Answer to Question #221987 in Mechanical Engineering for Omelatum

Question #221987
With 3% moisture, 2.5 kg of steam has an enthalpy of 6600 kJ. It is heated at constant pressure to a final condition of 90
1
Expert's answer
2021-08-05T03:19:26-0400

3%3 \% of moisture, x=10.03=0.97h=Hm=66002.5=2640kJ/kgAt x=0.97,h=2640kJ/kg;P=0.2MPax= 1-0.03=0.97 \\ h= \frac{H}{m}=\frac{6600}{2.5}=2640 kJ/kg\\ At \space x = 0.97, h= 2640 kJ/kg; P=0.2MPa\\

From the steam table

P=0.2MPaTsat=120.210Chf=504.7kJ/kghfg=2201.5kJ/kgh=hf+xhfg=504.7+0.972201.5=2640.155kJ/kgP= 0.2 MPa\\ T_{sat}=120.21^0C\\ h_f= 504.7 kJ/kg\\ h_{fg}=2201.5 kJ/kg\\ h= h_f+ xh_{fg}= 504.7+0.97*2201.5 =2640.155 kJ/kg\\

Degree of super heat =T2Tsat=T_2-T_{sat}

800C=T2120.210C    T2=200.210C80^0C=T_2 -120.21 ^0C \implies T_2 = 200.21^0C

At P=0.2MPaT=200.120Ch=2871.2kJ/kgH2=mh2=2.52871.2=7178kJAt \space P= 0.2MPa\\ T= 200.12 ^0C\\ h= 2871.2 kJ/kg\\ H_2= m*h_2= 2.5*2871.2 = 7178 kJ\\

Heat received is given as H2-H1

71786600=578kJ7178-6600=578 kJ


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