Question

A Carnot heat engine receives heat from a reservoir at \(900^{\circ} \mathrm{C}\) at a rate of \(800 \mathrm{kJ} / \mathrm{min}\) and rejects the waste heat to the ambient air at \(27^{\circ} \mathrm{C}\). The entire work output of the heat engine is used to drive a refrigerator that removes heat from the refrigerated space at \(-5^{\circ} \mathrm{C}\) and transfers it to the same ambient air at \(27^{\circ} \mathrm{C}\). 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

Based on the given information and calculations: (a) The maximum rate of heat removal from the refrigerated space is 1774.26 W. (b) The total rate of heat rejection to the ambient air is 5183.87 W.

Step-by-step solution

Convert given temperatures to Kelvin scale

To use the temperatures in calculations, we need to convert them from Celsius to Kelvin. Convert hot reservoir temperature, cold reservoir temperature, and refrigerated space temperature to Kelvin: \(T_{H} = 900 + 273.15 = 1173.15\,\mathrm{K}\) \(T_{C} = 27 + 273.15 = 300.15\,\mathrm{K}\) \(T_{R} = -5 + 273.15 = 268.15\,\mathrm{K}\)

Calculate the Carnot heat engine's efficiency

The efficiency of a Carnot heat engine is given by the following equation: \(\eta = 1 - \frac{T_C}{T_H}\) \(\eta = 1 - \frac{300.15}{1173.15}\) \(\eta = 0.7440\)

Calculate the rate of heat input in watts

Since we are given the rate of energy transfer in kJ per minute, we need to convert it to watts. \(\dot{Q_H} = \frac{800\,\mathrm{kJ/min}}{60\,\mathrm{s/min}} \times 1000\,\frac{\mathrm{J}}{\mathrm{kJ}} = 13333.33\,\mathrm{W}\)

Calculate the work output of the heat engine

Work output of the heat engine can be calculated using the efficiency and heat input: \(\dot{W} = \eta \times \dot{Q_H}\) \(\dot{W} = 0.7440 \times 13333.33\) \(\dot{W} = 9923.72\,\mathrm{W}\)

Calculate the coefficient of performance (COP) of the refrigerator

The COP of a Carnot refrigerator is given by the following equation: \(\mathrm{COP} = \frac{T_R}{T_C - T_R}\) \(\mathrm{COP} = \frac{268.15}{300.15 - 268.15}\) \(\mathrm{COP} = 5.5951\)

Determine the maximum rate of heat removal from the refrigerated space

The maximum rate of heat removal from the refrigerated space can be calculated using the work output of the heat engine and the COP of the refrigerator: \(\dot{Q_R} = \frac{\dot{W}}{\mathrm{COP}}\) \(\dot{Q_R} = \frac{9923.72}{5.5951}\) \(\dot{Q_R} = 1774.26\,\mathrm{W}\) (a) The maximum rate of heat removal from the refrigerated space is \(1774.26\,\mathrm{W}\).

Determine the total rate of heat rejection to the ambient air

The total rate of heat rejection to the ambient air can be calculated using the heat removal from the refrigerated space, work input to the refrigerator, and the heat input to the engine: \(\dot{Q_C} = \dot{Q_H} - \dot{W} + \dot{Q_R}\) \(\dot{Q_C} = 13333.33 - 9923.72 + 1774.26\) \(\dot{Q_C} = 5183.87\,\mathrm{W}\) (b) The total rate of heat rejection to the ambient air is \(5183.87\,\mathrm{W}\).

Key concepts

Carnot Cycle Efficiency

Understanding the efficiency of a Carnot heat engine is essential when studying thermodynamics. The Carnot efficiency represents the maximum possible efficiency that any heat engine can achieve operating between two temperatures. It is determined by the difference in temperature between the hot and cold reservoirs.

In our example, the Carnot heat engine operates between a hot reservoir at a high temperature and a cold reservoir, which is the ambient air. The efficiency equation, \(\eta = 1 - \frac{T_C}{T_H}\), where \(T_H\) and \(T_C\) are the absolute temperatures of the hot and cold reservoirs respectively, reflects this. The closer the temperatures of the reservoirs, the lower the efficiency, and this is why high-temperature sources are sought for to maximize the efficiency. This concept is critical, as it sets the upper limit for the energy conversion efficiency that can be reached.

Coefficient of Performance (COP)

The coefficient of performance (COP) is a measure of the efficiency of a heat pump or refrigerator. It is defined as the ratio of the heat removed to the work input. For a refrigerator, the COP is given by the formula \(\mathrm{COP} = \frac{T_R}{T_C - T_R}\), with \(T_R\) representing the temperature of the refrigerated space and \(T_C\) the temperature at which heat is rejected to the environment.

The higher the COP, the more efficient the refrigerator is. In our case, we calculated the COP using the given temperatures, converted to Kelvin. Notice how the efficiency of a refrigerator is dependent on the temperature difference between the space to be cooled and the environment. For practical reasons, refrigerators are designed to maximize this value while ensuring reliability and sustainability.

Heat Transfer

Heat transfer is a fundamental concept in thermodynamics that involves the movement of thermal energy from one object or material to another. It is the process which underpins the operation of heat engines and refrigerators. There are three modes of heat transfer: conduction, convection, and radiation. In the context of our Carnot engine and refrigerator system, we are interested in the transfer of heat energy from a high-temperature reservoir to the engine and then from the refrigerator to a lower temperature reservoir.

In the steps provided, we calculate the rate of heat removal from the refrigerated space, which is one aspect of heat transfer. There is also the waste heat that the engine rejects to the ambient air, which we considered to determine the overall heat transfer within the system. By calculating the heat input and the heat rejection rates, we can better understand the flow of energy and thus the efficiency of both the heat engine and the refrigerator.

Thermodynamic Temperature Conversion

Temperature conversion from Celsius to Kelvin is a fundamental step in solving thermodynamic problems because thermodynamic equations require absolute temperatures. Absolute temperature is measured in Kelvin, where 0 K is absolute zero, the point at which no further thermal energy can be removed from a substance.

To convert from Celsius to Kelvin, add 273.15 to the Celsius temperature. In our solution steps, this conversion was necessary to correctly apply the Carnot efficiency and COP formulae. Proper temperature conversion is crucial for accuracy in calculations, as using temperatures in Celsius or Fahrenheit could lead to incorrect interpretations of the thermodynamic properties and processes being analyzed.