Question

Air enters a 200 mm diameter adiabatic nozzle at 195 deg C, 500 kPa and 100 m/s. It exits at 85 kPa. If the exit diameter is 158 mm, what are the temperature and velocity at the exit?

Answer 1

Answer:

$v_2 = 160.23 m/s$

$T_2 = 475.797 k$

Explanation:

given data:

Diameter =$d_1 = 200mm$

$t_1 =195 degree$

$p_1 =500 kPa$

$v_1 = 100m/s$

$p_2 = 85kPa$

$d_2 = 158mm$

from continuity equation

$A_1v_1 = A_2v_2$

$v_2 = \frac{\frac{\pi}{4}d_1^2 v_1^2}{\frac{\pi}{4}d_2^2}$

$v_2 = \frac{d_2v_1}{d_2^2}$

$v_2 = [\frac{d_1}{d_2}]^2 v_1$

      $= [\frac{0.200}{0.158}]^2 \times 100$

$v_2 = 160.23 m/s$

by energy flow equation

$h_1 + \frac{v_1^2}{2} +gz_1 +q =h_2 + \frac{v_2^2}{2} +gz_2 +w$

$z_1 =z_2$ and q =0, w =0 for nozzle

therefore we have

$h_1 -h_2 =\frac{v_1^2}{2} -\frac{v_2^2}{2}$

$dh = \frac{1}{2} (v_1^2 -v_2^2)$

but we know dh = Cp dt

hence our equation become

$Cp(T_2 -T_1) = \frac{1}{2} (v_1^2 -v_2^2)$

$Cp (T_2 -T_1) = 7836.94$

$(T_2 -T_1) = \frac{7836.94}{1.005*10^3}$

$(T_2 -T_1) = 7.797$

$T_2 = 7.797 +468 = 475.797 k$