Question

A heat pump supplies heat energy to a house at the rate of \(140,000 \mathrm{kJ} / \mathrm{h}\) when the house is maintained at \(25^{\circ} \mathrm{C} .\) Over a period of one month, the heat pump operates for 100 hours to transfer energy from a heat source outside the house to inside the house. Consider a heat pump receiving heat from two different outside energy sources. In one application the heat pump receives heat from the outside air at \(0^{\circ} \mathrm{C} .\) In a second application the heat pump receives heat from a lake having a water temperature of \(10^{\circ} \mathrm{C}\). If electricity costs \(\$ 0.105 / \mathrm{kWh}\), determine the maximum money saved by using the lake water rather than the outside air as the outside energy source.

Short answer

Answer: The maximum money saved is $44.39.

Step-by-step solution

Calculate the heat pump's coefficient of performance (COP)

To calculate the COP, we will use the equation: \(\mathrm{COP} = \frac{Q_H}{W_{in}}\), where \(Q_H\) is the heat transferred to the house per hour, and \(W_{in}\) is the required work input per hour. First, we need to convert the given heating rate from kJ/h to kW. So, \(Q_H = 140,000 \,\text{kJ/h} × \left(\frac{1 \,\text{kW}}{1,000 \,\text{kJ}}\right) = 140 \,\text{kW}\). Next, we need to find the temperatures in Kelvin for inside, outside air, and lake water. Inside temperature, \(T_H = 25 + 273 = 298 \,\text{K}\) Outside air temperature, \(T_C1 = 0 + 273 = 273 \,\text{K}\) Lake water temperature, \(T_C2 = 10 + 273 = 283 \,\text{K}\). Now, we can calculate the COP for both outside air and lake water. For outside air: \(\mathrm{COP_1} = \frac{Q_H}{W_{in1}} = \frac{T_H}{T_H - T_C1} = \frac{298}{298 - 273} = 12.42\) For lake water: \(\mathrm{COP_2} = \frac{Q_H}{W_{in2}} = \frac{T_H}{T_H - T_C2} = \frac{298}{298 - 283} = 19.87\)

Calculate the work input for both outside sources

Now, we need to find the work input required for both outside energy sources. Using the COP values, we can find \(W_{in1}\) and \(W_{in2}\) as follows: For outside air: \(W_{in1} = \frac{Q_H}{\mathrm{COP_1}} = \frac{140}{12.42} = 11.27 \,\text{kW}\) For lake water: \(W_{in2} = \frac{Q_H}{\mathrm{COP_2}} = \frac{140}{19.87} = 7.05 \,\text{kW}\)

Calculate the cost difference for using lake water instead of outside air

To find the maximum money saved, we need to calculate the cost difference by using the given electricity costs. Since the heat pump operates for 100 hours over one month, we can multiply the work input and operation hours to find the monthly cost. For outside air: \(Cost_1 = W_{in1} × 100\,\text{h} × \frac{\$ 0.105}{\,\text{kWh}} = 11.27 \,\text{kW} × 100\,\text{h} × \$ 0.105/\,\text{kWh} = \$ 118.31\) For lake water: \(Cost_2 = W_{in2} × 100\,\text{h} × \frac{\$ 0.105}{\,\text{kWh}} = 7.05 \,\text{kW} × 100\,\text{h} × \$ 0.105/\,\text{kWh} = \$ 73.92\) Now, we can calculate the maximum money saved by subtracting the cost of using lake water from the cost of using outside air. Money saved = \(Cost_1 - Cost_2 = \$ 118.31 - \$ 73.92 = \$ 44.39\) Therefore, the maximum money saved by using the lake water rather than the outside air as the outside energy source is \( \$ 44.39\).

Key concepts

Coefficient of Performance (COP)

The Coefficient of Performance (COP) is a key concept when discussing the efficiency of heat pumps. In simple terms, it represents the ratio of heat energy transferred to a space to the electrical energy input required to transfer that heat. It's calculated using the formula \( \mathrm{COP} = \frac{Q_H}{W_{in}} \), where \( Q_H \) is the heat output and \( W_{in} \) is the electrical work input.

The higher the COP, the more efficient the heat pump. Efficiency is greatly influenced by the temperature difference between the heat source (such as outside air or lake water) and the destination (the inside of a house). The closer these temperatures are, the less work the heat pump needs to do to transfer the same amount of heat, and thus, the higher the COP will be. This is why in our exercise, the heat pump has a higher COP when using lake water (\( T_C=10^\circ C \) ) compared to outside air (\( T_C=0^\circ C \) ). The increase in COP can lead to significant energy savings over time.

Heat Transfer

Heat transfer is a fundamental concept in thermodynamics that involves the movement of thermal energy from one place to another. In the case of our heat pump example, heat is being transferred from the outside environment (air or lake water) into a house to maintain a comfortable temperature.

There are three modes of heat transfer: conduction, convection, and radiation. However, heat pumps mainly utilize the principles of convection — where heat is moved by the flow of a fluid, which in most heat pumps is a refrigerant — and then released to warm the inside of the house. The efficiency of this process is influenced by many factors including the temperatures of the heat sources, the temperature inside the house, and the qualities of the heat pump system itself.

Energy Cost Savings

Energy cost savings are an essential consideration for many homeowners when they evaluate the efficiency of their heating systems. By choosing a more efficient heat source or improving the efficiency of a heat pump, substantial savings on electricity bills can be achieved over time.

The exercise we are considering demonstrates how the COP affects the energy costs by comparing the cost-effectiveness of using an alternative heat source (lake water) instead of a less efficient source (outside air). The cost of electricity used by the heat pump over 100 hours of operation is calculated, and the result shows a clear saving when using the more efficient lake water source. This underscores how important it is to select the right heat pump and source temperatures to minimize operational costs.