Rate of change of Viscous Force in Couette Flow

[Soumalya]

For a situation where we have two infinite flat parallel plates with a viscous fluid in between if the upper plate if acted upon by a constant force then the plate should accelerate. The fluid layer adjacently below the upper plate should have the same acceleration as this plate (due to no-slip condition for fluids in contact with a stationary or moving surface). This layer causes a shear stress to act on the fluid layer adjacently below it so that every layer is dragged by the layer right above it. This results in a velocity gradient for the fluid layers in between the plates.

Since the plate is accelerating so there is a rate of change of velocity of the plate with time. It implies that the fluid layer right below the plate should have the same rate of change of velocity. The layer adjacently below this fluid layer moves with a velocity such that the difference in velocities between these layers is very small but constant. This is only possible if successive fluid layers are accelerating at the same pace but with a constant difference in velocities at any particular instant of time. The situation can be visualized as the slope of the velocity profile (assuming a liner velocity profile for simplicity) changing (decreasing) with time. As we know the shear stress is proportional to the velocity gradient (slope of the velocity profile) the viscous force should continually decrease with acceleration of the upper plate.

In most textbooks I have seen they assume the velocity for the upper plate to be constant such that the velocity gradient is constant (again assuming a liner velocity profile).But for the upper plate to have a constant velocity the net force on the plate should be zero. The forces acting on the plate during motion are the external force applied to the plate and viscous force on the plate due to fluid layer below it (actually the fluid layer below the adjacent fluid layer to the plate).Now when the fluid was at rest the upper plate must have been at rest. For the upper plate to reach the constant velocity when the external force on the upper plate is balanced by the viscous force due to fluid below it the plate must have passed a transient phase where it accelerated from rest to that constant velocity. Now as already established the shear stress due to viscosity and hence the viscous force on the upper plate keeps on decreasing as the plate accelerates. So the net force on the plate should actually increase with time rather than decreasing. Hence the resultant force on the plate can never approach zero.

Can anyone explain this correctly?

 

[Soumalya]

Well I have a slight correction to make here the subject should have been "Rate of change of Viscous force with time in a Couette Flow" and not viscocity:D

 

[mikeph]

As we know the shear stress is proportional to the velocity gradient (slope of the velocity profile) the viscous force should continually decrease with acceleration of the upper plate.

Surely the opposite will happen?

 

[Chestermiller]

Hi Soumalya,

You make a valiant effort to articulate what is happening, but I'm afraid your descriptions are way off base.

Paragraph 1 considers the situation where the upper plate is moved with a constant force, and paragraph 2 covers the situation where the upper plate is moved with a constant velocity. Analyzing the situation with constant velocity is easier, so let's start with that.

If the upper plate is moving with constant velocity, then if we do a force balance on the plate (free body diagram), the shear force exerted by the fluid on the bottom of the plate is going to be equal to the force we have to apply to the plate to keep it moving at constant velocity; the resultant force will always be equal to zero. So, if the shear force is decreasing with time, the force we apply to the plate must also be decreasing with time (until steady state is attained). Of course, if the plate has mass, to get it started initially, we may have to apply an impulsive force to it at time zero; but, if the plate has no mass, then that won't be necessary, although, from the solution to the viscous flow problem, we find that the initial force will still be infinite.

For the situation in paragraph 1, it is best to begin by assuming that the upper plate has no mass, so that the shear force exerted on the fluid at the base of the plate is constant. Why? If we can't solve this problem, we certainly will never be able to analyze the problem where the plate has mass. In this case, the net force on the plate is again zero. The shear stress at the top of the liquid is constant. We will have to increase the velocity of the plate in order to maintain the constant shear stress at the interface. If the slab of viscous fluid were infinitely thick, we would never reach steady state (and the plate velocity would have to increase forever to maintain constant shear stress at the top plate). However, once the effects of the bottom plate begin to kick in (i.e., the velocity boundary layer reaches the bottom plate), the top plate velocity can begin to level off, and eventually reach a steady state value.

Chet

 

[Soumalya]

If the upper plate is moving with constant velocity, then if we do a force balance on the plate (free body diagram), the shear force exerted by the fluid on the bottom of the plate is going to be equal to the force we have to apply to the plate to keep it moving at constant velocity; the resultant force will always be equal to zero. So, if the shear force is decreasing with time, the force we apply to the plate must also be decreasing with time (until steady state is attained).

Agreed.This is true only after the plate has achieved constant velocity.

Of course, if the plate has mass, to get it started initially, we may have to apply an impulsive force to it at time zero; but, if the plate has no mass, then that won't be necessary, although, from the solution to the viscous flow problem, we find that the initial force will still be infinite.

This is the part which needs more clarification.Suppose that the plate has mass and is initially at rest.There is no viscous force on the bottom of the plate at this stage.Now if we apply a constant force to the plate the plate starts accelerating.Why?Because at the very instant the plate starts it's motion there is only the applied force on it and no viscous force.After the plate is in motion there would be viscous force acting at its bottom trying to retard it's motion.What do you think is the magnitude of the viscous force now at the bottom of the plate?Remember the plate is currently in a transient motion (accelerating) and had not achieved a uniform motion yet.Assuming a liner velocity profile for the fluid in between plates at any instant of time the distribution looks like an inverted right angled triangle with the base at top of magnitude equal to the velocity of the upper plate.The velocity of the upper plate is increasing with time so you might visualize the change in shape of the velocity profile as an increase in the length of the base of the right angled triangle as time passes on.At this stage what do you think is the shear force due to viscosity at the bottom of this plate?It's certainly not constant (unlike it would be when the plate will achieve constant velocity) as the shear stress due to viscosity varies directly with the slope of the velocity profile which is decreasing as the velocity of the upper plate is increasing with time.

