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

[burian]

Sure, that was only the ideal-fluid case. For real fluids you have to take into account "friction", i.e., dissipation, and that's of course also part of the stress. There you need a difference in speed between the fluid in the volume and outside of it along the boundary of the volume. In the next approximation thus the corresponding stress from friction must be of first order in the gradient of $\vec{v}$. So it should be proportional to the components $\partial_j v_k$. Now the antisymmetric part of this tensor refers to a rotation of the fluid element as a whole and thus doesn't describe true velocity differences. So it's the symmetric part which must enter the first-order contribution to the stress. Since in a liquid or gas locally no direction is preferred there can only be two pieces, corresponding to the irreducible representations of the rotation group. For a symmetric 2nd rank tensor this are the five components of the trace-less part as well as the trace part. Thus one has
$$\sigma_{jk}^{(1)}=\eta (\partial_j v_k + \partial_k v_j - \frac{2}{3} \partial_{i} v_i \delta_{jk} )+\zeta \partial_i v_i \delta_{jk}.$$
As it must be the tensor only depends on two scalar quantities $\eta$, and $\zeta$, the shear and bulk viscosities.

Then you end up with the Navier-Stokes equation,
$$\rho (\partial_t \vec{v} + \vec{v} \cdot \vec{\nabla} \vec{v})=-\vec{\nabla} P + \eta \Delta \vec{v} + \left (\frac{\eta}{3}+\zeta \right) \vec{\nabla}(\vec{\nabla} \cdot \vec{v}),$$
where I have assumed the $\eta$ and $\zeta$ can be considered constant within the fluid (though they depend on temperature and density and thus are not strictly constant).

I am thrown of by the tensor notation xD. I'll learn it and reattend your reply some day.

 

[sophiecentaur]

How did you conclude that speed of pressure transmission is same as speed of gas?

I assume you mean speed of sound?? Well, the speed of sound in a substance is the speed at which disturbances travel through it. What other speed would be involved?
The piston moves and causes a difference in pressure at its face. The change in pressure is transferred to the rest of the gas. It's easier to consider it happening 'slowly enough' but there is a change and there is a time interval.

work will have been done.

Yes, but I said no work is being done (when the piston is no longer moving and its KE is now zero).

I can't believe the last equality, then what about the energy which goes into accelerating the piston?

By the time the piston reaches its final position the KE will be zero and will have ben transferred to Potential Energy change / Work. This sort of confusion happens all over the place when Potential Energy differences are discussed as things move from A to B. As with Elevators etc, the KE is zero at A and B (but not at a mid point C in the building).

 

[Chestermiller]

I can't believe the last equality, then what about the energy which goes into accelerating the piston?

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.

 

[Chestermiller]

I assume you mean speed of sound?? Well, the speed of sound in a substance is the speed at which disturbances travel through it. What other speed would be involved?
The piston moves and causes a difference in pressure at its face. The change in pressure is transferred to the rest of the gas. It's easier to consider it happening 'slowly enough' but there is a change and there is a time interval.

Yes, but I said no work is being done (when the piston is no longer moving and its KE is now zero).

By the time the piston reaches its final position the KE will be zero and will have ben transferred to Potential Energy change / Work. This sort of confusion happens all over the place when Potential Energy differences are discussed as things move from A to B. As with Elevators etc, the KE is zero at A and B (but not at a mid point C in the building).

I'm thinking of a horizontal cylinder and piston, without potential energy effect. The piston stops moving because its motion is damped by viscous stresses in the gas.

 

[jack action]

I can't believe the last equality, then what about the energy which goes into accelerating the piston?

It is dissipated by the viscous behavior of the gas.

I don't like that answer. What if there are no viscous behaviors?

I would prefer to say that the energy required to accelerate the piston is stored as kinetic energy in the piston, which can be released by decelerating the piston. Granted it can be a viscous force but it could be some other opposing force too.

In the problem I was referring to in post #6 (a piston held in a cylinder with both ends closed and different pressures on each side), if we assume no friction or viscous behaviors, when the piston is release we would end up with a never ending harmonic motion, just like a mass stuck between two springs. That is possible because the energy is continuously stored to, and retrieved from the moving mass.

 

[Chestermiller]

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).

 

[sophiecentaur]

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).

 

[Twigg]

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:

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.

 

[sophiecentaur]

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.

 

[Chestermiller]

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.

 

[burian]

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

 

[Chestermiller]

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.

 

[sophiecentaur]

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.

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.

 

[Chestermiller]

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.

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.

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.

 

[sophiecentaur]

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.

 

[cianfa72]

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).