Calculating Efficiency of a Heat Engine: Converting Celsius to Kelvin

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SUMMARY

The discussion focuses on calculating the efficiency of a heat engine by converting Celsius to Kelvin using the formula eff = [1 - (T cold / T hot)] * 100. The example provided illustrates the conversion of 320°C to 593K and 610°C to 883K, resulting in an efficiency calculation of approximately 32.8%. Participants clarified the correct conversion and formula application, emphasizing that real engines will exhibit lower efficiency than the ideal calculations suggest.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically heat engines.
  • Familiarity with temperature conversion from Celsius to Kelvin.
  • Knowledge of efficiency calculations in thermodynamic systems.
  • Basic mathematical skills for applying formulas.
NEXT STEPS
  • Research the Second Law of Thermodynamics and its implications on heat engine efficiency.
  • Learn about real-world factors affecting heat engine performance.
  • Explore different types of heat engines, such as Carnot and Rankine cycles.
  • Study advanced efficiency calculations for non-ideal heat engines.
USEFUL FOR

Students studying thermodynamics, engineers designing heat engines, and anyone interested in the principles of energy conversion and efficiency analysis.

cowgiljl
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when i am calculating the eggiciency of a heat engine i have to convert the celicus to kelvis and used this formula

eff = 1-(T cold/T hot)*100

fro example 302C = 593K 610C = 883K

eff = 1-593/883 = about 32.8% efficient

right ?

thanks joe
 
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Originally posted by cowgiljl
eff = 1-(T cold/T hot)*100
You mean: eff = [1-(T cold/T hot)]*100.

This is the efficiency of an ideal (reversible) heat engine; real engines will be less efficient.
...
eff = 1-593/883 = about 32.8% efficient
Right.
 
Originally posted by cowgiljl
fro example 302C = 593K

Nope, 302+273=575.
 
it was 320 not 302 but thanks
 


Originally posted by Tom
Nope, 302+273=575.
Good catch, Tom. :smile:
 

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