For those wondering, A1 = pi r ^2, and r is D/2 so when you square D/2 you get D^2/4. Same for A2. Now if you divide A1 by A2, you will get the ratio of A1 to A2, which you can then make A1 = (D1/D2)^2 x A2. Took me a while to figure it out and I saw it here, so hope it helps!
@@theartofproblemsolving1386 şu an avustralya açıklarında bi gemideyim. Bunu yazdığım zaman makine mühendisliği okuyordum. Kocaeli Üniversitesi Makine mühendisliğini bitirdim. Bir süre çalıştım, corporate hayattan istediğimi bulamadım, benim hayattan beklentilerimi karşılamadı daha doğrusu. Öyle olunca Piri reis üniversitesi gemi makineleri mühendisliği bitirdim ara dersleri vesaire verip. Şimdi devasa gemilerde yedi denizde geziyorum. Okyanusta giden gemide gemiyi tamir ediyoruz :D
@@theartofproblemsolving1386 yani bu konuda sana sadece şunu söyleyeyim, insanların sana lanse ettiği kadar önemli bi konu değil hangi üniversitede ne okuduğun. Bir işi yapmak için yeteri kadar merakın hevesin varsa o işi yapmak için gerekli olan bilgilere ulaşmanın bir yolunu bulursun. O yüzden kendini boş şeylere üzme, sadece değiştirebileceğin şeylere odaklan ve devamlı ileriye bak.
why they say that the squared diameter is proportional to the Area? I have understood everything else but that. As if you would explain it to a 7grader. Thanks.
Wow! It'd be hard to fail with a teacher like this. I've never heard of bernoullis equation or studied fluid dynamics, yet this made total sense to me!
Trust me, it's easier to understand if you haven't learned fluid dynamics, at least when your book has THE WRONG EQUATION LISTED AS BERNOULLI'S EQUATION
I encountered this equation in Continuum Mechanics a couple years ago, and the lecture notes for that course really sucked - they always made everything sound much more confusing than it needed to be, and these videos were much more to the point and much easier to understand.
Mr. Michel during @5:00 , the v² should just square the num. only not for the unit and the diff. in velocity is exactly = acceleration. so therefore when u times it, it will automatically came out N/m².
Yeah.. I attended an intro to mechanical engineering lecture for 2 hours and didn't have a take away. I watch this up to 5:32 and I understand it so clearly, unbelievable the standards we place on engineering "professors". Guess it has allot to do with having english as a second language, tenure status, and simply not working in a linear fashion such as this video. Thank you for the lecture! I appreciate your lecture style, straight forward explanation.👍
I have learned more from you in one hour than my entire physics courses. Thank you for explaining difficult concepts in a clear and concise manner. Just one question...why is the height taken from the middle of the tube and not the top?
That is where you find the average pressure. Typically the height difference due to the diameter of the pipe plays only a insignificant role, compared to the elevation gain or loss of the pipe.
This is a true gem. I recently enjoyed a similar book, and it was a true hidden gem. "The Art of Meaningful Relationships in the 21st Century" by Leo Flint
I usually use KhanAcademy as my extra learning tool, but it doesn't have a good playlist on Fluid. However, this video has helped me tremendously. Thank you!
Morning, thank you for your manner of presentiing Benoulli's. I.m teacher in French speaking country precisely Cote d'ivoire in west Africa. But i speak engliish very well. Thank you
This is glorious work. Doing my introduction to fluid mechanics, and this is great for extra studying and understanding. Thanks very much! P.S I love the bow tie!
yardy lad, Questions are never stupid. That equation is an equation of proportionality. Since the area of a circle is: A = (pi) * R^2 = (pi) * D^2/4 (D being the diameter) we can conclude that the area is proportional to the diameter squared.
the v² should just square the num. only not for the unit and the diff. in velocity is exactly = acceleration. so therefore when u times it, it will automatically came out N/m². *unit cannot be simply add it into the equation
At 5:33, units were confusing. A term 500kg/m2 would be 5000newton/m2. And a term 15m/sec2 would be newton. Multiplying these terms would end up 7500N squared/ m-sec2?.. Could you clarify sir? Thanks...
Michel van Biezen I apologize. I am so confused. I am definitely Missing something. How can you just change the operation like that? When you cancel m^2/m^3 wouldnt you get just 1/m? Wondering if you can break it down bc I have done this problem in the pass but never heard or saw it explained that way relative to the conversions your stating. Thanks in advance
Just like with any equation, as long as you do the same on the left side and the right side nothing changes. Here we are doing the same to the denominator and the numerator, so nothing changes. You can multiply the numerator by m^17 and multiply the denominator by m^17 and it would still be correct. What we did here is have a little foresight. Since we want to show that the units equal N/m^2 = Pa, we just wanted to divide the numerator and denominator by m (and not m^2)
Ignoring the viscosity of the fluid and the friction within the pipes, you must find the total cross sectional area by summing up the 2 pipes and then use the same equations as shown.
Sir thanks again for great explainations, i think that no one on a youtube does it in such a clear and accurate way as you do. With your permission' let me ask you a very simple question. All those examples take the Q as given parameter... My question' is actually simple a Yes\ No Question: example: I have 2 horizontal pipes of same lenght, which are NOT connected to same source of water (like the pipes are located in another countries or another rooms) pipe A has cross section 20 cm pipe B starts with cross section 20 cm and changes own cross section to 10 cm on a half of its lenght If i am given that Q in bouth pipes is same(for example 1 liter per second) , i offcorse now calculate the preasure, velosity of water, etc' using Bernulli' equasion. This part i understand very well and it is clear... , but i ask something different: the Q will not be equal unless pump in pipe B works "harder"' than the pump in pipe A. Am i right?
