Answer:
(a) The power input to the compressor: [tex]\dot{W}=73.07 kJ/s = 73.07 kW[/tex]
(b) The volume flow rate of the refrigerant at the compressor inlet: [tex]\dot{v}=0.209 m^{3}/s[/tex]
Explanation:
(a)
We need to check the values of enthalpy (as we have an open system) for both states, being the inlet, state 1 and the outlet, state 2. We will know these values by checking the vapor charts of R134a, I used the ones found in Thermodynamics of Cengel, 7th edition.
Then, our values are:
[tex]h_{1}=235.92kJ/kg\\h_{2}=296.81kJkg[/tex]
Now we proceed to know the work with the following expression:
[tex]\dot{W}=\dot{m}(h_{2}-h_{1})[/tex]
Now we replace values and our result is:
[tex]\dot{W}=73.07 kJ/s = 73.07 kW[/tex]
(b)
To know the volume rate at the compressor inlet, we need to know the specific volume in that phase, as we have that is saturated and at -24°C, we can read our table:
[tex]\nu=0.1739m^{3}/kg[/tex]
With our specific volume and the mass rate, we can calculate the volume rate:
[tex]\dot{v}=\nu * \dot{m}\\\dot{v}=0.209 m^{3}/s[/tex]
