Rate of change of Viscous Force in Couette Flow

[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.

 

[Soumalya]

I hope this is correct.

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

⇔$[\frac{\delta^2}{6}-L(\delta-1)](\frac{d\delta}{dt})=\nu\delta$

⇔$[\frac{\delta}{6}-L\left(1-\frac{1}{\delta}\right)](\frac{d\delta}{dt})=\nu$

 

[Chestermiller]

This is closer, but still not correct. The first term in brackets is correct, but the second term is not; its units don't even match those of the first term.

Chet

 

[Soumalya]

This is definitely correct.:p

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

⇔$(\frac{\delta^2}{6})(\frac{d\delta}{dt})=\nu\delta$

⇔$(\frac{\delta}{6})(\frac{d\delta}{dt})=\nu$

⇔$(\frac{d\delta}{dt})=\frac{6\nu}{\delta}$

 

[Chestermiller]

This is definitely correct.:p

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

⇔$(\frac{\delta^2}{6})(\frac{d\delta}{dt})=\nu\delta$

⇔$(\frac{\delta}{6})(\frac{d\delta}{dt})=\nu$

⇔$(\frac{d\delta}{dt})=\frac{6\nu}{\delta}$

Yes. Now solve this differential equation for δ(t), subject to the initial condition δ(0)=0. What do you get?

Chet

 

[Soumalya]

$(\frac{d\delta}{dt})=\frac{6\nu}{\delta}$

⇔$\int_{\delta_0}^{\delta_t}\delta d\delta=6\nu\int_{t_0}^{t}dt$

⇔$\left(\frac{\delta_t^2}{2}-\frac{\delta_0^2}{2}\right)=6\nu(t-t_0)$

Since at t=0, δ(0)=δ₀=0

⇔$\frac{\delta_t^2}{2}=6\nu t$

⇔$\delta_t=2\sqrt{3\nu t}$

⇔$\delta(t)=2\sqrt{3\nu t}$

 

[Chestermiller]

$(\frac{d\delta}{dt})=\frac{6\nu}{\delta}$

⇔$\int_{\delta_0}^{\delta_t}\delta d\delta=6\nu\int_{t_0}^{t}dt$

⇔$\left(\frac{\delta_t^2}{2}-\frac{\delta_0^2}{2}\right)=6\nu(t-t_0)$

Since at t=0, δ(0)=δ₀=0

⇔$\frac{\delta_t^2}{2}=6\nu t$

⇔$\delta_t=2\sqrt{3\nu t}$

⇔$\delta(t)=2\sqrt{3\nu t}$

Nicely done. Take a bow.

We're almost done now. Now go back to post #14 and use your equation for the boundary layer thickness δ to determine the velocity gradient at the wall (aka the wall shear rate) at time t. Then, use this value for the velocity gradient to obtain the shear stress at the wall τ at time t.

Chet

 

[Soumalya]

Nicely done. Take a bow.

:DThe velocity gradient at the wall i.e, at y=L is,\frac{\partial v}{\partial y}=\frac{2V}{\delta}

Since, $\delta(t)=2\sqrt{3\nu t}$

⇔\frac{\partial v}{\partial y}=\frac{2V}{2\sqrt{3\nu t}} (at y=L)

⇔\frac{\partial v}{\partial y}=\frac{V}{\sqrt{3\nu t}}

The shear stress at the wall at a particular time 't' is given by,

$\left(\tau_{shear}\right)_{wall}=\mu\left(\frac{∂v}{∂y}\right)_{y=L}$

⇔$\left(\tau_{shear}\right)_{wall}=\mu\frac{V}{\sqrt{3\nu t}}$

 

[Chestermiller]

Yes. Good job. Now, I can tell you that the exact solution to the differential equation at short times (i.e., when δ << L) is given by:

\frac{\partial v}{\partial y}=\frac{V}{\sqrt{\pi \nu t}}

So you can see that our approximate solution is only slightly different from the exact solution.

Now I have a special surprise for you. Boneh3ad and I have been in contact about this problem, and Boneh3ad has been nice enough to develop an exact solution to the differential equation and boundary conditions, including at longer times where the solution approaches steady state. In his next post, he will be presenting that solution for you to see. Enjoy!

Chet

 

[Soumalya]

Yes. Good job. Now, I can tell you that the exact solution to the differential equation at short times (i.e., when δ << L) is given by:

\frac{\partial v}{\partial y}=\frac{V}{\sqrt{\pi \nu t}}

So you can see that our approximate solution is only slightly different from the exact solution.

Now I have a special surprise for you. Boneh3ad and I have been in contact about this problem, and Boneh3ad has been nice enough to develop an exact solution to the differential equation and boundary conditions, including at longer times where the solution approaches steady state. In his next post, he will be presenting that solution for you to see. Enjoy!

Chet

This is such a special treat for me!:w

I feel immensely honored to have both yours and boneh3ad's helping hands exclusively for my concern.This is incredible:)

 

[boneh3ad]

So sometimes working it out by hand can give you immense insight into the problem. At other times, it may be advantageous to solve it numerically (or maybe you are just bored one day and feel like coding it up). Either way, if you start with the Navier-Stokes equations and make the relevant assumptions about the flow to remove most of the terms, you get the following:
\dfrac{\partial u}{\partial t} = \dfrac{1}{Re}\dfrac{\partial ^2 u}{\partial y^2},
where $Re$ is the Reynolds number and this entire equation has been made dimensionless (e.g. the $u$ here is actually the flow velocity normalized by the plate velocity $U$).

