Physics 34 Fluid Dynamics (7 of 7) Bernoulli's Equation
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- Опубліковано 22 сер 2024
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In this video I will show you how to use Bernoulli's equation to find the force that lifts an airplane off the ground.
First video in this series can be seen at:
• Physics 34 Fluid Dyna...
Michael. Excellent explanation of what happens to the air molecules as they pass around the airplane wing (as well as around the sail of a sail boat). The principle of lift of an airplane wing and the propulsion of a sail boat is still the difference in pressure between one side and the other, due to differential speed of the air molecules. However you are correct in saying that it is an oversimplification. But for the purpose of understanding physics at this level it makes it easier to understand. Fluid dynamics is a very complicated area of physics and very difficult to model correctly mathematically. That is why we use wind tunnels and smoke to help us understand the air flow dynamics. In the future when I complete all the basic sets of physics, math, chemistry, and astronomy, I will begin to put out videos at higher levels regarding a variety of topics.
sir i am with you. ...
How many kilograms does a fligth has which accomodate four passegers
Sir i need your idea kindly.how many kilogram does a flight has to make it fly. Am lauching my local made flight
It makes no sense how the Bernulli principles for the water in the tube can be applied to an open spaced area of gas. Obviously, the calculations cannot be correct because the density of the air near the wing will not remain constant.
Amazing, I really enjoyed watch all 7 videos about Bernoulli's equation and now I am looking to watching all of your videos. Many thanks Michel
Hi Michel, I love your style and presentation but I must take issue with you over the misplaced application of Bernouli's theorem to aerodynamic lift. I agreed with this idea for 30 years until I bought a sailing boat and discovered that the lift for a boat is the same as that for an aeroplane wing, yet there is no significant difference in the distance on either side of a sail. The aeroplane wing is generally speaking almost symetrical as are helicopter rotors. It is assumed that the air molecules must take the same time to travel round the upper surface as the lower surface which is not true (the "equal transit time fallacy") Even NASA now agrees that Newtonian forces (action and reaction) are more probable for aerodynamic lift but it seems from my research that the jury is still out for any absolute case for lift, although it does contain an element of angle of attack, in my opinion. But your series of videos is very useful and accurate, thanks and no disrespect meant herein.
What actually the point? Are you saying that Bernoulli's principle is not working to lift the airplane?
I believe you still get the correct lift force from Bernoulli equation given that you account for the actual speed over the top surface which is faster than equal transit time
@@Toast934 Yes, correct. The surface of an airfoil will typically be a streamline. You can find the velocity along that streamline, convert it to pressure using Bernoulli, then use the pressure distribution to get lift. This obviously is not a good explanation for lift because it doesn't tell you why the velocity is different over the upper and lower surfaces of the airfoil, but Bernoulli is correct.
You would get the same lift by computing the change in momentum of the lift, typically by putting a (metaphorical) box around the airfoil and measuring the momentum flux through the surfaces of the box. Those are identical; if there's a pressure and velocity distribution that implies a net upward force on the wing, there must also be a change in momentum of the air so that the air is moving a bit more downward as it exits.
>Even NASA now agrees that Newtonian forces (action and reaction) are more probable for aerodynamic lift
do you have a source for that buckeroo?
@@StrsAmbrg What he is saying is that the difference in pressure does not accurately describe lift. Some planes can fly inverted. Yet they are not pushed to the ground as this video stated.
I used to hate fluid dynamics, but after watching this, I'm falling in love with this stuff XD, well done sir!
Hello, I'd like to thank you for making these really useful videos. They help a lot for finals!
This video starts with a notorious fallacy, namely ‘the equal transit time theory’. It is well known in aeronautics and has been verified by direct observation that in fact the upper flow arrives at the trailing edge well ahead of the lower flow. For that reason I would not recommend the contents of this video to any student.
This video is literally made for high school physics. Those things are negligible for now.
@@chl0yx180 the video is literally wrong, so I don't care if it was made for kindergarten, such fallacies have no place in education at any level.
@@XPLAlN I told him that years ago. He doesn't care to be correct.
@@Observ45er let it go... 🙄
Sir, Excellent lectures and tutorials..you are a better teacher than most of our college professors..
For those who are preparing to go into aerodynamics or fluid dynamics, please don't try to understand wing aerodynamics from here. This whole explanation is completely wrong. There is absolutely no reason why the air above the wing and below the wing should reach the trailing edge of the wing at the same instant (in fact, they don't), and therefore the basic assumption about the wind speeds above and below the wing is wrong. This is a wrong explanation of lift that has been long debunked and should not be used anywhere in education anymore.
