Best lecturer on UA-cam. Currently *Reviewing* for my Engineering exam. Thank you for these videos, they're helping me a ton. And your website's great, too.
Appreciate how this professor teaches/instructs. Wish I had had him for fluid mechanics/dynamics. He teaches “practical” real world problems and not theory. Bet this professor knows how to install sprinkler systems ( this is a complement). Good job professor!
well done MICHEL!!.... by the way, it's amazing the difference in your Videos now. I'm referring to the HIGH DEF and the Brighter Lighting.... Your videos are so CLEAR!!.. Congrats!!.. Currently I'm following your Videos on the LINE INTEGRAL... I've always wanted a clear understanding of that. So far I'm through video #4, your latest... Thanks for all you do!!.. I trust your summer is going well and you're enjoying some Fun Time.... now it's back to more learning.. .. ....
Hi Philip, Yes, we have learned a few tricks of the trait over the years. We really didn't know what we were doing in the beginning. The series on line integrals were fun to make. Believe it or not, we are learning as well. I remember many years ago doing these line integrals without actually knowing what they were or what the represented. That is why we made video # 4 to contrast it to what people think they are. There is some interesting revelations on line integrals coming up later in the series.
Thank you Mr. Biezen. I Favorited this video. Now I just need to figure out how to incorporate 2-90deg bends. It is just like the diagram you have shown here except those two curves are 90 degree elbows. I'm going to watch the rest of your videos.
this man is master didact, i just understood constraint based solvers by watching this (at least i think so😅) thanks for your work if i may suggest something, maybe make the connection to conservation of energy and or kinematics? you have two very nice terms in your equation which correspond to kinetic and potential energy 🤔 Edit: he made the connection in a later video
7:49 but from where does this additional pressure comes from ? If we had 1 bar at point 1 this could get the water in the vertical pipe to approximately 10m, can we now take 7000 times wider pipe and get water to 1 kilometer in vertical pipe and still have 1 bar at the point 1 ? Or in the previous example, if pipe is constant diameter and height difference is 1000 meters and pressure at the point 1 is 2 atmospheres, I bet there would not be any water at the point 2 ??
kya atmospheric pressure sey energy bana saktey hain, tum burnouli equation ko jantey hain ho to heron's fountain and siphon and heron's fountain formula and inverted siphon and siphon and boyle flasks equation ko bhee compare karo hindi language means hindi film dekhtey ho to samajh jaogey
Does the pressure drop of the system matters? let say for the same problem in the video, assuming fluid transfer is via hose and it is given that the hose pressure drop is 0.1 bar (or any value).
Yes, the pressure drop matters, and if the drop is due to frictional drag, then it must be added to Bernoulli's equation. There is a playlist that includes additional terms to Bernoulli's equation: PHYSICS 34.1 BERNOULLI'S EQUATION & FLOW IN PIPES
Hello Michel, thanks for this I really liked it. However, it doesn't seems to work when P1 = P2 = Patm i.e. both ends of the tube are open to the atmosphere. Let's say 1 is higher than 2 and fluid still flow from 1 to 2 and let's say I am trying to determine the velocity at point 2 The Bernoulli equation you get when removing P1 and P2, and multiplying by 2*rho is 0 = 2g(h1-h2) + v1^2 - v2^2 If we input v1 = v2 (A2/A1) we get 0 = 2g(h1-h2) + v2^2 (A2^2/A1^2 -1) V2^2 = 2g(h1-h2)/ (A2^2/A1^2 -1) I have a problem here when A2 = A1 --> A2^2/A1^2 -1 = 0 and I end up dividing by 0. Also I get infinite speed if A1 and A2 are very close. There is definitely an issue somewhere but where is it !?
