Solutions to Navier-Stokes: Poiseuille and Couette Flow
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- Опубліковано 31 гру 2020
- MEC516/BME516 Fluid Mechanics, Chapter 4 Differential Relations for Fluid Flow, Part 5: Two exact solutions to the incompressible continuity and Navier-Stokes equations. One solution is for laminar steady flow between fixed parallel plates, where the flow is driven by a pressure gradient (Poiseuille Flow). The other solution is for laminar steady flow between parallel plates where the flow is driven only by the motion of the upper plate (Couette Flow).
All of the videos in this Introductory Fluid Mechanics course, sample exams (with solutions), and a copy (pdf) of this presentation can be downloaded at:
www.drdavidnaylor.net
Course Textbook: F.M. White and H. Xue, Fluid Mechanics, 9th Edition, McGraw-Hill, New York, 2021.
#fluidmechanics #fluiddynamics
![[CFD] Conservative, Advective & Material Derivative forms of the Navier-Stokes Equations](http://i.ytimg.com/vi/ljdv4T2U464/mqdefault.jpg)
All the videos (and pdf downloads) for this introductory Fluid Mechanics course are available at: www.drdavidnaylor.net/
this video alongside your excellent commentary is absolutely a gem to Mechanical Engineering education. much Love
Glad to hear you found this helpful. Bes of luck with your studies,
Oh yes, i have to say the same. Tomorrow i will write the fluid mechanics 1 exam, and this video is a really good explanation to improve my formula collection and my knowledge 🙂 Thank you!
Greetings from Turkey, this material helped me a lot. Thanks professor.
Thanks for the explanation. That was easier to understand than my textbook!
Sir This is absolutely incredible! You make an excellent relation between differential equations and the course material. I'm hoping you make a fluid dynamics course in its entirety. Thank you so much!
Thanks for the kind words.
This is very much appreciated, very clear and concise explanation.
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I wish you good health.
Mongolia! Hope to visit one day. Glad to hear the video was helpful.
thank you so much for your excellent explain
Thank you so much for these videos. They make my course much easier.
Thanks for the nice comments. Best of luck with your studies.
incredibly helpful
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Thanks for the kind words. Glad to hear the videos are helpful.
great vid broski
Thanks so much sir
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Sir thankyou for uploading this, can we have a video of generalized couette flow?
He saved my final exam😭😭😭
Outstanding
Thank you!
Great stuff, for the last problem i see we have our origin placed midway between the plates, I would therefore expect the y at the bottom to be y = -h
Agree. The axis in the graphic (that I stole from the book publisher) is in the wrong place. But I think it's totally clear in the presentation that y=0 is at the bottom. At some point, a long time ago, I fixed it in the pdf download.
Wow BLESS YOU I was about to give up
Thanks. Glad it helped.
Your videos are so helpful! What happened to dynamic viscosity in the Couette flow example around 17:53 of the video?
The assumption is that dynamic viscosity is constant, which is true for a Newtonian, isothermal flow. So, you can divide both sides by dynamic viscosity, and it goes away. This tells you that the form of velocity profile does not depend on the fluid viscosity for Couette flow.
@@FluidMatters thank you! I should have picked Ryerson. You have a real gift for explaining.
Really like your videos. Will just point out that slide at 20:11 should say "linear" instead of "parabolic" cor Couette flow. Although you do correct it about 10 seconds later.
Yeh, Sorry. These videos are recorded in one "session" without a script. So, misspeaking is going to happen now and again.
Why is d2u/dy2 not cancelled in one dimensional flow? Isnt the flow only in the x-direction
Your help is highly appreciated
"u" IS the x-component of velocity, which is not zero, except at the wall. Thus, d^2u/dy^2 is the curvature of the u-component of velocity, which is certainly no zero for Poiseuille flow. If you don't get this, review your basic calculus of the meanings of derivatives.
can you show an example of a case where the top plate is fixed and the pressure gradient is moving the bottom plate please
Just reverse the boundary conditions., With appropriate signs
What are the limitations of annalytic solution of fluid flow equations ?
For this to be an exact solution the flow has to be isothermal because the fluid properties (that vary with temperature ) are assumed to be constant. Also, flow must be laminar (and incompressible). So, Reynolds number must be ~
Sir What if we do not neglect the gravity effect (acceleration g for - y direction)? Does the calculation change?
If you include g in the y-direction you get a hydrostatic pressure gradient in the y-direction, but this has no influence on the flow (in the x-direction).
@@FluidMatters Thank you so much!
At 15:00 is the volumetric flow rate, Q= Vdot/w(width?)
No. The flow rate per unit depth (into the page) is Q=V_avg*A=V_avg*2h, where 2h is the flow cross sectional area per unit depth into the page.
The units are still the same as Q/w right? m^2/sec?@@FluidMatters
@@SumanthPhaniVarmaPenmetcha Yes. The units are flow rate (m^3/s) per unit depth (m). So, m^3/(s m)=m^2/s. That should make complete sense, if you think about it.
Pleas sir ...I want the reference for this subject ...thanks for you
It is in the video description: Course Textbook: F.M. White and H. Xue, Fluid Mechanics, 9th Edition, McGraw-Hill, New York, 2021.
If the distance apart is 2H, why do you use h for your boundary condition instead of 2h at upper wall?
The coordinate system is in the middle of the channel . Wall are y= +h, - h.
@@FluidMatters Yes that,why -h for a BC, instead of zero? How does -h and h sum up to '2H'?
@@nzearimo I don't think I use 2H in the problem, I use 2h. All I can say is: Look at the problem diagram more closely. Deta_y=h-(-h)=2h. I cannot help you beyond this.
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Thank you sir
Glad to hear the video was helpful. Good luck with your studies.