TR0LLREIGN hello i would Like to know more about aerodynamics(potentially about F1(thats a sport that has to do with Aero))
so do u have any videos where i could watch a vid that will teach me better and thats not longer over 20 mins
THANK YOU!! God, this is one serious pet peeve of mine. Bernoulli's explanation of flight is simply disproved by the fact that planes can fly upside down.
penguin rick
They can fly upside down only by the force of their engines. In rhis case the wings dont generate any lift that is why a boing 737 cannt fligh upside down but F16 whos engine trust is more greater then the weight of the plane can
@@iliag1 No, you are wrong. If that were the case, then airplanes would have to fly like helicopters when their airfoils are upside down relative to their flight direction. This would also mean that symmetrical airfoils would not produce any lift, let alone super critical airfoils used on modern jet airliners, which are _less_ curved at the top than the bottom.
And I know this comment is old, but I'm just leaving this here for the few people who will still watch this video and read your comment.
Sir, thanks to your passion for these subjects and your thorough methods of teaching, I have learned so much and found it enjoyable, all at the same time!
Thank you and thank you again!
Thank you. Glad you found our videos.
Michael,
This was never meant to be a "realistic" aeronautical problem. Just one to show the concept of Bernoulli's equation and how to apply it.
+Michel van Biezen Except this most certainly NOT how to apply Bernoulli's Equation.. It is two years later and it is still just as incorrect.
I understand the intent, but it perpetuates a false understanding of flight. Other than this, I love these videos and they've really helped my understanding of fluid physics!
?? Other than the fact it's wrong its a good video??? I think I know what you mean, but worded that way it looks strange.
Cheers
dude can i use this example for a school project.
Bernoulli's equation is for incompressible fluid but here the air is compressible .....I am having a doubt regarding this so pls help me to understand
It's a breakthrough game changer research.
Wat happens is moving airfoil hits the still air and deflects molecules away upwards, creating a partial vacuum or suction force over the wingtop. Various pressure zones are created around the wing in the process. Ambient atmospheric pressure below pushes wing up.
Book explains several aspects eg slat actio, stall, energy consumption, etc.
Nice input. Thank you. 🙂
Yup... enough to play with now. Thank You! Gonna do the otherseries when this is 'just over' logical in my head. So i'll never loose it again. Man... You should get paid a lot for making it logical.
Glad you found this helpful.
All 7 of these videos were great! Thanks for the help :)
Syed,
P = F/A therefore F = P * A = (4000 N/m^2) * (120 m^2) = 480,000 N
(assuming that 120m^2 is the total area of both wings)
w = mg therefore m = w/g = 480,000N / 9.8 m/sec^2 = 48,980 kg
You need to double everything if that was the area of just one wing.
You are truly a talented educator! I just watched your entire series on Bernoulli's equation, since I have an exam coming up on it, and I must say, I now feel confident about this test! What took my professor weeks to teach, you were able to do within an hour.
It's true that air above the wing travels faster than air below and that the pressure below the wing is greater than the pressure above. It's also true that Bernoulli's Principal is relevant.
However, the air above and below the wing do not have to meet up at the same point in time and the increased distance it has to travel above the wing does not cause lift.
Excellent teaching skills...Bernoulli eq. will stay with people watching this series forever!
Iman,
Thanks for the feedback. Much appreciated.
A wrong interpretation of it, that is. Both this example and the hurricane example are plain wrong. Not oversimplified, simply wrong.
These were awesome, thank you.
You are welcome. Glad you like them! 🙂
Michael, I must say your lectures are great and saw them all. Would love to have all my teachers just like you! Could you make another video relating the equations to a motor boat?
By witch physical law you had decided that the air molecules that flows up and down the wing must reach the end of the wing at the same time?? Couldn’t it be that they reach to the wing’s end at different times ?? So then bernoulli’s equation will not be the right way to explain the force lift.. Thanks!
The equal time argument (particles flowing above and below the wing have to reunite) is a beautyful way to explain lift but it's completely wrong. Moreover, Bernoulli's equation applies to the same streamline, not to different ones. Watch Prof Babinsky video instead.