@@MichelvanBiezen I am trying to solve this problem www.chegg.com/homework-help/questions-and-answers/water-tank-water-level-height-h-exposed-atmospheric-pressure-bottom-tank-nozzle-angle-thet-q18940888 For this, I want to know the speed of the water at the bottom of the tank. I thought this could be solved with Bernoulli's equation by considering a point A at the top of the tank (a large section pipe) and point B at the bottom of the tank (a small section pipe) both are at the atmospheric pressure
This problem is actually closer to what I am after www.chegg.com/homework-help/questions-and-answers/problem-3-water-tank-cross-sectional-area-a1-075-m2-open-atmosphere-outflow-pipe-bottom-cr-q25219741
how about, i have a waterpump which pumps 15 liters/min. with a pressure of 0.35atm. the length of the pipe is 2.5m and the height difference is 1m. i have to split the pipe at 0.5m height into two pipes. the total length will be 3.5m now. the diameter of the tubes is 10mm. how do i calculate the speed of the water at the top? is it just the total surface to double. or does the fact that 2 pipes instead of 1 pipe to go up that last 0.5m also has effect?
Assuming we can ignore the effect of viscosity and the effect at the junction, it is simply the total height difference (which causes a shift in pressure) and the total cross sectional area (which has now doubled and thus the velocity is half at the top.)
I have a problem with a bent in the pipe. When I checked online for tutorials, most problems with bends were solved by Bernoulli's principle assuming that the inlet and outlet heights are equal. And then forces were resolved at x and y directions at the bend. Can you explain why it is assumed? Is this method correct?
That is done for simplicity. Just like we ignore wind resistance in projectile motion, to make it easier. But with more advanced fluid motion problem, the loss due to bends in a pipe are included. We plan on adding those types of fluid dynamic theory and problems, but we don't know when we'll have time.
What if I am given both pressures, radius for both ends, and the vertical distance between point one and two, but no velocities? This is a problem on one of my worksheets and I dont know how to solve it.
Since P1, P2, rho * g * h1, and rho * g * h2 are known you can move all those terms to the left side and set it equal to C1 (a constant) Then move (1/2) * rho * v1^2 to the right side, multiply both sides by 2 and divide both sides by rho and you get a constant (C) = v2^2 - v1^2 on the right side. And then r1^2 * v1= r2^2 * v2.
Soooi... i kinda get it. But how does that work with an upgoing velocity but the pipe is narrower in the middle? I mean the velocity has to come down at the moment it leaves the narrow part, not? (So why doesn't that work on a universal scale? The only thing i can come up with is that the narrowest part of such a pipe is not completely filled with matter, so you can stretch and compress without any real problems. And that then leads me too.... how much would be enough too compress at the smallest part where something before that smallest part has got to give! Am i wrong, and if yes then why. (I get the equasions still it would be nice to have a video. But if not, equasions are fine. Thank you for all you already thought me as well
Sir i have a great video for you , how to calculate flow velocity when you no the flow rate and diameter of the orifice or pipe. Im having trouble finding a good answer
Hello Professor Van Biezen! thank you for all your effort and hard work, it's appreciated!! There's one topic that I've been kind of struggling with for awhile and I really hope you can clear it up for me. I always see the continuity equation used to solve for an unknown velocity under the assumption that flow rate is constant. That makes sense to me but the part that I find strange is that when using the Bernoulli equation, if you push the fluid to a high enough elevation, won't the flow rate be zero at some point? If the water does enough work against gravity, won't the velocity go to zero where nothing will flow out of the pipe at the "shutoff head"? Sorry for the long question, I just can't reconcile how the continuity equation can give the velocity at some height above the datum if eventually the velocity can be zero according to Bernoulli. Thanks Professor!
You are correct. If the pipe goes high enough, the force at the bottom creating the pressure may not be enough to push the water to that height and the flow will stop.
Thanks professor! So does that mean that the continuity equation isn't accurate if used on any significant elevations, like for example, my basement to the second floor sink? Something in there has me confused lol.
What would happen if the pipe creates negative pressure means (for this video) that the reference height is so high it creates negative pressure. What would be its physical mean?
What is the pressure of flowing water in pipe....has cross section area A and velocity of flow is v... density is R...... There is no elevation in pipe...
@@MichelvanBiezen yeah i watched that now a minutes ago... Thank you very much sir ... You are simply amazing .... I am working on these parameters to be calculated with the help of water trajectory method .... calculation will be done Via DESMOS. Graphing calculator .....
