Question about Flow between Parallel Plates

[Red_CCF]

Hello

With regards to a common example in most fluid mechanics books where there exists fluid between two stationary parallel plates and the top plate is pulled until it has a constant velocity U with force F, the result is \tau=F/A = μU/L. Although this is probably obvious to most, I am wondering if there is a formal proof of this equation and why a linear velocity profile makes sense physically.

I am also wondering the same for the velocity profile of a laminar flow through tube. I understand its derivation from the Navier Stokes equation, but physically why does the parabolic velocity profile and linear shear stress profile make sense?

Thanks very much

 

[Chestermiller]

Hey Red_CCF,

You are trying to develop an intuitive feel for this, so maybe this can help (or not).

In a solid, the stresses are related linearly to the strains, and, in a liquid, the stresses are related linearly to the rates of strain. So, let's see what would happen with a solid being deformed in your parallel plate geometry.

In both cases, the surfaces of the solid are glued to the flat plates to simulate the no-slip boundary condition for the fluid.

If the top plate is moved slightly sideways relative to the bottom plate, you expect the sideways displacement to be a linear function of distance from the bottom plate. This is just shearing of the solid between the plates. The stress is equal to the shear modulus G times the shear strain Δu/Δz. For the solid, the sideways displacement is analogous to the velocity.

In the case of a pressure difference between one end of the solid and the other, by symmetry, you expect the sideways displacement to be maximum at the center of the channel and zero at the wall. So you can see why you would expect the sideways displacement to be parabolic in shape. This is analogous to a parabolic velocity profile for a fluid. If the displacement is parabolic, then its derivative, which is the shear strain, will be linear. This would result in a linear shear stress profile.

Hope this analogy helps.

Chet

 

[Red_CCF]

Hi Chet

Hey Red_CCF,

You are trying to develop an intuitive feel for this, so maybe this can help (or not).

In a solid, the stresses are related linearly to the strains, and, in a liquid, the stresses are related linearly to the rates of strain. So, let's see what would happen with a solid being deformed in your parallel plate geometry.

I recall reading from somewhere that this was the definition of a liquid; is this originated from Newton's Law of Viscosity (strain rate = du/dy)?

How does one derive τ = μU/L from Navier Stokes? I ended up with all zero's for the x component of the equation. Is dp/dx = 0 in this case?

In both cases, the surfaces of the solid are glued to the flat plates to simulate the no-slip boundary condition for the fluid.

If the top plate is moved slightly sideways relative to the bottom plate, you expect the sideways displacement to be a linear function of distance from the bottom plate. This is just shearing of the solid between the plates. The stress is equal to the shear modulus G times the shear strain Δu/Δz. For the solid, the sideways displacement is analogous to the velocity.

So in both cases, shear stress is constant? If displacement analogous to velocity what is du/dy analogous to?

In the case of a pressure difference between one end of the solid and the other, by symmetry, you expect the sideways displacement to be maximum at the center of the channel and zero at the wall. So you can see why you would expect the sideways displacement to be parabolic in shape. This is analogous to a parabolic velocity profile for a fluid. If the displacement is parabolic, then its derivative, which is the shear strain, will be linear. This would result in a linear shear stress profile.

In this case, am I "pushing" the solid forward while fixing the boundaries? I can see that the final profile should be symmetrical and should be somewhat "curved", but what dictates that it must be parabolic?

Per your recommendation I began to read some sections of BSL on Newton's Law of Viscosity and had a couple of questions related to this topic.

According to Wikipedia, Newton's Law of Viscosity is a constitutive equation. Does this mean that the common form we see τ = μ∂u/∂x is a first order approximation and exact solution is given by an infinite series?

With regards to pressure, since P is always perpendicular the surface it applies on, when a gas molecule hits a surface at some angle, is only the perpendicular component of the force the gas applies considered pressure? Does the parallel (to the surface) component of the applied force not have any significance even if it is non-zero?

Lastly, the normal viscous stress τ_xx is described as a flux of x momentum in x-dir. If x-dir is perpendicular to a piston face in a piston-cylinder assembly, I'm having trouble seeing how this flux occurs physically.

Thank you very much

 

[boneh3ad]

I recall reading from somewhere that this was the definition of a liquid; is this originated from Newton's Law of Viscosity (strain rate = du/dy)?

No, that's the definition of a fluid. Gases and plasmas follow the same pattern. They needn't be Newtonian, though. Non-Newtonian fluids follow this pattern just as well as Newtonian fluids do, they simply have a different constitutive equation describing this relationship.

How does one derive τ = μU/L from Navier Stokes? I ended up with all zero's for the x component of the equation. Is dp/dx = 0 in this case?

Where exactly are you having trouble? You ought to be able to find this problem, Couette flow, in just about any fluids textbook as an example. Since this is a steady, incompressible flow in two dimensions, you have the continuity equation and the x- and y-momentum equations:
\dfrac{\partial u}{\partial x} + \dfrac{\partial v}{\partial y} = 0,
\rho\left(u\dfrac{\partial u}{\partial x} + v \dfrac{\partial u}{\partial y}\right) = -\dfrac{\partial p}{\partial x} + \mu \left(\dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2}\right),
\rho\left(u\dfrac{\partial v}{\partial x} + v \dfrac{\partial v}{\partial y}\right) = -\dfrac{\partial p}{\partial y} + \mu \left(\dfrac{\partial^2 v}{\partial x^2} + \dfrac{\partial^2 v}{\partial y^2}\right).

Couette flow assumes the domain is infinite and steady in x, so all the \partial/\partial x terms are going to drop out. It also assumes that the vertical velocity, v, is zero everywhere, so all of the v terms drop out and the continuity equation is therefore eliminated. The momentum equations reduce to
\dfrac{\partial^2 u}{\partial y^2} = 0,
\dfrac{\partial p}{\partial y} = 0.

Since this has reduced to a flow where the velocity only varies with one spatial variable, it is an ODE rather than a PDE, so you have
\dfrac{d^2 u}{d y^2} = 0.

So that leaves you with a differential equation describing the velocity profile as a function of y, which you should be able to easily integrate and use the boundary conditions to derive the solution you are seeking. Integrating it twice gives you
= u(y) = C_1 y + C_2
with boundary conditions
u(0) = 0, u(h) = U,
so C_2 = 0 and C_1 = U/h. Therefore,
u(y) = \dfrac{Uy}{h}.
Differentiate that ate your leisure to get the shear stress.

Anyway, back to your original question: you would expect this to be real because, for a Newtonian fluid, the shear stress is related linearly to the velocity gradient. Mathematically, the above shows how that results in the linear velocity profile.

According to Wikipedia, Newton's Law of Viscosity is a constitutive equation. Does this mean that the common form we see τ = μ∂u/∂x is a first order approximation and exact solution is given by an infinite series?

Ultimately, Newton's law of viscosity is an proportionality law that holds for a certain subset of fluids (Newtonian fluids) under the assumption that the flow is a continuum. It doesn't work for all fluids (non-Newtonian fluids have much more complicated constitutive equations), so it isn't any kind of universal law, but one that describes only a certain type of fluid. There may well be some larger constitutive equation that could describe literally every fluid, but as far as I knew, nothing of that sort exists and you just use the constitutive equation for the stress tensor that applies to the physics of the fluid in question.

With regards to pressure, since P is always perpendicular the surface it applies on, when a gas molecule hits a surface at some angle, is only the perpendicular component of the force the gas applies considered pressure? Does the parallel (to the surface) component of the applied force not have any significance even if it is non-zero?

Yes, only the wall-normal component of that collision contributes to pressure. Think of it this way, the molecule has a certain momentum, and when it collides obliquely with a surface, the only portion of that momentum that is affected (ignoring viscosity) is the wall normal portion. This is, in fact, one way to think about why the Bernoulli equation makes sense: the total pressure of relevant flows is constant and so the molecules have a constant energy, and as the velocity increases, those collisions get increasingly oblique, meaning less static pressure on the surface.

Lastly, the normal viscous stress τxx is described as a flux of x momentum in x-dir. If x-dir is perpendicular to a piston face in a piston-cylinder assembly, I'm having trouble seeing how this flux occurs physically.

This will be frame-dependent, but if you look in the inertial frame with a moving piston, the flux occurs because as the face of the piston passes through a plane, so, too, does the fluid on that face, so that is a flux. It certainly moves goes discontinuously to or from zero fluid flux at that instant. If you are in the frame of the piston, there is no flux there. The velocity is zero anyway at that point in that frame so those terms are zero.

 

[Chestermiller]

Ultimately, Newton's law of viscosity is an proportionality law that holds for a certain subset of fluids (Newtonian fluids) under the assumption that the flow is a continuum. It doesn't work for all fluids (non-Newtonian fluids have much more complicated constitutive equations), so it isn't any kind of universal law, but one that describes only a certain type of fluid. There may well be some larger constitutive equation that could describe literally every fluid, but as far as I knew, nothing of that sort exists and you just use the constitutive equation for the stress tensor that applies to the physics of the fluid in question.

To expand on what bondh3ad is saying here, I might add that there are a huge number of liquids and gases for which the Newtonian fluid model provides an excellent approximation to the actual constitutive behavior. That's why the model has such broad applicability.

Quote by RedCCF:
Lastly, the normal viscous stress τxx is described as a flux of x momentum in x-dir. If x-dir is perpendicular to a piston face in a piston-cylinder assembly, I'm having trouble seeing how this flux occurs physically.

The viscous stress is determined by the rate of deformation dv_x/dx. A positive value for this means that the material planes are getting further apart (compared to if the fluid or gas were not deforming). So the rate at which molecules are hitting the material planes is less, and this translates into lower compressive stress on the planes.

Chet

 

[Red_CCF]

Ultimately, Newton's law of viscosity is an proportionality law that holds for a certain subset of fluids (Newtonian fluids) under the assumption that the flow is a continuum. It doesn't work for all fluids (non-Newtonian fluids have much more complicated constitutive equations), so it isn't any kind of universal law, but one that describes only a certain type of fluid. There may well be some larger constitutive equation that could describe literally every fluid, but as far as I knew, nothing of that sort exists and you just use the constitutive equation for the stress tensor that applies to the physics of the fluid in question.

For Newtonian fluids only, is the linear relationship between stress and velocity gradient exact or also an approximation?

Yes, only the wall-normal component of that collision contributes to pressure. Think of it this way, the molecule has a certain momentum, and when it collides obliquely with a surface, the only portion of that momentum that is affected (ignoring viscosity) is the wall normal portion. This is, in fact, one way to think about why the Bernoulli equation makes sense: the total pressure of relevant flows is constant and so the molecules have a constant energy, and as the velocity increases, those collisions get increasingly oblique, meaning less static pressure on the surface.

Is the assumption that the momentum component parallel to the surface of the collision is (generally) conserved, and any frictional effects during the contact negligible?

This will be frame-dependent, but if you look in the inertial frame with a moving piston, the flux occurs because as the face of the piston passes through a plane, so, too, does the fluid on that face, so that is a flux. It certainly moves goes discontinuously to or from zero fluid flux at that instant. If you are in the frame of the piston, there is no flux there. The velocity is zero anyway at that point in that frame so those terms are zero.

The viscous stress is determined by the rate of deformation dv_x/dx. A positive value for this means that the material planes are getting further apart (compared to if the fluid or gas were not deforming). So the rate at which molecules are hitting the material planes is less, and this translates into lower compressive stress on the planes.

