Expansion in non-adiabatic Otto / Diesel cycles

[racctor]

TL;DR Summary

How can temperature be conserved while pressure drops in non adiabatic expansions/compressions?

Hello!

I have a question that has been bothering me for a while now. If we look at the expansion step of a real otto or diesel cycle we see that while the pressure drops to near surrounding levels the temperature remains relatively high ( high T of the exhaust gas). How is that possible? How can energy enter the system that increases the temperature, but not the pressure? Or in other words, how can energy dissipate into kinetic energy of the gas but not increase the pressure?

I think the same problem occurs for me in the compression step of real heat pumps: If i add work to a real system to reach a certain pressure, how come the temperature of real systems is higher than in the ideal isentropic compression?

If we take into account that in those examples we experience some heat loss to the surroundings, wouldn't that just decrease the mechanical Work i can get out of the system, but since heat is lost, the temperature drop should be the same as the pressure drop?

Your help is very much appreciated
kind regards

 

[jack action]

Or in other words, how can energy dissipate into kinetic energy of the gas but not increase the pressure?

Because the density has also changed. $P = \rho RT$, there are 3 variables in the ideal gas law.

If i add work to a real system to reach a certain pressure, how come the temperature of real systems is higher than in the ideal isentropic compression?

Because of heat added by friction or other inefficiencies.

wouldn't that just decrease the mechanical Work i can get out of the system

It does.

but since heat is lost, the temperature drop should be the same as the pressure drop?

Again, there is a density change to consider as well.

 

[PeterDonis]

How can energy enter the system that increases the temperature, but not the pressure?

Energy isn't entering the system during the expansion phase. It's leaving the system, because the system is doing work.

 

[PeterDonis]

how can energy dissipate into kinetic energy of the gas but not increase the pressure?

Energy isn't dissipating into kinetic energy of the gas during the expansion phase (except for a small amount due to friction losses, but in the idealized case these are zero). It's being used to do work.

 

[racctor]

Because the density has also changed. $P = \rho RT$, there are 3 variables in the ideal gas law.

Because of heat added by friction or other inefficiencies.

It does.

Again, there is a density change to consider as well.

Thanks for your answer
How can the density change if the number of particles stays the same?
Friction or heat loss to surroundings would only increase the T of the surroundings though?

kind regards

 

[PeterDonis]

How can the density change if the number of particles stays the same?

Um, because the volume is changing due to expansion?

 

[racctor]

Energy isn't dissipating into kinetic energy of the gas during the expansion phase (except for a small amount due to friction losses, but in the idealized case these are zero). It's being used to do work.

Let me rephrase it: How can the temperature still be so high although the work has already been done ( and pressure has dropped to near atm level)? High temperature should still cause high pressure?

 

[racctor]

Um, because the volume is changing due to expansion?

wouldn't that also happen in an ideal process?

 

[PeterDonis]

High temperature should still cause high pressure?

Not if the density has decreased. You are simply waving your hands instead of doing the math. Try looking up some actual values for all three variables--pressure, temperature, and density--before expansion and after expansion, plug them into the ideal gas law, and see how they match up. That is going to be a much better strategy than trying to guess based on your intuition.

wouldn't that also happen in an ideal process?

Of course it would. So what? @jack action wasn't arguing that the density change only happens in a non-ideal process. He was arguing that you are simply ignoring density change, even in the ideal case, and acting as though temperature and pressure are the only things that can vary, even in the ideal case. They're not.

Again, don't wave your hands, do the math.

 

[racctor]

Not if the density has decreased. You are simply waving your hands instead of doing the math. Try looking up some actual values for all three variables--pressure, temperature, and density--before expansion and after expansion, plug them into the ideal gas law, and see how they match up. That is going to be a much better strategy than trying to guess based on your intuition.



Of course it would. So what? @jack action wasn't arguing that the density change only happens in a non-ideal process. He was arguing that you are simply ignoring density change, even in the ideal case, and acting as though temperature and pressure are the only things that can vary, even in the ideal case. They're not.

Again, don't wave your hands, do the math.

okay, will try. I hope it isn't too complicated. I am actually studying something completely different.
Thanks for the input!

 

[Chestermiller]

Let me rephrase it: How can the temperature still be so high although the work has already been done ( and pressure has dropped to near atm level)? High temperature should still cause high pressure?

Who says that the temperature would stay high? It would be lower at the end.

 

[racctor]

If anyone else is looking for an answer to my question; the solution apparently can be found in "polytropic processes"

kind regards

 

[PeterDonis]

the solution apparently can be found in "polytropic processes"

Please elaborate.

 

[cjl]

It's worth mentioning that at the bottom of the expansion stroke (power stroke) in an engine, before the exhaust valve opens, the pressure is still substantially above ambient. It doesn't drop to ambient until it is allowed to expand further in the exhaust system, often through a turbine that extracts yet more energy from it.