Water emerges from a horizontal pipe at a rate of 0.5x10^ 3m^ 3/s. Find a velocity of water at a point in the pipe where the area of cross section is 12.6m^ 2.
Best lecturer on UA-cam.
Currently *Reviewing* for my Engineering exam.
Thank you for these videos, they're helping me a ton.
And your website's great, too.
I am not english native but your diction is so clear that it's a pleasure to hear you. Thank you.
Glad you are enjoying the videos.
If I had this guy as a prof, I don't think I would struggle learning this material. Well done.
Appreciate how this professor teaches/instructs. Wish I had had him for fluid mechanics/dynamics. He teaches “practical” real world problems and not theory. Bet this professor knows how to install sprinkler systems ( this is a complement). Good job professor!
Thank you for these free videos, you've helped alot of students!
We are glad these videos are helping students around the world
well done MICHEL!!.... by the way, it's amazing the difference in your Videos now. I'm referring to the HIGH DEF and the Brighter Lighting.... Your videos are so CLEAR!!.. Congrats!!.. Currently I'm following your Videos on the LINE INTEGRAL... I've always wanted a clear understanding of that. So far I'm through video #4, your latest... Thanks for all you do!!.. I trust your summer is going well and you're enjoying some Fun Time.... now it's back to more learning.. .. ....
Hi Philip,
Yes, we have learned a few tricks of the trait over the years. We really didn't know what we were doing in the beginning. The series on line integrals were fun to make. Believe it or not, we are learning as well. I remember many years ago doing these line integrals without actually knowing what they were or what the represented. That is why we made video # 4 to contrast it to what people think they are. There is some interesting revelations on line integrals coming up later in the series.
The explanation is clear and precise. New subscriber here
Thank you and welcome aboard!
I used flowrate formula to find the v2
got the same answer and way easier I guess
love your videos.
You are a great teacher you explained it very well
Lots of respect to you from India 🇮🇳
Thank you. Welcome to the channel!
You make Bernoulli's equation simple to understand.
Thank you Mr. Biezen. I Favorited this video. Now I just need to figure out how to incorporate 2-90deg bends. It is just like the diagram you have shown here except those two curves are 90 degree elbows. I'm going to watch the rest of your videos.
Sir, what if there is no given velocity, how would you calculate the velocities in the narrow part and in the other part of the pipe? Ty
Flow meter
yeah, I appreciate your teaching method. We learned
Glad it was helpful!
it's cool. I got the understanding perfectly. thanks
These videos helped me so so much! Thank you :)
we are glad you found them. 🙂
great explanations !!!! helped a lot, thanks!!
Great lecture. Thanks so much!
this man is master didact, i just understood constraint based solvers by watching this (at least i think so😅)
thanks for your work
if i may suggest something, maybe make the connection to conservation of energy and or kinematics? you have two very nice terms in your equation which correspond to kinetic and potential energy 🤔
Edit: he made the connection in a later video
Happy to help!
Thank you Mr.Biezen
7:49 but from where does this additional pressure comes from ?
If we had 1 bar at point 1 this could get the water in the vertical pipe to approximately 10m, can we now take 7000 times wider pipe and get water to 1 kilometer in vertical pipe and still have 1 bar at the point 1 ?
Or in the previous example, if pipe is constant diameter and height difference is 1000 meters and pressure at the point 1 is 2 atmospheres, I bet there would not be any water at the point 2 ??
Making the pipe wider will not cause the water to go higher with a particular pressure.
@@MichelvanBiezen so pressure is not higher in wider pipe ?
The Best Ever Thank You...
Thank you! 😃
kya atmospheric pressure sey energy bana saktey hain, tum burnouli equation ko jantey hain ho to heron's fountain and siphon and heron's fountain formula and inverted siphon and siphon and boyle flasks equation ko bhee compare karo
hindi language means hindi film dekhtey ho to samajh jaogey
Thanks alot sir i learned many more things from your videos 👍👍
Thanks you so much for Good presentation
You are welcome.
Best explanation
Glad you think so! 🙂
Does the pressure drop of the system matters? let say for the same problem in the video, assuming fluid transfer is via hose and it is given that the hose pressure drop is 0.1 bar (or any value).
