Thermodynamics energy efficiency

Therefore, the efficiency of the plant is 700/2000 = 0.35, or 35%. In summary, a nuclear power plant generates 2000 MJ of thermal energy each second, which is used to produce 700 MJ of electric power with a maximum possible efficiency of 53% and an actual efficiency of 35%.
  • #1
donkey11
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Homework Statement


Each second, a nuclear power plant generates 2000 MJ of thermal energy from nuclear reactions in the reactors core. This energy is used to boil water and produce high pressure steam at 573 Kelvins, the steam spins a turbine, which produces 700 MJ of electric power, then the steam is condensed and water is cooled to 303 kelvins before starting cycle again.
a) maximum possible efficiency of plant?
b) plant's actual efficiancy?



Homework Equations


emax= 1 - (Tc/Th)
e= what you get/ what you pay



The Attempt at a Solution



a)emax= 1- (303/573) = 0.53
b) e= work out/Qh = Qh-Qc/Qh = 1300/2000 = 0.65
but this is larger then my max!
 
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  • #2
Shouldn't the plant's efficiency be: 700/2000 ?

The electric power produced is 700MJ. That's what you get. What you pay is 2000 MJ.
 
  • #3


I would like to clarify a few things about the efficiency of a power plant. First, the maximum possible efficiency of any power plant is determined by the laws of thermodynamics and is known as the Carnot efficiency, which is given by the equation emax= 1 - (Tc/Th), where Tc is the temperature of the cold reservoir (in this case, 303 K) and Th is the temperature of the hot reservoir (573 K). This means that the maximum possible efficiency of the power plant in this scenario is 1- (303/573) = 0.47 or 47%.

Secondly, the actual efficiency of a power plant is typically lower than the maximum possible efficiency due to various factors such as energy losses, heat transfer inefficiencies, and mechanical losses. In this case, the plant's actual efficiency can be calculated by using the equation e= work out/Qh, where work out is the work output (700 MJ) and Qh is the heat input (2000 MJ). This gives an efficiency of 700/2000 = 0.35 or 35%.

It is important to note that the actual efficiency of a power plant can vary depending on several factors such as the design, maintenance, and operating conditions. Therefore, it is crucial to continuously monitor and improve the efficiency of power plants to ensure maximum energy efficiency and minimize energy wastage.
 

FAQ: Thermodynamics energy efficiency

What is thermodynamics energy efficiency?

Thermodynamics energy efficiency is the ratio of the output energy produced by a system to the input energy it receives. It is a measure of how well a system converts energy into useful work.

How is thermodynamics energy efficiency calculated?

Thermodynamics energy efficiency is calculated by dividing the output energy by the input energy and multiplying by 100 to get a percentage. It is important to note that this calculation is only valid for closed systems, where no energy is lost to the surroundings.

Why is thermodynamics energy efficiency important?

Thermodynamics energy efficiency is important because it allows us to understand and improve the performance of various systems, such as engines, power plants, and refrigerators. It also helps us conserve energy and reduce waste, which is crucial for sustainable development.

How can thermodynamics energy efficiency be improved?

Thermodynamics energy efficiency can be improved by minimizing energy losses in a system, such as reducing friction and heat transfer. It can also be improved by using more efficient materials and technologies, and by optimizing the design and operation of a system.

Is it possible to have 100% thermodynamics energy efficiency?

No, it is not possible to have 100% thermodynamics energy efficiency. This is because some energy will always be lost to the surroundings, usually in the form of heat. However, engineers and scientists continue to strive towards achieving higher and higher levels of energy efficiency in various systems.

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