Question

Air enters an adiabatic nozzle steadily at 300 kPa, 200°C and 20 m/s and leaves at 100 kPa and 200 m/s. The inlet area of the nozzle is 100 $c{m}^2$. The exit area of the nozzle is closest to (Assume $C_p$=1 kJ/kgK of air)

Answer 1

$m={\rho }_1{A}_1{V}_1=\frac{P_1}{R{T}_1}\times A_1\times V_1$

$=\frac{300}{0.287\times (273+200)}\times 100\times {10}^{-4}\times 20$

$m=0.442kg/s$

By applying steady flow energy equation between state 1 and state 2

$h_1+\frac{V_1^2}{2000}=h_2+\frac{V_2^2}{2000}$

$1\times 200+\frac{{20}^2}{2000}=1\times T_2+\frac{{200}^2}{2000}$

$T_2=180.2{}^{\circ }C$

Now,

$0.442=\frac{P_2}{R{T}_2}\times A_2\times V_2$

$=\frac{100}{0.287\times (273+180.2)}\times A_2\times 200$

$A_2=28.74c{m}^2$