Hi, How the Bernoulli’s equation can be applied when you have one input flow and multiple outlet flows? and what about including head loss in the balance? I hope you can help me to figure these extra challenges out Thanks Best Carlo
@@MichelvanBiezen Along this same question, can one assume that, given multiple outflows, the total volume of the multiple outflows would always equal the volume of the inflow? It would seem that for multiple outflows, of different diameters, the total pressure of all outflows would equal the inflow pressure, and the total velocity of all outflows would equal the inflow velocity. Is that a fair assumption? Could that be expressed that P for outflow 3 of 5 = Ptotal - (P1+P2+P4+P5) and V for outflow 3 of 5 = Vtotal - (V1+V2+V4+V5)? I apologize but I am not a mathematician. Great lecture though.
If a tube is conected to a compressor on one end and opened to the atmosphere on the other end, and the compressor injects a fluid at 40 psi, will the tube walls be subjected to the same pressure along the whole length of the tube? or is this pressure just to move the fluid? will the walls expand due to the 40 psi? Michel van Biezen Thanks for the help
How about if the diameter of the inlet is much larger than that of the second diameter? How do we express that to an equation? Thank you for your response.
Couldn't help but notice the classic opener, "welcome to ilectureonline" was missing ... instead we get a cold open ... love your videos but this one had a different wavelength to it....... i hope u can revise this video Michel.... (yeah i called him by his first name, we tight like that)
A 300 mm diameter pipe discharges water at the rate of 200 liters per second. Point A on the pipe has a pressure of 280kPa and 3.4 m below A is point B with a pressure of 300kPa. Compute the head lose between point A and B. Solve pls
We are currently working on a playlist of videos that solve these types of problems. We will begin publishing them soon. Essentially the head loss is an extra term on the right side of Bernoulli's equation. P1 + rho g h1 + (1/2) rho v1^2 = P2 + rho g h2 + (1/2) rho v2^2 + head loss rho g. Since the velocity appears constant the velocity terms drop out. Therefore your equation becomes: 280,000 + (1000 x 9.8 x 3.4) = 300,000 + 0 + head lossxrhoxg Then head loss x rho x g = 280,000 + 33,320 - 300,000 = 13,320 Pa Then head loss = 13,320 / 1000 / 9.8 = 1.36 m
Hi Michel how did you factor out one rho and a half and leave just 1 rho and half still in the equation ? i thought they would coucil each other out ? thanks josh
Hi Teacher, as far as I know the pressure goes from high to low; please tell me, if in the figure the flow goes from 1 to 2, why is P2 higher than P1?. Tks
@@MichelvanBiezen Because kg is mass, better to follow this : 1kgf=1kg*9,81m/s^2 and 1kgf=9,81N -> 9,81N=1kg*9,81m/s^2, dividing both parts by 9,81 -> 1N=1kg*m/s^2, correct ? Yorgos / GREECE
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.
@@MichelvanBiezen Sir, what is your opinion on motorized pumps with varying flow rates as well? Would they follow the same principle? Is Bernoulli's principle limited only to freely flowing fluids or would it include motorized pumps? P.S.: I'm sorry for the multiple comments, I had to get your attention sir. I appreciate your help! Thank you!
The pump will create a higher pressure on one side. Once you know that pressure, you can apply Bernoulli's principle. Off course in real pipes there is internal viscosity of the fluid, friction against the walls etc.
@@MichelvanBiezen Sir, will the overall flow rate between the two would equal to their sum despite pressure difference, viscosity, friction, etc? Thank you in advance!
Dear Professor ,Many thanks for your video I just have a quick question about the Bernoulli equation in this video. I tried to find my final pressure at p2 by assuming the flow rate of 2 m3/s..so lets say the cross section of our pipe is 5 cm at point A. and at point B the cross section change to 3 cm. as a result we expect more velocity throuth the pipe. However, if i want to assume the pressure at pint B it gives me negative pressure.what does the negative pressure means here.Does this mean the water cant go through the pipe?. Many thanks for your videos .I appreceat your help in advanced.
so it means that when the cross-section of the pipe become tighter...the area become smaller and the velocity increases. However, due to the Bernoulli equation ..v1 to power of 2 - v2 to power of 2 which means ...it always negative...can you kindly suggest me how should water through a water on the condition that the A1 is bigger than the A2
Many thanks. Just one more question .if for example there is a pipe of 5 cm diameter ..we put it on to the river with the velocity of 2 m3/sec ..the velocity inside the pipe is Q=AV which is gonna be 256.64 m/sec ..which means in each second we have 256.64 meter movement inside the pipe?? I believe something is wrong . I appreciate if you can help me regards that.
The continuity equation can be applied for an incompressible fluid even if the flow is not steady. But whenever there are bends, valves, corners, etc. we have to account for those in determining the fluid flow. We plan on making more videos on that topic in the future.
It is better to think of it in terms of the potential and kinetic energy of the fluid. As it moves from a higher pressure region to a lower pressure region the fluid must do work to overcome the pressure difference (W = F * d) and thus it loses kinetic energy which exhibits itself in a lower velocity.
Your Waifu Sucks the continuity equation and Bernoullis are equal to constants, so when one variable, lets say v2 is higher than v1, the increase on the right of the equation must mean the decrease of another variable on that side of the equation to balance the equations. That probably sounds convoluted but follow the orogibal equations and rationalize what would happen if you change a variable in order to keep the entire equation balanced.
Well in terms of natural phenomena. 1.this is the reason when high speep winds, lift sometimes the roof of a house. 2.See venturi principle in a car carburator, where the incoming air from air filter, inserting in the narow passage of carburator pipe (increasing air speed) , drops down the pressure and as result the fuel commes inside with speed (to balance the loss of pressure) and converts in spray with the air which commes from the air filter. 3. About on the same principle work all sorts mecanism which make spray from anticorrossion spray till parfume ones. Also many other aplies.
