One of the equations is wrong at, 5:58 the equation - v = (P1 - P2)/2uL* r^2 + C should be -v = (P1 - P2)/2uL* (r^2 + C), which gives a constant term (P1 - P2)/2uL*C, which is very confusing. I checked at another website www.insula.com.au/physics/1250/poiseuille.html and the correct way to do the integration is to integrate dv from v to 0 and integrate rdr from r to R to directly get v(r) = (P1-P2)/4uL *(R^2 - r^2) without using a constant term.
Thank you Michael I am so grateful to you. Secondly we can derive for volume flow rate from here For a cylinder of thickness dr the volume flow rate is v(r)×2πrdr integrate this from 0 to R we get V/t = (∆pπR^4)/8nl
A Fluid Mechanics problem that I'm trying to work out. A viscometer of the Redwood type has an oil-containing cylinder 4.75cm in diam and an agate tube 0.17 cm in diameter and 1.2 cm long. The oil surface, when flow starts, is 9 cm above the outlet from the agate tube. To allow for the sudden contractions at entry to the tube, the effective length of the tube may be taken as the actual length plus the tube radius. Making allowance for the decreasing head of oil, the viscous resistance through the tube, and the KE (kinetic energy) of discharge, calculate using arithmetical integration, the time required for 50 cm3 of an oil of viscosity 0.5 poise and specific gravity 0.92 to flow through the viscometer. How do you find the time? From this question, I have tried to use quadratic equation to solve for v, but it is proving more difficult than I thought
Is it correct to say that the negative sign comes from the direction in which the viscous forces act? For me it's easier to understand that friction forces are opposed to the movement. Great explanation!
For finding Fd, shouldn't you have used 'R' instead of 'r' ? How could you cancel the radius on both sides of equation before integrating it? Or is it possible to cancel variables like this in maths?
We were only considering a small section of the cross sectional area stopping at r. If the same variable exists on both sides of the equation you can cancel them out.
The traditional form of Poiseuille's law is written in terms of the volume per unit time passing by a particular point. Since dV/dt = A v if you take the equation in the video (which is expressed in terms of the velocity, v) and you divide both sides by the cross sectional area and then divide by 2 to get the average velocity for a cross section, you will get the equation you are familiar with.
Excuse me but I don't understand why the frictional force is calculated like that I know the formula but I fail do understand why we only use that force at that radius when the rest of the section that we selected to study also experiences friction at each radius...can you please clarify this thank you!
Hello, first of all thanks for your useful videos. But I had a question about the poiseuille's law equation. Many sources show the denominator part of the equation as 8μL instead of the 4μL you wrote. I'm confused about this. I would be very grateful if you could explain the reason for this situation.
You are correct that the correct equation has an "8" in the denominator instead of a "4". I need to revisit the derivation I put on the board to determine where I lost a factor of 2.
Watch video 13 in this series where it is explained why the shear strain on a portion of fluid is v/l (l is height of portion of fluid). dv/dr is in differential form because that force equation at that part in the video is itself describing the shear force upon a tiny little section of fluid. The shear strain over any tiny individual portion (or cube) of fluid is equal to the difference in velocity from one side of the portion to other, divided by the height of that tiny portion. Mathematically, that is written as dv/dr (r, since we're dealing with circular cross section). The integration later sums all the tiny velocity differences (dv) to describe the total accumulated velocity at any given r value.
Hello, I'm a high school student in Korea. Thank you for your useful video. However, I can't understand the reason for the differential coefficient in FsubR's formula. I'll appreciate it if you explain it to me.
Hi and welcome to the channel! If you watch the other videos in the playlist you'll see where that equation comes from. The retarding force in fluid flow is proportional to how fast the velocity changes from the center of the flow to the flow next to the pipe wall.
driving force is derived as the difference in pressure across two cross sections. That's why it uses the surface area of the "caps" of the tube. The resisting force calculates the area of the tube part, not the caps
Great lecture, but for an experiment, how would one know if the liquid is in a steady-state (i.e. the driving force equals the retarding force). How would the formula apply in that case?
Dear Sir, thank you for your nice lecture. It's pretty clear to me, considered as a layman in fluid mechanics. However, I have a basic question, what is R and r? I meant they are two difference radius. The R is indicating the outer radius, but I could not get point of small r! Thanks in advance.
Thank you very much. It was derived from the law of conservation of momentum, in a textbook titled 'Principles of biomedical engineering. (Engineering in medicine and biology)' in chapter 4. Could you please do a video to show the derivation. Though, yours is much faster and simpler to understand. Thanks.