When we put our hand out of a moving car, she lift up/down after a little angle turn from horizontal. This angle off attack subjects the hand to the moving air Momentum down incidence generation of force(law of Newton). This is the main explanation of lift, the other contributions better seen with CFD. In search engines writing "Bernoulli airplane" appear articles explaining as momentum and angle generate lift, and others why Bernoulli is incorrect theory to apply here. Van Biesen also show here why is Bernoulli wrong: 134m/s down and plus 45m/s to the speed above wing. If the wing have 5m length front-back, this 45m perimeter difference do a mountain shape to the wing, wich is not a real plane to flight. If the shape above is soft curvature like real plane wings, then the speed difference is maybe like 2m/s, and the plane again do not have lift to up by Bernoulli difference of pressure. The real difference of speeds of 45m/s and more are achieved by the angle of attack(air Momentum-Newton) ,not by Bernoulli. The helicopter rotor and the wind generators are other users of the angle of attack. Other commentators before me tried to said this; please Van Biesen change the title to "why a real plane cannot lift by Bernoulli".
Bernoulli can apply to all streamlines in this flow because they all go into the far field with constant pressure and velocity (thus same constant)
The basic application is a start of understanding more complicated topics in fluid dynamics. I enjoy the basics and expect more video on the following advanced topics.
Best aviation teacher ever
best vids ever thumbs up. now I can progress the basics are clear
This is called the equal transit theory and it is incorrect. The air above the wing does not need to reach the end at the same time as the air below the wing.
I agree that no mandatory for the air to be transit equally. But now question is, what is actually causing the huge 50 tons airplane lifted? Some say due to Coanda's effect. But this also seems not the reason as it only explain one phenomena rather than causing another action or reaction.
Assume the horizontal velocity along the x axis Vx be constant and it is the same velocity the air is moving below the wing and since the upper part of the wing is curved this forces the air to move upward with a velocity component Vy along y axis--- you can see that the air at the upper part of the wing travels with two velocities components vx and vy, taking vector sum of the components the resultant will be greater than the magnitude of each velocity component and therefore this is a proof that the air at the upper part of the wing travels faster than the air at the bottom of the wing.
@@mohamedkazema6381 That is a proof only if you assume the horizontal velocity along the x axis is constant. Unfortunately, that assumption is completely false.
@G1408 I agreee
This series was so helpful to me,, I was starting to hate this concept because I just couldn't get it, thanks for this series which brought some light. Be blessed.
wrong description on how a plane stays in the air. many planes can fly inverted. If your explanation was accurate then the plane could not fly inverted as it would be pushed to the ground. Lift is more than the air above the wing going faster than the air below it. I have enjoyed your lectures as you are a great teacher.
I told him that years ago. He doesn't care to be correct.
@@Observ45er let it go... 🙄
This Man you are literally the best.
Thanks Professor you have helped me a lot with your physic posted videos . God bless you.
sir mujhe bernaullis theorem bilkul samjh nhi aa raha thaa but aapki 1 hi video dekh kar aadha se jada samjh aa gaya
Thank you so much sir.I am teacher from Indonesia.God bless you.You are wonderful.
Thank you and great to make your acquaintance.
Thank you so much sir.God bless you
Except the upper and lower do not meet:
The non equal transit time (only) observation video by Holger Babinsky, Cambridge University.
Note this is in a wind tunnel which restricts the downwash and the wing is at an extreme angle of Attack to make the effect easier to see.
ua-cam.com/video/UqBmdZ-BNig/v-deo.html Narrated, no title at start Cambridge Univ.
Hi sir, I want to ask. Why does the air have to arrive at the end of the wing at the same time?
It doesn’t.
I think there is no valid reason to expect the air flow on the upper and lower side of the airfoil to meet after the same time at the tailing edge, and thus deduce an higher air speed on the upper side. Imagine a flat plate with a zig-zag channel on the upper side. Still think that the air on the upper side has to move faster due to the longer path?
Prof. Holger Babinsky shows a better explanation for lift, and also proves that the particles don't meet at the tailing edge (indeed, the air flow on the upper side is much faster)
It is true that there is more to this situation than a simple Bernoulli effect. But it is established that the pressure above the wing is lower than the pressure below the wing and that much of the lift is due to that pressure difference. The pressure difference is due to the difference in wind speed (higher above the wing as compared to lower below the wing). The Bernoulli equations is a good approximation and is therefore used at the elementary levels in physics text books.
Equal time is wrong. Visit nasa website to find why.
