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A refrigerator is removing heat from a cold medium at \(3^{\circ} \mathrm{C}\) at a rate of \(7200 \mathrm{kJ} / \mathrm{h}\) and rejecting the waste heat to a medium at \(30^{\circ} \mathrm{C}\). If the coefficient of performance of the refrigerator is \(2,\) the power consumed by the refrigerator is \((a) 0.1 \mathrm{kW}\) (b) \(0.5 \mathrm{kW}\) \((c) 1.0 \mathrm{kW}\) \((d) 2.0 \mathrm{kW}\) \((e) 5.0 \mathrm{kW}\)

Short Answer

Expert verified
Answer: The power consumed by the refrigerator is 1.0 kW.

Step by step solution

01

Write down the definition of the coefficient of performance (COP)

The COP is a measure of the efficiency of a refrigerator, defined as the ratio of the heat removed from the cold medium (Q_C) to the work input (W) required to remove that heat. Mathematically, it's written as: COP = \frac{Q_C}{W}
02

Use the given COP to rearrange the equation for the work input

We're given a COP of 2. Let's manipulate the equation to find the work input W: W = \frac{Q_C}{\text{COP}}
03

Convert the heat transfer rate to the appropriate units

We know that the heat is removed at a rate of 7200 kJ/h. First, we should convert the rate to kJ/s. There are 3600 seconds in an hour, so Q_C = \frac{7200 \text{ kJ}}{3600 \text{ s}} = 2 \text{ kJ/s}
04

Calculate the power consumed by the refrigerator

Now, we can substitute the values of Q_C and the COP into the equation from step 2 to find the work input (power consumed): W = \frac{2 \text{ kJ/s}}{2} = 1 \text{ kJ/s} Since 1 kJ/s is equivalent to 1 kW, the power consumed by the refrigerator is 1.0 kW, which corresponds to answer choice (c).

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Key Concepts

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

Refrigeration Cycle
Understanding the refrigeration cycle is fundamental when analyzing the workings of refrigerators, air conditioners, and any appliance that is used to cool a space or substance. The refrigeration cycle is a series of steps that involve manipulating a refrigerant to absorb heat from a cooled environment and then release it to the surroundings.

The process consists primarily of four stages: evaporation, compression, condensation, and expansion. During the evaporation stage, the refrigerant absorbs heat from the inside of a refrigerator and evaporates at a low pressure. This results in cooling the inside area. Then, the vapor is compressed, which increases its temperature and pressure. Following compression, the high-pressure hot vapor moves through a condenser where it releases heat to the outside environment and condenses into a liquid. Finally, the liquid passes through an expansion device that reduces its pressure and temperature, and the cycle begins again.

These mechanisms rely on the principles of heat transfer and thermodynamics, and the refrigeration cycle is designed to maximize energy efficiency through careful control of the refrigerant's physical states and thermodynamic properties.
Energy Efficiency
Energy efficiency in the context of refrigeration is a measure of how effectively a refrigerator uses electricity to remove heat from its interior. A more energy-efficient refrigerator consumes less power to perform the same amount of cooling compared to a less efficient one.

One way to quantify energy efficiency is through the coefficient of performance (COP). The COP is the ratio of the heat removed from a refrigerator's interior (useful cooling provided) to the electrical energy (work) it consumes to drive the cycle. A higher COP value indicates a more energy-efficient system. Energy efficiency is not just beneficial for reducing operational costs; it is also critical for reducing the environmental impact by lowering greenhouse gas emissions associated with electricity production.

Factors affecting the energy efficiency of a refrigeration cycle include the type and quality of insulation, the efficiency of the compressor, the properties of the refrigerant, and the design of the heat exchangers. Optimizing these factors can lead to significant energy savings.
Heat Transfer
Heat transfer is the process of thermal energy moving from a hotter object to a colder one. In the context of refrigeration, heat transfer occurs at both the internal and external interfaces of the refrigerator. Internally, heat is absorbed from the food compartments and transferred to the refrigerant circulating within the refrigerator. Externally, heat is released from the refrigerant to the environment when it passes through the condenser coils.

