Informally, viscosity is the quantity that describes a fluid’s resistance to flow. Fluids resist the relative motion of immersed objects through them as well as to the motion of layers with differing velocities within them.

Let us consider a fluid like oil enclosed between two glass plates as shown in Fig. 10.14 (a). The bottom plate is fixed while the top plate is moved with a constant velocity \(\mathbf{v}\) relative to the fixed plate. Hence, the layer of the liquid in contact with the top surface moves with a velocity \(\mathbf{v}\) and the layer of the liquid in contact with the fixed surface is stationary. The velocities of layers increase uniformly from the bottom (zero velocity) to the top layer (velocity \(\mathbf{v}\)). 

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. 10.14 (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 a short interval of time \((\Delta t)\). During this time interval, the liquid has undergone a shear strain of \(\Delta x / l\). 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. \(\Delta x /(l \Delta t)\) 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.
\(
\eta=\frac{F / A}{v / l}=\frac{F l}{v A} \dots(10.18)
\)

The SI unit of viscosity is poiseiulle (Pl). Its other units are \(\mathrm{N} \mathrm{s} \mathrm{m}^{-2}\) or Pa s. The dimensions of viscosity are \(\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]\). The coefficients of viscosity for some common fluids are listed in Table 10.2.

We point out two facts about blood and water that you may find interesting. As Table \(10.2\) indicates, blood is ‘thicker’ (more viscous) than water. Further, the relative viscosity \(\left(\eta \eta_{\eta_{\text {water }}}\right)\) of blood remains constant between \(0^{\circ} \mathrm{C}\) and \(37^{\circ} \mathrm{C}\).

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

Flow Through a Narrow Tube (Poiseuille’s Equation)

Suppose a fluid flows through a narrow tube in steady flow. Because of viscosity, the layer in contact with the wall of the tube remains at rest and the layers away from the wall move fast. Poiseuille derived a formula for the rate of flow of viscous fluid through a cylindrical tube.

Suppose a fluid having coefficient of viscosity \(\eta\) and density \(\rho\) is flowing through a cylindrical tube of radius \(r\) and length \(l\). Let \(P\) be the pressure difference in the liquid at the two ends. It is found that the volume of the liquid flowing per unit time through the tube depends on the pressure gradient \(P / l\), the coefficient of viscosity \(\eta\) and the radius \(r\). If \({V}\) be the volume flowing in time \(t\), we guess that
\(
\frac{V}{t}=k\left(\frac{P}{l}\right)^a \eta^b r^c
\)
where \(k\) is a dimensionless constant.
Taking dimensions,
\(
\begin{array}{rlrl}
\mathrm{L}^3 \mathrm{~T}^{-1}=\left(\frac{\mathrm{ML}^{-1} \mathrm{~T}^{-2}}{\mathrm{~L}}\right)^a\left(\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right)^b \mathrm{~L}^c \\
\text { or, } \quad \mathrm{L}^3 \mathrm{~T}^{-1} =\mathrm{M}^{a+b} \mathrm{~L}^{-2 a-b+c} \mathrm{~T}^{-2 a-b}
\end{array}
\)
Equating the exponents of M, L, and T we get,
\(
\begin{aligned}
0 & =a+b \\
3 & =-2 a-b+c \\
-1 & =-2 a-b .
\end{aligned}
\)
Solving these equations,
Thus,
\(
a=1, b=-1 \text { and } c=4 .
\)
\(
\frac{V}{t}=k \frac{\operatorname{Pr}^4}{\eta l} .
\)
The dimensionless constant \(k\) is equal to \(\pi / 8\) and hence the rate of flow is
\(
\frac{V}{t}=\frac{\pi \operatorname{Pr}^4}{8 \eta l}
\)
This is Poiseuille’s formula.

Example 1: A metal block of area \(0.10 \mathrm{~m}^2\) is connected to a \(0.010 \mathrm{~kg}\) mass via a string that passes over an ideal pulley (considered massless and frictionless), as in Fig. 10.15. A liquid with a film thickness of \(0.30 \mathrm{~mm}\) is placed between the block and the table. When released the block moves to the right with a constant speed of \(0.085 \mathrm{~m} \mathrm{~s}^{-1}\). Find the coefficient of viscosity of the liquid.

