Question

Derive the Poiseuille's Equation for steady flow of a liquid.​

Answer 1

Explanation:

Poiseuille analyzed the steady flow of liquid through a capillary tube. Let v=(V/t) be the volume of the liquid flowing out per second through a capillary tube. ... It depends on (1) coefficient of viscosity (η) of the liquid, (2) radius of the tube (r), and (3) the pressure gradient (P/l)

Answer 2

We can derive Poiseuille’s equation using dimensional analysis. Consider a liquid flowing steadily through a horizontal capillary tube. Let v=(V/t) be the volume of  the liquid flowing out per second through a capillary tube. It depends on (1) coefficient of viscosity (η) of the liquid, (2) radius of the tube (r), and (3) the pressure gradient (P/l) .

Then,

where, k is a dimensionless constant. Therefore,

So, equating the powers of M, L, and T on both sides, we get

a + c = 0, −a + b −2c =3, and −a −2c = −1

We have three unknowns a, b, and c. We have three equations, on solving, we get

a = −1, b = 4, and c = 1

Therefore, equation (7.24) becomes,

Experimentally, the value of k is shown to be π/8 , we have

The above equation is known as Poiseuille’s equation for the flow of liquid through a narrow tube or a capillary tube. This relation holds good for the fluids whose velocities are lesser than the critical velocity (vc).