Question

. A Carnot heat pump is to be used to heat a house and maintain it at 22 °C in winter. When the outdoor temperature remains at 3 °C, the house is estimated to lose heat at a rate of 76,000 kJ/h. If the heat pump consumes 9 kW of power, how long does it need to run in a single day to keep the temperature constant inside the house?

Answer 1

Answer:

The Carnot heat pump must work 3.624 hours per day to keep the temperature constant inside the house.

Explanation:

The net heat daily loss of the house is:

$Q_{losses} = \left(76000\,\frac{kJ}{h}\right)\cdot (24\,h)$

$Q_{losses} = 1.824\times 10^{6}\,kJ$

In order to keep the house warm, given heat must be equal to heat losses:

$Q_{H} = Q_{losses}$

Besides, the Coefficient of Performance for a Carnot heat pump is:

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

Where,

$T_{L}$ - Temperature of the cold reservoir (Outdoors), measured in Kelvin.

$T_{H}$ - Temperature of the hot reservoir (House), measured in Kelvin.

Given that $T_{L} = 276.15\,K$ and $T_{H} = 295.15\,K$, the Coefficient of Performance is:

$COP_{HP} = \frac{295.15\,K}{295.15\,K-276.15\,K}$

$COP_{HP} = 15.534$

For a real heat machine, the Coefficient of Performance is determined by the following expression:

$COP_{HP} = \frac{Q_{H}}{W}$

Where:

$Q_{H}$ - Heat received by the house, measured in kilojoules.

$W$ - Work consumed by the Carnot heat pump, measured in kilojoules.

The daily work consumed is now cleared in the previous expression:

$W = \frac{Q_{H}}{COP_{HP}}$

$W = \frac{1.824\times 10^{6}\,kJ}{15.534}$

$W = 117419.853\,kJ$

The working time is calculated by dividing this result by input power. That is:

$\Delta t = \frac{W}{\dot W}$

$\Delta t = \left(\frac{117419.853\,kJ}{9\,kW} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)$

$\Delta t = 3.624\,h$

The Carnot heat pump must work 3.624 hours per day to keep the temperature constant inside the house.