I think you are referring to the equation P = F / A That equation doesn't apply to the internal pressure of a fluid in motion. Based on Bernoulli's principle, the faster a fluid flows, the lower the pressure.
can someone explain to me why the pressure increases? I get the same result using the equation, but when the diameter increases and velocity slows down i would have thought the pressure would decrease as well
Think of it this way. If you have a box of marbles that are not moving, the pressure between the marbles will be at a maximum. Now when the marbles are rolling over one another, they will not be settled in and will push on one another with less force causing a lower pressure.
@@MichelvanBiezen thank you for responding! I'm fairly out of practice with these concepts, and in my planned project everything is happening on a very small scale. I also need to review how turbulence impacts flow. Would your content on dams be the best place to find that, or do you suggest other videos?
@@MichelvanBiezen I'm trying to approach redesigning under gravel filters for aquariums with fluid dynamics in mind. Turbulence and right angles result in dead spots and impacts flow
Hi, your lectures are helpful but i would like to check, If i have a tank of water 13,000 us gallons, at a height of 35 meters connected to a pipe descending back to ground which the same height 35 meters. question what will be the the velocity of water and the flaw rate of the water if i have a 1.2 dia pipe for 18 meter and a pipe 0.94 dia for the rest 18 meter to reach ground? thank you
Dear Sir............Example Problem : One hydraulic cylinder, the end is connected to the Threaded rod so that it can be rotated. Cylinders with a bore size of 18 mm, 10 mm rod and a stroke of 300 mm single acting in a condition that is filled with oil. If the cylinder is rotated 10 mm. a. How to calculate it b. What is the gauge pressure
please write down the equation which satisfied heron's fountain so we got that upper limit which can give us eternal source of energy please discussed it with me or show graphical representation of heron's fountain formula equation it become a great source of energy from atmospheric pressure forever
Dear Mr. Michel van Biezen, please I need your help to understand this, maybe I didn't get it well! As per continuity equation, increase in diameter leads to decrease in velocity, and as per Bernoulli`s equation, decrease in velocity leads to increase in pressure. But, I can't understand how fluid flows from low pressure point to high pressure point! I think that if we'll consider the friction loss across the pipeline, the pressure P2 will never be higher then pressure P1. Please, help me with a good explanation about, let's say, this misunderstanding of mine.
Lev A Yes, the continuity equation shows that when the diameter increases, the velocity decreases. You also know that when the velocity increases, the pressure decreases. That is all correct. The confusion comes in with the mis-interpretation of Bernoulli's equation. The equation simply states that the sum of the three terms: P + (1/2) * rho * v^2 + rho * g * h = constant Therefore you can go to any point inside the pipe and the sum of these 3 terms will always be the same. (Note that the units of each term equals pressure) Bernoulli's equation doesn't give you any indication as to why the fluid travels through the pipe. That is where you need principles of fluid flow. (I will cover that later this year in a new set of videos).
Michel van Biezen thank you so much for your quick answer. I saw almost all your videos, especially Thermodynamics (2nd Law with Carnot Cycle and heat engines) and fluid mechanics! I really want to thank you so much for this extremely important information that you shared for young engineers. Regarding the original misunderstanding of mine, I have an old book (more then 30 years published before) and there is Bernoulli`s Equation with this form: P1 + rho * g * h1 + (1/2) * rho * v^2 - Sum f * l/D * 1/2 * rho * v^2 - Sum * K sub L * 1/2 * rho * v^2 = P2 + rho * g * h2 + (1/2) * rho * v^2 First term is the Sum of major loss regarding friction loss and second term is the Sum of minor loss which has to do with all joints and connections between the pipes. I need for a confirmation from your part. So, if we'll consider this two terms, maybe the value of pressure in P2 is lower then value of pressure from P1. I will wait with great interest your new videos!
Lev A You are correct. If you include friction and joints and connections, the pressure will drop. (It is similar to the energy equation where friction will lower the energy.) With fluid flow, friction and joints will lower the pressure. (P2)
Michel van Biezen I thank you so much for help. My life would be very easy if you were my physics teacher, for sure. Now everything is clear for me! I will wait with great interest your new set of videos!
anybody can help out this to clarify Pressure Vs Area relationship was Inversally proportional, but in this case when area increase, pressure also increased HOW?
Good question. Even though pressure decreases, if the pump generates a lot of pressure to begin with, the decrease in pressure will be RELATIVELY small and there will be plenty of pressure left to do the job.
+Jinnie Tinianow There is some more information in these videos: Physics -Fluid Dynamics (1 of 2) Fluid Flow and Physics -Fluid Dynamics (2 of 2) Fluid Flow
The flow rate (dV/dt) which is the amount of fluid flowing through the tube per unit time is equal to the product of the cross-sectional area and the velocity of the fluid. That product must be the same everywhere because the flow rate must be the same everywhere
+Michel van Biezen , for me calculations that you do in the end is not clear enough. if don't care about numbers, we have N/m2 is equal to kg/m2*(m/sec2) , so what we have to do with this m/sec2 ?
My entire day has been spent watching your videos, I've come to appreciate the little things about ur videos! Why can't you be my physics professor?! Haha
I understated this vide thanks, but what would happened in a network of pipes (with pressures and flow rates) with two or three spray nozzles (all the same flow rate discharge of: 2.27 L/mint) and pressure in is just 5 bars...? Can anyone advise how to do it ?
The amount of fluid flow per unit time (dV/dt) is defined as: dV/dt = A*v where A is the cross sectional area and v is velocity of the fluid. Since dV/dt must be a constant, A1*v1 = A2*v2 = constant
For those wondering, A1 = pi r ^2, and r is D/2 so when you square D/2 you get D^2/4. Same for A2. Now if you divide A1 by A2, you will get the ratio of A1 to A2, which you can then make A1 = (D1/D2)^2 x A2. Took me a while to figure it out and I saw it here, so hope it helps!
