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Design a thermoelectric refrigerator that is capable of cooling a canned drink in a car. The refrigerator is to be powered by the cigarette lighter of the car. Draw a sketch of your design. Semiconductor components for building thermoelectric power generators or refrigerators are available from several manufacturers. Using data from one of these manufacturers, determine how many of these components you need in your design, and estimate the coefficient of performance of your system. A critical problem in the design of thermoelectric refrigerators is the effective rejection of waste heat. Discuss how you can enhance the rate of heat rejection without using any devices with moving parts such as a fan.

Short Answer

Expert verified
Answer: The heat rejection rate can be enhanced without using any moving parts by incorporating fins or heat sinks on the hot side of the Peltier elements, applying thermally conductive materials or compounds, and optimizing the design of the refrigerator to encourage natural convection and airflow around the hot side.

Step by step solution

01

Understand Thermoelectric Refrigeration

Thermoelectric refrigeration uses the Peltier effect, which occurs when current is passed through two different types of conductors or semiconductors that are connected in series. The current causes a temperature difference across the junctions, and thus creates a cooling effect. In order to generate sufficient cooling for the canned drink, we will need to use multiple thermoelectric semiconductor components.
02

Determine Required Power

First, we need to find the required power to cool the canned drink. To do so, we need to figure out the desired temperature drop, the heat capacity of the canned drink, and the time within which we want the drink to be cooled. The required power can be calculated using the formula: Power (W) = (Q * dT) / t Where Q is the heat capacity of the drink, dT is the desired temperature drop in K, and t is the cooling time in seconds.
03

Design the Refrigerator Sktech

Draw a simple sketch of the refrigerator design, showing the position of the cigarette lighter, thermoelectric elements, and the canned drink. Make sure to highlight the flow of current through the thermoelectric elements, and the cold and hot sides of these elements.
04

Determine Number of Components and Coefficient of Performance

Refer to the data provided by the semiconductor manufacturer to determine the cooling capacity of individual thermoelectric elements. Divide the required power by the cooling capacity of a single element to get the total number of elements needed. Next, calculate the coefficient of performance (COP) using the formula: COP = Q_cool / W_electric Where Q_cool is the cooling capacity (in watts) and W_electric is the electrical power input (in watts).
05

Enhance Heat Rejection Rate

To improve the heat rejection rate without using any moving parts, consider using passive methods, such as the following: 1. Incorporating fins or heat sinks on the hot side of the Peltier elements to increase the surface area and enhance heat dissipation. 2. Applying thermally conductive materials or compounds to improve the heat transfer between the hot side and surrounding environment. 3. Optimizing the design of the refrigerator to encourage natural convection and airflow around the hot side. Finally, revise your sketch to include these heat rejection enhancements in your refrigerator design.

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Key Concepts

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

Peltier effect
Understanding the Peltier effect is crucial when designing thermoelectric devices, including refrigerators like the one to cool a canned drink in a car. The Peltier effect occurs when an electric current is passed through two different types of materials that are connected. It results in a temperature difference; one junction absorbs heat, becoming colder, while the other releases heat, becoming hot. The effect is named after Jean Charles Athanase Peltier, who discovered it in 1834. It is this very principle that underpins the operation of a thermoelectric refrigerator.

For a practical application, multiple thermoelectric semiconductors are connected in series to amplify this effect and achieve the desired cooling. These semiconductor components, typically made of bismuth telluride, are chosen for their ability to produce a significant temperature difference and their efficiency in converting electrical energy into a temperature gradient.
Coefficient of performance
The coefficient of performance (COP) is a key parameter in assessing the efficiency of thermoelectric refrigeration systems. It is defined as the ratio of the amount of heat removed from the cold side (Q_cool) to the electrical power consumed (W_electric). Mathematically, it is expressed as:
\[\begin{equation}COP = \frac{Q_{cool}}{W_{electric}}\end{equation}\]The higher the COP, the more efficient the refrigerator. For the design task of cooling a canned drink using a car's cigarette lighter, estimating COP helps you determine how effectively the thermoelectric elements use the electrical power to cool the drink. In context, achieving a high COP is critical for a car refrigerator due to limited available power from the cigarette lighter and for energy conservation.
Heat rejection
Heat rejection is a critical aspect of any refrigeration system, including thermoelectric varieties. The objective here is to expel the heat absorbed from the cold side and generated by the Peltier elements to the surroundings as efficiently as possible. Without effective heat rejection, the heat can flow back into the cold side, diminishing the refrigeration effect.

In the context of the design challenge, it is essential to tackle heat rejection without moving parts to maintain simplicity and reliability. Solutions include the use of heat sinks or fins to increase the surface area for dissipating heat. Another recommendation is to enhance natural convection, possibly by strategically placing hot-side elements to drive air flow due to temperature-driven density gradients. Additionally, using materials with high thermal conductivity can significantly improve the effectiveness of the heat rejection strategy.
Thermal conductivity
Thermal conductivity is an intrinsic property of a material that quantifies how well it conducts heat. In a thermoelectric refrigerator, materials with high thermal conductivity are essential on the heat-rejecting side of the Peltier elements. This property ensures that the heat absorbed from the cooled space and generated by the Peltier effect is swiftly transferred to the heatsink or surrounding air.

For an enhanced heat transfer without active components, materials or compounds with high thermal conductivity can be applied between the hot side of the Peltier elements and the heat sinks to minimize thermal resistance. These can include metallic foams, paste, or even aerogels, depending on the design needs and constraints. Overall, selecting appropriate materials for their thermal properties is key to designing an efficient thermoelectric refrigeration system.

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Most popular questions from this chapter

Can a vapor-compression refrigeration system with a single compressor handle several evaporators operating at different pressures? How?

A heat pump operates on the ideal vapor compression refrigeration cycle with \(\mathrm{R}-134 \mathrm{a}\) as the working fluid between the pressure limits of 0.32 and \(1.2 \mathrm{MPa}\). If the mass flow rate of the refrigerant is \(0.193 \mathrm{kg} / \mathrm{s}\), the rate of heat supply by the heat pump to the heated space is \((a) 3.3 \mathrm{kW}\) (b) \(23 \mathrm{kW}\) \((c) 26 \mathrm{kW}\) \((d) 31 \mathrm{kW}\) \((e) 45 \mathrm{kW}\)

The COP of vapor-compression refrigeration cycles improves when the refrigerant is subcooled before it enters the throttling valve. Can the refrigerant be subcooled indefinitely to maximize this effect, or is there a lower limit? Explain

A refrigerator uses refrigerant-134a as the working fluid and operates on the ideal vapor-compression refrigeration cycle except for the compression process. The refrigerant enters the evaporator at \(120 \mathrm{kPa}\) with a quality of 34 percent and leaves the compressor at \(70^{\circ} \mathrm{C}\). If the compressor consumes \(450 \mathrm{W}\) of power, determine \((a)\) the mass flow rate of the refrigerant, ( \(b\) ) the condenser pressure, and ( \(c\) ) the COP of the refrigerator.

An absorption refrigeration system that receives heat from a source at \(95^{\circ} \mathrm{C}\) and maintains the refrigerated space at \(0^{\circ} \mathrm{C}\) is claimed to have a COP of \(3.1 .\) If the environmental temperature is \(19^{\circ} \mathrm{C}\), can this claim be valid? Justify your answer.

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