Yes, the pressure drop matters, and if the drop is due to frictional drag, then it must be added to Bernoulli's equation. There is a playlist that includes additional terms to Bernoulli's equation: PHYSICS 34.1 BERNOULLI'S EQUATION & FLOW IN PIPES
Thank you!
would P2 be 1 atm for most cases? 1 atm resisting the pressure coming out of the tube?
Normally while inside the pipe, the pressure will be greater than 1 ATM, and after it leaves the pipe, the pressure will then be 1 ATM
Thank you very much for the series.
Hello Michel, thanks for this I really liked it. However, it doesn't seems to work when P1 = P2 = Patm i.e. both ends of the tube are open to the atmosphere.
Let's say 1 is higher than 2 and fluid still flow from 1 to 2 and let's say I am trying to determine the velocity at point 2
The Bernoulli equation you get when removing P1 and P2, and multiplying by 2*rho is
0 = 2g(h1-h2) + v1^2 - v2^2
If we input v1 = v2 (A2/A1) we get
0 = 2g(h1-h2) + v2^2 (A2^2/A1^2 -1)
V2^2 = 2g(h1-h2)/ (A2^2/A1^2 -1)
I have a problem here when A2 = A1 --> A2^2/A1^2 -1 = 0 and I end up dividing by 0. Also I get infinite speed if A1 and A2 are very close.
There is definitely an issue somewhere but where is it !?
You may be setting up a hypothetical situation which violates the laws of physics.
@@MichelvanBiezen I am trying to solve this problem www.chegg.com/homework-help/questions-and-answers/water-tank-water-level-height-h-exposed-atmospheric-pressure-bottom-tank-nozzle-angle-thet-q18940888
For this, I want to know the speed of the water at the bottom of the tank. I thought this could be solved with Bernoulli's equation by considering a point A at the top of the tank (a large section pipe) and point B at the bottom of the tank (a small section pipe) both are at the atmospheric pressure
This problem is actually closer to what I am after www.chegg.com/homework-help/questions-and-answers/problem-3-water-tank-cross-sectional-area-a1-075-m2-open-atmosphere-outflow-pipe-bottom-cr-q25219741
God bless u sir
Thank u prof Michel 100%
You're welcome 100%.
how about, i have a waterpump which pumps 15 liters/min. with a pressure of 0.35atm.
the length of the pipe is 2.5m and the height difference is 1m. i have to split the pipe at 0.5m height into two pipes. the total length will be 3.5m now. the diameter of the tubes is 10mm.
how do i calculate the speed of the water at the top? is it just the total surface to double. or does the fact that 2 pipes instead of 1 pipe to go up that last 0.5m also has effect?
Assuming we can ignore the effect of viscosity and the effect at the junction, it is simply the total height difference (which causes a shift in pressure) and the total cross sectional area (which has now doubled and thus the velocity is half at the top.)
thank you a lot ,@@MichelvanBiezen.
the total height difference is that with or without the double height from the extra pipe, so 1.5m instead of 1m
It depends on where you are determining the velocity and the pressure. It would be the same point.
Where and in which direction does the P2 exactly act?
Thank you in advance.
Using Pascal's principle the pressure acts in all directions.
thank you, great lecture
You are welcome!
Work done by an aneroid barometer is what per milibar of atmospheric pressure ??
Thank you.
I have a problem with a bent in the pipe. When I checked online for tutorials, most problems with bends were solved by Bernoulli's principle assuming that the inlet and outlet heights are equal. And then forces were resolved at x and y directions at the bend. Can you explain why it is assumed? Is this method correct?
That is done for simplicity. Just like we ignore wind resistance in projectile motion, to make it easier. But with more advanced fluid motion problem, the loss due to bends in a pipe are included. We plan on adding those types of fluid dynamic theory and problems, but we don't know when we'll have time.
@@MichelvanBiezen Thank you!
What if I am given both pressures, radius for both ends, and the vertical distance between point one and two, but no velocities? This is a problem on one of my worksheets and I dont know how to solve it.
