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I'm essentially describing a fallacy due to a flaw in my understanding, and I'm trying to understand what the flaw is.

I know that if a pipe narrows, the fluid moving through it will move faster to preserve the same volumetric flow rate.

But, by that logic, an hourglass should not work, nor any type of funnel, nor should drains ever clog, because fluids should just flow faster through the narrower hole and ultimately pass through the opening in the same amount of time as if it were a wider opening.

What is the flaw in my reasoning?

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I think you're considering this in reverse.

Yes; the flow does speed up in the area of a smaller cross section to remain constant volumetric flow (if incompressible). If I understand your question right; you're wondering why this would cause a slower flow, if having a large diameter tube with no bottleneck would have to have the same volumetric flow rate.

The thing is; the overall volumetric flow rate depends on what is downstream of the system as well. If you're familiar with electric circuits; a good way to picture it is through the electronic-hydraulic analogy.

You have a potential difference of energy between the top of the fluid and the exit after the flow restriction. This is analogous to a voltage source in a circuit. Adding a flow restriction is like adding another resistor in series of the circuit.

The potential energy difference (voltage) is still the same; but now the flow everywhere in the circuit is lower because of this restriction. The volumetric flow rate is analogous to the amperage of the circuit; adding resistance lowers the flow at every point in this system.

Essentially; any flow restriction will decrease the overall volumetric flow rate of the hydraulic system. I didn't get too in depth in the actual fluid dynamics in this answer; but hopefully this gives an intuitive/familiar way to see it.

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As mentioned in other answers, the fluid doesn't flow on its own.

Think of your problem this way:
1. Gravity pulls the fluid through the opening. Only the fluid column above the opening is moving down as the other parts are restricted by the container. Larger the opening more fluid can be pulled down per unit time. The volume left void is just replaced by the remaining fluid in the top half.

2.

I know that if a pipe narrows, the fluid moving through it will move faster to preserve the same volumetric flow rate.

Yes, in case of steady flow of incompresisble fluid. The hourglass still passes criteria for constant volumetric flow. By definition, during a steady-flow process, the total amount of mass (=volume when incompressible) contained within a control volume does not change with time. The control volume here is the opening, where Volume in = Volume out.

3. What would make it flow faster?
A larger acceleration (due to gravity here) would make it flow faster. Its independent of the size of the opening. The same applies in other situations. If the flow is also steady and incompressible then constant volumetric flow applies.

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This is a very common fallacy, yes!

So when we say that "fluid flows faster as the pipe gets narrower" we mean within the same pipe. We do not mean across all circumstances. The cause of the increased fluid flow is that water is a highly incompressible fluid. Because of this, any mass that flows into a box must also flow out of that box, and any volume that flows into a box must also flow out of that box: otherwise we'd be compressing the fluid within that box.

It's worth taking a second to understand that point better, so let me recommend this experiment: in lots of places you can find thin plastic drink bottles in a 1.5L or 2L size, and generally they have screw-on caps which form a nice airtight seal. Take an empty one (containing only air) and put the top on it, and then squeeze it. Try to estimate how much you're actually compressing this volume—10%? 20%?—at different amounts of pressure from your fingertips. The point is that there is a curve: the more pressure you apply, the more you compress this thing. Now if you fill it with water, you will notice that whatever this curve is it is very steep: it takes immense amounts of pressure to compress water even the barest of degrees. So this is not an absolute rule and in fact it will be violated at the sub-microscopic scale in non-steady-states all the time, but the point is that it works well to first approximation because any significant volume change would require much more pressure than you're typically dealing with.

Now when you compare two different pipes, say with the same pressure across them, usually the fluid flows slower as the pipe gets narrower. One can get a rough idea of what this should be with a scaling argument and dimensional analysis: it should go proportional to pressure and inversely with viscosity; those together form $[[p/\mu]] = \text s^{-1}$ and one immediately recognizes that whatever we multiply this by must be a length having units of $\text m$ so that we get $\text m\cdot\text s^{-1}$ and we have a velocity. There are two length scales $D,L$ at play though -- the diameter of the flow and the length. But we expect that this sort of friction would cause twice the pressure drop over twice the $L$ at constant $v$, introducing a unit of $\text m^{-1}.$ So the full equation would be $v \propto p D^2/(\mu L).$ So at constant pressure with twice the diameter we can expect four times faster flow.