Where do you think I am wrong until now?

For the situation in paragraph 1, it is best to begin by assuming that the upper plate has no mass, so that the shear force exerted on the fluid at the base of the plate is constant.

Do you mean if the plate has mass the shear force exerted on the fluid at the bottom of the plate will vary? I can't think about the reason.Could you clarify more?

For the situation in paragraph 1, it is best to begin by assuming that the upper plate has no mass, so that the shear force exerted on the fluid at the base of the plate is constant. Why? If we can't solve this problem, we certainly will never be able to analyze the problem where the plate has mass. In this case, the net force on the plate is again zero. The shear stress at the top of the liquid is constant.

Again this is only true after the plate had achieved a constant velocity.I am focusing on the transient motion of the plate from rest to the point it attains this constant velocity and the variation of viscous force and consequently the net force acting on the upper plate for a fixed external force applied on it.

We will have to increase the velocity of the plate in order to maintain the constant shear stress at the interface.

I couldn't understand this part either.According to Newton's law of Viscosity,

τ_shear=μ.(du/dy)

To maintain a constant shear stress at the interface we need the velocity gradient to be constant as viscosity is always constant for a fluid.The velocity gradient is constant only after the plate has achieved a constant velocity.During its transient motion the velocity gradient is likely to change as explained earlier.

 

[Soumalya]

For an easy understanding in the picture above we see the velocity profile for the case under consideration.Since we have already discussed for the transient part of motion the velocity of the upper plate increases with time (plate is accelerating) so the velocity profile must changes it's shape with time.In the figure above the velocity profile has changed it's shape to the one shown in red extension of the profile.One may clearly see the gradient which is the slope ∂u/∂y varies with time as u=u(y,t).The slope increases as the velocity of the upper plate increases with time.So the shear stress increases with time until the shear force balances the force on the upper plate when the plate keeps moving with a constant velocity.This was my precise understanding I wanted to share.

But I will tell you where I went wrong!The slope ∂u/∂y in the figure above is the angle the velocity profile makes with the y-axis.So the angle increases as the velocity of the upper plate increases with time while in transient motion and so does the shear stress.I was considering the angle the velocity profile makes with the u-axis which was decreasing as the upper plate was accelerating so I thought the shear stress would have continually decreased with the acceleration of the plate and the stage would never come when it would balance the constant external force on the plate!

I know I am stupid at times but I am glad I finally arrive at the meaningful conclusion:D

 

[Soumalya]

Surely the opposite will happen?

Were you asking me or you were ascertain that I was mistakingly anticipating the opposite result?

 

[Chestermiller]

Agreed.This is true only after the plate has achieved constant velocity.

Didn't I say that we make the plate have constant velocity from time t = 0 onward? The plate doesn't have to "achieve" constant velocity. We force it to have constant velocity for all time t >0. I can force either the velocity or the shear stress at the upper boundary to be whatever function of time I want. In fact, I can do this even if the upper plate has mass. If you don't think I can do that, then you're wrong. All I need is a motor with enough "oomph" and an automatic control system. So let's stop focusing on the plate and turn our attention to where it should be, which is on the fluid.

This is the part which needs more clarification.Suppose that the plate has mass and is initially at rest.

We're not going to be focusing on the plate any more.

Assuming a liner velocity profile for the fluid in between plates at any instant of time the distribution looks like an inverted right angled triangle with the base at top of magnitude equal to the velocity of the upper plate.

The velocity profile is not linear during the transient period.

Where do you think I am wrong until now?

You're going wrong in many ways.

τ_shear=μ.(du/dy)

To maintain a constant shear stress at the interface we need the velocity gradient to be constant as viscosity is always constant for a fluid.The velocity gradient is constant only after the plate has achieved a constant velocity.During its transient motion the velocity gradient is likely to change as explained earlier.

No. This is only true if the velocity profile is linear. To maintain a constant shear stress at the interface, we only need the velocity gradient at that boundary to be constant (not throughout the fluid). During the transient period, the velocity gradient varies with depth through the fluid.

I don't know where to begin. Have you had partial differential equations yet? If so, we can derive the fluid mechanics equations for the problem, and look at the solution. Have you heard the term "boundary layer" yet, and do you know what it means?

Chet

 

[Soumalya]

I don't know where to begin. Have you had partial differential equations yet? If so, we can derive the fluid mechanics equations for the problem, and look at the solution.

Well yes I have gone through partial differential equations.But I must say I don't have it in total grip.I know how to approach them analytically but visualizing physical interpretations for problems using PDEs seems difficult as I am yet to improve much on calculus of several variables.Will a basic understanding of PDEs do at this stage?

Have you heard the term "boundary layer" yet, and do you know what it means?

