Question

Two streams of air, one at 1 bar, ${27}^{o}C$ and velocity of 30 m/ s and the other at 5 bar, 227°C, and velocity of 50 m/s, are mixed in equal proportion in a chamber from which heat at the rate of 100 kJ/ kg is removed. The mixture is then passed through an adiabatic nozzle. If the temperature of the air leaving the nozzle is 27°C and its $C_p=1.005$ kJ/ kg-K then, the velocity of steam issuing out of the nozzle is _______ m/s.

• A. 50.05
• B. 58.76
• C. 51.96
• D. 41.96

Answer 1

The correct option is C 51.96

As per steady flow energy equation,
$\Rightarrow$ $Q+{\dot{m}}_1[h_1+\frac{V_1^2}{2}]+{\dot{m}}_2[h_2+\frac{V_2^2}{2}]={\dot{m}}_3[h_3+\frac{V_3^2}{2}]+\dot{W}$

$\Rightarrow$ $h_1=c_p{T}_1=1.005\times 300=301.5kJ/kg$

$\Rightarrow$ $h_2=c_p{T}_2=1.005\times 500=502.5kJ/kg$


$V_1=30m/s$

$V_2=50m/s$

$\dot{W}=0$

$Now,{\dot{m}}_1={\dot{m}}_2\left(given\right)$

$Let{\dot{m}}_1={\dot{m}}_2=\dot{m}$

$Now{\dot{m}}_3=[{\dot{m}}_1+{\dot{m}}_2]=2\dot{m}$

$\dot{Q}=-[{\dot{m}}_1+{\dot{m}}_2]\times 100=-2\dot{m}100kJ/s$

Putting values in the equation (i), we get

$-2\dot{m}100+\dot{m}[301.5+\frac{{30}^2}{2}\times {10}^{-3}]+\dot{m}[502.5+\frac{{50}^2}{2}\times {10}^{-3}]$

$=2\dot{m}[1.005\times 300+\frac{V_3^2}{2}\times {10}^{-3}]$
$605.7=2[301.5+\frac{V_3^2}{2}\times {10}^{-3}]$

$V_3=51.96m/s$