Fluids - Squeezing flow - How to find force, work, and flux

[OldStudent0382]

Homework Statement

The initial separation of the pistons is 2cm. They are submerged in water at room temperature.

I need to calculate the work required to have two pistons touch, find the volumetric flux of the water as a function of time, and the force at a given time

Homework Equations

The Attempt at a Solution

F= \int p(r,z) 2 \pi r dr (from 0 to R)

F=(3/2) (pi u *V)/(H^3) * [z^2*R^2+R^4/2-R^4/4+Po*R^2/2][/B]

This is as far as I got with it.

Is my solution for force correct? I'm not sure of the integration I need to do for work and flux from here. How do I find the force as a function of time?

I tried googling for lectures on axisymmetric squeezing flow, but I only found research papers on the topic that did not help me in solving this problem. What other keywords should I be searching for?

 

[Chestermiller]

Homework Statement

View attachment 82605
The initial separation of the pistons is 2cm. They are submerged in water at room temperature.

I need to calculate the work required to have two pistons touch, find the volumetric flux of the water as a function of time, and the force at a given time

Homework Equations

View attachment 82602
View attachment 82603
View attachment 82604

The Attempt at a Solution

F= \int p(r,z) 2 \pi r dr (from 0 to R)

F=(3/2) (pi u *V)/(H^3) * [z^2*R^2+R^4/2-R^4/4+Po*R^2/2][/B]

This is as far as I got with it.

Is my solution for force correct?

No. The p₀ should not be inside the parenthesis. In fact, it should not be in there at all; you forgot to include the force of the water pushing down on the piston from the top. Also, you should have used the pressure at z = H to get the force on the piston, and not at arbitrary z. I didn't check the rest of your integration. Also, your equations are hard to read because you don't use Latex.

What is H as a function of time?

Chet

 

[OldStudent0382]

Here's what i meant to enter earlier:
F=\frac{3\pi \mu V}{2H^{3}}*[z^2R^2+\frac{R^4}{2}-\frac{R^4}{4}]+\pi P_{o}*R^2

Besides replacing "z" with H, Is Po the only correction I need to make? Just dropping it? In thinking about the problem, I had a feeling it would cancel out.

For work, would I just integrate the force over the distance "z" the pistons move?

The initial gap between the two pistons is 2cm and the velocity of each piston is 0.1cm/s, so I'm taking H(t) to be
H(t)=2cm-2*0.1cm/s*t

Thank you!

 

[Chestermiller]

Here's what i meant to enter earlier:
F=\frac{3\pi \mu V}{2H^{3}}*[z^2R^2+\frac{R^4}{2}-\frac{R^4}{4}]+\pi P_{o}*R^2

Besides replacing "z" with H, Is Po the only correction I need to make?

No. I think the R⁴ term should have an 8 in the denominator, not a 4.

Just dropping it?

As I said, you neglected to include the downward force from the water on the top of the piston, which exactly cancels out the pressure term that you got in the integration.

In thinking about the problem, I had a feeling it would cancel out.

Your intuition was correct.

For work, would I just integrate the force over the distance "z" the pistons move?

Sure, but I don't know whether they want you to include the work on both the pistons or on just one of them.

The initial gap between the two pistons is 2cm and the velocity of each piston is 0.1cm/s, so I'm taking H(t) to be
H(t)=2cm-2*0.1cm/s*t

The diagram looks incorrect. From your equations for the velocities, it is obvious that H is half the distance between the pistons (and z is measured upward from the center-plane). So, H = 1 - (0.1) t

Chet

 

[OldStudent0382]

No. I think the R⁴ term should have an 8 in the denominator, not a 4.

As I said, you neglected to include the downward force from the air on the top of the piston, which exactly cancels out the pressure term that you got in the integration.Your intuition was correct.Sure, but I don't know whether they want you to include the work on both the pistons or on just one of them.

