fluid dynamics - I disagree with the definition of Reynolds number. Why am I wrong?

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?

Answer

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.