Expansion or Compression Work by Gas ##=\int{P_{ext}dV}##

In summary, @burian and I have been discussing why, in compression and expansion of a gas, the work can be calculated from ##W=\int{P_{ext}dV}##. We thought it would be of value to open the discussion to the full membership.
  • #36
jack action said:
I don't like that answer. What if there are no viscous behaviors?
All gases are Newtonian fluids and exhibit viscous behavior. Bird, Stewart, and Lightfoot, Transport Phenomena, Section 1.4 MOLECULAR THEORY OF THE VISCOSITY OF GASES AT LOW DENSITY (i.e., ideal gas limit).
 
Science news on Phys.org
  • #37
Chestermiller said:
without potential energy effect.
I didn't say Gravitational Potential Energy. I was referring to the potential energy of changing the volume of a gas / spring etc..

I don't know why the mass of the piston is of interest here, in any case. That's a bit like wanting to know the sort of battery or PSU that's connected to a theoretical circuit. We (more than often) assume the supply is constant voltage and I would say that the power for the piston is a high impedance source ( 'constant displacement' could be the name).
 
  • #38
First off, this thread is a great resource. If it's possible to pin it to the thermodynamics subforum, it has my vote.

Beyond that, I have a very nitpicky question (sorry in advance). Regarding this statement:
sophiecentaur said:
The speed at which the effect of applied pressure transmits through the gas is basically the speed of sound in the gas. There will be a delay as with any force deforming anything.
I was under the impression that this was only true for perfectly inviscuous fluids (or viscous fluids in very high pressure gradients or flowing in a converging nozzle). For viscous fluids at lower pressure gradients, I thought it was flow instability (turbulence) that limits the speed at which energy can be transported? Am I wrong about this? Sorry again, not trying to nitpick, just testing my understanding.
 
  • #39
Twigg said:
Am I wrong about this?
The context of the thread is a gas, which is why I introduced the speed of sound. I have no strong feelings about this though.
 
  • #40
sophiecentaur said:
I didn't say Gravitational Potential Energy. I was referring to the potential energy of changing the volume of a gas / spring etc..

I don't know why the mass of the piston is of interest here, in any case. That's a bit like wanting to know the sort of battery or PSU that's connected to a theoretical circuit. We (more than often) assume the supply is constant voltage and I would say that the power for the piston is a high impedance source ( 'constant displacement' could be the name).
The mass of the piston is important only in so far as some members thought that it might affect the total amount of work that is done by the gas between the initial and final thermodynamic equilibrium states. The analysis I have done has shown that, because of the viscous damping of the piston motion, even though the piston mass affects the amount of work done by the gas during the process, in the end it has no effect on the ultimate work done by the gas from initial to final thermodynamic equilibrium states.
 
  • Like
Likes sophiecentaur
  • #41
Chestermiller said:
It is dissipated by the viscous behavior of the gas. Think of the gas as if it is a pre-compressed spring in parallel with a viscous damper. The spring mimics the equilibrium P-V behavior of the gas and the damper comes into play in rapid irreversible deformations in mimicking the viscous effect. So, think of mean compressive stress in the rapidly deforming gas as being approximated by the linearized relationship $$P_{int}=\bar{\sigma}_z=P_{int,init}-C_1(V-V_{init})-C_2\frac{dV}{dt}$$where C2 is proportional to the gas viscosity. So, from our force balance on the piston, we have the linearized differential equation: $$\frac{m}{A}\frac{d^2V}{dt^2}+C_1A(V-V_{init})+C_2A\frac{dV}{dt}$$$$+
(P_0(t)-P_{int,init})A=0$$This equation describes a forced oscillation on the piston. Basically, what will happen is that the piston will oscillate about the final equilibrium position until the amplitude of its oscillation is damped to zero by viscous stresses in the gas.
In this case, why didn't you include the viscous foces in the energy balance in the post where you introduced the Newton's second law and integrate it to get the energy balance
 
  • #42
burian said:
In this case, why didn't you include the viscous foces in the energy balance in the post where you introduced the Newton's second law and integrate it to get the energy balance
The viscous forces are implicitly included in ##P_{int}## in the energy balance I introduced and in its integration. To complete the derivation, we needed to assume that at infinite time, the piston is no longer moving.

