Derive Poiseuille's formula for the volume of a liquid flowing per second through a pipe under streamlined flow.

Explanation

Consider a liquid flowing steadily through a horizontal capaillary tube. Let `v = (V/t)` be the volume of the liquid flowing out per second through a capillary tube. It depends on (1) coeffiecient of viscosity `(eta`) of the liquid, (2) radius of the tube (r ), and (3) the pressure gradient `(P/l)` .
Then , `" " v prop eta^(a)r^(b)(P/l)^(c )`
`" " v = k eta^a r^b (P/l)^c " "......(1)`
Where, k is a dimensionless constant.
Therefore, `[v] = ("Volume")/("Time") = [L^3T^(-1)] , [(dP)/(dx)] = ("Pressure")/("Distance") = [ML^(-2)T^(-2)]`
`[eta] = [ML^(-1)T^(-1)] and [r ] = [L]`
Substituting in equation (1)
`[L^3T^(-1)] = [ML^(-1)T^(-1)]^(a)[L]^(b)[ML^(-2)T^(-2)]^(c )`
`M^0L^3T^(-1) = M^(a+ c) L^(-a + b - 2c) T^(-a-2c)`
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
Therefore, equation (1) become
`v = k eta^(-1)r^(4)(P/l)^(1)`
Experimentally, the value of k is shown to be `pi/8`, we have
`v = (pi r^4P)/(8 eta l)`
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 `(v_c)`.