MMAE 320Thermodynamics6/11/1811st Midterm ExamProblem 1:Solution:#1. GivenTmax = 140oFP1 = 50 psiaP2 = 14.7 psiaMass flow rate overall = 1 lb/sRequired of the problem:Mass flow rate of the steamWhen tap is opened water flows from 50 psia to 14.7 psia. This process can be assumed to beisentropic process ( It is throttling process as there is no enthalpy loss with decrease in pressure).Knowing that:In State 1 =>1 = 50 , 1 140In State 2 =>2 = 14.7 50 PsiaT14.7 Psia12SMMAE 320Thermodynamics6/11/182h12S1 (50 , 140 ) = 1 = 2 (14.7) = Hence, (. ) is enthalpy of saturated liquid at 14.7 .We observe based on the result that h2 is smaller than hsl ( 14.7 psia ). In this case, the State 2 i.e iswater at 14.7 will remain fully as liquid without water vapor. Therefore, the mass flow rateof water vapor is equal to .MMAE 320Thermodynamics6/11/18Problem 2:Solution:Constant Volume: (Process 1- 2)Given that:1 = 10 1 = 2 0.05 103 3 = 01 = 3002 = 600Consider that:1 1 = 1=1 1 10 100 0.5 103== . 10. ...
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