How to show that Carnot engine is more efficient?

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SUMMARY

The discussion focuses on comparing the efficiency of an Ideal gas engine with that of a Carnot engine. The efficiency of the Ideal gas engine is defined by the equation $$e=1-\frac{T_2 - T_L}{T_H - T_1}$$, while the Carnot engine's efficiency is given by $$e=1-\frac{T_L}{T_H}$$. The analysis reveals that as the temperatures approach each other (T1 ~ TH and TL ~ T2), the Ideal gas engine's efficiency approaches an indefinite form, indicating that the Carnot engine is indeed more efficient under these conditions.

PREREQUISITES
  • Understanding of thermodynamic cycles, specifically isobaric and adiabatic processes.
  • Familiarity with the concepts of temperature and efficiency in thermodynamics.
  • Knowledge of the Carnot theorem and its implications for engine efficiency.
  • Basic mathematical skills to manipulate and analyze equations related to thermodynamics.
NEXT STEPS
  • Study the derivation of the Carnot efficiency equation in detail.
  • Explore the implications of adiabatic processes in thermodynamic cycles.
  • Investigate real-world applications of Carnot engines and their efficiency limits.
  • Learn about the differences between Ideal gas engines and real engines, including losses and inefficiencies.
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Students of thermodynamics, mechanical engineers, and anyone interested in understanding the principles of engine efficiency and thermodynamic cycles.

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Homework Statement


Consider an Ideal gas engine with the following cycle:
i. Isobaric expansion (T1 -> Th)
ii. adiabatic expansion (Th -> T2)
iii. Isobaric compression (T2 -> T_L)
iv. adiabatic compression (T_L -> T1)
a. Find its efficiency
b. When operated in two temperature, show that Carnot engine is more efficient

Homework Equations



Efficiency of the Ideal gas engine: $$e=1-\frac{T_2 - T_L}{T_H - T_1}$$

Efficiency of the Carnot Engine: $$e=1-\frac{T_L}{T_H}$$

The Attempt at a Solution



I try to approximate the efficiency of the Ideal gas when T1 is very close to TH and TL is very close to T2, T1~TH, TL~T2, then subtract it to the Carnot Efficiency, if it turns out to become positive then Carnot Engine is more efficient. However I get an Indefinite for the Ideal gas efficiency since the denominator becomes zero.
 
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Useful information: two legs of the cycle for the ideal gas are adiabatic processes between the same pressures.
 
Last edited:

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