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

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

answer; i believe that the correct answer is (d) a very small amount of mass

a lot of it means it can be "burned" for energy and sold at one price. if it is scarce, the price will go up