A wave front has the form of a
1 answer:
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Answer:
Work out = 28.27 kJ/kg
Explanation:
For R-134a, from the saturated tables at 800 kPa, we get
= 171.82 kJ/kg
Therefore, at saturation pressure 140 kPa, saturation temperature is
= -18.77°C = 254.23 K
At saturation pressure 800 kPa, the saturation temperature is
= 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, =
= 171.82 kJ/kg
We know COP of heat pump
COP =
=
= 6.076
Therefore, Work out put, W =
= 171.82 / 6.076
= 28.27 kJ/kg