i am an engineering student and this is the best explanation of flow i have seen...doesnt go deep enough for what I need but it got me on the right path....
@Amr Elsayed Mohammed Abdulhady: In the barrel example, you are correct that V1 does not precisely equal zero. However, depending upon the exact dimensions of the barrel, including the dimensions of the hole at the bottom, if the hole's diameter is substantially smaller than the barrel diameter, the velocity will be negligible. For example, if the barrel's diameter is 50cm, and the hole's diameter is 5mm, the magnitude of the velocity of the stream coming out the bottom is 10,000x as great as the speed with which the water line is descending in the barrel. If we were using mathematical analysis to place a man on the moon, or build a nuclear power plant, then of course we wouldn't ignore 1 part in 10,000. However, if we are measuring the speed with which water shoots out of a barrel, it's highly unlikely such a small error will be even measurable, let alone significant.
This was so helpful. Your explanations of the derivations are spot on. Thank you! One thing I do that helps me learn: once everything is derived, ask 'If I increased A2 by a factor of 2, what changes and by how much?'. Do it for several of the variables and I feel like it drives the concept home.
That looks like the equation for energy conservation in Thermodynamics with Work and Heat transfer remains unchanged and everything divided by Volume. So if I multiply by volume to everything in the Bernoulli's equation, P1*V1 would just be the flow energy, I think.
I would like to see a video regarding vestibular neurosis. The reasearch I have been able to find really only identifies physical therapy as a means of preventing the vertigo that develops with this condition. I feel that there must be more information on this condition and would really like you to explain to us what this condition is all about. Thank you in advance.
What if we have a huge barrel with a pipe at the bottom with the same diameter as the opening on top? If we then assume that P is equal on both sides of the equation (Patm), this pressure cancels out on both sides.. Then we would have 1/2ρv1^2 + ρgy1 = 1/2 ρv2^2 + ρgy2.. The potential energy on the left side will be much higher than on the right (0) and thus the kinetic energy on the right side would have to be higher.. But that would mean that the velocity at the hole (which has the same diameter as the top opening of the barrel in this case) is higher.. But that's not possible due to the continuity equation saying that with same cross-sectional areas we have to have the same velocity?!
If bernoulli equation states that velocity and internal pressure have inverse relationships and if the speed increases, then the pressure inside the tube has decreased. Also, Venturi effect states that when there is a constriction in a pipe your pressure is lower. my question is how does this apply to blood pressure? when there is vasoconstriction you have an increase in blood pressure. this is bugging me a lot, I'm studying for the MCAT and it seemes as if physics is going agains what i have already learned in the circulatory system.
Firstly, thanks for making the video, but I am still unsure about what I've seen. So in the barrel example, first you ignored v1 (velocity "in"), because we ASSUME the area of the top is much much larger than the area out, but this assumption is artificial, so practically, in real world situations we would not be able to use such an assumption I assume? So we wouldn't be able to actually find v2 (velocity "out")? Additionally, why is it that we do not include the pressure exerted by the column of water above point #2 (where the water is flowing out), and only consider the atmospheric pressure? Shouldn't the pressure exerted by the fluid be taken into consideration?
It is not the height as the radius of the tube, it is about the y axis reference. You can put the 0 of the y axis as the reference wherever you want to simplify the problem. So if in this one, if we put the reference in the middle of the tube, the stream line will go in a straight line, at the same height through the tube, thus cancelling out the two height terms of the equation
as a main principle of physics problem solving, the answer will be the same independant of where you put the reference. But, depending where you put it, this allow you to cancel some terms that will end up being equal to zero. To put into an analogy, imagine yourself climbing some stairs. If you start at the middle, and go until the end, it will be the same distance walked as if you started in the bottom and went to the middle. The difference is that, if the stairs has 2m height, in the first scenario you would calculate as 2m - 1m = 1m, and in the second one the math would be 1m - 0m = 1m. Doesnt look like a great difference, but when you are working with complex equations, you can avoid a lot of algebra by setting the right reference in the problem.