There are three fundamental modes of heat transfer: conduction, convection, and radiation. In a standard refrigerator, conduction occurs within the walls and shelves, convection occurs as the air inside circulates over these surfaces, and radiation can happen with the heat exchange between the coils and the surrounding air.

Improving heat transfer efficiency can also improve the overall energy efficiency of a refrigeration system. This can be achieved through the use of better heat-conducting materials, increasing surface area for heat exchange, and ensuring proper airflow around the condenser coils to disperse heat more effectively.
Thermodynamic Principles
Thermodynamics is a branch of physics concerned with heat, work, and temperature, and their relation to energy, radiation, and physical properties of matter. The thermodynamic principles govern the operation of refrigeration cycles and determine their performance and efficiency.

The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed in an isolated system. The energy going into a refrigerator (electrical work) must equal the sum of the energy removed (cooling effect) and the energy losses like waste heat. The second law of thermodynamics introduces the concept of entropy and states that in all energy exchanges, if no energy enters or leaves the system, the potential energy of the state will always be less than that of the initial state. This principle dictates that some energy will be lost as waste heat in any real system, making 100% efficiency impossible.

Understanding these principles helps in identifying the optimal conditions for the refrigeration cycle to obtain maximum efficiency. It explains why a refrigerator is more efficient if the difference between the temperature of the refrigerated space and the outside environment is minimized, and why refrigerators must release waste heat to function properly.

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Most popular questions from this chapter

Design a hydrocooling unit that can cool fruits and vegetables from 30 to \(5^{\circ} \mathrm{C}\) at a rate of \(20,000 \mathrm{kg} / \mathrm{h}\) under the following conditions: The unit will be of flood type, which will cool the products as they are conveyed into the channel filled with water. The products will be dropped into the channel filled with water at one end and be picked up at the other end. The channel can be as wide as \(3 \mathrm{m}\) and as high as \(90 \mathrm{cm} .\) The water is to be circulated and cooled by the evaporator section of a refrigeration system. The refrigerant temperature inside the coils is to be \(-2^{\circ} \mathrm{C}\), and the water temperature is not to drop below \(1^{\circ} \mathrm{C}\) and not to exceed \(6^{\circ} \mathrm{C}\) Assuming reasonable values for the average product density, specific heat, and porosity (the fraction of air volume in a box), recommend reasonable values for \((a)\) the water velocity through the channel and ( \(b\) ) the refrigeration capacity of the refrigeration system.

A cold canned drink is left in a warmer room where its temperature rises as a result of heat transfer. Is this a reversible process? Explain.

Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquid-vapor mixture region using \(0.96 \mathrm{kg}\) of refrigerant-134a as the working fluid. It is known that the maximum absolute temperature in the cycle is 1.2 times the minimum absolute temperature, and the net work input to the cycle is \(22 \mathrm{kJ}\). If the refrigerant changes from saturated vapor to saturated liquid during the heat rejection process, determine the minimum pressure in the cycle.

Using a timer (or watch) and a thermometer, conduct the following experiment to determine the rate of heat gain of your refrigerator. First make sure that the door of the refrigerator is not opened for at least a few hours so that steady operating conditions are established. Start the timer when the refrigerator stops running and measure the time \(\Delta t_{1}\) it stays off before it kicks in. Then, measure the time \(\Delta t_{2}\) it stays on. Noting that the heat removed during \(\Delta t_{2}\) is equal to the heat gain of the refrigerator during \(\Delta t_{1}+\Delta t_{2}\) and using the power consumed by the refrigerator when it is running, determine the average rate of heat gain for your refrigerator, in \(\mathrm{W}\). Take the COP (coefficient of performance) of your refrigerator to be 1.3 if it is not available.

A heat engine operates between a source at \(477^{\circ} \mathrm{C}\) and a \(\operatorname{sink}\) at \(25^{\circ} \mathrm{C}\). If heat is supplied to the heat engine at a steady rate of \(65,000 \mathrm{kJ} / \mathrm{min},\) determine the maximum power output of this heat engine.

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