I can't possibly tell how much these videos help me with my studies. I hope I will have the chance to thank Prof. Biddle in person one day. His way of teaching makes all these complex topics very simple to understand.
At 15:23 there's no U but only 0.332.U^(3/2)(rho.mu/x)^(1/2) in the expression of the shear stress. Also 23:50 , the integral of the shear stress should provide us with: Dx=0.664.b.x^(1/2).(rho.mu)^(1/2).U^(3/2) Then when equating with DL, Cd will have the right answer Cd=1.328/Re^(1/2).
This is a super excellent lecture! I just want to point out that there is an extra U in the Blasius equation that makes the exponent of the velocity 5/2. It should be 3/2 total.
I think you've made a mistake in formula for sheer stress at 15:07. You wrote U in front of the square root that shouldn't be there. Also I think you're missing the x under the square root at 24:00
There's no U power of 1 there in the shear stress Equation keep just the one U with power of 3/2 alone, he forgot the square X too in the drag force D. Overall it was so helpful thanks !
I'm not sure why an equation around 9:30 is called "x-momentum" equation. Is momentum not defined to be mass (density) times velocity? du/dx and du/dy aren't mass though so I'm confused.
Correct me if I'm wrong, but I believe he is talking about Chap. 9... not chap.7 (from Munson, Young, and Okiishi's Fundamentals of Fluids book) and he's referring to Table 9.1 (not 7.1)!
The authors have two wrong scientific approaches: researching the creation of Lift force and Low pressure at upper side of the wing, relative to the ground surface and Earth. I explain the aerodynamic cavitation and existence of Lee side aerocavern, and creation of Aerodynamic force.
Please can you help me? I commented here weeks ago and am still really struggling with the derivation of the boundary layer equations. I don't understand why we can get rid of the terms you mentioned in the video. If you could explain I would be extremely grateful. I found a video (link below) that had an explanation without going in depth about order of magnitude which was good for me as I'm a beginner in this area but it's reasoning for dp/dy = 0 makes no sense. Also if you could explain why the pressure gradient is zero that would be helpful too. This other video gave an explanation but I can't follow his reasoning. (his reasoning is at 2:28 of this video link if you are interested in helping a stressed student th-cam.com/video/h3RuIhR6TTU/w-d-xo.html).
the analysis of order of magnitude in that video is totally valid. as the whole point of that analysis is to non-dimensioalize the navier-stokes equation to actually obtain the order of magnitude of the pertaining values. and if you are still confused about what is the order of magnitude is all about, the reasoning goes something like this: first we analyze the navier-stokes for the x component which is the most prominent id we are dealing with a uniform fluid velocity (Uo) then we can obtain through non-dimensionlaization that the partial second order derivative of the velocity (u in respect to x is 1:1) which means that suppose that the velocity is 2m/s then the change of x is in the same order of magnitude (ie something like 5m or something) however, when we analyze the partial second order derivative of velocity u in respect to the horizontal height of the boundary layer, then through the nondimeonalized equation, we can observe that the previously assumed velocity in x (Uo) is 5m/s but the change in boundary layer is something like 0.000001 for example which gives us a way larger value than the (delsquared(u)/del(x)squared) and thus we can ignore it accordingly. the same thing goes with the navier-stokes in the y direction, you can see from the nondimenosionalized equation that (dp over the change in y) is by sandwiching the order by the previous analysis that have been done in the video is going to be something like 0.00001 and thus we can ignore the change in the pressure over the vertical displacement overall as it shows no significant change overall. and if the change in pressure over the vertical is almost insignificant then we can assume that navier-stokes is only viable for the horizontal direction only
@@NostalgiaGames_Gamer Many many thanks for exaplining that. Two questions: 1. Do you need a pressure change to make NS equation viable? 2. Now that we have the boundary layer equation, how do I go about solving it? I'm relatively basic level and the PDE looks very complex.
