So going further it is necessary to understand the underlying concept of poiseullie’s laws in very simplest form.

In poiseullie’s laws fluid motion should be streamline, through any pipe.

So let’s starts:

We are starting the concept with a question that in what manner a viscous fluid flows through a cylindrical pipe?

In any cylindrical pipe when a viscous fluid flows then they follow a convex meniscus,but why? It is because, we know that a streamlined fluid has several stream of fluid.

And we know that the pipe is cylindrical then the shape of water which are flowing through it, must be in the shape of cylinder.

Then we can conclude that every fluid lamina in the pipe are in the shape of cylinder.there are infinite laminae between the centre of the pipe and it’s wall.

And friction occurred between the curved surface area the pipe and fluid lamina.

Due to this friction a opposing forces generated and this force is termed as laminar force or viscous force.And in the pipe each lamina opposes the velocity of each lamina, first lamina opposes second lamina, and second opposes the third and so on. And this opposition started from the wall, so the stream which is nearer the wall has minimum velocity, and as distance increases from the wall velocity of the fluid also increases, and become maximum at the centre, it means it’s velocity is effected by the change of radius ml and same occurred from the other sides of the wall, so this form the convex meniscus of the fluid in the pipe.

And this is not only for cylindrical body, this laminar flow held in all types body in which fluid is flowing. This gives a convex form as given below.

So now it’s time to talk about the poiseullie’s laws:

First of all we have to know that poiseullie’s laws measures what?

Poiseullie’s laws measures only the rate of flow of volume of fluid per second in the pipe.

Going further let’s give a sight on, at what factors can volume rate depends:

And third is flow rate is inversely proportional to the viscosity, larger viscosity larger friction then larger opposition and then smaller velocity then smaller flow rate.And fourth is it is inversely proportional to the length of the pipe, how? If length increases then time of friction to the wall also increases, and friction increases then same as above.

So now we are going to derive a formula for rate of flow of fluid volume per second.

## DERIVATION

Let’s take a cylinder of radius R and length L as shown below,

And flowing a fluid of viscosity η , and dr is the small distance between two cylindrical lamina. For understanding see diagram below.

When a fluid is in the pipe then the force of viscosity generated due to the friction between the two lamina is given as below:

F = -ηA(dv/dr)

Where A is the curved surface of the lamina with radius r and length L that is;

A = -2πrL

And dv/dr is the velocity gradient, means velocity changes with change in radius.

Here sign is negative because this force opposes the velocity of the fluid.

Then putting all the values F become

F= -2πηrL(dv/dr)

And we know that fluid is flowing through the pipe due to pressure difference;

P = P1-P2

This pressure can be expressed in external force per unit area, then force is given as;

F = Pπr^2

In equilibrium condition, when external force became equal to the laminar frictional force that is viscous force.

Equating both forces we get

✓ Here we take only magnitude

Pπr^2 = 2πηrL(dv/dr)

After simplification we get,

dv/dr = Pr/2ηL

Then. dv = (Pr/2ηL)dr , now integrating

Both sides from r to R, we get

Int(dv) = (P/2ηL).int(r to R) rdr

On solving we get

v = [P/4ηL](R^2-r^2)

Now applying equation of continuity, we get

V = A1v1=A2v2. Where V is volume

Then for a small cross sectional area dA we get small volume of fluid that is dV

dV = dA×v. Putting all values, we get

dV = 2πrdr× P(R^2-r^2)/4ηL

On solving we get

dV = Pπ(r.R^2-r^3)dr/2ηL

Again integrating both sides from 0 to R because we want to find whole volume of fluid in the pipe. Then we get

Int(dV) = (Pπ/2ηL).int(0 to R)[r.R^2-r^3]dr

On solving we get

V = (Pπ/2ηL)[(R^2.R^2)/2 -(R^4/4)]

V= (Pπ/2ηL)[(R^4/2) – (R^4/4)]

On solving we get

V = PπR^4/8ηL

This is the fluid which are flowing per second through the pipe.