Solutions to Navier-Stokes: Poiseuille and Couette Flow

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  • čas přidán 31. 12. 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

Komentáře • 57

  • @FluidMatters
    @FluidMatters  Před rokem +9

    All the videos (and pdf downloads) for this introductory Fluid Mechanics course are available at: www.drdavidnaylor.net/

  • @vor6126
    @vor6126 Před 10 měsíci +11

    this video alongside your excellent commentary is absolutely a gem to Mechanical Engineering education. much Love

    • @FluidMatters
      @FluidMatters  Před 10 měsíci

      Glad to hear you found this helpful. Bes of luck with your studies,

  • @StressedHades37
    @StressedHades37 Před 3 lety +7

    Thanks for the explanation. That was easier to understand than my textbook!

  • @abdofast5
    @abdofast5 Před 3 lety +15

    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!

  • @PsychicJaguar19
    @PsychicJaguar19 Před 2 lety +2

    This is very much appreciated, very clear and concise explanation.

  • @majroi
    @majroi Před 2 měsíci +1

    Greetings from Turkey, this material helped me a lot. Thanks professor.

  • @portreemathstutor
    @portreemathstutor Před 4 měsíci +1

    Thank you so much for these videos. They make my course much easier.

    • @FluidMatters
      @FluidMatters  Před 4 měsíci

      Thanks for the nice comments. Best of luck with your studies.

  • @batbayarbatsukh6393
    @batbayarbatsukh6393 Před 3 měsíci +1

    Greetings from Mongolia. Thanks for the clear explanation. Тhis helped me a lot
    I wish you good health.

    • @FluidMatters
      @FluidMatters  Před 3 měsíci

      Mongolia! Hope to visit one day. Glad to hear the video was helpful.

  • @javijavi6166
    @javijavi6166 Před 2 lety +1

    Sir thankyou for uploading this, can we have a video of generalized couette flow?

  • @vidurapaathukkalam..8744
    @vidurapaathukkalam..8744 Před 2 lety +1

    God gave me the opportunity to see ur video..thank god..thank u for ur explanation..i really Adore 🥰

  • @user-safaa334
    @user-safaa334 Před 3 lety +2

    thank you so much for your excellent explain

  • @specter1001
    @specter1001 Před 2 lety +1

    incredibly helpful

  • @requiem-ph5xx
    @requiem-ph5xx Před rokem +2

    great vid broski

  • @baaa-ej7xj
    @baaa-ej7xj Před měsícem +1

    He saved my final exam😭😭😭

  • @masoudadli108
    @masoudadli108 Před 3 měsíci +1

    the best of the best,, you make life easy..thanks from Sudan Africa

    • @FluidMatters
      @FluidMatters  Před 3 měsíci

      Thanks for the kind words. Glad to hear the videos are helpful.

  • @meraihianouar9022
    @meraihianouar9022 Před 2 lety +1

    Im so grateful nd thankful to u sir, very useful this vidéo

  • @luisenriquejaracieza6498

    Thank you so much, sir.

  • @Frostbyte-Game-Studio
    @Frostbyte-Game-Studio Před 2 lety

    good sir thank you for your video, helped out a lot

  • @ThePerks.
    @ThePerks. Před 6 měsíci

    Outstanding

  • @ahmedmokkedem7525
    @ahmedmokkedem7525 Před 3 lety +1

    Thanks so much sir

  • @user-qy7gg6hg5b
    @user-qy7gg6hg5b Před 2 lety

    Thank you!

  • @RGCAD
    @RGCAD Před měsícem

    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

    • @FluidMatters
      @FluidMatters  Před měsícem

      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.

  • @alexklaver9936
    @alexklaver9936 Před 2 lety +1

    Your videos are so helpful! What happened to dynamic viscosity in the Couette flow example around 17:53 of the video?

    • @FluidMatters
      @FluidMatters  Před 2 lety +2

      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.

    • @alexklaver9936
      @alexklaver9936 Před 2 lety

      @@FluidMatters thank you! I should have picked Ryerson. You have a real gift for explaining.

  • @splendidteaching
    @splendidteaching Před 2 lety +1

    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.

    • @FluidMatters
      @FluidMatters  Před 2 lety +1

      Yeh, Sorry. These videos are recorded in one "session" without a script. So, misspeaking is going to happen now and again.

  • @soso-zz9qf
    @soso-zz9qf Před 4 měsíci +1

    Wow BLESS YOU I was about to give up

  • @francisbacor3214
    @francisbacor3214 Před 2 lety +1

    Why is d2u/dy2 not cancelled in one dimensional flow? Isnt the flow only in the x-direction
    Your help is highly appreciated

    • @FluidMatters
      @FluidMatters  Před 2 lety +1

      "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.

  • @jeanbedessounda3571
    @jeanbedessounda3571 Před 2 lety

    can you show an example of a case where the top plate is fixed and the pressure gradient is moving the bottom plate please

    • @snehajain4334
      @snehajain4334 Před 2 lety

      Just reverse the boundary conditions., With appropriate signs

  • @sajadadnan9555
    @sajadadnan9555 Před 3 lety

    What are the limitations of annalytic solution of fluid flow equations ?

    • @FluidMatters
      @FluidMatters  Před 3 lety +1

      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 ~

  • @user-xz7xt5cl1m
    @user-xz7xt5cl1m Před 2 lety

    Sir What if we do not neglect the gravity effect (acceleration g for - y direction)? Does the calculation change?

    • @FluidMatters
      @FluidMatters  Před 2 lety +1

      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).

    • @user-xz7xt5cl1m
      @user-xz7xt5cl1m Před 2 lety

      @@FluidMatters Thank you so much!

  • @nzearimo
    @nzearimo Před rokem

    If the distance apart is 2H, why do you use h for your boundary condition instead of 2h at upper wall?

    • @FluidMatters
      @FluidMatters  Před rokem

      The coordinate system is in the middle of the channel . Wall are y= +h, - h.

    • @nzearimo
      @nzearimo Před rokem

      @@FluidMatters Yes that,why -h for a BC, instead of zero? How does -h and h sum up to '2H'?

    • @FluidMatters
      @FluidMatters  Před rokem

      @@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.

  • @SumanthPhaniVarmaPenmetcha
    @SumanthPhaniVarmaPenmetcha Před 9 měsíci

    At 15:00 is the volumetric flow rate, Q= Vdot/w(width?)

    • @FluidMatters
      @FluidMatters  Před 9 měsíci

      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.

    • @SumanthPhaniVarmaPenmetcha
      @SumanthPhaniVarmaPenmetcha Před 9 měsíci

      The units are still the same as Q/w right? m^2/sec?@@FluidMatters

    • @FluidMatters
      @FluidMatters  Před 9 měsíci

      @@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.

  • @mrsengineering3663
    @mrsengineering3663 Před 2 lety

    Pleas sir ...I want the reference for this subject ...thanks for you

    • @FluidMatters
      @FluidMatters  Před 2 lety +1

      It is in the video description: Course Textbook: F.M. White and H. Xue, Fluid Mechanics, 9th Edition, McGraw-Hill, New York, 2021.

  • @DrDerivative
    @DrDerivative Před 8 měsíci +1

    🐐🐐🐐🐐🐐🐐🐐🐐 You sir are the GOAT.

  • @abdulhakeemshehu3559
    @abdulhakeemshehu3559 Před rokem +1

    Thank you sir

    • @FluidMatters
      @FluidMatters  Před rokem +1

      Glad to hear the video was helpful. Good luck with your studies.