Answer to Question #263460 in Molecular Physics | Thermodynamics for den

Question

Steam with a flow rate of 1360 kg/h enters an adiabatic nozzle (adiabatic, Q=0) at 1378

kPa, specific volume of 0.147 m3

/kg, and specific internal energy of 2510 kJ/kg. The exit

conditions are 137.8 kPa, 1.099 m3

/kg specific volume, and 2263 kJ/kg of specific internal

energy. Find the exit velocity.

Answer 1

Solution;

For an adiabatic nozzle;

W=0

Q=0

Given;

$\dot{m}=1360kg/h=0.3778kg/s$

$P_1=1378kPa$

$v=0.147m^3/kg$

$u_1=2510kJ/kg$

$P_2=137.8kPa$

$v_2=1.099m^3/kg$

$u_2=2263kJ/kg$

$V_1=3.05m/s$

Energy balance equation;

$h_1+\frac{v_1^2}{2000}=h_2+\frac{v_2^2}{2000}$

But h=u+PV

$u_1+P_1v_1+\frac{V_1^2}{2000}=u_2+P_1v_1+\frac{V_2^2}{2000}$

By direct substitution;

$2510+1378×0.147+\frac{3.05^2}{2000}=2263+137.8×1.099+\frac{V_2^2}{2000}$

$V_2^2=621409.3025$

$V_2=788.3m/s$