Example 37.1: Air CM Given: V1=0.4m3 Rigid tank Environment p1=10 bars @T。=300K,andp。=1bari T =400K Find:Exergy of the air ● Assumptions:》△KE=△PE=0》Air is ideal gas》Closed system Solution: Φar,1=ma(u1-uo)+Po(V-Vo)-mmrT.(s,-s。) 》Mass of air: PiV 10.105N/20.4m31J m mair =3.484kg RT 287 K400K INm From air tables:u=uair (T=400K)=286.16 kJ/kg u。=uair(T。=300K)=214.07kJ/kg 上游通大学 May2,2018 11 SHANGHAI JLAO TONG UNIVERSITYMay 2, 2018 11  Given:  Find: Exergy of the air  Assumptions: Air V1 = 0.4 m3 p1 = 10 bars T1 = 400 K CM Rigid tank Environment @ To = 300K, and po= 1 bar Example 37.1: » ΔKE = ΔPE = 0 » Air is ideal gas » Closed system  Solution: » Mass of air: » From air tables: F       air,1 o 1 o m u u air 1 o   po V T 1 o V s s  mair   2 5 3 1 1 air 1 N m J kg K 10 10 0.4m 1J p V m 3.484kg RT 287 400K 1Nm          1 air 1 o air o u u T 400K 286.16 kJ / kg u u T 300K 214.07 kJ / kg      