Solution:

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 \(\mathrm{m}\). Thus, the shear force \(F\) is
\(
F=T=m g=0.010 \mathrm{~kg} \times 9.8 \mathrm{~m} \mathrm{~s}^{-2}=9.8 \times 10^{-2} \mathrm{~N}
\)
Shear stress on the fluid \(=F / A=\frac{9.8 \times 10^{-2}}{0.10} \mathrm{~N} / \mathrm{m}^2\)
\(
\begin{aligned}
& \text { Strain rate }=\frac{v}{l}=\frac{0.085}{0.30 \times 10^{-3}} \\
& \eta=\frac{\text { stress }}{\text { strain rate }} \mathrm{s}^{-1} \\
& =\frac{\left(9.8 \times 10^{-2} \mathrm{~N}\right)\left(0.30 \times 10^{-3} \mathrm{~m}\right)}{\left(0.085 \mathrm{~m} \mathrm{~s}^{-1}\right)\left(0.10 \mathrm{~m}^2\right)} \\
& =3.46 \times 10^{-3} \mathrm{Pas} \\
&
\end{aligned}
\)

Stokes’ Law

The Stokes law states when a body falls through a fluid, it drags the layer of the fluid in contact with it. A relative motion is developed between the different layers of the fluid, and as a result, the body experiences a retarding force. This retarding force acting on a body as it passes through a viscous fluid is directly proportional to the sphere’s velocity, radius, and fluid viscosity.

A few common examples of such a motion include a person falling with a parachute, pouring raindrops, and swinging a pendulum bob. It is seen that the viscous force is proportional to the velocity of the object and is opposite to the direction of motion.

To derive the Stokes Law Formula, he performed several experiments to understand the motion of small spherical bodies in different fluids, and he concluded that the viscous force \(F\) acting on a spherical body of radius \(r\) depends directly on the following quantities, the radius \((r)\) of the sphere falling through the liquid, the velocity \((v)\) of the sphere falling through the liquid, and the coefficient of viscosity \((\eta)\) of the liquid.

Thus, The stokes Law formula states that the drag force \(F\) acting upward in resistance to the fall can be given as:
\(
F=6 \pi r \eta v \dots(10.19)
\)

This law is an interesting example of retarding force, which is proportional to velocity. 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 \(v_{\mathrm{t}}\) is given by
\(
6 \pi \eta a v_{\mathrm{t}}=(4 \pi / 3) a^3(\rho-\sigma) g
\)

where \(\rho\) and \(\sigma\) are mass densities of sphere and the fluid, respectively. We obtain
\(
v_{\mathrm{t}}=d / t =2 a^2(\rho-\sigma) g /(9 \eta) \dots(10.20)
\)
So the terminal velocity \(v_{\mathrm{t}}\) depends on the square of the radius of the sphere and inversely on the viscosity of the medium.

Measuring coefficient of Viscocity by Stokes’ Method

The viscosity of a liquid may be determined by measuring the terminal velocity of a solid sphere in it. The figure below shows the apparatus. A test tube \(A\) contains the experimental liquid and is fitted into a water bath \(B\). A thermometer \(T\) measures the temperature of the bath. A tube \(C\) is fitted in the cork of the test tube \(A\). There are three equidistant marks \(P, Q\), and \(R\) on the test tube well below the tube \(C\).

A spherical metal ball is dropped into tube \(C\). The time interval taken by the ball to pass through the length \(P Q\) and through the length \(Q R\) are noted with the help of a stop watch. If these two are not equal, a smaller metal ball is tried. The process is repeated till the two-time intervals are the same. In this case the ball has achieved its terminal velocity before passing through the mark \(P\). The radius of the ball is determined by a screw gauge. It’s mass \(m\) is determined by weighing it. The length \(P Q=Q R\) is measured with a scale.
Let \(r=\) radius of the spherical ball
\(m=\) mass of the ball
\(t=\) time interval in passing through the length
\(P Q\) or \(Q R\)
\(d=\) length \(P Q=Q R\)
\(\eta=\) coefficient of viscosity of the liquid
\(\sigma=\) density of the liquid.
The density of the solid is \(\rho=\frac{m}{\frac{4}{3} \pi r^3}\) and the terminal velocity is \(v_0=d / t\). Using the equation of terminal velocity (10.20, radius is taken as \(a=r\))
\(
\eta=\frac{2}{9} \frac{(\rho-\sigma) g r^2}{d / t} \text {. }
\)
This method is useful for a highly viscous liquid such as Castor oil.

