How Do You Calculate Qh for a Non-Carnot Heat Engine?

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SUMMARY

The calculation of Qh for a non-Carnot heat engine involves understanding the relationship between work output and thermal energy extracted. Given a hot reservoir temperature (Thot) of 600K and a cold reservoir temperature (Tcold) of 300K, the efficiency is defined as W/Qh. The discussion confirms that the Carnot efficiency is 0.50, but emphasizes that the engine in question is not a Carnot engine. The relevant equations include Qhot = W + Qcold and efficiency = 1 - (Qcold/Qhot), which can be combined to derive Qh.

PREREQUISITES
  • Understanding of thermodynamic laws, specifically the First and Second Laws of Thermodynamics.
  • Familiarity with the concept of efficiency in thermodynamic systems.
  • Knowledge of heat engine operation and the distinction between Carnot and non-Carnot engines.
  • Ability to manipulate equations involving thermal energy and work output.
NEXT STEPS
  • Study the First Law of Thermodynamics and its application in cyclic processes.
  • Explore the Second Law of Thermodynamics and its implications for heat flow.
  • Learn about the efficiency calculations for various types of heat engines, including non-Carnot engines.
  • Investigate practical examples of non-Carnot heat engines and their performance metrics.
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Students and professionals in thermodynamics, mechanical engineers, and anyone involved in the design or analysis of heat engines seeking to deepen their understanding of efficiency calculations and thermodynamic principles.

physics.stu
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1. Thot reservoir= 600K, Tcold reservoir= 300K, Output= 0.5 J of work for every J of thermal energy extracted from the hot reservoir. What is Qh?



2. efficiency= W/Qh; Carnot efficiency= 1- (Tcold/Thot);



3. I have found the Carnot efficiency to be 0.50. I don't know where to go after this.
 
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physics.stu said:
1. Thot reservoir= 600K, Tcold reservoir= 300K, Output= 0.5 J of work for every J of thermal energy extracted from the hot reservoir. What is Qh?



2. efficiency= W/Qh; Carnot efficiency= 1- (Tcold/Thot);



3. I have found the Carnot efficiency to be 0.50. I don't know where to go after this.

1. what is the 1st law?
2. what is the 2nd law? (the assumption seems to be that you're running a Carnot engine.)
3. what is the definition of efficiency? oops, you have that equation already I see ...

3 equations, 3 unknowns.
 
First Law: dU = Q-W; dU=0.0J because it returns to initial state (cyclic engine)

I am NOT running a Carnot engine, it's the only thing that I could solve for.
Second Law: thermal energy will flow spontaneously from a hot object to cold object but cannot spontaneously flow from cold to hot object.

More on efficiency:
Qhot= W + Qcold
Efficiency= 1-(Qcold/Qhot)
 
physics.stu said:
Output= 0.5 J of work for every J of thermal energy extracted from the hot reservoir
So what is the efficiency?
 
physics.stu said:
First Law: dU = Q-W; dU=0.0J because it returns to initial state (cyclic engine)

I am NOT running a Carnot engine, it's the only thing that I could solve for.
Second Law: thermal energy will flow spontaneously from a hot object to cold object but cannot spontaneously flow from cold to hot object.

More on efficiency:
Qhot= W + Qcold
Efficiency= 1-(Qcold/Qhot)

You're right - you don't need to make the Carnot assumption. Answer Dr Claude's question!
Combine your two equations above and you have the answer!
 
Last edited:

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