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I was browsing PhysicsForums and I came across an interesting fluids question. I understood the method suggested to solve it, but I wanted to see if I could solve it through integral calculus.

[Question source - original method.]

A tube of uniform cross-section $a$ is bent to form a three-quarter circular arc of radius R, and a liquid is forced through the tube with a linear speed $v$. The liquid enters the tube and exits with the same speed.

I'm having some difficulty setting up the equations for this. I'm taking the basic element as $dm$ which is a tiny slice of the fluid with thickness $dx$, so $dm = \rho adx$.

This element moves with constant speed in a circular path, so if we measure angular displacement of the element from the initial position, we get $x=R\theta$, so $x$ varies from $0$ to $\frac{3\pi}{2}$. I also think that the angular velocity of $dm$ which is $\omega = \frac{d\theta}{dt} = \frac{v}{r}$ should be constant because speed doesn't change?

The basic idea is to use the fact that force is the rate of change of momentum. I wanted to find the net horizontal and vertical forces exerted on the fluid element throughout its range of motion and the force on the tube would be the corresponding net reaction force in the opposite direction.

So I wrote out the components of velocity (with constant magnitude $v$) as trigonometric functions of $\theta$. Assuming the ring to be observed symmetrically (the angle marked is $\theta$)

 [diagram

The horizontal and vertical components of velocity are: $v_x = v\sin(\frac{\pi}{4}-\theta)$ and $v_x = v\cos(\frac{\pi}{4}-\theta)$.

So, integrating this for $F_x$:

$$F_x = \frac{dm}{dt}v_x + m\frac{dv_x}{dt}$$ $$=\rho a \frac{dx}{dt}\cdot v\sin(\frac{\pi}{4}-\theta) + \rho adx \cdot v\cos(\frac{\pi}{4}-\theta)\cdot (-\frac{d\theta}{dt})$$

Now I have no idea how to carry this further. So I guess my question is, are these equations accurate (i.e. can they be solved to give the correct answer - net force is $\sqrt{2}\rho a v^2$) or if these equations are wrong, could you please help me find the correct equations to integrate?

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    $\begingroup$ See the section on Macroscopic Momentum Balances in Transport Phenomena by Bird, Stewart, and Lightfoot. $\endgroup$ Commented Jan 16 at 11:50

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I don't understand your calculation method. It should be noted that even for an open system, of possibly variable mass, Newton's law is not written $ \dfrac{d(m\vec{v})}{dt}=\vec{F}$

This type of exercise is usually handled by considering the overall closed system formed by the water in the tube at a given moment. The reasoning is classical and we arrive at Euler's theorem $D_m(\vec{v_s}- \vec{v_e})=\vec{F}=\vec{F_{tube \rightarrow water}}+ \vec{F_{pressure}}=\vec{F_{tube \rightarrow water}}+P_e S_e \vec{t_e}- P_s S_s \vec{t_s}$

If you wish to analyze in detail what is happening within the tube, we can reason simply by labeling with an index $i$ each identical elementary mass $dm$ of velocity $\vec{v_i}$. Every $dt$, the particle $i$ takes the place of the particle $i+1$. Newton's law is written :

$ dm \dfrac{d\vec{v_i}}{dt}=d\vec{F_i}$ with $d\vec{F_i}= d\vec{F_{i_{tube\rightarrow water} }}+ d\vec{F_{i_{pressure}}}$

The term $ d\vec{F_{i_{pressure}}}=\vec{F_{i_{pressure_{left}}}}-\vec{F_{i_{pressure_{right}}}}$ is related to the pressure forces exerted by adjacent elements, on either side except at the entrance and exit.

To find the force on the tube, we must add up all the forces on the tube: $\vec{F_{tube\rightarrow water}}=\Sigma d\vec{F_{i_{tube\rightarrow water}}}$

By replacing: $\vec{F_{tube\rightarrow water}}=\Sigma dm \dfrac{d\vec{v_i}}{dt} - \Sigma d\vec{F_{i_{pressure}}} $

The pressure forces are compensated 2 to 2 except at the inlet and outlet: $-\Sigma d\vec{F_{i_{pressure}}} =\vec{F_{i_{pressure_{s}}}}-\vec{F_{i_{pressure_{e}}}}= - P_s S_s \vec{t_s}+P_e S_e \vec{t_e}$

It remains to calculate:

$\Sigma dm \dfrac{d\vec{v_i}}{dt}=\dfrac{dm }{dt}\Sigma \dfrac{\vec{v_i}(t+dt) - \vec{ v_i}(t)}{dt}$

But, by construction $\vec{v_i}(t+dt)= \vec{v_{i+1}}(t)$ so that: $\Sigma dm \dfrac{d\vec{v_i}}{dt}=D_m\Sigma (\vec{v_{i+1}}(t) - \vec{v_i}(t))$

Here again, everything is simplified except for input and output: $\Sigma dm \dfrac{d\vec{v_i}}{dt}=D_m(\vec{v_s}-\vec{v_e})$

We obviously find Euler's theorem. Hope it can help. I'm not sure it's any more enlightening than the usual method !

EDIT :

For an open system, the total momentum is $\vec{P}=\iiint_\Sigma \vec{v} \,dm$, the integral relating to the control surface $\Sigma$.

Newton's law is not written $ \dfrac{d\vec{P}}{dt}=\vec{F}$. This is easily seen in your example in which the total momentum is constant since we are in steady state. Even in the case of the rocket, we do not find a correct result by writing $ \dfrac{d(m\vec{v})}{dt}=\vec{F}$.

In reality, the correct result is very simple. Just remember that for an open system, there are two ways to change the momentum: by exerting a force (source term) but also by bringing the momentum into the system. Newton's law is therefore written:

$ \dfrac{d\vec{P}}{dt}=\vec{F}+ \vec{\Phi}$ with $\vec{\Phi}$ the momentum entering the system per unit of time . For a fixed control surface, with a single entry point and a single exit point, this incoming flow is written: $D_{me} \vec{v_e}- D_{ms} \vec{v_s}$ with $D_{me}$ and $ D_{ms}$ the incoming and outgoing mass flow rates.

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  • $\begingroup$ Huh, so basically, you're saying that even when we take differential fluid elements, there's no need to integrate, because a net force sum already gets rid of almost all the terms in between. So we skipped the integration by directly adding the force on the tube by each $i$th element ... though could you please elaborate on why we do not use $F = \frac{d(mv)}{dt}$ in such scenarios? $\endgroup$
    – zxayn
    Commented Jan 15 at 16:52
  • $\begingroup$ Sorry for the delay. I was very unavailable today. I will add a comment as soon as possible. $\endgroup$ Commented Jan 16 at 17:57
  • $\begingroup$ I added a complement about Newton's law for an open system. $\endgroup$ Commented Jan 18 at 10:54
  • $\begingroup$ Got it, thanks a lot! $\endgroup$
    – zxayn
    Commented Jan 19 at 18:18

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