Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Physics
Classical Physics
Thermodynamics
Temperatures of the hot source and cold sink in a heat engine
Reply to thread
Message
[QUOTE="FranzDiCoccio, post: 6267562, member: 141434"] Hi, I was just wondering about the efficiency of a cycle that is not Carnot cycle. In that case one should use [itex]\eta = 1-\left|\frac{Q_{\rm out}}{Q_{\rm in}}\right|[/itex], where [itex]Q_{\rm in}[/itex] and [itex]Q_{\rm out}[/itex] are the amounts of heat absorbed and released during the cycle. For instance, I guess that in an (ideal) Stirling cycle (isochoric - isothermal expansion - isochoric - isothermal compression) [itex]Q_{\rm in}[/itex] involves the isochoric transformation with increasing temperature and the isothermal expansion, while [itex]Q_{\rm out}[/itex] involves the remaining two transformations. I quickly browsed online and I actually found something like that. Now the usual diagram for a generic machine (at least in my textbook) seems to suggest that [itex]Q_{\rm in}[/itex] is absorbed from a heat resevoir at some temperature [itex]T_h[/itex] and [itex]Q_{\rm out}[/itex] released into a heat sink at a lower temperature [itex]T_c<T_h[/itex]. See e.g. this one I found on the web, which is similar to the ones in my textbook [CENTER][ATTACH=full]253391[/ATTACH][/CENTER]But, in the case of the Stirling cycle, what does it really mean that [itex]Q_{\rm in}=Q_{\rm isothermal}+Q_{\rm isochoric}[/itex] is absorbed "at [itex]T_h[/itex]"? It seems to me that this could apply to the isothermal part only, because the heat intake in the isochoric process does not happen at a fixed temperature...Are all of these diagrams actually referring to a Carnot engine? Or perhaps this has to do with the fact that any cycle can be decomposed into many Carnot cycles, so ultimately the diagrams assume that the heat exchanges happen at fixed temperatures? Thanks a lot for any insight. Franz [/QUOTE]
Insert quotes…
Post reply
Forums
Physics
Classical Physics
Thermodynamics
Temperatures of the hot source and cold sink in a heat engine
Back
Top