Expansion in non-adiabatic Otto / Diesel cycles

In summary, the conversation discusses the concept of energy and heat transfer in a real otto or diesel cycle, as well as in the compression step of real heat pumps. It is explained that during the expansion phase, energy is leaving the system and being used to do work, and that the temperature can remain high while the pressure decreases due to changes in density. The effects of friction and heat loss to the surroundings are also mentioned. It is emphasized that instead of relying on intuition, it is important to use the ideal gas law and actual values to understand the relationship between pressure, temperature, and density in these processes.
  • #1
racctor
7
0
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
 
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  • #2
racctor said:
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.
racctor said:
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.
racctor said:
wouldn't that just decrease the mechanical Work i can get out of the system
It does.
racctor said:
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.
 
  • #3
racctor said:
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.
 
  • #4
racctor said:
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.
 
  • #5
jack action said:
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
 
  • #6
racctor said:
How can the density change if the number of particles stays the same?

Um, because the volume is changing due to expansion?
 
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  • #7
PeterDonis said:
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?
 
  • #8
PeterDonis said:
Um, because the volume is changing due to expansion?
wouldn't that also happen in an ideal process?
 
  • #9
racctor said:
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.

racctor said:
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.
 
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  • #10
PeterDonis said:
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!
 
  • #11
racctor said:
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.
 
Last edited:
  • #12
If anyone else is looking for an answer to my question; the solution apparently can be found in "polytropic processes"

kind regards
 
  • #13
racctor said:
the solution apparently can be found in "polytropic processes"

Please elaborate.
 
  • #14
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.
 

Related to Expansion in non-adiabatic Otto / Diesel cycles

1. What is non-adiabatic expansion in Otto/Diesel cycles?

Non-adiabatic expansion refers to the process in which the working fluid in an Otto or Diesel cycle expands while exchanging heat with its surroundings. This is in contrast to adiabatic expansion, where no heat is exchanged during the expansion process.

2. How does non-adiabatic expansion affect the efficiency of an Otto/Diesel cycle?

Non-adiabatic expansion results in a decrease in the efficiency of an Otto or Diesel cycle. This is because some of the energy that could have been used to do work is instead lost to the surroundings as heat.

3. What factors influence the amount of non-adiabatic expansion in an Otto/Diesel cycle?

The amount of non-adiabatic expansion in an Otto or Diesel cycle is influenced by several factors, including the design of the engine, the properties of the working fluid, and the operating conditions.

4. How can non-adiabatic expansion be minimized in an Otto/Diesel cycle?

Non-adiabatic expansion can be minimized by improving the design of the engine, using a more efficient working fluid, and operating the engine at optimal conditions. Additionally, insulation can be used to reduce heat exchange with the surroundings.

5. What are the practical implications of non-adiabatic expansion in Otto/Diesel cycles?

The practical implications of non-adiabatic expansion in Otto and Diesel cycles include a decrease in engine efficiency, which can lead to lower power output and increased fuel consumption. It also contributes to the overall heat loss of the engine, which can impact its durability and lifespan.

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