As it turns out, this is just a form of the heat equation, and that shouldn't really surprise you. After all, viscosity is simply a diffusion coefficient for momentum much like thermal conductivity is a diffusion coefficient for heat. Mathematically, then, this problem is identical to a bar initially at a constant temperature with both ends fixed at that temperature, then instantly increasing the temperature at one end to some higher temperature. Physics is filled with neat parallels like that.

Anyway, I went ahead and solved the equation numerically for $Re = 1000$. All of the values in these plots have been made dimensionless so that they apply more easily to many different physical problems as long as $Re$ is the same.

If you happen to be interested in the numerical solution method, it's just a relatively simple finite difference scheme called Crank-Nicolson. Let's be a little careful about what we call an "exact solution" though. This is not exact, as it is still a numerical approximation subject to finite spatial resolution, finite time-stepping, and the vast but limited numerical precision of a computer which introduces error. It should be very close, though.

 

[Chestermiller]

In what Boneh3ad did, he called U what we have been calling V and he called h what we have been calling L. His Reynolds number is defined as:

Re = \frac{VL}{\nu}

I might have done this a little differently by lumping the Reynolds Number into the dimensionless time. In that case, the dimensionless time would be:
\frac{\nu t}{L^2}. For the case that Boneh3ad considered, my dimensionless times would be 1000 times smaller than his, but then the graphs would have applied to all Reynolds numbers, and not just Re = 1000.

Chet

 

[Soumalya]

So sometimes working it out by hand can give you immense insight into the problem. At other times, it may be advantageous to solve it numerically (or maybe you are just bored one day and feel like coding it up). Either way, if you start with the Navier-Stokes equations and make the relevant assumptions about the flow to remove most of the terms, you get the following:
\dfrac{\partial u}{\partial t} = \dfrac{1}{Re}\dfrac{\partial ^2 u}{\partial y^2},
where $Re$ is the Reynolds number and this entire equation has been made dimensionless (e.g. the $u$ here is actually the flow velocity normalized by the plate velocity $U$).

As it turns out, this is just a form of the heat equation, and that shouldn't really surprise you. After all, viscosity is simply a diffusion coefficient for momentum much like thermal conductivity is a diffusion coefficient for heat. Mathematically, then, this problem is identical to a bar initially at a constant temperature with both ends fixed at that temperature, then instantly increasing the temperature at one end to some higher temperature. Physics is filled with neat parallels like that.

Anyway, I went ahead and solved the equation numerically for $Re = 1000$. All of the values in these plots have been made dimensionless so that they apply more easily to many different physical problems as long as $Re$ is the same.

If you happen to be interested in the numerical solution method, it's just a relatively simple finite difference scheme called Crank-Nicolson. Let's be a little careful about what we call an "exact solution" though. This is not exact, as it is still a numerical approximation subject to finite spatial resolution, finite time-stepping, and the vast but limited numerical precision of a computer which introduces error. It should be very close, though.

I studied the graphs carefully and your approach to the solution and I have a better picture in my mind about the situation now.:)

 

[Soumalya]

Now, I can tell you that the exact solution to the differential equation at short times (i.e., when δ << L) is given by:

\frac{\partial v}{\partial y}=\frac{V}{\sqrt{\pi \nu t}}

So you can see that our approximate solution is only slightly different from the exact solution

From the exact solution of the differential force balance equation it's evident that the velocity gradient at any 'y' decreases with time so it implies the shear stress and hence the shear force say at y=L (at the upper plate fluid interface) decreases with time.Since we studied the situation of the upper plate moving with constant velocity so for the plate to sustain constant velocity(zero resultant force) the external force has to be decreased constantly with time in equation to the shear force at the plate.So is this the reason why you were indicating about the need of a motor with an automatic control system?

Also, you mentioned that from the solution we need an infinite amount of force to impulsively begin the motion of the plate (post #4)! Theoretically it's the case but we don't have to apply an infinite force to start the motion of the plate at t=0.What does that mean?

What would have been the situation if we were to apply a constant external force this time to the upper plate enough to impulsively start the motion of the plate?

I am trying to focus my attention to the plate this time

 

[Chestermiller]

From the exact solution of the differential force balance equation it's evident that the velocity gradient at any 'y' decreases with time so it implies the shear stress and hence the shear force say at y=L (at the upper plate fluid interface) decreases with time.Since we studied the situation of the upper plate moving with constant velocity so for the plate to sustain constant velocity(zero resultant force) the external force has to be decreased constantly with time in equation to the shear force at the plate.So is this the reason why you were indicating about the need of a motor with an automatic control system?

I said that because I wanted you to realize that we can move the plate in any way we want, with whatever force history we want.

Also, you mentioned that from the solution we need an infinite amount of force to impulsively begin the motion of the plate (post #4)! Theoretically it's the case but we don't have to apply an infinite force to start the motion of the plate at t=0.What does that mean?

It means that the force we have to apply is very high to begin with and then decreases with time.

What would have been the situation if we were to apply a constant external force this time to the upper plate enough to impulsively start the motion of the plate?

If the plate has no mass, then all we need to do is apply a constant force (equal to the constant shear force). In this case the velocity will start out from zero and increase continuously with time until it approaches a constant value. So, in this case, we would not be impulsively starting the motion of the plate.

Chet