law of thermo dynamics - 80 = 80 + sw = Mc_v(T_2 - T_i) + sw = 0.8 times .718 times (369.3 - 300) - 53.070 theta implies - 13.24 KJ heat is negative, that means heat is rejected from the system