Question

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

Answer 1

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, `"v" ∝ η^"a""r"^"b"("P"/"l")^"c"`

v = `"k"η^"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], [η] = ["M"^-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" + "c") "L"^(-"a" + "b" - 2"c") "T"^(-"a" - 2"c")`

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 = `"k"η^-1"r"^4("P"/"l")^1`

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

v = `(π"r"^4"P")/(8η"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).