There's no magic behind it. It was done by non-dimensionalizing the momentum equation in the Navier-Stokes equations.
Starting with:
$$\frac{\partial u_i}{\partial t} + u_j\frac{\partial u_i}{\partial x_j} = -\frac{1}{\rho}\frac{\partial P}{\partial x_i} + \nu \frac{\partial^2 u_i}{\partial x_j x_j}$$
which is the momentum equation for an incompressible flow. Now you non-dimensionalize things by choosing some appropriate scaling values. Let's look at just the X-direction equation and assume it's 1D for simplicity. Introduce $\overline{x} = x/L$, $\overline{u} = u/U_\infty$, $\tau = tU_\infty/L$, $\overline{P} = P/(\rho U_\infty^2)$ and then substitute those into the equation. You get:
$$ \frac{\partial U_\infty \overline{u}}{\partial \tau L/U_\infty} + U_\infty\overline{u}\frac{\partial U_\infty \overline{u}}{\partial L \overline{x}} = - \frac{1}{\rho}\frac{\partial \overline{P}\rho U_\infty^2}{\partial L\overline{x}} + \nu \frac{\partial^2 U_\infty \overline{u}}{\partial L^2 \overline{x}^2} $$
So now, you collect terms and divide both sides by $U_\infty^2/L$ and you get:
$$ \frac{\partial \overline{u}}{\partial \tau} + \overline{u}\frac{\partial \overline{u}}{\partial \overline{x}} = -\frac{\partial \overline{P}}{\partial \overline{x}} + \frac{\nu}{U_\infty L}\frac{\partial^2 \overline{u}}{\partial \overline{x}^2}$$
Where now you should see that the parameter on the viscous term is $\frac{1}{Re}$. Therefore, it falls out naturally from the definitions of the non-dimensional parameters.
The intuition
There's some other ways to come up with it. The Buckingham Pi theorem is a popular way (demonstrated in Floris' answer) where you collect all of the units in your problem in this case $L, T, M$ and find a way to combine them into a number without dimension. There is one way to do that, which ends up being the Reynolds number.
The interpretation of inertial to viscous forces comes from looking at the non-dimensional equation. If you inspect the magnitude of the terms, namely the convective (or inertial term) and the viscous term, the role of the number should be obvious. As $Re \rightarrow 0$, the magnitude of the viscous term $\rightarrow \infty$, meaning the viscous term dominates. As $Re \rightarrow \infty$, the viscous term $\rightarrow 0$ and so the inertial terms dominates. Therefore, one can say that the Reynolds number is a measure of the ratio of inertial forces to viscous forces in a flow.
Viscous stress is proportional to $\frac{du}{dy}$
This is actually only partly true; not all fluids are Newtonian fluids and non-Newtonian fluids behave differently from Newtonian fluids.
There are many models for viscosity as a function of shear:
- Newtonian: $\tau=\mu\frac{du}{dy}$
- Bingham: $\tau = \tau_0+\mu_{\infty}\frac{du}{dy}$ for $\tau>\tau_0$ else $\frac{du}{dy}=0$
- Ostwald–de Waele: $\tau = K(\frac{du}{dy})^n$
- others like pseudoplastic, carreau-yasuda, etc
Basically in general (very simplified):
$$\tau = \tau_0 + K(\frac{du}{dy})^n$$ for $\tau>\tau_0$ else $\frac{du}{dy}=0$
Here $\tau_0$ is known as the yield stress and it is basically the equivalent of static friction. It is the stress that must be applied before the fluid even starts to move and only then will viscosity (or dynamic friction) play a roll.
Best Answer
Inertial force, as the name implies is the force due to the momentum of the fluid. This is usually expressed in the momentum equation by the term $(\rho v)v$. So, the denser a fluid is, and the higher its velocity, the more momentum (inertia) it has. As in classical mechanics, a force that can counteract or counterbalance this inertial force is the force of friction (shear stress). In the case of fluid flow, this is represented by Newtons law, $\tau_x = \mu \frac{dv}{dy}$. This is only dependent on the viscosity and gradient of velocity. Then, $Re = \frac{\rho v L}{\mu}$, is a measure of which force dominates for a particular flow condition.
The inertial forces are what gives rise to the dynamic pressure. Another way to look at the Reynolds Number is by the ratio of dynamic pressure $\rho u^2$ and shearing stress $μ v/ L$ and can be expressed as $$Re =\frac{\rho u^2} {μ v/ L} = \frac{ u L} {\nu} $$
At very high Reynolds numbers, the motion of the fluid causes eddies to form and give rise to the phenomena of turbulence.