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 |