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

Question #267545

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

1
Expert's answer
2021-11-17T18:30:42-0500

Solution;

Give;

h1=2850kJ/kgh_1=2850kJ/kg

v1=50m/sv_1=50m/s

T1=200°cT_1=200°c

P1=1.5barP_1=1.5bar

m˙=1500kg/h\dot{m}=1500kg/h

v2=650m/sv_2=650m/s

Q=1200kJ/hrQ=1200kJ/hr

Find h2h_2 ;

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

m˙[h1+v122+z1g]+Q=m˙[h2+v222+z2g]+W\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;

m˙[h1+v122]+Q=m˙[h2+v222]\dot{m}[h_1+\frac{v_1^2}{2}]+Q=\dot{m}[h_2+\frac{v_2^2}{2}]

m˙h2=m˙[h1+v12v222]+Q\dot{m}h_2=\dot{m}[h_1+\frac{v_1^2-v_2^2}{2}]+Q

By direct substitution;

1500h2=1500[2850+50265022000]+120001500h_2=1500[2850+\frac{50^2-650^2}{2000}]+12000

1500h2=39720001500h_2=3972000

h2=2648kJ/kgh_2=2648kJ/kg




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