Mark Palmisano Saw it asked here a couple times**
true, but he has A1=(D2/D1) not (D1/D2). ????
I keep getting: A1/D1=A2/D2 --> A2=A1(D2/D1). I still have no idea where the square comes from. It's driving me insane.
A1=r1^2pi , A2=r2^2pi, r1=D1/2 , r2=D2/2, pi=pi A1/r1*2=A2/r2*2 insert D1/2 and D2/2 insted r1,r2
and the end resault is A2=A1 D1*2/D2*2
It was just A1V1=A2V2. Just substitute the given and solve for V2 which is 1m/s.
You are better teacher than most of our University Professors sir.
th-cam.com/channels/gnq8tH5o-X3byKaMsuwqHA.htmlvideos
doğukan abi tam olarak 8 yıl geçmiş nerelerdesin şuan hangi üniversiteden mezun oldun :)))
@@theartofproblemsolving1386 şu an avustralya açıklarında bi gemideyim. Bunu yazdığım zaman makine mühendisliği okuyordum. Kocaeli Üniversitesi Makine mühendisliğini bitirdim. Bir süre çalıştım, corporate hayattan istediğimi bulamadım, benim hayattan beklentilerimi karşılamadı daha doğrusu. Öyle olunca Piri reis üniversitesi gemi makineleri mühendisliği bitirdim ara dersleri vesaire verip. Şimdi devasa gemilerde yedi denizde geziyorum. Okyanusta giden gemide gemiyi tamir ediyoruz :D
@@dogukanturgal6758 Vay be. Yolun açık olsun abi bende bilgisayar mühendisliği düşünüyorum, bakalım neler çıkarsa karşımıza.
@@theartofproblemsolving1386 yani bu konuda sana sadece şunu söyleyeyim, insanların sana lanse ettiği kadar önemli bi konu değil hangi üniversitede ne okuduğun. Bir işi yapmak için yeteri kadar merakın hevesin varsa o işi yapmak için gerekli olan bilgilere ulaşmanın bir yolunu bulursun. O yüzden kendini boş şeylere üzme, sadece değiştirebileceğin şeylere odaklan ve devamlı ileriye bak.
I love how this teacher doesn't rush the calculations, or take massive leaps of logic. It makes it much easier to learn from scratch.
The area is proportional to the radius or to the diameter.
why they say that the squared diameter is proportional to the Area? I have understood everything else but that. As if you would explain it to a 7grader.
Thanks.
Wow! It'd be hard to fail with a teacher like this. I've never heard of bernoullis equation or studied fluid dynamics, yet this made total sense to me!
***** LOL! Reading this reminds me of some of the tweets on "celebrities reading mean tweets"!
Trust me, it's easier to understand if you haven't learned fluid dynamics, at least when your book has THE WRONG EQUATION LISTED AS BERNOULLI'S EQUATION
I encountered this equation in Continuum Mechanics a couple years ago, and the lecture notes for that course really sucked - they always made everything sound much more confusing than it needed to be, and these videos were much more to the point and much easier to understand.
th-cam.com/channels/gnq8tH5o-X3byKaMsuwqHA.htmlvideos
Mr. Michel during @5:00 , the v² should just square the num. only not for the unit and the diff. in velocity is exactly = acceleration. so therefore when u times it, it will automatically came out N/m².
*unit cannot be simply add it into the equation. if needed there must be same for both left and right side equation
The units must be squared as well.
Yeah.. I attended an intro to mechanical engineering lecture for 2 hours and didn't have a take away. I watch this up to 5:32 and I understand it so clearly, unbelievable the standards we place on engineering "professors". Guess it has allot to do with having english as a second language, tenure status, and simply not working in a linear fashion such as this video. Thank you for the lecture! I appreciate your lecture style, straight forward explanation.👍
Glad the video helped. Thanks for sharing.
The "Bernoulli's equation" written in my hydraulics book is entirely different. Thank you for making an impossible problem possible.
th-cam.com/channels/gnq8tH5o-X3byKaMsuwqHA.htmlvideos
I was feeling scared about physics but now I'm actually looking forward to it!
That is great. It is fun when we begin to understand the concepts.
th-cam.com/channels/gnq8tH5o-X3byKaMsuwqHA.htmlvideos
I have learned more from you in one hour than my entire physics courses. Thank you for explaining difficult concepts in a clear and concise manner. Just one question...why is the height taken from the middle of the tube and not the top?
That is where you find the average pressure. Typically the height difference due to the diameter of the pipe plays only a insignificant role, compared to the elevation gain or loss of the pipe.
th-cam.com/channels/gnq8tH5o-X3byKaMsuwqHA.htmlvideos
This is a true gem. I recently enjoyed a similar book, and it was a true hidden gem. "The Art of Meaningful Relationships in the 21st Century" by Leo Flint
You refernce seems like an interesting book. The description definitely seems like something we can all learn from "mindful communication".
Thank you Michel, you are the reason im still alive in physics :)
th-cam.com/channels/gnq8tH5o-X3byKaMsuwqHA.htmlvideos
I usually use KhanAcademy as my extra learning tool, but it doesn't have a good playlist on Fluid. However, this video has helped me tremendously.
Thank you!
th-cam.com/channels/gnq8tH5o-X3byKaMsuwqHA.htmlvideos
Great Videos!! I appreciate how you explain the general concept before going into all of the details of the equation.
th-cam.com/channels/gnq8tH5o-X3byKaMsuwqHA.htmlvideos
ohh thank you your last words about pressure increasing because the fluid slows down were gold for me
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You are a good teacher i am following you from algeria even we study physics in french but i could understand thank you so much sir
Thank you! 😃 Welcome to the channel!
In 5:06 , how did he added m for the two numbers. I just didn’t understand. I know that he wanted to coverts into Pa. But how ???
