Why can we neglect the inertial (but not viscous) term in Navier-Stokes at low flow and high viscosity?

Complete Navier-Stokes: $\rho \frac{D\vec{v}}{Dt}=\rho g - \nabla P+ \mu \nabla ^2 \vec{v}$

Inertial term: $\frac{D\vec{v}}{Dt}= \frac{\partial\vec{v}}{\partial t}+ \frac{\partial\vec{v}}{\partial x}v_x+ \frac{\partial\vec{v}}{\partial y}v_y+ \frac{\partial\vec{v}}{\partial z}v_z$.

And as we assume a stationary flow and low rate: $ \frac{\partial\vec{v}}{\partial t}=0, \frac{\partial\vec{v}}{\partial x}\approx0, \frac{\partial\vec{v}}{\partial y}\approx0, \frac{\partial\vec{v}}{\partial z}\approx0$. And so it follows that the inertial term can be ignored.

However in my material it is also stated that the $\mu \nabla ^2 \vec{v}$ will be the dominating term during these circumstances. Why will it not be so that $\nabla ^2 \vec{v} \rightarrow \frac{\partial^2\vec{v}}{\partial x^2}\approx0, \frac{\partial^2\vec{v}}{\partial y^2}\approx0, \frac{\partial^2\vec{v}}{\partial z^2}\approx0 \rightarrow \mu \nabla ^2 \vec{v} \approx 0$ as well?

  • $\begingroup$ Can you cite the book/document/etc. that made this claim? It would help to see it in context. $\endgroup$
    – Carlton
    Nov 25, 2015 at 17:29
  • $\begingroup$ The course book is "Fundamentals of Momentum, Heat and Mass Transfer" by Welty, Rorrer and Foster. Unfortunately this is a problem in a practice problem document in Swedish, so I am not sure if it will be of much help. $\endgroup$
    – Raoul
    Nov 25, 2015 at 17:32
  • $\begingroup$ I have the 4th edition of that book (English version). I will look and see if they explain more. $\endgroup$
    – Carlton
    Nov 25, 2015 at 17:36
  • $\begingroup$ It is quite a significant mistake/misprint to state $\partial_\beta u_\alpha \approx 0$ and i can understand your confusion concerning the second derivatives! If the practice problem is made by a TA/professor (s)he should really review the basics of fluid mechanics... $\endgroup$
    – nluigi
    Nov 25, 2015 at 22:50
  • $\begingroup$ @nluigi I don't see where you're getting the $\partial_\beta u_\alpha \approx 0$ from the OP's question, or even what that expression means. What are $\beta$ and $\alpha$, and what is the differential wrt? $\endgroup$ Nov 26, 2015 at 9:33

2 Answers 2


Usually what is implied by low flowrate and high viscosity is that we are dealing with a so-called low Reynolds number flow. The Reynolds number is dimensionless number which is the ratio of inertial forces ($\rho U U$) and viscous force ($\mu U/L$): $$\mathrm{Re}=\frac{\rho U U}{\mu U/L}=\frac{\rho U L}{\mu}$$ For low $\mathrm{Re}$ viscous forces dominate (laminar regime) and for high $\mathrm{Re}$ inertial forces dominate (turbulent regime). Dimensionless numbers like $\mathrm{Re}$ show up naturally through a process known as 'scaling' in which the equations are made non-dimensional; through this process it is then possible to say which terms are negligible based on the values of relevant dimensionless numbers. For more information check my answer to this question.

Technically, saying 'low flow and high viscosity' is not sufficient to say we are dealing with a low $\mathrm{Re}$ flow because it also depends on the length scale $L$ (usually a pipe diameter, etc) and the density $\rho$ (of air or water), but it is usually implied that is the case.

Now saying for a low flowrate that $\partial_\beta u_\alpha \approx 0$ is incorrect; what you probably mean is that $u_\beta\partial_\beta u_\alpha \ll \mu\partial_\beta^2u_\alpha$. This justifies the simplification of the equations by saying that $u_\beta\partial_\beta u_\alpha\approx 0$ which physically means that inertial terms are completely negligible compared to viscous terms. $u_\beta\partial_\beta u_\alpha\approx 0$ does not imply $\partial_\beta u_\alpha \approx 0$ rather the low flowrate implies $u_\beta\approx0$ while $\partial_\beta u_\alpha$ can be significant. Consider the order-of-magnitude estimation of $\partial_\beta u_\alpha \sim U/L$; for small values of $L$ (contributes to low $\mathrm{Re}$) it may be much larger than order $O(U)$. A similar order-of-magnitude analysis of the viscous terms $\partial_\beta^2 u_\alpha \sim U/L^2$ shows that these will be even more significant. Hence, the reason why the inertial terms are negligible but the viscous terms are not.


Since the final term is proportional to $\nabla ^2 \vec{v}$, and not $|\vec{v}|$, the term can be large even if the velocity magnitude is small. Consider the simple case of no-slip laminar flow in an x-oriented pipe. This is a unidirectional flow, so we can discard $v$ and $w$ and focus on $u$. We'll take as given that the flow speed is small.

Just because $u$ is generally small, however, doesn't mean we can conclude that $\nabla u$ is also generally small. In fact, the narrower the pipe, the larger the magnitude of $\nabla u$. Looking at the gradient field of the horizontal velocity, we see it tends to point inwards, towards the center of the pipe, where the velocity is at a maximum. This means we have an negative divergence, for which the magnitude is dependent on the sharpness of the variation in the velocity, and not its overall magnitude.

Hence $\nabla ^2 \vec{v} = \{-|\nabla ^2 u|, 0, 0\}$ is not insignificant, which of course is just Navier Stokes accurately predicting the tendency of viscous forces to cause a pressure drop in steady state, horizontal pipe flow (or you could call it the necessity of a pressure drop to force flow against viscous forces, as you please). Cancel out the gravitational and time varying terms and see for yourself.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.