Sir, in case the pipe is moving to the opposite direction of the flow, how can I find the velocity equation of the flow? could you please help me with this problem?
Don't you mean mu, not pit? Look up online, 'mean value of paraboloid'. A paraboloid is a 3d parabola (sort of), where a parabola is rotated about its axis of symmetry, this is pertinent here since we're looking at the cross section of a pipe. Basically, that formula comes from finding the mean velocity by adding up the velocities of all the tiny portions of fluid in any cross section of the pipe, then dividing that sum by the quantity of portions added (and that quantity is just the area of the cross section). This is from the mean value theorem.
They're just two things I don't understand ,first: I prefer working with definite integrals so if I do that what range of integration would I take for the (r) and second: if we're going to take a slice of thickness dr from r why did you ignore that at this video? sorry for the long comment 😙☺️
@@MichelvanBiezen There are two professors from IIT(Indian institute of technology) Prof.v balakrishnan, quantum physics and Prof suman chakraborty, fluid mechanics. They are very good at delivering lectures. You can watch their lectures.
R is the radius of the cylinder. (It is a constant and it is a fixed value). r is a variable, it represents any distance from the center of the cylinder to the edge of the cylinder.
I can't thank this guy enough. He always help me understand the fluids dynamics concepts I'm struggling with. Thanks a lot.
WOW! I have been staring at my lecture notes for a while now. And your explanation just made it sooooo easy to understand. Thank you sir!
Thanks for uploading these types of videos! Fluids may have been much easier if I had this channel years ago! :D
One of the equations is wrong at, 5:58 the equation - v = (P1 - P2)/2uL* r^2 + C should be -v = (P1 - P2)/2uL* (r^2 + C), which gives a constant term (P1 - P2)/2uL*C, which is very confusing. I checked at another website www.insula.com.au/physics/1250/poiseuille.html and the correct way to do the integration is to integrate dv from v to 0 and integrate rdr from r to R to directly get v(r) = (P1-P2)/4uL *(R^2 - r^2) without using a constant term.
In my study group we noticed the same thing. I don't understand why it wouldn't be -v = (P1 - P2)/2uL* (r^2 + C) making C = -(R^2)/2 ???
Thank you Michael I am so grateful to you.
Secondly we can derive for volume flow rate from here
For a cylinder of thickness dr the volume flow rate is v(r)×2πrdr integrate this from 0 to R we get
V/t = (∆pπR^4)/8nl
A Fluid Mechanics problem that I'm trying to work out. A viscometer of the Redwood type has an oil-containing cylinder 4.75cm in diam and an agate tube 0.17 cm in diameter and 1.2 cm long. The oil surface, when flow starts, is 9 cm above the outlet from the agate tube. To allow for the sudden contractions at entry to the tube, the effective length of the tube may be taken as the actual length plus the tube radius. Making allowance for the decreasing head of oil, the viscous resistance through the tube, and the KE (kinetic energy) of discharge, calculate using arithmetical integration, the time required for 50 cm3 of an oil of viscosity 0.5 poise and specific gravity 0.92 to flow through the viscometer. How do you find the time? From this question, I have tried to use quadratic equation to solve for v, but it is proving more difficult than I thought
You saved me from a hard question on my exam. Hero!
Thank you sir 😊.Your way of teaching and your language both are fabulous.
Thank you. 🙂
if only you could be my lecturer! thanks !!!
I know how you feel.
Very nice and simple explanation. My college notes needs 5 pages to arrive the final eqn . I quickly understood the concept. Many thanks
Is it correct to say that the negative sign comes from the direction in which the viscous forces act? For me it's easier to understand that friction forces are opposed to the movement. Great explanation!
The viscosity will cause the fluid to slow down, the greater the viscosity the smaller the velocity so the negative sign makes sense in that context.
@@MichelvanBiezen Thank you, sir. New suscriber over here! Congrats on your great channel!
For finding Fd, shouldn't you have used 'R' instead of 'r' ? How could you cancel the radius on both sides of equation before integrating it? Or is it possible to cancel variables like this in maths?
We were only considering a small section of the cross sectional area stopping at r. If the same variable exists on both sides of the equation you can cancel them out.
@@MichelvanBiezen thank you sir
Thank u sir. I am from India.after watching this it's become so easy to understand
Welcome to the channel!
I hate the Indian syllabus for not teaching the derivation. What's so hard in it!!!
