Physics 34 Fluid Dynamics (5 of 7) Bernoulli's Equation
HTML-код
- Опубликовано: 9 сен 2024
- Visit ilectureonline.com for more math and science lectures!
In this video I will show you how to use Bernoulli's equation to find Visit ilectureonline.com for more math and science lectures!
In this video I will show you how to use Bernoulli's equation to find the force needed to for a hurricane to rip the roof off a house.
Next video in this series can be seen at:
• Physics 34 Fluid Dyna...
Amazing. I've never seen this principal explained so eloquently and so perfectly. You deserve a medal. Thank you.
Jellevanderidder,
Thank you for the feedback.
Actually my name is from Belgium. I used to live in a small town 4 km from the border with the Netherlands.
Groetjes van America.
+Michel van Biezen heel erg bedankt en fijne dag nog ; )
Dat was niet verwacht! Hartelijk bedankt voor alle video's!
This example is a bit more confusing because, unlike in previous examples, the fluir (air) in the house is not connected to the air outside the house - and yet, the outside air can exert such a large force on the surface area of the roof.
Gotta say your lectures are the best I've found covering this material in Fluids, thanks! 👌
Bernoulli's Principle is the single principle that helps explain how heavier-than-air objects can fly. Bernoulli's Principle states that faster moving air has low air pressure and slower moving air has high air pressure.
This is a great lecture. I'm just was watching out of curiosity and now I'm more interested in studying this.
Technically Bernoulli's principle cannot even be applied to this particular problem. Bernoulli's equation is a form of the general principle of the conservation of energy. Since the two systems (inside the house and above the roof) are in no way connected in the example, the energy is not conserved between the two systems. Bernoulli's principle can only be applied through a streamline where conservation of energy is valid.
While the final result is correct, the application of Bernoulli's principle is theoretically and physically wrong. If the inside of my house was sealed (air tight) and I would have a compressor installed that increases the pressure to 3ATM inside the house, then suddenly the outside pressure would also increase, because your equation states that p1 = p2 - 1/2*rho*v^2. However, the outside pressure should be completely independent of the pressure inside, which makes the equation wrong.
yes and it can only be applied at 2 points along a streamline. Great videos but this is fundamentally flawed
But you can think of the wind blowing from the left side to the right side and compare a point 1 from the right side, and then a point 2 on the roof which will be the same as point 1 and then a point 3 inside the room, and think of point 3 is on the streamline of point 1 but with same height and 0 velocity. So comparing point 1 with point 3 is basically the same as comparing point 2 with point 3.
I'm just starting to learn this now. Could Bernoulli's principle be applied assuming that the windows are blown off and wind has complete access to the inside of the house? Thanks
Yet he proved mathematically the difference of force by using Bernoulli equation. 😎
Actually what one should do is compere point 1 that is outside the house on the roof with point 2 that is also outside, on the same streamline but so far away that is not in the storm anymore and the speed is 0. Then you can use bernoullise and the point 2 is getting same values as a point inside the house. You example where the pressure inside the house is 3ATM doesn’t work since such a point doesn’t exist on the streamline.
I cannot thank Michael van Biezen enough for his videos. He makes learning so easy and fun. He explains everything but somehow makes it seem simple!
U r connecting the day today life with physics thank u sir, u r great, i don't know how to show my gratitude for ur help here in srilanka teacher are teach bernoulli therom for 1 month but u cleverly and clearly explain within 10 minutes, im soo happy to learn from u sir 💙💜💙💜💙💜💙💜
i believe my god jesus show ur channel, and im really blessed by my god
You are a hero! Thank you so much for posting these amazing lectures online. I wish you were my professor!
Veronica,
Thanks for the feedback. Much appreciated.
Actually , Bernoulli's equation is only applied along streamline , so u cant apply Bernoulli equation between outside and inside the house , the term P2 u just calculated is called the stagnation pressure ,so if u put ur hand against the airspeed u will feel the pressure p2 , so the pressure inside the house stay atmospheric pressure (101325pa for ex ) and above the house roof the static pressure(p1)= Pinfinty-0.5*rho*v^2 , u can put Pinfinty=Patm as assumption,hence the house roof will experience a pressure= patm-p1, u will notice that u will get the same answer cuz u put Pinfinty=Patm,but the concept is not right ,btw the outside house walls will experience a stagnation pressure (p2) .
