Answer :
Answer:
[tex]\dot m = 0.726\,\frac{kg}{s}[/tex]
Explanation:
The process made by the compressor at steady-state is modelled after the First Principle of Thermodynamics:
[tex]-\dot Q_{out} + \dot W_{in} + \dot m \cdot (h_{in} - h_{out}) = 0[/tex]
The mass flow rate is:
[tex]\dot m = \frac{\dot Q_{out}-\dot W_{in}}{h_{in}-h_{out}}[/tex]
For ideal gases, specific enthalpies depends on temperature only. Properties at inlet and outlet are, respectively:
[tex]h_{in} = 300.19\,\frac{kJ}{kg}[/tex]
[tex]h_{out} = 503.02\,\frac{kJ}{kg}[/tex]
The mass flow rate of air is:
[tex]\dot m = \frac{22.67\,kW-170\,kW}{300.19\,\frac{kJ}{kg}-503.02\,\frac{kJ}{kg} }[/tex]
[tex]\dot m = 0.726\,\frac{kg}{s}[/tex]