Calculating Heat Exhaust from a Heat Engine

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SUMMARY

The discussion focuses on calculating the heat exhaust from a heat engine operating between 40°C and 380°C, with an efficiency of 60% of the theoretical Carnot efficiency. The ideal Carnot efficiency is calculated as 0.52, leading to an operational efficiency of 0.312. The heat exhaust rate is determined to be 41.3 kJ/s, derived from the equation 0.312 = (Qh - Ql) / Qh, where Qh is the heat absorbed at 60 kJ/s.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically heat engines
  • Knowledge of Carnot efficiency calculations
  • Familiarity with the first law of thermodynamics
  • Basic proficiency in unit conversions (kW to kJ/s)
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  • Study the derivation of Carnot efficiency for various temperature ranges
  • Learn about real-world factors affecting heat engine efficiency
  • Explore advanced thermodynamic cycles beyond the Carnot cycle
  • Investigate methods for improving heat recovery in heat engines
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Engineers, thermodynamics students, and professionals involved in energy systems and heat engine design will benefit from this discussion.

flower76
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I'm not sure if I'm interpreting this question correctly

A heat engine operated between 40C and 380C. Being a real engine, it efficiency is only 60% of that theoretically possible for a Carnot engine at these temperatures. If it absorbs heat at a rate of 60kW at what rate does it exhaust heat?

So should I be finding the ideal (Carnot engine) and then taking 60% of that?

ideal = 1-Tl/Th = 1- (313K/653K) = 0.52

so then I figured it runs at 60% of ideal so e should be 0.312

Is this right or is e just 60% ??
 
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You are almost there. You did not answer what the question asked of you, it said what rate does it exhaust heat, not the efficiency.
 
If e is 0.312

then 0.312 = (Qh-Ql)/Qh
0.312 = (60 kJ/s - Ql)/60 kj/s

Ql = 41.3 kj/s So it exhausts at a rate of 41.3 kj/s

Does this look right?
 
seems reasonable.
 

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