I have a very little understanding on the subject of "boundary layer theory" and I am yet to study it throughly.Until now by boundary layer I understand it to be the layer of fluid in the immediate vicinity of a bounding surface where the effects of viscosity are significant.

Is it necessary to have a in-depth study on the boundary layer theory before I proceed to understand the problem?

 

[Chestermiller]

I think you have enough background for us to proceed.

Let's start out by just talking about what's happening physically at short times. We'll consider the case of constant plate velocity V. At short times, there develops a thin region of fluid of nominal thickness δ immediately adjacent to the moving plate in which the velocity is varying very rapidly with distance from the plate. The fluid velocity at the plate is V, and the velocity at y = L - δ is essentially equal to zero. At distances greater than δ from the plate, the fluid velocity is zero. As time progresses, the thickness of this boundary layer increases until the boundary layer penetrates to the lower plate at y = 0. So the boundary layer thickness is a function of time δ(t). (This is only an approximate picture of the velocity profile, but it is adequate for our purposes).

As a first approximation to the velocity profile (before the boundary layer penetrates to y = 0), we can write:

$v=V\left(1-\frac{(L-y)}{δ}\right)^2$ for (L-δ) < y < L

$v=0$ for y < (L-δ)

From these equations,

What is the velocity at y = L?
What is the velocity at y = (L-δ)
What is the equation for the velocity gradient dv/dy?
What is the velocity gradient at y = L?
What is the velocity gradient at y = (L-δ)?
What is the velocity gradient below y = (L-δ)?

According to this description, the fluid below y = (L-δ) does not even know that the plate is moving yet. Can you see that? Why do you think it is that the velocity profile is curved? Which part of the fluid has been accelerated, and which part has not been accelerated yet?

I'll continue after I hear back from you.

Chet

 

[Soumalya]

http://[PLAIN]http://i58.tinypic.com/2ag7zfd.jpg

What is the velocity at y = L?

At y=L, v=V (The velocity of the fluid at the plate is V)

What is the velocity at y = (L-δ)

At y=L-δ, v=0

What is the equation for the velocity gradient dv/dy?

For (L-δ) < y < L, dv/dy=2V.(1−(L−y)/δ).(1/δ)=(2V/δ).(1−(L−y)/δ)

y < (L-δ), dv/dy=0

What is the velocity gradient at y = L?

At y=L, dv/dy=0

What is the velocity gradient at y = (L-δ)?

At y = (L-δ), dv/dy=0

What is the velocity gradient below y = (L-δ)?

Below y = (L-δ), dv/dy=0

According to this description, the fluid below y = (L-δ) does not even know that the plate is moving yet. Can you see that?

Obviously Yes

Why do you think it is that the velocity profile is curved?

Because the velocity varies non uniformly with y i.e, dv/dy=f(y).

Which part of the fluid has been accelerated, and which part has not been accelerated yet?

The part of fluid within the boundary layer thickness has been accelerated while the part of fluid between the lower plate and the boundary layer i.e, below below y = (L-δ) remains at rest.

 

[Chestermiller]

http://[PLAIN]http://i58.tinypic.com/2ag7zfd.jpg

This figure does not represent what is happening in our specific physical situation. In our situation, the fluid stream lines remain perfectly horizontal (i.e., constant y). Think of it as a deck of cards that is being sheared.

At y=L, dv/dy=0

No. This one is not correct. Please try again.

Because the velocity varies non uniformly with y i.e, dv/dy=f(y).

Sorry. I didn't mean mathematically. I meant, physically, why do you think that the velocity profile is curved?

The next thing I'd like to do is derive the differential force balance on the thin slab of fluid between y and y + Δy. Let A represent the horizontal area that we are focusing on. The volume of the fluid in the region under consideration is AΔy. The mass of this fluid is ρAΔy. The force at the top of the slab, exerted by the fluid above, is given by: $Aμ(∂v/∂y)_{y+Δy}$. The force exerted on the bottom of the slab by the fluid below is given by: $-Aμ(∂v/∂y)_y$. The mass of the slab times its acceleration is given by: $ρAΔy(∂v/∂t)$.

So the force balance on the slab gives us:
$$ρAΔy\frac{∂v}{∂t}=Aμ\left(\frac{∂v}{∂y}\right)_{y+Δy}-Aμ\left(\frac{∂v}{∂y}\right)_{y}$$
If we divide this equation by AΔy, and take the limit as Δy →0, we obtain:
$$ρ\frac{∂v}{∂t}=μ\frac{∂^2 v}{∂y^2}$$
Unlike the approximate equations I have written in my previous post, this equation is exact. We will be using it in our solution to the problem at hand for short times.

Please tell me if you are comfortable with this development so far.

Chet

 

[Soumalya]

This figure does not represent what is happening in our specific physical situation. In our situation, the fluid stream lines remain perfectly horizontal (i.e., constant y). Think of it as a deck of cards that is being sheared.

Could you draw the velocity profile for the approximation?

It would give me a better pictorial understanding.

No. This one is not correct. Please try again.

In your previous post you didn't mention what is represented by "L".Assuming that the distance between the two plates is denoted by "L" and y is measured from the lower plate vertically upwards, the velocity gradient at y=L should be dv/dy at y=L where v=V.(V is the upper plate velocity)

So, dv/dy at y=L ⇔ dV/dy=0 (since the plate velocity is constant and the fluid layer immediately adjacent to the plate should have the same velocity as the plate.)

