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[[Special:Contributions/86.3.108.41|86.3.108.41]] ([[User talk:86.3.108.41|talk]]) 15:30, 28 September 2011 (UTC)
[[Special:Contributions/86.3.108.41|86.3.108.41]] ([[User talk:86.3.108.41|talk]]) 15:30, 28 September 2011 (UTC)

I believe that the author of the article sees this cycle as an ideal because it represent the maximum amount of work for a piston steam engine given a boiler pressure and fixed stroke. Steam locomotives use this cycle when starting from a standstill in the station giving the greatest accelerating force to get the train rolling. However it is horribly inefficient as the expansion is done at constant volume which means that at the end of the stroke the high pressure steam is throttled to the atmosphere. As the locomotive approaches its design speed, the full pressure is only applied for part of the stroke and the steam is left to expand to atmospheric pressure within the cylinder in an (ideally) adiabatic process. When the exhaust valve open atmospheric pressure steam is evacuated from the cylinder during the exhaust stroke. This greatly increases the efficiency of the locomotive. The inefficient cycle is only used for a small fraction of the whole journey to quickly get the train up to speed.

Revision as of 03:52, 15 December 2011

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Gravity chimney based cycle

Does anyone know the name of the gravity based cycle used in things like solar chimneys, thunderstorms and very large airconditioning systems? The cycle heats a gas at a constant pressure, expands the gas in a tall chimney (i.e. hundreds of meters tall), cools at a constant pressure and compresses down another tall chimney. I read a paper on it years ago and I can't find any references to it now. njh 00:37, 28 January 2006 (UTC)[reply]


Removed some spam. 76.184.116.146 08:10, 28 April 2007 (UTC)[reply]

Entropy equation untrue!!

There is a severe failure in this article

The integral (integral( dQ/T))over a hole cycle is only zero if it is an ideal cycle but for any other cycle the so called Clausius inequality is true!! where the Integral is smaller or equal zero! —Preceding unsigned comment added by 83.27.25.158 (talk) 11:05, 8 January 2008 (UTC)[reply]

Absorption

"The absorption refrigeration cycle is an alternative that absorbs the refrigerant in a liquid solution rather than evaporating it."

The main difference between the absorption and vapor-compression cycles is the method by which the cycle is driven: vapor-compression cycles are work-driven and absorption cycles are heat-driven. The COMPRESSOR of the vapor-compression cycle is replaced by the absorption components, NOT the evaporator. Both cycles evaporate the refrigerant to provide a cooling heat sink. The article statement quoted above is simply untrue.

Removed material

I removed the 'Demonstration' section which was a highly confusing 'demonstration' that the delta S over a cycle is zero. Looked like it was copied from a textbook or course notes.Cerireid (talk) 20:04, 14 August 2008 (UTC)[reply]

FrauEngineer (talk) 19:04, 19 June 2008 (UTC)[reply]

Help! I'm confused about the Stoddard cycle!

This page says that the Stoddard cycle has 4 stages: adiabatic, isometric, adiabatic, isometric. But the Stoddard engine page says this cycle is adiabatic, *isobaric*, adiabatic, *isobaric*. Can some expert fix the page that's wrong?

Stoddard's 1933 patent --- you can see it on the Stoddard engine page --- says: "It is to be noted that the increase in pressure at the end of the up stroke of the piston is due to the rise in temperature and not to mechanical work". That sounds like an isometric stage, not an isobaric one. John Baez (talk) 17:42, 13 February 2010 (UTC)[reply]

Use of the term "isometric" in the "table of thermodynamic cycles"

Wouldn't it be better to substitute the word "isometric" by "isochoric", since the only occurence of this term is in the said table, while the rest of the article uses the term "isochoric" for processes with constant volume? I couldn't access the table myself, since I'm not used to edit pages on wikipedia, but would somebody mind to change it?

Rankine cycle in the "Some example thermodynamic cycles"

The table of thermodynamic cycles (Some example thermodynamic cycles) in this page, as error the Rankine cycle isn't cycle with internal combustion. This cycle has an external source of heat (external combustion). I would correct the mistake but I don´t how. Apologies for my mistakes.Gonçalo Quintal (talk) 11:51, 3 May 2011 (UTC)[reply]


--79.208.112.246 (talk) 08:59, 22 September 2010 (UTC)[reply]

Agreed, the article should be internally consistent. FYI this table is in Template:Table of thermodynamic cycles; the change was not a problem for any of the other articles it is used in.
 Done: used "isochoric" everywhere David Hollman (Talk) 10:36, 22 September 2010 (UTC)[reply]

Ideal cycle

Why is it this cycle that is chosen to be ideal: that is, what about it is 'ideal'?

(There is no 'ideal cycle' article, which suggests that it is not all that notable).

86.3.108.41 (talk) 15:30, 28 September 2011 (UTC)[reply]

I believe that the author of the article sees this cycle as an ideal because it represent the maximum amount of work for a piston steam engine given a boiler pressure and fixed stroke. Steam locomotives use this cycle when starting from a standstill in the station giving the greatest accelerating force to get the train rolling. However it is horribly inefficient as the expansion is done at constant volume which means that at the end of the stroke the high pressure steam is throttled to the atmosphere. As the locomotive approaches its design speed, the full pressure is only applied for part of the stroke and the steam is left to expand to atmospheric pressure within the cylinder in an (ideally) adiabatic process. When the exhaust valve open atmospheric pressure steam is evacuated from the cylinder during the exhaust stroke. This greatly increases the efficiency of the locomotive. The inefficient cycle is only used for a small fraction of the whole journey to quickly get the train up to speed.