Thank you Mr Van Biesen! love from a classified location in south east asia! (seriously)
Welcome to the channel!
Equal transit theory is incorrect. The air molecules do not in fact travel across the top and bottom of the wing in the same amount of time
Good input.
thank you very much i hope that i become good teacher like you best regards
Thank you Michel van Biezen !! Your videos are pretty well explained and easy to understand. I wish my professor was as good as you are. Good job ! :D
Finally I`ve got the understanding of how the plane flies! All I had to do is to consider this from the point of fluid mechanics!
Thanks a lot! Wish you all the best and hope that you will continue lectures on Fluid flow
Love you man
This isn't how planes fly! www.scienceeducationreview.com/open_access/eastwell-bernoulli.pdf
how can you use Bernoulli equation at differents streamlines?
Gino Andrade you can’t XD
u can if the flow is non-rotational, check out the Bernouilli's demonstration and it will seem obvious
can u solve my problem please really appreciate it :)
Question: The uplift on an aeroplane is a consequence of the faster moving air at A having a lower pressure than that at B. For Boeing 737 cruising at 270 ms-1, this pressure difference is about 4*10^3.
a) Calculate the uplift if the underwing area is 120m^2.
b) Estimate the mass of the plaine and its load.
thanks in advance.
I am just 13 but i know the law of aerodynamics
I have interest in aerodynamics
I want to become aeronautic engineer
Thanks Sir Michel van Biezen. It sure helps me a lot !
thanks for the videos. correct me if i am wrong. bernoulli states the total energy per unit volume is constant anywhere along the line of fluid flow. but in this example, point1 & point2 are not lying simultaneously along the flow. i mean line joining point1 & point2 should represent line of fluid flow (here it is perpendicular to line of fluid flow). my second question is does the air flowing over wing should create suction on wing (rather than pressure shown in figure)? please correct me if I am wrong.
for my 1st question, may be we should split the fluid mass into 2 parts. one flowing over wing. other flowing bottom of wing. now for portion of fluid mass flowing over wing, select 2 points (one point at leading edge, other point between leading & trailing edge) & then apply bernoulli equation to these 2 points. similarly, for portion of fluid mass flowing under the wing. now calculate the pres diff between 2 points (mid point at top of wing & mid point at bottom of wing) which shall create lift. i am not expert at this concept. but i would assume like this for perfect application of bernouli equation.
You're right, but all lines of fluid flow (streamlines) go off to infinity where the velocity and pressure is the same everywhere. So any two points can be equated because all streamlines ultimately reach points with the same property. For an accurate treatment, the velocity and pressure continuously changes over the surface of the airfoil, so you actually need to do a sum over infinitesimal segments of the airfoil - so really you need calculus.
Your videos are a godsend, sir. Thank you so much for them!
Why did you add the surface area of both wings to calculate the lift force?
Each wing produces 1/2 the lift.
There is an equation called "Lift Force Equation" and in that equation, there is no velocity differential. Instead, there is only one velocity, which is aircraft velocity. Why are they ignoring V2-V1 and write only V ? Additionally, we know that lift force increases as aircraft speed increases. How can I understand it through this equation you wrote? V2 will increase to the same extend V1 does. Doesn't it?
Lift on an airplane wing is very complicated and can be more easily understood with the use of wind tunnels. One of the principles is indeed Bernoulli's principles (but not the only one). Using Bernoulli's principle is a simplistic method to get into the ballpark of the wing lift.
@@MichelvanBiezen Would you be able to do a video to explain the cloud we see over the wing when an aircraft lands in humid air conditions. My understanding is the humid ait at atmospheric pressure hits the upper wing of lower pressure and adiabatic expansion and cooling occurs as the air expands and does work. In doing work it loses energy and its temperature drops to the dew point and condenses forming the cloud droplets. Would this be correct. Thanks
It would be nice to see something more advanced explaining the compresibility problems, influecne of angle of attack and the swept angle for example. Where can i find a book that shows something like that? Any idea?
Hi Michel,
I'm trying to model the effect of sea swelling on a floating object.
Would Bernoulli's principle appliable by chosing 2 points on the top and bottom surface of the object? Knowing the airspeed above it and waterspeed below it, the thickness of the object (y1-y2) and its area, would the resultant force be a reasonable approximation?
way better than my lecturer
awesome video....Salute to you Sir
What if the airplane is moving? We still have to use the air velocities itself in the equation or the relative velocities of air with respect to airplane?
