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The thermal characteristics of a small, dormitory refrigerator are determined by performing two separate experiments, each with the door closed and the refrigerator placed in ambient air at \(T_{\infty}=25^{\circ} \mathrm{C}\). In one case, an electric heater is suspended in the refrigerator cavity, while the refrigerator is unplugged. With the heater dissipating \(20 \mathrm{~W}\), a steady-state temperature of \(90^{\circ} \mathrm{C}\) is recorded within the cavity. With the heater removed and the refrigerator now in operation, the second experiment involves maintaining a steady-state cavity temperature of \(5^{\circ} \mathrm{C}\) for a fixed time interval and recording the electrical energy required to operate the refrigerator. In such an experiment for which steady operation is maintained over a 12-hour period, the input electrical energy is \(125,000 \mathrm{~J}\). Determine the refrigerator's coefficient of performance (COP).

Short Answer

Expert verified
The coefficient of performance (COP) of the refrigerator is approximately 6.92.

Step by step solution

01

Find Heat Absorbed by Refrigerator from Surroundings

First, we need to find the heat absorbed by the refrigerator from the surroundings when its steady-state temperature is maintained at \(5^{\circ} \mathrm{C}\). Since the electric heater is dissipating \(20 \mathrm{~W}\) when the steady-state temperature reaches \(90^{\circ} \mathrm{C}\), it means that the heat extracted by the refrigerator from the surroundings is \(20 \mathrm{~W}\).
02

Calculate Work Input Rate to the Refrigerator

Now, let's find the work input rate in watts to maintain the steady-state temperature at \(5^{\circ} \mathrm{C}\). We are given the electrical energy input \(125,000 \mathrm{~J}\) in a 12-hour period. To find the work input rate, we can divide the total energy input by the time it was consumed: \[W_{input} = \frac{Total~Energy~Input}{Time}\] We need to convert the time to seconds. 12 hours is equivalent to \(12 \times 3600 = 43200\) seconds. Thus, \[W_{input} = \frac{125,000 \mathrm{~J}}{43,200 \mathrm{~s}} \approx 2.89 \mathrm{~W}\]
03

Calculate Coefficient of Performance (COP)

Finally, we can calculate the COP using the formula mentioned in the analysis: \[COP = \frac{Q_{absorbed}}{W_{input}}\] Substituting the values we calculated: \[ COP = \frac{20 \mathrm{~W}}{2.89 \mathrm{~W}} \approx 6.92\] The COP of the refrigerator is approximately 6.92.

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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 a key metric used to quantify the efficiency of refrigeration and heat pump systems. In simple terms, it measures how well a refrigeration cycle can perform in terms of energy transfer. For refrigerators, COP is defined as the ratio of the heat energy absorbed from the inside (or the refrigerated space) to the work or electrical energy input needed to maintain the cycle:

\[COP = \frac{Q_{absorbed}}{W_{input}}\]
- \(Q_{absorbed}\) is the heat extracted from the refrigerated space, measured in watts (or joules per second).- \(W_{input}\) is the electrical work input to the refrigerator, also in watts.
A higher COP indicates a more efficient refrigeration cycle, meaning that less energy is required to extract a given amount of heat from the refrigerating space. For example, in the solution provided, we calculated a COP of approximately 6.92. This means for every unit of electrical energy used, the refrigerator is able to extract nearly 7 units of thermal energy from its interior.
Understanding COP is crucial as it allows engineers and consumers to compare the energy efficiency of different refrigeration appliances, aiding in selecting more energy-efficient and cost-effective options.
Refrigeration Cycle
The refrigeration cycle is at the heart of any refrigerator or air conditioning unit. It is a process that transfers heat from the interior of the refrigerator to its exterior. The fundamental mechanism of the refrigeration cycle involves several key stages:

