FRICTION IN FLUIDS – VISCOSITY

PHYSICS

CLASS: SS1

LESSON NOTE: 3

VISCOSITY – FRICTION IN FLUIDS

Frictional force exists between different layers of liquid surfaces such a frictional force is referred to as Viscosity. The viscosity of a liquid is low if the frictional force is low and vice versa. For instance, the viscosity of water is comparatively lower than that of palm oil. This is because an object moving through water will experience less resistance (frictional force) than when it moves through palm oil. The attraction between molecule in neighbouring area is responsible for the viscosity of the liquid. The shear stress (F/A) is thus needed to facilitate uniform flow of the liquid. Between two solid surfaces, the coefficient of friction, μ = F_f/R. where F_f = Frictional force, R = Reaction.

            Hence, the coefficient of friction depends on the nature of the solid surface and is independent of the area in contact and the relative velocity of the solid surface.

            There is also coefficient of friction between two liquid surfaces.

            Coefficient of viscosity, η = F/ (A X velocity gradient)

            F = Frictional force, A = Area of liquid surface considered.

            Here, the coefficient of viscosity depends on the area of contact and the relative velocity between the liquid surfaces.

            The coefficient of viscosity is defined as the frictional force per unit area of a liquid when it is in a region of unit velocity gradient. The coefficient of viscosity of a liquid is a measure of the resistance of the liquid in orderly flow.

            The unit of coefficient of viscosity is given by Nsm^-2 or Poise.

VARIATION OF VISCOSITY WITH TEMPERATURE AND PRESSURE

The variation of viscosity with temperature is such that for liquids, the coefficient of viscosity decreases as the temperature increases. However, some liquids (e.g. motor engine) are described as viscostatic such that decrease in their coefficient of viscosity hardly change with temperature change.

For most gases, the coefficient of viscosity increases with increase in temperature. However, the coefficient of viscosity is independent of pressure.

SIMILARITIES BETWEEN VISCOSITY AND FRICTION

1. Bothe forces oppose relative motion between surfaces.
2. Both depend on the nature of the materials in contact.

DIFFERENCES BETWEEN VISCOSITY AND FRICTION

1. Friction does not depend on areas of surfaces in contact while viscosity depends on the area of surfaces in contact.
2. Friction is dependent on normal reaction while viscosity does not.
3. Friction occurs in solids while viscosity takes place in both liquids and gases.
4. Friction does not depend on relative velocity between two layers while viscosity depends on the relative velocity between two layers. 

APPLICATIONS OF VISCOSITY

1. Viscous liquids are used as lubricants
2. The knowledge of viscous drag is applied in the design of ships, aircrafts and vehicles.
3. In doors, it is used as damper.
4. It is used in hydraulic press or filter pump.

DETERMINATION OF COEFFICIENT OF VISCOSITY

Figure: Determination of coefficient of viscosity

At Equilibrium,

The Net Force = 0.

This occurs when Terminal Velocity is attained.

The total force acting on the ball = Net Force, i.e.

W – V – U = ma, a = 0 (Terminal Velocity)

Making V the subject of the formula,

V = W – U

But, W = m_s x g

         M_s = ρ_s x v_s

              V_s = 4/3 x πr³

         r = d/2

         V_s = 4/3 x π(d/2)³

          V_s = πd³/6

         M_s = ρ_sπd³/6

        W = ρ_sπd³g/6

Also,

       U = ρ_lπd3g/6

For V, From Stoke’s law

       V = 3πηdu

Where u = Terminal velocity, η = coefficient of viscosity, d = diameter of the sphere.

Hence,

W = U + V

ρ_sπd³g/6 = ρ_lπd3g/6 + 3πηdu

Therefore,

                 η = d²g/18u (ρ_s – ρ_l)

Example: Calculate coefficient of viscosity of a liquid whose density is 1020kgm^-3. If the radius of a sphere which falls inside a liquid is given as 50cm and density 1250 kgm^-3, given that the sphere terminates at a uniform speed of 350cms^-1. (Take g= 9.81ms^-2).

Solution

             η = d²g/18u (ρ_s – ρ_l)

             Given: d = 2r = 2 x 50 = 100cm = 1m;

                         g = 9.81ms^-2

                         ρ_l = 1020 kgm^-3

                         ρ_s = 1250 kgm^-3

                                       ^­u = 350 cms^-1 = 3.5ms^-1

           Required: i. η = ?

           Analysis:

          η = d²g/18u (ρ_s – ρ_l)

          η = 1 x 1 x 9.81 x (1250 – 1020)/18 x 3.5 = 35.8 kgm^-1s^-1.