How this argument fails

This argument is not fully rigorous and I am pulling a fast one on you by requiring $p\propto v$, which only applies in the laminar-flow regime. A fully rigorous dimensional-analysis argument could focus on the pressure loss $p$, finding a fully correct expression would be something more like $$p = \frac{\mu v L}{D^2}~f\left({\rho~v~D\over\mu},~{L\over D},~\frac\epsilon D\right),$$where an arbitrary unknown function $f$ is being taken of all of the dimensionless coefficients of the flow and $\epsilon$ is the typical length scale of surface imperfections on the sides. The argument that $L$ should not enter into this function $f$ works for both turbulent and laminar flow.

Now this claim that $f$ is just a constant is true only for small velocities where we expect $p\propto v:$ but at larger velocities this ceases to be true and instead, at some laminar-to-turbulent flow regime, the flow velocity begins to depend very strongly on the exact pressure between these two, almost discontinuously jumping higher as one puts more pressure and more vortices can form in the fluid. Finally when turbulence has fully taken hold, the pressure instead hits a limit where $p\propto v^2.$ The above expression still holds but it would be more appropriate to write it as $(\rho~v^2~L/D)~\alpha(\epsilon/D)$ for some undetermined function $\alpha$. The sudden lack of importance of the viscosity is a huge clue about what's going on in this expression; momentum is now being lost by fluid molecules smacking into these surface imperfections $\epsilon$ and getting scattered in a random direction in this boundary layer of the flow.

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The flaw in your reasoning is that you're presuming that the fluid is being forced to flow faster.

Yes, if you have a situation where you want the same volume of fluid to flow through a thinner pipe, the fluid must be made to "flow faster". However, to say that the fluid "will move faster" somehow implies that the fluid will just do that without any particular reason to do it. If that's what you think, you're under a misapprehension.

Short answer: Fluids should not just "flow faster through the narrower". This is a misunderstanding.

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  • $\begingroup$ Can you give an example of what would make it flow faster? This is what confuses me. I was picturing constant pressure at the pipe entrance. What conditions would induce constant volumetric flowrate? $\endgroup$ – temporary_user_name Oct 22 '17 at 2:16
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If you were dealing with an ideal fluid then the speed $v$ at which the fluid emerged from an orifice would be $\sqrt{2gh}$ where $h$ is the height of the fluid above the orifice.
[As pointed out by @LucJ.Bourhis such an analysis assumes that the surface speed is very much smaller than that at the orifice which is reasonable when the area of the surface is much larger than the area of the orifice.]
The speed at which the fluid emerged from the orifice does not depend on the area of the orifice $A$ but the volume flux $vA$ does.
Such flow under the force of gravity is discussed here.

In non-ideal situations then for liquid viscosity would play a part and reduce the size of the orifice would make the viscous nature of the liquid have an effect on the flow rate as would the non-laminar flow.
So for a funnel or a drain with a given head of liquid one would expect the flow rate to decrease even more than foe an ideal fluid as the aperture decreased.

The hour glass example is entirely different in that the particles of sand do not behave in the same way as the molecules in a liquid and the pressure at the bottom of the sand does not depend on the height of the sand above the orifice if the height of the sand is greater than the diameter of the container noting that an hour glass has tapered sides.
For a liquid one has that the pressure is proportional to the height of the liquid.

So above a certain height of sand in the hour glass the rate at which sand flows through the orifice is independent of the height of sand in the sand glass.
This independence of the pressure is called the Janssen effect and is due to the grains of sand forming bridges which help transmit the weight of the sand grains to the container walls were upward frictional forces are acting.
One notices this effect when pouring rice, bird seed etc from a plastic bag with a small in it even to the extent that the flow actually stops if the area of the orifice is too small.
The flow of particles though an orifice is the subject of much research as is shown in this article and also in this article in which the importance of the Janssen effect is disputed.

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  • $\begingroup$ You need to neglect the velocity of the fluid on the surface to reach the conclusion in your first paragraph. $\endgroup$ – user154997 Oct 22 '17 at 8:56
  • $\begingroup$ @LucJ.Bourhis Thank you for your comment. I have added a note to this effect in my answer. $\endgroup$ – Farcher Oct 22 '17 at 9:02

protected by Qmechanic Oct 22 '17 at 8:52

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