What is the efficiency of this reversible engine?

AI Thread Summary
The discussion revolves around calculating the efficiency of a reversible engine operating with an ideal monatomic gas through a defined cycle. The efficiency was determined to be 13.8%, calculated using the formula efficiency = work(out)/Q(in). There was confusion regarding the work done by the engine in one cycle, as the user initially equated it to work(out) used for efficiency but found it was not accepted as the correct answer. The output work was calculated as 0.0036499, with Q(in) being 0.0265285. Clarification on the distinction between work done and work output is sought to resolve the misunderstanding.
GeorgeCostanz
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Homework Statement



A reversible engine contains 0.5 mol of an ideal monatomic gas and operates using a cycle, similar to that shown in Fig. 19.21 on page 324 of the textbook, consisting of constant volume and constant pressure segments. (You will need to draw your own diagram using the data that follow.)

Starting at the lower left corner, the first part of the cycle goes from 1.62 atm to 2.75 atm at a constant volume of 0.00255 m3. Next, we have an isobaric expansion (at 2.75 atm) from 0.00255 m3 to 0.00578 m3. In the third segment of the cycle, the pressure falls from 2.75 atm to 1.62 atm at a constant volume of 0.00578 m3, while the final part of the cycle is an isobaric compression (at 1.62 atm) to 0.00255 m3.

(a) Determine the efficiency for this cycle. Give your answer as a percentage (do not enter units).

(b) How much work is done by this engine in one cycle?

Homework Equations



efficiency = work(out)/Q(in)

The Attempt at a Solution



I sketched my PV diagram and worked the problem out completely. I got e = 13.8% which was right.

Part b asked for work done by the engine in one cycle. Isn't that just work(out) which I used to find efficiency? Apparently not because it's not accepting that answer.

Any help would be appreciated, thanks
 
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GeorgeCostanz said:
Part b asked for work done by the engine in one cycle. Isn't that just work(out) which I used to find efficiency? Apparently not because it's not accepting that answer.

Any help would be appreciated, thanks
What was your answer for the output work?

AM
 
w(out) was .0036499

Q(in) was .0265285
 
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