Example 15.1 Solution P s bars 22C pul3 =constant w-fow W R(T2-F 1bar m 1-31.3 T2 from the process equation (n-1)/m =295K (1.3-1)/1.3 =428K D W= R(T2-T) 8.314kJ 428K-295K =-127.2kJ/kg 1-n 1-1.3 Table A-17 Energy balance △u=9-1w→9=w+(4-4)=-127.2+(306.53-210.49)=-31.16kkg 上游充通大 Mar/15,Wed,2017 9 SHANGHAI JIAO TONG UNIVERSITYMar/15, Wed, 2017 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