Respuesta :
Answer:
Work out = 28.27 kJ/kg
Explanation:
For R-134a, from the saturated tables at 800 kPa, we get
[tex]h_{fg}[/tex] = 171.82 kJ/kg
Therefore, at saturation pressure 140 kPa, saturation temperature is
[tex]T_{L}[/tex] = -18.77°C = 254.23 K
At saturation pressure 800 kPa, the saturation temperature is
[tex]T_{H}[/tex] = 31.31°C = 304.31 K
Now heat rejected will be same as enthalpy during vaporization since heat is rejected from saturated vapour state to saturated liquid state.
Thus, [tex]q_{reject}[/tex] = [tex]h_{fg}[/tex] = 171.82 kJ/kg
We know COP of heat pump
COP = [tex]\frac{T_{H}}{T_{H}-T_{L}}[/tex]
= [tex]\frac{304.31}{304.31-254.23}[/tex]
= 6.076
Therefore, Work out put, W = [tex]\frac{q_{reject}}{COP}[/tex]
= 171.82 / 6.076
= 28.27 kJ/kg
