Thermal Efficiency and Heat Extraction in Heat Engines: Figure Analysis

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Homework Help Overview

The discussion revolves around analyzing the thermal efficiency and heat extraction of a heat engine, as depicted in a provided figure. Participants are exploring the relationships between work output, heat extracted from the hot reservoir, and the cold reservoir in the context of thermodynamics.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to calculate the thermal efficiency and the heat extracted from the hot reservoir. Questions arise regarding the significance of heat flows and how to correctly determine the values for Qc and Qh. There is also discussion about whether to sum the heat leaving the engine.

Discussion Status

The conversation is ongoing, with some participants providing hints and guidance regarding the relationships between the quantities involved. There is an exploration of different interpretations of the heat flows, and while some calculations have been proposed, no consensus has been reached on the correct approach yet.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available for discussion. There is a mention of a note regarding the sum of heat leaving the engine, which is being questioned.

Ravey
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1. What is the thermal efficiency for the heat engine shown in the figure?
What is the heat extracted from the hot reservoir for the heat engine shown in the figure?t

Figure shown here:
knight_Figure_19_16.jpg




2.Wout = 1/2 B*H Also, Qc=Qh-Wout, efficiency: Wout/Qh



3. Wout = .5 * 200KPA* (400x10^-6m^3). I have no idea how to obtain Qc (cold reservoir) or Qh (hot)
 
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Hi Ravey, welcome to PF. What's the significance of the heat flows marked on the image?
 
Thanks!

I believe that is heat leaving the engine*, but what do I do? Add them? According to some note Qc is the sum of the heat leaving the engine. So I assume so. However, qh is not 280-40 = 240. What's going wrong?
 
Last edited:
Stumped too I see ;)
 
Hint: No energy is stored in the engine from cycle to cycle.
 
Got it!


Qh=Q_c+W_out = 280J + 40J = 320J.

40/320 = efficiency.
 
Sounds good!
 

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