Poiseuille formula

Flow of viscous liquid through a capillary tube :

The velocity v at a distance y from the capillary axis for a flow of liquid of viscosity η in a capillary tube of length L and radius r under a pressure difference p across it is given by

$ \displaystyle v = \frac{P (r^2 – y^2)}{4 \eta L} $

and the volume of liquid flowing per second is given by

$ \displaystyle V = \frac{\pi P r^4}{8 \eta L} $

Illustration : A metal plate 0.04 m2 in area is lying on a liquid layer of thickness 10-3m and coefficient of viscosity 140 poise. Calculate the horizontal force needed to move the plate with a speed of 0.040 m/s.

Solution : Area of the place A = 0.04 m2

Thickness Δx = 10-3m

Δx is the distance of the free surface with respect to the fixed surface.

Velocity gradient,

$ \displaystyle \frac{\Delta v}{\Delta x} = \frac{0.04}{10^{-3}}= 40 s^{-1} $

Coefficient of viscosity, η = 14kg/ms-1

Let F be the required force,

Then, $ \displaystyle F = \eta A \frac{\Delta v}{\Delta x} $

= 22.4 N

Illustration : A liquid flows through a pipe of 1.5 mm radius and 15 cm length under a pressure of 15,000 dyne/cm2. Calculate the rate of flow and the speed of the liquid coming out of the pipe. The coefficient of viscosity of the liquid is 1.40 centipoise.

Solution: Radius r = 1.5mm = 0.15 cm

Length L = 15 cm

Pressure difference, p = 15×103 dyne/cm2

Coefficient of viscosity, η = 1.40 centipoise = 0.0140 poise

Rate of flow,

$ \displaystyle V = \frac{\pi P r^4}{8 \eta L} $

= 14.19 cm3/s

Velocity,

$ \displaystyle v = \frac{V}{cross-sectional \, area} $

= 200.84 cm/s

Also Read:

Surface Tension
Surface Energy
Excess Pressure inside a soap bubble
Angle of contact
Capillarity & Ascent Formula
Viscosity , Stoke’s Law & Terminal Velocity

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