Hi, would you help me? I have had a troubling wondering on one airflow situation for a long time, my question is perhaps best shown by an example: A straight horizontal wind flows parallel and just over a flat roof of building . This wind DOES NOT hit the side of the building, the wind is just flowing very close above the roof. Will the roof feel an uplift from the pressure differential ? , so my question is about an airflow running just above a surface. Your help is so appreciated. Thanks
+Kaledius in my opinion the roof will experience an uplift force,as above the roof velocity of fluid(air) is high so presure will be less (by bernaulli equation) and at bottom pressure will be more so pressure difference is generated just take your copy and blow as hard as you can over the pages u will be surprised that pages will rise up...same principle... hope it helped allah hafiz
+LONE FALCON This conclusion is incorrect and is based on the all too popular misconception that Bernoulli's Principle is that "Fast-air-has-lower-pressure". The pressure along a straight (constant speed) flow is the same inside the flow as it is outside the flow. This is well known and can be measured. ... As you describe it, the roof will experience NO LIFT force. ... You are stating a misconception about Bernoulli's Equation. The pressure will not be lower above the roof, because velocity alone does not determine pressure. Bernoulli speaks of Velocity *_CHANGES_* along a path, not absolute speeds. Look carefully at the way the Equation is used along any of the pipes. This is CHANGES along one path of flow, NOT comparing two completely different flows.
because bernoulli equation is for invisid flow, if you consider p2=r0*g*h , systems seems like a closed one, so we have to consider p2=atm which makes a invisid flow
i am an engineering student and this is the best explanation of flow i have seen...doesnt go deep enough for what I need but it got me on the right path....
@Amr Elsayed Mohammed Abdulhady: In the barrel example, you are correct that V1 does not precisely equal zero. However, depending upon the exact dimensions of the barrel, including the dimensions of the hole at the bottom, if the hole's diameter is substantially smaller than the barrel diameter, the velocity will be negligible. For example, if the barrel's diameter is 50cm, and the hole's diameter is 5mm, the magnitude of the velocity of the stream coming out the bottom is 10,000x as great as the speed with which the water line is descending in the barrel. If we were using mathematical analysis to place a man on the moon, or build a nuclear power plant, then of course we wouldn't ignore 1 part in 10,000. However, if we are measuring the speed with which water shoots out of a barrel, it's highly unlikely such a small error will be even measurable, let alone significant.
This was so helpful. Your explanations of the derivations are spot on. Thank you!
One thing I do that helps me learn: once everything is derived, ask 'If I increased A2 by a factor of 2, what changes and by how much?'. Do it for several of the variables and I feel like it drives the concept home.
That looks like the equation for energy conservation in Thermodynamics with Work and Heat transfer remains unchanged and everything divided by Volume. So if I multiply by volume to everything in the Bernoulli's equation, P1*V1 would just be the flow energy, I think.
what a beautiful way to explain this equation, thanks!!
Outstanding presentation and appreciated!
I would like to see a video regarding vestibular neurosis. The reasearch I have been able to find really only identifies physical therapy as a means of preventing the vertigo that develops with this condition. I feel that there must be more information on this condition and would really like you to explain to us what this condition is all about.
Thank you in advance.
What if we have a huge barrel with a pipe at the bottom with the same diameter as the opening on top? If we then assume that P is equal on both sides of the equation (Patm), this pressure cancels out on both sides.. Then we would have 1/2ρv1^2 + ρgy1 = 1/2 ρv2^2 + ρgy2.. The potential energy on the left side will be much higher than on the right (0) and thus the kinetic energy on the right side would have to be higher.. But that would mean that the velocity at the hole (which has the same diameter as the top opening of the barrel in this case) is higher.. But that's not possible due to the continuity equation saying that with same cross-sectional areas we have to have the same velocity?!
Can you please explain flow meter calculations. I hopefully get on that (single phase and multiphase calculations. Thank you for your information
why omit Patm tong when barometric pressure is equal to 1 and above altitude difference
If bernoulli equation states that velocity and internal pressure have inverse relationships and if the speed increases, then the pressure inside the tube has decreased. Also, Venturi effect states that when there is a constriction in a pipe your pressure is lower. my question is how does this apply to blood pressure? when there is vasoconstriction you have an increase in blood pressure. this is bugging me a lot, I'm studying for the MCAT and it seemes as if physics is going agains what i have already learned in the circulatory system.
