Thermodynamics problem, Heat Engine with two heat sinks

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SUMMARY

The discussion revolves around calculating the maximum thermal efficiency of a heat engine with one heat source at 1000 K and two heat sinks at 200 K and 300 K, respectively. The thermal efficiency is derived using the Carnot efficiency formula: η = 1 - TC/TH. The user successfully applied the entropy balance equation and acknowledged the importance of considering all three temperatures in the calculations. The final solution involves determining the heat rejected by each sink and using these values to compute the overall maximum efficiency.

PREREQUISITES
  • Understanding of Carnot efficiency and its formula
  • Familiarity with the principles of thermodynamics, specifically the first and second laws
  • Knowledge of entropy balance equations in thermodynamic systems
  • Ability to perform calculations involving multiple heat sinks and sources
NEXT STEPS
  • Study the derivation and application of the Carnot efficiency formula in various thermodynamic scenarios
  • Learn about entropy balance equations and their role in thermodynamic analysis
  • Explore the implications of multiple heat sinks in heat engine design
  • Investigate real-world applications of thermodynamic principles in heat engines
USEFUL FOR

This discussion is beneficial for thermodynamics students, engineers designing heat engines, and anyone interested in optimizing thermal efficiency in engineering applications.

TeddyLu
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I'm having a hard time setting up an equation for a heat engine problem with one heat source and two heat sinks given only the temperature of the heat source and temperature of the two heat sinks as:

TH = 1000 K
TC1 = 200 K
TC2 = 300 K

It is given that the two heat rejected are of equal value.

determine the maximum thermal efficiency.

Homework Equations


thermal efficiency (carnot) = 1 - TC/TH = 1 - QC/QH

The Attempt at a Solution


Since there was two heat sinks with an equal value of heat rejected at each:
QC1 = QC2 = QC

therefore, QH = QC1 + QC2 will turn into
QH = 2QC

I took the entropy balance equation to solve for QC:

0 = QH/1000 - QC/200 - QC/300

but I don't have a constant on the other side to figure out for QC to plug back into find QH and then solve for thermal efficiency.

any help please?
 
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Consider 2 systems First one with heat source at 1000 K and heat sink at 300 K. Find out the maximum thermal efficiency. For the second one, consider the source at 300 K and sink at 200 K. Calculate the maximum efficiency. Multiply the two to get the overall maximum efficiency.

I kept wondering why the equal rejected fact is mentioned. Couldn't figure it out. I'm not too sure this method is correct. Verify it from someone.
 
TeddyLu said:
therefore, QH = QC1 + QC2 will turn into
QH = 2QC
How did you get this? What happened to W?

You can use the 1st and 2nd laws to solve this problem. The answer is a function of all three temperatures, as you might guess.
I would start from 1st principles rather than attempt to squeeze the Carnot law into the problem. A Carnot cycle runs between 2 temperatures only, by definition.
 
@siddharth23
Thank you for your reply! I found out that rude man's suggestion below helped with finding the answer to my question.

@rude man
Thank you also for your reply. To answer your question, I used the entropy balance equation to comes to that conclusion. There was no W stated so I excluded it from the problem.
I followed your instructions and was able to formulate an equation to find out the maximum carnot efficiency. The use of all three temperatures and made sure to take half of the heat rejected (QC1 & QC2) and solved for QC/QH to plug back into my equation for the carnot efficiency.

Appreciate all the help for this problem!
 
TeddyLu said:
@siddharth23
Thank you for your reply! I found out that rude man's suggestion below helped with finding the answer to my question.

@rude man
Thank you also for your reply. To answer your question, I used the entropy balance equation to comes to that conclusion. There was no W stated so I excluded it from the problem.
I followed your instructions and was able to formulate an equation to find out the maximum carnot efficiency. The use of all three temperatures and made sure to take half of the heat rejected (QC1 & QC2) and solved for QC/QH to plug back into my equation for the carnot efficiency.

Appreciate all the help for this problem!
Good. So what was your answer?
 

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