Currently, I am visualizing this as, as a piston is compressed, the gas velocity decreases away from the piston face, and the high and low velocity molecules "mixes", leading to an extra load for the piston to keep moving at its velocity. However, I have trouble seeing the necessity of viscosity as it appears that inertia is the reason for the viscous normal stress on the piston. I'm also having trouble seeing how normal stresses are induced within the fluid itself.

Where exactly are you having trouble? You ought to be able to find this problem, Couette flow, in just about any fluids textbook as an example. Since this is a steady, incompressible flow in two dimensions, you have the continuity equation and the x- and y-momentum equations:
\dfrac{\partial u}{\partial x} + \dfrac{\partial v}{\partial y} = 0,
\rho\left(u\dfrac{\partial u}{\partial x} + v \dfrac{\partial u}{\partial y}\right) = -\dfrac{\partial p}{\partial x} + \mu \left(\dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2}\right),
\rho\left(u\dfrac{\partial v}{\partial x} + v \dfrac{\partial v}{\partial y}\right) = -\dfrac{\partial p}{\partial y} + \mu \left(\dfrac{\partial^2 v}{\partial x^2} + \dfrac{\partial^2 v}{\partial y^2}\right).

Couette flow assumes the domain is infinite and steady in x, so all the \partial/\partial x terms are going to drop out. It also assumes that the vertical velocity, v, is zero everywhere, so all of the v terms drop out and the continuity equation is therefore eliminated. The momentum equations reduce to
\dfrac{\partial^2 u}{\partial y^2} = 0,
\dfrac{\partial p}{\partial y} = 0.

Since this has reduced to a flow where the velocity only varies with one spatial variable, it is an ODE rather than a PDE, so you have
\dfrac{d^2 u}{d y^2} = 0.

So that leaves you with a differential equation describing the velocity profile as a function of y, which you should be able to easily integrate and use the boundary conditions to derive the solution you are seeking. Integrating it twice gives you
= u(y) = C_1 y + C_2
with boundary conditions
u(0) = 0, u(h) = U,
so C_2 = 0 and C_1 = U/h. Therefore,
u(y) = \dfrac{Uy}{h}.
Differentiate that ate your leisure to get the shear stress.

Anyway, back to your original question: you would expect this to be real because, for a Newtonian fluid, the shear stress is related linearly to the velocity gradient. Mathematically, the above shows how that results in the linear velocity profile.

Thank you for the complete derivation. I was attempting to derive the shear - velocity gradient relationship directly using this stranger form of the Navier Stokes equation. I thought that the shear stress in this case should be \tau_yx (if x is parallel to the flow and y perpendicular) but did not find this anywhere and ended up with ρ∂u²/∂x = 0 (x-component).

How come the knowledge of linearity between shear and velocity gradient dictate that the velocity profile itself be linear?

Thank you very much

 

[Chestermiller]

For Newtonian fluids only, is the linear relationship between stress and velocity gradient exact or also an approximation?

A Newtonian Fluid is defined as one for which the stress tensor is a linear function of the velocity gradient tensor. Fortunately for us, a huge number of fluids satisfy this requirement.

Is the assumption that the momentum component parallel to the surface of the collision is (generally) conserved, and any frictional effects during the contact negligible?

No. The assumption for this kind of deformation is that changes in momentum tangent to the surface statistically cancel out.

Currently, I am visualizing this as, as a piston is compressed, the gas velocity decreases away from the piston face, and the high and low velocity molecules "mixes", leading to an extra load for the piston to keep moving at its velocity. However, I have trouble seeing the necessity of viscosity as it appears that inertia is the reason for the viscous normal stress on the piston. I'm also having trouble seeing how normal stresses are induced within the fluid itself.

As the gas is getting compressed and the gas (average molecular) velocity (in the x-direction) decreases (with distance x) away from the piston face, the material planes within the gas are getting closer together, and gas molecules are striking them more frequently. This results in a higher force per unit area. See Bird et al for really good discussions of the molecular interpretation of viscosity. You are having trouble seeing the necessity of viscosity because you are forgetting that you can't track the momentum changes of every single molecule and, in practice, you need to consider the statistical average of the momentum changes. This is where viscosity comes in.

Thank you for the complete derivation. I was attempting to derive the shear - velocity gradient relationship directly using this stranger form of the Navier Stokes equation. I thought that the shear stress in this case should be \tau_yx (if x is parallel to the flow and y perpendicular) but did not find this anywhere and ended up with ρ∂u²/∂x = 0 (x-component).

How come the knowledge of linearity between shear and velocity gradient dictate that the velocity profile itself be linear?

Thank you very much

From the conservation of momentum equations that you presented, and boneh3ad's discussion of the flow and boundary conditions, you are left with:
\frac{dτ_{xy}}{dy}=0
This gives: τ_{xy}=Const.
From the Newtonian constitutive equation for this situation (using BSL notation), τ_{xy}=-μ\frac{du}{dy}
Therefore, du/dy = C

Chet

 

[Red_CCF]

From the conservation of momentum equations that you presented, and boneh3ad's discussion of the flow and boundary conditions, you are left with:
\frac{dτ_{xy}}{dy}=0
This gives: τ_{xy}=Const.
From the Newtonian constitutive equation for this situation (using BSL notation), τ_{xy}=-μ\frac{du}{dy}
Therefore, du/dy = C

Chet

If x is parallel to the flow and y is perpendicular, is the correct notation \tau_xy or \tau_yx? Eq. 1.1-2 in BSL had \tau_yx which is why I though ∂\tau_xy/∂y = 0 in the equation.

This is related a bit to our previous discussion. If the additional normal stress from viscous dissipation theoretically didn't exist in a insulated piston-cylinder compression/expansion cycle but wall friction did (i.e. between the piston and cylinder or shear stress from gas-cylinder even though viscosity is required for the latter), should I expect P_ext during both expansion and compression both be higher than that from the reversible case? I know that the correct P-v curve shows a higher pressure for compression and lower pressure for expansion compared to a reversible case, but my current logic is that friction -> heat -> higher T -> higher P so I'm definitely missing something.

Thank you very much

 

[Chestermiller]

If x is parallel to the flow and y is perpendicular, is the correct notation \tau_xy or \tau_yx? Eq. 1.1-2 in BSL had \tau_yx which is why I though ∂\tau_xy/∂y = 0 in the equation.

Well, the stress tensor is symmetric (τ_xy=τ_yx), so we don't have to pay too much attention to the order of the subscripts. I never do.

This is related a bit to our previous discussion. If the additional normal stress from viscous dissipation theoretically didn't exist in a insulated piston-cylinder compression/expansion cycle but wall friction did (i.e. between the piston and cylinder or shear stress from gas-cylinder even though viscosity is required for the latter), should I expect P_ext during both expansion and compression both be higher than that from the reversible case? I know that the correct P-v curve shows a higher pressure for compression and lower pressure for expansion compared to a reversible case, but my current logic is that friction -> heat -> higher T -> higher P so I'm definitely missing something.

Well, we can speculate about this forever, or we can remove all the uncertainty by just modelling the doggone thing. This is a very straightforward problem to model. Are you up for it?

Chet

 

[Red_CCF]

Well, the stress tensor is symmetric (τ_xy=τ_yx), so we don't have to pay too much attention to the order of the subscripts. I never do.

Under what circumstances would the pressure gradient be 0?

Well, we can speculate about this forever, or we can remove all the uncertainty by just modelling the doggone thing. This is a very straightforward problem to model. Are you up for it?

Chet

Yes.

Thank you very much

 

[Red_CCF]

Well, the stress tensor is symmetric (τ_xy=τ_yx), so we don't have to pay too much attention to the order of the subscripts. I never do.

I read τ_xy as a force acting on a face perpendicular to x-axis and in the y-dir and thus the stress is perpendicular to the flow, but I read τ_yx as acting on a face perpendicular to the y-axis and in the x-dir which would be parallel to the flow. Is it the magnitude of the two that are the same?

Thanks very much

 

[Chestermiller]

I read τ_xy as a force acting on a face perpendicular to x-axis and in the y-dir and thus the stress is perpendicular to the flow, but I read τ_yx as acting on a face perpendicular to the y-axis and in the x-dir which would be parallel to the flow. Is it the magnitude of the two that are the same?

Thanks very much

Yes. Check out the Newtonian fluid relationships between stress components and velocity derivatives.

 

[Chestermiller]

Under what circumstances would the pressure gradient be 0?

This is too general a question. After you've worked a few problems, you will recognize it right away.

Yes.

OK. Here we go.

Here are the assumptions I'm proposing to use:

1. During the expansion or compression, the friction force on the piston is constant at F.
2. The piston is massless, and has zero heat capacity
3. All the heat generated between the wall and the piston ends up in the gas
4. The expansion and compression are carried out quasi-statically

Are you in agreement with these assumptions?
Are there any others that you think we should be using?

Which do you want to do first, expansion or compression?
What to you want included in the system: (a) just the gas or (b) the gas plus the piston? (I'm suggesting we eventually do it both ways to see how the two analyses compare).

Chet

 

[Red_CCF]

Yes. Check out the Newtonian fluid relationships between stress components and velocity derivatives.

If τ_xy is a shear stress applied to a face parallel to y and perpendicular to x, what is causing this stress? Is it to cancel out the moment caused by τ_yx?

For the Couette flow case I can see that τ_xy = τ_yx symmetry since shear stress is constant throughout. However, is this true in any case? In BSL I only see the symmetry property referenced under fluid in a pure rotation.

OK. Here we go.

Here are the assumptions I'm proposing to use:

1. During the expansion or compression, the friction force on the piston is constant at F.
2. The piston is massless, and has zero heat capacity
3. All the heat generated between the wall and the piston ends up in the gas
4. The expansion and compression are carried out quasi-statically

Are you in agreement with these assumptions?
Are there any others that you think we should be using?

Which do you want to do first, expansion or compression?
What to you want included in the system: (a) just the gas or (b) the gas plus the piston? (I'm suggesting we eventually do it both ways to see how the two analyses compare).

Chet

Yes the assumptions are the same as what I had in mind. I am hoping that we start with expansion as we didn't really discuss it and gas plus piston as the system first so the friction is occurring at the system boundary.

Thank you very much

 

[Chestermiller]

If τ_xy is a shear stress applied to a face parallel to y and perpendicular to x, what is causing this stress? Is it to cancel out the moment caused by τ_yx?

If I remember correctly, yes.

For the Couette flow case I can see that τ_xy = τ_yx symmetry since shear stress is constant throughout. However, is this true in any case? In BSL I only see the symmetry property referenced under fluid in a pure rotation.

The Newtonian fluid constitutive equation not only applies to homogeneous deformations; it applies locally to all deformations, including non-homogeneous deformations. Check out those partial derivatives in the general Newtonian fluid equations. The validity of applying the Newtonian fluid model locally to non-homogeneous deformations has been demonstrated by hundreds of years of success.

Yes the assumptions are the same as what I had in mind. I am hoping that we start with expansion as we didn't really discuss it and gas plus piston as the system first so the friction is occurring at the system boundary.

OK. Let's get started.

