Answer
$\dot{m}=100.0\ kg/min$
Work Step by Step
From tables A-11 to A-13:
Inlet ($P_3=1\ MPa, T_3=90°C$): $h_3=324.66\ kJ/kg$
Outlet ($P_4=1\ MPa, T_4=30°C$): $h_4=93.58\ kJ/kg$
For the air:
$\dot{m}_1=\frac{P_1\dot{V}_1}{RT_1}$
Since $P_1=100\ kPa,\ R=0.287\ kJ/kg.K,\ T_1=300\ K,\ \dot{V}_1=600\ m³/min$
$\dot{m}_1=696.9\ kg/min$
Energy balance for the air ($c_{p_1}=1.005\ kJ/kg.K,\ T_2=60°C,\ T_1=27°C$):
$\dot{Q}=\dot{m}_1c_{p_1}(T_2-T_1)$
$\dot{Q}=23112.69\ kJ/min$
For the R-134a:
$\dot{Q}=-\dot{m}_3(h_4-h_3)$
So $\dot{m}=100.0\ kg/min$
