Example 37.2:Continued... Po=100 kPa T0=25C Exergy at Initial state Φ1=m[(41-4)-Ts1-so)+P(1-Vo] Steam P=1 MPa P2=200 kPa =(0.05kg){(2793.7-104.83)kJ/kg T=300C T2=150°C 2kJ -(298K)[(7.1246-0.3672)kJ/kgK State I State 2 +(100kPa)[(0.25799-0.00103)m3/kg]}(kJ/kPam3) =35.0kJ Exergy at Final state 2 ml(uz-uo)-To(s2 -so)Po(v2-vo)] =(0.05kg){(2577.1-104.83)kJ/g -(298K)[(7.2810-0.3672)kJ/kgK +(100kPa)[(0.95986-0.00103)m3/kg}kJ/kPam3) =25.4kJ (b)The exergy change 7Φ=中2-Φ1=25.4-35.0=-9.6kJ 上游文通大学 May2,2018 15 SHANGHAI JLAO TONG UNIVERSITYMay 2, 2018 15 Example 37.2: Continued… Exergy at Initial state Exergy at Final state (b) The exergy change 𝛻𝛷 = 𝛷2 - 𝛷1= 25.4 - 35.0 = -9.6 kJ