The first step is to write down the force balance on the piston. In terms of the friction force F, the piston area A, the externally applied pressure P_ext, and the gas normal stress at the piston interface τ_I, what is the force balance on the piston if we are considering expansion? (Of course, for quasi-static deformation, τ_I is going to be equal to p, the ideal gas law pressure within the cylinder).

Chet

 

[Red_CCF]

If I remember correctly, yes.

The Newtonian fluid constitutive equation not only applies to homogeneous deformations; it applies locally to all deformations, including non-homogeneous deformations. Check out those partial derivatives in the general Newtonian fluid equations. The validity of applying the Newtonian fluid model locally to non-homogeneous deformations has been demonstrated by hundreds of years of success.

What are homogeneous and non-homogeneous deformations?

From looking at the general equation, what I get from it is that viscous shear stresses acting on some plane are functions of velocity gradients for velocities that may or may not be in the same direction, such as \tau_xy = μ\frac{∂v_x}{∂y} in this example where velocity in y-dir is 0 yet shear stress in the y-dir is non-zero as velocity in x-dir is nonzero. Is this correct?

OK. Let's get started.

The first step is to write down the force balance on the piston. In terms of the friction force F, the piston area A, the externally applied pressure P_ext, and the gas normal stress at the piston interface τ_I, what is the force balance on the piston if we are considering expansion? (Of course, for quasi-static deformation, τ_I is going to be equal to p, the ideal gas law pressure within the cylinder).

Chet

P_I = RT/v and P_I = P_ext + F/A_piston if the piston is expanding.

Thanks very much

 

[Chestermiller]

What are homogeneous and non-homogeneous deformations?

A homogeneous deformation is one for which the velocity gradients are the same at all locations throughout the flow field. A non-homogeneous deformation is one for which the velocity gradients are not the same at all locations throughout the flow field.

From looking at the general equation, what I get from it is that viscous shear stresses acting on some plane are functions of velocity gradients for velocities that may or may not be in the same direction, such as \tau_xy = μ\frac{∂v_x}{∂y} in this example where velocity in y-dir is 0 yet shear stress in the y-dir is non-zero as velocity in x-dir is nonzero. Is this correct?

Yes.

P_I = RT/v and P_I = P_ext + F/A_piston if the piston is expanding.

Good. So, if we combine these equations, we have:
\frac{nRT}{V}-\frac{F}{A}=P_{ext}
We next use this equation to get the work done by the system on the surroundings when the volume of gas increases by dV:
dW=P_{ext}dV=\frac{nRT}{V}dV-\frac{F}{A}dV
We're next going to apply the 1st Law to the system. Let dU = nC_vdT be the change in internal energy of the system when the volume of the gas increases quasistatically by dV. Please write that 1st law equation.

Chet

 

[Red_CCF]

A homogeneous deformation is one for which the velocity gradients are the same at all locations throughout the flow field. A non-homogeneous deformation is one for which the velocity gradients are not the same at all locations throughout the flow field.

So a homogeneous deformation would be the Couette flow example, and a non homogeneous deformation would be like a laminar flow in circular pipe?

Good. So, if we combine these equations, we have:
\frac{nRT}{V}-\frac{F}{A}=P_{ext}
We next use this equation to get the work done by the system on the surroundings when the volume of gas increases by dV:
dW=P_{ext}dV=\frac{nRT}{V}dV-\frac{F}{A}dV
We're next going to apply the 1st Law to the system. Let dU = nC_vdT be the change in internal energy of the system when the volume of the gas increases quasistatically by dV. Please write that 1st law equation.

This is where I was confused. I get:

δQ - δW = dU

I am assuming that the work done to overcome friction is equivalent to the heat addition

δQ = \frac{F}{A}dV

but then I end up with which doesn't look right:

-\frac{nRT}{V}dV = nC_v dT

Unless the friction heat addition doesn't come into first law because it occurs at the system boundary and not added via finite temperature gradient. If this is the case δQ = 0 and I get this:

-\frac{nRT}{V}dV +\frac{F}{A}dV = nC_v dT

Thanks very much

 

[Chestermiller]

So a homogeneous deformation would be the Couette flow example, and a non homogeneous deformation would be like a laminar flow in circular pipe?

Yes.

This is where I was confused.

I suspected that this would be the trouble spot.

I get:

δQ - δW = dU

I am assuming that the work done to overcome friction is equivalent to the heat addition

δQ = \frac{F}{A}dV

but then I end up with which doesn't look right:

-\frac{nRT}{V}dV = nC_v dT

Unless the friction heat addition doesn't come into first law because it occurs at the system boundary and not added via finite temperature gradient.

Not exactly. We said that the cylinder is insulated, so no heat leaves or enters through there. And we said that all the "heat generated" stays in the system, so none leaves or enters through the top of the piston. Since δQ represents the heat entering through the interface between the system and the surroundings, δQ = 0.

We also said that the piston has zero heat capacity, so that the change in internal energy for the system is purely the result of the change in internal energy of the gas (without any contribution from the piston).

Incidentally, unlike the gas dynamics analysis which was rather complicated in the irreversible problems we discussed in the other thread, the heat transfer analysis for the heat generated between the cylinder and the piston, and its conduction through the piston to the gas below is rather easy to set up. This might give you some additional insight into what is happening. If you're interested, just say so.

If this is the case δQ = 0 and I get this:
-\frac{nRT}{V}dV +\frac{F}{A}dV = nC_v dT

Yes, this is the correct result. So now, rearranging this into the form of a first order ordinary differential equation, we get:

nC_v \frac{dT}{dV}+\frac{nRT}{V} =\frac{F}{A}

The initial condition on this differential equation is T = T₀ at V = V₀. Do you remember how to solve a differential equation of this form? If so, please proceed.

Chet

 

[Red_CCF]

Not exactly. We said that the cylinder is insulated, so no heat leaves or enters through there. And we said that all the "heat generated" stays in the system, so none leaves or enters through the top of the piston. Since δQ represents the heat entering through the interface between the system and the surroundings, δQ = 0.

We also said that the piston has zero heat capacity, so that the change in internal energy for the system is purely the result of the change in internal energy of the gas (without any contribution from the piston).

Since heat is generated at and parallel to the system boundary, it doesn't satisfy the definition of Q in first law as the heat isn't technically entering by conduction from the surroundings?

How is the effect of friction heat addition on system T and P compared to the reversible case reflected in just the modified work equation. To me it only shows that applied work is lower and nothing about how the friction energy going back to the system is affecting T and P, or is the effect shown implicitly in the equation below?

Incidentally, unlike the gas dynamics analysis which was rather complicated in the irreversible problems we discussed in the other thread, the heat transfer analysis for the heat generated between the cylinder and the piston, and its conduction through the piston to the gas below is rather easy to set up. This might give you some additional insight into what is happening. If you're interested, just say so.

Some insight would be appreciated.

Yes, this is the correct result. So now, rearranging this into the form of a first order ordinary differential equation, we get:

nC_v \frac{dT}{dV}+\frac{nRT}{V} =\frac{F}{A}

The initial condition on this differential equation is T = T₀ at V = V₀. Do you remember how to solve a differential equation of this form? If so, please proceed.

Chet

From solving the ODE I got:

T(V) = \frac{Const}{V^\frac{R}{C_v}}+\frac{F}{nA(R+C_v)}V

Const = V_o^\frac{R}{C_v}(T_o - \frac{FV_o}{nA(R+C_v)})

T(V) = T_o(\frac{V_o}{V})^\frac{R}{C_v}+\frac{FV_o}{nA(R+C_v)}[(\frac{V_o}{V})^\frac{R}{C_v}-\frac{V}{V_o}]

Thanks very much

 

[Chestermiller]

Since heat is generated at and parallel to the system boundary, it doesn't satisfy the definition of Q in first law as the heat isn't technically entering by conduction from the surroundings?

Basically, yes. The heat is generated just inside the "system boundary," especially if the system also includes the cylinder wall (assumed to have zero heat capacity and zero thermal conductivity). Then the heat is definitely generated inside the system boundary. Think of the cylinder as a black box with insulation. No heat enters or leaves through its wall and no heat enters or leaves through the top of the piston.

How is the effect of friction heat addition on system T and P compared to the reversible case reflected in just the modified work equation. To me it only shows that applied work is lower and nothing about how the friction energy going back to the system is affecting T and P, or is the effect shown implicitly in the equation below?

It is shown explicitly in what we are going to do below.

Some insight would be appreciated.

I think what I'll do is write up the heat transfer analysis in a Word Document, and email it to you. That way I can include figures if I want.

From solving the ODE I got:

T(V) = \frac{Const}{V^\frac{R}{C_v}}+\frac{F}{nA(R+C_v)}V

Const = V_o^\frac{R}{C_v}(T_o - \frac{FV_o}{nA(R+C_v)})

T(V) = T_o(\frac{V_o}{V})^\frac{R}{C_v}+\frac{FV_o}{nA(R+C_v)}[(\frac{V_o}{V})^\frac{R}{C_v}-\frac{V}{V_o}]

My solution doesn't match yours. I got a minus sign in from of the last term. The friction term should raise the temperature, not lower it.

T(V) = T_o(\frac{V_o}{V})^\frac{R}{C_v}-\frac{FV_o}{nA(R+C_v)}[(\frac{V_o}{V})^\frac{R}{C_v}-\frac{V}{V_o}] = T_o(\frac{V_o}{V})^\frac{R}{C_v}+\frac{FV}{nAC_p}[1-(\frac{V_o}{V})^\frac{C_p}{C_v}]
So,

T(V) = T_o\left(\frac{V_o}{V}\right)^{γ-1}+\frac{FV}{nAC_p}\left[1-\left(\frac{V_o}{V}\right)^γ\right]
Note that the first term is the temperature change if there were no friction, and the second term is the temperature rise resulting from the friction.

Now, please use the ideal gas law to determine the gas pressure P when the gas has expanded to volume V (in terms of the initial pressure P₀), and then use the calculated temperature T to determine the total change in internal energy when the gas expands to volume V. From this, and the first law, you can also get the total amount of work done. What are your conclusions from these results?

Chet

 

[Red_CCF]

I think what I'll do is write up the heat transfer analysis in a Word Document, and email it to you. That way I can include figures if I want.

My solution doesn't match yours. I got a minus sign in from of the last term. The friction term should raise the temperature, not lower it.

T(V) = T_o(\frac{V_o}{V})^\frac{R}{C_v}-\frac{FV_o}{nA(R+C_v)}[(\frac{V_o}{V})^\frac{R}{C_v}-\frac{V}{V_o}] = T_o(\frac{V_o}{V})^\frac{R}{C_v}+\frac{FV}{nAC_p}[1-(\frac{V_o}{V})^\frac{C_p}{C_v}]
So,

T(V) = T_o\left(\frac{V_o}{V}\right)^{γ-1}+\frac{FV}{nAC_p}\left[1-\left(\frac{V_o}{V}\right)^γ\right]
Note that the first term is the temperature change if there were no friction, and the second term is the temperature rise resulting from the friction.

Now, please use the ideal gas law to determine the gas pressure P when the gas has expanded to volume V (in terms of the initial pressure P₀), and then use the calculated temperature T to determine the total change in internal energy when the gas expands to volume V. From this, and the first law, you can also get the total amount of work done. What are your conclusions from these results?

Chet

My bad, I had a typo in the last equation and the plus on the last term should be a minus.

