Try another method: 15m 38C F2850 -055 MPa- 4名-0.7MP Energy balance for open system: 0.74 Process 2-3:inlet-exit increment δQp-h.δm。=dU--→δQp=h.δm。+a(mu) =hδm。+mdu+udm Conservation of mass: δm。=-dmAtequlibr state:h。=h ,-md(h)dm=mic,dT-prdm m Integration: T m % m T m2 = T RT, R 上游气通大粤 March 31,2018 8 SHANGHAI JLAO TONG UNIVERSITYMarch 31, 2018 8 Energy balance for open system: Process 23: inlet – exit = increment δ δ d Q h m U p e e   Conservation of mass: δ d m m e   At equilibrium state: he =h δ d d d d Q m u h u m mc T pv m p V        d d V pV pV c T m RT m   δ δ d Q h m mu p e e     δ d d e e    h m m u u m Try another method: 3 3 2 2 3 3 2 2 d d ln ln T m V p V T m pV T m c T m Q c pV pV R T m R T m             3 3 3 p V m RT  3 3 2 ln p p c p V T Q R T  3 2 2 3 m T m T  2 2 2 p V m RT  3 2 p p p   Integration: