Question

A Carnot heat engine receives heat from a reservoir at \(1700^{\circ} \mathrm{F}\) at a rate of \(700 \mathrm{Btu} / \mathrm{min}\) and rejects the waste heat to the ambient air at \(80^{\circ} \mathrm{F}\). The entire work output of the heat engine is used to drive a refrigerator that removes heat from the refrigerated space at \(20^{\circ} \mathrm{F}\) and transfers it to the same ambicnt air at \(80^{\circ} \mathrm{F}\). Determine \((a)\) the maximum rate of heat removal from the refrigerated space and ( \(b\) ) the total rate of heat rejection to the ambient air.

Short answer

(a) Maximum rate of heat removal from the refrigerated space: 52.58 Btu/s (b) Total rate of heat rejection to the ambient air: 64.25 Btu/s

Step-by-step solution

1. Convert temperature to the absolute scale (Kelvin)

First, we need to convert the given temperatures from Fahrenheit to the absolute temperature scale (Kelvin). To do this, we will use the following conversion equation: \(T_K = \frac{5}{9}(T_F - 32) + 273.15\) Now, we convert the temperatures: \(T_{H} = \frac{5}{9}(1700 - 32) + 273.15 = 925.93 K\) \(T_{L,engine} = T_{H,refrigerator} = \frac{5}{9}(80 - 32) + 273.15 = 299.82 K\) \(T_{L,refrigerator} = \frac{5}{9}(20 - 32) + 273.15 = 266.48 K\)

2. Determine the efficiency of Carnot heat engine and the coefficient of performance for refrigerator

Next, we need to find the efficiency of the heat engine and the coefficient of performance of the refrigerator. The formulas for these, given that we are working with Carnot cycles, are: \(\eta_{engine} = 1 - \frac{T_{L,engine}}{T_H}\) \(COP_{refrigerator} = \frac{T_{L,refrigerator}}{T_{H,refrigerator} - T_{L,refrigerator}}\) Now, we calculate the values for the heat engine: \(\eta_{engine} = 1 - \frac{299.82}{925.93} = 0.676\) And for the refrigerator: \(COP_{refrigerator} = \frac{266.48}{299.82 - 266.48} = 6.657\)

3. Calculate the work output of the heat engine and the rate of heat removal from the refrigerator

The heat engine takes in heat from the hot reservoir at a rate of \(700 \mathrm{Btu} / \mathrm{min}\). Using the efficiency of the heat engine, we can find the work output and the waste heat rejected to the ambient air. The formula for the work output is: \(W_{engine} = \eta_{engine} \cdot Q_{H,engine}\) Converting the heat rate to Btu/s: \(\dot{Q}_{H,engine} = \frac{700 \mathrm{Btu}}{60 \mathrm{s}} = 11.67 \mathrm{Btu/s}\) Next, we calculate the work output: \(\dot{W}_{engine} = 0.676 \cdot 11.67 \mathrm{Btu/s} = 7.895 \mathrm{Btu/s}\) This work output is used to drive the refrigerator, so the rate of heat removal from the refrigerator can be determined using its coefficient of performance and the formula: \(\dot{Q}_{L,refrigerator} = COP_{refrigerator} \cdot \dot{W}_{engine}\) Now, we compute the rate of heat removal from the refrigerator: \(\dot{Q}_{L,refrigerator} = 6.657 \cdot 7.895 \mathrm{Btu/s} = 52.58 \mathrm{Btu/s}\)

