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Bad White [126]
3 years ago
5

Consider a heat pump that operates on the reversed Carnot cycle with R-134a as the working fluid executed under the saturation d

ome between the pressure limits of 140 and 800 kPa. R-134a changes from saturated vapor to saturated liquid during the heat-rejection process. The net work input for this cycle is ____
Physics
1 answer:
Schach [20]3 years ago
8 0

Answer:

Work out = 28.27 kJ/kg

Explanation:

For R-134a, from the saturated tables at 800 kPa, we get

h_{fg} = 171.82 kJ/kg

Therefore, at saturation pressure 140 kPa, saturation temperature is

T_{L} = -18.77°C = 254.23 K

At saturation pressure  800 kPa, the saturation temperature is

T_{H} = 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, q_{reject} = h_{fg} = 171.82 kJ/kg

We know COP of heat pump

COP = \frac{T_{H}}{T_{H}-T_{L}}

        = \frac{304.31}{304.31-254.23}

         = 6.076

Therefore, Work out put, W = \frac{q_{reject}}{COP}

                                              = 171.82 / 6.076

                                              = 28.27 kJ/kg

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<u>We are Given:</u>

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