@@marquez2390 no worries, 1) a pressure change is negligible in the x direction because of the constant uniform flow velocity U, an thus can be ignored in the xmomentum equation of NS, thus we will have the equation that the professor wrote down, and also the pressure change in the y direction is similarly neglected when we talk about the y-momentum in general, and thus pressure change in the BL theorem equations are mostly neglected in laminar flow, however, in general (not related to BLT) we of course have to take the NS equation with all it's glory, however solving it requires computational power and it is beyond our ability without numerical approximations 2) Don't worry about the partial differentials for now in the BLT, blasius has already done the job by using a similarity solution (i advise you to watch (Mod-01 Lec-12 Laminar External flow past flat plate (Blasius Similarity Solution) ) video, very informative, and yielded a beautiful simple approximate solution for laminar BLs , but as a general note please review the partial differential equation solving methods as they will be helpful later on. Greetings .
Big "thumb's up" to the cameraman. Kept the prof and whiteboard within perfect frame/zoom.
I wish I could, one day, meet this professor just to tell him how good of a job he's doing. Thank you from DOWN UNDER
I can't possibly tell how much these videos help me with my studies. I hope I will have the chance to thank Prof. Biddle in person one day. His way of teaching makes all these complex topics very simple to understand.
At 15:23 there's no U but only 0.332.U^(3/2)(rho.mu/x)^(1/2)
in the expression of the shear stress.
Also 23:50 , the integral of the shear stress should provide us with: Dx=0.664.b.x^(1/2).(rho.mu)^(1/2).U^(3/2)
Then when equating with DL, Cd will have the right answer Cd=1.328/Re^(1/2).
A man of culture as well
I wish all professors could be like this guy! Thank you sir! From Portugal
This is a super excellent lecture! I just want to point out that there is an extra U in the Blasius equation that makes the exponent of the velocity 5/2. It should be 3/2 total.
His delivery is amazing
Thank you so much for the video! He is so detailed in his explanation. I wish he was my fluid prof
online sucks. Reading off slide shows and going 10x faster . Not even answering questions properly 😒. Thank you for your videos they helped a lot!
Thank you very much Dr. Biddle!!!
Very good explanation of the topic, useful for a quick study session before the final
you're my hero...
Saya mewakili Griffon Engine 21 mengucapkan terima kasih banyak doktorr Biddle. Yellboysss!!!
I think you've made a mistake in formula for sheer stress at 15:07. You wrote U in front of the square root that shouldn't be there.
Also I think you're missing the x under the square root at 24:00
good stuff, i still have trouble viualizing the boundary layer depth, is there maybe a rule of thumb i could use for fast approximation?
wait you guys have fluids 1? this is my first fluids class...
Need more videos on laminar and turbulent flow
Very good
How can be get the values in the tabular form
There's no U power of 1 there in the shear stress Equation keep just the one U with power of 3/2 alone, he forgot the square X too in the drag force D.
Overall it was so helpful thanks !
this video is very important,how can i gate the full lecture
Very good. No need to go to the book now
Reading the book adds a lot of additional information and context... Dr. Biddle certainly read many fluid mechanics textbooks over the years. :)
Super helpful video!
We're glad it was helpful!
There is an error to write the Froude number as V/(gL) in the last similarity problem. The dimensions do not match, should be v^2/(gL) or v/sqrt(gL).
What book is used for this course?
I'm not sure why an equation around 9:30 is called "x-momentum" equation. Is momentum not defined to be mass (density) times velocity? du/dx and du/dy aren't mass though so I'm confused.
go back to navier-stokes equation, it is just an algebraic manipulation of the equation nothing else
great lecture
Thanks
Useful videos 👌👌👌
why don't mention the Blasius equation?
Thanks
Correct me if I'm wrong, but I believe he is talking about Chap. 9... not chap.7 (from Munson, Young, and Okiishi's Fundamentals of Fluids book) and he's referring to Table 9.1 (not 7.1)!
The book is Fluid Mechanics (8th) by Frank M. White.
I wish I had you as my professor
Ric Flair teaches fluid mechanics?
Where do I get the notes for this so I can follow this better where you leave gaps.
Sorry, Dr. Biddle's lecture notes are not available.
I love you sir
Table 7.1 from which book ?
Why does the height of boundary layer increases as we move along x axis?
More fluid is in contact with wall.....so there is more effect of wall stress
As you move along the wall, the influence of the wall (no slip boundary condition) has more time to impact the flow near the wall.
