Question

A Carnot heat pump is to be used to heat a house and maintain it at \(25^{\circ} \mathrm{C}\) in winter. On a day when the average outdoor temperature remains at about \(2^{\circ} \mathrm{C}\), the house is estimated to lose heat at a rate of \(55,000 \mathrm{kJ} / \mathrm{h}\). If the heat pump consumes \(4.8 \mathrm{kW}\) of power while operating, determine \((a)\) how long the heat pump ran on that day; ( \(b\) ) the total heating costs, assuming an average price of \(11 \mathrm{e} / \mathrm{kWh}\) for electricity; and \((c)\) the heating cost for the same day if resistance heating is used instead of a heat pump.

Short answer

(a) The heat pump ran for approximately 0.509 hours on that day. (b) The total heating costs using the heat pump are approximately 26.84€. (c) The heating cost for the same day using resistance heating would be approximately 168.08€, which is significantly higher than using a heat pump.

Step-by-step solution

Calculate the COP of the Carnot heat pump

The Carnot COP for a heat pump is given by the formula: COP_carnot = \(\frac{T_h}{T_h-T_c}\) where \(T_h\) is the indoor temperature (in Kelvin) and \(T_c\) is the outdoor temperature (in Kelvin). First, let's convert the temperatures to Kelvin: \(T_h = 25^{\circ} \mathrm{C} + 273.15 \mathrm{K} = 298.15 \mathrm{K}\) \(T_c = 2^{\circ} \mathrm{C} + 273.15 \mathrm{K} = 275.15 \mathrm{K}\) Now, substitute these temperatures into the formula and find COP_carnot: COP_carnot \(\approx \frac{298.15}{298.15 - 275.15} \approx 6.25\)

Calculate the heat delivered by the heat pump per second

The actual heat delivered by the heat pump (Q_hp) can be calculated as: \(Q_{hp} = COP \times W\) where \(W\) is the power consumed by the heat pump. Since the heat pump consumes 4.8 kW of power, we have: \(Q_{hp} = 6.25 \times 4.8 \mathrm{kW} = 30 \mathrm{kW}\)

Determine the heat pump's operation time

The heat pump operation time can be calculated by dividing the total heat loss of the house by the heat delivered per second (Q_hp): operation_time = \(\frac{55,000 \mathrm{kJ}}{30 \mathrm{kW} \times 3600 \mathrm{s}}\) operation_time = \(\frac{55,000}{108,000} \mathrm{h} \approx 0.509 \mathrm{h}\)

Calculate the electricity consumed by the heat pump

The electricity consumed by the heat pump can be calculated as: consumed_electricity = operation_time \(\times\) power_consumed consumed_electricity = \(0.509 \mathrm{h} \times 4.8 \mathrm{kW}\) consumed_electricity = \(2.44 \mathrm{kWh}\)

Calculate the total heating costs

The total heating costs can be calculated by multiplying the electricity consumed (in kWh) by the cost of electricity (11€/kWh): heating_cost = consumed_electricity \(\times\) cost_of_electricity heating_cost = \(2.44 \mathrm{kWh} \times 11 \mathrm{e/kWh}\) heating_cost = \(26.84 \mathrm{e}\)

Calculate the heating cost if resistance heating were used

For resistance heating, the COP equals 1. Therefore, the electricity consumed by the resistance heater is equal to the total heat loss of the house, which is 55,000 kJ or 15.28 kWh (because 1 kWh = 3600 kJ). Now, let's calculate the heating cost for resistance heating: resistance_heating_cost = \(15.28 \mathrm{kWh} \times 11 \mathrm{e/kWh} = 168.08 \mathrm{e}\) The answers to the exercise are: (a) The heat pump ran for approximately 0.509 hours on that day. (b) The total heating costs using the heat pump are approximately 26.84€. (c) The heating cost for the same day using resistance heating would be approximately 168.08€, which is significantly higher than using a heat pump.

Key concepts

COP (Coefficient of Performance)

The Coefficient of Performance (COP) is a critical measure of the effectiveness of heat pumps and refrigeration devices. It is defined as the ratio of heat output to the energy input, for a heat pump, or the ratio of heat removed to the energy input, for a refrigeration system. In simpler terms, it tells us how many units of heat are moved by the device per unit of work or energy consumed.

For a Carnot heat pump, which is an idealized heat pump operating on the Carnot cycle, the COP can be calculated using the formula: \[COP_{carnot} = \frac{T_h}{T_h - T_c}\]where \(T_h\) is the temperature at which heat is delivered (the indoor temperature, in our case) measured in Kelvin, and \(T_c\) is the temperature from which heat is extracted (the outdoor temperature) also in Kelvin.

The COP provides an upper limit for the efficiency of real-world heat pumps. A higher COP indicates a more efficient device, meaning less electricity is needed for the same amount of heating. In our example, the Carnot heat pump has a COP of 6.25, which means it's quite efficient compared to traditional heating methods like electric resistance heaters, which have a COP of 1.

Energy Consumption Calculation

Understanding energy consumption is vital when assessing the performance and sustainability of heating systems. To calculate the energy consumed by a heat pump, we apply the formula:
\[Q_{hp} = COP \times W \]where \(Q_{hp}\) is the heat delivered by the heat pump, and \(W\) is the electrical power consumed by the heat pump in kilowatts (kW).

In the example, the heat pump operates at 4.8 kW of power and has a COP calculated as 6.25. Multiplying these numbers gives us the rate at which the heat pump delivers heat, which is 30 kW. Knowing the rate of heat delivery and the total heat loss of the house, we can determine the operation time and subsequently calculate the total energy consumption.

Practical Tips for Calculating Energy Consumption

For students aiming to master this concept, it's crucial to convert all given units accordingly and ensure consistency across calculations. By doing so, the derived energy consumption figures will accurately represent the real-world operating conditions of heating systems, enabling effective assessments and comparisons.

Heating Cost Comparison

Making a heating cost comparison is crucial for economic considerations, especially when one has the option to choose between different heating methods. The cost comparison is directly influenced by both the efficiency, as indicated by the COP, and the energy price.

To find the heating costs, we multiply the energy consumption by the unit cost of energy. For a heat pump, the costs are generally lower due to higher efficiency compared to resistance heating, where the COP is 1, making it less efficient.

Using the example provided, we calculated the heat pump's total heating costs to be around 26.84€ for a day's usage. Comparatively, the costs using resistance heating for the same period would amount to 168.08€ – a significant difference.

Saving Potential

Using the COP and energy price, individuals can decide on the most cost-effective heating method. The example demonstrates the potential savings heat pumps offer over traditional resistance heating methods. This comparison not only benefits the budget but also encourages the adoption of energy-efficient technologies for environmental sustainability.