I might be wrong in my understanding but unable to see where.Please clarify.

Sorry. I didn't mean mathematically. I meant, physically, why do you think that the velocity profile is curved?

Each layer of the fluid is acted upon by a net forward force which is different for the fluid layers resulting in different accelerations for the fluid layers.

The next thing I'd like to do is derive the differential force balance on the thin slab of fluid between y and y + Δy. Let A represent the horizontal area that we are focusing on. The volume of the fluid in the region under consideration is AΔy. The mass of this fluid is ρAΔy. The force at the top of the slab, exerted by the fluid above, is given by: Aμ(∂v/∂y)y+ΔyAμ(∂v/∂y)_{y+Δy}. The force exerted on the bottom of the slab by the fluid below is given by: −Aμ(∂v/∂y)y-Aμ(∂v/∂y)_y. The mass of the slab times its acceleration is given by: ρAΔy(∂v/∂t)ρAΔy(∂v/∂t).

So the force balance on the slab gives us:

ρAΔy∂v∂t=Aμ(∂v∂y)y+Δy−Aμ(∂v∂y)y​

ρAΔy\frac{∂v}{∂t}=Aμ\left(\frac{∂v}{∂y}\right)_{y+Δy}-Aμ\left(\frac{∂v}{∂y}\right)_{y}
If we divide this equation by AΔy, and take the limit as Δy →0, we obtain:

ρ∂v∂t=μ∂2v∂y2​

ρ\frac{∂v}{∂t}=μ\frac{∂^2 v}{∂y^2}
Unlike the approximate equations I have written in my previous post, this equation is exact. We will be using it in our solution to the problem at hand for short times.

Please tell me if you are comfortable with this development so far.

Chet

Understood but I would like to proceed further after I have cleared my preliminary doubts.

 

[Chestermiller]

Could you draw the velocity profile for the approximation?

It would give me a better pictorial understanding.

I would like you (rather than me) to draw a graph of the velocity profile. Make y the vertical axis, and make v the horizontal axis. Do it at one value of time t.

In your previous post you didn't mention what is represented by "L".Assuming that the distance between the two plates is denoted by "L" and y is measured from the lower plate vertically upwards, the velocity gradient at y=L should be dv/dy at y=L where v=V.(V is the upper plate velocity)

So, dv/dy at y=L ⇔ dV/dy=0 (since the plate velocity is constant and the fluid layer immediately adjacent to the plate should have the same velocity as the plate.)

I might be wrong in my understanding but unable to see where.Please clarify.

Yes, L is the gap between the plates. In your previous post, you correctly wrote:
$$\frac{\partial v}{\partial y}=\frac{2V}{\delta}\left(1-\frac{(L-y)}{\delta}\right)$$
From this, at y = L, I get:
$$\frac{\partial v}{\partial y}=\frac{2V}{\delta}$$ at y =L

Each layer of the fluid is acted upon by a net forward force which is different for the fluid layers resulting in different accelerations for the fluid layers.

Excellent.

 

[Soumalya]

As a first approximation to the velocity profile (before the boundary layer penetrates to y = 0), we can write:

$v=V\left(1-\frac{(L-y)}{δ}\right)^2$ for (L-δ) < y < L

$v=0$ for y < (L-δ)

From these equations,

What is the velocity gradient at y = L?

For the velocity distribution formulated as an approximation,

$v=V\left(1-\frac{(L-y)}{δ}\right)^2$ for (L-δ) < y < L

$v=0$ for y < (L-δ)Thus,
$$\frac{\partial v}{\partial y}=\frac{2V}{\delta}\left(1-\frac{(L-y)}{\delta}\right)$$

can be used for (L-δ) < y < L but not at y=L.(that is y cannot touch the value L for the velocity gradient to be approximated by the formula.)

 

[Soumalya]

The next thing I'd like to do is derive the differential force balance on the thin slab of fluid between y and y + Δy. Let A represent the horizontal area that we are focusing on. The volume of the fluid in the region under consideration is AΔy. The mass of this fluid is ρAΔy. The force at the top of the slab, exerted by the fluid above, is given by: $Aμ(∂v/∂y)_{y+Δy}$. The force exerted on the bottom of the slab by the fluid below is given by: $-Aμ(∂v/∂y)_y$. The mass of the slab times its acceleration is given by: $ρAΔy(∂v/∂t)$.

So the force balance on the slab gives us:
$$ρAΔy\frac{∂v}{∂t}=Aμ\left(\frac{∂v}{∂y}\right)_{y+Δy}-Aμ\left(\frac{∂v}{∂y}\right)_{y}$$
If we divide this equation by AΔy, and take the limit as Δy →0, we obtain:
$$ρ\frac{∂v}{∂t}=μ\frac{∂^2 v}{∂y^2}$$
Unlike the approximate equations I have written in my previous post, this equation is exact. We will be using it in our solution to the problem at hand for short times.

Please tell me if you are comfortable with this development so far.

Chet

Yes I would like to proceed with the analysis now.

 

[Chestermiller]

Yes I would like to proceed with the analysis now.

