Navier-Stokes Equation Final Exam Question
ฝัง
- เผยแพร่เมื่อ 10 ก.ค. 2024
- MEC516/BME516 Fluid Mechanics I: A Fluid Mechanics Final Exam question on solving the Navier-Stokes equations (Chapter 4). Calculation of the fully-developed velocity field between two parallel plates, with a pressure gradient and one plate moving. (This is a classical exact solution that dates back 100+ years: Couette flow with a pressure gradient.)
All the videos for this course and a copy (pdf) of this fluid mechanics presentation can be downloaded at: www.drdavidnaylor.net
This website also includes an exam bank and solved problem sets.
Course Textbook: F.M. White and H. Xue, Fluid Mechanics, 9th Edition, McGraw-Hill, New York, 2021.
#fluidmechanics #fluiddynamics #Navier-Stokes
All the videos (and pdf downloads) for this introductory Fluid Mechanics course are available at: www.drdavidnaylor.net/
I need the PDF Sir
@@badejosamuel-qk5ud I've fixed the file error on my website. It is here: www.drdavidnaylor.net/exam-review-questions.html
@@FluidMatterscan u send more application to understund the low of navier ?
a full quarter of fluids and it only makes sense now after watching these videos!
Glad to hear it was helpful.
Best video for navier stokes example in the whole youtube
You are right, explaining all the simplified terms and the logic of solving the exercise.
15 minutes of pure gold, thank you so much!
Good luck at 8am. See you on zoom.
I watched the entire play list of Navier Stokes equation. It was very helpful. Thank you so much!
A very nice example and explanation that can rarely be found on youtube.
The Simplest, mlst understandable explanation.
Thank you !
Thank you so much, I finally understand the individual terms of the N.S. equaitons. It has been hard to get a detailed answer on this, so I'm grateful for this video🙏
i agree.. everyone just leaves it in 3 dimensions and never breaks it down. I do wish.. and am curious could he provided a value for U .. and b.. could we solve the velocity?
Nice clarification in approaching N-S equation problems!
It's so easy to have the equation from your lecture sir ...Thanks a lot sir.
thank you so much for the simplicity
You made me gain confidence in my knowledge thanks for your hard work
Where was this channel 😭,how nicely he explained 🙏
Best video on navier stokes for sure
Right to the point, thanks!!
the best explanation on TH-cam 😍
Very well presented example. Thanks!
Thank you so much for making this video. You're a great teacher. Wish I went to Ryerson.
THIS VIDEO IS SO GOOD
Very useful and well prepared example, it helped a lot. Thanks!
Thank you really really much! This saved my life! I searched for something like this a week long. Thank you again!!
Glad it helped!
I really enjoyed this exercise thank you
after this video, I got Navier stokes equetions. thanks
thank you so much.i needed this🌹
Glad you found it helpful.
Thank you for the good explaining. This video was very helpful for me to understand the Navier-Stokes equation.
Glad to hear it was helpful. Thanks for the kind words. Best of luck with your studies.
Excellent!
Wow, thank you Sir!
Excellent video!!
Thanks. Glad to hear it was helpful. Good luck with your studies!
Very helpful!
Vielen dank, sehr sehr hilfreich
Wonderful explanation. He made an impossibly difficult problem into an easy one to understand and solve
Thanks so much for kind words! Best of luck with your studies.
I came just to check for a concept, then proceed to finish the whole series.
Glad to be able to help. Best of luck with your studies.
Thank you so much for sharing. Best regards from Panama 🇵🇦
Glad to hear it was helpful.
Very helpful video sir thanks
This was my quiz question
Thank you sir
Its perfect, right from the start to the very end
Thanks. Glad to hear it was helpful.
super helpful! explained more than I learned in an entire fluids class. awesome!
Thanks. Glad I could help.
Thanks a lot
thanks a lot
Thank you so much sir for all the effort you put in this work❤, it really helped me with my studies 😊.
Glad to hear it was helpful. Good luck with your studies.
thank you
Thanks So much, Sr. good lecture.
Thanks for the kind words.
invaluable
That elimination of possiblity of u being a function of x(fully developed flow) using contuinity equation was sick. Now I always use Navier stokes and contuinity equation together. 😊
That's a good mathematical insight! You are really understanding the details. Sick! Ha Ha.
cristal clear lecture
Could you please give me the references for these info in the video
Thank you, I just clapped at the end of the video
Thanks! Glad that it helped.
