Navier Stokes Equation

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The Navier Stokes equation is one of the most important topics that we come across in fluid mechanics. The Navier stokes equation in fluid mechanics describes the dynamic motion of incompressible fluids. Finding the solution of the Navier stokes equation was really challenging because the motion of fluids is highly unpredictable. This equation can predict the motion of every fluid like it might be the motion of water while pouring into a container, motion of smoke of match, etc.


Stokes Equation

Leonard Euler first derived an equation of motion for incompressible and frictionless fluids in the 18th century, later in the early 1800s Claude- Louis Navier came up with a correction to the Euler equation by including the viscosity term. In the 19th century, sir Gabriel George Stoke worked on these equations and finally concluded with a set of equations that will describe the motion of fluid available in the world.

The main concepts involved in deriving Navier stokes equations was Newton’s second law of motion, conservation of mass, conservation of momentum. The Navier stokes equation represents the conservation of momentum.

So, Euler gave the equation of motion for incompressible and frictionless fluids as:

⇒ \[\frac{\partial u}{\partial t}\] + u.\[\bigtriangledown\]u = - \[\frac{\bigtriangledown P}{\rho }\] 

This equation could not explain the fluid dynamics of incompressible viscous fluids, thus Navier- stokes equation was derived with major correction by introducing viscosity term in the Euler equation:

⇒ \[\frac{\partial u}{\partial t}\] + u.\[\bigtriangledown\]u = -\[\frac{\bigtriangledown P}{\rho }\] + v\[\bigtriangledown\]\[^{2}\]u 

Where u is the velocity of the liquid, is the density of equation, v is the kinematic viscosity and P is the pressure. The equation can be written by using either cartesian coordinates or cylindrical coordinates. 


Navier Equation in Cylindrical Coordinates is Given by:

Navier Stokes in cylindrical coordinates is as given below, it is considered to be one of the most tedious equations to solve.


Incompressible Continuity Equation: 

\[\frac{1}{r}\] \[\frac{\partial (ru_{r})}{\partial r}\] + \[\frac{1}{r}\] \[\frac{\partial (u_{\theta})}{\partial \theta}\] + \[\frac{\partial u_{z}}{\partial z}\] = 0    eq a)

r - component

\[\rho\] [\[\frac{\partial u_{r}}{\partial t}\] + u\[_{r}\] \[\frac{\partial u_{r}}{\partial r}\] + \[\frac{u_{\theta }}{r}\] [\[\frac{\partial u_{r}}{\partial\theta}\] - \[\frac{u_{\theta }^{2}}{r}\] + u\[_{z}\]  [\[\frac{\partial u_{r}}{\partial z}\]]

= - \[\frac{\partial P}{\partial r}\] + \[\rho\]g\[_{r}\] + μ[\[\frac{1}{r}\] [\[\frac{\partial }{\partial r}\][r[\[\frac{\partial u_{r}}{\partial r}\]] - \[\frac{u_{r}}{r^{2}}\] + [\[\frac{1}{r^{2}}\] [\[\frac{\partial ^{2} u_{r}}{\partial \theta^{2}}\] - [\[\frac{2}{r^{2}}\] [\[\frac{\partial u_{\theta}}{\partial \theta}\] + [\[\frac{\partial^{2} u_{r}}{\partial z^{2}}\]]    eq b)

\[\theta\] - component:

\[\rho\] [\[\frac{\partial u_{\theta}}{\partial t}\] + u\[_{r}\] \[\frac{\partial u_{\theta}}{\partial r}\] + \[\frac{u_{\theta }}{r}\] [\[\frac{\partial u_{\theta}}{\partial\theta}\] - \[\frac{u_{r}u_{\theta}}{r}\] + u\[_{z}\]  [\[\frac{\partial u_{\theta}}{\partial z}\]]

= - \[\frac{1}{r}\] \[\frac{\partial P}{\partial \theta}\] + \[\rho\]g\[_{\theta}\] + μ[\[\frac{1}{r}\] [\[\frac{\partial }{\partial r}\][r[\[\frac{\partial u_{\theta}}{\partial r}\]] - \[\frac{u_{\theta}}{r^{2}}\] + [\[\frac{1}{r^{2}}\] [\[\frac{\partial ^{2} u_{\theta}}{\partial \theta^{2}}\] + [\[\frac{2}{r^{2}}\] [\[\frac{\partial u_{r}}{\partial \theta}\] + [\[\frac{\partial^{2} u_{\theta}}{\partial z^{2}}\]]   eq c) 

z - component: 

 \[\rho\] [\[\frac{\partial u_{z}}{\partial t}\] + u\[_{r}\] \[\frac{\partial u_{z}}{\partial r}\] + \[\frac{u_{\theta }}{r}\] [\[\frac{\partial u_{z}}{\partial\theta}\] + u\[_{z}\] [\[\frac{\partial u_{z}}{\partial z}\]]

= - \[\frac{\partial P}{\partial z}\] + \[\rho\]g\[_{z}\] + μ[\[\frac{1}{r}\] [\[\frac{\partial }{\partial r}\][r[\[\frac{\partial u_{z}}{\partial r}\]] + [\[\frac{1}{r^{2}}\] [\[\frac{\partial ^{2} u_{z}}{\partial \theta^{2}}\] + \[\frac{\partial^{2} u_{z}}{\partial z^{2}}\]]    eq d)


Navier Stokes Problem

The incompressible Navier Stokes equations play a major role in fluid dynamics. The terms that made Navier stokes equation unique are the diffusion term and the convection term.

Where, 

The diffusion term is = v\[\bigtriangledown\]\[^{2}\]u

The convection term is = u.\[\bigtriangledown\]u

  • If u.\[\bigtriangledown\]u < v\[\bigtriangledown\]\[^{2}\]u we will get a smooth flow of the fluid.

[Image will be Uploaded Soon]

  • At the same time if u.\[\bigtriangledown\]u > v\[\bigtriangledown\]\[^{2}\]u we will get rapid turbulence flow. 

[Image will be Uploaded Soon]

Due to these conventions, the solution for the Navier stokes equation is the most difficult one and it was considered as one of the millennium questions. We can solve the equation if ignore the convection term but that is not possible.

In physics in order to eliminate unwanted terms will just assume that the term is negligibly small, we can do the same process here, we will assume that the convection term is negligibly small and it can be ignored. This can be done by considering the following cases:

  1. Assume that the flow of fluid is very slow. This was done by two japan based mathematicians Hiroshi Fujita and Tosio Kato in their groundbreaking paper- on the Navier stokes initial value problem.

  2. The fluid has turbulent only at a small scale. This was brought into the limelight by french mathematicians in 1994.

The Navier stokes equation or Navier Stokes theorem is so dynamic in fluid mechanics it explains the motion of every possible fluid existing in the universe. It is always been challenging to solve million-dollar questions and the solution for the Navier Stokes equation is one among them.


Did You Know?

  • There are seven million-dollar mathematical questions selected by the clay mathematics institute of Cambridge, Massachusetts, US. They are considered to be the hardest ones to resolve and hence they are considered for the special rewards.

  • The Reynolds averaged Navier stokes equation is derived for equation motion of fluids over averaged time. This equation was mainly discussed for the turbulence flow.

FAQ (Frequently Asked Questions)

1: Is the Navier Stokes Equation Available in Cylindrical Coordinates?

Ans: Yes. The Navier stokes equation cylindrical equation is written by considering the cylindrical coordinates, i.e., u = u(r, θ, z, t).

2: What is the Cauchy Momentum Equation?

Ans: The Cauchy momentum equation is given by:

⇒ Du/Dt = 1/ρ ▽.σ + g