Thank you for the clear presentation. But I read Calculus: early transcendentals, and it said that this equation is called the law of laminar flow.
Same thing
The traditional form of Poiseuille's law is written in terms of the volume per unit time passing by a particular point. Since dV/dt = A v if you take the equation in the video (which is expressed in terms of the velocity, v) and you divide both sides by the cross sectional area and then divide by 2 to get the average velocity for a cross section, you will get the equation you are familiar with.
i'm a korean high school student and this helped me very much. thank you!
Welcome to the channel!
sir your video is a lifesaver
Great explanations and derivation!
Very helpful for 3hours before exam 🥲
Glad you find the videos helpful. All the best on your exam! 🙂
is it 1/4 or 1/8 in the end
Excuse me but I don't understand why the frictional force is calculated like that I know the formula but I fail do understand why we only use that force at that radius when the rest of the section that we selected to study also experiences friction at each radius...can you please clarify this thank you!
you explain things really well keep doing what you are doing!!
You are so imaginative person,
I learnt it✌
Hello, first of all thanks for your useful videos. But I had a question about the poiseuille's law equation. Many sources show the denominator part of the equation as 8μL instead of the 4μL you wrote. I'm confused about this. I would be very grateful if you could explain the reason for this situation.
You are correct that the correct equation has an "8" in the denominator instead of a "4". I need to revisit the derivation I put on the board to determine where I lost a factor of 2.
no dude , he didnt wrote the poisuelle eqn , he
wrote the value of v only
@@MichelvanBiezen you wrote it correctly sir , this the value of v , not Q
@@None0fY0urConcern wait can u pls explain this
can this work with deriving V(p)
There are usually multiple ways to get to the same solution. I would say: "Give it a try and see what happens".
wow that's easy to understand to me. thank you keep it !
Insightful explanation !!!
Thank you. Glad you liked it. 🙂
why does the dv/dr term appear in the force equation?....any theoretical explanation pls...
Watch video 13 in this series where it is explained why the shear strain on a portion of fluid is v/l (l is height of portion of fluid). dv/dr is in differential form because that force equation at that part in the video is itself describing the shear force upon a tiny little section of fluid. The shear strain over any tiny individual portion (or cube) of fluid is equal to the difference in velocity from one side of the portion to other, divided by the height of that tiny portion. Mathematically, that is written as dv/dr (r, since we're dealing with circular cross section). The integration later sums all the tiny velocity differences (dv) to describe the total accumulated velocity at any given r value.
why area is nt equal to 2*pi*r*dr while deriving the velocity
The area here is the area of the surface of the cylinder (not including the top and bottom)
Hello, I'm a high school student in Korea. Thank you for your useful video. However, I can't understand the reason for the differential coefficient in FsubR's formula. I'll appreciate it if you explain it to me.
Hi and welcome to the channel! If you watch the other videos in the playlist you'll see where that equation comes from. The retarding force in fluid flow is proportional to how fast the velocity changes from the center of the flow to the flow next to the pipe wall.
How do we know that the viscous force is equal to the driving force? Wouldn't there be acceleration due to a pressure difference?
At steady state, the velocity will be constant, therefore there is no acceleration, therefore there is no net force.
@@MichelvanBiezen Sorry for my ignorance, but would this occur when the fluid is fully developed (therefore viscous forces equal driving forces)?
why when little r is equal to R the velocity is zero ?
+steve khan
This is based on a model. The friction force between the fluid and the wall would make the velocity there near zero.
+Michel van Biezen So basically the friction between the fluid and wall will be so great it will cause the fluid to be static ?
Down at the microscopic level, that is essentially correct.
why there it is πx^2 in driving force and in resisting force it is 2πxl
driving force is derived as the difference in pressure across two cross sections. That's why it uses the surface area of the "caps" of the tube. The resisting force calculates the area of the tube part, not the caps
Hello! I am struggling to understand why does the driving force is equal to the recurring force to have a constant flow
Great lecture, but for an experiment, how would one know if the liquid is in a steady-state (i.e. the driving force equals the retarding force). How would the formula apply in that case?
Did you find the answer
thank you you're great teacher
You are welcome! Glad you think so.
thank you for this great efforts
Awesome ! Appreciate your work !
Thanks a lot!
Dear Sir, thank you for your nice lecture. It's pretty clear to me, considered as a layman in fluid mechanics. However, I have a basic question, what is R and r? I meant they are two difference radius. The R is indicating the outer radius, but I could not get point of small r! Thanks in advance.
