Question

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

Answer 1

Work out = 28.27 kJ/kg

Step-by-step 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 =
`(T_(H))/(T_(H)-T_(L))`

=
`(304.31)/(304.31-254.23)`

= 6.076

Therefore, Work out put, W =
`(q_(reject))/(COP)`

= 171.82 / 6.076

= 28.27 kJ/kg