Po=100 kPa Example 37.2:Continued. T0=25C (c)The total exergy destroyed Steam P =I MPa P2=200 kPa 中destrayed=-￥2-化se T1=300C T2=150°C 2 kJ State I State 2 Energy balance equation △U=Q-Wb,out→Wb,out=Q-△U=Q-m(u2-u1) =(-2k)-(0.05kg)(2577.1-2793.7)k/kg =8.8k (Total boundary work) Useful work Wu =W-Wsurr =Wb.out -Po(V2-V1)=Wb.out -Pom(v2-v1) =8.8J-100kPax0.05kg[0.9599-0.2579)mKgl1kPam 1kJ =5.3kJ Φdestroyed=中1-中2-Wse=35.0-25.4-5.3=4.3kJ 上游充通大 May2,2018 16 SHANGHAI JLAO TONG UNIVERSITYMay 2, 2018 16 Example 37.2: Continued… (c) The total exergy destroyed 𝛷destroyed = 𝛷1 − 𝛷2 − 𝑊use √ √ ?? Energy balance equation Δ𝑈 = 𝑄 − 𝑊b,out 𝑊b,out = 𝑄 − Δ𝑈 = 𝑄 − 𝑚(𝑢2 − 𝑢1) = −2kJ − 0.05kg 2577.1 − 2793.7 kJ/kg = 8.8 kJ (Total boundary work) Useful work: 𝛷destroyed= 𝛷1 − 𝛷2 − 𝑊use = 35.0 − 25.4 − 5.3 = 4.3 kJ