Answer to Question #285792 in Mechanical Engineering for Tan

Question #285792

A rigid vessel contains 6 kg of wet steam at 0.4 MPa with a quality of 75% until the steam


has a temperature of 300C. Determine the initial internal energy and the final pressure.

1
Expert's answer
2022-02-08T02:10:01-0500

at wet steam at 0.4 MPa with a quality of 75% 

 Final Temperature T2= 4600C from steam table

Internal Energy

δE=m×Cpdt=6×1×(733573)=960kJkg\delta E= m\times C_p dt =6\times 1\times (733-573) =960\frac{kJ}{kg}

P2=P1×0.75=0.4×0.75=0.3MpaP2 =P1\times 0.75 =0.4\times 0.75 = 0.3 Mpa


as we konw,


P1T1=P2T2P_1T_1=P_2T_2


p2=0.4×573733=0.3Mpap_2= \frac{0.4\times573}{733}=0.3Mpa



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