Thermal Efficiency and Heat Extraction in Heat Engines: Figure Analysis

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The discussion focuses on calculating the thermal efficiency and heat extraction for a specific heat engine. Participants analyze the relationship between work output (Wout), heat extracted from the hot reservoir (Qh), and heat rejected to the cold reservoir (Qc). The formula Wout = 1/2 B*H is used to derive Wout, while the total heat flow is clarified as Qh = Qc + Wout. Ultimately, the efficiency is calculated as the ratio of Wout to Qh, resulting in a final efficiency value. The conversation emphasizes the importance of understanding energy flow in heat engines for accurate calculations.
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?
 
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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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