1 N = kg *m/sec^2 Area = m^2 Pressure = Force / Area = N/m^2
Morning, thank you for your manner of presentiing Benoulli's. I.m teacher in French speaking country precisely Cote d'ivoire in west Africa. But i speak engliish very well. Thank you
That is great! Thanks for the comment.
th-cam.com/channels/gnq8tH5o-X3byKaMsuwqHA.htmlvideos
These lessons are sooo gooooood, thank you very much!!!!!!!!!
Glad you like them! 🙂
This is glorious work. Doing my introduction to fluid mechanics, and this is great for extra studying and understanding. Thanks very much! P.S I love the bow tie!
th-cam.com/channels/gnq8tH5o-X3byKaMsuwqHA.htmlvideos
Sorry for the stupid question, but what equation is the A2=A1(D2/D1)^2 I haven't seen it before and where can I know more about it?.
yardy lad,
Questions are never stupid.
That equation is an equation of proportionality.
Since the area of a circle is: A = (pi) * R^2 = (pi) * D^2/4 (D being the diameter) we can conclude that the area is proportional to the diameter squared.
Thank you very much , I will look more into it.
Oh I got it thank you.
+Michel van Biezen why not radius?
I guess it's because if D/2=R, then D^2/2^2 =R
on 5:00 why do we need to add m, because 1 Pa = 1 kg/(m·s^2) ?
Shalva Abramia allready I mean
+Shalva Abramia I was thinking the same...
the v² should just square the num. only not for the unit and the diff.
in velocity is exactly = acceleration. so therefore when u times it, it
will automatically came out N/m². *unit cannot be simply add it into the equation
Great series of videos. Well presented and brilliantly explained. Thanks
At 5:33, units were confusing. A term 500kg/m2 would be 5000newton/m2. And a term 15m/sec2 would be newton. Multiplying these terms would end up 7500N squared/ m-sec2?.. Could you clarify sir? Thanks...
The units are: kg /m sec^2 which is the same as N/m^2 = Pa
Very well explained the Bernoulli's equation
@ 7:00 How does the units work out? The m^2 cancels out how can you just put a “m”
Wouldn’t it be kg/m sec^2 ?
we changed m^2 / m^3 to m / m^2 that is how we went from kg m^2 / m^3 sec^2 to kg m / m^2 sec^2 to N / m^2 to Pa (units of pressure)
Michel van Biezen I apologize. I am so confused. I am definitely
Missing something. How can you just change the operation like that? When you cancel m^2/m^3 wouldnt you get just 1/m? Wondering if you can break it down bc I have done this problem in the pass but never heard or saw it explained that way relative to the conversions your stating. Thanks in advance
Just like with any equation, as long as you do the same on the left side and the right side nothing changes. Here we are doing the same to the denominator and the numerator, so nothing changes. You can multiply the numerator by m^17 and multiply the denominator by m^17 and it would still be correct. What we did here is have a little foresight. Since we want to show that the units equal N/m^2 = Pa, we just wanted to divide the numerator and denominator by m (and not m^2)
Michel van Biezen got it now. Didn’t know that’s what you were doing. Thanks
Incredible 👏👏👏👏 amazing ❤ 👏 😍
This gentleman wonderful teacher is my favourite lovely teacher, he studies so fluent
Thank you! 😃 For your kind comment.
Hard subject made easy with brilliant teaching. Thank you!
th-cam.com/channels/gnq8tH5o-X3byKaMsuwqHA.htmlvideos
Thank you ! The person like you we need..who dedicates the time for the others.sory for my english. Good look!
What if this pipe splits off into two pipes with different diameters (instead of the diameter of the pipe just changing like it is here ? )
Ignoring the viscosity of the fluid and the friction within the pipes, you must find the total cross sectional area by summing up the 2 pipes and then use the same equations as shown.
Sir thanks again for great explainations, i think that no one on a youtube does it in such a clear and accurate way as you do.
With your permission' let me ask you a very simple question.
All those examples take the Q as given parameter...
My question' is actually simple a Yes\ No Question:
example:
I have 2 horizontal pipes of same lenght, which are NOT connected to same source of water (like the pipes are located in another countries or another rooms)
pipe A has cross section 20 cm
pipe B starts with cross section 20 cm and changes own cross section to 10 cm on a half of its lenght
If i am given that Q in bouth pipes is same(for example 1 liter per second) , i offcorse now calculate the preasure, velosity of water, etc' using Bernulli' equasion. This part i understand very well and it is clear...
, but i ask something different: the Q will not be equal unless pump in pipe B works "harder"' than the pump in pipe A.
Am i right?
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@@engineeringsimulationstudi1404
what?
you are an excellent teacher/professor.
th-cam.com/channels/gnq8tH5o-X3byKaMsuwqHA.htmlvideos
Hi,
How the Bernoulli’s equation can be applied when you have one input flow and multiple outlet flows? and what about including head loss in the balance?
I hope you can help me to figure these extra challenges out
Thanks
Best
Carlo
We still need to make additional videos on fluid flow besides the simple applications of Bernoulli's equation. (Something for the future).
@@MichelvanBiezen Along this same question, can one assume that, given multiple outflows, the total volume of the multiple outflows would always equal the volume of the inflow? It would seem that for multiple outflows, of different diameters, the total pressure of all outflows would equal the inflow pressure, and the total velocity of all outflows would equal the inflow velocity. Is that a fair assumption? Could that be expressed that P for outflow 3 of 5 = Ptotal - (P1+P2+P4+P5) and V for outflow 3 of 5 = Vtotal - (V1+V2+V4+V5)? I apologize but I am not a mathematician. Great lecture though.