Since P1, P2, rho * g * h1, and rho * g * h2 are known you can move all those terms to the left side and set it equal to C1 (a constant) Then move (1/2) * rho * v1^2 to the right side, multiply both sides by 2 and divide both sides by rho and you get a constant (C) = v2^2 - v1^2 on the right side. And then r1^2 * v1= r2^2 * v2.
Soooi... i kinda get it. But how does that work with an upgoing velocity but the pipe is narrower in the middle? I mean the velocity has to come down at the moment it leaves the narrow part, not? (So why doesn't that work on a universal scale? The only thing i can come up with is that the narrowest part of such a pipe is not completely filled with matter, so you can stretch and compress without any real problems. And that then leads me too.... how much would be enough too compress at the smallest part where something before that smallest part has got to give! Am i wrong, and if yes then why. (I get the equasions still it would be nice to have a video. But if not, equasions are fine.
Thank you for all you already thought me as well
The pipe is always filled everywhere and the pipe can narrow and widen in different places and the speed and pressure will vary throughout the pipe.
@@MichelvanBiezen that part i get, it is even true on niversal scale but...
Thanks sir 😊
You are welcome.
Sir i have a great video for you , how to calculate flow velocity when you no the flow rate and diameter of the orifice or pipe. Im having trouble finding a good answer
Flow rate = dV/dt = v x A where A is the cross sectional area.
Thanks, Thanks, Thanks
welcome. Glad you find the videos helpful. 🙂
what I can't understand in a two hours class I understand every single word here
Hello Professor Van Biezen! thank you for all your effort and hard work, it's appreciated!! There's one topic that I've been kind of struggling with for awhile and I really hope you can clear it up for me. I always see the continuity equation used to solve for an unknown velocity under the assumption that flow rate is constant. That makes sense to me but the part that I find strange is that when using the Bernoulli equation, if you push the fluid to a high enough elevation, won't the flow rate be zero at some point? If the water does enough work against gravity, won't the velocity go to zero where nothing will flow out of the pipe at the "shutoff head"? Sorry for the long question, I just can't reconcile how the continuity equation can give the velocity at some height above the datum if eventually the velocity can be zero according to Bernoulli. Thanks Professor!
You are correct. If the pipe goes high enough, the force at the bottom creating the pressure may not be enough to push the water to that height and the flow will stop.
Thanks professor! So does that mean that the continuity equation isn't accurate if used on any significant elevations, like for example, my basement to the second floor sink? Something in there has me confused lol.
The equation is accurate in all circumstances. Simply use v = 0.
Thank you Professor Van Biezen!
What would happen if the pipe creates negative pressure means (for this video) that the reference height is so high it creates negative pressure. What would be its physical mean?
If the pipe goes to high, the water won't reach there, (unless there is siphoning action).
What is the pressure of flowing water in pipe....has cross section area A and velocity of flow is v... density is R......
There is no elevation in pipe...
Have you seen this playlist: PHYSICS 34.1 BERNOULLI'S EQUATION & FLOW IN PIPES
@@MichelvanBiezen yeah i watched that now a minutes ago... Thank you very much sir ... You are simply amazing ....
I am working on these parameters to be calculated with the help of water trajectory method .... calculation will be done Via DESMOS. Graphing calculator .....
But what if v1 is different from V2
If the diameter of the pipe changes, as it does in this example, v1 and v2 will be different.
Great !!!
Thank you. 🙂
Good ❤
Thank you. 🙂
thanks to you
Thannk you
Sir i do have questions.....
Water emerges from a horizontal pipe at a rate of 0.5x10^ 3m^ 3/s. Find a velocity of water at a point in the pipe where the area of cross section is 12.6m^ 2.
The amount of water flowing through a pipe ever second = A x v (A is the cross sectional area and v is the velocity of the flow). v = A / flow rate
Thank you sir you did helped a great 🥰
computing is easy.. the hardest part is how to derive the formulas
ottimo
=202.600-49000-24000
=-72797.4
Shouldnt that be the result?
In the US a comma "," is just a separator --> 200,000 = 200000 and a decimal point "." is used for decimal places --> 200.000 = 200
ur my physics teacher dont u know sir ?
7:19 lol
it's cool. I got the understanding perfectly. thanks