This graph is not correct. The velocity and the velocity gradient are continuous at y = L-δ, and your curvature is flipped. Just assume some values for L and δ, and plot the points.

For the velocity distribution formulated as an approximation,

$v=V\left(1-\frac{(L-y)}{δ}\right)^2$ for (L-δ) < y < L

$v=0$ for y < (L-δ)Thus,
$$\frac{\partial v}{\partial y}=\frac{2V}{\delta}\left(1-\frac{(L-y)}{\delta}\right)$$

can be used for (L-δ) < y < L but not at y=L.(that is y cannot touch the value L for the velocity gradient to be approximated by the formula.)

This is not correct. The equation actually applies right up to y = L. The no-slip boundary condition requires only that the fluid velocity matches the boundary velocity. It does not require that the velocity gradient be zero. Certainly, in the steady state solution, the velocity gradient is not zero. Otherwise the shear stress at the boundary would be zero, which it obviously is not.

Chet

 

[Chestermiller]

Yes I would like to proceed with the analysis now.

I want to wait until you get the issues in the previous post straightened out. It's important that you understand what the velocity profile looks like.

Chet

 

[boneh3ad]

http://[PLAIN]http://i58.tinypic.com/2ag7zfd.jpg

I think I see where you confusing yourself here based on these two images. You are viewing the $\delta$ in this case as if it was a boundary layer developing over a semi-infinite flat plate (i.e. with a leading edge) and have therefore drawn each figure sort of like you would see a Blasius boundary layer on a flat plate develop rather than what it would look like with an infinite flat plate moving over the liquid where this $\delta$ will be the same for any value of $x$ (or $z$ for that matter if this was three-dimensional).

What @Chestermiller is asking you to do is plot the velocity profile $v = v(y)$ of the fluid between the two plates. This profile should be some shape of line stretching from one plate all the way to the other for a given instant in time. He seems to want you to just pick some instant in time after the plate has accelerated and before the system reaches steady state and try to draw it.

Thus,
$$\frac{\partial v}{\partial y}=\frac{2V}{\delta}\left(1-\frac{(L-y)}{\delta}\right)$$

can be used for (L-δ) < y < L but not at y=L.(that is y cannot touch the value L for the velocity gradient to be approximated by the formula.)

As you approach the plate from below, the velocity is changing, so it clearly has a finite value of $dv/dy$ at all of those points and continues to change right up until it hits the plate. After all, the fluid at the plate is moving with $v=V$ but the very nearest parcel of fluid is moving with $v < V$. For calculating shear stress at the wall (which is ultimately what the goal here is), you are therefore finding $dv/dy$ at the wall as the wall is approached from below. Just because the velocity at the wall is constant in time does not mean it is constant in space through the fluid.

 

[Soumalya]

I think I see where you confusing yourself here based on these two images. You are viewing the δ\delta in this case as if it was a boundary layer developing over a semi-infinite flat plate (i.e. with a leading edge) and have therefore drawn each figure sort of like you would see a Blasius boundary layer on a flat plate develop rather than what it would look like with an infinite flat plate moving over the liquid where this δ\delta will be the same for any value of xx (or zz for that matter if this was three-dimensional).

Yes you are correct.I was visualizing the situation incorrectly.In fact the figures were nowhere near the actual situation.I am glad you pointed out the mistake so precisely:)

What @Chestermiller is asking you to do is plot the velocity profile v=v(y)v = v(y) of the fluid between the two plates. This profile should be some shape of line stretching from one plate all the way to the other for a given instant in time. He seems to want you to just pick some instant in time after the plate has accelerated and before the system reaches steady state and try to draw it.

Yes I noticed in the figures above the velocity profile was discontinuous at y=L-δ whereas it should have been continuous at all values of 'y' and monotonically increasing from y=0 to y=L.

As you approach the plate from below, the velocity is changing, so it clearly has a finite value of dv/dydv/dy at all of those points and continues to change right up until it hits the plate. After all, the fluid at the plate is moving with v=Vv=V but the very nearest parcel of fluid is moving with v<Vv < V. For calculating shear stress at the wall (which is ultimately what the goal here is), you are therefore finding dv/dydv/dy at the wall as the wall is approached from below. Just because the velocity at the wall is constant in time does not mean it is constant in space through the fluid.

Yes correct.But Chestermiller formulated the equation as,

$v=V\left(1-\frac{(L-y)}{δ}\right)^2$ for (L-δ) < y < L

$v=0$ for y < (L-δ)

So according to the equation mathematically at y=L the velocity is not properly defined.The equation is perhaps incomplete in it's formulation as at y=L we mean the velocity of fluid layer immediately adjacent to the moving plate which should be same as the plate i.e, V (no-slip condition).Thus as a complete approximation of the velocity profile the piecewise defined velocity distribution function should have been,

$v=V\left(1-\frac{(L-y)}{δ}\right)^2$ for (L-δ) < y ≤ L

$v=0$ for y < (L-δ)

$v=0$ for y = (L-δ)It is understood that the velocity of the fluid layer not directly adjacent to the moving plate but the fluid layer adjacent to the fluid layer immediately adjacent to the moving plate is of course less than V as you pointed out.This layer is again not at a distance of 'L' from the lower plate but at a distance infinitesimally less than 'L' from the lower plate and hence withing the region L-δ<y<L.

