Question

A heat engine operates between two reservoirs at 800 and \(20^{\circ} \mathrm{C} .\) One-half of the work output of the heat engine is used to drive a Carnot heat pump that removes heat from the cold surroundings at \(2^{\circ} \mathrm{C}\) and transfers it to a house maintained at \(22^{\circ} \mathrm{C}\). If the house is losing heat at a rate of \(62,000 \mathrm{kJ} / \mathrm{h}\) determine the minimum rate of heat supply to the heat engine required to keep the house at \(22^{\circ} \mathrm{C}\).

Short answer

The minimum rate of heat supply required for the heat engine is approximately 170,566 kJ/h.

Step-by-step solution

Calculate the efficiency of the heat engine

The efficiency of a heat engine can be calculated using the following formula: \(\eta = 1 - \frac{T_L}{T_H}\), where \(T_L\) is the temperature of the low-temperature reservoir, and \(T_H\) is the temperature of the high-temperature reservoir. To use this formula, we need to convert the temperatures from Celsius to Kelvin: \(T_H = 800 + 273.15 = 1073.15\,\text{K}\) and \(T_L = 20 + 273.15 = 293.15\,\text{K}\). Now, we calculate the efficiency: \(\eta = 1 - \frac{293.15}{1073.15} \approx 0.727\)

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

For a Carnot heat pump, the coefficient of performance is given by: \(\text{COP} = \frac{1}{1 - \frac{T_C}{T_H}}\), where \(T_C\) is the temperature at which heat is extracted from the surroundings, and \(T_H\) is the temperature at which heat is delivered. To use this formula, we need to convert the temperatures from Celsius to Kelvin: \(T_C = 2 + 273.15 = 275.15\,\text{K}\) and \(T_H = 22 + 273.15 = 295.15\,\text{K}\). Now, we calculate the COP: \(\text{COP} = \frac{1}{1 - \frac{275.15}{295.15}} \approx 19.159\)

Calculate the work output of the heat engine and the heat output of the heat pump

We know that half of the heat engine's work output is used to power the heat pump. Let's denote \(W\) as the work output of the heat engine. Thus, the heat output of the heat pump, \(Q_{out}\), can be expressed as: \(Q_{out} = \frac{1}{2} W\). The house loses heat at a rate of 62,000 kJ/h, meaning the heat pump should deliver 62,000 kJ/h to maintain a constant temperature. Therefore, \(Q_{out}\) = 62,000 kJ/h. Now, we find the work output of the heat engine: \(W = 2 \times Q_{out} = 2 \times 62{,}000 \,\text{kJ/h} = 124{,}000 \,\text{kJ/h}\)

Calculate the heat input to the heat engine, \(Q_{in}\)

Using the efficiency equation, we can calculate the heat input to the heat engine, \(Q_{in}\), as follows: \(W = \eta \times Q_{in} \Rightarrow Q_{in} = \frac{W}{\eta}\) Substituting the values, we get: \(Q_{in} = \frac{124000}{0.727} \approx 170{,}566 \, \text{kJ/h}\)

Determine the minimum rate of heat supply to the heat engine

The minimum rate of heat supply required to keep the house at 22°C is equal to the heat input to the heat engine, \(Q_{in}\). So, the minimum rate of heat supply to the heat engine is: \(Q_{in} \approx 170{,}566 \, \text{kJ/h}\)

Key concepts

Carnot Heat Pump

Just like the Carnot engine is a model for the most efficient heat engine, the Carnot heat pump represents the idealized version of a heat pump operating between two thermal reservoirs. Such a pump absorbs heat from a cooler place and releases it to a warmer area, which aligns with our everyday use of heat pumps for heating our homes.

By definition, a Carnot heat pump is a reversible cycle, which means it can achieve the maximum theoretical efficiency. No actual heat pump can fully reach Carnot efficiency, but it provides a benchmark to measure real systems against. The thermodynamic temperatures in Kelvin are crucial in determining its performance, emphasizing the significance of temperature conversions between Kelvin and Celsius, especially when dealing with heat pump calculations.

The performance of a Carnot heat pump is evaluated using the Coefficient of Performance (COP). COP is a ratio that compares the heat delivered to the warm reservoir to the work input required to transfer that heat. The importance of COP lies in its ability to quantify the energy efficiency of heat pump systems. The higher the COP, the more efficient the heat pump is, as it can transfer more heat per unit of work.

Coefficient of Performance (COP)

Understanding the Coefficient of Performance, or COP, is critical when evaluating the performance of heating and cooling systems, including the Carnot heat pump described earlier. The COP is simply a measure of efficiency used for heat pumps and refrigeration systems.

The COP is described by the ratio of the useful heating or cooling provided to the work required to produce that heating or cooling. For a heat pump, the COP represents the ratio of the heat output (what is delivered to warm up your house, for example) to the work input (the electrical energy used to run the pump). As temperatures and operation conditions change, so does the COP. This makes it a vital, real-time indicator of a system’s efficiency.

In the example from the exercise, we calculated a COP around 19.159, indicating that for every one unit of energy put into running the pump, just over 19 units of heat energy are moved from the colder outside to the warmer inside. It is essential to note that the COP would vary if the temperature conditions changed, which highlights how temperature affects heat pump efficiency.

Temperature Conversion: Kelvin to Celsius

Temperature conversions are a fundamental part of physics and engineering, as all thermal processes are influenced by temperature. The Kelvin and Celsius scales are closely related -- both increments are the same size, so the conversion is relatively straightforward.

To convert from Celsius to Kelvin, you add 273.15 to the Celsius temperature. This is because 0 degrees Celsius is equivalent to 273.15 Kelvin, the temperature at which water freezes under standard atmospheric conditions. Conversely, to convert from Kelvin to Celsius, you subtract 273.15 from the Kelvin temperature.

For example, the room temperature of 22 degrees Celsius that we want to maintain in the heat pump exercise is equivalent to 295.15 Kelvin. This conversion is vital when using the formulas for efficiency and COP since these require absolute temperatures in Kelvin. When managing thermal systems like heat engines and heat pumps, an accurate temperature conversion ensures proper calculations and system design.