Question

4.15. Carbon dioxide flows in a horizontal adiabatic, no-work system. Pressure and temper- ature at section 1 are 7 atm and 600 K. At a downstream section, p2 = 4 atm., T2 = 550 K, and the Mach number is M2 = 0.90. (a) Compute the velocity at the upstream location. (b) What is the entropy change? (e) Determine the area ratio A2/A.

Answers

a. 156.8 m/s

b. 32.5 J/kg K

c. 0.763

Answer 1

Given Pressure at section 1, P1 = 7 atm Temperature at at section 1, T = 600 K Pressure at section 2, P2 = 7 atm Temperature at at section 2, T, = 550 K Mach number at exit, M2 = 0.9 (a) Velocity,V, of the gas at exit V2 = M, X sonic vel coity at exit temperature V2 = M, VYRT, = 0.9 1.29 x 189 550 = 329.57 m/s Since the process is adiabatic, there's no heat transfer between system and surroundings. Also there's no work done. Since the flow take place in horizontal pipe no potential energy change to the flow. Now writing energy balance at section 1 and section 2 tha – h) + = co(P2 –16)+" 329.572 = 841(550 - 600) + - = 12258.19 solving for V V1 = 156.577 m/s

(b) Entropy change for an ideal gas is given by 52 - = cp in (62) – R in (6) substituting the known values sz-sı = 841 In (500) – 189 in (7) = 32.5J/kg.K S2-S1 = 32.5J/kg.K (c) Since the mass flow rate remains constant P1A2V1 = P2A, V2 A2_AV (RT)v ( PTV A P2 V2 substituting the known values 42 = 0.763