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 (l) coefficient of viscosity `(eta)` of the liquid, (2) radius of the tube (r), and (3) the pressure gradient `(P/l)`
Then, `v propeta^(a)r^(b)(P/l)^(c)`
v = `keta^(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 (1)
`[L^(3)T^(-1)] = [ML^(-1)T^(-1)]^(a) [L]^(b) [ML^(-2)T^(-2)]^(c)`
`M^(0)L^(3)T^(-1) = M^(a+b)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
a = -1, b = 4 and c = 1
Therefore, equation (1) becomes,
v = `keta^(-1) r^(4) ((P)/(l))^(1)`
Experimentally, the value of k is shown to be `pi/8` we have
`v = (pir^(4)P)/(8etal)`
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))`.