The shear stress at the moving plate is not due to the fluid layer directly below it because the relative velocity of the fluid layer immediately adjacent to the moving plate to the plate is zero.Hence, the shear stress is due to the fluid layer next to the fluid layer sticking to the moving plate (adhesion) due to the existence of relative velocity between them along with the property of viscosity.

 

[Soumalya]

This graph is not correct. The velocity and the velocity gradient are continuous at y = L-δ, and your curvature is flipped. Just assume some values for L and δ, and plot the points.

Is this one correct?

This is not correct. The equation actually applies right up to y = L. The no-slip boundary condition requires only that the fluid velocity matches the boundary velocity. It does not require that the velocity gradient be zero. Certainly, in the steady state solution, the velocity gradient is not zero. Otherwise the shear stress at the boundary would be zero, which it obviously is not.

I suppose at y=L it is meant the fluid layer directly below the moving plate which should have the velocity exactly same as the plate i.e, V.Thus v=V at y=L. I suggest a more complete formulation of the approximation for the velocity profile as,

$v=V\left(1-\frac{(L-y)}{δ}\right)^2$ for (L-δ) < y ≤ L

$v=0$ for y < (L-δ)

$v=0$ for y = (L-δ)


Earlier I calculated the derivate of the function v(y) at y=L taking the functional value at y=L which is constant at y=L.

But again I realized that one needs to find out the rate of change of v w.r.t y at y=L approaching from values of v(y) lower than that at y=L i.e, using the functional expression for values of y between L-δ and L.So you were correct when you wrote:

Yes, L is the gap between the plates. In your previous post, you correctly wrote:
$$\frac{\partial v}{\partial y}=\frac{2V}{\delta}\left(1-\frac{(L-y)}{\delta}\right)$$
From this, at y = L, I get:
$$\frac{\partial v}{\partial y}=\frac{2V}{\delta}$$ at y =L

I was wrong of course. :D

 

[Soumalya]

If I am correct this time we could proceed with the analysis further.

 

[Chestermiller]

Is this one correct?

Yes. This is correct. But, what is that partial with respect to t doing as a label on the figure?

The differential force balance equation that I derived earlier is given by:
$$\frac{\partial v}{\partial t}= \nu \frac{\partial ^2 v}{\partial y^2}$$
where $\nu=\mu/\rho$ is the so-called kinematic viscosity.

What I would like you to do next is to substitute our approximate velocity profile into this equation and report back what you obtain.

After that, I'll tell you what the next step is for determining the boundary layer thickness as a function of time.

Chet

 

[Soumalya]

But, what is that partial with respect to t doing as a label on the figure?

I am sorry but that should have been 'δt' instead of '∂t' which denotes a very small change in time.I have considered the velocity profile at an instant δt after the motion of the plate was set.I thought as δ was already used to denote boundary layer thickness in the figure it would be confusing to use it again as a very small change in time 't'.

The differential force balance equation that I derived earlier is given by:
$$\frac{\partial v}{\partial t}= \nu \frac{\partial ^2 v}{\partial y^2}$$
where $\nu=\mu/\rho$ is the so-called kinematic viscosity.

What I would like you to do next is to substitute our approximate velocity profile into this equation and report back what you obtain.

After that, I'll tell you what the next step is for determining the boundary layer thickness as a function of time.

Chet

From the differential force balance equation :

$$\frac{\partial v}{\partial t}= \nu \frac{\partial ^2 v}{\partial y^2}$$

⇔(-2.V.y/δ²).{1-(L-y)/δ}.(∂δ/∂t)=ϑ.(2.V/δ²) (ϑ→Kinematic viscosity)

⇔ (∂δ/∂t)=- ϑ/y.[1/{1-(L-y)/δ}]

Sorry I couldn't find the way to write those equations in proper form as you do and also the symbol for 'nu'.Could you teach me how do you use the panel to write the equations?

 

[boneh3ad]

So according to the equation mathematically at y=L the velocity is not properly defined.The equation is perhaps incomplete in it's formulation as at y=L we mean the velocity of fluid layer immediately adjacent to the moving plate which should be same as the plate i.e, V (no-slip condition).Thus as a complete approximation of the velocity profile the piecewise defined velocity distribution function should have been,

$v=V\left(1-\frac{(L-y)}{δ}\right)^2$ for (L-δ) < y ≤ L

$v=0$ for y < (L-δ)

$v=0$ for y = (L-δ)It is understood that the velocity of the fluid layer not directly adjacent to the moving plate but the fluid layer adjacent to the fluid layer immediately adjacent to the moving plate is of course less than V as you pointed out.This layer is again not at a distance of 'L' from the lower plate but at a distance infinitesimally less than 'L' from the lower plate and hence withing the region L-δ<y<L.

That's not true, though. If you plug $y=L$ directly into the equation, the result is that $v=V$, which is both defined and exactly as it should be. The velocity at the wall should match the plate. Any layer of fluid infinitesimally less than $L$ will have a velocity infinitesimally less than $V$, which is not what you are looking for in this case.

 

[Chestermiller]

I am sorry but that should have been 'δt' instead of '∂t' which denotes a very small change in time.I have considered the velocity profile at an instant δt after the motion of the plate was set.I thought as δ was already used to denote boundary layer thickness in the figure it would be confusing to use it again as a very small change in time 't'.