Yes, the relative velocities to the wing is what matters.
This is not how flight works. This is a short article which explains it better. www.explainthatstuff.com/howplaneswork.html
One Question - how does the surface area of the fuselage get considered in a real life analysis of this situation? Does it not add to the wing area since the air moves over it also?
It is usually minor in relation to the wing, but it is included in the final design for sure.
Awesome series! Awesome work.
1 and 2 are not in the same streamline, why can use the Bernoulli’s equation?
Air flow over object are very complicated and usually cannot be accurately calculated which is why engineers use wind tunnels to determine wind flow. Bernoulli's equation does offer a good approximation.
So which theory account for the lift force and how?
It is a combination of things. As viewers indicated in their comments, Bernoulli's principle is not the whole story, but it definitely plays an important role in the lift obtained by airplanes.
In one of the question in my book the velocity is not given bit the pressure on lower suface of area is given
So how to find it sir.. Plzz help
Force upward = pressure on the lower surface x area (But you need to know more information such as the pressure on the upper surface).
Michel van Biezen thank you sir
And what if i have more doubts i cant always comment. Is there any other mean to contact to you so to clear my doubts
This is the best way. (We get many many requests and this is the easiest way for us to respond)
the change from 300 to 400 is 33% but the upper surface of wing is usually 2 to 4 % cambered. on which aircraft the wing is so much cambered ????
thank u for this video...
actually I hav a doubt regarding our product..
we r manufacturing ceiling fans....with blades of airfoil shape...
we can calculate V2 from anemometer. can u suggest a way to find out V1 ?? so that we can calculate the forces.
Do not use this simplistic explanation for your technical work. Use this: en.wikipedia.org/wiki/Affinity_laws
how did you apply bernouilli's equation on the airplane example? isn't an equation for fluids?
Air is also a fluid.
Thankyou professor !!I wish you can help me getting rid of all my problems in physics ❤️
Happy to help
Can you help by posting explanation about angular momentum and mechanics , hard applications because I've finals so near🥺
We cannot turn around requests like that so quickly. We have a lot of other jobs and endeavors that keep us occupied and squeeze making these videos in wherever we can.
When we put our hand out of a moving car, she lift up/down after a little angle turn from horizontal. This angle of attack subjects the hand to the moving air Momentum down face incidence generation of force(law of Newton). This is the main explanation of lift, the other contributions better seen with CFD. In search engines writing "Bernoulli airplane" appear articles explaining as momentum and wing angle of attack generate lift, and others why Bernoulli is incorrect theory to apply here. Van Biesen also show here why is Bernoulli wrong: 134m/s down and plus 45m/s to the speed above wing. If the wing have 5m length front-back, this 45m perimeter difference do a mountain shape to the wing, wich is not a real plane to flight. If the shape above is soft curvature like real plane wings, then the speed difference is maybe like 2m/s, and the plane again do not have lift to up by Bernoulli this small difference of pressure. The real difference of speeds of 45m/s and more are achieved by the wing angle of attack(air Momentum-Newton) ,not by Bernoulli. The helicopter rotor ,propellers and the wind generators are other users of the angle of attack. Other commentators before me tried to said this; please Mr Van Biesen change the title to "why a real plane cannot lift by Bernoulli", and with some adjustments all will be ok.
No, Bernoulli is completely correct but this is a misapplication of Bernoulli. You're thinking of equal transit, which is correct. Momentum is also true, but leads to the bad misinterpretation that the air simply deflects off the bottom of the wing (when the top of the wing is more important).
Exactly is a misapplication of Bernoulli, because to test Bernoulli alone due to the shape of wing we must put angle of attack zero. That we see the difference of speed is so small wich only generate lift to a "paper plane" not to a 70 tons plane. The shape of wing I understand is maximized to the Coanda effect. Well this is all working until the angle of attack of 20°, after that the wing stall, finishing suddenly the lift, all Newton, coanda and bernoulli disappear.
@@RF-fi2pt That makes no sense. First, wing stall does not lead to no lift. The lift generally decreases somewhat. "Bernoulli alone" still doesn't make sense; you're actually just measuring lift at zero angle of attack. Obviously, in the approximation of thin airfoil theory, lift is linear with angle of attack, but that's neither physical nor allows you to break anything down into "Bernoulli" or "Newton" components. And Coanda effect is completely irrelevant
Ansys saf1 lesson 2, ua-cam.com/video/Xi7ERZD_m9g/v-deo.html have the CFD to airfoil WITH AOA. At time 14:40 we see velocities: 112m/s incidence air. At first 1/3 down velocity half (56m/s)(big deceleration against wing), next same streamline increase to again 112 during 1/3lenght down and in the last 1/3 accelerate to 170m/s following always the down shape. Above the air doubles from 112 to 224 during de first 1/3, but leaving the shape staying almost vacuum there.
