Answer to Question #257941 in Mechanics | Relativity for giel

Question

A nozzle is designed to expand air from 689 𝑘𝑃𝑎𝑎 and 32℃ to 138 𝑘𝑃𝑎𝑎. Assume an isentropic expansion and negligible approach velocity. The airflow rate is 1.36 kg/s. Calculate (a) the exit velocity; (b) the proper exit cross-sectional area.

Answer 1

(a) Since the expansion is isentropic

$\frac{T_2}{T_1} = (\frac{p_2}{p_1})^{γ-1/γ} \\ \frac{T_2}{305} = (\frac{138}{689})^{1.4-1/1.4} \\ T_2 = 192.625 \;K$

Applying steam flow energy equation to nozzle

$m(h_1 + \frac{v^2_1}{2000} + \frac{z_1g}{1000}) + Q = m(h_2 + \frac{v^2_2}{2000} + \frac{z_2g}{1000}) + w \\ m(h_1 + 0 +0) + 0 = m(h_2 + \frac{v^2_2}{2000} + 0) + 0$

Neglecting potential energy change

$mh_1 = m(h_2 + \frac{v^2_2}{2000}) \\ mc_pT_1 = m(c_pT_2 + \frac{v^2_2}{2000}) \\ 1.36 \times 1.005 \times 305 = 1.36(1.005 \times 192.652 + \frac{v^2_2}{2000}) \\ v_2 = 475.205 \;m/s$

(b) Mass flow rate $m =ρ_1A_1v_1 = ρ_2A_2v_2$

From ideal gas equation

$p_2 = ρ_2RT_2 \\ ρ_2 = \frac{p_2}{RT_2} = \frac{138}{0.287 \times 192.652} = 2.4958 \;kg/m^3 \\ m=ρ_2A_2V_2 \\ 1.36 = 2.4958 \times A_2 \times 475.205 \\ A_2 = 1.1467 \times 10^{-3} \;m^2$