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 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 `(eta)` of the liquid, (2) radius of the tube (r), (3) the pressure gradient `((P)/(l))`.
Then,
`v alpha 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^(3)T^(-1)]`.
`[(dP)/(dx)] = ("Pressure")/("distance") = [ML^(-2)T^(-2)]`.
`[eta] = [ML^(-1)T^(-1)]` and `[r] = [L]`
Substituting in equation `L^(3)T^(-1) = [ML^(-1)T^(-1)][L]^(b)[ML^(-2)T^(-2)]^(c )`
So, equating the powers of M, L, and T on both sides, we get `m^(0)L^(-3)T^(-1) = M^(a+c)L^(-a+b-2c)T^(-a-2c)`
`a + b =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 (1) becomes,
`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^(4)P)/(8 eta l)`