The increase in pressure occurs ahead of the constriction, no? So it is still lower within the constriction?
Firstly, thanks for making the video, but I am still unsure about what I've seen. So in the barrel example, first you ignored v1 (velocity "in"), because we ASSUME the area of the top is much much larger than the area out, but this assumption is artificial, so practically, in real world situations we would not be able to use such an assumption I assume? So we wouldn't be able to actually find v2 (velocity "out")?
Additionally, why is it that we do not include the pressure exerted by the column of water above point #2 (where the water is flowing out), and only consider the atmospheric pressure? Shouldn't the pressure exerted by the fluid be taken into consideration?
Simply and perfectly explained! Thank you!
This is great. Thanks a lot for these videos!
why is y1 = y2 ?Aren't we were to consider the height at the middle where the diameter is least?If yes then y1 will not be the same as y2....
It is not the height as the radius of the tube, it is about the y axis reference. You can put the 0 of the y axis as the reference wherever you want to simplify the problem. So if in this one, if we put the reference in the middle of the tube, the stream line will go in a straight line, at the same height through the tube, thus cancelling out the two height terms of the equation
as a main principle of physics problem solving, the answer will be the same independant of where you put the reference. But, depending where you put it, this allow you to cancel some terms that will end up being equal to zero. To put into an analogy, imagine yourself climbing some stairs. If you start at the middle, and go until the end, it will be the same distance walked as if you started in the bottom and went to the middle. The difference is that, if the stairs has 2m height, in the first scenario you would calculate as 2m - 1m = 1m, and in the second one the math would be 1m - 0m = 1m. Doesnt look like a great difference, but when you are working with complex equations, you can avoid a lot of algebra by setting the right reference in the problem.
This was very helpful, thank you for posting.
Thank you very much for your great video.
why P2 is equal to atmosphere pressure?
I think cause p2 is just the moment it leaves the drum
I don't get how the height is the same.
I dont understand about how y1 and y2 equal
could you help me ?
It's the pressure head, velocity head and elevation head.
Very descriptive material
Hi, would you help me? I have had a troubling wondering on one airflow situation for a long time, my question is perhaps best shown by an example:
A straight horizontal wind flows parallel and just over a flat roof of building . This wind DOES NOT hit the side of the building, the wind is just flowing very close above the roof. Will the roof feel an uplift from the pressure differential ? , so my question is about an airflow running just above a surface. Your help is so appreciated. Thanks
+Kaledius in my opinion the roof will experience an uplift force,as above the roof velocity of fluid(air) is high so presure will be less (by bernaulli equation) and at bottom pressure will be more so pressure difference is generated just take your copy and blow as hard as you can over the pages u will be surprised that pages will rise up...same principle...
hope it helped
allah hafiz
+LONE FALCON This conclusion is incorrect and is based on the all too popular misconception that Bernoulli's Principle is that "Fast-air-has-lower-pressure". The pressure along a straight (constant speed) flow is the same inside the flow as it is outside the flow. This is well known and can be measured.
...
As you describe it, the roof will experience NO LIFT force.
...
You are stating a misconception about Bernoulli's Equation. The pressure will not be lower above the roof, because velocity alone does not determine pressure. Bernoulli speaks of Velocity *_CHANGES_* along a path, not absolute speeds. Look carefully at the way the Equation is used along any of the pipes. This is CHANGES along one path of flow, NOT comparing two completely different flows.
Very nice, well done, thank you
why is the height of hole 2 zero?
great
video!
Why cant we use P2=ro*g*h. and why P2=atm pressure.
because bernoulli equation is for invisid flow, if you consider p2=r0*g*h , systems seems like a closed one, so we have to consider p2=atm which makes a invisid flow
Manoj Kumar thanks a lot man
But what is the relation between a closed system and inviscid fluid?
Thank you
Dude thank you!
Great vid !! thanks
thanks really helpful
thank you for the explanation
thanks alot
🙏
I'm a curious kitty
oiuk
good thank you