Comparing Carnot efficiency to Stirling efficiency

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SUMMARY

The discussion focuses on comparing the efficiencies of a Stirling cycle and a Carnot cycle operating between temperatures Th = 700K and Tc = 400K. The efficiency of the Stirling cycle is defined as η = W/Qpos, where W is the work done and Qpos is the total positive heat flow. The Carnot efficiency is calculated using the formula 1 - Tc/Th. The participant encountered difficulties in deriving the correct efficiency, indicating potential miscalculations related to the molar heat capacity (Cv = 5/2 * R) and the number of moles of gas involved in the process.

PREREQUISITES
  • Understanding of thermodynamic cycles, specifically Stirling and Carnot cycles.
  • Familiarity with the concepts of work and heat in thermodynamics.
  • Knowledge of the ideal gas law and molar heat capacities.
  • Ability to perform integrals and logarithmic calculations in thermodynamic equations.
NEXT STEPS
  • Research the derivation of work done in thermodynamic cycles using the integral PdV.
  • Study the implications of the number of moles on heat capacity calculations in thermodynamics.
  • Learn about the efficiency calculations for different thermodynamic cycles, including real-world applications.
  • Explore the differences between monatomic and diatomic gases in the context of heat capacity and efficiency.
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Students and professionals in thermodynamics, mechanical engineers, and anyone interested in the efficiency of thermal engines and cycles.

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Homework Statement



In the problem, we were to derive work and heat for each leg of the cycle. Then we are asked:

In the definition of the efficiency of this cycle, η =W Qpos , where Qpos is the total positive heat flow to the engine, what is the efficiency of the cycle when Th = 700K and
Tc = 400K? V1 = 0.5 L and V2 = 1.5 L. Compare this efficiency to the efficiency of a Carnot cycle that operates between the same temperature extremes.

Homework Equations



Work = integral from Vi to Vf of PdV
W done during cycle = nr(Th-Tc)ln(vf/vi) Derivation here:

http://www.pha.jhu.edu/~broholm/l39/node5.html

Q = Cv(Th -Tc) - nRT ln (Vf/Vi)

We are told Cv = 5/2 * R

efficiency of a carnot cycle = 1- Tc/Th

The Attempt at a Solution



I have attempted to get the solution but have come up with a result that is about an order of magnitude too small that i am unable to rectify. It is part c in the attachment.
 

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Last edited by a moderator:
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CV = 5R/2 is the molar heat capacity at constant volume for a diatomic gas. You need to take account the number of moles when using CV. (You didn't specify if the number of moles is given or whether the gas is monatomic or diatomic.)
 

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