Question

A heat pump is to be used for heating a house in winter. The house is to be maintained at \(78^{\circ} \mathrm{F}\) at all times. When the temperature outdoors drops to \(25^{\circ} \mathrm{F}\), the heat losses from the house are estimated to be \(70,000 \mathrm{Btu} / \mathrm{h}\). Determine the minimum power required to run this heat pump if heat is extracted from \((a)\) the outdoor air at \(25^{\circ} \mathrm{F}\) and \((b)\) the well water at \(50^{\circ} \mathrm{F}\)

Short answer

Answer: (a) The minimum power required for extracting heat from outdoor air at 25°F is 4810.65 Btu/h. (b) The minimum power required for extracting heat from well water at 50°F is 2122.61 Btu/h.

Step-by-step solution

Scenario (a) - Extracting heat from outdoor air at 25°F#

First, let us convert the temperatures to Rankine. $$ T_{outdoor} = 25 + 459.67 = 484.67\,^{\circ} \mathrm{R} $$ To maintain the house at a temperature of \(78^{\circ} \mathrm{F}\), the heat pump must deliver the same \(70,000 \mathrm{Btu} / \mathrm{h}\) as the heat losses. For a maximum COP, we will assume that the heat pump works in a reversible Carnot process, and the COP for a reversible Carnot heat pump is given by: $$ COP_{HP}^{Carnot} = \frac{T_{house}}{T_{house} - T_{outdoor}} $$ We first need to convert T_house to Rankine: $$ T_{house} = 78 + 459.67 = 537.67\,^{\circ} \mathrm{R} $$ Now, we can calculate the Carnot COP: $$ COP_{HP}^{Carnot} = \frac{537.67}{537.67 - 484.67} = 14.55 $$ Finally, we can find the minimum power required to run the heat pump: $$ W_{cycle} = \frac{Q_{in}}{COP_{HP}^{Carnot}} = \frac{70,000}{14.55} = 4810.65 \,\mathrm{Btu} / \mathrm{h} $$

Scenario (b) - Extracting heat from well water at 50°F#

Let us convert the temperature to Rankine. $$ T_{well} = 50 + 459.67 = 509.67\,^{\circ} \mathrm{R} $$ The COP for a reversible Carnot heat pump is: $$ COP_{HP}^{Carnot} = \frac{T_{house}}{T_{house} - T_{well}} $$ $$ COP_{HP}^{Carnot} = \frac{537.67}{537.67 - 509.67} = 32.97 $$ Now, we can find the minimum power required to run the heat pump: $$ W_{cycle} = \frac{Q_{in}}{COP_{HP}^{Carnot}} = \frac{70,000}{32.97} = 2122.61 \,\mathrm{Btu} / \mathrm{h} $$ In conclusion, the minimum power required to run the heat pump while extracting heat from (a) the outdoor air at \(25^{\circ} \mathrm{F}\) is \(4810.65 \,\mathrm{Btu} / \mathrm{h}\) and (b) the well water at \(50^{\circ} \mathrm{F}\) is \(2122.61 \,\mathrm{Btu} / \mathrm{h}\).

Key concepts

Carnot Cycle

The Carnot cycle is a theoretical thermodynamic cycle proposed by Nicolas Léonard Sadi Carnot in 1824. It represents the most efficient possible cycle for converting thermal energy into work or, inversely, using work to produce refrigeration. The cycle is comprised of two isothermal processes where heat is transferred while the working fluid is at a constant temperature, and two adiabatic processes where the system does not exchange heat with its surroundings and the temperature of the working fluid changes.

When applied to heat pumps, the Carnot cycle provides a model for understanding the upper limit of efficiency. A heat pump operating in a reversible Carnot cycle absorbs heat from a low-temperature reservoir ({T_{low}}) and discharges it into a high-temperature reservoir ({T_{high}}). We use this principle to establish the Coefficient of Performance (COP) for an ideal heat pump.

Coefficient of Performance (COP)

The Coefficient of Performance (COP) is a measure of a heat pump's efficiency. It is defined as the ratio of the heat delivered to the desired reservoir (for example, a house interior during winter) to the work input required to transfer that heat.

In the context of heat pumps operating on the Carnot cycle, the COP can be expressed mathematically as
\[COP_{HP}^{Carnot} = \frac{T_{house}}{T_{house} - T_{source}}\]

Where {T_{house}} is the inside temperature of the house and {T_{source}} is the temperature of the outdoor air or well water from which heat is extracted. As seen in the exercise, the efficiency hugely depends on the difference in temperature between the heat source and the destination.

Thermodynamic Temperature Conversion

Thermodynamic temperature conversion is necessary for calculating COP and efficiency because the operations of heat pumps are based on absolute temperature scales. In our exercise, we worked with Rankine (°R), which is the imperial counterpart to Kelvin (K).

To convert Fahrenheit to Rankine, we use the formula:
\[T_{Rankine} = T_{Fahrenheit} + 459.67\].

Correct temperature conversion is crucial for accurate COP calculation in the Carnot cycle. Misinterpreting temperatures can lead to significant errors in the assessment of the heat pump's performance.

Heating Load Calculation

Heating load calculation is a critical aspect of determining the power required by a heat pump to maintain comfortable living conditions in a building. The heating load is the amount of heat energy that must be added to a space to maintain a set temperature. This value is influenced by factors like outdoor temperature, building insulation, and the properties of the construction materials.

For the given exercise, the heat losses from the house were estimated at 70,000 BTU/h. This heating load must be matched or exceeded by the heat pump's output to maintain the set indoor temperature. Ignoring detailed load calculations can result in an inefficiently sized heat pump, leading to increased energy costs and wear on the pump.