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A refrigerator used to cool a computer requires \(1.2 \mathrm{kW}\) of electrical power and has a COP of \(1.8 .\) Calculate the cooling effect of this refrigerator, in \(\mathrm{kW}\)

Short Answer

Expert verified
Answer: The cooling effect of the refrigerator is 2.16 kW.

Step by step solution

01

Understand the Coefficient of Performance (COP)

The COP of a refrigerator is defined as the ratio of the cooling effect (output) to the electrical power input. Mathematically, it can be represented as: COP = Cooling effect (Q) / Power input (W) We are given the power input and COP, and we need to find the cooling effect.
02

Rearrange the formula and plug in the given values

We can rearrange the formula to solve for the cooling effect, Q: Q = COP × W Now plug in the given values: COP = 1.8 W = 1.2 kW Q = 1.8 × 1.2 kW
03

Calculate the cooling effect

Now, calculate the cooling effect by multiplying the COP by the power input: Q = 1.8 × 1.2 kW Q = 2.16 kW Therefore, the cooling effect of the refrigerator is 2.16 kW.

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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 essential when dealing with cooling systems like refrigerators and air conditioners. The cycle illustrates the path of a refrigerant as it goes through phase changes to absorb and dissipate heat, providing the desired cooling effect. In essence, the refrigeration cycle includes four main stages: evaporation, compression, condensation, and expansion.

During the evaporation stage, the refrigerant absorbs heat from the inside of a refrigerator or other enclosed space and evaporates. As a low-pressure vapor, it moves to the compressor, where its pressure and temperature are increased. Next, in the condenser, the high-pressure hot vapor releases heat to the outside and condenses into a liquid. Finally, through the expansion valve, the liquid refrigerant undergoes a pressure drop, cools down, and returns to the evaporator to start the cycle anew.

Comprehending this cycle is beneficial in identifying how the COP, a measure of a refrigerator's efficiency, is influenced by various stages. For instance, if a component of the cycle, such as the compressor, is not working efficiently, this can adversely affect the COP.
Thermal Efficiency
Thermal efficiency is a crucial concept in understanding how well a system converts energy into work or, in the case of refrigeration systems, into a cooling effect. It is a ratio of the useful output of a process to the input of energy, represented as a percentage. In thermal systems, this efficiency is often limited by the second law of thermodynamics, which states that not all supplied heat can be converted into work due to losses, most often as waste heat.

In the context of a refrigerator, thermal efficiency is closely related to the Coefficient of Performance (COP), which we have already explored. However, it's important to differentiate that COP is specific to refrigeration and heat pumps and is not expressed as a percentage but rather as a ratio. Higher COP values indicate a more efficient refrigerator, meaning it provides more cooling for the same amount of electric power input.

Improving the thermal efficiency of refrigeration affects both performance and energy consumption. Enhancing components such as insulation, compressor, and refrigerants can lead to better efficiency and lower operating costs. These improvements directly contribute to a higher COP, ensuring that more of the input electrical energy contributes to cooling rather than being lost.
Heat Transfer
Heat transfer is a fundamental concept in refrigeration as it is the process that allows refrigerators to remove heat from the space being cooled. It involves the movement of thermal energy from one place to another and is driven by temperature differences. The three primary mechanisms of heat transfer are conduction, convection, and radiation.

In refrigeration,
  • Conduction occurs within the walls of the refrigerator, where heat is transferred from the external warmer environment to the cooler interior.
  • Convection is responsible for the movement of heat within the refrigerant and the air inside the fridge. This process circulates the cooler air, ensuring even temperatures throughout.
  • Radiation, although less involved in refrigeration, can affect the external temperature of the unit.
Understanding how these heat transfer mechanisms function together helps clarify how refrigeration systems achieve the cooling effect. The COP measure, in the exercise provided, is a direct reflection of how effectively this heat transfer is achieved with the given amount of electrical power. Optimizing heat transfer processes within the refrigerator can lead to a better COP, meaning the system requires less electrical power to achieve the same level of cooling.

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Most popular questions from this chapter

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 \(22^{\circ} \mathrm{C}\).

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 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.

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.

Why does a nonquasi-equilibrium compression process require a larger work input than the corresponding quasiequilibrium one?

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