If a tube is conected to a compressor on one end and opened to the atmosphere on the other end, and the compressor injects a fluid at 40 psi, will the tube walls be subjected to the same pressure along the whole length of the tube? or is this pressure just to move the fluid? will the walls expand due to the 40 psi? Michel van Biezen
Thanks for the help
How about if the diameter of the inlet is much larger than that of the second diameter? How do we express that to an equation? Thank you for your response.
+lj vailoces
Just change point 1 for point 2 and visa versa and then work out the problem in the other direction.
+Michel van Biezen Thank you! It helps :)
Couldn't help but notice the classic opener, "welcome to ilectureonline" was missing ...
instead we get a cold open ... love your videos but this one had a different wavelength to it....... i hope u can revise this video Michel.... (yeah i called him by his first name, we tight like that)
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A 300 mm diameter pipe discharges water at the rate of 200 liters per second. Point A on the pipe has a pressure of 280kPa and 3.4 m below A is point B with a pressure of 300kPa. Compute the head lose between point A and B. Solve pls
We are currently working on a playlist of videos that solve these types of problems. We will begin publishing them soon. Essentially the head loss is an extra term on the right side of Bernoulli's equation. P1 + rho g h1 + (1/2) rho v1^2 = P2 + rho g h2 + (1/2) rho v2^2 + head loss rho g. Since the velocity appears constant the velocity terms drop out. Therefore your equation becomes: 280,000 + (1000 x 9.8 x 3.4) = 300,000 + 0 + head lossxrhoxg Then head loss x rho x g = 280,000 + 33,320 - 300,000 = 13,320 Pa Then head loss = 13,320 / 1000 / 9.8 = 1.36 m
Very detailed and understandable. Thank you very much.
how to compute the P1 if it is not given? Thanks
Then you must be given a hint, or other factors are given.
Hi Michel how did you factor out one rho and a half and leave just 1 rho and half still in the equation ? i thought they would coucil each other out ?
thanks josh
We factored out (1/2) rho from the last two terms. (algebra)
Hello that was good. How fluid flows when downstream pressure is high?
Depends on the circumstances and conditions. The pressure at a lower elevation does not need to be higher.
@@MichelvanBiezen plz elaborate
Hi Teacher, as far as I know the pressure goes from high to low; please tell me, if in the figure the flow goes from 1 to 2, why is P2 higher than P1?. Tks
If water is delivered to a house on a hill, the pressure at the house will be much lower than at the bottom of the hill.
Good explain sir thank you.
Sir . Can you tell me how boiler feed pump impeller change velocity of water in pressure.
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Can someone explain how he added a "meter" when multiplying the density and velocity to get Pascal??
multiplying the numerator and denominator by "m" you end up with kg * m / sec^2 in the numerator and m^2 in the denominator.
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@@MichelvanBiezen
Because kg is mass, better to follow this : 1kgf=1kg*9,81m/s^2 and 1kgf=9,81N -> 9,81N=1kg*9,81m/s^2, dividing both parts by 9,81 -> 1N=1kg*m/s^2, correct ?
Yorgos / GREECE
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.
@@MichelvanBiezen Sir, what is your opinion on motorized pumps with varying flow rates as well? Would they follow the same principle?
Is Bernoulli's principle limited only to freely flowing fluids or would it include motorized pumps?
P.S.: I'm sorry for the multiple comments, I had to get your attention sir. I appreciate your help! Thank you!
The pump will create a higher pressure on one side. Once you know that pressure, you can apply Bernoulli's principle. Off course in real pipes there is internal viscosity of the fluid, friction against the walls etc.
@@MichelvanBiezen Sir, will the overall flow rate between the two would equal to their sum despite pressure difference, viscosity, friction, etc? Thank you in advance!
Dear Professor ,Many thanks for your video I just have a quick question about the Bernoulli equation in this video. I tried to find my final pressure at p2 by assuming the flow rate of 2 m3/s..so lets say the cross section of our pipe is 5 cm at point A. and at point B the cross section change to 3 cm. as a result we expect more velocity throuth the pipe. However, if i want to assume the pressure at pint B it gives me negative pressure.what does the negative pressure means here.Does this mean the water cant go through the pipe?. Many thanks for your videos .I appreceat your help in advanced.
A negative pressure at B means there is not enough pressure at A to push the water through.
so it means that when the cross-section of the pipe become tighter...the area become smaller and the velocity increases. However, due to the Bernoulli equation ..v1 to power of 2 - v2 to power of 2 which means ...it always negative...can you kindly suggest me how should water through a water on the condition that the A1 is bigger than the A2
We can say that when the pipe becomes smaller the velocity increases and the pressure decreases. (The pressure is therefore not necessarily negative).
Many thanks. Just one more question .if for example there is a pipe of 5 cm diameter ..we put it on to the river with the velocity of 2 m3/sec ..the velocity inside the pipe is Q=AV which is gonna be 256.64 m/sec ..which means in each second we have 256.64 meter movement inside the pipe?? I believe something is wrong . I appreciate if you can help me regards that.
hi sir micheal... does incompressible continuity equation still applicable in pipes with valve not fully open?....hoping for your answer thanks
The continuity equation can be applied for an incompressible fluid even if the flow is not steady. But whenever there are bends, valves, corners, etc. we have to account for those in determining the fluid flow. We plan on making more videos on that topic in the future.
What if I need to calculate the v2 when the diameter change and h2 too ? What’s the equation
We have examples just like that in the playlist.
I'm gonna be starting uni in September I can't wait!
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Engineering Simulation Studio : Numerical Approach I just graduated in May
sir michael thank you for making all of these videos!
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If the fluid is flowing from a1 to a2 but the psi is higher in a2 wouldn't their be a net force backwards from a2 to a1?
It is better to think of it in terms of the potential and kinetic energy of the fluid. As it moves from a higher pressure region to a lower pressure region the fluid must do work to overcome the pressure difference (W = F * d) and thus it loses kinetic energy which exhibits itself in a lower velocity.
also the fluid is incompressible.
