Heat Cycles, Work, and Efficiency

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Homework Help Overview

The discussion revolves around heat cycles, work, and efficiency in thermodynamic processes. Participants are examining the equations related to work done in a system and the efficiency of that work in relation to heat transfer.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to apply the work equation W=-∫PdV and are questioning the correctness of their integration methods. There are inquiries about alternative equations for calculating work based on the provided diagram.

Discussion Status

Some participants have pointed out errors in the integration process and are seeking clarification on the appropriate equations to use. Guidance has been offered regarding the integral of PdV, with suggestions to align calculations with the diagram presented.

Contextual Notes

There is an emphasis on ensuring that the calculations are consistent with the visual representation of the process, and participants are working within the constraints of the problem as defined by the homework assignment.

Marcin H
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Homework Statement


New Doc 13_1.jpg


Homework Equations


W=-∫PdV
ε=W/Q
PV=nRT
COP=Q/W


The Attempt at a Solution


Using the first equation for work, I got W=-nrT*ln(3V1/V1) but I don't think that is right. Is this the wrong formula to use for this problem? My answer should be in terms of P1 and V1.
 
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That is not the correct integration of PdV for the process shown in the figure.
 
Chestermiller said:
That is not the correct integration of PdV for the process shown in the figure.
What equation can I use to find the work then?
 
Marcin H said:
What equation can I use to find the work then?
You can use the integral of PdV. Just do it in a way that is consistent with what you see in the diagram.
 
Chestermiller said:
You can use the integral of PdV. Just do it in a way that is consistent with what you see in the diagram.
I just took the area of the rectangle. So W=2P1V1. I got an efficiency of 17% with that.
 

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