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CR Drost
CR Drost
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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 kinetic 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.

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 kinetic 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.

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.

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CR Drost
CR Drost
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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 kinetic 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.