How do I find the area and velocity of the outlet if I knew the P1, P2, v1 and A1?
If the pipe does not change height, then use Bernoulli's equation to find v2. Then use the flow rate equation to find A2.
Thank you very much! This is very helpful, please continue teaching us! :D
I don't get it. How does increasing the speed decreases the pressure?
Your Waifu Sucks the continuity equation and Bernoullis are equal to constants, so when one variable, lets say v2 is higher than v1, the increase on the right of the equation must mean the decrease of another variable on that side of the equation to balance the equations. That probably sounds convoluted but follow the orogibal equations and rationalize what would happen if you change a variable in order to keep the entire equation balanced.
Well in terms of natural phenomena. 1.this is the reason when high speep winds, lift sometimes the roof of a house. 2.See venturi principle in a car carburator, where the incoming air from air filter, inserting in the narow passage of carburator pipe (increasing air speed) , drops down the pressure and as result the fuel commes inside with speed (to balance the loss of pressure) and converts in spray with the air which commes from the air filter.
3. About on the same principle work all sorts mecanism which make spray from anticorrossion spray till parfume ones. Also many other aplies.
How did you get A2 = A1(D2/D1)2
thank you
i was lost where the part (500kg/m^2)(15m/s^2) became N/m^2 ? how did the unit changed to N/m^2?
1 N = 1 kg m / sec^2
Sir I thought the smaller the area the greater the pressure? Why the opposite in this issue
I think you are referring to the equation P = F / A That equation doesn't apply to the internal pressure of a fluid in motion. Based on Bernoulli's principle, the faster a fluid flows, the lower the pressure.
Dear Mr. Michel van Biezen,
How did you get A1v1=A2v2?
Thank you.
oh.i got.
volumetric flow rate.)
what if the pipe A1 = 10cm and A2= 5cm?
what if the height and diameter both changing ; how it will be balanced?
We have an example of that in the playlist.
@@MichelvanBiezen yes sir I have seen. Thank you. Your videos are super useful
can someone explain to me why the pressure increases? I get the same result using the equation, but when the diameter increases and velocity slows down i would have thought the pressure would decrease as well
Think of it this way. If you have a box of marbles that are not moving, the pressure between the marbles will be at a maximum. Now when the marbles are rolling over one another, they will not be settled in and will push on one another with less force causing a lower pressure.
Question: Can v2 be increased by an increase of temperature at that end of the pipe?
An increase in temperature would cause a change in density (although slight) which would have an effect on Bernoulli's equation.
@@MichelvanBiezen thank you for responding! I'm fairly out of practice with these concepts, and in my planned project everything is happening on a very small scale.
I also need to review how turbulence impacts flow. Would your content on dams be the best place to find that, or do you suggest other videos?
I would not expect for you to have turbulent flow, unless the pipes are too narrow or the velocity is too high.
@@MichelvanBiezen I'm trying to approach redesigning under gravel filters for aquariums with fluid dynamics in mind. Turbulence and right angles result in dead spots and impacts flow
as simple as your explanation
wow double pipe diameter but only .074 atm gained... thnks
how to calculate the both viscosities when they are both unknown in Bernoulli's Equation
Benroulli's equation typically do not include viscosities. Instead they include the density of the fluid.
what is P here? is it pressure exerted by air?
Hi, your lectures are helpful but i would like to check, If i have a tank of water 13,000 us gallons, at a height of 35 meters connected to a pipe descending back to ground which the same height 35 meters. question what will be the the velocity of water and the flaw rate of the water if i have a 1.2 dia pipe for 18 meter and a pipe 0.94 dia for the rest 18 meter to reach ground?
thank you
You really help me
Glad we did. Thanks for sharing.
Dear Sir............Example Problem : One hydraulic cylinder, the end is connected to the Threaded rod so that it can be rotated. Cylinders with a bore size of 18 mm, 10 mm rod and a stroke of 300 mm single acting in a condition that is filled with oil. If the cylinder is rotated 10 mm.
a. How to calculate it
b. What is the gauge pressure
Sir im always gratitude for ur help thank u very much 🙏🙏
Excellent 😊
Thank you. Glad you liked it.
why do you square the diameters when determining the ratio of A2 with respect to A1
What if the given is pressure difference together with the radii of the sections. But you have to find the water speed in the two sections?
Then use the equation: A1* v1 = A2 * v2
Along with Bernoulli's equation.
Can you tell why are you Squaring d2/d1
The cross-sectional area is proportional to the radius squared or the diameter squared.
Michel sir Always it happen or some special cases it happen????? plz clarify me
please write down the equation which satisfied heron's fountain so we got that upper limit which can give us eternal source of energy
please discussed it with me or show graphical representation of heron's fountain formula equation
it become a great source of energy from atmospheric pressure forever
Pls help guys how did he get that 0.07 coz when I press the calculator I don't get it...??
Dear Mr. Michel van Biezen, please I need your help to understand this, maybe I didn't get it well!
As per continuity equation, increase in diameter leads to decrease in velocity, and as per Bernoulli`s equation, decrease in velocity leads to increase in pressure.
But, I can't understand how fluid flows from low pressure point to high pressure point!
I think that if we'll consider the friction loss across the pipeline, the pressure P2 will never be higher then pressure P1.
Please, help me with a good explanation about, let's say, this misunderstanding of mine.
Lev A
Yes, the continuity equation shows that when the diameter increases, the velocity decreases.
You also know that when the velocity increases, the pressure decreases.
That is all correct.
The confusion comes in with the mis-interpretation of Bernoulli's equation.
The equation simply states that the sum of the three terms:
P + (1/2) * rho * v^2 + rho * g * h = constant
Therefore you can go to any point inside the pipe and the sum of these 3 terms will always be the same. (Note that the units of each term equals pressure)
Bernoulli's equation doesn't give you any indication as to why the fluid travels through the pipe.