The cylinder pressure:

P(V) = P_o\left(\frac{V_o}{V}\right)^{γ}+\frac{FR}{AC_p}\left[1-\left(\frac{V_o}{V}\right)^γ\right]

The gas/system internal energy change:

ΔU = nC_v(T(V) - T_o) where T(V) is as before

Since this is expansion, V_o/V < 1, so the last term in the P(V) and T(V) equation is always positive. This tells me that the gas pressure at any point during an expansion when there is piston-cylinder friction will always be higher compared to that of a reversible expansion at the same V. The same can be said for gas temperature/internal energy. Is this correct?

Thanks very much

 

[Chestermiller]

My bad, I had a typo in the last equation and the plus on the last term should be a minus.

The cylinder pressure:

P(V) = P_o\left(\frac{V_o}{V}\right)^{γ}+\frac{FR}{AC_p}\left[1-\left(\frac{V_o}{V}\right)^γ\right]

The gas/system internal energy change:

ΔU = nC_v(T(V) - T_o) where T(V) is as before

Since this is expansion, V_o/V < 1, so the last term in the P(V) and T(V) equation is always positive. This tells me that the gas pressure at any point during an expansion when there is piston-cylinder friction will always be higher compared to that of a reversible expansion at the same V. The same can be said for gas temperature/internal energy. Is this correct?

Yes. I was hoping that you would substitute your equation for T(V) into the equation for ΔU. This will then also tell you the total amount of work done for the expansion.

What we are looking at here is an irreversible process (even though the deformation is quasi static). We can confirm this by calculating the change in entropy ΔS for the gas. Since the process path is adiabatic, from Cauchy's inequality, the entropy change for the system should be positive if the process is irreversible (since ∫dQ/T_I is equal to zero). Please determine the change in entropy.
ΔS=n(C_pln(T(V)/T_0)+Rln(P_0/P(V)))

Also, please start thinking about how we can modify these results if the deformation is compression, rather than expansion (without doing the entire analysis over again).

Chet

 

[Red_CCF]

Yes. I was hoping that you would substitute your equation for T(V) into the equation for ΔU. This will then also tell you the total amount of work done for the expansion.

ΔU = nC_v(\frac{FV}{nAC_p}\left[1-\left(\frac{V_o}{V}\right)^γ\right]- T_o\left[1-\left(\frac{V_o}{V}\right)^{γ-1}\right])

From the above equation I find that |ΔU| decreases with increasing F, correlating to less work done with greater friction as expected.

Also, please start thinking about how we can modify these results if the deformation is compression, rather than expansion (without doing the entire analysis over again).

Chet

I believe that for compression I can sub F as -F, recalculate the arbitrary constant, and every term in the T(V) equation becomes a plus.

T(V) = T_o\left(\frac{V_o}{V}\right)^{γ-1}+\frac{FV}{nAC_p}\left[1+\left(\frac{V_o}{V}\right)^γ\right]

In this case, the temperature of the system is also higher compared to that of the reversible case.

What we are looking at here is an irreversible process (even though the deformation is quasi static). We can confirm this by calculating the change in entropy ΔS for the gas. Since the process path is adiabatic, from Cauchy's inequality, the entropy change for the system should be positive if the process is irreversible (since ∫dQ/T_I is equal to zero). Please determine the change in entropy.
ΔS=n(C_pln(T(V)/T_0)+Rln(P_0/P(V)))
Chet

ΔS=n(C_pln(\left(\frac{V_o}{V}\right)^{γ-1}+\frac{FV}{nAC_pT_o}\left[1-\left(\frac{V_o}{V}\right)^γ\right])-Rln(\left(\frac{V_o}{V}\right)^{γ}+\frac{FR}{AC_pP_o}\left[1-\left(\frac{V_o}{V}\right)^γ\right]) &gt; 0

I am unsure how to prove that the components inside the natural logarithm is greater than 1 or the possibility that one of the logarithms are negative and the other one positive with a greater magnitude. I have always seen the second law imposed on situations like these to find the relationship between state properties.

Slightly off topic, but in our previous discussion it was mentioned that the TdS equations were only valid on a differential basis for irreversible processes. However, it is used in this case and also in a combustion text I've been reading (entropy change for carbon-oxygen combustion in an isolated system). How come this would be valid?

Thanks very much

 

[Chestermiller]

ΔU = nC_v(\frac{FV}{nAC_p}\left[1-\left(\frac{V_o}{V}\right)^γ\right]- T_o\left[1-\left(\frac{V_o}{V}\right)^{γ-1}\right])

From the above equation I find that |ΔU| decreases with increasing F, correlating to less work done with greater friction as expected.



I believe that for compression I can sub F as -F, recalculate the arbitrary constant, and every term in the T(V) equation becomes a plus.

T(V) = T_o\left(\frac{V_o}{V}\right)^{γ-1}+\frac{FV}{nAC_p}\left[1+\left(\frac{V_o}{V}\right)^γ\right]

In this case, the temperature of the system is also higher compared to that of the reversible case.



ΔS=n(C_pln(\left(\frac{V_o}{V}\right)^{γ-1}+\frac{FV}{nAC_pT_o}\left[1-\left(\frac{V_o}{V}\right)^γ\right])-Rln(\left(\frac{V_o}{V}\right)^{γ}+\frac{FR}{AC_pP_o}\left[1-\left(\frac{V_o}{V}\right)^γ\right]) &gt; 0

I am unsure how to prove that the components inside the natural logarithm is greater than 1 or the possibility that one of the logarithms are negative and the other one positive with a greater magnitude. I have always seen the second law imposed on situations like these to find the relationship between state properties.

Slightly off topic, but in our previous discussion it was mentioned that the TdS equations were only valid on a differential basis for irreversible processes. However, it is used in this case and also in a combustion text I've been reading (entropy change for carbon-oxygen combustion in an isolated system). How come this would be valid?

Thanks very much

Hi. I don't have time to address everything right now, because we are on our way out to adopt our new dog. But I'll be back later.

Your equation for ΔS looks very daunting, but it can be simplified tremendously. In the first ln expression, use the ideal gas law to express the "F coefficient" in terms of P₀. Note how it then compares with the "F coefficient" in the second ln term. Then you can also make use of alnx=ln(x^a) and also lnx + lny = ln(xy). Play around with the equation, and see what you can come up with.

Chet

 

[Chestermiller]

<br /> <br /> Slightly off topic, but in our previous discussion it was mentioned that the TdS equations were only valid on a differential basis for irreversible processes. However, it is used in this case and also in a combustion text I&#039;ve been reading (entropy change for carbon-oxygen combustion in an isolated system). How come this would be valid?<br />

<br /> I don&#039;t remember saying that the &quot;TdS equations&quot; were only valid on a differential basis for <b>irreversible</b> processes. If I did say that, I didn&#039;t mean to. I meant reversible. And, of course, no matter how the change between two differentially separated equilibrium states takes place, the TdS equations always describes the relationship between dS, dU, and dV between these two closely neighboring equilibrium states.<br /> <br /> In the present problem (which is irreversible), we did not use the &quot;TdS equations.&quot; The relationship we used for ΔS was derived in general for any ideal gas reversible path between any two equilibrium end states, including our irreversible process.<br /> <br /> Chet

 

[Red_CCF]

Your equation for ΔS looks very daunting, but it can be simplified tremendously. In the first ln expression, use the ideal gas law to express the "F coefficient" in terms of P₀. Note how it then compares with the "F coefficient" in the second ln term. Then you can also make use of alnx=ln(x^a) and also lnx + lny = ln(xy). Play around with the equation, and see what you can come up with.

Chet

After playing around with the equation I got:

ΔS=nln\left(\frac{\left[\left(\frac{V_o}{V}\right)^{γ}\left[1-\left(\frac{FR}{AC_pP_o}\right)\right]+ \frac{FR}{AC_pP_o}\right]^{C_v}}{\left(\frac{V_o}{V}\right)^{C_p}}\right)<br />

It seems unlikely that the numerator be smaller than the denominator, though I'm not sure if there is a way to prove this.

I don't remember saying that the "TdS equations" were only valid on a differential basis for irreversible processes. If I did say that, I didn't mean to. I meant reversible. And, of course, no matter how the change between two differentially separated equilibrium states takes place, the TdS equations always describes the relationship between dS, dU, and dV between these two closely neighboring equilibrium states.

So is the integral of the TdS equation between any two equilibrium states valid for any process, including irreversible and why? Is it only valid on a differential basis for reversible processes because they traverses through differentially separated equilibrium states?

In the present problem (which is irreversible), we did not use the "TdS equations." The relationship we used for ΔS was derived in general for any ideal gas reversible path between any two equilibrium end states, including our irreversible process.

Chet

From my recollection, the ΔS was derived from the TdS equations via substitution of the ideal gas law and constant specific heats, so would their underlying assumptions/range of applications be the same or are their additional/different constraints?

Thank you very much

 

[Chestermiller]

After playing around with the equation I got:

ΔS=nln\left(\frac{\left[\left(\frac{V_o}{V}\right)^{γ}\left[1-\left(\frac{FR}{AC_pP_o}\right)\right]+ \frac{FR}{AC_pP_o}\right]^{C_v}}{\left(\frac{V_o}{V}\right)^{C_p}}\right)<br />

It seems unlikely that the numerator be smaller than the denominator, though I'm not sure if there is a way to prove this.

Hi Red_CCF,

Here's what I came up with regarding ΔS, starting with your original equation and using the ideal gas law:

ΔS=n(C_pln(\left(\frac{V_o}{V}\right)^{γ-1}+\frac{FRV}{AC_pP_oV_0}\left[1-\left(\frac{V_o}{V}\right)^γ\right])-Rln(\left(\frac{V_o}{V}\right)^{γ}+\frac{FR}{AC_pP_o}\left[1-\left(\frac{V_o}{V}\right)^γ\right])

If I factor V/V₀ out of the first log term, I get:

ΔS=n C_pln\left(\frac{V}{V_0}\right)+<br /> nC_pln(\left(\frac{V_o}{V}\right)^{γ}+\frac{FR}{AC_pP_o}\left[1-\left(\frac{V_o}{V}\right)^γ\right])-nRln(\left(\frac{V_o}{V}\right)^{γ}+\frac{FR}{AC_pP_o}\left[1-\left(\frac{V_o}{V}\right)^γ\right])

If I now combine the two log terms, I get:

ΔS=n C_pln\left(\frac{V}{V_0}\right)+<br /> n(C_p-R)ln\left(\left(\frac{V_o}{V}\right)^{γ}+\frac{FR}{AC_pP_o}\left[1-\left(\frac{V_o}{V}\right)^γ\right]\right)
If I factor out (V_0/V)^γ from the log term, I get:

ΔS=n C_pln\left(\frac{V}{V_0}\right)+n(C_p-R)γln\left(\frac{V_0}{V}\right)+<br /> n(C_p-R)ln\left(1+\frac{FR}{AC_pP_o}\left[\left(\frac{V}{V_0}\right)^γ-1\right]\right)

The first two terms exactly cancel one another, so we are left with:

ΔS=nC_vln\left(1+\frac{FR}{AC_pP_o}\left[\left(\frac{V}{V_0}\right)^γ-1\right]\right)
For expansion, the second term in parenthesis is clearly greater than zero, so ΔS>0.

So is the integral of the TdS equation between any two equilibrium states valid for any process, including irreversible and why?

