Answer to Question #267545 in Molecular Physics | Thermodynamics for Little D

Question

Steam enters a nozzle at pressure of 6 bar and 2000c with enthalpy 2850 kJ/kg and leaves at a pressure of 1.5 bar. The initial velocity of steam is 50m/s and the exit velocity from nozzle is 650 m/s. The mass flow rate through the nozzle is 1500kg/h, the heat loss from the nozzle is 12000 kJ/hr. Determine the final enthalpy of steam

Answer 1

Solution;

Give;

$h_1=2850kJ/kg$

$v_1=50m/s$

$T_1=200°c$

$P_1=1.5bar$

$\dot{m}=1500kg/h$

$v_2=650m/s$

$Q=1200kJ/hr$

Find $h_2$ ;

Assume the Nozzle undergoes a steady flow ,we apply the steady flow energy equation;

$\dot{m}[h_1+\frac{v_1^2}{2}+z_1g]+Q=\dot{m}[h_2+\frac{v_2^2}{2}+z_2g]+W$

W=0

The equation reduces to;

$\dot{m}[h_1+\frac{v_1^2}{2}]+Q=\dot{m}[h_2+\frac{v_2^2}{2}]$

$\dot{m}h_2=\dot{m}[h_1+\frac{v_1^2-v_2^2}{2}]+Q$

By direct substitution;

$1500h_2=1500[2850+\frac{50^2-650^2}{2000}]+12000$

$1500h_2=3972000$

$h_2=2648kJ/kg$