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Is the efficiency of a thermoelectric generator limited by the Carnot efficiency? Why?

Short Answer

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Answer: The efficiency of a thermoelectric generator is mainly limited by material properties (such as the Seebeck coefficient, electrical conductivity, and thermal conductivity), Joule heating, and irreversible heat transfer. The efficiency is not limited by the Carnot efficiency, since thermoelectric generators are usually less efficient due to the limitations mentioned above.

Step by step solution

01

Define Carnot efficiency

The Carnot efficiency is the maximum efficiency that any heat engine can achieve when operating between two thermal reservoirs at constant temperatures T_hot and T_cold. It is given by the formula: Carnot efficiency = (T_hot - T_cold) / T_hot where T_hot and T_cold are the temperatures of the hot and cold reservoirs, respectively. This efficiency is an upper limit that can be achieved only by an ideal, reversible engine. In practice, all real heat engines are irreversible and thus have efficiencies lower than the Carnot efficiency.
02

Thermoelectric generator

A thermoelectric generator is a type of heat engine that converts heat directly into electrical energy through the Seebeck effect. It is made up of thermoelectric materials, which are characterized by their thermoelectric properties, such as the Seebeck coefficient, electrical conductivity, and thermal conductivity.
03

Factors limiting efficiency for thermoelectric generators

The efficiency of a thermoelectric generator is determined by the properties of the thermoelectric material and the operating conditions (temperature difference and heat transfer rates). Some major factors that limit the efficiency of a thermoelectric generator are: 1. Material properties: The efficiency of a thermoelectric generator is dependent on the thermoelectric properties of the material, which include the Seebeck coefficient, electrical conductivity, and thermal conductivity. It is challenging to find materials with the right combination of these properties to achieve high efficiency. 2. Irreversible heat transfer: In practical applications, there will be losses in the heat transfer process due to irreversibilities, such as conduction, convection, and radiation losses. 3. Joule heating: When an electric current flows through a thermoelectric material, it generates heat due to the resistance of the material. This Joule heating effect reduces the efficiency of the thermoelectric generator.
04

Comparison with Carnot efficiency

While the Carnot efficiency sets an upper limit for the efficiency of heat engines, the efficiency of thermoelectric generators is primarily limited by the material properties and their interplay, which usually results in lower efficiencies than the Carnot efficiency. Therefore, the efficiency of a thermoelectric generator is not limited by the Carnot efficiency, but rather by the factors mentioned above. In summary, the efficiency of a thermoelectric generator is not limited by the Carnot efficiency, but by material properties, Joule heating, and other factors that contribute to its operation. Despite these limitations, thermoelectric generators still offer unique advantages, such as compactness, no moving parts, and the ability to operate on small temperature differences, which makes them ideal for various applications.

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Most popular questions from this chapter

A vapor-compression refrigeration system absorbs heat from a space at \(0^{\circ} \mathrm{C}\) at a rate of \(24,000 \mathrm{Btu} / \mathrm{h}\) and rejects heat to water in the condenser. The water experiences a temperature rise of \(12^{\circ} \mathrm{C}\) in the condenser. The COP of the system is estimated to be \(2.05 .\) Determine \((a)\) the power input to the system, in \(\mathrm{kW},(b)\) the mass flow rate of water through the condenser, and \((c)\) the second-law efficiency and the exergy destruction for the refrigerator. Take \(T_{0}=20^{\circ} \mathrm{C}\) and \(c_{p, \text { water }}=4.18 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\).

A space is kept at \(-15^{\circ} \mathrm{C}\) by a vapor-compression refrigeration system in an ambient at \(25^{\circ} \mathrm{C}\). The space gains heat steadily at a rate of \(3500 \mathrm{kJ} / \mathrm{h}\) and the rate of heat rejection in the condenser is \(5500 \mathrm{kJ} / \mathrm{h}\). Determine the power input, in \(\mathrm{kW}\), the COP of the cycle and the second-law efficiency of the system.

A reversible absorption refrigerator consists of a reversible heat engine and a reversible refrigerator. The system removes heat from a cooled space at \(-15^{\circ} \mathrm{C}\) at a rate of \(70 \mathrm{kW}\) The refrigerator operates in an environment at \(25^{\circ} \mathrm{C}\). If the heat is supplied to the cycle by condensing saturated steam at \(150^{\circ} \mathrm{C},\) determine \((a)\) the rate at which the steam condenses, and (b) the power input to the reversible refrigerator. (c) If the COP of an actual absorption chiller at the same temperature limits has a COP of \(0.8,\) determine the second-law efficiency of this chiller.

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.

Consider a two-stage cascade refrigeration system operating between the pressure limits of 0.8 and 0.14 MPa. Each stage operates on the ideal vapor- compression refrigeration cycle with refrigerant-134a as the working fluid. Heat rejection from the lower cycle to the upper cycle takes place in an adiabatic counterflow heat exchanger where both streams enter at about 0.4 MPa. If the mass flow rate of the refrigerant through the upper cycle is \(0.24 \mathrm{kg} / \mathrm{s}\), determine (a) the mass flow rate of the refrigerant through the lower cycle, \((b)\) the rate of heat removal from the refrigerated space and the power input to the compressor, and \((c)\) the coefficient of performance of this cascade refrigerator.

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