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A refrigerator in a laboratory uses \(R-22\) as the working substance. The high pressure is 1200 \(\mathrm{kPa},\) the low pressure is \(201 \mathrm{kPa},\) and the compressor is reversible. It should remove \(500 \mathrm{W}\) from a specimen currently at \(-20^{\circ} \mathrm{C}\) (not equal to \(T_{L}\) in the cycle) that is inside the refrigerated space. Find the cycle COP and the electrical power required.

Short Answer

Expert verified
The coefficient of performance (COP) of the refrigeration cycle is around 4.94 and the electrical power required by the compressor is about 101.21\, \mathrm{W}.

Step by step solution

01

Understand COP and its Calculation

The Coefficient of Performance (COP) for a refrigeration cycle is defined as the ratio of the desired output (the amount of heat being removed, \(Q_{L}\)) to the required input (the work done by the system, \(W_{cycle}\)). Mathematically, it can be expressed as \(COP = \frac{Q_{L}}{W_{cycle}}\). In this case, we don't have direct values of \(Q_{L}\) and \(W_{cycle}\), but we can calculate the COP through the Carnot cycle equation, which is \(\frac{1}{T_{H}/T_{L}-1}\), where \(T_{H}\) is the high temperature and \(T_{L}\) is the low temperature. But these temperatures are in Kelvin.
02

Convert Pressures to Temperatures

We have been provided with the high and low pressures, which we need to convert into equivalent temperatures. The temperature-pressure relationship is given by the properties of \(R-22\). For \(R-22\), at 1200 kPa, \(T_{H} = 40.16^{\circ}\mathrm{C} = 313.31\, \mathrm{K}\) and at 201 kPa, \(T_{L} = -52.72^{\circ}\mathrm{C} = 220.43\, \mathrm{K}\).
03

Calculate the COP

Now, we can substitute the values of \(T_{H}\) and \(T_{L}\) from step 2 into the COP equation in step 1, thus obtaining the COP value. So, the COP = \(\frac{1}{313.31/220.43 - 1} = 4.94\).
04

Calculate the Required Electrical Power

From the definition of COP, it's known that the refrigeration effect or the heat removal from the cold reservoir \(Q_{L}\) is equal to the product of the work done \(W_{cycle}\) and the COP. We know that the refrigerator is removing \(500\, \mathrm{W}\) from the specimen. So, \(W_{cycle} = \frac{Q_{L}}{COP} = \frac{500}{4.94} = 101.21\, \mathrm{W}\). This is the work done by the refrigerator or the electrical power required.

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Key Concepts

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

Coefficient of Performance (COP)
The Coefficient of Performance, often abbreviated as COP, is a key concept in understanding how efficient a refrigeration system is. It measures how effectively the system uses energy to transfer heat. In refrigeration, COP is defined as the ratio of the heat removed from the refrigerated space (i.e., the output we desire) to the work input required by the system (i.e., the electrical energy it consumes). In equation form, it looks like this: \[ \text{COP} = \frac{Q_{L}}{W_{cycle}} \]Where:
  • \( Q_{L} \) is the heat removed from the refrigerated space.
  • \( W_{cycle} \) is the work input needed for the cycle to remove \( Q_{L} \).
A high COP value indicates a more efficient refrigeration cycle, as you get more cooling effect for every unit of work input. It’s like getting more bang for your buck but for energy. Understanding COP helps in designing energy-efficient refrigeration systems and is critical in energy management.
Carnot Cycle
The Carnot Cycle represents an idealized cycle that offers the maximum possible efficiency for any given refrigeration or heat pump cycle operating between two temperature limits. Named after the French physicist Sadi Carnot, this theoretical cycle serves as a benchmark to compare real-world systems.In the Carnot cycle, the efficiency is dependent solely on the temperature of the two reservoirs involved, the hot (\( T_{H} \)) and cold (\( T_{L} \)) reservoirs. The COP for a refrigeration cycle using a Carnot Cycle is given by the formula:\[ \text{COP}_{\text{Carnot}} = \frac{1}{\frac{T_{H}}{T_{L}} - 1} \]Where:
  • \( T_{H} \) is the temperature of the hot reservoir, in Kelvin.
  • \( T_{L} \) is the temperature of the cold reservoir, in Kelvin.
The ideal COP is a useful tool in determining how close a real system is to the optimal. While no real refrigeration cycle can achieve this ideal value due to unavoidable inefficiencies and non-idealities, the Carnot cycle sets a standard for designing more efficient refrigeration systems.
Temperature-Pressure Relationship
In refrigeration systems, the temperature and pressure of the working fluid (often a refrigerant like R-22) are closely linked. As the pressure of the refrigerant changes, so does its temperature, thanks to the principles of thermodynamics governing liquid-vapor equilibrium. This relationship is crucial in predicting how the system behaves at various stages of the refrigeration cycle. For refrigerant R-22, specific pressure values correspond to specific temperatures, which are often provided in thermodynamic property tables. For instance, in the given exercise:
  • At a high pressure of 1200 kPa, R-22 has a corresponding temperature of 40.16°C, which is 313.31 K.
  • At a low pressure of 201 kPa, the corresponding temperature is -52.72°C, equivalent to 220.43 K.
Understanding this relationship allows engineers to design effective refrigeration cycles, ensuring that the system operates safely and efficiently within its specified temperature and pressure limits. Temperature-pressure data is essential for accurately calculating other cycle parameters and for controlling the process in practice.

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Most popular questions from this chapter

To approximate an actual spark-ignition engine, consider an air-standard Otto cycle that has a heat addition of \(1800 \mathrm{kJ} / \mathrm{kg}\) of air, a compression ratio of \(7,\) and a pressure and temperature at the beginning of the compression process of \(90 \mathrm{kPa}\) and \(10^{\circ} \mathrm{C}\). Assuming constant specific heat, with the value from Table A.5, determine the maximum pressure and temperature of the cycle, the thermal efficiency of the cycle, and the mean effective pressure.

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