3% of moisture, x=1−0.03=0.97h=mH=2.56600=2640kJ/kgAt x=0.97,h=2640kJ/kg;P=0.2MPa
From the steam table
P=0.2MPaTsat=120.210Chf=504.7kJ/kghfg=2201.5kJ/kgh=hf+xhfg=504.7+0.97∗2201.5=2640.155kJ/kg
Degree of super heat =T2−Tsat
800C=T2−120.210C⟹T2=200.210C
At P=0.2MPaT=200.120Ch=2871.2kJ/kgH2=m∗h2=2.5∗2871.2=7178kJ
Heat received is given as H2-H1
7178−6600=578kJ
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