Solving Heat Engine & Refrigerator Problems

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SUMMARY

The discussion focuses on solving problems related to a Carnot heat engine operating between temperatures T1=400K and T2=300K, with a heat input (Qin) of 1200 J. Key calculations include determining the heat output (Qout) to the low-temperature reservoir, the work done by the engine, and the performance coefficient of the refrigerator when operated in reverse. The efficiency of the engine is calculated using the formula e=W/QH, while the performance coefficient (COP) for the refrigerator is derived from the Carnot cycle equations. The correct interpretation of sign conventions and equations is emphasized to avoid confusion in calculations.

PREREQUISITES
  • Understanding of Carnot cycle principles
  • Familiarity with thermodynamic equations, specifically QH + QC + W = 0
  • Knowledge of efficiency calculations for heat engines
  • Concept of performance coefficient (COP) for refrigerators
NEXT STEPS
  • Study the derivation of Carnot cycle efficiency
  • Learn about the implications of the second law of thermodynamics
  • Explore real-world applications of heat engines and refrigerators
  • Investigate the impact of temperature differences on engine performance
USEFUL FOR

Students studying thermodynamics, engineers working with heat engines and refrigeration systems, and anyone interested in the practical applications of the Carnot cycle in energy efficiency.

Ki-nana18
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Homework Statement


A Carnot heat engine operates between temperatures T1= 400K and T2= 300K. In each cycle the engine receives Qin=1200 J from the high temperature reservoir.
a) Calculate the heat Qout delivered to the low temperature reservoir.
b) Suppose the engine is operated in reverse as a refrigerator. Now the engine receives Qin= 1200 J from the low temperature reservoir. Determine the heat delivered to the high temperature reservoir.
c) Calculate the work done by the engine in part a, and the work done on the refrigerator in part b. Note that both answers should be positive.
d) Find the efficiency of the engine in part a, and the performance coefficient of the refrigerator in part b.

Homework Equations


QH+QC+W=0
##\frac{Q_H} {T_H}##+##\frac{Q_C} {T_C}##=0
e=W/QH
performance coefficient=##\frac{|Q_c|} {W}##

The Attempt at a Solution


[/B]
My confusion arises when I get negative values for part c. Mathematically they should be negative, I suspect that I am not grasping a concept somewhere.
20141004_152957.jpg
 
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Let's take the case for the engine first. Look at the diagram you drew at the top. Heat flows from the hot reservoir and a part of it is converted into work done by the engine, while the rest is dumped into the cold reservoir. This statement, translated into mathematical language says ##Q_h=W+Q_c## where ##Q_h,~Q_c,~W## are defined analogously from the above sentence with positive values corresponding to the bolded words. Does this equation make sense to you in terms of the language that it is translated to? Does the sign convention make sense to you?

Now look at the equation you provided in your "relevant equations". You have the quite different equation ##Q_h+Q_c+W=0##, so again, looking at the picture you drew, what English sentence do you think this equation gets translated to?
 
Your value for COP, .3333, is not correct. Since it is a Carnot cycle, COP = |Qc/W| = |Qc/(Qh-Qc)| = |Tc/(Th-Tc)|

AM
 

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