Due to Viscosity
What assumption are involved in the derivation of momentum eq and energy eq for laminar boundary layer?
Thnks sir..
You're welcome.
Owsome
The authors have two wrong scientific approaches: researching the creation of Lift force and Low pressure at upper side of the wing, relative to the ground surface and Earth. I explain the aerodynamic cavitation and existence of Lee side aerocavern, and creation of Aerodynamic force.
What does the letter "nu" represent
Kinematic viscosity... it is (absolute viscosity / density)
@@CPPMechEngTutorials Thank you very much.
why is D-x not a function of x! another typo or what!
Please can you help me? I commented here weeks ago and am still really struggling with the derivation of the boundary layer equations. I don't understand why we can get rid of the terms you mentioned in the video. If you could explain I would be extremely grateful. I found a video (link below) that had an explanation without going in depth about order of magnitude which was good for me as I'm a beginner in this area but it's reasoning for dp/dy = 0 makes no sense.
Also if you could explain why the pressure gradient is zero that would be helpful too. This other video gave an explanation but I can't follow his reasoning. (his reasoning is at 2:28 of this video link if you are interested in helping a stressed student th-cam.com/video/h3RuIhR6TTU/w-d-xo.html).
the analysis of order of magnitude in that video is totally valid. as the whole point of that analysis is to non-dimensioalize the navier-stokes equation to actually obtain the order of magnitude of the pertaining values. and if you are still confused about what is the order of magnitude is all about, the reasoning goes something like this: first we analyze the navier-stokes for the x component which is the most prominent id we are dealing with a uniform fluid velocity (Uo) then we can obtain through non-dimensionlaization that the partial second order derivative of the velocity (u in respect to x is 1:1) which means that suppose that the velocity is 2m/s then the change of x is in the same order of magnitude (ie something like 5m or something) however, when we analyze the partial second order derivative of velocity u in respect to the horizontal height of the boundary layer, then through the nondimeonalized equation, we can observe that the previously assumed velocity in x (Uo) is 5m/s but the change in boundary layer is something like 0.000001 for example which gives us a way larger value than the (delsquared(u)/del(x)squared) and thus we can ignore it accordingly. the same thing goes with the navier-stokes in the y direction, you can see from the nondimenosionalized equation that (dp over the change in y) is by sandwiching the order by the previous analysis that have been done in the video is going to be something like 0.00001 and thus we can ignore the change in the pressure over the vertical displacement overall as it shows no significant change overall. and if the change in pressure over the vertical is almost insignificant then we can assume that navier-stokes is only viable for the horizontal direction only
@@NostalgiaGames_Gamer Many many thanks for exaplining that. Two questions: 1. Do you need a pressure change to make NS equation viable?
2. Now that we have the boundary layer equation, how do I go about solving it? I'm relatively basic level and the PDE looks very complex.
@@marquez2390 no worries, 1) a pressure change is negligible in the x direction because of the constant uniform flow velocity U, an thus can be ignored in the xmomentum equation of NS, thus we will have the equation that the professor wrote down, and also the pressure change in the y direction is similarly neglected when we talk about the y-momentum in general, and thus pressure change in the BL theorem equations are mostly neglected in laminar flow, however, in general (not related to BLT) we of course have to take the NS equation with all it's glory, however solving it requires computational power and it is beyond our ability without numerical approximations
2) Don't worry about the partial differentials for now in the BLT, blasius has already done the job by using a similarity solution (i advise you to watch (Mod-01 Lec-12 Laminar External flow past flat plate (Blasius Similarity Solution)
) video, very informative, and yielded a beautiful simple approximate solution for laminar BLs , but as a general note please review the partial differential equation solving methods as they will be helpful later on.
Greetings
.
@@NostalgiaGames_Gamer Thank you a lot.
Who's here from The University of Sydney CIVL3612 2020 cohort?
Me.
Froude number equations are wrong
Can you provide a timestamp?
@@CPPMechEngTutorials 53:08
@@CPPMechEngTutorials Fr=v/(Lg)^0.5
@@tuanle4958 You are correct. We will place a correction in the video comments. Thanks for catching this!
Great lecture