Chapter 6: Problem 43
A food department is kept at \(-12^{\circ} \mathrm{C}\) by a refrigerator in an environment at \(30^{\circ} \mathrm{C}\). The total heat gain to the food department is estimated to be \(3300 \mathrm{kJ} / \mathrm{h}\) and the heat rejection in the condenser is \(4800 \mathrm{kJ} / \mathrm{h}\). Determine the power input to the compressor, in \(\mathrm{kW}\) and the COP of the refrigerator.
Short Answer
Expert verified
Answer: The power input to the compressor is 0.4167 kW, and the COP of the refrigerator is approximately 2.20.
Step by step solution
01
Calculate the heat transfer in the evaporator (Q1)
We are given that the total heat gain to the food department is 3300 kJ/h. This is equal to the heat transfer in the evaporator (Q1).
So, \(Q_1 = 3300 \, \mathrm{kJ/h}\).
02
Convert Q1 to Watts (W)
Since we need to find the power input to the compressor in Watts, we need to convert \(Q_1\) from kJ/h to W (1 W = 1 J/s).
\(Q_1 = \frac{3300 \, \mathrm{kJ/h}}{3.6 \, \mathrm{MJ/h}} \times 1000 = 916.67 \, \mathrm{W}\)
03
Calculate the heat transfer in the condenser (Q2)
The heat rejection in the condenser is given as 4800 kJ/h. This is equal to the heat transfer in the condenser (Q2).
So, \(Q_2 = 4800 \, \mathrm{kJ/h}\).
04
Convert Q2 to Watts (W)
Similar to step 2, convert \(Q_2\) from kJ/h to W (1 W = 1 J/s).
\(Q_2 = \frac{4800 \, \mathrm{kJ/h}}{3.6 \, \mathrm{MJ/h}} \times 1000 = 1333.33 \, \mathrm{W}\)
05
Calculate the power input to the compressor (W_comp)
According to conservation of energy principle, the power input to the compressor (\(W_\mathrm{comp}\)) is the difference between the heat transfer in the condenser (Q2) and the heat transfer in the evaporator (Q1).
\(W_\mathrm{comp} = Q_2 - Q_1 = 1333.33 \, \mathrm{W} - 916.67 \, \mathrm{W} = 416.66 \, \mathrm{W}\)
06
Convert W_comp to kW
Since we need to find the power input to the compressor in kW, we need to convert \(W_\mathrm{comp}\) from W to kW (1 kW = 1000 W).
\(W_\mathrm{comp} = 416.66 \, \mathrm{W} \times \frac{1 \, \mathrm{kW}}{1000 \, \mathrm{W}} = 0.4167 \, \mathrm{kW}\)
07
Calculate the Coefficient of Performance (COP)
The Coefficient of Performance (COP) is defined as the ratio of the heat transfer in the evaporator (Q1) to the power input to the compressor (W_comp).
\(\mathrm{COP} = \frac{Q_1}{W_\mathrm{comp}} = \frac{916.67 \, \mathrm{W}}{416.66 \, \mathrm{W}} \approx 2.20\)
The power input to the compressor is \(0.4167 \, \mathrm{kW}\) and the COP of the refrigerator is approximately \(2.20\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Heat Transfer in Refrigeration
In the context of refrigeration, heat transfer is a key concept that explains how thermal energy is moved from one area to another. Specifically, it refers to the process by which the refrigeration system removes heat from the food department (inside) and releases it into the environment (outside).
In our exercise, the food department is kept at \( -12^\circ \mathrm{C} \) while the environment is at \( 30^\circ \mathrm{C} \). The 'heat transfer' to the food department, or the 'heat load', is given as 3300 kJ/h, indicating the amount of thermal energy entering the refrigerator's cooled space. This heat is subsequently absorbed by the evaporator, a component of the refrigeration cycle that typically contains a cold refrigerant which absorbs heat as it vaporizes.
