Solution;
Given;
m=5.0kg
QA=9585kJ
P1=0.4MPa
Vessel is rigid, meaning;
V1=V2
P2=2.0MPa
T2=700°c=973K
Heat added;
QA=m(h2−h1)
From steam tables,at 700°c and 2.0MPa;
h2=3923.81kJ/kg
Therefore;
9585=5×(3923.81−h1)
h1=519619.05−9585
h1=2006.81kJ/kg
From steam tables ,at 0.4MPa and h1=2006.81kJ/kg;
hf=604.37kJ/kg
hg=2737.68kJ/kg
vf=0.001084m3/kg
vg=0.4624m3/kg
ug=2552.74kJ/kg
uf=603.93kJ/kg
h1=hf1+xhg1 =hf+x(hg−hf)
2006.81=604.37+x(2737.68−604.37)
x=0.657
Take;
v1=vf+x(vg−vg)
v1=0.001084+0.657(0.4624−0.001084)
Hence,the specific volume is;
v1=0.3043m3/kg
To calculate internal energy;
u=uf+x(ug−uf)
u=603.93+0.657(2552.74−603.93)
Hence the internal energy is;
u=1884.29kJ/kg
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