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

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

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.



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

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]

[tex]

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?

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.

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

 

[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:

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

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:

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

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:

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

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

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

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

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 $\Sigma$μ_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 $\Sigma$μ_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