Question

A completely reversible heat pump produces heat at a rate of 300 kW to warm a house maintained at 24 °C. The exterior air, which is at 7 °C, serves as the source. Calculate the rate of entropy change of the two reservoirs and determine if this heat pump satisfies the second law according to the increase of entropy principle.

Answer 1

Answer:

No.

Explanation:

The Coefficient of Performance of the reversible heat pump is determined by the Carnot's cycle:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

$COP_{HP} = \frac{297.15\,K}{297.15\,K-280.15\,K}$

$COP_{HP} = 3.339$

The power required to make the heat pump working is:

$\dot W = \frac{300\,kW}{3.339}$

$\dot W = 89.847\,kW$

The heat absorbed from the exterior air is:

$\dot Q_{L} = 300\,kW - 89.847\,kW$

$\dot Q_{L} = 210.153\,kW$

According to the Second Law of Thermodynamics, the entropy generation rate in a reversible cycle must be zero. The formula for the heat pump is:

$\frac{\dot Q_{L}}{T_{L}} - \frac{\dot Q_{H}}{T_{H}} + \dot S_{gen} = 0$

$\dot S_{gen} = \frac{\dot Q_{H}}{T_{H}} - \frac{\dot Q_{L}}{T_{L}}$

$\dot S_{gen} = \frac{300\,kW}{297.15\,K}-\frac{210.153\,kW}{280.15\,K}$

$\dot S_{gen} = 0.259\,\frac{kW}{K}$

Which contradicts the reversibility criterion according to the Second Law of Thermodynamics.