The integral of the TdS equation between any two equilibrium states is independent of the process. It only depends on the initial and final equilibrium states.

Is it only valid on a differential basis for reversible processes because they traverses through differentially separated equilibrium states?

As I said, it is independent of the process. Since a reversible path must be used to calculate the change in entropy between any two equilibrium states, the TdS equation is the perfect vehicle for doing this because it automatically gives the same result as integrating along any arbitrary reversible path, even if the reversible path is split into several different segments (for example, adiabatic and isothermal).

From my recollection, the ΔS was derived from the TdS equations via substitution of the ideal gas law and constant specific heats, so would their underlying assumptions/range of applications be the same or are their additional/different constraints?

The particular relationship we used for determining ΔS was obtained by integrating the TdS equations specifically for the case of an ideal gas with constant heat capacity, so it doesn't apply outside the ideal gas region. However, the TdS equations can also be integrated to get ΔS for substances that are not ideal gases.

Chet

 

[Red_CCF]

The first two terms exactly cancel one another, so we are left with:

ΔS=nC_vln\left(1+\frac{FR}{AC_pP_o}\left[\left(\frac{V}{V_0}\right)^γ-1\right]\right)
For expansion, the second term in parenthesis is clearly greater than zero, so ΔS>0.

It looks like I did not simplify my equation far enough. After factoring out (V_o/V)^γ from the numerator I got the same result. Thank you for the full derivation.

From the result, my impression is that friction causes the system T and P to be higher than that of its reversible counterpart for the same volume change in both compression and expansion and at the same time the work extract during expansion is lower and work applied during compression is higher (for the same ΔV), is this correct?

The integral of the TdS equation between any two equilibrium states is independent of the process. It only depends on the initial and final equilibrium states.

As I said, it is independent of the process. Since a reversible path must be used to calculate the change in entropy between any two equilibrium states, the TdS equation is the perfect vehicle for doing this because it automatically gives the same result as integrating along any arbitrary reversible path, even if the reversible path is split into several different segments (for example, adiabatic and isothermal).

I accept that the TdS equation is valid for any process reversible or not, but I'm just wondering if there is a proof or argument that shows why this is so. For a general irreversible process, I do not see how one can integrate TdS since the system isn't necessarily in equilibrium during the process. So is the argument that, since the TdS equation is a function of state properties only, if we are interested in Δs for some irreversible process, we can evaluate the integral for a reversible process between the same end states and the numerical value of Δs would be the same?

Thank you very much

 

[Chestermiller]

It looks like I did not simplify my equation far enough. After factoring out (V_o/V)^γ from the numerator I got the same result. Thank you for the full derivation.

From the result, my impression is that friction causes the system T and P to be higher than that of its reversible counterpart for the same volume change in both compression and expansion and at the same time the work extract during expansion is lower and work applied during compression is higher (for the same ΔV), is this correct?

Yes.

I accept that the TdS equation is valid for any process reversible or not, but I'm just wondering if there is a proof or argument that shows why this is so.

The integral of the TdS equation is valid for any process, reversible or not because the changes in S, U, and V depend only on the beginning and end equilibrium states. Since the TdS equation describes differential changes between closely neighboring equilibrium states, the integral implies moving along a reversible path.

For a general irreversible process, I do not see how one can integrate TdS since the system isn't necessarily in equilibrium during the process. So is the argument that, since the TdS equation is a function of state properties only, if we are interested in Δs for some irreversible process, we can evaluate the integral for a reversible process between the same end states and the numerical value of Δs would be the same?

Yes, that's exactly what I've been trying to say, although you articulated it much better.

Chet

 

[Red_CCF]

Yes.

My original confusion arises from this. I have seen graphs in books such at the one attached, where the P-v curve for an irreversible expansion lies below that of the reversible (in the attached graph the dotted line is the reversible process). Is there something I'm missing, because the derivations above challenges this.

Also, if wall friction was neglected but viscous dissipation is now the only irreversibility, qualitatively would the trend in the attached graph be accurate since τ_xx is negative during expansion such that σ_I decreases compared to the reversible case?

The integral of the TdS equation is valid for any process, reversible or not because the changes in S, U, and V depend only on the beginning and end equilibrium states. Since the TdS equation describes differential changes between closely neighboring equilibrium states, the integral implies moving along a reversible path.

Yes, that's exactly what I've been trying to say, although you articulated it much better.
Chet

With regards to the ΔS equation for ideal gases:

ΔS=n(C_pln(T(V)/T_0)+Rln(P_0/P(V)))

For our example, since the process is quasistatic, a path of equilibrium states exists. Would integrating dS = c_p dT/T - R dp/p directly as opposed to using the above equation give the same result? Does the assumption of δQ = TdS in the equation (as opposed to δQ ≤ TdS in reality) impact this in anyway?

Is P(V) here supposed to be the interface pressure? In my substitution I used the gas pressure, which is < P_ext; should I have added F/A to P(V)?

Is the equation in this form valid for combustion in an isolated system, or must \Sigmaμ_iN_i be appended to the end of this equation?

Thank you very much

 

[rkmurtyp]

I was referred to this thread by Chet.

This is too general a question. After you've worked a few problems, you will recognize it right away.

OK. Here we go.

Here are the assumptions I'm proposing to use:

1. During the expansion or compression, the friction force on the piston is constant at F.
2. The piston is massless, and has zero heat capacity
3. All the heat generated between the wall and the piston ends up in the gas
4. The expansion and compression are carried out quasi-statically

Are you in agreement with these assumptions?
Are there any others that you think we should be using?

Which do you want to do first, expansion or compression?
What to you want included in the system: (a) just the gas or (b) the gas plus the piston? (I'm suggesting we eventually do it both ways to see how the two analyses compare).

Chet

The points 1 and 4 above need a comment.

1. Since we are interested in a process of irreversible adiabatic expansion (or compression) of an ideal gas, it is not necessary that the external pressure (Force on the pistion) be constsnt. The necessary and sufficient condition is that 1. for expansion the external pressure, P_ext plus F/A be less than the pressure of the gas, P_sym and 2. for compression the external pressure, P_extis greater than the pressure of the gas, P_symplus F/A. F is the frictional force and A the area of the pistion over which F acts.

4. Irreversible adiabatic expansion (or compression) involving friction and quasistatic process don't go together - they are mutually exclusive. If we assume presence of friction, then quasistatic process is impossible; similarly, if we assume quasistatic process absence of friction is implied.

With these in mind we discuss below, the calculation of entropy change for an irreversible adiabatic expansion (or compression) process AB of an ideal gas.

Irreversible adiabatic expansion process AB of an ideal gas.

Let the system (a fixed amount of ideal gas) go from an initial equilibrium state A (P_A, V_A, T_A) to a final equilibrium state B (P_B, V_B, T_B) by an irreversible adiabatic process involving friction. Note that presence of friction is not a necessary condition for irreversible expansion of an ideal gas. The entropy of the gas changes from S_A to S_B, The change in the entropy of the system, ΔS_sym =S_B-S_A. During this change the surroundings suffer no entropy change. ΔS_sur=0. ∴ ΔS_sym = ΔS_AB = ΔS_univ>0.

If the gas had expanded revrsibly (without friction) from state A through the same volume change, it would have reached equilibrium state C (P_C, V_C, T_C), where V_B=V_C and T_B>T_C.

From state B let us cool the system reversibly at constant volume to state C. This process requires a regenerator - an ideal device of a set of heat reservoirs (HRs) of continuously varying temperature between any two desired values of temperatures. Since this process BC is reversible ΔS_univ=0. ∴ ΔS_BC=-ΔS_CB= - ΔS_sur= -∑(Q_i/T_i), where Q_i is the heat change suffered (absorbed) by HR at temperature T_i.

Since states A and C lie on a reversible adiabat, entropy values of the system in these two states is the same ie S_A=S_C. ∴ ΔS_BA=-ΔS_AB =ΔS_BC= -∑(Q_i/T_i) <0. ∴ ΔS_AB ∑(Q_i/T_i > 0.

An alternate way of seeing this is: When the system reached state B from state A by a spontaneous (irreversible) process, we take the system back from state B to state A by a reversible path (process) BCA, BC being a reversible cooling process and CA being a reversible adiabatic compression process, thus completing a cycle. For such a cyclic process ΔS_sym=0. ∴ ΔS_AB=-ΔS_BCA=ΔS_sur= ∑(Q_i/T_i>0.

Calculation of ΔS for an irreversible adiabatic compression process AB of an ideal gas can be carried out on similar lines.

 

[Red_CCF]

For a quasistatic (reversible) process we need to satisfy the condition P_ext= P_sym. When friction is present it is impossible to satisfy this condition.

This was the first time I came across the fact that processes with friction cannot be quasistatic as everything I've read up to this point (such as the sources below and other threads) including has indicated that it is possible to have a quasistatic process with friction, so I'm having a tough time accepting this. If the argument for friction applies, what about a process with heat transfer with finite temperature gradient; can those be quasistatic then?

From Wikipedia: http://en.wikipedia.org/wiki/Quasistatic_process

An example of a quasistatic process that is not reversible is a compression against a system with a piston subject to friction — although the system is always in thermal equilibrium, the friction ensures the generation of dissipative entropy, which directly goes against the definition of reversible.

From a journal article (I only have access to the abstract unfortunately): http://adsabs.harvard.edu/abs/1960AmJPh..28..119T

Quasi-static processes are not reversible when sliding friction forces are present. An example is considered consisting of a cylinder containing a gas and equipped with a piston for which sliding friction forces are significant.

The speed (or time rate of change) of a process (slow enough of fast enough etc) does not and should not enter thermodynamic arguments - time has no role to play in (equilibrium) thermodynamics.

A process that takes the system through a series of continuous set of equilibrium states has no chance to be out of equilibrium ever, consequently be irreversible ever!

I've always thought of the slow enough argument to be qualitative to mean that the rate in which the process proceeds << the rate in which equilibrium is established during the process after an infinitessimal disturbance (an infinitessimal movement of the cylinder). Quantitatively I do not believe time ever factors into any analysis.

Irreversible adiabatic expansion process AB of an ideal gas.

Let the system (a fixed amount of ideal gas) go from an initial equilibrium state A (P_A, V_A, T_A) to a final equilibrium state B (P_B, V_B, T_B) by an irreversible adiabatic process involving friction. Note that presence of friction is not a necessary condition for irreversible expansion of an ideal gas. The entropy of the gas changes from S_A to S_B, The change in the entropy of the system, ΔS_sym =S_B-S_A. During this change the surroundings suffer no entropy change. ΔS_sur=0. ∴ ΔS_sym = ΔS_AB = ΔS_univ>0.

If the gas had expanded revrsibly (without friction) from state A through the same volume change, it would have reached equilibrium state C (P_C, V_C, T_C), where V_B=V_C and T_B>T_C.

From state B let us cool the system reversibly at constant volume to state C. This process requires a regenerator - an ideal device of a set of heat reservoirs (HRs) of continuously varying temperature between any two desired values of temperatures. Since this process BC is reversible ΔS_univ=0. ∴ ΔS_BC=-ΔS_CB= - ΔS_sur= -∑(Q_i/T_i), where Q_i is the heat change suffered (absorbed) by HR at temperature T_i.

