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Prove that the COP of all completely reversible refrigerators must be the same when the reservoir temperatures are the same.

Short Answer

Expert verified
Prove it. #Answer# Yes, when the reservoir temperatures are the same, all completely reversible refrigerators have the same Coefficient of Performance (COP). This can be proved by analyzing the relationship between the heat transfers in the refrigeration system and the reservoir temperatures. For completely reversible refrigerators, we can express the COP as a function of reservoir temperatures: COP = \(\frac{1}{\frac{T_H}{T_C} - 1}\), where \(T_H\) and \(T_C\) are the temperatures of the hot and cold reservoirs, respectively. Therefore, when the reservoir temperatures are the same, the COP of all completely reversible refrigerators must be the same.

Step by step solution

01

Define the COP for a refrigerator

The Coefficient of Performance (COP) is a measure of the efficiency of a refrigeration system. It is defined as the cooling effect (the heat removed from the cold reservoir) divided by the work input required to achieve this cooling effect: COP = \frac{Q_C}{W} Here, \(Q_C\) represents the heat absorbed from the cold reservoir and \(W\) is the work input required for the refrigerator to operate.
02

Analyze the role of heat transfers and reservoir temperatures in the system's efficiency

We would need to relate the two heat reservoirs and work input of the process with the temperatures of the reservoirs. To do this, we use the first and second laws of thermodynamics. According to the first law, energy is conserved: \(Q_H = Q_C + W\) Here, \(Q_H\) is the heat released to the hot reservoir. The second law of thermodynamics states that the heat exchanged between two reservoirs is proportional to their temperatures, which can be expressed in terms of absolute temperatures using the Carnot efficiency: \(\eta = \frac{T_H - T_C}{T_H} = \frac{W}{Q_H}\)
03

Consider the definition of completely reversible refrigerators

A completely reversible refrigerator is one whose operations can be reversed without any changes in efficiency. In other words, if the refrigerator were operated as a heat engine, it would follow the same efficiency relationships from the second law of thermodynamics. For a fully reversible refrigerator, we have the following relation: \(\frac{W}{Q_H} = \frac{T_C}{T_H - T_C}\)
04

Show that when reservoir temperatures are the same, the COPs are equal for all completely reversible refrigerators

Let's now express the COP as a ratio of the temperatures of the reservoirs. From the first law relationship, we can substitute \(Q_H\) in terms of \(Q_C\) and W: \(Q_H = Q_C(1 + \frac{W}{Q_C})\) Now substitute the relation found in Step 3: \(Q_H = Q_C\left(1 + \frac{T_C}{T_H - T_C}\right)\) And now we use this relationship to express the COP as a function of reservoir temperatures: COP = \(\frac{Q_C}{W} = \frac{Q_C}{Q_H - Q_C} = \frac{1}{\frac{Q_H}{Q_C} - 1} = \frac{1}{\frac{T_H}{T_C} - 1}\) As we can see, when the reservoir temperatures \(T_H\) and \(T_C\) are the same for completely reversible refrigerators, the Coefficient of Performance (COP) will be the same for all of them since COP is a function of only those two temperatures: COP = \(\frac{1}{\frac{T_H}{T_C} - 1}\) Thus, we have proved that the COP of all completely reversible refrigerators must be the same when the reservoir temperatures are the same.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Thermodynamics in Refrigeration
Thermodynamics plays a pivotal role in refrigeration, which is essentially about moving heat from one place to another. It's guided by two main laws. The first law, also known as the law of energy conservation, indicates that energy cannot be created or destroyed, only transferred or changed in form. In the context of refrigeration, this law ensures that the heat absorbed from a cold space (\(Q_C\)) plus the work input (\(W\) by the refrigerator equals the heat released to the warmer space (\(Q_H\)..

The second law of thermodynamics introduces the concept of entropy, a measure of system disorder. It states that energy naturally disperses from being localized to becoming spread out if it is not hindered from doing so. For refrigerators, this means that they need work to move heat from a cooler to a warmer place, thereby decreasing entropy locally while overall, the universe's entropy increases. This law also indicates that no process is completely efficient, as some energy is always dispersed as waste heat.

