Holooly Plus Logo
Holooly Plus Logo

Question 6.4: Heating a House with a Heat Pump A heat pump is used to meet...

Heating a House with a Heat Pump

A heat pump is used to meet the heating requirements of a house and maintain it at 20°C. On a day when the outdoor air temperature drops to –2°C, the house is estimated to lose heat at a rate of 80,000 kJ/h. If the heat pump under these conditions has a COP of 2.5, determine (a) the power consumed by the heat pump and (b) the rate at which heat is absorbed from the cold outdoor air.

Question Data is a breakdown of the data given in the question above.
  • Outdoor air temperature: -2°C
  • Heat loss rate of the house: 80,000 kJ/h
  • COP of the heat pump: 2.5
  • Desired indoor temperature: 20°C
The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.
Step 1:
The power consumed by the heat pump is determined using the definition of the coefficient of performance (COP). The COP is a measure of the efficiency of the heat pump and is defined as the ratio of the heat delivered to the heat absorbed. In this case, the COP of the heat pump is given as 2.5.
Step 2:
To find the power consumed, we divide the heat absorbed (given as 80,000 kJ/h) by the COP. This gives us a value of 32,000 kJ/h or 8.9 kW.
Step 3:
The house is losing heat at a rate of 80,000 kJ/h. In order to maintain a constant temperature of 20°C, the heat pump must deliver heat to the house at the same rate. Therefore, the rate of heat transfer from the outdoor becomes the difference between the heat absorbed (80,000 kJ/h) and the power consumed (32,000 kJ/h), which is equal to 48,000 kJ/h.
Step 4:
It is important to note that out of the 80,000 kJ/h heat delivered to the house, 48,000 kJ/h is actually extracted from the cold outdoor air. This means that we are only paying for the 32,000 kJ/h energy that is supplied as electrical work to the heat pump. If we were to use an electric resistance heater instead, we would have to supply the entire 80,000 kJ/h to the heater as electric energy. This would result in a heating bill that is 2.5 times higher. This is why heat pumps are preferred as heating systems despite their higher initial cost.

Final Answer

The COP of a heat pump is given. The power consumption and the rate of heat absorption are to be determined.

Assumptions     Steady operating conditions exist.

Analysis     (a) The power consumed by this heat pump, shown in Fig. 6–24, is determined from the definition of the coefficient of performance to be

\dot{W}_{\mathrm{net}, \mathrm{in}}=\frac{\dot{Q}_{H}}{\mathrm{COP}_{\mathrm{HP}}}=\frac{80,000  \mathrm{~kJ} / \mathrm{h}}{2.5}=32,000  \mathrm{k} \mathrm{J} / \mathrm{h}(\text { or } 8.9  \mathrm{~kW})

(b) The house is losing heat at a rate of 80,000 \mathrm{~kJ} / \mathrm{h}. If the house is to be maintained at a constant temperature of 20^{\circ} \mathrm{C}, the heat pump must deliver heat to the house at the same rate, that is, at a rate of 80,000 \mathrm{~kJ} / \mathrm{h}. Then the rate of heat transfer from the outdoor becomes

\dot{Q}_{L}=\dot{Q}_{H}-\dot{W}_{\text {net,in }}=(80,000-32,000)  \mathrm{kJ} / \mathrm{h}=48,000  \mathrm{k} \mathrm{J} / \mathrm{h}

Discussion     Note that 48,000 of the 80,000 \mathrm{~kJ} / \mathrm{h} heat delivered to the house is actually extracted from the cold outdoor air. Therefore, we are paying only for the 32,000 \mathrm{~kJ} / \mathrm{h} energy that is supplied as electrical work to the heat pump. If we were to use an electric resistance heater instead, we would have to supply the entire 80,000 \mathrm{~kJ} / \mathrm{h} to the resistance heater as electric energy. This would mean a heating bill that is 2.5 times higher. This explains the popularity of heat pumps as heating systems and why they are preferred to simple electric resistance heaters despite their considerably higher initial cost.

6.24

Related Answered Questions

Malfunction of a Refrigerator Light Switch The interior lighting of refrigerators is provided by incandescent lamps whose switches are actuated by the opening of the refrigerator door. Consider a refrigerator whose 40-W lightbulb remains on continuously as a result of a malfunction of the switch

Verified Answer:
The lightbulb of a refrigerator malfunctions and r...

Heating a House with a Carnot Heat Pump A heat pump is to be used to heat a house during the winter, as shown in Fig. 6–52. The house is to be maintained at 21°C at all times. The house is estimated to be losing heat at a rate of 135,000 kJ/h when the outside temperature drops to –5°C. Determine

Verified Answer:
A heat pump maintains a house at a constant temper...

A Carnot Refrigeration Cycle Operating in the Saturation Dome A Carnot refrigeration cycle is executed in a closed system in the saturated liquid– vapor mixture region using 0.8 kg of refrigerant-134a as the working fluid (Fig. 6–51). The maximum and the minimum temperatures in the cycle are 20 and

Verified Answer:
A Carnot refrigeration cycle is executed in a clos...

Analysis of a Carnot Heat Engine A Carnot heat engine, shown in Fig. 6–47, receives 500 kJ of heat per cycle from a high-temperature source at 652°C and rejects heat to a low-temperature sink at 30°C. Determine (a) the thermal efficiency of this Carnot engine and (b) the amount of heat rejected to

Verified Answer:
The heat supplied to a Carnot heat engine is given...

Analysis of a Household Refrigerator A household refrigerator with a COP of 1.2 removes heat from the refrigerated space at a rate of 60 kJ/min (Fig. 6-23). Determine (a) the electric power consumed by the refrigerator and (b) the rate of heat transfer to the kitchen air.

Verified Answer:
The COP and the refrigeration rate of a refrigerat...

Fuel Consumption Rate of a Car A car engine with a power output of 65 hp has a thermal efficiency of 24 percent. Determine the fuel consumption rate of this car if the fuel has a heating value of 19,000 Btu/lbm (that is, 19,000 Btu of energy is released for each lbm of fuel burned).

Verified Answer:
The power output and the efficiency of a car engin...

Net Power Production of a Heat Engine Heat is transferred to a heat engine from a furnace at a rate of 80 MW. If the rate of waste heat rejection to a nearby river is 50 MW, determine the net power output and the thermal efficiency for this heat engine.

Verified Answer:
The rates of heat transfer to and from a heat engi...