Since states A and C lie on a reversible adiabat, entropy values of the system in these two states is the same ie S_A=S_C. ∴ ΔS_BA=-ΔS_AB =ΔS_BC= -∑(Q_i/T_i) <0. ∴ ΔS_AB ∑(Q_i/T_i > 0.

An alternate way of seeing this is: When the system reached state B from state A by a spontaneous (irreversible) process, we take the system back from state B to state A by a reversible path (process) BCA, BC being a reversible cooling process and CA being a reversible adiabatic compression process, thus completing a cycle. For such a cyclic process ΔS_sym=0. ∴ ΔS_AB=-ΔS_BCA=ΔS_sur= ∑(Q_i/T_i>0.

Calculation of ΔS for an irreversible adiabatic compression process AB of an ideal gas can be carried out on similar lines.

With regards to ΔS_BC=-ΔS_CB= - ΔS_sur= -∑(Qi/Ti); Qi here is w.r.t. the surroundings and hence positive? Other than that everything makes sense.

Thanks very much

 

[Chestermiller]

I was referred to this thread by Chet.


Thanks rkmurtyp.

The points 1 and 4 above need a comment.

1. Since we are interested in a process of irreversible adiabatic expansion (or compression) of an ideal gas, it is not necessary that the external pressure (Force on the pistion) be constsnt. The necessary and sufficient condition is that 1. for expansion the external pressure, P_ext plus F/A be less than the pressure of the gas, P_sym and 2. for compression the external pressure, P_extis greater than the pressure of the gas, P_symplus F/A. F is the frictional force and A the area of the pistion over which F acts.

In the statement of our problem, we were very specific about what we were solving. In particular, we were carrying out the expansion or compression under conditions where the piston is moved extremely slowly (quasi statically), in which case the > and < in the above comment become = signs.

4. Irreversible adiabatic expansion (or compression) involving friction and quasistatic process don't go together - they are mutually exclusive. If we assume presence of friction, then quasistatic process is impossible; similarly, if we assume quasistatic process absence of friction is implied.

We were also very specific about what we considered our system. We chose to regard the system as the combination of cylinder, piston, and gas. For this system, the expansion is adiabatic and quasistatic, since we also assume that all the heat generated at the interface between the piston and the cylinder remains within the system, and goes into the gas.

If was had chosen as our system the gas alone, then the situation would have been quite different. For this system, the process would have been non-adiabatic and non-isothermal, but reversible.

With these in mind we discuss below, the calculation of entropy change for an irreversible adiabatic expansion (or compression) process AB of an ideal gas.

Irreversible adiabatic expansion process AB of an ideal gas.

Let the system (a fixed amount of ideal gas) go from an initial equilibrium state A (P_A, V_A, T_A) to a final equilibrium state B (P_B, V_B, T_B) by an irreversible adiabatic process involving friction. Note that presence of friction is not a necessary condition for irreversible expansion of an ideal gas. The entropy of the gas changes from S_A to S_B, The change in the entropy of the system, ΔS_sym =S_B-S_A. During this change the surroundings suffer no entropy change. ΔS_sur=0. ∴ ΔS_sym = ΔS_AB = ΔS_univ>0.

If the gas had expanded revrsibly (without friction) from state A through the same volume change, it would have reached equilibrium state C (P_C, V_C, T_C), where V_B=V_C and T_B>T_C.

From state B let us cool the system reversibly at constant volume to state C. This process requires a regenerator - an ideal device of a set of heat reservoirs (HRs) of continuously varying temperature between any two desired values of temperatures. Since this process BC is reversible ΔS_univ=0. ∴ ΔS_BC=-ΔS_CB= - ΔS_sur= -∑(Q_i/T_i), where Q_i is the heat change suffered (absorbed) by HR at temperature T_i.

Since states A and C lie on a reversible adiabat, entropy values of the system in these two states is the same ie S_A=S_C. ∴ ΔS_BA=-ΔS_AB =ΔS_BC= -∑(Q_i/T_i) <0. ∴ ΔS_AB ∑(Q_i/T_i > 0.

An alternate way of seeing this is: When the system reached state B from state A by a spontaneous (irreversible) process, we take the system back from state B to state A by a reversible path (process) BCA, BC being a reversible cooling process and CA being a reversible adiabatic compression process, thus completing a cycle. For such a cyclic process ΔS_sym=0. ∴ ΔS_AB=-ΔS_BCA=ΔS_sur= ∑(Q_i/T_i>0.

Calculation of ΔS for an irreversible adiabatic compression process AB of an ideal gas can be carried out on similar lines.

In my judgement, all this is entirely equivalent to what we did in determining the change in entropy of the system. However, it is of value in providing additional information for Red_CCF of how what he calls the "TdS equations" were arrived at.

Chet

 

[Chestermiller]

My original confusion arises from this. I have seen graphs in books such at the one attached, where the P-v curve for an irreversible expansion lies below that of the reversible (in the attached graph the dotted line is the reversible process). Is there something I'm missing, because the derivations above challenges this.

I don't see an attached graph. But, as we discussed previously, there is considerable ambiguity as to what the P-v curve for an irreversible expansion lies, since, typically, in an irreversible expansion, the pressure varies with position throughout the system during an irreversible path.

Also, if wall friction was neglected but viscous dissipation is now the only irreversibility, qualitatively would the trend in the attached graph be accurate since τ_xx is negative during expansion such that σ_I decreases compared to the reversible case?

Again, I don't see the graph.

With regards to the ΔS equation for ideal gases:

ΔS=n(C_pln(T(V)/T_0)+Rln(P_0/P(V)))

For our example, since the process is quasistatic, a path of equilibrium states exists. Would integrating dS = c_p dT/T - R dp/p directly as opposed to using the above equation give the same result?

Those are exact differentials on the right hand side of the equation. So, if you integrate the dS equation between the initial and final states, you end up with the ΔS equation.

Does the assumption of δQ = TdS in the equation (as opposed to δQ ≤ TdS in reality) impact this in anyway?

Well, for the combined system of cylinder, piston, and gas, δQ is equal to zero. So certainly, that wouldn't give us the correct ΔS. See my next answer below.

Is P(V) here supposed to be the interface pressure? In my substitution I used the gas pressure, which is < P_ext; should I have added F/A to P(V)?

The above equations apply exclusively to an ideal gas between equilibrium states. Since there is no change in the piston entropy, we were able to get the change in entropy of the combined system by focusing exclusively on the gas. So for that calculation, the gas pressure would be the correct one to use. If we had used the gas as our system in the first place, then there would have been heat transferred from the piston to the gas, and the gas expansion in our process could have been considered reversible, since the heat was added in vanishingly small increments. So, if we integrated over the exact path of the gas, rather than any other arbitrary path, we would have ended up with the exact same change in entropy.

Is the equation in this form valid for combustion in an isolated system, or must \Sigmaμ_iN_i be appended to the end of this equation?

No. The equation in this form is applicable to an ideal gas without reaction. The case with reaction must take into account the changes in entropy as a result of the reaction. That's beyond what we are talking about here.

Chet

 

[rkmurtyp]

This was the first time I came across the fact that processes with friction cannot be quasistatic as everything I've read up to this point (such as the sources below and other threads) including has indicated that it is possible to have a quasistatic process with friction, so I'm having a tough time accepting this. If the argument for friction applies, what about a process with heat transfer with finite temperature gradient; can those be quasistatic then?

From Wikipedia: http://en.wikipedia.org/wiki/Quasistatic_process


From a journal article (I only have access to the abstract unfortunately): http://adsabs.harvard.edu/abs/1960AmJPh..28..119T


I've always thought of the slow enough argument to be qualitative to mean that the rate in which the process proceeds << the rate in which equilibrium is established during the process after an infinitessimal disturbance (an infinitessimal movement of the cylinder). Quantitatively I do not believe time ever factors into any analysis.


With regards to ΔS_BC=-ΔS_CB= - ΔS_sur= -∑(Qi/Ti); Qi here is w.r.t. the surroundings and hence positive? Other than that everything makes sense.

Thanks very much

The references imply this: For quasistatic processes the difference in the values of the intensive properties of the system and surrounding differ by infinitesimal amounts. To that infinitesimal extent the process would be irreversible; only in the case the differences are zero reversibility is attained - when not zero no reversibility (one may coin hundred names such as quasistatic etc, they all continue to remain irreversible). I have used in my arguments reversibility and quasistatic synonimously and ΔP to be zero ( not tending to zero!).

'Rverersibility' concept is a beatiful mental costruct that aids in simplifying the concepts and helps understanding - just as massless, frictionless pistons and pulleys help understanding the concepts easily.


I've always thought of the slow enough argument to be qualitative to mean that the rate in which the process proceeds << the rate in which equilibrium is established during the process after an infinitessimal disturbance (an infinitessimal movement of the cylinder). Quantitatively I do not believe time ever factors into any analysis.

True, time plays no role in thermodynamic processes.


With regards to ΔS_BC=-ΔS_CB= - ΔS_sur= -∑(Qi/Ti); Qi here is w.r.t. the surroundings and hence positive? Other than that everything makes sense.

You are right; Qi here is w.r.t. the surroundings and hence positive - remember when HRs receive heat Qi is positive, and negative when they lose heat.

 

[Red_CCF]

The references imply this: For quasistatic processes the difference in the values of the intensive properties of the system and surrounding differ by infinitesimal amounts. To that infinitesimal extent the process would be irreversible; only in the case the differences are zero reversibility is attained - when not zero no reversibility (one may coin hundred names such as quasistatic etc, they all continue to remain irreversible). I have used in my arguments reversibility and quasistatic synonimously and ΔP to be zero ( not tending to zero!).

'Rverersibility' concept is a beatiful mental costruct that aids in simplifying the concepts and helps understanding - just as massless, frictionless pistons and pulleys help understanding the concepts easily.

I have always thought that quasistatic processes are one where the perturbations are infinitessimal (i.e. an infinitessimally higher P during compression or a temperature gradient where the temperature differences are infinitessimal). If, say for piston-cylinder compression, ΔP = 0 and Pext and the system pressure equal exactly, I am not sure physically how the piston would move to begin with, since the net force would be exactly zero on a still piston.

Thanks very much

 

[Red_CCF]

I don't see an attached graph. But, as we discussed previously, there is considerable ambiguity as to what the P-v curve for an irreversible expansion lies, since, typically, in an irreversible expansion, the pressure varies with position throughout the system during an irreversible path.

Again, I don't see the graph.

My apologies, the graph should (hopefully) be visible now.

Those are exact differentials on the right hand side of the equation. So, if you integrate the dS equation between the initial and final states, you end up with the ΔS equation.

Well, for the combined system of cylinder, piston, and gas, δQ is equal to zero. So certainly, that wouldn't give us the correct ΔS. See my next answer below.

So to summarize, the integral of dS = c_p dT/T - Rdp/p to give the ΔS equation assumes a reversible path between initial and final states. Had I substituted p = P(V) and T = T(V) such that dS = c_p dT/T(V) - Rdp/p(V) and then integrated, I should not expect to get the same result as substituting T(V) and P(V) into the ΔS equation (as we had done earlier) as dS ≠ δQ/T?

The above equations apply exclusively to an ideal gas between equilibrium states. Since there is no change in the piston entropy, we were able to get the change in entropy of the combined system by focusing exclusively on the gas. So for that calculation, the gas pressure would be the correct one to use. If we had used the gas as our system in the first place, then there would have been heat transferred from the piston to the gas, and the gas expansion in our process could have been considered reversible, since the heat was added in vanishingly small increments. So, if we integrated over the exact path of the gas, rather than any other arbitrary path, we would have ended up with the exact same change in entropy.

