Thermodynamics energy efficiency

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SUMMARY

The discussion centers on calculating the energy efficiency of a nuclear power plant that generates 2000 MJ of thermal energy and produces 700 MJ of electric power. The maximum possible efficiency, calculated using the formula emax = 1 - (Tc/Th), is determined to be 0.53. The actual efficiency is calculated as e = (Qh - Qc) / Qh, resulting in an efficiency of 0.65. This value exceeds the maximum efficiency, leading to confusion regarding the correct interpretation of the efficiency metrics.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically the laws of thermodynamics.
  • Familiarity with efficiency calculations in thermal systems.
  • Knowledge of the Carnot efficiency formula and its application.
  • Basic concepts of nuclear energy generation and power conversion processes.
NEXT STEPS
  • Study the Carnot efficiency and its implications in real-world systems.
  • Research the differences between thermal efficiency and overall efficiency in power plants.
  • Explore the impact of temperature differences on efficiency in thermodynamic cycles.
  • Learn about the specific operational parameters of nuclear power plants and their efficiency metrics.
USEFUL FOR

Students studying thermodynamics, engineers involved in energy systems, and professionals in the nuclear energy sector seeking to understand efficiency calculations and thermodynamic principles.

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!
 
Physics news on Phys.org
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.
 

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