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I understand that the reynolds number is given by the expression $Re=\frac{\rho v L}{\mu}$, where $\rho$ is the density, $v$ is the fluid velocity and $\mu$ is the dynamic viscosity. For any given fluid dynamics problem, $\rho$, $v$, and $\mu$ are trivially given. But what exactly is the characteristic length $L$? How exactly do I calculate it? What can I use from a given problem to determine the characteristic length automatically?

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  • $\begingroup$ Could you explain why the Reynoldsnumber is the similarity which describes your flow problem? $\endgroup$ – rul30 Oct 10 '15 at 15:09
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I would like to approach this question from a mathematical perspective which can be fruitful as discussed in some of the comments and answers. The given answers are useful, however i would like to add:

  1. In general the smallest available length scale is the characteristic length scale.
  2. Sometimes (e.g. in dynamic systems) there is no fixed length scale to choose as a characteristic length scale. In such cases often a dynamic length scale can be found.

Characteristic length scales:

TL;DWTR: for $R/L\ll1$, $R$ is the characteristic length scale; for $R/L\gg1$, $L$ is the characteristic length scale. This implies that the smaller length scale is (usually) the characteristic length scale.

Consider the pipe flow case discussed in the other answers; there is the radius $R$ but also the length $L$ of the pipe. Usually we take the pipe diameter to be the characteristic length scale but is this always the case? Well, lets look at this from a mathematical perspective; let's define the dimensionless coordinates: $$\bar{x}=\frac{x}{L} \quad \bar{y}=\frac{y}{R} \quad \bar{u}=\frac{u}{U} \quad \bar{v}=\frac{v}{V} \quad \bar{p}=\frac{p}{\rho U^2}$$

Here, $L$, $R$, $U$, $V$ are $x$-$y$ coordinate and velocity scales but not necessarily their characteristic scales. Note that the choice of the pressure scale $P=\rho U^2$ is only valid for $\mathrm{Re}\gg1$. The case $\mathrm{Re}\ll1$ requires a rescaling.

Transforming the continuity equation to dimensionless quantities:
$$\boldsymbol{\nabla}\cdot\boldsymbol{u}=0 \rightarrow \partial_{\bar{x}}\bar{u}+\partial_{\bar{y}}\bar{v}=0$$

which can only be the case when we assume $\frac{U}{V}\frac{R}{L}\sim1$ or $\frac{V}{U}\sim\frac{R}{L}$. Knowing this, the Reynolds number may be redefined:

$$\mathrm{Re}=\frac{UR}{\nu}=\frac{U}{V}\frac{R}{L}\frac{VL}{\nu}=\frac{VL}{\nu}=\hat{\mathrm{Re}}$$

Similarly, let's transform the Navier-Stokes equations ($x$-component only to keep it short): $$\boldsymbol{u}\cdot\boldsymbol{\nabla u}=-\frac{1}{\rho}\boldsymbol{\nabla}p+\nu\triangle\boldsymbol{u}$$ $$\bar{u}\partial_{\bar{x}}\bar{u}+\bar{v}\partial_{\bar{y}}\bar{u}=-\partial_{\bar{x}}\bar{p}+\frac{1}{\mathrm{Re}}\left[\frac{R}{L}\partial_{\bar{x}}^{2}\bar{u}+\frac{L}{R}\partial_{\bar{y}}^{2}\bar{u}\right]$$ We see here the Reynolds number occuring naturally as part of the scaling process. However, depending on the geometric ratio $R/L$, the equations may require rescaling. Consider the two cases:

  • The pipe radius is much smaller than the pipe length (i.e. $R/L\ll1$):

    The transformed equation then reads: $$\bar{u}\partial_{\bar{x}}\bar{u}+\bar{v}\partial_{\bar{y}}\bar{u}=-\partial_{\bar{x}}\bar{p}+\frac{1}{\mathrm{Re}}\frac{L}{R}\partial_{\bar{y}}^{2}\bar{u}$$ Here we have a problem because the term $\frac{1}{\mathrm{Re}}\frac{L}{R}$ could be very large and a properly scaled equation only has coefficients $O(1)$ or smaller. So we require a rescaling of the $\bar{x}$ coordinate, $\bar{v}$ velocity and $\bar{p}$ pressure: $$\hat{x}=\bar{x}\left(\frac{R}{L}\right)^{\alpha}\quad\hat{v}=\bar{v}\left(\frac{R}{L}\right)^{-\alpha}\quad\hat{p}=\bar{p}\left(\frac{R}{L}\right)^{\beta}$$ This choice of rescaled quantities ensures that the continuity equation remains of the form: $$\partial_{\hat{x}}\bar{u}+\partial_{\bar{y}}\hat{v}=0$$ The Navier-Stokes equations in terms of the rescaled quantities yields: $$\bar{u}\partial_{\hat{x}}\bar{u}+\hat{v}\partial_{\bar{y}}\bar{u}=-\partial_{\hat{x}}\hat{p}+\frac{1}{\mathrm{Re}}\partial_{\bar{y}}^{2}\bar{u}$$ which is properly scaled with coefficients of $O(1)$ or smaller when we take the values $\alpha=-1,\,\beta=0$. This indicates the pressure scale didn't need any rescaling but the length and velocities scales have been redefined: $$\hat{x}=\bar{x}\frac{L}{R}=\frac{x}{R}\quad\hat{v}=\bar{v}\frac{R}{L}=\bar{v}\frac{V}{U}=\frac{v}{U}\quad\hat{p}=\bar{p}=\frac{p}{\rho U^{2}}$$ and we see that the characteristic length and velocity scale for respectively $x$ and $v$ isn't $L$ and $V$ as assumed at the beginning but $R$ and $U$.