Since the piston ΔS = 0 anyways, does this mean that had I chosen to add F/A to P(V) (essentially use P_ext instead), I should still get the same change in entropy using the ΔS equation, or does that equation assume the entire system is an ideal gas?

Thanks very much

 

[rkmurtyp]

I have always thought that quasistatic processes are one where the perturbations are infinitessimal (i.e. an infinitessimally higher P during compression or a temperature gradient where the temperature differences are infinitessimal). If, say for piston-cylinder compression, ΔP = 0 and Pext and the system pressure equal exactly, I am not sure physically how the piston would move to begin with, since the net force would be exactly zero on a still piston.

Thanks very much

This is the reason why reversibility is a mental construct, to imagine it in practice we invoke that quasistatic process and use all the jargon of infinitesimal differeneces etc.

Think for yourself how one can have massless frictionless pulleys, pistons etc to carry out expts in mechanics, then you will be able to appreciate the concept of reversibility in thermodynamics.

 

[rkmurtyp]

In the statement of our problem, we were very specific about what we were solving. In particular, we were carrying out the expansion or compression under conditions where the piston is moved extremely slowly (quasi statically), in which case the > and < in the above comment become = signs.

The 'extreamly slowly' doesn't qualify quasistatic condition; the inequalities define conditions for irreversible and equality sign defines conditions for reversible process


We were also very specific about what we considered our system. We chose to regard the system as the combination of cylinder, piston, and gas. For this system, the expansion is adiabatic and quasistatic, since we also assume that all the heat generated at the interface between the piston and the cylinder remains within the system, and goes into the gas.

By choosing the combination of cylinder, piston and gas as the system we sufficiently ill define the system (its properties) we end up no where in the analysis!

If was had chosen as our system the gas alone, then the situation would have been quite different. For this system, the process would have been non-adiabatic and non-isothermal, but reversible.

It is not clear to me what you are saying


In my judgement, all this is entirely equivalent to what we did in determining the change in entropy of the system. However, it is of value in providing additional information for Red_CCF of how what he calls the "TdS equations" were arrived at.

So far so good.

Chet

 

[Chestermiller]

My apologies, the graph should (hopefully) be visible now.

My original confusion arises from this. I have seen graphs in books such at the one attached, where the P-v curve for an irreversible expansion lies below that of the reversible (in the attached graph the dotted line is the reversible process). Is there something I'm missing, because the derivations above challenges this.

Not really. P in the figure is what we have been calling P_ext, which inherently assumes that the gas, cylinder, and piston constitute the system. On expansion, P_ext is less than the gas pressure would have been if the gas expansion were reversible. (See if you can verify this.) On compression, P_ext is greater than the gas pressure would have been if the gas expansion were reversible. (See if you can verify this.)

Also, if wall friction was neglected but viscous dissipation is now the only irreversibility, qualitatively would the trend in the attached graph be accurate since τ_xx is negative during expansion such that σ_I decreases compared to the reversible case?

In an irreversible expansion, P_ext is less than the gas pressure would have been if the gas expansion were reversible. Your interpretation of why is probably correct.

So to summarize, the integral of dS = c_p dT/T - Rdp/p to give the ΔS equation assumes a reversible path between initial and final states.

No, the ΔS equation applies to an ideal gas between two equilibrium states, irrespective of whether the actual process path was reversible or not.

Had I substituted p = P(V) and T = T(V) such that dS = c_p dT/T(V) - Rdp/p(V) and then integrated, I should not expect to get the same result as substituting T(V) and P(V) into the ΔS equation (as we had done earlier) as dS ≠ δQ/T?

If the system is the gas alone, then substituting the initial and final states into the ΔS equation gives the correct result, because, for the gas alone, δQ was not equal to zero. And, in our problem, with the system taken as the gas only, the expansion is reversible. And for that path for the gas, dS=δQ/T. However, if we took the system as the cylinder, piston, and gas, since this system is insulated, and δQ for that system = 0 for the entire path. As we've said earlier, it all depends on what you take as your system.

Since the piston ΔS = 0 anyways, does this mean that had I chosen to add F/A to P(V) (essentially use P_ext instead), I should still get the same change in entropy using the ΔS equation, or does that equation assume the entire system is an ideal gas?

That would assume that the entire system is an ideal gas, which it isn't. You can only use the ΔS equation we employed for an ideal gas.

Try to think of the entropy as a physical property of the material (in this case an ideal gas) rather than something associated with the specific nature of the process path.Chet

 

[Red_CCF]

Not really. P in the figure is what we have been calling P_ext, which inherently assumes that the gas, cylinder, and piston constitute the system. On expansion, P_ext is less than the gas pressure would have been if the gas expansion were reversible. (See if you can verify this.) On compression, P_ext is greater than the gas pressure would have been if the gas expansion were reversible. (See if you can verify this.)

Pext= P_o\left(\frac{V_o}{V}\right)^{γ}+\frac{FR}{AC_p}\left[1-\left(\frac{V_o}{V}\right)^γ\right] - F/A

Pext= P_o\left(\frac{V_o}{V}\right)^{γ}+\frac{F}{A}\left[\frac{R}{c_p}(1-\left(\frac{V_o}{V}\right)^γ)-1\right]

R/c_p = 1 - c_v/c_p &lt; 1
1-(\frac{V_o}{V})^γ &lt; 1

Hence the second term in the Pext equation is negative, and Pext is always lower than that from the reversible case (where Pext = P(V)). So the gas pressure/temperature (P(V), T(V)) always increases with friction regardless of expansion or compression, but the applied pressure to the surroundings is lower in expansion but higher in compression?

No, the ΔS equation applies to an ideal gas between two equilibrium states, irrespective of whether the actual process path was reversible or not.

I think I worded my question poorly. I meant to ask, although the final ΔS equation is valid for any process, my understanding is that the differential form dS = c_p dT/T - R dp/p was constructed for a reversible process (as it assumes TdS = Q) and integrated along a reversible path (due to the substitution of the ideal gas law)?

That would assume that the entire system is an ideal gas, which it isn't. You can only use the ΔS equation we employed for an ideal gas.

Try to think of the entropy as a physical property of the material (in this case an ideal gas) rather than something associated with the specific nature of the process path.

Would this be due to the fact that c_p comes into play (as it is a material property)?

Thanks

 

[Chestermiller]

Pext= P_o\left(\frac{V_o}{V}\right)^{γ}+\frac{FR}{AC_p}\left[1-\left(\frac{V_o}{V}\right)^γ\right] - F/A

Pext= P_o\left(\frac{V_o}{V}\right)^{γ}+\frac{F}{A}\left[\frac{R}{c_p}(1-\left(\frac{V_o}{V}\right)^γ)-1\right]

R/c_p = 1 - c_v/c_p &lt; 1
1-(\frac{V_o}{V})^γ &lt; 1

Hence the second term in the Pext equation is negative, and Pext is always lower than that from the reversible case (where Pext = P(V)). So the gas pressure/temperature (P(V), T(V)) always increases with friction regardless of expansion or compression, but the applied pressure to the surroundings is lower in expansion but higher in compression?

Yes.

I think I worded my question poorly. I meant to ask, although the final ΔS equation is valid for any process, my understanding is that the differential form dS = c_p dT/T - R dp/p was constructed for a reversible process (as it assumes TdS = Q) and integrated along a reversible path (due to the substitution of the ideal gas law)?

The only way I know of for getting the change in entropy between two equilibrium states of a system is to follow a reversible path between the two equilibrium states. The TdS equations provide an automatic way of doing this. But in our problem, we could just as easily have followed a path in which we first adiabatically and reversibly expanded the gas to its final volume, and then reversibly transferred heat at constant volume to arrive at the final temperature. Or we could have expanded it isothermally and reversibly to its final pressure and then transferred heat reversibly at constant pressure to bring it to its final temperature.

Would this be due to the fact that c_p comes into play (as it is a material property)?

Not really. The entropy per unit mass is a function of state of the material, every bit as much as internal energy, enthalpy, and heat capacity. For a single phase material, it is a function of temperature and pressure (or any other pair of intensive properties). Don't forget to consider the statistical thermo interpretation of entropy, which is related to the number of quantum mechanical states that the material exhibits.

Hey Red_CCF, remember when we were talking about the analogy between a gas undergoing an irreversible expansion (e.g., say free expansion) and the response of a multiple spring-mass system that is suddenly released from being compressed. Well the following link has an analysis and results for this problem (see post #16) in the limit of a continuous spring:
https://www.physicsforums.com/showthread.php?t=752813

Look it over, especially the results. Note the non-uniform nature of the spring compression with spatial position during the expansion.

Chet

 

[Red_CCF]

With regards to processes involving a massless piston, I am wondering, in these hypothetical scenarios, are these processes completed instantaneously? Since the piston is massless, it cannot accelerate, thus if beginning at its initial equilibrium state where P_{ext, i} = P_I, and suddenly dP is added (infinitessimal disturbance for quasistatic) or ΔP (finite disturbance for non-quasistatic) to P_{ext, i}, by Newton's third law the P_I must balance P_{ext, i} + dP or ΔP; does this mean that P_I should balance the disturbance at immediately as it occurs?

Not really. The entropy per unit mass is a function of state of the material, every bit as much as internal energy, enthalpy, and heat capacity. For a single phase material, it is a function of temperature and pressure (or any other pair of intensive properties).

I can kind of see the analogy between internal energy and entropy, but I'm having trouble interpreting it as material specific; what if the system is pure but is multiphased, or the system undergoes a chemical reaction (such that the "material" changes), or the system has components of different substances and phases (maybe solid metal with gas like we have), or even gases like air with multiple species; how does one generalize entropy to describe that of a whole system if the system itself has multiple components?

Thanks very much

 

[Chestermiller]

With regards to processes involving a massless piston, I am wondering, in these hypothetical scenarios, are these processes completed instantaneously? Since the piston is massless, it cannot accelerate, thus if beginning at its initial equilibrium state where P_{ext, i} = P_I, and suddenly dP is added (infinitessimal disturbance for quasistatic) or ΔP (finite disturbance for non-quasistatic) to P_{ext, i}, by Newton's third law the P_I must balance P_{ext, i} + dP or ΔP; does this mean that P_I should balance the disturbance at immediately as it occurs?

Sure. That's what Newton's second law tells us. Just draw a free body diagram of the piston. If the piston is massless, then ma = 0.

I can kind of see the analogy between internal energy and entropy, but I'm having trouble interpreting it as material specific; what if the system is pure but is multiphased, or the system undergoes a chemical reaction (such that the "material" changes), or the system has components of different substances and phases (maybe solid metal with gas like we have), or even gases like air with multiple species; how does one generalize entropy to describe that of a whole system if the system itself has multiple components?

If you have multiple phases of a single substance, then, no problem. You just get the entropy of the material in each phase, taking into account the change in entropy for the phase transition.

In the case of mixtures or solutions (chemical reaction, multiple components), J. Williard Gibbs figured out how to extend the theory to include this back in the early 1900's. For ideal gas mixtures, this is presented very nicely in Chapter 10 of Smith and Van Ness. What you end up doing is determining the entropy per unit mass of each individual species in the mixture. This requires the inclusion of an additional term in the equation for the specific entropy of each species (over and above the total pressure ratio term and the temperature ratio term).