TBNR mido Couldn't you consider the air going over the house as a streamline?
Max Lightning air stream coming from far field point (1)where pressure =Patm and continue flowing over house where u can consider point (2) so u can consider line connecting those two points as stream line then u can simply apply BERNOULLI equation to calculate the pressure over the Roof ,,,But u cant apply the principle directly between inside the house and Roof top simply cuz there is no air stream passing from inside to upward penetrating the roof.
+TBNR mido was just about to ask him this, thanks for explaining it
Thanks for giving me peace of mind ! before watching the video, this's exactly the way I thought it should be solved but then I saw the video's solution and got me scratching my head ... I just hate how a lot of resources regarding Bernoulli's equation completely forget about the fact that this equation is only valid for two points in the same streamline and try to apply it incorrectly to all sorts of things, such as in the case of lift in airplanes
One of the best. Hats off to you. Well explained. Your effort is incredible and highly apprecitaed
There is absolutely no lifting force on the roof of a house under high wind loads. The force of the prevailing winds will apply positive pressure to the left side of the roof and the trusses will distribute that force as a clockwise rotational torque to the right side, not lifting it up. High pressure will build on the left side of your house diagram and low pressure pockets will develop on the right. The sum of all force vectors is more sideways left to right, preferring to shear the house from it's foundation.
Thank you sir ! Practical example with excellent explanation.
GOAT Michel!
Thank you. Glad you like our videos.
You are genuinely the best!!!
Thank you. Glad you found our videos. 🙂
Fascinating! I wonder what would be the drop in force if a builder would install one or more sort of trap doors that would "relieve" the inside positive pressure saving the roof from totally lifting off? Just a thought.
Fascinating example. Thanks for these videos.
Glad you like them! 🙂
Thank you very much! Without the help from this video, I'd have struggled to complete my physics assignment. Greatly Helpful!
really good lessons! thank you
saludos from argentina!
+Elizabeth Perez
Great to hear from people around the world!
Thank you very much Dr. Please correct me if i'm wrong, because I got confused, the atm pressure is equal to 101325 pa which is too much higher than 1288 pa, not a slight difference as you explained. (101325-1288=100037 pa). I'm very grateful for your lectures.
Inside the house, the air pressure is atomospheric pressure = 101,325 Pa. The pressure outside the house is less due to the high wind velocity. The difference in pressure is 1288 Pa.
If you draw a window on both sides so you have your streamline then you should be fine applying Bernoulli
Berrnoulli’s equation was basically modelled and established on a single streamline, someone using this equation should consider this streamline assumption.
This example is indeed an oversimplification. We have a video on that in the review videos. But it does offer us a good approximation.
Sir, I found your videos to be very comprehensive & effective. That's why I have a request. Please, make a video on the derivation of Sabine's formula. It will be very helpful.
That is an interesting formula indeed. We added it to the list of requests. (Don't know when we can get to it).
Hi,
Thank you for the videos, they are amazing! just a quick question about the example 5. Shouldn't you use that Bernoulli equation through a streamline? The calculation seems alright but the explanation is questionable i think. How can you connect the points inside the house and point on the roof? i think it would be clearer to assume a point far from the house with zero velocity at ATM pressure. Use it as the first point (1) and and the second point should be on the roof (2). Assuming the air inside the house is with zero velocity at ATM pressure, the calculation will give the same result but at least it will prevent further miss understandings.
Ali,
Good suggestion. However during a hurricane (or very strong winds), the air will be moving quickly anywhere outside the house, thus it would be difficult to make such a comparison. That is why I picked a point inside the house. Bernoulli's equation works as well in this situation (and others) where there is no direct flow relationship.
Hi Michel,
Thank you for your answer. I didn't say it doesn't work, it works because the air inside the house is at 1 ATM with zero velocity (which is same with the assumption I wrote earlier. A point far from the house). the way you wrote that Bernoulli equation doesn't make sense to me. Bernoulli equation can be derived from the conservation of energy and it says that sum of all mechanical energy is constant on any point in streamline.
If we use your calculations, the pressure inside the house does not effect the pressure difference which is used later to calculate the force on the roof. Lets assume that the pressure inside the house is 2 ATM, what will be the answer? From you calculation, the answer will be same. if the answer is same that means the pressure outside the house is changed which should not be related with the pressure inside the house.
