Why in a pressure driven flow are we able to neglect gravity? Wouldn't gravity slightly change the magnitudes of the vectors in z (smaller for the bottom half and larger for the top half) due to gravity pulling particles down onto a surface we assumed to have 0 velocity? Or are we assuming that this problem is in a 0 gravity environment so the orientation of the pipe makes no difference. It seems like gravity would almost create a shift of the Vz(r) that was calculated
If the pipe is held horizontal ,then gravity can be neglected . And if its vertical or diagonal , gravity would contribute in it. As you see there is a "-" before pressure gradient,if the pipe is horizontal, the gravitational force will be only in y direction which would be zero ,by equating the pressure gradient with gravitational force which are opposite to each other. Just the terms should be converted x,y,z from r,Q(theta),z to get this
Thanks for explaining. It is very interesting and challenging for me. I have question. In applying the conservation laws how we got 1/r before the partial derivatives? Sorry if it is too easy
This is actually quite a difficult question, and it relates to the derivation from first principles around circular or cylindrical coordinates. A fundamental level physics/engineering book (maybe Maxwells) will contain the solution from the initial volume equations in the cylinder. An easier way to think of it is in terms of units; as the theta term is dimensionless, the r variable adds the required dimensional element to the term. Check the units on every term to see how it balances out.
Hello Victor- Wouldn't it be more appropriate to say that d(vz)/dz = 0 is an outcome of continuity equation (at 3.4 mins), rather than saying that we know in advance v does not change with z? Then this outcome of the continuity equation i.e. d(vz)/dz =0, is adopted later in momentum equation at 8.43 mins and 9.33 mins for d2(vz)/dz2.
How is it possible for p to vary with z, if v_z doesn't. I mean, you eliminated ∂v_z/∂z, but kept ∂p/∂z. It doesn't make sense to me (I'm starting to believe that this pressure is not the same pressure that we use in other equations, such as Bernoulli's equation) D;
stay at home mom stumbled upon this & fascinated. stuff just jumps out at me & it seems obvious. I never took calc & please, please don't think i'm here to talk trash. i have a formula but don't know how to write it out. lol Just have to say , thrilled i found this & curious to see if my entire "formula" continues to be correct. now i know why teachers want you to show your work, lol. Also I cant help but notice & seems obvious ...a lot of Nothing-from Nothing Equals Nothing. I assume all the incorrect formulas are for the process of elimination.
Great video. You actually make this exciting. I have a question though; how would this change if this was a gravity driven flow, and not pressure? I simplified it to be rho*g= 0. However, wouldn't that make my entire equation go to zero?
>an 11 minute long video on youtube is more valuable than 4 hours of lecture at my uni
really makes you think...
I'm an engineering student from Brazil. I'm grateful for the explanations.
Thank you! You explain it like you are genuinely amazed that all this stuff works out nicely and you want to let others in on the secret. Appreciated.
Thank you a lot I really needed some slowly explained example. Cheers!
u dumb?
Yes ;)
me too :'(
Thanks! Much appreciated for explaining this. Keep up the good work
Why in a pressure driven flow are we able to neglect gravity? Wouldn't gravity slightly change the magnitudes of the vectors in z (smaller for the bottom half and larger for the top half) due to gravity pulling particles down onto a surface we assumed to have 0 velocity? Or are we assuming that this problem is in a 0 gravity environment so the orientation of the pipe makes no difference. It seems like gravity would almost create a shift of the Vz(r) that was calculated
If the pipe is held horizontal ,then gravity can be neglected . And if its vertical or diagonal , gravity would contribute in it. As you see there is a "-" before pressure gradient,if the pipe is horizontal, the gravitational force will be only in y direction which would be zero ,by equating the pressure gradient with gravitational force which are opposite to each other. Just the terms should be converted x,y,z from r,Q(theta),z to get this
amazing video, explained everything. I was confused by what terms need to be cancelled out for a while
wonderfully explained, thank you.
You are my savior.
Thank you. This video was really helpful
Thanks for explaining. It is very interesting and challenging for me.
I have question. In applying the conservation laws how we got 1/r before the partial derivatives? Sorry if it is too easy
This is actually quite a difficult question, and it relates to the derivation from first principles around circular or cylindrical coordinates. A fundamental level physics/engineering book (maybe Maxwells) will contain the solution from the initial volume equations in the cylinder. An easier way to think of it is in terms of units; as the theta term is dimensionless, the r variable adds the required dimensional element to the term. Check the units on every term to see how it balances out.
@@thefish1711 Thank you very much for the precise explanation!
In the Navier-Stokes how do you derive the cylindrical coordinate form for the laplacian * the velocity vector?
What are Vr and Vtheta? Is Vtheta angular velocity? What direction are they going in?
Hello Victor- Wouldn't it be more appropriate to say that d(vz)/dz = 0 is an outcome of continuity equation (at 3.4 mins), rather than saying that we know in advance v does not change with z?
Then this outcome of the continuity equation i.e. d(vz)/dz =0, is adopted later in momentum equation at 8.43 mins and 9.33 mins for d2(vz)/dz2.
it is nice everyone should look it &understand the real world
Very complicated problem for me but the uploader tamed it i must admit
Motion towards taking another hit
Awesome sir
You are my hero😭😭😭 gracias :D
How is it possible for p to vary with z, if v_z doesn't. I mean, you eliminated ∂v_z/∂z, but kept ∂p/∂z. It doesn't make sense to me (I'm starting to believe that this pressure is not the same pressure that we use in other equations, such as Bernoulli's equation) D;
Thanks alot ❤️
When velocity varies in r-direction then how we have v subscript r=0????
What varies in the r-direction is the z-component of the velocity (v subscript r, in turn, remains at 0)
i dont get vz(r) ? how r is another direction and z is another direction?
Imagine having a pipe of water where the velocity on the direction Z varies depending on the position in the R axis.
any reference book ?
stay at home mom stumbled upon this & fascinated. stuff just jumps out at me & it seems obvious. I never took calc & please, please don't think i'm here to talk trash. i have a formula but don't know how to write it out. lol Just have to say , thrilled i found this & curious to see if my entire "formula" continues to be correct. now i know why teachers want you to show your work, lol. Also I cant help but notice & seems obvious ...a lot of Nothing-from Nothing Equals Nothing. I assume all the incorrect formulas are for the process of elimination.
Great video. You actually make this exciting. I have a question though; how would this change if this was a gravity driven flow, and not pressure? I simplified it to be rho*g= 0. However, wouldn't that make my entire equation go to zero?
All things are destiny!
Gravity in the r direction shouldn't be neglected according to my textbook.
awesome
Thank you!
Remember in high school when I said weed never again?
At this point, one would hope that the Navier-Stokes equations would be open source
Some day later in life sam revolinski and I will smoke weed together again.
Mark Yeri sent me this
I am on last stage I solved navier Stokes equation
Donald trump was a modern day tea party member.
My Indian neighbor in kansas supported donald trumps manufacture in America first and impose high tariffs and taxes on wine imports policy.