- **Evaporation:** A refrigerant absorbs heat from the refrigerator's interior, turning from a liquid into a vapor.- **Compression:** The refrigerant vapor is compressed by a compressor, increasing its pressure and temperature.- **Condensation:** The hot, high-pressure refrigerant vapor releases heat to the surrounding environment through condenser coils, turning back into a liquid.- **Expansion:** The refrigerant now at a lower temperature and pressure, passes through an expansion valve or capillary tube, completing the cycle.
This cycle maintains the refrigerated space at a desired lower temperature. In the original exercise, the refrigeration cycle effectively maintains a steady-state temperature of \(5^{\circ} \mathrm{C}\) inside the fridge despite the external ambient temperature of \(25^{\circ} \mathrm{C}\). By removing heat from the inside and releasing it outside, it keeps the interior cooler than its surroundings.The efficiency of this cycle can be assessed through the COP, revealing how effectively it converts electrical energy into thermal energy transfer.
Thermal Analysis
Thermal analysis evaluates the behavior of materials or systems under thermal conditions by observing how heat moves and is managed in such systems. In the context of refrigeration, thermal analysis helps understand how effectively the refrigerator compartment is maintained at a desired temperature despite outside influences.

In the original exercise, thermal analysis was performed through controlled experiments: 1. Using a heater to maintain a consistent temperature of the refrigerator's cavity and observing the heat transfer. 2. Measuring the electrical energy required to maintain the refrigerator's operating temperature over a set period.
This analysis involves: - **Temperature Measurements:** Observing changes in temperature to evaluate how the refrigerator maintains low temperatures. - **Heat Transfer Coefficients:** Calculating how much heat is being transferred internally and externally during operation.
Effective thermal management is crucial to maximizing energy efficiency and maintaining refrigeration performance. By understanding heat transfer processes, we can better design and operate refrigeration systems to be both cost-effective and environmentally friendly.

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Most popular questions from this chapter

Finned passages are frequently formed between parallel plates to enhance convection heat transfer in compact heat exchanger cores. An important application is in electronic equipment cooling, where one or more air-cooled stacks are placed between heat-dissipating electrical components. Consider a single stack of rectangular fins of length \(L\) and thickness \(t\), with convection conditions corresponding to \(h\) and \(T_{\infty}\). (a) Obtain expressions for the fin heat transfer rates, \(q_{f, o}\) and \(q_{f, L}\), in terms of the base temperatures, \(T_{o}\) and \(T_{L}\). (b) In a specific application, a stack that is \(200 \mathrm{~mm}\) wide and \(100 \mathrm{~mm}\) deep contains 50 fins, each of length \(L=12 \mathrm{~mm}\). The entire stack is made from aluminum, which is everywhere \(1.0 \mathrm{~mm}\) thick. If temperature limitations associated with electrical components joined to opposite plates dictate maximum allowable plate temperatures of \(T_{o}=400 \mathrm{~K}\) and \(T_{L}=350 \mathrm{~K}\), what are the corresponding maximum power dissipations if \(h=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=300 \mathrm{~K} ?\)

A plane wall of thickness \(2 L\) and thermal conductivity \(k\) experiences a uniform volumetric generation rate \(\dot{q}\). As shown in the sketch for Case 1 , the surface at \(x=-L\) is perfectly insulated, while the other surface is maintained at a uniform, constant temperature \(T_{o}\). For Case 2 , a very thin dielectric strip is inserted at the midpoint of the wall \((x=0)\) in order to electrically isolate the two sections, \(\mathrm{A}\) and \(\mathrm{B}\). The thermal resistance of the strip is \(R_{t}^{\prime \prime}=0.0005 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The parameters associated with the wall are \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L=\) \(20 \mathrm{~mm}, \dot{q}=5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\), and \(T_{o}=50^{\circ} \mathrm{C}\). (a) Sketch the temperature distribution for Case 1 on \(T-x\) coordinates. Describe the key features of this distribution. Identify the location of the maximum temperature in the wall and calculate this temperature. (b) Sketch the temperature distribution for Case 2 on the same \(T-x\) coordinates. Describe the key features of this distribution. (c) What is the temperature difference between the two walls at \(x=0\) for Case 2 ? (d) What is the location of the maximum temperature in the composite wall of Case 2 ? Calculate this temperature.

A spherical tank for storing liquid oxygen on the space shuttle is to be made from stainless steel of \(0.80-\mathrm{m}\) outer diameter and 5 -mm wall thickness. The boiling point and latent heat of vaporization of liquid oxygen are \(90 \mathrm{~K}\) and \(213 \mathrm{~kJ} / \mathrm{kg}\), respectively. The tank is to be installed in a large compartment whose temperature is to be maintained at \(240 \mathrm{~K}\). Design a thermal insulation system that will maintain oxygen losses due to boiling below \(1 \mathrm{~kg} /\) day.

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