# Viscosity and Stokes’ Law

Most of the fluids are not ideal ones and offer some resistance to motion. This resistance to fluid motion is like an internal friction analogous to friction when a solid moves on a surface. It is called viscosity. This force exists when there is relative motion between layers of the liquid. Suppose we consider a fluid like oil enclosed between two glass plates as shown in Fig. 1 (a).

Fig 1 (a) A layer of liquid sandwiched between two parallel glass plates, in which the lower plate is fixed and the upper one is moving to the right with velocity v (b) velocity distribution for viscous flow in a pipe.The bottom plate is fixed while the top plate is moved with a constant velocity v relative to the fixed plate. If oil is replaced by honey, a greater force is required to move the plate with the same velocity. Hence we say that honey is more viscous than oil. The fluid in contact with a surface has the same velocity as that of the surfaces. Hence, the layer of the liquid in contact with top surface moves with a velocity v and the layer of the liquid in contact with the fixed surface is stationary. The velocities of layers increase uniformly from bottom (zero velocity) to the top layer (velocity v).

For any layer of liquid, its upper layer pulls it forward while lower layer pulls it backward. This results in force between the layers. This type of flow is known as laminar. The layers of liquid slide over one another as the pages of a book do when it is placed flat on a table and a horizontal force is applied to the top cover. When a fluid is flowing in a pipe or a tube, then velocity of the liquid layer along the axis of the tube is maximum and decreases gradually as we move towards the walls where it becomes zero, Fig. 1(b). The velocity on a cylindrical surface in a tube is constant.

On account of this motion, a portion of liquid, which at some instant has the shape ABCD, take the shape of AEFD after short interval of time (∆t). During this time interval the liquid has undergone a shear strain of ∆x/l. Since, the strain in a flowing fluid increases with time continuously. Unlike a solid, here the stress is found experimentally to depend on ‘rate of change of strain’ or ‘strain rate’ i.e. ∆x/(l ∆t) or v/l instead of strain itself. The coefficient of viscosity (pronounced ‘eta’) for a fluid is defined as the ratio of shearing stress to the strain rate.

$\eta = \frac{F/A}{v/l}=\frac{Fl}{vA}$

The SI unit of viscosity is poiseiulle (Pl). Its other units are N.s.m$^{-2}$ or Pa.s. The dimensions of viscosity are [$ML^{-1}T^{-1}]. Generally, thin liquids, like water, alcohol, etc., are less viscous than thick liquids, like coal tar, blood, glycerine, etc. The coefficients of viscosity for some common fluids are listed in Table 1.

Table 1 The viscosities of some fluids |

We point out two facts about blood and water that you may find interesting. As Table 1 indicates, blood is ‘thicker’ (more viscous) than water. Further, the relative viscosity (η/η$_{water}$) of blood remains constant between 0 $^o$C and 37$^o$C.

The viscosity of liquids decreases with temperature, while it increases in the case of gases.

Example 1

A metal block of area 0.10 m$^2$ is connected to a 0.010 kg mass via a string that passes over an ideal pulley (considered massless and frictionless), as in Fig. 2. A liquid with a film thickness of 0.30 mm is placed between the block and the table. When released the block moves to the right with a constant speed of 0.085 m.s$^{-1}$. Find the coefficient of viscosity of the liquid.

Fig. 2 Measurement of the coefficient of viscosity of a liquid.

Answer

The metal block moves to the right because of the tension in the string. The tension T is equal in magnitude to the weight of the suspended mass m. Thus, the shear force F is

F = T = mg = 0.010 kg × 9.8 m.s$^{–2}$ = 9.8 × 10$^{-2}$ N

Shear stress on the fluid = F/A = $\frac{9.8 \times 10^{-2}}{0.10}$ N/m$^2$

Strain rate = $\frac{v}{l}=\frac{0.085}{0.30 \times 10^{-3}}$

$\eta$ = (stress/strain rate) s$^{-1}$

$\eta=\frac{(9.8 \times 10^{-2} \ N)(0.30 \times 10^{-2} \ m)}{(0.085 \ m.s^{-1})(0.010 \ m^2)}$

= 3.46 × 10$^{-3}$ Pa.s

** Stokes’ Law**

When a body falls through a fluid it drags the layer of the fluid in contact with it. A relative motion between the different layers of the fluid is set and, as a result, the body experiences a retarding force. Falling of a raindrop and swinging of a pendulum bob are some common examples of such motion. It is seen that the viscous force is proportional to the velocity of the object and is opposite to the direction of motion.

The other quantities on which the force F depends are viscosity η of the fluid and radius a of the sphere. Sir George G. Stokes (1819– 1903), an English scientist enunciated clearly the viscous drag force F as

F = 6$\pi \eta av$

This is known as Stokes’ law. We shall not derive Stokes’ law.

This law is an interesting example of retarding force, which is proportional to velocity. We can study its consequences on an object falling through a viscous medium. We consider a raindrop in air. It accelerates initially due to gravity. As the velocity increases, the retarding force also increases. Finally, when viscous force plus buoyant force becomes equal to the force due to gravity, the net force becomes zero and so does the acceleration. The sphere (raindrop) then descends with a constant velocity.

Thus, in equilibrium, this terminal velocity v$_t$ is given by

6$\pi \eta av_t=\frac{4\pi}{3}a^3(\rho-\sigma)g$

where ρ and σ are mass densities of sphere and the fluid, respectively. We obtain

$v_t=\frac{2a^2 (\rho-\sigma)g}{9\eta}$

So the terminal velocity vt depends on the square of the radius of the sphere and inversely on the viscosity of the medium.

**Example 1**

The terminal velocity of a copper ball of radius 2.0 mm falling through a tank of oil at 20$^o$C is 6.5 cm.s$^{-1}$. Compute the viscosity of the oil at 20$^o$C. Density of oil is 1.5 ×10$^3$ kg,m$^{-3}$, density of copper is 8.9 × 10$^3$ kg m$^{-3}$ .

Answer

We have $v_t$ = 6.5 × 10$^{-2}$ m.s$^{-1}$,

a = 2 × 10$^{-3}$ m,

g = 9.8 m.s$^{-2}$,

ρ = 8.9 × 10$^3$ kg.m$^{-3}$,

σ = 1.5 × 10$^3$ kg.m$^{-3}$.

From Eq. (10.20)

$\eta=\frac{2(2\times 10^{-3} \ m)^2 (7.4 \times 10^{-3} \ kg.m^3) \times 9.8 \ m.s^{-2}}{9\times 6.5 \times 10^{-2} \ m.s^{-1} }$

$\eta$ = 9.9 × 10$^{-1}$ kg.m$^{–1}$.s$^{–1}$

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