That is where you need principles of fluid flow. (I will cover that later this year in a new set of videos).
Michel van Biezen thank you so much for your quick answer.
I saw almost all your videos, especially Thermodynamics (2nd Law with Carnot Cycle and heat engines) and fluid mechanics! I really want to thank you so much for this extremely important information that you shared for young engineers.
Regarding the original misunderstanding of mine, I have an old book (more then 30 years published before) and there is Bernoulli`s Equation with this form:
P1 + rho * g * h1 + (1/2) * rho * v^2 - Sum f * l/D * 1/2 * rho * v^2 - Sum * K sub L * 1/2 * rho * v^2 = P2 + rho * g * h2 + (1/2) * rho * v^2
First term is the Sum of major loss regarding friction loss and second term is the Sum of minor loss which has to do with all joints and connections between the pipes.
I need for a confirmation from your part. So, if we'll consider this two terms, maybe the value of pressure in P2 is lower then value of pressure from P1.
I will wait with great interest your new videos!
Lev A
You are correct. If you include friction and joints and connections, the pressure will drop.
(It is similar to the energy equation where friction will lower the energy.) With fluid flow, friction and joints will lower the pressure. (P2)
Michel van Biezen I thank you so much for help.
My life would be very easy if you were my physics teacher, for sure.
Now everything is clear for me!
I will wait with great interest your new set of videos!
Michel van Biezen this means that by neglecting the friction loss we can never use the Bernoulli equation in fluid dynamics(oil industry)?
can someone tell me why did he do ( d2/d1 )? isn't it supposed to be ( d1/d2 )?
The video is correct.
thanks :)
Is this for high school physics? because this is still coming up so clutch and I'm in my 3rd year of college.
This is more geared towards college physics. Although you will find this sometimes in high school as well.
In a way yes, because its a derivation of the law of conservation of energy, which is taught in grade 12
good explanation.
anybody can help out this to clarify Pressure Vs Area relationship was Inversally proportional, but in this case when area increase, pressure also increased HOW?
Why the unit of diameter not converted from cm to m?
Since we use the values as ratios the units cancel out so we don't have to convert them.
I agree this guy is good!
I don't understand, pressure decreases if pipe gets thinner, then how water flowing through thin pipe use to cut stones or hard substances?
Good question. Even though pressure decreases, if the pump generates a lot of pressure to begin with, the decrease in pressure will be RELATIVELY small and there will be plenty of pressure left to do the job.
So for A1V1=A2V2... i don't think we cover that in class, could you refer me to more information about the equation?
+Jinnie Tinianow There is some more information in these videos: Physics -Fluid Dynamics (1 of 2) Fluid Flow and
Physics -Fluid Dynamics (2 of 2) Fluid Flow
Great thanks!
+Michel van Biezen Are these videos on your channel I cant seem to find them
I really don't get where does the A1V1=A2V2 ? Can someone please explain to me.
The flow rate (dV/dt) which is the amount of fluid flowing through the tube per unit time is equal to the product of the cross-sectional area and the velocity of the fluid. That product must be the same everywhere because the flow rate must be the same everywhere
could you clarify what you do with sec2 ?
+Константин Валеев Instead of using s for seconds I use "sec" for seconds.
+Michel van Biezen , for me calculations that you do in the end is not clear enough. if don't care about numbers, we have N/m2 is equal to kg/m2*(m/sec2) , so what we have to do with this m/sec2 ?
+Константин Валеев Note that in Bernoulli's equation, every term has the units of pressure: N/m^2 that can be written as: N/m^2 = kg/(m*sec^2)
i like the my little pony in the corner
Maryjoy,
You are the first person to comment on the little figures in the corner.
My entire day has been spent watching your videos, I've come to appreciate the little things about ur videos! Why can't you be my physics professor?! Haha
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thank you teacher
Welcome!!
Does anyone know what will happen if P2 is smaller than P1
Then the fluid would flow in the opposite direction.
@@MichelvanBiezen Thanks, but won’t P2-P1 be negative while 1/2p(v1^2-v2^2) is still positive
Sure, but that doesn't matter. It all works out in the end.
@@MichelvanBiezen ok thanks a lot
I understated this vide thanks, but what would happened in a network of pipes (with pressures and flow rates) with two or three spray nozzles (all the same flow rate discharge of: 2.27 L/mint) and pressure in is just 5 bars...?
Can anyone advise how to do it ?
333geoalv
My plan is to place those type of videos on line in the near future.
Why was the ratio D2/D1 become squared?
+Rhehan Pamlian Because the cross sectional area is a function of the diameter squared.
Because the suface of cycle is S=π.r^2 or S=π. (D\2)^2 hence S=π(D^2\4), since
r=D\2
7:02 you pulled a Michael Scott
How do you know A1V1=A2V2 ?
That is explained in this video: Physics -Fluid Dynamics (1 of 2) Fluid Flow th-cam.com/video/Cdpoo2XM6Hg/w-d-xo.html
h1 & h2 are not equal na.. thn y sir ignore changes and ?? Someone. Plz help me..
Sorry, I didn't understand your question. If you look at the other examples you may get the answer.
@@MichelvanBiezen okay.. thank you sir..
Best ever
Thank you. Glad you found our videos.
can you help me with some questions?
We respond to questions the best we can as time permits
Michel van Biezen would I be able to email you the question?
Great explanations!!! Thanks :)
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why A1V1=A2V2? SAME FLOW?
The amount of fluid flow per unit time (dV/dt) is defined as: dV/dt = A*v where A is the cross sectional area and v is velocity of the fluid. Since dV/dt must be a constant, A1*v1 = A2*v2 = constant
sir thank you i want geotecnic -1 all topic please sir
Thank you sir
You are welcome.