The figure you have drawn is not the velocity profile at the very instant after the motion of the plate was set. It is at a significant time t after time t = 0, after the boundary layer has had time to grow a little.

From the differential force balance equation :

$$\frac{\partial v}{\partial t}= \nu \frac{\partial ^2 v}{\partial y^2}$$

⇔(-2.V.y/δ²).{1-(L-y)/δ}.(∂δ/∂t)=ϑ.(2.V/δ²) (ϑ→Kinematic viscosity)

⇔ (∂δ/∂t)=- ϑ/y.[1/{1-(L-y)/δ}]

This is not quite correct. The equation I get is:

$$\frac{2V(L-y)}{\delta ^2}\left(1-\frac{(L-y)}{\delta}\right)\frac{d\delta}{dt}=2\nu\frac{V}{\delta^2}$$
or equivalently,
$$(L-y)\left(1-\frac{(L-y)}{\delta}\right)\frac{d\delta}{dt}=\nu$$
Please see if you can get the algebra correct, and end up with what I got. Also note that the time derivative in the equation is not a partial derivative, but an ordinary time derivative, since δ is a function only of time.

You will note from this equation that, as might be expected, our approximate velocity profile does not satisfy the differential force balance equation exactly. (There are y's on the left side of the equation, and an ordinary time derivative of δ). However, we can satisfy the differential force balance approximately by averaging over the boundary layer thickness. This is done by integrating over the boundary layer with respect to y, from y = (L-δ) to y = L. Such an approach is called a "momentum integral technique." Please carry out this integration and report back to me what you get.

Sorry I couldn't find the way to write those equations in proper form as you do and also the symbol for 'nu'.Could you teach me how do you use the panel to write the equations?

You have been with Physics Forums a long time now, and it is now time for you to "bite the bullet" and learn how to use the LaTex equation editor. I learned it as a 70 year old, so I'm sure you can. It doesn't take long. Physics Forums has an excellent tutorial on LaTex that is available. This is something that I'm not about to teach you.

Chet

 

[Soumalya]

The figure you have drawn is not the velocity profile at the very instant after the motion of the plate was set. It is at a significant time t after time t = 0, after the boundary layer has had time to grow a little.

Understood.Let it be at an instant t+δt after the plate was set to motion so that the boundary layer had enough time to develop.

T

This is not quite correct. The equation I get is:

$$\frac{2V(L-y)}{\delta ^2}\left(1-\frac{(L-y)}{\delta}\right)\frac{d\delta}{dt}=2\nu\frac{V}{\delta^2}$$
or equivalently,
$$(L-y)\left(1-\frac{(L-y)}{\delta}\right)\frac{d\delta}{dt}=\nu$$
Please see if you can get the algebra correct, and end up with what I got.

Sorry I made a mistake while evaluating the differential $\frac{\partial v}{\partial t}$ on the left hand side.I re-evaluated the whole equation and it matches with the one you wrote.

Also note that the time derivative in the equation is not a partial derivative, but an ordinary time derivative, since δ is a function only of time.

Yes I made a mistake again.The boundary layer thickness is only a function of time and hence an ordinary derivative is applicable.

You will note from this equation that, as might be expected, our approximate velocity profile does not satisfy the differential force balance equation exactly. (There are y's on the left side of the equation, and an ordinary time derivative of δ).

I am looking into the result of the equation as something like this:

We have $\nu$ in the right hand side of the equation which is constant throughout thought the fluid as it's a property of the fluid and the thermodynamic state of the fluid doesn't change during our observation.The left hand side of the equation is a function of both 'y' and 't'(both δ and $\frac{d\delta}{dt}$ are functions of only time).Thus, for a constant value of 'y' i.e, for a particular fluid layer the value of the expression in left hand side of the equation changes with time which doesn't make sense and effectively violates the equation as $\nu$ must be constant with time.Again at a particular instant of time i.e, a fixed 't' the value of the expression on the left hand side of the equation changes with 'y' i.e, from layer to layer which doesn't make any sense either and again violates the equation as $\nu$ must be constant with spatial location as well.So, our approximation of the velocity profile does not satisfy the force balance equation well.

This is what I understand from the interpretation of the solution obtained.

However, we can satisfy the differential force balance approximately by averaging over the boundary layer thickness. This is done by integrating over the boundary layer with respect to y, from y = (L-δ) to y = L. Such an approach is called a "momentum integral technique." Please carry out this integration and report back to me what you get.

I will get back with the result of the integral soon.

 

[Chestermiller]

Understood.Let it be at an instant t+δt after the plate was set to motion so that the boundary layer had enough time to develop.

Why not just time t? Why do you feel that you need the δt in there?

We have $\nu$ in the right hand side of the equation which is constant throughout thought the fluid as it's a property of the fluid and the thermodynamic state of the fluid doesn't change during our observation.The left hand side of the equation is a function of both 'y' and 't'(both δ and $\frac{d\delta}{dt}$ are functions of only time).Thus, for a constant value of 'y' i.e, for a particular fluid layer the value of the expression in left hand side of the equation changes with time which doesn't make sense and effectively violates the equation as $\nu$ must be constant with time.Again at a particular instant of time i.e, a fixed 't' the value of the expression on the left hand side of the equation changes with 'y' i.e, from layer to layer which doesn't make any sense either and again violates the equation as $\nu$ must be constant with spatial location as well.So, our approximation of the velocity profile does not satisfy the force balance equation well.