4. Calculate the total rate of heat rejection to the ambient air

Finally, we need to find the total rate of heat rejection to the ambient air. This will consist of two parts: the heat rejected by the heat engine and the heat rejected by the refrigerator. The formulas are: \(\dot{Q}_{H,refrigerator} = \dot{Q}_{L,refrigerator} + \dot{W}_{engine}\) And for the heat engine: \(\dot{Q}_{L,engine} = \dot{Q}_{H,engine} - \dot{W}_{engine}\) Now, we compute the heat rejected by the heat engine: \(\dot{Q}_{L,engine} = 11.67 \mathrm{Btu/s} - 7.895 \mathrm{Btu/s} = 3.775 \mathrm{Btu/s}\) And the heat rejected by the refrigerator: \(\dot{Q}_{H,refrigerator} = 52.58 \mathrm{Btu/s} + 7.895 \mathrm{Btu/s} = 60.475 \mathrm{Btu/s}\) Therefore, the total rate of heat rejection to the ambient air is: \(\dot{Q}_{total} = \dot{Q}_{L,engine} + \dot{Q}_{H,refrigerator} = 3.775 \mathrm{Btu/s} + 60.475 \mathrm{Btu/s} = 64.25 \mathrm{Btu/s}\)

5. Results

(a) The maximum rate of heat removal from the refrigerated space is \(\dot{Q}_{L,refrigerator} = 52.58 \mathrm{Btu/s}\). (b) The total rate of heat rejection to the ambient air is \(\dot{Q}_{total} = 64.25 \mathrm{Btu/s}\).

Key concepts

Understanding Thermodynamic Cycles

Thermodynamic cycles play a crucial role in understanding how engines and refrigerators operate. In a thermodynamic cycle, a working substance undergoes a series of transformations through different phases of pressure, temperature, and volume, which can result in work being done on or by the system. One of the most idealized and efficient cycles is the Carnot cycle, which consists of four reversible processes: two isothermal (constant temperature) and two adiabatic (no heat transfer).

In a Carnot heat engine, which the exercise describes, heat is absorbed from a high-temperature reservoir, partially converted to work, and the remainder is rejected to a low-temperature reservoir. The efficiency of a Carnot heat engine is determined by the temperatures of the hot and cold reservoirs and is always less than 1, meaning that no heat engine can be perfectly efficient due to the second law of thermodynamics.

Notably, for any engine or cycle, the work output or input can be calculated by integrating the area under the process curves on a pressure-volume (PV) diagram. This helps visualize the process but is simplified through the use of efficiency or coefficient of performance (COP) calculations in our exercise.

Coefficient of Performance (COP) of Refrigerators

The Coefficient of Performance (COP) is a measure of the effectiveness of a refrigerator or heat pump. It represents the ratio of heat transfer from the cold reservoir (refrigerated space) to the work input required to transfer that heat to the hot reservoir (ambient air). In the case of the refrigerator mentioned in the exercise, the COP is high, which indicates a more effective refrigeration process.

The higher the COP of a refrigerator, the more efficient it is at transferring heat from the inside to the outside while using less work. In our problem, the COP also tells us how effectively the heat engine's work output is being used to drive the refrigeration cycle. The calculation in the exercise uses the temperatures of the cold and hot reservoirs to find the COP. This demonstrates that the efficiency of a refrigeration cycle is greatly dependent on the temperature difference between the cold and hot reservoirs; a smaller difference yields a higher COP.

Heat Transfer in Carnot Engines and Refrigerators

Heat transfer is an essential concept in thermodynamics and is especially pertinent in the operation of heat engines and refrigerators. In a Carnot engine, heat is absorbed from a hot reservoir, work is done as it operates, and finally, heat is rejected to a cold reservoir. For the system described in our exercise, the engine absorbs heat at a high temperature, performs work, and the waste heat is rejected to ambient air.

When examining heat transfer in refrigeration, the reverse process occurs. Heat is removed from a cold space, and in the process of refrigeration, work is applied, causing heat to be released into a warmer environment. The total heat rejected includes both the heat extracted from the refrigerated space and the input work from the heat engine.

Moreover, in our scenario, the refrigerator is driven by the work output of the Carnot heat engine. This linkage showcases a real-world application of energy transfer where the waste output of one system becomes the input for another, emphasizing the importance of understanding the intricacies of heat transfer for energy efficiency and practical design.