Trying to understand quasi-static processes

  • Context: Graduate 
  • Thread starter Thread starter johng23
  • Start date Start date
  • Tags Tags
    Quasi-static
Click For Summary
SUMMARY

This discussion clarifies the distinction between quasi-static and reversible processes in thermodynamics, emphasizing that quasi-static processes occur through equilibrium states without friction or dissipative effects. The example of a paddle wheel in an adiabatic container illustrates how energy transfer can be quasi-static if the paddle is turned slowly, maintaining equilibrium among gas molecules. However, rapid turning introduces pressure variations and turbulence, leading to non-equilibrium states and energy dissipation. The conversation also addresses the challenges of avoiding friction in quasi-static processes, concluding that while friction can be minimized, it cannot be entirely eliminated.

PREREQUISITES
  • Understanding of thermodynamic concepts, specifically quasi-static and reversible processes.
  • Familiarity with the principles of thermal and mechanical equilibrium.
  • Knowledge of adiabatic processes and their implications in thermodynamics.
  • Basic comprehension of kinetic energy and its relationship to temperature in gases.
NEXT STEPS
  • Study the principles of thermal equilibrium and its mathematical formulations.
  • Explore the concept of adiabatic processes in greater detail, focusing on their applications in real-world systems.
  • Investigate the effects of friction in mechanical systems and methods to minimize it.
  • Learn about turbulence in fluid dynamics and its impact on energy transfer and thermodynamic processes.
USEFUL FOR

Students and professionals in thermodynamics, mechanical engineers, and anyone interested in the principles of energy transfer and system equilibrium.

johng23
Messages
292
Reaction score
1
In my thermo course, we made a distinction between quasi-static and reversible transitions (which apparently some don't make). Reversible means both system and surroundings need to be able to return to their initial state. Quasi-static only means that the system transitions to a new state by going through only equilibrium states. There is no friction or dissipative effects.

Intuitively, this makes sense but I'm not sure precisely how to differentiate an equilibrium state from a non-equilibrium state. The case I want to use to illustrate is an adiabatic container with rigid walls, with a paddle wheel going into it.

If I turn the paddle wheel, I can bring the system to any isometric state of higher energy by increasing the entropy. Is this a quasi-static process? Let's assume there is no friction between the paddle wheel and the wall. Then the general picture I have is that if I turn the paddle wheel slowly enough, the process is quasi-static since the distribution of gas molecules never appears macroscopically different than it otherwise would, even though you are gradually transferring energy to the system. But can I really reach an arbitrarily high energy this way? If I move the paddle wheel fast enough that there are pressure variations or heterogeneous velocity distributions in the container, at that point I am no longer going through equilibrium states. But why does it matter if energy is dissipated in these ways between molecules? I'm still transferring the same amount of energy to the system.

As another (maybe clearer) question, how do we avoid friction in a quasi-static process? If I slide two plates across each other, is it even conceptually possible to avoid friction? As I pull two atoms away from each other which are attracted by van der Waals forces or otherwise, don't they increase distance up to a point and then reach some threshold where they pull apart and start to oscillate? That's how it seems like it would happen in my mind.
 
Science news on Phys.org
johng23 said:
In my thermo course, we made a distinction between quasi-static and reversible transitions (which apparently some don't make). Reversible means both system and surroundings need to be able to return to their initial state. Quasi-static only means that the system transitions to a new state by going through only equilibrium states. There is no friction or dissipative effects.

Intuitively, this makes sense but I'm not sure precisely how to differentiate an equilibrium state from a non-equilibrium state.
Here is my definition of a quasi-static process involving a system and its surroundings: it is a process in which the changes occur slowly enough so that there is always:
1. thermal equilibrium within in all system components,
2. thermal equilibrium between all system components and surroundings across whose boundaries heat flows and
3. mechanical equilibrium between all system components and surroundings between which mechanical work/energy is transferred.​
So, in the case of a quasi-static heat flow from a reservoir to a gas, the heat flow occurs due to infinitessimal temperature difference between the reservoir and gas. In the case of a gas doing work on its surroundings, the gas pressure is infinitessimally higher than the surroundings at all times. All components are always in equilibrium (ie. out of equilbrium by an infinitessimal amount).

If there is a finite difference in temperature between two components in thermal contact, these components are not in equilibrium. If there is a finite pressure difference between two components in mechanical contact, these components are not in equilibrium. These processes are, therefore, not quasi-static.

The case I want to use to illustrate is an adiabatic container with rigid walls, with a paddle wheel going into it.

If I turn the paddle wheel, I can bring the system to any isometric state of higher energy by increasing the entropy. Is this a quasi-static process? Let's assume there is no friction between the paddle wheel and the wall. Then the general picture I have is that if I turn the paddle wheel slowly enough, the process is quasi-static since the distribution of gas molecules never appears macroscopically different than it otherwise would, even though you are gradually transferring energy to the system. But can I really reach an arbitrarily high energy this way? If I move the paddle wheel fast enough that there are pressure variations or heterogeneous velocity distributions in the container, at that point I am no longer going through equilibrium states. But why does it matter if energy is dissipated in these ways between molecules? I'm still transferring the same amount of energy to the system.
If you accelerate the paddle slowly (quasi-statically), you will do some work on the gasbut once you reach a constant speed no more work will be done (ignoring friction). Because the paddle is rigid, you are not changing the volume of the gas between the paddles. So a slow constant speed turning of the paddle does no work. Ignoring friction, the paddle should keep turning without any addition of energy. All the work done on the paddle results in uniform kinetic energy of the gas and does not cause the temperature of the gas to increase. That kinetic energy can be fully recovered as work.

If you accelerate the paddle quickly, you will be compressing the gas immediately in front of the turning paddle and expanding the parts behind the paddle. Ultimately what you are doing is adding non-uniform kinetic energy to the gas. After you stop accelerating the paddle and the gas reaches kinetic equilibrium, its temperature will be higher than when you started.
As another (maybe clearer) question, how do we avoid friction in a quasi-static process? If I slide two plates across each other, is it even conceptually possible to avoid friction? As I pull two atoms away from each other which are attracted by van der Waals forces or otherwise, don't they increase distance up to a point and then reach some threshold where they pull apart and start to oscillate? That's how it seems like it would happen in my mind.
You can minimize friction but you can't really avoid it. Friction losses are ultimately turned into heat.

For your paddle example, if you turn the paddle quickly, you create turbulence which is kinetic energy that does not follow a Maxwell-Boltzmann distribution (temperature distribution). As turbulence settles down, the molecular motion will follow the Maxwell-Boltzmann distribution and, therefore, appears as heat. You can think of the turbulence as "friction".

However, if you accelerate the paddle slowly so that there is not turbulence created (ie quasi-statically), the energy transferred from the paddle to the fluid is retained in the fluid as mechanical energy rather than heat. That mechanical energy can be recovered fully from the gas (ie. by putting a load on the paddle the turning gas does work on the paddle until the turning ends).

AM
 
Last edited:

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 135 ·
5
Replies
135
Views
9K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K