In the video of Mechatech ua-cam.com/video/l2W-J34YwkQ/v-deo.html is airfoil and AOA Zero. Here at time 22:30 a 45m/s air incidence half to 25 in the edge between top and down and above increase to 85 during 2/3 and down to 75 also during 2/3 following almost parallel to the shapes, not impacting and not creating vacuum buble, no useful lift force.
About AOA vs Lift graphic one at aviationchief.com/angle-of-attack.html and exactly after the maximum lift it drops suddenly after that Angle, as in this page said "lift is destroyed"(Lift=CL.ρ.v^2.S/2).We see many airline plane crashes by stall in program "Mayday air disasters", one of them Air France 447 Rio de Janeiro to Paris.
Breaking in Bernoulli only is what van Biesen do and as we said is misapplication.
Thank you sir..for ur presentation..
good explanation. will you make a video of how to find the minimum speed for the plane in your explanation to fly?
Why don’t the air molecules at the top of the wing maintain their initial velocity? Why is the time constant?
Thankyou so much!
Glad it helped!
I am surprised that Leonardo da vinchi didn't discovered aerodynamic by the way he had alot of free time
The (two) wings of an aeroplane are both 6.00m long with an average width of 1.50m. The wind is designed so that the average air-path length over the upper surface of the wing is 1.80m and 1.60m for the lower surface. Estimate the lift when the wings are horizontal and the air speed is 150km/hr. (Ignore any difference in potential pressure due to the thickness of the wing.) any solutions for this question
+Roshaan Prabakaran Assume the v (of the air) below the wing to be 41.7 m/sec (150 km/hr) and assume the v above the wing to be (1.8/1.6) x 41.7 m/sec = 46.9 m/sec The difference in pressure will be (1/2) * density * 46.9^2 - (1/2) * density * 41.7^2 The lift force will be F = delta P * Area of the wing.
Thank You
Good explanation
My mind is blown.
Videos on engineering mechanics
Are the lift forces then equivalent torques (w/respect to the plane’s center of mass) on both wings?
Thank you Michel great vid.. I heard equal time is wrong theory? then what is correct how lift is really generated?
Fluid flow can become very complicated very quickly when diverting from the most basic cases. That said the Bernoulli equation approach is a good approximation.
Yup, equal time theory is wrong. There is a reduction in pressure above the wing, but it is much faster than equal transit, resulting in more air further above the wing sucked downward. The curved shape further helps direct the air downward due to the Coandă effect. If the mass of the air going DOWN does not equal the mass of the plane, it can't stay in the air. The angle of attack is also vital. Many good resources out there on this.
Thank you @penguin rick
Great knowledge of the equation. Totally incorrect explanation of lift. NASA also simplify their explanation without getting it wrong (they even helpfully call this out as incorrect theory of lift, and definitely not a good approximation).
They also explain that it gets complicated. Check their site. Fluids simply do not behave over a wing form as you describe. Check any wind tunnel experiment. It is not a simplification, just an incorrect observation, and application of the equation.
Like 'simplifying' gravity by saying that objects fall upward to explain wind resistance and terminal velocity.
Why include this as an example? Why not correct the video, and explain that fluids do not actually meet back up at the trailing edge of the wing?
Why not explain that this is totally incorrect, but that for some reason you have assumed that fluids behave differently way over a wing to how they actually do? That would save some confusion.
@@pngnrick Nothing at all to do with the Coanda effect. Not sure why everyone seems to bring up viscosity and the Coanda effect when an entirely inviscid treatment of airflow still generates lift.
Why this explanation is wrong:
1. The air above and under the airfoil don’t meet at the same time.
2. Bernoulli’s equation doesn’t apply for gas ! (The fluid should be incompressible)
3. You can’t use Bernoulli’s equations for two streamlines !
It’s only valid along a single streamline. Once the the stream line has split, the air above and the air under the airfoil doesn’t have the same total energy, thus the sum of pressures is different ! You can’t compare the two with the standard Bernoulli’s equation.