Big R is a constant. It is the radius of the tube. Little r is a variable, it represents any point of consideration from the center to the edge.
Yes, I got the point. Thank you very much.
I thought there was an r to the 4th power in pouiselles law?
That is a different form of the equation.
@@MichelvanBiezen ok I will delve for the other forms. thanks
shouldn't be fr the opposite of fd?
thank's for teaching❤️❤️👌🏻👌🏻
You are welcome.
thank's for good video. I have a question. why does V=0 when the small r becomes R?
Friction between the fluid and the wall prevents the layer of fluid closest to the wall from moving.
does that means that fluid which is in the ring located in r=R and have thickness of dr has the 0 velocity?
That is correct. (from a theoretical perspective according to Poisseuille's law)
This is brilliant! Thank you so much
You're very welcome! Glad you found our videos! 🙂
I didnt understand the r=R step. How's the velocity zero there?
The assumption is that the molecules next to the edge have zero velocity (that is the model used)
6:08 why is v=0 when r=R? thanks in advance sir.
no moving if no pressure difference right.
if r=R they are the same thing
Thank you very much. It was derived from the law of conservation of momentum, in a textbook titled 'Principles of biomedical engineering. (Engineering in medicine and biology)' in chapter 4. Could you please do a video to show the derivation. Though, yours is much faster and simpler to understand. Thanks.
tusm sir
truly respected
Why isn't the area equal to 2*π*r*dr while calculating driving force?
What area are you referring to?
@@MichelvanBiezen I am saying that why don't you take another r+dr surface above the cylinder, and use the face area of that cylindrical shell?
Can you recommend a textbook that includes Naviet Stokes Equations?
I haven't found a good book yet.
Sir, in case the pipe is moving to the opposite direction of the flow, how can I find the velocity equation of the flow? could you please help me with this problem?
There's a subtle detail that is not clear at first. At least for me. It cost me 2 hours of thorough mental processing :)
But how is
Velocity (avg.) = (∆P)R^2/8πL
Don't you mean mu, not pit? Look up online, 'mean value of paraboloid'. A paraboloid is a 3d parabola (sort of), where a parabola is rotated about its axis of symmetry, this is pertinent here since we're looking at the cross section of a pipe. Basically, that formula comes from finding the mean velocity by adding up the velocities of all the tiny portions of fluid in any cross section of the pipe, then dividing that sum by the quantity of portions added (and that quantity is just the area of the cross section). This is from the mean value theorem.
Thanks a lot, Great job!
you're awesome mate
Thank You.
sir am unable to find 18th video of this series
Yes, we still have to make more videos in this playlist.
Love this
Thank you. 🙂
They're just two things I don't understand ,first: I prefer working with definite integrals so if I do that what range of integration would I take for the (r)
and second: if we're going to take a slice of thickness dr from r why did you ignore that at this video?
sorry for the long comment 😙☺️
Nicely explained
Thank you so much 🙂
Sir your way of teaching is very nice. I want to know about your educational qualification. Are you holding master or PhD in engineering?
I have a masters degree in physics (and 35 years of experience as an engineer)
@@MichelvanBiezen
Thanks for reply . Great work.
@@MichelvanBiezen
There are two professors from IIT(Indian institute of technology)
Prof.v balakrishnan, quantum physics and Prof suman chakraborty, fluid mechanics. They are very good at delivering lectures. You can watch their lectures.
beautiful. thank you so much
you are welcome.
This man here is a superman
Thanks a lot sir.
Most welcome
Amazing explanation, thank you so much!!!!
tqsm
great❤️❤️
Thank you.
Dear sir, where are the videos of part 17 to 25?
They still need to be made. (A work in progress)
I like to learn to wd uh....🙋👍
WOW Thank you so much.
why is it dv/dr
The flow rate is a function of the distance from the center of the pipe.
Respect.
I love youuuuu. Good explanation
Glad it helped!
How can r be R ?
R is the radius of the cylinder. (It is a constant and it is a fixed value). r is a variable, it represents any distance from the center of the cylinder to the edge of the cylinder.
Absolute Beast
Thanks sir
Thanks a lot!
Thank you sir thank you thank you
Most welcome
Thankyou so much!
You're welcome!
beautiful
Thank you! 😊
keep it up
you awesome! Cheers
❤
Thank you. Glad you liked it. 🙂
Anyone else think he kinda sounds like Gru from despicable me
You are not the first one to mention that.