Viscosity , Stoke’s Law , Terminal Velocity

Viscosity : When a fluid flows such that a velocity gradient is set up within it, forces act within the fluid so as to prevent the velocity gradient from existing. This force is due to a property called viscosity.

Suppose that a glass plate in contact with a water column of height h is moved with constant velocity v. Forces of viscosity appear between the solid surface and the layer in contact.

$ \displaystyle F = -\eta A \frac{dv}{dz} $

where η is a constant called co-efficient of viscosity, its cgs unit is poise. Dimension is ML-1T-1 . The SI unit of viscosity equals 10 poise.

Stoke’s Law & Terminal Velocity

When a sphere of radius r moves with a velocity v through a fluid of viscosity η, the viscous force opposing the motion of the sphere is

F = 6πη r v

If for a sphere viscous force become equal to the net weight acting downward, the velocity of the body become constant and is known as terminal velocity.

$ \displaystyle 6\pi \eta r v = \frac{4}{3}\pi r^3 (\rho -\sigma)g $

$ \displaystyle v = \frac{2 r^2 (\rho – \sigma) g}{9 \eta} $

Exercise : Raindrops fall from a very great height. As we see the speed with which it falls is not that large. Why is this so ?

Illustration : A spherical ball of radius 1×10-4 m and density 104 kg/m3 falls freely under gravity through a distance h before entering a tank of water. If after entering the water the velocity of the ball does not change, find h. The viscosity of water is 9.8 × 10-6N -m2

Solution : After falling through a height h, the velocity of the ball becomes

$ \displaystyle v = \sqrt{2 g h} $

After entering water, this velocity does not change, this velocity is equal to the terminal velocity,

$ \displaystyle \sqrt{2 g h} = \frac{2 r^2 (\rho – \sigma) g}{9 \eta} $

On putting given values , we get

or,  h = 20.41 m

Illustration : Two identical drops of water are falling through air with a steady velocity v. If the drops coalesced, what will be the new velocity?

Solution : Let r be the radius of each drop. The terminal velocity vT of a drop of radius r is given by

$ \displaystyle v = \frac{2 r^2 (\rho – \sigma) g}{9 \eta} $   . . . (i)

Now when two drops each of radius r coalesce to form a new drop, the volume of coalesced drop will be given by

$ \displaystyle \frac{4}{3}\pi R^3 = 2\times \frac{}{}\pi r^3 $

The radius of the coalesced drop will be

R = (2)1/3 r

$ \displaystyle \frac{v’}{v} = (\frac{R}{r})^2 = 2^{2/3} $

Hence, the new terminal velocity of the coalesced drop is

or v’ = (2)2/3 v    [as vT = v]

Also Read:

→ Surface Tension
→ Surface Energy
→ Excess Pressure inside a soap bubble
→ Angle of contact
→ Capillarity & Ascent Formula
→ Poiseuille’s formula

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