What is the efficiency of this reversible engine?

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SUMMARY

The discussion centers on calculating the efficiency and work output of a reversible engine operating with 0.5 mol of an ideal monatomic gas. The engine cycle consists of constant volume and constant pressure segments, resulting in an efficiency of 13.8%. The work done by the engine in one cycle was calculated as 0.0036499 J, while the heat input (Q(in)) was determined to be 0.0265285 J. The calculations were verified through the efficiency formula: efficiency = work(out)/Q(in).

PREREQUISITES
  • Understanding of ideal gas laws and monatomic gases
  • Familiarity with thermodynamic cycles and PV diagrams
  • Knowledge of efficiency calculations in thermodynamics
  • Ability to perform calculations involving work and heat transfer
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  • Study the derivation of the efficiency formula for thermodynamic cycles
  • Learn about the characteristics of ideal monatomic gases
  • Explore the implications of isobaric and isochoric processes in thermodynamics
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Students studying thermodynamics, engineers working with heat engines, and anyone interested in the principles of energy conversion and efficiency in mechanical systems.

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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