The more recent equation in which the viscous forces are explicitly included in the analysis uses only a crude approximation to ##P_{int}## (in terms of gas volume V and its time derivative) to provide us with a rough qualitative understanding of how the viscous forces come into play in damping the piston motion.
 
Last edited:
  • Like
Likes burian
  • #43
Chestermiller said:
some members thought that it might affect the total amount of work that is done by the gas between the initial and final thermodynamic equilibrium states.
If you take my point about the way people approach electric circuit theory then there's definitely a different attitude to the behaviour of gases. How many people even stop to think about what happens in an electrical supply when they are given a value for the supply voltage? Does the battery charge or discharge when a Potentiometer is moved?
I suppose there is one big difference in that Heat Engines involve Work Done on a piston but the behaviour of gases under pressure is the first thing that we're taught.

Chestermiller said:
The piston stops moving because its motion is damped by viscous stresses in the gas.
In the basic model, it's the piston that is being moved (by an irresistible applied force) so doesn't the piston stops moving when you choose to make it stop?

PS I find you a terrific source of help with Thermodynamics.
 
  • #44
sophiecentaur said:
I suppose there is one big difference in that Heat Engines involve Work Done on a piston but the behaviour of gases under pressure is the first thing that we're taught.
It's the first thing we're taught for gases at thermodynamic equilibrium. It isn't until we take a course in fluid mechanics that we learn about the irreversible deformational behavior of gases at finite deformation rates; that's when we learn about the viscous aspect of their behavior.
sophiecentaur said:
In the basic model, it's the piston that is being moved (by an irresistible applied force) so doesn't the piston stops moving when you choose to make it stop?
In homework problems in thermodynamics, there are cases where the piston is stopped manually or with a detent within the cylinders, but in most of the cases, the gas, piston, and other surroundings are allow to continue moving spontaneously until they finally re-equilibrate on their own.
sophiecentaur said:
PS I find you a terrific source of help with Thermodynamics.
Thank you. You are very kind, and it's a pleasure interacting with you.

I'm a chemical engineer, and thermodynamics is our bread and butter. ChE is founded on the application of thermodynamics in the design and operation of chemical plant processing equipment.
 
  • Like
Likes burian, Twigg, vanhees71 and 1 other person
  • #45
Chestermiller said:
I'm a chemical engineer, and thermodynamics is our bread and butter.
yeah well - my thermodynamics extends to the basics of A level. Not surprisingly. I can easily get hold of the wrong end of the stick. But it's all grist to the mill.
 
  • #46
Chestermiller said:
I'm going to continue what I was discussing in post #7 for an irreversible deformation of the gas. If ##\bar{\sigma_z}=P_{int}## is the average force per unit area exerted by the gas on the inside face of the piston, and we choose the gas alone as our system, then, calling the average external force per unit area exerted by the piston on the gas ##P_{ext}##, we must have that $$P_{ext}=P_{int}=\bar{\sigma_z}$$
I would like to add that it is always true in force of 3rd Newton law: the average force per unit area the gas exerts on the inside face of the piston is the same as the average force the inside face of the piston exerts on the gas (here we are assuming the gas inside the cylinder is the system).
 

Similar threads

Replies
5
Views
1K
Replies
4
Views
881
  • Thermodynamics
Replies
8
Views
691
  • Thermodynamics
Replies
20
Views
2K
Replies
22
Views
2K
Replies
13
Views
2K
  • Thermodynamics
Replies
17
Views
1K
Replies
65
Views
75K
Replies
5
Views
627
  • Thermodynamics
Replies
11
Views
5K
Back
Top