In essence, the efficiency of a refrigerator is all about how effectively it can transfer heat with minimal work input. This is where the Coefficient of Performance (COP) comes into play, showcasing how much cooling effect can be achieved per unit of work. By understanding these thermodynamic principles, one can appreciate how even small differences in temperature between the cold and warm reservoirs can significantly affect the COP and the overall efficiency of the refrigeration system.
Exploring the Refrigeration Cycle
The refrigeration cycle is a practical application of thermodynamics that keeps our food fresh and our rooms cool. Fundamentally, it involves a repeated cycle of compression, condensation, expansion, and evaporation of a refrigerant that absorbs and releases heat where needed.

During this cycle:
  • The refrigerant is compressed, raising its pressure and temperature.
  • It then flows through the condenser where it loses heat to the surroundings and turns into a liquid.
  • After passing through an expansion device, the refrigerant cools down and moves to the evaporator.
  • In the evaporator, it absorbs heat from the cold space, causing it to vaporize back into a gas. This completes the cycle as the gas returns to the compressor.

A key aspect of the refrigeration cycle is that it's driven by the transfer of heat, and not the creation of cold. Refrigerators do not generate 'coldness' but rather remove heat from a space. The efficiency of this cycle is often measured by the Coefficient of Performance (COP), which quantifies how effectively a refrigerator transfers heat relative to the work needed to run the cycle.
The Principle of Reversible Processes
Reversible processes are idealized concepts in thermodynamics where a system can transition between states without leaving any net change in the environment. Such processes are particularly important in theoretical discussions about the maximization of energy efficiency.

In a reversible process, every single step can be undone perfectly without any sign that the process took place. This means no energy has been wasted on friction, unwanted heat transfer, or other inefficiencies. While completely reversible processes don't occur in the real world as they require infinite time, they serve as an important benchmark.

When it comes to refrigerators, if they operated under fully reversible conditions, their COP would be tied solely to the temperatures of the cold and hot reservoirs, devoid of any other inefficiencies. This is why reversible refrigerators, operating between the same two temperatures, would all have the same COP, illustrating the deep relationship between thermodynamics and the fundamentals of cooling technology.

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Most popular questions from this chapter

The cargo space of a refrigerated truck whose inner dimensions are \(12 \mathrm{m} \times 2.3 \mathrm{m} \times 3.5 \mathrm{m}\) is to be precooled from \(25^{\circ} \mathrm{C}\) to an average temperature of \(5^{\circ} \mathrm{C}\). The construction of the truck is such that a transmission heat gain occurs at a rate of \(120 \mathrm{W} /^{\circ} \mathrm{C}\). If the ambient temperature is \(25^{\circ} \mathrm{C}\) determine how long it will take for a system with a refrigeration capacity of \(11 \mathrm{kW}\) to precool this truck.

Is it possible to develop \((a)\) an actual and \((b)\) a reversible heat-engine cycle that is more efficient than a Carnot cycle operating between the same temperature limits? Explain.

A homeowner buys a new refrigerator with no freezer compartment and a deep freezer for the new kitchen. Which of these devices would you expect to have a lower COP? Why?

A heat pump with refrigerant-134a as the working fluid is used to keep a space at \(25^{\circ} \mathrm{C}\) by absorbing heat from geothermal water that enters the evaporator at \(60^{\circ} \mathrm{C}\) at a rate of \(0.065 \mathrm{kg} / \mathrm{s}\) and leaves at \(40^{\circ} \mathrm{C}\). Refrigerant enters the evaporator at \(12^{\circ} \mathrm{C}\) with a quality of 15 percent and leaves at the same pressure as saturated vapor. If the compressor consumes \(1.6 \mathrm{kW}\) of power, determine \((a)\) the mass flow rate of the refrigerant, \((b)\) the rate of heat supply, \((c)\) the \(\mathrm{COP}\), and \((d)\) the minimum power input to the compressor for the same rate of heat supply.

A heat engine receives heat from a heat source at \(1200^{\circ} \mathrm{C}\) and rejects heat to a heat \(\operatorname{sink}\) at \(50^{\circ} \mathrm{C}\). The heat engine does maximum work equal to \(500 \mathrm{kJ}\). Determine the heat supplied to the heat engine by the heat source and the heat rejected to the heat sink.

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