Yes. Good analysis. The only change I would make would be to say that it doesn't satisfy the force balance "exactly", rather than "well." On average, it satisfies the force balance well (as we'll see when you do the integration and compare our results for the shear stress at the boundary with the exact solution).

Chet

 

[Soumalya]

Why not just time t? Why do you feel that you need the δt in there?

Yes it should be at time 't' only as we are observing the situation after a certain time from the commencement of motion of the plate at t=0 and not over an infinitesimal interval of time.

The average value of the velocity distribution for the boundary layer as obtained by integrating our velocity approximation over the boundary layer is as follows,

$\int_{y=L-δ}^L v \ dy= \int_{y=L-δ}^L [V \left(1-\frac{(L-y)}{δ}\right)^2] dy=\frac{Vδ}{3}$

The average velocity turns out to be a function of time only and thus varies with time.Please do check if my result matches with yours as obtained in your calculations.

 

[Chestermiller]

The average value of the velocity distribution for the boundary layer as obtained by integrating our velocity approximation over the boundary layer is as follows,

$\int_{y=L-δ}^L v \ dy= \int_{y=L-δ}^L [V \left(1-\frac{(L-y)}{δ}\right)^2] dy=\frac{Vδ}{3}$

The average velocity turns out to be a function of time only and thus varies with time.Please do check if my result matches with yours as obtained in your calculations.

I didn't ask you to integrate the velocity distribution over the boundary layer; I asked you to integrate our final version of the differential force balance equation over the boundary layer.

Chet

 

[Soumalya]

I didn't ask you to integrate the velocity distribution over the boundary layer; I asked you to integrate our final version of the differential force balance equation over the boundary layer.

Chet

The expression on the left hand side of our final version of the differential force balance equation is a function of both 'y' and 't'.When you say about integrating the equation over the boundary layer I suppose you mean to treat 't' and $\frac {d\delta}{dt}$(a function of time) as constant i.e, carry out the integration over the boundary layer at a particular instant of time.

Thus,

The figure you have drawn is not the velocity profile at the very instant after the motion of the plate was set. It is at a significant time t after time t = 0, after the boundary layer has had time to grow a little.This is not quite correct. The equation I get is:

$$\frac{2V(L-y)}{\delta ^2}\left(1-\frac{(L-y)}{\delta}\right)\frac{d\delta}{dt}=2\nu\frac{V}{\delta^2}$$
or equivalently,
$$(L-y)\left(1-\frac{(L-y)}{\delta}\right)\frac{d\delta}{dt}=\nu$$
Please see if you can get the algebra correct, and end up with what I got. Also note that the time derivative in the equation is not a partial derivative, but an ordinary time derivative, since δ is a function only of time.

You will note from this equation that, as might be expected, our approximate velocity profile does not satisfy the differential force balance equation exactly. (There are y's on the left side of the equation, and an ordinary time derivative of δ). However, we can satisfy the differential force balance approximately by averaging over the boundary layer thickness. This is done by integrating over the boundary layer with respect to y, from y = (L-δ) to y = L. Such an approach is called a "momentum integral technique." Please carry out this integration and report back to me what you get.

You have been with Physics Forums a long time now, and it is now time for you to "bite the bullet" and learn how to use the LaTex equation editor. I learned it as a 70 year old, so I'm sure you can. It doesn't take long. Physics Forums has an excellent tutorial on LaTex that is available. This is something that I'm not about to teach you.

Chet

 

[Soumalya]

I didn't ask you to integrate the velocity distribution over the boundary layer; I asked you to integrate our final version of the differential force balance equation over the boundary layer.

Chet

The expression on the left hand side of our final version of the differential force balance equation is a function in both 'y' and 't'.When you say about integrating the equation over the boundary layer I suppose you mean to treat 't' and $\frac {d\delta}{dt}$(a function of time) as constant i.e, carry out the integration over the boundary layer at a particular instant of time.

Thus, $\int_{L-δ}^{L} [(L-y)\left(1-\frac{(L-y)}{\delta}\right)](\frac{d\delta}{dt}) dy=\int_{L-δ}^{L}\nu dy$

⇔$[\frac{\delta^2}{2}+L^2\left(1-\frac{1}{\delta}\right)-L(\delta-1)-\frac{\delta}{3}](\frac{d\delta}{dt})=\nu\delta$

 

[Chestermiller]

The integration was done incorrectly. Please try again.

Chet

 

[Soumalya]

Is this the integration to be evaluated at all?

$\int_{L-δ}^{L} [(L-y)\left(1-\frac{(L-y)}{\delta}\right)](\frac{d\delta}{dt}) dy=\int_{L-δ}^{L}\nu dy$

Or the step is itself incorrect?

 

[Chestermiller]

Is this the integration to be evaluated at all?

$\int_{L-δ}^{L} [(L-y)\left(1-\frac{(L-y)}{\delta}\right)](\frac{d\delta}{dt}) dy=\int_{L-δ}^{L}\nu dy$

Or the step is itself incorrect?

This is the integration to be evaluated.