Answer:
[tex]\dot V_{out} = 0.061\,\frac{m^{3}}{s}[/tex], [tex]\dot W = 108.875\,kW[/tex]
Explanation:
The process in the compressor is modelled after the First Law of Thermodynamics:
[tex]\dot W + \dot m \cdot (h_{1}-h_{2}) = 0[/tex]
The specific enthalpies of the refrigerant at inlet and outlet are, respectively:
Inlet (Saturated vapor)
[tex]\nu = 0.10744\,\frac{m^{3}}{kg}[/tex]
[tex]h = 243.34\,\frac{kJ}{kg}[/tex]
Outlet (Superheated Vapor)
[tex]\nu = 0.036373\,\frac{m^{3}}{kg}[/tex]
[tex]h = 308.34\,\frac{kJ}{kg}[/tex]
The mass flow is:
[tex]\dot m = \frac{\left(0.18\,\frac{m^{3}}{s} \right)}{0.10744\,\frac{m^{3}}{kg} } \co[/tex]
[tex]\dot m = 1.675\,\frac{kg}{s}[/tex]
The volumetric flow rate at the exit is:
[tex]\dot V_{out} = \left(1.675\,\frac{kg}{s} \right)\cdot \left(0.036373\,\frac{m^{3}}{kg} \right)[/tex]
[tex]\dot V_{out} = 0.061\,\frac{m^{3}}{s}[/tex]
The power needed to make the compressor work is:
[tex]\dot W = \dot m \cdot (h_{2}-h_{1})[/tex]
[tex]\dot W = \left(1.675\,\frac{kg}{s} \right)\cdot \left(308.34\,\frac{kJ}{kg}-243.34\,\frac{kJ}{kg} \right)[/tex]
[tex]\dot W = 108.875\,kW[/tex]
