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A thermoelectric generator receives heat from a source at \(340^{\circ} \mathrm{F}\) and rejects the waste heat to the environment at \(90^{\circ} \mathrm{F}\). What is the maximum thermal efficiency this thermoelectric generator can have?

Short Answer

Expert verified
Answer: To calculate the maximum thermal efficiency, use the Carnot efficiency formula: efficiency = \(1 - \frac{T_{2K}}{T_{1K}}\), where \(T_{1K}\) is the heat source temperature in Kelvin, and \(T_{2K}\) is the environment temperature in Kelvin. First, convert the temperatures to Kelvin: \(T_{1K} = (340 - 32) \times \frac{5}{9} + 273.15\) \(T_{2K} = (90 - 32) \times \frac{5}{9} + 273.15\) Next, substitute the values and solve for the efficiency: Efficiency = \(1 - \frac{(90 - 32) \times \frac{5}{9} + 273.15}{(340 - 32) \times \frac{5}{9} + 273.15}\) Calculate the efficiency and express the result as a percentage to find the maximum thermal efficiency of the thermoelectric generator.

Step by step solution

01

Convert temperatures to Kelvin

To calculate the Carnot efficiency, we need the temperatures in Kelvin. We can convert Fahrenheit to Kelvin using the following formula: \(T_{K} = (T_{F} - 32) \times \frac{5}{9} + 273.15\). Heat source temperature in Fahrenheit T1 = \(340^{\circ} \mathrm{F}\) Environment temperature in Fahrenheit T2 = \(90^{\circ} \mathrm{F}\) Convert both to Kelvin: \(T_{1K} = (340 - 32) \times \frac{5}{9} + 273.15\) \(T_{2K} = (90 - 32) \times \frac{5}{9} + 273.15\)
02

Calculate Carnot efficiency

Now, we can calculate the maximum thermal efficiency using the Carnot efficiency formula, which is: efficiency = \(1 - \frac{T_{2K}}{T_{1K}}\), where \(T_{1K}\) is the heat source temperature in Kelvin and \(T_{2K}\) is the environment temperature in Kelvin. Efficiency = \(1 - \frac{T_{2K}}{T_{1K}}\)
03

Substitute the values and solve

Substitute the values of \(T_{1K}\) and \(T_{2K}\) that we calculated in step 1, and solve for the maximum thermal efficiency. Efficiency = \(1 - \frac{(90 - 32) \times \frac{5}{9} + 273.15}{(340 - 32) \times \frac{5}{9} + 273.15}\) Calculate the efficiency and express the result as a percentage.
04

Interpret the result

The resulting value represents the maximum thermal efficiency that the thermoelectric generator can achieve. This means that it is the highest possible percentage of heat energy from the heat source that can be converted into useful work, while the remaining energy will be rejected to the environment as waste heat.

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Most popular questions from this chapter

An ideal gas refrigeration cycle using air as the working fluid operates between the pressure limits of 80 and \(280 \mathrm{kPa} .\) Air is cooled to \(35^{\circ} \mathrm{C}\) before entering the turbine. The lowest temperature of this cycle is \((a)-58^{\circ} \mathrm{C}\) \((b)-26^{\circ} \mathrm{C}\) \((c) 5^{\circ} \mathrm{C}\) \((d) 11^{\circ} \mathrm{C}\) \((e) 24^{\circ} \mathrm{C}\)

Consider isentropic compressor of a vaporcompression refrigeration cycle. What are the isentropic efficiency and second-law efficiency of this compressor? Justify your answers. Is the second-law efficiency of a compressor necessarily equal to its isentropic efficiency? Explain.

A heat pump using refrigerant-134a heats a house by using underground water at \(8^{\circ} \mathrm{C}\) as the heat source. The house is losing heat at a rate of \(60,000 \mathrm{kJ} / \mathrm{h}\). The refrigerant enters the compressor at \(280 \mathrm{kPa}\) and \(0^{\circ} \mathrm{C}\), and it leaves at \(1 \mathrm{MPa}\) and \(60^{\circ} \mathrm{C}\). The refrigerant exits the condenser at \(30^{\circ} \mathrm{C}\). Determine \((a)\) the power input to the heat pump, (b) the rate of heat absorption from the water, and (c) the increase in electric power input if an electric resistance heater is used instead of a heat pump.

Why is the reversed Carnot cycle executed within the saturation dome not a realistic model for refrigeration cycles?

Consider a heat pump that 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 \(1.2 \mathrm{MPa}\). The coefficient of performance of this heat pump is \((a) 0.17\) \((b) 1.2\) \((c) 3.1\) \((d) 4.9\) \((e) 5.9\)

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