  • The pipe radius is much larger than the pipe length (i.e. $R/L\gg1$):

    The transformed equation then reads: $$\bar{u}\partial_{\bar{x}}\bar{u}+\bar{v}\partial_{\bar{y}}\bar{u}=-\partial_{\bar{x}}\bar{p}+\frac{1}{\mathrm{Re}}\frac{R}{L}\partial_{\bar{x}}^{2}\bar{u}$$ Likewise to the previous case, $\frac{1}{\mathrm{Re}}\frac{R}{L}$ could be very large and requires a rescaling. Except this time we require a rescaling of the $\bar{y}$ coordinate, $\bar{u}$ velocity and $\bar{p}$ pressure: $$\hat{y}=\bar{y}\left(\frac{R}{L}\right)^{\alpha}=\frac{y}{L}\quad\hat{u}=\bar{u}\left(\frac{R}{L}\right)^{-\alpha}\quad\hat{p}=\bar{p}\left(\frac{R}{L}\right)^{\beta}$$ This choice of rescaled quantities again ensures that the continuity equation remains of the form: $$\partial_{\bar{x}}\hat{u}+\partial_{\hat{y}}\bar{v}=0$$ The Navier-Stokes equations in terms of the rescaled quantities yields: $$\hat{u}\partial_{\bar{x}}\hat{u}+\bar{v}\partial_{\hat{y}}\hat{u}=-\partial_{\bar{x}}\hat{p}+\frac{1}{\mathrm{\hat{\mathrm{Re}}}}\partial_{\bar{x}}^{2}\hat{u}$$ which is properly scaled with coefficients of $O(1)$ or smaller when we take the values $\alpha=1\,\beta=-2$. This indicates the length, velocities and pressure scales have been redefined: $$\hat{y}=\bar{y}\frac{R}{L}=\frac{y}{L}\quad\hat{u}=\bar{u}\frac{L}{R}=\bar{u}\frac{U}{V}=\frac{u}{V}\quad\hat{p}=\bar{p}\left(\frac{L}{R}\right)^{2}=\bar{p}\left(\frac{U}{V}\right)^{2}=\frac{p}{\rho V^{2}}$$ and we see that the characteristic length, velocity and pressure scales for respectively $x$, $v$ and $p$ isn't $R$, $U$, $\rho U^{2}$ as assumed at the beginning but $L$, $V$ and $\rho V^{2}$.

In case you had forgotten the point of this all: for $R/L\ll1$, $R$ is the characteristic length scale; for $R/L\gg1$, $L$ is the characteristic length scale. This implies that the smaller length scale is (usually) the characteristic length scale.

Dynamic length scales:

Consider diffusion of a species into semi-infinite domain. As it is infinite in one direction, it does not have a fixed length scale. Instead a length scale is established by the 'boundary layer' slowly penetrating into the domain. This 'penetration length' as the characteristic length scale is sometimes called is given as: $$\delta\left(t\right) = \sqrt{\pi D t}$$

where $D$ is the diffusion coefficient and $t$ is the time. As seen, there is no length scale $L$ involved as it is determined completely by the diffusion dynamics of the system. For an example of such a system see my answer to this question.

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  • $\begingroup$ What exactly do you mean by available when you say " smallest available length scale"? What exactly determines what is available and what isn't? $\endgroup$ – Paul Oct 15 '15 at 18:05
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    $\begingroup$ @Paul 'available' was meant in relation to obvious geometric length scales like length, height, width, diameter, etc. This in contrast to dynamic length scales which are much less obvious and are determined by the dynamics of the system. $\endgroup$ – nluigi Oct 15 '15 at 19:59
  • $\begingroup$ Is there any particular justification for generally using the "smallest available length" as opposed to any other length available? $\endgroup$ – Paul Feb 26 '17 at 2:59
  • $\begingroup$ @Paul The gradients are generally the largest there so most of the transport occurs at the small length scales $\endgroup$ – nluigi Feb 26 '17 at 11:28
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This is a practical, empirical question, not a theoretical one that can be "solved" by mathematics. One way to answer it is to start from what Reynolds number means physically: it represents the ratio between "typical" inertia forces and viscous forces in the flow field.

So, you look at a typical flow pattern, and choose the best length measurement to represent that ratio of forces.