This professor is a great UA-cam teacher, but this video is misleading and should be removed so others won’t be confused.
We could say that almost every example in every physics book is wrong since they don't take everything into account. No wind resistance for projectile motion, not energy loss in collisions, no friction motion examples, etc. That said, Bernoulli's principle can be applied here, knowing that it doesn't explain the full example. This example can be found in many physics text books, just like projectile motion problems without wind resistance.
Thank you for the videos
Absolutly amazing, but I think that in aviation, we use the nautical mile (1852 m)
You are correct.
Thanks you sir. Very interesting video.
May I know while calculating the final force (which is F=Delta*P*A) why area of only wings(A=50+50) is considered ? Why you neglected the area of hull of plane ?
The lift is primarily provided by the wings, (and in particular, how the wings are shaped).
PLEASE use metric units, thank you.
It's hard to understand with ur metric units
Very Cool!!!
Glad you liked it. 🙂
Question: bernaulli equation can be apllied to 2 points which they are at the same streamline. In this problem, 1 and 2 is not at the same streamline. Any explanations ? Do I know wrong ?
Technically, you are correct, that Bernoulli's equation was developed by assuming the same streamline in a pipe. However it turns out that it offers a very good approximation for an example like this where the pressure (relative to still air) can be calculated above the wing and the pressure (relative to still air) can be calculated below the wing, and since they have the same reference, we can calculate the difference in pressure.
The correct way to look at is that the two parcels of air ahead of the wing have t5he same energy level to start. It turns out that as the travel above and below that Bernoulli CAN be used because of this.
.
He is WRONG about the equal transit time.
..
See Charles Eastlake. I talked with Prof Eastlake about this very thing
..
Charles Eastlake's paper may seem a bit heavy and, I think, rather disorganized in the beginning, but it is clear toward the end that he describes that there are two ways to CALCULATE the magnitude of the lift force, not describe the phenomenon completely. One method (using Bernoulli's Equation) calculates the lift force directly by examining the velocities and pressures around the wing to get the upward pressure difference that the air has on the wing. The other (using the mass of the air continually deflected/turned downward) calculates the Equal and Opposite force on the air (downwash). Each is a distinct part of the whole story, not two equivalent ways of describing the phenomenon fully..:
wiki.mattrude.com/images/4/44/Bernoulli_Newton_Lift.pdf
the explantion for the difference in pressure of the airfoil is not true!!!
It is an approximation used in most physics text books when applying Bernoulli's equation. At the introductory level we use a lot of approximations, such as ignoring wind resistance when introducing projectile motion. The real life scenarios of the aerodynamics around objects such as wings, cars, etc. are so complicated that wind tunnels are used rather than trying to calculate it.
sir micheal van your all 7 videos are so helpfull for me with all concept are there but i need bernoulli eqution derivation if possible and within one or two days ..
You can find that video here: Physics: Fluid Dynamics: Fluid Flow (1.6 of 7) Bernoulli's Equation Derived
thanku so muchhhh😊😊😊👍
Why do the molecules of air traveling over the wing have to reach the other end of the wing at the exact same time as the molecules of air traveling below the wing?
Once air is flowing over the wings and steady state is reached, air will be moving across the upper and across the lower surfaces. If it took longer for the air to travel past the wing on one side, then air would bunch up on one side, which is not possible.
@@MichelvanBiezen But it's not a confined space. Air can also move laterally over the wing (certain old soviet jets had "walls" on wings). And bunched up air is possible to some degree since it's compressable, whereas in the examples with water, the area was confined and water is virtually incompressible.
That's my reasoning at least.
Thank you so much for these videos btw!
Excellent explanations.
@@MichelvanBiezen That is a valid argument that the velocity at the trailing edge must be equal, but that doesn't mean the same air molecules must meet.
I don't understand why the air above the wing has to be faster ( I know it travels more space in the same amount of time but this is only the definition of 'faster' )... Why doesn't it simply go at the same speed?
If you travel farther in the same amount of time, then you are traveling faster. (Note that this is a simplified, but workable, representation of what actually takes place above the airplane wing).
But this (equal transit time) has been proven wrong a long time ago. See for yourself:
The non equal transit time (only) observation video by Holger Babinsky, Cambridge University.
Note this is in a wind tunnel which restricts the downwash and the wing is at an extreme angle of Attack to make the effect easier to see.
ua-cam.com/video/UqBmdZ-BNig/v-deo.html Narrated, no title at start Cambridge Univ.
