Question

Air flows adiabatically with friction through a long, constant area pipe. At the upstream section, p₁=60 kPa (abs), T₁ = 60 ^oC and V₁=200 m/s. At the downstream section, T₂ = 30 ^oC. Determine p₂, V₂ and the stagnation pressure ratio p_o2/p_o1.

Answer 1

P_01 - P_1 (1 + 0.2 m_a1^2)^3.5 = 60 (1 + 0.2 times 0.54^2)^3.5 p_01 = 73.16 kPa since the flow is adiabatic stagnation temperature is constant T_01 = T_02 = 352.9 k isentropic stagnation pressure at point p_02 = p_2 (T_02/T_2) r/v - 1 T_2 = 303 K = 43.11 (352.9/303) 1.4/1.4 - 1 P_02 = 73.50 stagnation pressure ratio = P_02/P_01 = 73.50/73.16 = 1.004 \r\n let air be approximated as ideal gas with constant k R = 0.287 Kgk, k = 1.4, C_p = 1.005 151 kgk h_1 + v^2_12000 = h_2 + v^2_2/2000 C_p T_1 + v^2_1/2000 = C_P T_2 + v^2_2/2000 1005 times 333 + 200^22000 = 1.005 times 303 + v^2_2/2000 50.15 = v^2_2/2000 v_2 = 316.70 m/s T_1 = 333 k T_2 = 303 k air flow T_2/T_1 = (P_2/P_1) r - 1/y 303/333 = (P_2/60) 1.4 - 11.4 P_2 = 43.11 kPa once the match number is found, local stagnation pressure follows m_a1 = squareroot 5 (T_01/T_1 - 1)