For example, in flow through a circular pipe, the viscous (shear) forces depend on the velocity profile from the axis of the pipe to the walls. If the velocity along the axis of the pipe remains the same, doubling the radius will (roughly) halve the rate of shear between the axis and the walls (where the velocity is zero). So the radius, or the diameter, are a good choices for the characteristic length.

Obviously Re will be different (by a factor of 2) if you choose the radius or the diameter, so in practice everybody makes the same choice and everybody uses the same critical value of Re for the transition from laminar to turbulent flow. From a practical engineering point of view, the size of a pipe is specified by its diameter since that is what is easy to measure, so you might as well use the diameter for Re also.

For a pipe that is approximately circular, you might decide (by a similar sort of physical argument) that the circumference of the pipe is really the most important length, and therefore compare the results with circular pipes by using an "equivalent diameter" defined as (circumference / pi).

On the other hand, the length of the pipe doesn't have much influence on the fluid flow pattern, so for most purposes that would be a poor choice of characteristic length for Re. But if you are considering flow in a very short "pipe" where the length is much less than the diameter, the length might be the best number to use as the parameter describing the flow.

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  • $\begingroup$ I disagree with your statement that math can not help here. The procedure you describe would be of no use in many cases with no obvious length scales, such as a boundary layer. That is the question at hand. Dimensional analysis of the governing equations has proved quite helpful in finding relevant length scales in laminar and turbulent boundary layers, e.g., the laminar boundary layer thickness scaling and viscous length scales, respectively. The far-field scaling of thermal plumes is another case where it's much less obvious how to do the analysis you suggest, but dimensional analysis helps. $\endgroup$ – Ben Trettel Oct 11 '15 at 19:52
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    $\begingroup$ @BenTrettel - I agree that a dimensional analysis can greatly help in determining the characteristic length scale. See my answer for a 'simple' example. $\endgroup$ – nluigi Oct 15 '15 at 9:51
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There are three main ways to determine which groups of terms (more general than just length or time scales) are relevant. The first is by math, which could involve solving a problem or an analogous or appropriate problem analytically and seeing which terms appear and making selections which simplify things as appropriate (more on this below). The second approach is by trial and error, more or less. The third is by precedent, usually when someone else in the past has already done some sort of the previously mentioned analysis in this problem or related ones.

There are a number of ways to do theoretical analysis, but one useful one in engineering is non-dimensionalizing governing equations. Sometimes, the characteristic length is obvious, as is the case in a pipe flow. But other times, there are no obvious characteristic lengths, as is the case in free shear flows, or a boundary layer. In these cases, you can make the characteristic length a free variable, and choose one which simplifies the problem. Here are some good notes on non-dimensionalization, which have the following suggestions for finding characteristic time and length scales:

  1. (always) Make as many nondimensional constants equal to one as possible.
  2. (usually) Make the constants that appear in the initial or boundary conditions equal to one.
  3. (usually) If there is a nondimensional constant that, if we were to set it equal to zero, would simplify the problem significantly, allow it to remain free and then see when we can make it small.

The other main approach is to solve a problem entirely and see which groups of terms appear. Generally the relevant length is obvious if you are grabbing the term from this type of theoretical analysis, though this sort of analysis is often easier said than done.

But how do you figure out a good length if you don't have a theoretical analysis to go off of? Often, it doesn't matter too much which length you pick. Some people seem to think this is confusing, because they were taught that turbulence transition occurs at $Re$ of 2300 (for a pipe), or 500,000 (for a flat plate). Recognize that in the pipe case, it doesn't matter if you pick the diameter or radius. That just scales the critical Reynolds number by a factor of two. What does matter is making sure that any criteria you use are consistent with the definition of the Reynolds number you use, and the problem you are studying. It's tradition that dictates that we use the diameter for pipe flows.

Also, to be general, analysis or experimentation could suggest another number, say the Biot number, which also has a "characteristic length" in it. The procedures in this case are identical to that already mentioned.

Sometimes you can make a heuristic analysis to determine the relevant length. In the Biot number example, this characteristic length is usually given as the volume of an object divided by its surface area, because this makes sense for heat transfer problems. (Larger volume = slower heat transfer to center and larger surface area = faster heat transfer to center.) But I suppose it's possible to derive this from certain approximations. You can make a similar argument justifying the hydraulic diameter.

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  • $\begingroup$ If I choose L arbitrarily and the problem is non-canonical such that the flow regimes and analytical solutions are not known a priori, then trial and error is really the only way? $\endgroup$ – Paul Oct 10 '15 at 2:01
  • $\begingroup$ I don't think so. You might be able to get something useful by non-dimensionalizing the relevant governing equations with arbitrary length and time scales. This is generally my first step when analyzing a problem with clear governing equations but no clear length or time scales. If you are confused about how to do this in your particular case, post it as a question on here and I'll give it a shot. $\endgroup$ – Ben Trettel Oct 10 '15 at 2:18

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