Thermodynamics and Steam Engine

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SUMMARY

The discussion focuses on calculating the heat discharged per second from a steam engine operating at half of its theoretical maximum efficiency (εmax) while performing work at a rate of W J/s. The key equations used include WMax = εMax * QH and W = QH - QC. The final expression derived for the heat discharged is QC = (2W (1 - (εMax / 2))) / εMax. The discussion highlights a potential misunderstanding regarding the expression of maximum efficiency in terms of temperatures TC and TH.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically the first and second laws of thermodynamics.
  • Familiarity with the concepts of heat transfer and work in thermodynamic systems.
  • Knowledge of efficiency calculations in heat engines, particularly the definition of maximum efficiency.
  • Ability to manipulate algebraic equations to solve for unknown variables in thermodynamic contexts.
NEXT STEPS
  • Research the derivation of maximum efficiency in steam engines using Carnot's theorem.
  • Learn about the relationship between temperatures TC and TH in the context of thermodynamic cycles.
  • Explore the implications of operating below maximum efficiency on thermal performance and energy loss.
  • Investigate real-world applications of steam engines and their efficiency metrics in industrial settings.
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Students studying thermodynamics, engineers working with steam engines, and anyone interested in the efficiency of thermal systems.

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Homework Statement


If a steam engine operates at half of its theoretical maximum efficiency (emax) and does work at a rate of W J/s, calculate how much heat is discharged per second.


Homework Equations


WMax = \epsilon_{Max}*Q_{H}
W = QH-QC


The Attempt at a Solution


W = \frac{1}{2}*\epsilon_{Max}*Q_{H}

\frac{2W}{\epsilon_{Max}} = Q_{H}

W = Q_{H} - Q_{C}

W = \frac{2W}{\epsilon_{Max}} - Q_{C}

\epsilon_{Max} *W = 2W - \epsilon_{Max} * Q_{C}

\epsilon_{Max} *W - 2W= - \epsilon_{Max} * Q_{C}

W(\epsilon_{Max} - 2) = - \epsilon_{Max} * Q_{C}

\frac{W(\epsilon_{Max} - 2)}{- \epsilon_{Max}} = Q_{C}

Answer = \frac{2W (1-\frac{\epsilon_{Max}}{2})}{ \epsilon_{Max}}

Where did i go wrong?
 
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What is the given answer?
 
Last edited:
Maybe they want you to express the maximum efficiency in terms of temperatures Tc and Th.

AM
 

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