The diagram looks incorrect. From your equations for the velocities, it is obvious that H is half the distance between the pistons (and z is measured upward from the center-plane). So, H = 1 - (0.1) t

Chet

Thanks - I wasn't too sure about H(t).

As for where you said the 4 should be an 8.

Here's how I got it:

F=\int_{0}^{R}[\frac{3 \mu V}{4H^{3}}*[2z^2+R^2-r^2]]*2\pi r dr

becomes:

F=\frac{3\pi \mu V}{2H^{3}}*[z^2 r^2+\frac{R^2*r^2}{2}-\frac{r^4}{4}]

Then substituting in R:
F=\frac{3\pi \mu V}{2H^{3}}*[z^2 R^2+\frac{R^4}{2}-\frac{R^4}{4}]

I don't think I'm making any mistakes -- please let me know if there is something you see that I'm overlooking!

Thanks again!

 

[OldStudent0382]

In addtion - for work, would taking the work done by 1 piston and multiplying it by 2 provide a valid answer? or even just integrating from -1cm to +1cm would provide the same result as integrate from 0 to +1 or -1 and multipling by two, I think. I always doubt myself on these situations and this is where I make my mistakes.

 

[OldStudent0382]

Should I be replacing "H" with either the function "H(t)" or z? At first I thought it should be the size of the initial gap or distance from z=0 (in this case, 1cm), but I'm starting to question if that assumption was correct.

For the volumetric flux Q(t), would it be the circumference of the pistons (2*pi*r) * u_r*H(t)?

Thanks!

 

[Chestermiller]

Thanks - I wasn't too sure about H(t).

As for where you said the 4 should be an 8.

Here's how I got it:

F=\int_{0}^{R}[\frac{3 \mu V}{4H^{3}}*[2z^2+R^2-r^2]]*2\pi r dr

becomes:

F=\frac{3\pi \mu V}{2H^{3}}*[z^2 r^2+\frac{R^2*r^2}{2}-\frac{r^4}{4}]

Then substituting in R:
F=\frac{3\pi \mu V}{2H^{3}}*[z^2 R^2+\frac{R^4}{2}-\frac{R^4}{4}]

I don't think I'm making any mistakes -- please let me know if there is something you see that I'm overlooking!

Thanks again!

Oooops. My mistake. Sorry. That's what I get for trying to do it in my head.

You should mention to your professors about that error in the figure regarding H. Another issue is that the platens are immersed in water. So, unless the platens are neutrally buoyant, they are going to contribute to the force, and the bottom force is going to differ from the upper force. So, I guess you need to assume that they are neutrally buoyant.

Chet

 

[Chestermiller]

In addtion - for work, would taking the work done by 1 piston and multiplying it by 2 provide a valid answer? or even just integrating from -1cm to +1cm would provide the same result as integrate from 0 to +1 or -1 and multipling by two, I think. I always doubt myself on these situations and this is where I make my mistakes.

You want to do the upper piston and integrate from 0 to H(0), then multiply by 2 to get the work by both the pistons. Or, do the lower piston and integrate from 0 to -H(0), then multiply by 2 to get the work by both pistons. But neither piston moves over the full gap, so, even though the result might give the right answer mathematically (I haven't tried it to make sure), it doesn't make sense physically.

Chet

 

[Chestermiller]

Should I be replacing "H" with either the function "H(t)" or z? At first I thought it should be the size of the initial gap or distance from z=0 (in this case, 1cm), but I'm starting to question if that assumption was correct.

It doesn't matter what you use. Either way, you are going to be integrating from 0 to H(0).

For the volumetric flux Q(t), would it be the circumference of the pistons (2*pi*r) * u_r*H(t)?

No. You have to integrate over z, from z = -H(t) to z = +H(t). Here's another (simpler) way to do it:

What is the volume of fluid between the plattens at time t when their separation is 2H(t)? What is the rate of change of that volume with t? That must be equal to minus the volumetric flow rate out. (It has nowhere else to go).

Chet