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Chapter 5: Problem 41

By removing energy by heat transfer from its freezer compartment at a rate of \(1.5 \mathrm{~kW}\), a refrigerator maintains the freezer at \(-22^{\circ} \mathrm{C}\) on a day when the temperature of the surroundings is \(28^{\circ} \mathrm{C}\). Determine the minimum theoretical power, in \(\mathrm{kW}\), required by the refrigerator at steady state.

Short Answer

Expert verified
0.2987 kW

Step by step solution

01

Identify the temperatures

Convert the given temperatures to Kelvin. Temperature of freezer, \( T_{L} = -22^{\textdegree} \text{C} + 273.15 = 251.15 \text{K} \) Temperature of surroundings, \( T_{H} = 28^{\textdegree} \text{C} + 273.15 = 301.15 \text{K} \)
02

Understand the coefficient of performance (COP) for a refrigerator

The coefficient of performance (COP) of an ideal refrigerator operating between two temperatures is given by: \( \text{COP} = \frac{T_{L}}{T_{H} - T_{L}} \)
03

Calculate the COP

Substitute the values of \( T_{L} \) and \( T_{H} \) into the COP formula: \( \text{COP} = \frac{251.15}{301.15 - 251.15} = \frac{251.15}{50} = 5.023 \)
04

Calculate the minimum theoretical power required

The power required by the refrigerator can be deduced from the relationship: \( Power (P) = \frac{Q}{\text{COP}} \) where \( Q \) is the rate of heat transfer (1.5 kW here). Thus, \( P = \frac{1.5 \text{ kW}}{5.023} = 0.2987 \text{ kW} \)

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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 (COP) is crucial in understanding refrigeration efficiency. It measures how well a refrigerator transfers heat relative to the work input. The higher the COP, the more efficient the system. For an ideal refrigerator working between two temperatures, the COP is given by the formula: \ \ \( \text{COP} = \frac{T_{L}}{T_{H} - T_{L}} \) \ \ Here, \(T_{L}\) is the lower temperature (inside the fridge in Kelvin), and \(T_{H}\) is the higher temperature (outside in Kelvin). A better grasp of COP helps in optimizing energy usage and ensuring economical operation.
Heat Transfer
Heat transfer is the process of removing heat from the refrigerator's freezer compartment to maintain a low temperature. In the given problem, the rate of heat transfer is specified as \(1.5 \mathrm{~kW}\). This energy represents the thermal power that needs to be continuously removed to keep the freezer at \(-22^{\circ} \mathrm{C}\). There are several methods of heat transfer, including conduction, convection, and radiation. In refrigeration, it is essential to minimize undesired heat gain to maintain efficiency. An effective insulation system is key to limiting the rate of heat flow into the cold compartment.
Power Requirement Calculation
Calculating the power requirement determines the minimum energy needed to keep the refrigerator running efficiently. Using the previously calculated COP, we can find the power using: \ \ \( P = \frac{Q}{\text{COP}} \) \ \ In this equation, \(Q\) is the heat transfer rate (1.5 kW in this case). For our specific problem, substituting the given values yields: \ \ \( P = \frac{1.5 \text{ kW}}{5.023} = 0.2987 \text{ kW} \) \ \ This calculation highlights the relationship between COP and power requirement, emphasizing the importance of optimizing both to reduce energy consumption and operational cost.
Temperature Conversion
Temperature conversion is essential for applying the thermodynamic equations correctly. Always ensure temperatures are in Kelvin when inserting into formulas. Conversions from Celsius to Kelvin are straightforward: add 273.15 to the Celsius temperature. For the problem at hand: \ \ - Freezer temperature \(-22^{\circ} \mathrm{C}\) converts to \(251.15 \mathrm{K}\). \ - Surrounding temperature \(28^{\circ} \mathrm{C}\) converts to \(301.15 \mathrm{K}\). \ \ This step is crucial as using the wrong units might result in incorrect calculations, undermining the entire process.

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Most popular questions from this chapter

A refrigeration cycle rejects \(Q_{H}=540 \mathrm{~kJ}\) per cycle to a hot reservoir at \(T_{\mathrm{H}}=500 \mathrm{~K}\), while receiving \(Q_{\mathrm{C}}=400 \mathrm{~kJ}\) per cycle from a cold reservoir at temperature \(T_{C}\). For 20 cycles of operation, determine (a) the net work input, in \(\mathrm{kJ}\), and (b) the minimum theoretical temperature \(T_{C}\), in \(\mathrm{K}\).

A reversible power cycle has the same thermal efficiency for hot and cold reservoirs at temperature \(T\) and \(600 \mathrm{~K}\), respectively, as for hot and cold reservoirs at 4000 and \(2000 \mathrm{~K}\), respectively. Determine \(T\), in \(\mathrm{K}\).

One \(\mathrm{kg}\) of air as an ideal gas executes a Carnot power cycle having a thermal efficiency of \(50 \%\). The heat transfer to the air during the isothermal expansion is \(50 \mathrm{~kJ}\). At the end of the isothermal expansion, the pressure is \(574 \mathrm{kPa}\) and the volume is \(0.3 \mathrm{~m}^{3}\). Determine (a) the maximum and minimum temperatures for the cycle, in \(\mathrm{K}\) (b) the pressure and volume at the beginning of the isothermal expansion in bar and \(\mathrm{m}^{3}\), respectively. (c) the work and heat transfer for each of the four processes, in \(\mathrm{kJ}\). (d) Sketch the cycle on \(p\)-v coordinates.

A reversible power cycle operating between hot and cold reservoirs at \(900 \mathrm{~K}\) and \(200 \mathrm{~K}\), respectively, receives \(120 \mathrm{~kJ}\) by heat transfer from the hot reservoir for each cycle of operation. Determine the net work developed in 12 cycles of operation, in \(\mathrm{kJ}\).

Ocean temperature energy conversion (OTEC) power plants generate power by utilizing the naturally occurring decrease with depth of the temperature of ocean water. Near Florida, the ocean surface temperature is \(27^{\circ} \mathrm{C}\), while at a depth of \(700 \mathrm{~m}\) the temperature is \(7^{\circ} \mathrm{C}\). (a) Determine the maximum thermal efficiency for any power cycle operating between these temperatures. (b) The thermal efficiency of existing OTEC plants is approximately \(2 \%\). Compare this with the result of part (a) and comment.
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