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?

  • $\begingroup$ Are you sure it's not because you're looking for steady state? $\endgroup$ – Kyle Kanos Nov 20 '16 at 23:37
  • $\begingroup$ You're correct that we are looking for the steady state. My professor doesn't indicate his reasoning towards his results - just the end result. In that case, I see what you're saying. $\endgroup$ – Austin Nov 20 '16 at 23:46
  • $\begingroup$ What is the value of telling us the flow is low reynolds number? When would we ever use Euler's approximation (high Re) and Stokes theorem (low Re)? $\endgroup$ – Austin Nov 20 '16 at 23:46
  • $\begingroup$ At low Reynolds numbers, the flow is laminar/smooth; I suspect your professor is being specific about that point (that there is no turbulent flow to worry about)--but you may want to ask your professor that question. $\endgroup$ – Kyle Kanos Nov 21 '16 at 4:02
  • $\begingroup$ All you did was multiply each term in the equation by a constant, namely, the dimensionless Reynolds number. This does not reduce the equation to dimensionless form, so you have no basis for comparing the magnitudes of the various terms. You need to compare the magnitudes of the terms in a fully dimensionless version of the equation. $\endgroup$ – Chet Miller Nov 21 '16 at 11:43

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.

  • $\begingroup$ +1 i like this approach to scaling. Note that depending on the situation the pressure gradient may not be negligible for $\mathrm{Re}\ll1$ and then requires a rescaling of the pressure. $\endgroup$ – nluigi Nov 24 '16 at 20:54

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).

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    $\begingroup$ Minor correction: Steady state only implies $\frac{\partial \textbf{u}}{\partial t}=0$. Total derivative $\frac{D\textbf{u}}{Dt}$ can be non-zero in a steady flow, due to presence of convective acceleration. $\endgroup$ – Deep Nov 22 '16 at 6:21

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.


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


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.

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    $\begingroup$ @KyleKanos No. He was wondering why the gravitational term does not disappear at low Re if the inertial term disappears at low Re. Austin, is that in fact what your were wondering? $\endgroup$ – Chet Miller Nov 21 '16 at 16:08
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    $\begingroup$ @KyleKanos Excellent. Then, in that case, the development I presented precisely addresses this issue. Note that it is critical in scaling situations like this to reduce the equations to dimensionless form. $\endgroup$ – Chet Miller Nov 21 '16 at 16:26
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    $\begingroup$ I guess we have different perspectives on this. $\endgroup$ – Chet Miller Nov 21 '16 at 17:14
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    $\begingroup$ Exactly. So, in the limit of low Re ( as specified in the problem statement), the first term vanishes, while the gravitiational term, according to the dimensional analysis I provided, does not have to vanish. $\endgroup$ – Chet Miller Nov 21 '16 at 17:19
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    $\begingroup$ That applies to the sample problem they were working on. But the question posed by the OP seemed to me to be more general than that. And, in the limit of Reynolds numbers approaching zero, even if the flow is not steady state, the leading term will still not contribute significantly. $\endgroup$ – Chet Miller Nov 21 '16 at 17:51

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