My question is about the intuition behind the Reynolds number. For example they say that if a fluid is moving

1 10 mph past a sphere of radius 10m with viscosity v

2 100 mph past a sphere of radius 1 m also with viscosity v

then they will have the same reynolds number.

The formula is Re = LU/v where L is characteristic length and U is characteristic velocity. They say that 2 flows are "similar" if they have the same reynolds number.

I would consider the following flows

1 100 mph past a sphere of radius 10 m

2 10 mph past a sphere of radius 1 m

with the same viscosity similar. So my equation would be Re = L/Uv

What am I missing here? Why should the first 2 flows above be similar but the bottom 2 are not?

  • 3
    $\begingroup$ Have you done experiments to verify that your two "similar" flows actually behave similarly? $\endgroup$ Commented Oct 13, 2015 at 3:56
  • $\begingroup$ no. but don't you think it seems reasonable? $\endgroup$
    – Joe
    Commented Oct 13, 2015 at 4:00
  • 3
    $\begingroup$ Not without experimental verification, no. The idea of the Reynolds number is that it's a quantity that has empirically been found to be useful in describing systems. It's not really relevant what you "want" it to be. $\endgroup$ Commented Oct 13, 2015 at 4:02
  • 1
    $\begingroup$ @Joe with your equation, changing the units you use to measure speed, length and viscosity would change the number you get, which is why having a dimensionless number is important. $\endgroup$
    – David
    Commented Oct 13, 2015 at 5:00
  • $\begingroup$ Probably useful reading: physics.stackexchange.com/q/131732, physics.stackexchange.com/q/143328 and most certainly this one on the Reynolds number, among others. $\endgroup$
    – Kyle Kanos
    Commented Oct 13, 2015 at 11:45

2 Answers 2



I'm not sure that I understand your question entirely and if I don't, please don't hesitate to further clarify. As you stated, the Reynolds number is as follows (here I use dynamic viscosity instead of kinematic like you did):

$Re=\frac{\rho U D}{\mu}$

which is a relationship between intertial and viscous forces (the numerator being inertial and denominator being viscous). The first example that you show,

  1. $U=10 mph$ and $D=10m$
  2. $U=100 mph$ and $D=1m$

do indeed have the same Reynolds number. This is due to the scaling of the problem remaining constant. Now I don't quite understand what you're trying to say with the second portion since the values you give would produce different values of Reynolds number if you use the above equation (note that in engineering applications, diameter is most often used for the characteristic size, not radius) with the same fluid parameters.

Maybe you're confusing the Reynolds number for some other equation in the second equation you type? I ask because the Reynolds number belongs to a class of parameters called dimensionless values and your second equation is not dimensionless. Here is a link to a Wikipedia page which has a good list of dimensionless numbers used in fluid dynamics (link). Note that Reynolds number is arguably the most important of the list since it dictates the flow regime which greatly changes the physics of the problem at hand.

I hope this clarified some things,


  • $\begingroup$ Thank you for your answer Patamoose. I think I can further clarify my question by stating that the reason I think the second pair of flows I gave should be similar is that it takes the same amount of time for the fluid to run past the sphere for each flow. For the first pair of flows it takes 100 times as long for one than it takes for the other. $\endgroup$
    – Joe
    Commented Oct 13, 2015 at 4:10
  • $\begingroup$ I can't see physically how those flows should be "similar" when it takes 100 times as long for the first one to pass the sphere than the second one. $\endgroup$
    – Joe
    Commented Oct 13, 2015 at 4:12
  • $\begingroup$ Reynolds number shouldn't really be compared with the time of a flow (at least not quantitatively, it does give a good approximation for "seeing" a flow). The first pair of flows might take 100 times as long for one versus the other since scale of one system is 10 times larger than the other. Take for example water and your first flow pair, this gives Re=~450 which for an external flow case is definitely laminar. In 1) the water creeps around a very large sphere while in 2) the water quickly goes past a very small sphere. In a physical sense, the water in 1) has to "crawl" more than in 2). $\endgroup$
    – Patamoose
    Commented Oct 13, 2015 at 4:27
  • $\begingroup$ Thank you for your help Patamoose. I think I understand the meaning now. As far as I understand it a fluid has a viscosity level which determines "how much Reynolds number" it can take before it gets turbulent. So for example, say water, if we push it through fast enough it will get turbulent or if we make the sphere big enough. My misunderstanding was in what the word similar meant. It has to do with how laminar/turbulent the flow will be, is that correct? $\endgroup$
    – Joe
    Commented Oct 13, 2015 at 4:50
  • $\begingroup$ Very roughly speaking, it may take 100x longer for the first one to pass the sphere, but the sphere is a much larger obstacle to negotiate, so there is still a large effect on the flow from the sphere. $\endgroup$
    – David
    Commented Oct 13, 2015 at 4:59

Maybe a theoretical treatment of the Reynolds number would be fruitful here. For some introduction see my answer to another question.

In a fluid flow there are typically two relevant characteristic forces present, i.e. inertial $\left(\rho UU\right)$ and viscous $\left(\mu U/L\right)$ forces. The ratio of these quantities $\left(\rho UU / \mu U/L= U L / \nu\right)$ is what is known as the Reynolds number and signifies the relative importance of inertial to viscous forces. When viscous forces are more important then $\mathrm{Re}\ll1$, known as the laminar regime. Vice versa, when inertial forces are more important the $\mathrm{Re}\gg1$, known as the turbulent regime.

A mathematical treatment of this is shown by transforming the Navier-Stokes equations which describe fluid flow:

$$\mathbf{u}\cdot\mathbf{\nabla}\mathbf{u}=-\frac{1}{\rho}\mathbf{\nabla}p + \nu\mathbf{\nabla}^2\mathbf{u}$$

to a non-dimensional form:

$$\bar{\mathbf{u}}\cdot\bar{\mathbf{\nabla}}\bar{\mathbf{u}}=-\bar{\mathbf{\nabla}}\bar{p} + \frac{1}{\text{Re}}\bar{\mathbf{\nabla}}^2\bar{\mathbf{u}}$$

Here we see that the Reynolds number occurs as a coefficient in front of the viscous term. When $\mathrm{Re}\gg1$, the viscous term becomes negligible compared to the pressure term and inertial term, which means inertial forces are more important than viscous forces. The same treatment can be done for $\mathrm{Re}\ll1$, but requires rescaling of the pressure term which is not really the point now.

The Reynolds number naturally occured as a result of non-dimensionalizing the Navier-Stokes equations. The equation went from being described by four parameters $\rho$, $\nu$, $U$ and $L$ to one degree of freedom $\mathrm{Re}$. Now two flows are called 'similar' when their Reynolds numbers are the same because it means that their flow is described by the exact same dimensionless differential equation which yields a solution only dependent on one degree of freedom, $\mathrm{Re}$.

The first two flows you indicate have the same Reynolds number (i.e $vL=10\cdot10=100\cdot1$ for the same viscosity) and are therefore described by exactly the same dimensionless equation and thus 'similar'. The bottom flows are not 'similar' as they have different Reynolds numbers (i.e. $vL=100\cdot10\neq10\cdot1$ for the same viscosity). Simply redefining Reynolds number is not physical as it then no longer is dimensionless.

Note that, in general, flows are 'similar' when all their dimensionless numbers are the same. If for example there is a body force included, this would introduce an additional dimensionless number known as the Froude number. Flows described by such an equation are only 'similar' when both the Reynolds and Froude numbers are the same.


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