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Yesterday I answered this question, using analysis of forces from this web page.

But today I started having doubts regarding the completeness of this analysis. It occurred to me this approach does not take into account any centripetal forces caused by the changing direction of the fluid flow. Simply put:

$\vec{F}=m\frac{d\vec{v}}{dt}$.

To analyse this, consider the following geometry for a $90^0$ bend in a pipe lying in a horizontal $(x,y)$ plane:

Bent pipe.

Assumptions:

  • turbulent plug flow with constant fluid speed $v$, fluid density $\rho$, cross-section of pipe $A$, bend radius $R$.
  • $\frac{R}{A}\gg 1$

Right now I am only concerned with centripetal forces needed to keep the flow on its rotational trajectory.

Take an infinitesimal element at $\theta$:

$dm=\rho A Rd\theta$.

The centripetal forces are:

$dF=-\frac{v^2 dm}{R}=-\rho A v^2 d\theta$

$dF_x=-\rho A v^2\cos\theta d\theta$.

$dF_y=-\rho A v^2\sin\theta d\theta$.

So these would be the forces the pipe needs to exert on the fluid:

$F_x=-\rho A v^2\int_0^{\frac{\pi}{2}}\cos\theta d\theta=-\rho A v^2$.

$F_y=-\rho A v^2\int_0^{\frac{\pi}{2}}\sin\theta d\theta=-\rho A v^2$.

If this is correct then I obviously need to amend my answer. So my question is: is this correct?

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I got the same answer using momentum considerations.

In time $dt$ the amount of mass hitting the side of the pipe is:

$$dm=\rho A v dt$$

Now, we know that the force must satisfy:

$$F_{x}dt=dp=0-vdm=-\rho A v^2 dt $$

Dividing both sides by $dt$ gives:

$$F_{x}=-\rho A v^2$$

The force in the $y$ direction is the same as here.

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  • $\begingroup$ Ah. That’s very interesting. I think the author of the second page I linked to also used momentum because $F_{x}=-\rho A v^2$ can be re-written as $F_{x}=-\dot{m}v$. That would mean my original answer was indeed correct. I will wait for any (?) more answers and then upvote. Many thanks! $\endgroup$
    – Gert
    Sep 22 '15 at 14:28

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