# Rotation of fluid like a rigid body in a cylindrical contain

1. Jul 19, 2015

### benny_91

Hello!

While studying from the book 'Fluid Mechanics' by Cengel I came across the section which explains the behavior of fluids acting like a rigid body when the cylindrical container which contains the liquid rotates with a uniform angular velocity. Without much explanation the author states the governing equations using cylindrical co-ordinate system which are as follows:
$$\frac{\partial P}{\partial r}=\rho r \omega^{2} , \frac{\partial P}{\partial \theta}=0 , \frac{\partial P}{\partial z}=-\rho g$$
But when I tried to derive the first equation myself I ended up with something else. Here is my attempt to derive it. I am not including the diagram of the container so I request you to imagine the shape of the fluid element while I explain it. Let us assume that the density of the liquid is constant throughout.
As stated above, the variation in pressure with respect to angle in cylindrical co-ordinate system is zero. Hence let us imagine a cylindrical fluid element of radius 'r', thickness 'dr' and height 'dz'. Now we know that the pressure is the same at regions having same elevation and radius. Let the pressure be denoted by 'P'. Now we can do the force balance considering the pressure forces and the centrifugal force.
$$(P+\frac{\partial P}{\partial r}\frac{dr}{2})2\pi(r+\frac{dr}{2})dz - (P-\frac{\partial P}{\partial r}\frac{dr}{2})2\pi(r-\frac{dr}{2})dz = 2\pi r dr dz \rho r \omega^2$$
On simplifying this we get,
$$P dr+r\frac{\partial P}{\partial r} dr = \rho r^2 \omega^2 dr$$
Further simplification yields,
$$P+r\frac{\partial P}{\partial r} = \rho r^2 \omega^2$$
Finally this simplifies to
$$\frac{\partial}{\partial r}(Pr) = \rho r^2 \omega^2$$
This final result that we get has considerable difference from the one that is given in the book.

Please help me to find out where I have gone wrong in the derivation.
Thank you!

Last edited: Jul 19, 2015
2. Jul 19, 2015

### Drakkith

Staff Emeritus
Did you have a question?

3. Jul 20, 2015

### benny_91

Yes. Please help me find the mistake in my derivation.

Last edited: Jul 21, 2015
4. Jul 20, 2015

### Drakkith

Staff Emeritus
Whoops! I guess I missed the beginning of the 2nd paragraph. My mistake.

5. Jul 21, 2015

### benny_91

No. It was my mistake actually. I uploaded the second paragraph later. Beginner you know. Anyway here's the question!

6. Jul 21, 2015

### Nathanael

I haven't checked your work in detail, but your term on the right, 2πρrdrdzω2 should have an extra factor of r.
One r comes from the centripetal acceleration (rω2) and one r comes from the volume (2πrdrdz)

Also you don't need to work from the middle and move dr/2 in both directions, you can just start from one side and move dr to the other side.
Your ("midpoint") method would be more accurate if we were working with finite approximations, but in the limit of dr→0 the result is the same.

Edit:
It isn't correct to consider an entire cylindrical shell of liquid. You have to remember you are dealing with vectors (dA and dF) and they cancel out when you consider an entire cylinder (the net force needed for a cylinder to rotate is zero...)
Instead just work with an element of liquid which has a surface area dA which is small enough that it is approximately flat (so none of the force from the pressure cancels out).

Last edited: Jul 21, 2015
7. Jul 21, 2015

### benny_91

Thank you for your response. Firstly, I think you have missed to notice that I have included the 'r' factor twice. It is just that the order in which I have written is different from yours. But the equation is correct to the best of my knowledge.
Thank you for your suggestion regarding the "midpoint" method. That was really informative. Appreciated.
Further, I think it doesn't make any difference whether we consider the entire cylindrical shell or just a tiny part of it. The mass of this element would be $r d\theta dr dz$. The factor $d\theta$ would then be cancelled since it would be common on both sides of the equation. If you check the first line of the derivation you will notice that the common factor $2\pi$ has been cancelled in the same way. So the final equation will remain the same independent of whether the entire cylindrical shell is considered or only a tiny part of it.
I do agree that no force is required to keep a cylindrical shell in rotational motion. But at the same time there should be some agent that provides the necessary centripetal force required for rotation. In case of a rigid cylinder made from a solid material the centripetal force is provided by the circumferential (or hoop) stresses that is produced in the walls of the cylindrical shell. The problem with fluids is that it cannot produce the required centripetal force by generating hoop or tensile stresses since they are readily deformed. In the case discussed above the centripetal force is provided by the walls of the container. This force is transferred through the liquid and appears as radial pressure forces in the liquid which has been used in the derivation.
Please do respond if you find something wrong in the above explanation or you find the error in my derivation.
Thank you!

8. Jul 21, 2015

### Nidum

Just for interest :

It takes a theoretically infinite time for all the fluid in a rotating cylinder to achieve the same angular velocity after initiating cylinder rotation . Consider the fluid as a very large number of nested cylindrical shells each one being 'driven' by fluid shear from the next one out .

9. Jul 21, 2015

### Staff: Mentor

The error of your derivation is related to something you said regarding a rigid cylinder. In the case of a rigid cylinder, there is a tensile hoop stress. In the case of your fluid, there is compressive stress by pressure in the hoop direction. You omitted that from your derivation. When you include it, you will find that it exactly cancels the P/r term.

Chet

10. Jul 21, 2015

### Nathanael

You are right, I missed the other r, sorry.
It is not the same You need to be careful: what is the shape of this element? Is the only radial pressure coming from those two sides?

This is what you are doing $(P+dP)(A+dA)-PA = rω^2ρAdr$ which leads to $\frac{dP}{dr}=ρrω^2-\frac{P}{A}\frac{dA}{dr}$ with the right-most term equal to P/r
But the left side of this equation should have another term (which causes the P/r to vanish). This term comes from the other two sides of the element:

See, when you take the two sides to have different areas, this necessarily implies there is another component of radial force on the other sides.
There is a simple way to take account of these sides, which will give you the right answer.
Or you can just consider an element which has two equal sides, so that the force on the other sides vanishes: $(P+dP)A-PA=rω^2ρAdr \Rightarrow \frac{dP}{dr}=ρrω^2$

Last edited: Jul 21, 2015
11. Jul 21, 2015

### benny_91

But Nidum, I believe that the derivation is done for the steady state when all the fluid elements have attained the same angular velocity, maybe after a very long time. As there is no rubbing between the adjacent cylindrical shells there is no shear stress.

12. Jul 21, 2015

### benny_91

Thank you Chester. That explains the error.

13. Jul 21, 2015

### benny_91

Thank you Nathaneal for the help. The basic error was in considering the entire cylindrical shell instead of the tiny element. Further by including the compressive stresses in the circumferential direction as suggested by chestermiller we get the exact same result as given in the textbook.