Thank you so much ! These are very helpful
interesting and fantastic way to explain.. thanks alot
Glad you liked it. 🙂
You are great. Great example
Amazing Sir. Thanking you from Bangladesh :)
Welcome to the channel!
I am sri lankan 🇱🇰. This is very usefull topic for us.Thank you very much sir.
Hi and welcome to the channel!
Ya ya it would come final
Thank you so much sir! super explanation
Best explanation ever
Are we wrong to apply Bernoulli's principle here? I am not arguing this method but am curious if this is correct. Thank you.
Technically, it is not a "pure" example of Bernoulli's principle. But it will give you the correct answer to the problem. This is a typical example found in many text books under Bernoulli's principle.
You are my hero
Correction @ 6:10
F = (1288 N/m^2)(200 m^2) = 257,600 N
What if you have 2 or 3 incoming and varying flow rates going to different pipes and meeting at one point?
1. A1V1
2. A2V2
3. A3V3
Is the total output flow rate (total AV) equal to = A1V1 + A2V2 + A3V3?
Yes, because the sum of the volumes per unit time flowing through each of the smaller pipes equals the volume per unit time flowing through the big pipe.
Thank you sir.
you are welcome.
Thank you so so much
Just out of curiosity, wouldn't the air flowing across the left face of the roof create a high pressure against that side of the roof, and a vacuum on the right side of the roof? I don't think it is correct to assume that the gain of pressure on the left side of the roof is equal to the loss of pressure on the right side, but I am not complete sure. If I am correct I believe the conservation of energy cannot be applied to this problem. In regards to that I do not believe that Bernoulli's equation can be applied either because it is a derivation of the conservation of energy.
This is off course a VERY oversimplified example. When it comes to fluid flow there are many other things we have to consider. But it does offer a good approximation of what happens.
So, say that a tornado is near the house. The tornado is strong enough to rip the roof off. If I open windows in the house, so now there is air flowing through the house, will that equalize the pressure enough to lessen the force pushing up against the roof?
+Zachary Bertrand A tornado is a very different "beast". Opening the windows will not save your house.
I pray god for sir.Michael van beizen
Thank you...🇺🇸 😎👍☕
You are welcome. 🙂
Wow, works for me!!
How come the difference in the pressure times the area gives you the force instead of just the pressure inside the house times the area giving you the force?
Because there is pressure inside the house pushing up against the roof the and there is pressure outside the house pushing down against the roof. In this example the pressure inside the house is greater than the pressure outside the house.
@@MichelvanBiezen ahhh, i see. Thank you!
it is working perfectly now :) thank you .... and about the corroborator it is okay .... I could know how ... because i am designing a familiar system :) is there examples for Cavitation ... have a nice day respects for you Majesty :) ... is there any contact info. if I am facing a case I need a help with ?
Thanx a lot sir
Can you please clarify: F = P /A..... I see you are putting F = PA?
P = F/A F = PA
I learned that Bernoulli's equation is only applicable along a pathline. Why can we still use the equation here?
It will not be exact, but it gives a good approximation. (A good example of Bernoulli;s principle).
@@MichelvanBiezen Ah I see, thank you very much for your reply!
i thought you use Bernoulli's formula for a the same steamline??? the top and bottom of the roof aren't on the same streamline????
Doesn't have to be.
Hi, I have asked you earlier, is this perhaps the video your are referring to for my answer? The house example here in your video is a peaked roof one. I was asking about a flat roof house where the wind flows just above the flat roof and parallel to the flat roof without hitting the sidewall, so would the roof feel an uplift? or is it important that the airflow of the wind finds an OBSTRUCTION like a side wall or a peaked roof to make the roof lift?
Only considering the basic principles, the shape of the roof doesn't matter. The roof would still feel the uplift, just like in the example.
When you get into obstructions, turbulence created by them, etc. the problem becomes very complicated very quickly and requires a much more advanced analysis, way beyond the scope of introductory physics.
Thank you Michel!, please bear with me, I think I see the point but, would the flat roof feel an uplift even if the horizontal wind portion actually ONLY flows just slightly above the roof surface and never originally hit the side wall of building as a constriction?
