Question #263460

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.


1
Expert's answer
2021-11-11T12:03:16-0500

Solution;

For an adiabatic nozzle;

W=0

Q=0

Given;

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

P1=1378kPaP_1=1378kPa

v=0.147m3/kgv=0.147m^3/kg

u1=2510kJ/kgu_1=2510kJ/kg

P2=137.8kPaP_2=137.8kPa

v2=1.099m3/kgv_2=1.099m^3/kg

u2=2263kJ/kgu_2=2263kJ/kg

V1=3.05m/sV_1=3.05m/s

Energy balance equation;

h1+v122000=h2+v222000h_1+\frac{v_1^2}{2000}=h_2+\frac{v_2^2}{2000}

But h=u+PV

u1+P1v1+V122000=u2+P1v1+V222000u_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+3.0522000=2263+137.8×1.099+V2220002510+1378×0.147+\frac{3.05^2}{2000}=2263+137.8×1.099+\frac{V_2^2}{2000}

V22=621409.3025V_2^2=621409.3025

V2=788.3m/sV_2=788.3m/s






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