I feel more confident on my aerospace test now.
+Say Holder You shouldn't because this is an incorrect application of Bernoulli's Equation.
Excellent lecture
made me laugh when i saw the error in the text at the bottom of the video (physicist consider WATER a fluid. well yes.. i think most people do.) great videos for bernoullis equation. thanks for making them
I thought Bernoulli‘s equation is for a single streamline you’re showing two stream lines?
Essentially it is used twice, once on the upper side and once on the lower side in order to compare the pressure difference. But since we are looking for the pressure difference, it can be done with a single equation.
@@MichelvanBiezen so the airplane air speed is 300 miles/hour?
The assumption in this problem is that the speed of the air rushing over the wing is different at the top and at the bottom of the wing. The speed of the plane is not given in this problem.
u cannot apply bernoulli's Eq btw to different currant line
Dear michel van Beizen, how can i calculate the speed of over and botton of the airfoil?
could you kindly explain.
thank you.
That would be very difficult to do. Typically this is measure in a wind tunnel, or using instruments on the flight.
@@MichelvanBiezen thank you for Quick response
There are computer simulations that are very good these days called CFD Computational fluid dynamics that do the calculations based on the known fluid properties.
@@Observ45er The simulations are very good if you know the physics behind them (particularly the assumptions). But also very easy for beginners (like me) to use a solver that makes an assumption that is completely false for the situation.
Yea. That's a problem, but I'm talking about what professionals use. I don't know what 'consumer' versions do.
im a 14 year old boy coming from Croatia that is really interested in aerodynamic
i understood everything till 2:31
than i realized oh here we go again with this shit long formulas
It is not because the air has to travel longer distance in the same amount of time. Please stop perpetuating wrong information. There are modern aircraft with longer underside and shorter upper side, and it still flys. Please do your research.
Thank you for the help
Is it A 100 m sqr? Not 50 m sqr?
It is 50 m^2 for each wing. (x 2 because there are 2 wings)
Is there a set of videos for Bernoulli's Equation where you factor in the friction, the work done on the pipes from the fluid, the work done by a pump and all that...
Could someone direct me to a video explaining how to solve this for example:
A pump takes water from a large reservoir and delivers it to the bottom of an open elevated tank 25ft above the reservoir surface through a 3inch inside diameter pipe. The pump is located 10ft below the water surface, and the water level in the tank is kept constant at 160ft above the reservoir surface. The pump delivers 150 gal/min. If the total energy loss due to friction in the piping system is 35lbf-ft/lb, calculate the pump power (in hp) required for this duty if overall efficiency of the pump is 55%.
aag
You don't need Bernoulli's equation for that problem. It is a work/energy type of problem (look in the work/energy playlist)
If you ignore the work needed to give the water kinetic energy you would solve it like this:
W + PEo +KEo = PEf + KEf + energy lost
For 1 second with 2.5 gallons pumped every second and 1 gallon having the weigth of 8.345 lbs and since the pump is only 55% efficient this becomes:
0.55*W + 0 + 0 = (2.5) * (8.345) * (195) + 0 + (2.5) * (8.345) * (35)
W = 9770 ft lbs
P = 9770 ft lbs/sec = 17.76 hp
Now when you also take into account the kinetic energy (KE)
Then KEf = (1/2) m v^2 = (1/2) * (2.5*8.345 / 32) * (2.5/7.48) /(pi*(1/4)^2/4) =
15.11 ft lbs which would only add 0.05 hp with the 55% efficiency
Thank you very much Sir, i will check the Playlist you've mentioned.
@@MichelvanBiezen Worth noting that the formula you effectively came up with is often called the extended or engineering Bernoulli equation. But obviously it's totally valid to look at it as conservation of energy directly as you did.
Thank you very much!
Life saver
Isn't the first part wrong??
He just assumed the airspeed at the top and down are the same and that is wrong!
NO, he did not assume both speeds are the same. Listen and read carefully. One is 400 and the other is 300 m/hr. What he did assume is that air molecules at the upper side of the wing reach the rear of the wing at the same time as the air molecules of the stream below the wing.
And that assumption is false, but he doesn't know why. The upper and lower airs don't meet, but you can measure the speeds and integrate over the area of the wing to get the actual pressure difference.
..
While this is using two different streamlines, this is valid because it is a uniform flow and they both have a common source far ahead of the wing.
Pretty good