I mean, in ordinary cases, the wind say flows from left to right on a flat roofed house, the wind first meets the house´s left SIDEWALL which is a constriction which makes the wind flow and deflect up faster for(conservation of energy and Bernoulli) and the wind now flows faster ie. -faster than before it hit the left wall- so the wind now flows faster yes above the flat roof which creates the true lower pressure and thus uplift.
My question was , would the uplift still exist even if the wind stream portion didnt hit the left sidewall(or any constriction) at all? Instead, this wind stream portion flowed as earlier from left to right as earlier BUT NOW at a height from ground as to never hit any sidewall or constriction. Thus the the wind just very simple flowed just exactly past and above the flat roof from start to finish without hitting any constriction or the side wall.
would there be still an uplift on the flat roof even if there was no constriction to the wind flow?
Please, you are perhaps my only hope in understanding this. I really appreciate your help as I find no help at all in this example and if I find anything, people are arguing with contradictions. Thanks a lot for you Time.
K.A
leviterande
As strange as it may seem. The answer is still yes.
The reduced pressure is simply a result of the speed of the flow of air.
The reduction in pressure = (1/2) * density * v**2
And I also suspected and was most sure that the uplift would still be there!, But, the problem is that, some educated people like on
physicsforums website, are saying that unless there is a constriction to airflow, there would be no uplift, and these are educated people... Thats why I am confused...
If we assume that the fluid density is high, would you take the height into account or would assume that diffrence in height is very small that you would cancel them out_ Thanks
With liquids the height in the fluid is important and we keep the term in Bernoulli's equation. But with air, a few meters makes a very small difference.
Sir, Bernolli’s equation looks almost exact to conservation of energy except for the P(2) what are your thoughts? It doesnt look like energy lost to overcome friction
Also why not build flat roofs?
Absolutely love your videos.
Take a look at this video, it will explain what you just discovered:
Physics: Fluid Dynamics: Bernoulli's & Flow in Pipes (10 of 38) Understanding Bernoulli's Equation
can you explain why bernoulli's equation is valid in that problem? we derived the equation from conservation of energy on a pipe, I personally would never think to use bernoulli's equation here
Bernoulli's equation works for fluids in motion. That includes gases which behave like liquids in this instance.
@@MichelvanBiezen but the air outside and the air inside are not linked, as I see it they are in 2 different "tubes", correct me if I'm wrong
We are comparing the pressure inside the house and the pressure outside (above the roof). The pressure difference is caused by the difference in height (which is minimal with air) and the difference in velocity.
@@MichelvanBiezen OK thank you for the clarification
Brilliant
can someone explain to me why is the pressure inside the house pushing up not down? thank you
+ANDY CHEN The pressure inside a fluid declines with increasing velocity of the fluid. (See Bernoulli's equation).That means the pressure inside the house is higher than the pressure outside above the roof , and thus the force pushes up.
In a fluid, the pressure is pushing all sides of the container. So if the pressure is assumed constant in a room, then it pushes the floor, the walls and the roof as well. If it's easier for you to imagine it that way, it's not the pressure inside the house that pushes the roof up, but the low pressure above the house is sucking it upwards.
thanks for your lecture Mr.Biezen. I dont think the Unit of pressure for the equation P2-P1=1/2*rho*(v2)2 are not matching. pls explain.
See video # 8 in this playlist: PHYSICS 0.5 STANDARD UNITS IN PHYSICS
thanks a bunch bruh
Hey Dr. van Biezen,
Can I use Pascal's principle on this type of problem ?
No, for Pascal's principle, the fluid needs to be in a container.
Just recalled that they are using Pounds in America!
sizwe mbokazi
Yes, we still do. They have been trying to change that but old habits are hard to break.
I am facing a problem with the website ... are you developing it .... ?
is there any example about how much is the amount of the gasoline will be provided to an engine when it starts to work at the corroborator please
+Vladislav Intaniosovic No, nothing like that on our channel.
How we made electricity by siphoning the water 🕉🕉
How we found the next date of earth quake since we know speeds pf tactonic plate and all the inclination of earth and combinations of temperature increasing globally 🕉🕉🕉
I'm not from america so I don't know much about imperal system but isn't lb used to measure mass? So how can you convert force in newtons to lbs?
i pound = 4.448 N
There is a unit of measure called lbf (pound force) and a unit of measure called lbm (pound mass). There is a factor to use to convert lbm to lbf. 1 lbf = 32.174 lbm-ft/sec^2.
