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Topic: Carnot Cycle  (Read 11366 times)

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Offline renai

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Carnot Cycle
« on: July 24, 2007, 10:18:32 PM »
Just sharpening up on my thermo again.  In the carnot cycle, since the first two stages involve successive pressure drops, 1st from isothermal expansion, 2nd from adiabatic expansion, am I to understand that at the end of those two steps, when the piston has stopped moving, i.e. expansion ends, that the pressure of the gas in the chamber is exactly equal to atmospheric pressure and that therefore at the beginning of the cycle the gas must first quickly be heated to a pressure above atmospheric.

If this is the case then I'm assuming that the work calculation done cannot involve an equation of state substitution since the pressure in the chamber IS NOT equal to the external pressure it's doing work against.  Unless I'm totally wrong.

Can anyone clarify?

Cheers,
Renai

Offline enahs

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Re: Carnot Cycle
« Reply #1 on: July 25, 2007, 10:28:34 AM »
You contradicted your self.

In your first paragraph you say
Quote
that the pressure of the gas in the chamber is exactly equal to atmospheric pressure
Which is not true.

But then in your second paragraph you say
Quote
since the pressure in the chamber IS NOT equal to the external pressure it's doing work against


I am not really sure what you are asking then. But the first compression stroke of the piston is isothermal, so no it does not heat up quickly first.


*edit* What ygg said sounds much better.
« Last Edit: July 25, 2007, 04:51:37 PM by enahs »

Offline Yggdrasil

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Re: Carnot Cycle
« Reply #2 on: July 25, 2007, 03:41:43 PM »
To extract the maximum efficiency from the Carnot cycle, every step must be reversible.  So, for an ideal carnot cycle, the external pressure is always equal to the internal pressure and you can use an equation of state to calcuclate the Pexternal in the work calculation.

Offline renai

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Re: Carnot Cycle
« Reply #3 on: July 26, 2007, 10:36:26 PM »
Sorry for not being clear.  What I meant to say was that since the Carnot process is reversible (i.e. internal and external pressure are always equal), and the pressure during the first two stages is dropping (first isothermally, then adiabatically) then that means that the external pressure is also dropping.  That's what I don't quite get.  I keep on thinking of the gas expanding against a constant pressure but clearly it isn't.  Can you then explain to me why the external pressure is dropping.  I'm probably just not visualising what's going on.  I keep on imagining a gas in a chamber expanding against the constant combined pressrue of a piston weight (if its vertical) and/or the atmosphere.

Renai

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