Take a look at Chapter 10, and carefully read over the section on ideal gas mixtures. I think you'll find it very worthwhile.

Chet

 

[Red_CCF]

Sure. That's what Newton's second law tells us. Just draw a free body diagram of the piston. If the piston is massless, then ma = 0.

Would one be able to analyze the effects of viscous dissipations with a massless piston involved and/or would there be an error if this is done? If P_I (or σ_I) equilibrates to P_ext + dP or ΔP instantaneously,, wouldn't that have a significant effect on du/dx required to find τ_xx?

I was also thinking a little more about static/kinetic friction effects. If the piston is stationary before a compression process such that P_I + F_static/A = P_ext, once dP to get the piston moving, friction becomes kinetic and instantaneously drops such that there is a finite pressure difference between P_I + F_kinetic/A and P_ext + dP. If the piston is massless, P_I would increase instantaneously such that P_I + F_kinetic/A = P_ext + dP and since the piston is massless it should stop. But a finite pressure must be added to P_ext in addition to dP to overcome the static friction to get the piston moving again. Is the above still considered quasistatic, or is there some assumption we can toss into ignore this (and which friction would we end up using)?

In the case of mixtures or solutions (chemical reaction, multiple components), J. Williard Gibbs figured out how to extend the theory to include this back in the early 1900's. For ideal gas mixtures, this is presented very nicely in Chapter 10 of Smith and Van Ness. What you end up doing is determining the entropy per unit mass of each individual species in the mixture. This requires the inclusion of an additional term in the equation for the specific entropy of each species (over and above the total pressure ratio term and the temperature ratio term).

Take a look at Chapter 10, and carefully read over the section on ideal gas mixtures. I think you'll find it very worthwhile.

Chet

I think this came up in the combustion book I am reading; they essentially applied the ΔS equation we used and calculating ∑ΔS_i where ΔS_i is the entropy for each gas species?

Thanks very much

 

[Chestermiller]

Would one be able to analyze the effects of viscous dissipations with a massless piston involved and/or would there be an error if this is done? If P_I (or σ_I) equilibrates to P_ext + dP or ΔP instantaneously,, wouldn't that have a significant effect on du/dx required to find τ_xx?

Sure. If you were doing a gas dyanmics analysis (compressible fluid mechanics), this would be handled in the boundary condition at the piston face.

I was also thinking a little more about static/kinetic friction effects. If the piston is stationary before a compression process such that P_I + F_static/A = P_ext, once dP to get the piston moving, friction becomes kinetic and instantaneously drops such that there is a finite pressure difference between P_I + F_kinetic/A and P_ext + dP. If the piston is massless, P_I would increase instantaneously such that P_I + F_kinetic/A = P_ext + dP and since the piston is massless it should stop. But a finite pressure must be added to P_ext in addition to dP to overcome the static friction to get the piston moving again. Is the above still considered quasistatic, or is there some assumption we can toss into ignore this (and which friction would we end up using)?

The coefficient of kinetic friction is less than the coefficient of static friction, so, if F decreases, σ(I) increases. The gas near the interface would start compressing at just the right rate so that the force balance is still satisfied. Also, again, if the piston is massless, equality of the forces does not mean that the piston is not moving or accelerating. So the piston won't stop moving once the coefficient of static friction is exceeded. However, in this deformation, you can be certain that the deformation will not be quasistatic.

I think this came up in the combustion book I am reading; they essentially applied the ΔS equation we used and calculating ∑ΔS_i where ΔS_i is the entropy for each gas species?

For this to be correct, they would have to use the partial pressure of the individual gas species in calculating the ΔS_i, rather than the total pressure of the combination of gases. Is this what they did?

Chet

 

[Red_CCF]

Sure. If you were doing a gas dyanmics analysis (compressible fluid mechanics), this would be handled in the boundary condition at the piston face.

What is the boundary condition at the piston-gas interface for a massless piston? Does the piston at any time have a measurable velocity if it cannot accelerate and equilibrium is instantaneous?

The coefficient of kinetic friction is less than the coefficient of static friction, so, if F decreases, σ(I) increases. The gas near the interface would start compressing at just the right rate so that the force balance is still satisfied. Also, again, if the piston is massless, equality of the forces does not mean that the piston is not moving or accelerating. So the piston won't stop moving once the coefficient of static friction is exceeded. However, in this deformation, you can be certain that the deformation will not be quasistatic.

So the piston will move Δx (instead of dx due to the finite pressure difference) instantaneously such that σ(I) can increase instantaneously as F changes from static to kinetic in order to satisfy Newton's Second Law for a massless piston? I assume that the pressure of the rest of the system cannot react as fast, thus the experiment we were analyzing even theoretically must be nonquasistatic?

Since the piston has no inertia, I assume that once σ(I) and kinetic friction balances the external forces the piston must stop, friction becomes static again, and the cycle continues?

For this to be correct, they would have to use the partial pressure of the individual gas species in calculating the ΔS_i, rather than the total pressure of the combination of gases. Is this what they did?

Chet

The exact equation for the entropy of the ith species of the combustion products mixture given was:

s_i = s_i^o (Tref) + ∫c_p\frac{dT}{T} - R_uln(\frac{P_i}{P^o})

where the integration was taken from some reference state T_ref to T_f; p_i is the species partial pressure.

Although it is not stated, I am assuming that total entropy change of the reaction is ∑s_i,products - ∑s_i,reactants.

Is the point of having a reference state just so that the lower limit of the integral is known and we only need to add s_i^o(T_ref) to find absolute entropy? Is there an issue of integrating from the state where s = 0 and if so how were the tabulated s values measured?

Thanks

 

[Chestermiller]

What is the boundary condition at the piston-gas interface for a massless piston?

Irrespective of whether the piston is massless, the boundary condition at the piston-gas interface (using BSL notation, in which compressive stresses are positive) is

F=σ_IA=(p-2μ\frac{∂v}{∂x})A

where F is the force that the gas exerts on the piston face, A is the area of the piston, p is the local thermodynamic pressure of the gas at the interface =ρRT, ρ is the local molar gas density at the interface, μ is the gas viscosity at the interface, and ∂v/∂x = the spatial axial velocity gradient of the gas in the vicinity of the interface.

Does the piston at any time have a measurable velocity if it cannot accelerate and equilibrium is instantaneous?

Who says the piston can't accelerate. Suppose the mass of the piston is 10^-15 grams. How much of a force difference between the two faces of the piston does it take in order to accelerate it?

So the piston will move Δx (instead of dx due to the finite pressure difference) instantaneously such that σ(I) can increase instantaneously as F changes from static to kinetic in order to satisfy Newton's Second Law for a massless piston?

No. The piston doesn't move Δx instantaneously. All that is required for σ(I) to increase. This will happen as soon as the piston starts moving. The viscous stress contribution causes σ(I) to increase. Even more importantly, the local gas density in close proximity to the interface will increase, resulting in a local increase in the thermodynamic pressure p. This is because, as the piston begins to move, the entirety of the gas cannot compress all at once because of its inertia. So a small region of higher density begins to develop adjacent to the piston face. This higher density region will grow in extent as time progresses.

I assume that the pressure of the rest of the system cannot react as fast, thus the experiment we were analyzing even theoretically must be nonquasistatic?

Correct.

Since the piston has no inertia, I assume that once σ(I) and kinetic friction balances the external forces the piston must stop, friction becomes static again, and the cycle continues?

No. As we said, the acceleration of the piston is not zero. Consider again the case where the mass of the piston is 10^-15 grams, rather than zero. The difference in forces on the two sides of the piston necessary to accelerate the piston and keep it movingwill be very close to zero.

The exact equation for the entropy of the ith species of the combustion products mixture given was:

s_i = s_i^o (Tref) + ∫c_p\frac{dT}{T} - R_uln(\frac{P_i}{P^o})

where the integration was taken from some reference state T_ref to T_f; p_i is the species partial pressure.

Although it is not stated, I am assuming that total entropy change of the reaction is ∑s_i,products - ∑s_i,reactants.

No. The entropies need to be weighted in terms of the mole fractions.

Is the point of having a reference state just so that the lower limit of the integral is known and we only need to add s_i^o(T_ref) to find absolute entropy? Is there an issue of integrating from the state where s = 0 and if so how were the tabulated s values measured?

No. The absolute entropy is never used. Everything is referenced to the entropy of each pure species in the reference state. The tabulated values of the pure species entropies in the reference state are determined from the free energies of formation and heats of formation of the pure species in the reference state. The free energies of formation are "backed out" from measurements of the equilibrium constants for various equilibrium reactions, and the heats of formation are "backed out" from the heats of reaction for these reactions. Once the free energies of formation and heats of formation of the species are established, these values can be used to determine the equilibrium constants and heats of reaction for any other reactions involving species included in the overall list of species. This is what makes the methodology so powerful.

Chet

 

[Red_CCF]

No. The piston doesn't move Δx instantaneously. All that is required for σ(I) to increase. This will happen as soon as the piston starts moving. The viscous stress contribution causes σ(I) to increase. Even more importantly, the local gas density in close proximity to the interface will increase, resulting in a local increase in the thermodynamic pressure p. This is because, as the piston begins to move, the entirety of the gas cannot compress all at once because of its inertia. So a small region of higher density begins to develop adjacent to the piston face. This higher density region will grow in extent as time progresses.

If the process was truly quasistatic, the system pressure = interface pressure = ideal gas pressure and thus pressure change is dependent on the volume change during compression. In reality, does the change in p(I) have little dependence on the volume change of the system and thus no longer quasistatic? Does σ(I) immediately compensate for the drop in friction from static to kinetic (σ(I) + F_kin/A = P_ext +dP) such that the net force on the massless piston is zero?

Who says the piston can't accelerate. Suppose the mass of the piston is 10^-15 grams. How much of a force difference between the two faces of the piston does it take in order to accelerate it?

No. As we said, the acceleration of the piston is not zero. Consider again the case where the mass of the piston is 10^-15 grams, rather than zero. The difference in forces on the two sides of the piston necessary to accelerate the piston and keep it movingwill be very close to zero.

My thought is that since net force must equal 0 per Newton's Second Law, essentially 0 = 0*a, but now a is undefined since there's an infinite number of solutions, so how do we know the kinematics of the piston (displacement, velocity, acceleration) for every dP addition?

No. The absolute entropy is never used. Everything is referenced to the entropy of each pure species in the reference state. The tabulated values of the pure species entropies in the reference state are determined from the free energies of formation and heats of formation of the pure species in the reference state. The free energies of formation are "backed out" from measurements of the equilibrium constants for various equilibrium reactions, and the heats of formation are "backed out" from the heats of reaction for these reactions. Once the free energies of formation and heats of formation of the species are established, these values can be used to determine the equilibrium constants and heats of reaction for any other reactions involving species included in the overall list of species. This is what makes the methodology so powerful.

Chet

Is absolute entropy used when finding ΔS (pre and post combustion), as the equation in my book seem to imply this by having the reference entropy on the right side?

Is the reference state entropy is analogous/equivalent to the enthalpy of formation component of a specie's absolute enthalpy? How come it is independent pressure unlike the enthalpy of formation?

Thank you very much