❤️❤️🖤
That house needs a ridge line - it's upsets my OCD! j/k
P atm. when do you use it? why does it affect the INSIDE of a house?
When you use Bernoulli's equation you must determine the pressure, velocity, and height of each of the two points in consideration. Since the air inside the house is not moving, we can assume that the air pressure inside the house is equal to atmospheric pressure. The pressure just above the house will be less than atmospheric pressure because it is moving. The decrease in pressure above the house can be determine by delta P = (1/2) x (density of air) x v^2
Now its actually getting C😎😎L
Glad you liked it.
but what about the angle of the roof?
Good question. We ignored the angle of the roof for the sake of simplicity.
what will be the equation if i want to put angle in the equation?
Fluid flow is a very complicated topic. We plan on making videos on this topic when we find the time to do so.
so what will be the angle set to default in the equation you showed in the video? is it when the surface is at 90 degrees? if so can i just time by sin theta ?
How we become consciousness as the natural selection of geometric consequences adapted by first molecules of life 🕉🕉🕉
👌
Sorry Sir ,you said 1mile=1609meters,did'nt you make a mistake about giving the answer as 47,7m/s which means 100 miles=47m/s.
No mistake. 100 miles/hour = 44.7 m/sec (you also have to convert hours to seconds)
Can u sometimes speak Dutch in your videos, say a few words :D
Sir Bernoulli equation applicable for incompressible fluid I think so
1288*200= 257,600
+St Lee You need to go back and carry the calculation from the beginning or you will get a rounding error. (The 1288 is not exact, see the previous calculation)
oh thank you :) I was puzzled, thank you Sir :)
No homo nigga but I love u you explained this shit so nicely
iit legends watching these vedios before exams
Good luck on your exams.
@@MichelvanBiezen 😂
Hi Michel, I had to check back on THIS particular problem because of a Tornado Video I saw. If you have 60 seconds then check out this video. You can see how the Roofs of these Houses are being sucked right off their foundation due to the Wind Over them. It actually is more like an Explosion. What I do wonder is what is all the WHITE color you see when that Low Pressure instantly arrives over a Roof. It may be Water Vapor due to the Lower Pressure and the Humidity surrounding that area. Anyway... I'm watching your Video now.. thank you. Oh, here's the Link to that Tornado... ruclips.net/video/lxdFh8nYMgM/видео.html
Hi Philip. Tornadoes have always fascinated me. They are powerful and their destructive power is almost beyond imagination. Yes the high wind speeds cause the pressure to drop and they almost act like huge vacuum cleaners, and the reduction in pressure pulls the roofs off the houses. Also higher up, the reductionin pressure allows the vapor in the air to condense, showing the white cloud like appearance of the funnel. A few years ago I actually saw one in the Mojave Desert in California. (A very rare sight in California).
@@MichelvanBiezen thanks Michel for your reply! Wow, congrats on having SEEN a real tornado.. What a thrill that must be, from a safe distance. :) I appreciate your response, especially in regards to the Condensation. Be safe. Thanks again. :)
Yes, it was about 5-10 miles away and it was very tall. Luckily in the desert there wasn't much to be destroyed. Tornadoes are very rare in California. I remember that about 41 years ago a tornado touched down in the town next to ours and damaged 18 houses. We drove by to take a look and it was unreal. But nothing like the damage the big tornados cause in the midwest.
@@MichelvanBiezen 41years ago? you must have been 5 years old!! :D but amazing to see something so Natural and still somewhat of a mystery. I do see a tornado in the sink whenever i drain the water. :) Vortex physics is angular velocity at it's best.. :)
Philip, Very flattering, but we are a lot older than you think. 🙂
This question technically wrong.
Yes, fluid flow is very complex, but using Bernoulli's equation is a good approximation. This is what is taught in all college textbooks, untill you reach higher levels of fluid flow. (Like we ignore wind resistance in mechanics, so you could call everthing wrong in a college physics book)
@@MichelvanBiezen This answer is technically true 😂. Thank you so much sir, I am appreciate..
W wW wey
Thank you so much! Your video's have been such a great help. And I like your bow tie.
Did you know the origin of your name is in the Netherlands (Holland)? Greetings from here.