Solution Wev=? P1=1 bar T1=290K Air 2 P2=7bar Mass flow rate VI=6 m/s compressor T,=450K A1=0.1m2 V2 =2 m/s AVI m Ideal gas EoS Qcy=-180 kJ/min 1(0.1m2)6m/s)(105N/m (R/M)T 2897gR)290 8314Nm 2=0.72kgs Given temperature,Table A-17 (ideal gas air )>h,h2 T1=290K,h1=290.16k/kg.T2=450K,h2=451.8k/kg. -e.+a--(] -(-1)l+02e016-451 kJ +(,ergdisoal -119.4kW 上游充通大学 negative March 28,2019 6 SHANGHAI JIAO TONG UNIVERSITYMarch 28, 2019 6 Solution Mass flow rate Ideal gas EoS Given temperature, Table A-17 (ideal gas air ) h1 , h2 T1 = 290 K, h1 = 290.16 kJ/kg. T2 = 450 K, h2 = 451.8 kJ/kg. negative