JEE MAIN & ADVANCE 11th PCM Physics-Real Fluids Demo Videos

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Hello students we have discussed about the different phenomena viscosity we have talked about how exactly the liquids flow and so on right let’s try to understand about that other aspects of viscosity this in this particular topic we are going to discuss about poiseuilles equation of poiseuilles formula this is very, very important because this will tell us how exactly the fluid rate changes are how exactly the fluids move let’s try to understand about it so in this case what exactly the point this equation let’s try to understand about so this gives us this . this equation gives the pressure drop in the fluid flowing through a long cylindrical pipe let me explain how so just take it very long cylindrical pipe ok and if I just assume that both the ends have a uniform area of cross section there is some distance between the two ends there is going to be some pressure difference right because the fluid is moving from one into the other obviously that is something which is going to be different as a result the fluid is moving is the fluid would have been rest right so that tells us that is some pressure difference which makes the fluid to move how much is it how do we find it out what is the register’s with the flow that’s what we are going to understand in this particular topic so there are a few assumptions that we need to take care of here assumptions of the equation our first assumption flow is laminar that is it steady flow that is all the layers would have tangents being parallel each other or tangents are not at different angles then it’s a viscous force viscous fluid so obviously that is going to be viscous force that is going to be acting between the different layers then the liquid is incompressible that is you need to take into account that the density should not change from one point with other institution means constant from one into the other so it’s incompressible fine going further the flow is through a constant cross section they should not be it change in the cross section because if there is a change obviously the liquid is going to accelerate at some point it’s going to deselect some other right so the area of cross section has to be the same it is assumed that there is no acceleration  of liquid taking place in the pipe it should not be accelerating there should be a constant velocity with which it should be moving so these are the assumptions that we are going to make you so let’s take it further from you so if we take the case of cylinder of ok the cylinder would have some length and there is a pressure difference between it so p1 is like that the pressure at one end and p2 is the pressure . in L be the length that liquid is entering all the fluid is entering from in and coming out from other r is the radius of the cylinder taking these things into account let’s go further and define everything so what is going to be the force the force due to the pressure is going to be defined as it’s going to be close to the pressure difference into the area, area of cross section that you have we are assuming that they remain same there’s also going to be a physical force because there are different layers of the liquid they are going to exert a force in the opposite direction so you have – ita  2piRL why 2pirl on it because it’s a cylinder so soon they will have a of cross section which is going to be given by a the circumference 2PI R into the length so 2pirL correct but we know that we assume that there is no acceleration right the net force has to be equal to zero so if I just add the pressure force and the viscous force we add the top so that gives the net force to be 0 so if you do that will get what – Delta P 2 pi r square is going to be equals to Ita2pir Dv divided by dr so this is what we have so from here dv by dr can be figured out and that comes out in the form of delta p divide 2ita nlr correct now we need to integrate this so using the empirical velocity gradient we can see that lets say from small are to capital R if you try to find out how exactly the velocity varies between the different radii so we get into it this simple looking expression here correct this expression is related in terms of the pressure difference between the two ends the coefficient of viscosity ita the length of the tube l and the different radii right let’s go further we are now going to talk about in terms of the equation of continuity so using the equation of continuity let’s try to talk about the volume flux so volume flux this is going to be given by DV by DT that is the rate at which the volume is changing with time so that is going to be equal to the velocity into the area we are going to talk about the equation of continuity later but i’m just using that ok so what basically does that tell us we have the expression of velocity if you use that here we’ll get this particular expression in that just need to integrate it so if we integrate from the center to the radius R we’re going to get this particular expression of the flow rate so Q which is equal to the flow rate is going to be given by DV by DT which is dependent on the pressure difference the radius R coefficient of viscosity ita and the length of the tube right let’s try to understand and try to modify this expression to this particular equation is what is known as a poiseuilles of this equation it should be using now instead of delta p happy if he effectively write it down in terms of the pressure difference p 1 and p 2 so p1 and p2 for write it down so we can say that the expression Q can be given by p1 minus p2 and rest Pir4 if we take it to the denominator so we can write it in a very simple-looking form equal to p1 minus p2 upon Rp so what is Rp this term RP is known as the resistance that resistance to the flow of fluid that is what exactly stops the fluid from moving freely so this is what is known as that resistance to the flow of fluid we are going to do a numerical based on this so hopefully you have understood the concept of it thank you

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