Deriving Poiseuille's Law from the Navier-Stokes Equations

In this video, I use the Navier-Stokes Equations to derive Poiseuille's Law (aka. The Hagen-Poiseuille Equation). This is a rather simple derivation carried out by simplifying Navier-Stokes in cylindrical coordinates, making some substitutions, and determining the solution of the resulting ODE. The end result is a parabolic velocity profile for laminar flow of an incompressible, Newtonian fluid in a cylindrical pipe.

This velocity profile is then used to deduce the relationship between the pressure difference and the radius of the cylindrical pipe (it turns out to be a 1/R^4 dependence). Once Poiseuille's Law is derived, I use it to discuss the significance of the pressure-flow rate-radius relationship in the clinical context.

Questions/requests? Let me know in the comments!

Prerequisites: Basic knowledge of Fluid Mechanics and the Navier-Stokes equations. I plan on adding to this playlist in the future so hopefully I can replace this line by actual video links some day.

Special thanks to my Patrons:
- Jennifer Helfman
- Justin Hill
- Jacob Soares
- Yenyo Pal
- Lisa Bouchard

[RajPatel-mjch]

Love the way you explain, it gives overall picture and why we are learning what we are learning! Please keep it up and bring more videos like this

[farooqfox]

Very well done, and nice to know about the medical application

[xiaodongchen]

Looking foreword to more Fluid Mechanics and the Navier-Stokes equations!!!^^

[akshaya]

very much understood...thnks from INDIA

[hfy]

Good one! I love the way you explain it

[monibniazi]

at 4:15, you get dP/d(theta). What happened to the -1/r

[jessicah.]

this video is fantastic, thank you so much

[niksapraljak]

Your videos are great. I would like to ask if you can introduce "Pulsatile Flow in a Rigid Tube" and "Pulsatile Flow in an Elastic Tube." This would be greatly appreciative!

[ostensiblyquerulous]

Geez you are a champion of college maths.

[АлександрКабанец-хз]

Why the pressure drop occurs itself? Is it derived from head loss or it's just an assumption based on empirical evidence?

[shreyasms]

where can i get the proof of those navier stokes in cylindrical coordinates.. You just stated those 3 equations.thanks in advance

[belfodilfarid]

Please I need to know How to get N-S eqns in cylindrical coordinates from cartisians. thanks

[thebluebeyond]

Our physics lecturer doesn't want us reciting equations, therefore in every test we need to draw the diagram and derive the equation before using it.

[ryanchatterjee]

I love your channel, but that audio editing can be super annoying. Maybe try doing more takes instead of splicing takes together?

[prashantdahiya]

Sorry to say but Hagen Poiseuille Formula tells the head loss, which is change in pressure upon density and acceleration due to gravity, and even after considering that you are just talking about the pressure change, then also the final output is incorrect as it's inversely proportional to radius square, not raise to the power 4, I have R.K Bansal, 9th Edition and it's proof is in page 390, go check it out for Yourself, for those who have different edition then it's in the chapter viscous flow.

[galinalyutskanova-zhekova]

There are some issues in the video. The first part with the derivation of the equations is fine. However, there a couple of problems solving the system. First of all, the reason why 1/eta dP/dz = 1/r d/dr(rdvz/dz) is equal to a constant is not clear. The reason is the following. dP/dr=0 and thus P is a function of z only. Analogiously, dv_z/dz=0 and thus v_z is a function of r only. The equation dP/dz = 1/r d/dr(rdvz/dz) means that a function of r is equal to a function of z. The only way this is possible if both sides of this equation are equal to certain constant. The size of this constant depends on the inlet and outlet boundary conditions (the conditions for the pressure), which are not included in the list with the boundary conditions (I really have no idea why you did not include them). The condition for the final velocity is not a boundary condition, it is not applied at the boundary but at the axis of symmetry. It is just common sense. In physical world, there is not a such thing as infinite velocity but we get one due to math...