Navier Stokes Approximation For Low Reynold's number

Question

The generic form of the Navier Stokes equation is (assuming incompressibility and Newtonian fluid):

$$\rho\cfrac{Dv}{Dt} = -\nabla P + \mu\nabla^2v + \rho g$$

This equation can be rearranged as follows with respect to the Reynold's number:

$$Re\cfrac{Dv}{Dt} = -\cfrac{vL}{\mu}\nabla P + vL\nabla^2v + Re* g$$

Notably, for small Re, the first and last terms disappear (according to my intuition).

However, when solving a problem, my professor only removes the first term based on this assumption. Note that the problem involves fluid flow down an inclined plane, and that the gravitational term is actually important, since this flow is gravity driven.

Why can we not negate the first and last terms, given that we are told that the flow is low Re?

Answer 1

When looking at what terms are important or not in an equation, you want to non-dimensionalize things and then see which terms are O(1) in magnitude. To make this easier, it's usually better to divide by Reynolds number, so you get:

$$ \frac{D\mathbf u}{Dt}=-\nabla p + \frac{1}{\mathrm{Re}}\nabla^2\mathbf u+\mathbf g $$

From here, it should be immediately obvious what happens as the Reynolds number changes between small values and large ones. At very small Re, the viscous term dominates every other term in the expression and so gravity and pressure gradients won't matter. Think of something like tree sap in cold weather -- it's still liquid, but it really doesn't want to move, even in gravity.

On the other hand, if Re is of O(1), then all the terms are important and each has an effect.

If I had a third hand, then on that one when Re is really big, the viscous term is negligibly small compared to the others and you won't see viscous effects in the solution.

Other than that, what KyleKanos said about the steady state solution is why the temporal derivative was neglected in this case.

Answer 2

The Navier-Stokes formula, when using Reynolds number $\mathrm{Re}=\rho uL\,/\,\mu$, is, $$ \mathrm{Re}\,\frac{D\mathbf u}{Dt}=-\mathrm{Re}\,\nabla p+\nabla^2\mathbf u+\mathrm{Re}\,\mathbf g $$ Your professor likely specified the Reynolds number as low to indicate that the flow you are studying is laminar (i.e., the viscous forces are dominant such that the flow is smooth).

For the case of the inclined plane, your professor then was looking for the steady-state case, so $D_t\mathbf u=0$, hence eliminating that first term and none of the latter terms (as explained in nluigi's answer to your other question).

Answer 3

The way to proceed in this problem is to first reduce the equation to dimensionless form. Start out by defining characteristic parameters for the flow as follows:

$v_0$ = characteristic velocity for the flow

$x_0$ = characteristic length for the flow

$p_0$ = characteristic pressure for the flow

$t_0$ = characteristic time

Then define dimensionless parameters for the flow as follows:

$\vec{v^*}=\frac{\vec{v}}{v_0}$ = dimensionless velocity

$x^*=\frac{x}{x_0}$ = dimensionless length

$p^*=\frac{p}{p_0}$ = dimensionless pressure

$t^*=\frac{t}{t_0}$

Substituting these into the differential equation yields: $$\left[\frac{\rho v_0}{t_0}\right]\frac{D\vec{v^*}}{Dt^*}=-\left[\frac{p_0}{x_0}\right]\nabla^* p^*+\left[\frac{\mu v_0}{x_0^2}\right](\nabla^*)^2\vec{v^*}+\rho g \vec{i_g}$$where $\nabla^*=x_0\nabla$ and $\vec{i_g}$is a unit vector in the direction of the gravity body force.

If we next choose $t_0=x_0/v_0$ and $p_0=\mu v_0/x_0$, and multiply the equation by $\frac{x_0^2}{\mu v_0}$, we obtain out dimensionless NS equation: $$Re\frac{D\vec{v^*}}{Dt^*}=-\nabla^* p^*+(\nabla^*)^2\vec{v^*}+\left[\frac{\rho g x_0^2}{\mu v_0}\right] \vec{i_g}$$ where $Re=\frac{\rho v_0 x_0}{\mu}$. At low Reynolds numbers, this reduces to $$0=-\nabla^* p^*+(\nabla^*)^2\vec{v^*}+\left[\frac{\rho g x_0^2}{\mu v_0}\right] \vec{i_g}$$The term in brackets represents the ratio of gravitational to viscous effects, and is not necessarily small at low Reynolds numbers.

Answer 4

As you must know $\frac{D\textbf{u}}{Dt}$ is the acceleration of a fluid element. At small $Re$ you may want to throw away all the terms accompanied by $Re$ so as to make your life simple, but you see reality gets in the way. In a sluggish flow, it is reasonable to say that accelerations are negligible and so throw that term away. But flow down the incline is caused solely by gravity, so stop, don't throw away the gravity term! The equation will cease to represent the actual flow at hand, if you were to ignore the very cause of the flow (remember we are doing dynamics and not just kinematics). This kind of physical reasoning is what makes you a physicist/engineer rather than a mathematician. The only justification you may provide for neglecting inertial but not the gravity term is that you want the equation to represent the actual flow you are studying.