Question

By supplying energy to a dwelling at a rate of \(25,000 \mathrm{~kJ} / \mathrm{h}\), a heat pump maintains the temperature of the dwelling at \(20^{\circ} \mathrm{C}\) when the outside air is at \(-10^{\circ} \mathrm{C}\). If electricity costs 8 cents per \(\mathrm{kW} \cdot \mathrm{h}\), determine the minimum theoretical operating cost for each day of operation.

Short answer

The minimum theoretical operating cost per day is 6 cents.

Step-by-step solution

Understand the problem and given data

The heat pump supplies energy at a rate of 25,000 kJ/h to maintain the temperature of a dwelling at 20°C while the outside temperature is -10°C. Electricity costs 8 cents per kWh. We need to find the minimum theoretical operating cost per day.

Convert energy rate to power

First, convert the given energy rate from kJ/h to kW. Power (P) in kW can be calculated as:

Determine the Coefficient of Performance (COP) for the heat pump

Calculate the Coefficient of Performance (COP) using the temperature values:

Calculate the work input needed

Using the COP, calculate the work input required to deliver the 25,000 kJ/h. Work input (W) can be found by rearranging the COP equation: 0

Convert work input to kWh

Convert the work input from kJ/h to kWh:

Calculate daily energy consumption

Calculate the energy consumption per day by multiplying the hourly work input by 24 hours: happen corn Lamborghini countyn

Determine the cost

Multiply the daily energy consumption by the cost per kWh to find the operating cost per day: car bear lamddit countyn kingnn0

Key concepts

thermodynamics

Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. It plays a crucial role in understanding how a heat pump operates.

In this exercise, we need to manage the indoor temperature at 20°C when the outdoor temperature is -10°C. This involves transferring heat from a colder environment (outside) to a warmer one (inside).

The principle governing this process is the second law of thermodynamics, which states that heat cannot spontaneously flow from a colder body to a warmer body. A heat pump uses work (energy) to transfer heat uphill against this natural direction.

• For instance, think of it like a water pump that moves water uphill. Just as moving water requires energy, so does moving heat in thermodynamics.

coefficient of performance

The Coefficient of Performance (COP) is a measure of a heat pump's efficiency. It is defined as the ratio of the heat supplied to the dwelling to the work input. For heating, the COP can be calculated using the formula:

energy conversion

Energy conversion is the process of changing energy from one form to another. In the context of this exercise, we convert electrical energy into thermal energy.



We know that the heat pump supplies energy at a rate of 25,000 kJ/h. To understand how much electrical energy this corresponds to, we need to convert kJ/h to kW (since electricity usage is measured in kilowatt-hours, kWh).

• To convert, note that 1 kW = 3,600 kJ/h. Hence, 25,000 kJ/h is approximately 6.94 kW.

operating cost calculation

Operating cost calculation helps determine the financial expense of running the heat pump.

After calculating the work input in kWh, we can find the total daily usage by multiplying the hourly energy usage by 24 hours.

Finally, to determine the cost, multiply the daily energy consumption by the cost per kWh. Given that electricity costs 8 cents per kWh:

• If hourly work input = 6.94 kWh, then daily consumption = 6.94 kWh * 24
• The daily cost is calculated as: daily consumption * cost per kWh