
#1
Apr813, 10:24 AM

P: 158

I am thinking about the adibatic compression when the gas gets back to Th from Tc... My question is how the p and V gets same as the initial pressure and temperature... Can you prove it?




#2
Apr813, 02:54 PM

P: 864

Your question is hard to understand. Could you please explain it more fully?




#3
Apr913, 05:40 AM

P: 158

Okay... I am changing the pressure of the gas using little pebbles...
The initial p,v is p1,v1... In the first cycle,we expand the gas isothermally at Th by getting rid of some pebbles ... Then we get to p2,v2... In this process Q1 heat is given to the system... The system then adiabatically expands by again losing some pebbles... It now gets to p3,v3 and Tc... Tc is less than Th as the internal energy is doing the work in this process... Now we add some pebbles back and the gas contracts isothermally at Tc... Q2 heat is given to the heat absorber at Tc... It now gets to p4,v4... Now we add some pebbles back and compress the gas adiabatically... We get back to p1 pressure... My ques is how do we get back to V1 volume and Th temperature... How does the three variables synchronize themselves to the initial state... They all get back to the initial state? Its pretty much clear that this happens... Can you prove why this happens?Or am I just missing sth? 



#4
Apr913, 06:38 AM

P: 864

carnot cycle
Suggested translation:
'getting rid of pebbles' = allowing gas to expand and do work. 'adding pebbles' = compressing the gas, so work is done on it. I'm sorry that my answer to your question is not going to be very profound. The experimenter can decide the endpoints (values of p and V) of any change, isothermal or adiabatic. Points 1 and 2 and 3 can be chosen arbitrarily. If you draw an isothermal curve through point 3, and an adiabatic curve through point 1, these curves will intersect at a point '4'. The third stage (isothermal) of the cycle, starting at 3, must be stopped at point 4, as just determined. The final stage, an adiabatic compression starting at 4, will then automatically go through 1. 



#5
Apr913, 07:39 AM

P: 158

You are saying if I determine a certain p1,v1 then p4,v4 will be determined too?
But I didn't say how much work was done on the gas in the isothermal compression... If I do a little less or more, it doesn't seem to matter... I am still confused... 



#6
Apr913, 11:29 AM

P: 864

Not exactly. There's are 'families' of possible Carnot cycles starting at point 1. We can vary the length of the first isothermal and first adiabatic to produce different family members. See thumbnail, which shows, I think, just four of the infinity of possible Carnot cycles starting at 1.




#7
Apr913, 02:32 PM

P: 158

So you are saying 1st and 2nd power stroke determines the cycle... It makes sense cause the isotherm curve at Tc and the adibatic curve from Tc and Th are not dependent on anything I guess except temperature... It also visualizes the equation for efficiency of a carnot engine,why it is only dependent on the Th and Tc...
Another ques... At the isothermal compression if I do not get to the adibatic curve, will I be able to complete a carnot cycle? I guess not... 



#8
Apr913, 02:52 PM

P: 864

Agreed. You won't, then, be able to complete the Carnot cycle starting at your first point.




#9
Apr913, 04:04 PM

P: 158

Thank you for the discussion... :)




#10
Apr913, 04:13 PM

P: 864

It was a pleasure. But I'm still a bit puzzled about where pebbles fit in...



Register to reply 
Related Discussions  
Carnot cycle  Classical Physics  1  
Carnot Cycle  Engineering, Comp Sci, & Technology Homework  3  
The Carnot Cycle  Classical Physics  4  
carnot cycle  Introductory Physics Homework  1  
carnot cycle  Introductory Physics Homework  3 