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!
The most general format of N-S equations is free of coordinate system (see Incompressible Navier-Stokes equations (convective form) in the link en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations ). You can then specify certain coordinate system and get the coordinate form of the equations. In order to do that, just use the coordinate form of the del operator in cylindrical coordinate (see en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates )
Yeah my apologies. I realized after posting this video that after around 6:30, the audio gets weird and starts cutting off a few words. I'll be more careful for future videos, since this hasn't really happened before as far as I can recall. I don't think it's an issue with editing though; it's more something that happened with my recording software/in processing the audio that some words got cut off.
Faculty of Khan Btw now that I got your attention, you should do a video about hearing the shape of a drum/Dirichlet problem for the laplacian. I never quite understood that and you do such a great job of explaining things.
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...
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.
Ok, so I checked out the book just because you asked. Neither of us are wrong; the Hagen-Poiseuille equation still has the pressure change inversely proportional to r^4 as I mentioned. It's just that in your book, the author's written: delta p = 8*mu*u^bar*L/(R^2). My equation doesn't contain u^bar, the average speed through the cross-section. Instead, it contains Q = pi*R^2*u^bar. You can plug in u^bar into the equation in your book in terms of Q and see that the result is equivalent to what I came up with here. Also, many resources online write the pressure change in terms of Q, making it inversely proportional to R^4, so you can confirm my results there as well.
Note that Khan's formula has "Q" in the numerator. Q (flowrate) = Area * Velocity = π * r^2 * V Rewriting his equation replacing Q with (π * r^2 * V) gives: ΔP = (8uL * [π r^2 V] ) / (π r^4) Cancelling the πs and rs gives: ΔP = (8uL * V) / (r^2) Or, if you prefer to work with diameter: ΔP = (32uL * V) / (d^2) So, ΔP is actually proportional to 1/r^2 in Khan's version too.
Very well done, and nice to know about the medical application
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
very much understood...thnks from INDIA
Looking foreword to more Fluid Mechanics and the Navier-Stokes equations!!!^^
Good one! I love the way you explain it
Geez you are a champion of college maths.
I appreciate the kind words!
this video is fantastic, thank you so much
just what i needed ty
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!
Thank you! Sure, I'll add your requests to my to-do list.
Why the pressure drop occurs itself? Is it derived from head loss or it's just an assumption based on empirical evidence?
at 4:15, you get dP/d(theta). What happened to the -1/r
where can i get the proof of those navier stokes in cylindrical coordinates.. You just stated those 3 equations.thanks in advance
Awesome vedio
Please I need to know How to get N-S eqns in cylindrical coordinates from cartisians. thanks
The most general format of N-S equations is free of coordinate system (see Incompressible Navier-Stokes equations (convective form) in the link en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations ). You can then specify certain coordinate system and get the coordinate form of the equations. In order to do that, just use the coordinate form of the del operator in cylindrical coordinate (see en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates )
hey!! u save me! thanks!!!
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.
It is actually a good approach tbh, It will help you in future and you will see!
I love your channel, but that audio editing can be super annoying. Maybe try doing more takes instead of splicing takes together?
Yeah my apologies. I realized after posting this video that after around 6:30, the audio gets weird and starts cutting off a few words. I'll be more careful for future videos, since this hasn't really happened before as far as I can recall. I don't think it's an issue with editing though; it's more something that happened with my recording software/in processing the audio that some words got cut off.
Faculty of Khan Btw now that I got your attention, you should do a video about hearing the shape of a drum/Dirichlet problem for the laplacian. I never quite understood that and you do such a great job of explaining things.
Thanks, and I'll add it to my to-do list!
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...
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.
Ok, so I checked out the book just because you asked. Neither of us are wrong; the Hagen-Poiseuille equation still has the pressure change inversely proportional to r^4 as I mentioned. It's just that in your book, the author's written: delta p = 8*mu*u^bar*L/(R^2). My equation doesn't contain u^bar, the average speed through the cross-section. Instead, it contains Q = pi*R^2*u^bar. You can plug in u^bar into the equation in your book in terms of Q and see that the result is equivalent to what I came up with here. Also, many resources online write the pressure change in terms of Q, making it inversely proportional to R^4, so you can confirm my results there as well.
Note that Khan's formula has "Q" in the numerator.
Q (flowrate) = Area * Velocity = π * r^2 * V
Rewriting his equation replacing Q with (π * r^2 * V) gives:
ΔP = (8uL * [π r^2 V] ) / (π r^4)
Cancelling the πs and rs gives:
ΔP = (8uL * V) / (r^2)
Or, if you prefer to work with diameter:
ΔP = (32uL * V) / (d^2)
So, ΔP is actually proportional to 1/r^2 in Khan's version too.