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Chapter 19: Problem 27

The maximum steam temperature in a nuclear power plant is \(570 \mathrm{K}\) The plant rejects heat to a river whose temperature is \(0^{\circ} \mathrm{C}\) in the winter and \(25^{\circ} \mathrm{C}\) in the summer. What are the maximum possible efficiencies for the plant during these seasons?

Short Answer

Expert verified
The maximum possible efficiencies for the nuclear power plant during winter and summer should be approximately 52.11% and 47.72% respectively.

Step by step solution

01

Conversion of temperatures to Kelvin

To compare the temperatures appropriately and to apply the formula of Carnot efficiency, all temperatures must be in the same unit. So first, convert the river temperatures in the winter and the summer from Celsius to Kelvin. The conversion from Celsius to Kelvin can be done by adding 273 to the given temperatures. Therefore, \(0^{\circ} \mathrm{C}=0+273=273 \mathrm{K}\) and \(25^{\circ} \mathrm{C}=25+273=298 \mathrm{K}\)
02

Calculation of Carnot Efficiency

Now, calculate the Carnot efficiency using the formula as \[\eta = 1 - \frac{Tc}{Th}\] where \( \eta \) is the efficiency, Tc is the temperature of the cold reservoir (the river), and Th is the temperature of the hot reservoir (the nuclear plant). Compute this for both winter and summer. For winter, \[ \eta_w = 1 - \frac{273}{570} \] and for summer, \[ \eta_s = 1 - \frac{298}{570} \]
03

Convert to percentage

As efficiency is often expressed as a percentage, convert the efficiency from step 2 to percentages by multiplying by 100. So, the efficiency for winter by be calculated as \[ \eta_w \% = \eta_w \times 100 \] and the efficiency for summer be calculated as \[ \eta_s \% = \eta_s \times 100 \]

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Most popular questions from this chapter

Refrigerators remain among the greatest consumers of electrical energy in most homes, although mandated efficiency standards have decreased their energy consumption by some \(80 \%\) in the past four decades. In the course of a day, one kitchen refrigerator removes \(30 \mathrm{MJ}\) of energy from its contents, in the process consuming \(10 \mathrm{MJ}\) of electrical energy. The electricity comes from a \(40 \%\) efficient coal-fired power plant. The total energy rejected to the surrounding kitchen during the course of the day is a. \(10 \mathrm{MJ}\). b. \(30 \mathrm{MJ}\). c. \(40 \mathrm{MJ}\). d. \(75 \mathrm{MJ}\).

Could you cool the kitchen by leaving the refrigerator open? Explain.

Refrigerators remain among the greatest consumers of electrical energy in most homes, although mandated efficiency standards have decreased their energy consumption by some \(80 \%\) in the past four decades. In the course of a day, one kitchen refrigerator removes \(30 \mathrm{MJ}\) of energy from its contents, in the process consuming \(10 \mathrm{MJ}\) of electrical energy. The electricity comes from a \(40 \%\) efficient coal-fired power plant. The electrical energy a. is used to run the light bulb inside the refrigerator. b. wouldn't be necessary if the refrigerator had enough insulation. c. retains its high-quality status after the refrigerator has used it. d. ends up as waste heat rejected to the kitchen environment.

Use energy-flow diagrams to show that the existence of a perfect heat engine would permit the construction of a perfect refrigerator, thus violating the Clausius statement of the second law.

Name some irreversible processes that occur in a real engine.
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