Bro love you 3000....I watched this video 20 mins. Before my final exxam from IIT.
I had the very same question in the exam for 20% marks.🎉
Glad it helped. Maybe your prof saw this video too! Ha Ha.
LEGENDDDDDDDDDD
Thanks!
Thank you so much ,do you have exam for boundary layer
The mathematics of viscous boundary layers are not part of this intro course. Sorry.
Thanks for the video. I have a question about the pressure profile. Should we continue and solve for the pressure profile in x direction or it is enough to stop at the final form demonstrated by the video?
The questions asks "Derive and expression for the velocity profile", not the pressure gradient dp/dx. So, you can stop where I did.
@@FluidMatters Good. In case, we want to continue, what is needed to solve the pressure gradient term?
Thanks.
@@tammammohammed4442 You would likely want dp/dx as a function of the volume flow rate, Q. To do this I would integrate the velocity profile across the channel to get the flow rate: Q=f(U,b, dp/dx). Then rearrange the expression to solve for dp/dx, which is a constant.
@@FluidMatters Thanks a lot!
thank you. but please how can we find the pressure gradient?
The pressure gradient (dp/dx) is supplied by the pump. In the final equation, the value of dp/dx is your choice (i.e. an input) and could be estimated from the pump head curve. The bigger the pump, the higher the flow rate, and a larger pressure loss per meter of channel length (dp/dx).
this seemed easy, but "under pressure" it is not. i feel like having a slightly over-tuned pressure gradient doing this in an exam again xD
😍
Hello , what about the unsteady case? how can we sole this problem for u(t,y)?
There are exact solutions for some simple unsteady problems, like an impulsively accelerated plane wall. This is beyond the undergrad level. See the classic book "Boundary-Layer Theory" by Schlichting, for example.
@@FluidMatters Ok,thank you very much🙏🙏
Why can't my university professors be like this, instead of being so hard to understand
Same…
🙏🙏
what happens when the plates are moving in the same direction
That's another possible exam variation. You can define U_1 for the bottom plate and U_2 for the upper plate. The solution is identical up to 12:44. The evaluation of C_1 (at 12:44) is different: u=U_2 at y=b. Results in a slightly different superposition of Poiseuille (pressure gradient-driven flow) and Couette flow (plate motion-driven flow).
At 13:29, shouldn’t it be C1 = -1/(2*mu*b)*dp/dx - U/b and not C1 = -b/(2*mu)*dp/dx - U/b ? Doing the algebra, I feel the b should be in the bottom of the fraction and not the top. Can someone explain please?
Before simplification, we have b^2 in the numerator (from y^2 in the original expression applied at b). So, when you divide by b to isolate C1, you end up with b in the numerator. I hope that helps.
@@FluidMattersThanks! I didn’t see that
How long roughly would students be expected to solve this in?
I recall this question was one of five on a three hour final exam. So, say, ~40 minutes.
@@FluidMatters thanks
@@FluidMatters oh, we took a similar question for couette flow (N. S. E. ) in maximum 15 minutes in the midterm exam
What will happen if the flow is unsteady
You'd need to specify initial conditions, like a still flow (u=v=0 everywhere) and perhaps an impulsively started plate at time zero. In this case you'd have the acceleration term du/dt (partials, of course) and it would be more difficult to solve, as u=u(x,t). These solutions are more advanced: See the classic book "Boundary Layer Theory by H. Schlichting", McGraw Hill.
so it wasn't my fault, apparently my professors can't explain anything. Thank you so much.
Thanks for the kind words.
@fluidmatters Your mathematical equation is incorrect. It is impossible to have a "No Slip" situation for the lower plate boundary. The formula should instead show that the lower and upper plates are directly related to each other. Instead of making the upper plate move and the lower plate stationary...you should instead halve the velocities of both plates. This is both physically and mathematically the only way this equation can be properly solved and I can prove it. This is works the same for all fluids, whether it be water flowing through a fully contained pipe or the Pacific ocean flowing over bedrock.
I disagree. Couette flow with a pressure gradient is classical exact solution, dating back 100+ years. I will not debate it here. If you feel all the experts and dozens of textbooks are in error, you should submit your new idea to a refereed journal, rather than post an inadequately explained claim in the comments on TH-cam. Best of luck.
this seemed easy, but "under pressure" it is not. i feel like having a slightly over-tuned pressure gradient doing this in an exam again xD
btw. this was helpful
btw. this was helpful