Solution cont. 75°C Condenser 14 bar 3 Entropy rate balance 28℃ 0=mef(S2-S3）+mir(s5-6）+cond 3.5 bar mair (no-hs) (AV)5 =(AV5 Ps (h2-h3) Us RTs (energy balance,1st law) (ag) (1 bar) 10N/m2 1kJ 8.314 =0.5 kg/s kJ 1 bar 103N·m 293K) 28.97kg·K (05 (323-293)K Indoor return air Qcv=0 Supply air =0.07kg/s T5=20C T6=50C (294.17-79.05)kJ/kg Ps 1 bar P6=1 bar (AV)5=0.42m3/s VWM 3 P2 =14 bar P3 14 bar Condenser T2=75C T3=28C 上游充通大 April 26,2018 16 SHANGHAI JIAO TONG UNIVERSITYApril 26, 2018 16 Solution cont. Condenser Entropy rate balance (energy balance, 1st law) Qcv=0