Example 2: The terminal velocity of a copper ball of radius \(2.0 \mathrm{~mm}\) falling through a tank of oll at \(20^{\circ} \mathrm{C}\) is \(6.5 \mathrm{~cm} \mathrm{~s}^{-1}\). Compute the viscosity of the oil at \(20^{\circ} \mathrm{C}\). Density of oil is \(1.5 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\), density of copper is \(8.9 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\)

Solution:

We have \(v_{\mathrm{t}}=6.5 \times 10^{-2} \mathrm{~ms}^{-1}, a=2 \times 10^{-3} \mathrm{~m}\), \(g=9.8 \mathrm{~ms}^{-2}, \rho=8.9 \times 10^3 \mathrm{~kg} \mathrm{~m}{ }^{-3}\),
\(\sigma=1.5 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\). From Eq. (10.20)
\(
\begin{aligned}
\eta & =\frac{2}{9} \times \frac{\left(2 \times 10^{-3}\right)^2 \mathrm{~m}^2 \times 9.8 \mathrm{~m} \mathrm{~s}^{-2}}{6.5 \times 10^{-2} \mathrm{~m} \mathrm{~s}^{-1}} \times 7.4 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3} \\
& =9.9 \times 10^{-1} \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-1}
\end{aligned}
\)

Example 3: An air bubble of diameter \(2 \mathrm{~mm}\) rises steadily through a solution of density \(1750 \mathrm{~kg} \mathrm{~m}^{-3}\) at the rate of \(0.35 \mathrm{~cm} \mathrm{~s}^{-1}\). Calculate the coefficient of viscosity of the solution. The density of air is negligible.

Solution:

The force of buoyancy \(B\) is equal to the weight of the displaced liquid. Thus,
\(
B=\frac{4}{3} \pi r^3 \sigma g .
\)
This force is upward. The viscous force acting downward is
\(
F=6 \pi \eta r v .
\)
The weight of the air bubble may be neglected as the density of air is small. For uniform velocity
\(
\begin{aligned}
F & =B \\
\text { or, } 6 \pi \eta r v & =\frac{4}{3} \pi r^3 \sigma g \\
\text { or, } \quad \eta & =\frac{2 r^2 \sigma g}{9 v} \\
& =\frac{2 \times\left(1 \times 10^{-3} \mathrm{~m}\right)^2 \times\left(1750 \mathrm{~kg} \mathrm{~m}^{-3}\right)\left(9 \cdot 8 \mathrm{~m} \mathrm{~s}^{-2}\right)}{9 \times\left(0.35 \times 10^{-2} \mathrm{~m} \mathrm{~s}^{-1}\right)} \\
& \approx 11 \text { poise. }
\end{aligned}
\)
This appears to be a highly viscous liquid.

Critical Velocity and Reynolds Number

When a fluid flows in a tube with small velocity, the flow is steady. As the velocity is gradually increased, at one stage the flow becomes turbulent. The largest velocity which allows a steady flow is called the critical velocity.

Whether the flow will be steady or turbulent mainly depends on the density, velocity, and coefficient of viscosity of the fluid as well as the diameter of the tube through which the fluid is flowing. The quantity

\(N=\frac{\rho v D}{\eta}\)

is called the Reynolds number and plays a key role in determining the nature of the flow. It is found that if the Reynolds number is less than 2000, the flow is steady. If it is greater than 3000, the flow is turbulent. If it is between 2000 and 3000, the flow is unstable. In this case it may be steady and may suddenly change to turbulent or it may be turbulent and may suddenly change to steady.