Example 15.1 Solution P s bars 22C pul3 =constant w-fow →w= W R(T2-F) 1 bar m -313 T2 from the process equation (n-1)/m (1.3-1)/1.3 =295K =428K W= R(T2-T) 8.314kJ 428K-295K =-127.2kJ/kg 1-n 1-1.3 Table A-17 Energy balance △2u=q-1w→9=w+(42-4)=-127.2+(306.53-210.49)=-31.16kg 上游充通大 March 25,2019 9 SHANGHAI JIAO TONG UNIVERSITYMarch 25, 2019 9 Example 15.1 Solution 22˚C 1.3 ? T2 from the process equation Energy balance    u q w w  w  2 1 q w u u     ( ) Table A-17