Answer to Question #213463 in Mechanical Engineering for Ajimm

Question

1 m3 of air at 8 bar and 120 oC is cooled at constant pressure process until the temperature drops to 27 oC. Given R = 0.287 kJ/kgK and Cp = 1.005 kJ/kgK, calculate the heat rejected during the process and the volume of the air after cooling.

Answer 1

Part a

Mass of the air

$m= \frac{PV}{RT}=\frac{800*1}{0.287*(120+273)}=7.093 kg$

$Q_R = m_{air}*C_p*(T_1-T_2) = 7.093* 1.005 ((273+120)-(273+27))=662.95 kW$

Part b

$V_2= \frac{mRT}{P}=\frac{7.093*0.287*(27+273)}{800}=0.7634 m^3$