Fluid mechanics and circular flow

[tomz]

Dear all

I am having a problem on circular flow of fluid. On all books I have read they say
\frac{dp}{dr}=\rhov^{2}/r

Which make sense by using infinitesimal square volume and take the force exert.

But if I use a circular infinitesimal volume (which is usually the case for circular things), i get a different answer because the side with greater pressure have greater area.

Why is this ?


Thank you

 

[Simon Bridge]

Please describe the situation you are dealing with.
What do you mean by "circular flow of fluid"?
To me that means the pipe goes in a circle, or that there is a vortex or whirlpool.
But I don't see how you can have an infinitesimal volume with different areas on different sides.

 

[tomz]

Please describe the situation you are dealing with.
What do you mean by "circular flow of fluid"?
To me that means the pipe goes in a circle, or that there is a vortex or whirlpool.
But I don't see how you can have an infinitesimal volume with different areas on different sides.

Thanks for the reply. Sorry I did not make it clear.

I mean for a vortex, or any flow that is not along a straight line.

Because from what I deal with viscid fluid/statics/pressure vessel.. when there is rotation or cylindrical coordinate, we tend to use a curved shape (sector) infinitesimal area with area rdrdθ.
But for this shape, the force from outside is (r+dr)dθ(p+dp/dr*dr), force from inside is rdθ*p. Giving a difference of (dp/dr*r+p)*dθ*dr rather than just dp/dr*r*dθ*dr as for a square infinitesimal area.

Please tell me if I still make it unclear. Thank you

 

[Simon Bridge]

In cylindrical coordinates, the force on the inner area of the infinitesimal volume $dV$ is $F_1=p(\rho)\rho d\theta dz$ while the outer area is $F_2=(\rho+d\rho)p(\rho+d\rho) d\theta dz$ and your assertion is that this means that $F_1<F_2$. Where $p(\rho)$ is the pressure at radius $\rho$.

Let's say there is no circular motion - the water is still.
Cylindrical coordinates have been chosen because of the cylindrical container or something.
Then $F_2>F_1$ would make the water tend to hump up in the middle wouldn't it?

Is that the jist of things?

 

[tomz]

In cylindrical coordinates, the force on the inner area of the infinitesimal volume $dV$ is $F_1=p(\rho)\rho d\theta dz$ while the outer area is $F_2=(\rho+d\rho)p(\rho+d\rho) d\theta dz$ and your assertion is that this means that $F_1<F_2$. Where $p(\rho)$ is the pressure at radius $\rho$.

Let's say there is no circular motion - the water is still.
Cylindrical coordinates have been chosen because of the cylindrical container or something.
Then $F_2>F_1$ would make the water tend to hump up in the middle wouldn't it?

Is that the jist of things?

Am I making the mistake of ignoring the tangential component of forces...

Thank you so much!

 

[Simon Bridge]

In my example, there are no tangential forces.

You also get a similar issue vertically - the pressure of the water is higher as you go deeper, so the force up from the bottom of an infinitesimal volume is greater than the force down from the top...

These are things you've dealt with before.

If the water is in uniform circular motion, then there must be an unbalanced force on each volume element that points to the center or the motion is not possible right?