The heat is then transported to the condenser, where it is released at a higher temperature to the outside environment. This process effectively keeps the food department cold and is the principle upon which refrigerators operate. When we talk about the 'heat rejection in the condenser' at 4800 kJ/h, we refer to the heat being expelled outside, typically more than the amount of heat absorbed due to the work done by the compressor, which is part of the energy that is being conserved and transferred during the refrigeration cycle.
In our exercise, the food department is kept at \( -12^\circ \mathrm{C} \) while the environment is at \( 30^\circ \mathrm{C} \). The 'heat transfer' to the food department, or the 'heat load', is given as 3300 kJ/h, indicating the amount of thermal energy entering the refrigerator's cooled space. This heat is subsequently absorbed by the evaporator, a component of the refrigeration cycle that typically contains a cold refrigerant which absorbs heat as it vaporizes.
The heat is then transported to the condenser, where it is released at a higher temperature to the outside environment. This process effectively keeps the food department cold and is the principle upon which refrigerators operate. When we talk about the 'heat rejection in the condenser' at 4800 kJ/h, we refer to the heat being expelled outside, typically more than the amount of heat absorbed due to the work done by the compressor, which is part of the energy that is being conserved and transferred during the refrigeration cycle.
Coefficient of Performance (COP) in Refrigerators
The Coefficient of Performance (COP) is an important measure in refrigeration and heat pump systems, representing the efficiency of the system's heat transfer process. It is calculated as the ratio of useful heating or cooling provided to the work required. In the language of our refrigerator example, COP is the ratio of the heat removed from the inside of the refrigerator to the work done by the compressor (electrical energy input).
To understand this better, let's take the final calculated COP from our exercise, which is around 2.20. This indicates that for every unit of work (energy) the compressor uses, it transfers approximately 2.20 units of heat from the food department. A higher COP means a more efficient refrigerator, as it requires less energy to transfer a certain amount of heat.
An important point to note for students is that COP is a unitless measure because it represents a ratio of energies (or powers, which are simply rates of energy transfer) that are typically measured in the same unit.
To understand this better, let's take the final calculated COP from our exercise, which is around 2.20. This indicates that for every unit of work (energy) the compressor uses, it transfers approximately 2.20 units of heat from the food department. A higher COP means a more efficient refrigerator, as it requires less energy to transfer a certain amount of heat.
An important point to note for students is that COP is a unitless measure because it represents a ratio of energies (or powers, which are simply rates of energy transfer) that are typically measured in the same unit.
Energy Conservation in Thermodynamics
The principle of energy conservation is fundamental to all areas of physics and is especially crucial in thermodynamics, which studies energy and its transformations. According to this principle, energy cannot be created or destroyed, but it can change forms, such as kinetic, potential, thermal, or work.
In our refrigerator scenario, the principle of energy conservation is applied when understanding how the power input to the compressor, heat transfer in the evaporator and condenser, and overall system performance relate to each other. The difference in heat transfer between the condenser and the evaporator quantifies the work done by the compressor. This reflects the conservation of energy within the system, as all the work that is put into the compressor (electrical energy) is converted into moving heat out of the refrigerator (heat energy), minus the losses due to inefficiencies.
Energy conservation in refrigeration ensures that the energy balance is maintained and helps to calculate the system's performance. The exercise clearly demonstrates how energy is conserved and transferred throughout the refrigeration cycle, ultimately enabling us to determine the necessary power input for the compressor in kilowatts.
In our refrigerator scenario, the principle of energy conservation is applied when understanding how the power input to the compressor, heat transfer in the evaporator and condenser, and overall system performance relate to each other. The difference in heat transfer between the condenser and the evaporator quantifies the work done by the compressor. This reflects the conservation of energy within the system, as all the work that is put into the compressor (electrical energy) is converted into moving heat out of the refrigerator (heat energy), minus the losses due to inefficiencies.
Energy conservation in refrigeration ensures that the energy balance is maintained and helps to calculate the system's performance. The exercise clearly demonstrates how energy is conserved and transferred throughout the refrigeration cycle, ultimately enabling us to determine the necessary power input for the compressor in kilowatts.
