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I've been searching yet haven't found any great intuition about what the Reynolds Number implies. Most people would characterize it as the measure of turbulence in a flow (usually defined by a specific value, a, where Re<a = laminar, Re>a = turbulent).

I am simply asking what about the increasing properties of the Reynolds Number makes the flow more turbulent. My intuition would be that for any flow gradient, any particle is attempting to rotate/curl due to a net moment about the particle's centroid. I suppose the intermolecular forces and density would be the forces generally restricting this rotation?

Any insights or more ac

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  • $\begingroup$ It's not really that higher Re = more turbulent. It's more like higher Re = more likely to be turbulent. As far as characteristics of the fluid go, it's not just about density--it's really the ratio of density to viscosity that matters. E.g., air has higher viscosity per unit mass than water, so all else being equal, the Reynolds number in air will be lower. $\endgroup$ Dec 27, 2023 at 6:03
  • $\begingroup$ Further to the first point: if the interior of the pipe is rougher, the flow will become turbulent at a lower Reynolds number. The usually quoted cutoff for pipes is Re=2300. So the 2000 shown in your figure is apparently based on a rougher surface. With a smoother surface, you might be able to maintain laminar flow at Re=2500, and with a really highly polished surface, maybe even up to Re = 3000 (I'm picking numbers out of the air here, but substantially higher, anyway). $\endgroup$ Dec 27, 2023 at 6:08
  • $\begingroup$ Flow can change to turbulent above about 1700. Which source did you use? You need to check out others - there are many that cover laminar to turbulent flow . Mechanics of Fluids by Massey is one. $\endgroup$
    – Solar Mike
    Dec 27, 2023 at 8:29

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Multiple the numerator and denominator by the cross sectional area $A$ to obtain

$$ N_{Re} = \frac{\rho\ v\ L\ A}{\mu\ A}$$

The numerator and denominator are momentum. The numerator is the momentum supplied by an external means to a packet (control volume) of the fluid. The denominator is the momentum transferred from a given packet to its surrounding fluid packets.

At low $N_{Re}$, a fluid packet behaves in cooperation with its surrounding packets. Momentum input is effectively transferred between all packets. This is plug flow or laminar flow conditions. As $N_{Re}$ increases, more momentum is supplied to any given packet than can be transferred effectively to its surrounding packets. Each fluid packet begins to "slip" over its surrounding packets; the packets begin moving independently of each other. The independent movement includes rotation about the center of mass, since each individual packet is less and less bound to the surrounding packets. This is the transition to turbulent flow conditions.

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Intuitive version: Re = ratio of inertia vs viscous forces

A denser or less viscous fluid will tend to keep swirling once you start it swirling, thus more turbulent etc.

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Starting from a laminar solution $u_0$ to the Navier-Stokes equations (e.g. Poiseuille flow in a pipe), add a small disturbance - a random fluctuation - $\Delta\! u$ onto the velocity field. Because the disturbance is small, it's safe to neglect terms in the Navier-Stokes equation quadratic in the disturbance, and terms independent of the disturbance cancel because the laminar velocity profile is a solution of the N-S equations. Hence, we need consider only terms linear in the disturbance to get an equation for the Eulerian time derivative of velocity: $$\frac{\partial\!\!\mathrm{\Delta}\! u}{\partial\! t} = -u_0\frac{\partial\!\!\mathrm{\Delta}\! u}{\partial\! x}+\nu\nabla^2\left(\mathrm{\Delta}\! u\right)$$

where $\nu$ is the kinematic viscosity of the fluid and $x$ is the direction co-ordinate along which the laminar flow is generally aligned. If there are no-slip boundary conditions on both sides of, e.g., a pipe or a Hele-Shaw cell, then $\nabla^2\left(\mathrm{\Delta}\! u\right)$ must (averaged over the space) have the opposite sign from $\mathrm{\Delta}\! u$. Hence, the $\nu\nabla^2\left(\mathrm{\Delta}\! u\right)$ term tends to make the disturbance die away, i.e. to prevent turbulence. The form of disturbance that is most inclined to grow, i.e. which will decide whether there is any turbulence, is the form where $-u_0\partial\!\mathrm{\Delta}\! u/\partial\! x$ most strongly has the same sign as $\mathrm{\Delta}\! u$. This requires the spatial fluctuations of the disturbance to be, in some sense, in phase quadrature with the spatial variation of the laminar solution, so it's a form of disturbance that varies on the same order-of-magnitude length scale $D$ as the laminar solution. Hence, if we denote the amplitude of the disturbance by $U$ and the mean velocity in the laminar solution by $\left\langle u\right\rangle$ , $-u_0\partial\!\mathrm{\Delta}\! u/\partial\! x$ has order-of-magnitude size $\left\langle u\right\rangle U/D$ and the same sign as $\mathrm{\Delta}\! u$, and $\nu\nabla^2\left(\mathrm{\Delta}\! u\right)$ has order-of-magnitude size $\nu U/D^2$ and the opposite sign from $\mathrm{\Delta}\! u$. The disturbance grows (i.e. there is turbulence) iff $\left\langle u\right\rangle U/D-\nu U/D^2$ has the same sign as $U$, i.e. iff $\left\langle u\right\rangle/D-\nu/D^2$ is positive: that is, if the Reynolds number $\left\langle u\right\rangle D/\nu$ is large.

(This argument makes it look like the critical Reynolds number is $1$, but you'll have noticed that this argument is very rough-and-ready at several points. A more detailed analysis would reveal the familiar critical Reynolds number of a couple of thousand, but not change the essential qualitative point that there is a critical Reynolds number.)

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