Solution;
Given;
m˙=1kg/s
T1=20oc=293
W=15kW
V1=100m/s
V2=150m/s
cp=1.00kJ/kgK
The process is adiabatic;
Q=0
Apply steady flow energy equation;
m˙(h1+2000V12)+W=m˙(h2+2000V22)
where;
h=cpT
By direct substitution;
1[(1.005×293)+20001002]+15=1[(1.005×T2)+20001502)
314.465=1.005T2+11.25
T2=301.7064K or 28.71oc
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