Chapter 10.13 - Viscosity

In the previous section, we saw Magnus effect. In this section we will see viscosity

Basic details about viscosity can be written in 9 steps:

1. Fig.10.42(a) shows some oil enclosed between two glass plates
    ♦ The bottom glass plate is fixed
    ♦ The top glass plate is movable


Fig.10.42
2. The oil between the two plates can be considered to be made up of layers
    ♦ This is shown in fig.b
3. Let a horizontal force F be applied on the top glass plate
• As a result, that glass plate move towards the right with a velocity v
(See fig.10.43 below)
4. Motion of layers:
• The top most layer-1 which is in contact with the top glass, will move with the same velocity v
    ♦ So we can write: v₁ = v
• The bottom most layer-8 which is in contact with the fixed glass will be stationary
    ♦ So we can write: v₈ = 0
• The layer-2 will move with a velocity lesser than v₁
    ♦ So we can write: v₂ < v₁
• The layer-3 will move with a velocity lesser than v₂
    ♦ So we can write: v₃ < v₂
so on…
5. Since the layers move with different velocities, the distances covered in a time Δt will be different for different layers
    ♦ Upper layers having greater velocities, will move greater distances in Δt
    ♦ Lower layers having lesser velocities, will move lesser distances in Δt
• So the layers will slide past one another
• This is shown in fig.10.43 below:


Fig.10.43
6. We saw that, there is a reduction in velocities from upper layers to lower layers. Such a 'reduction in velocities' occur in a uniform manner
• This can be explained in 5 steps:
(i) We can plot a graph with velocity along the horizontal axis and height along the vertical axis
This is shown in fig.10.43 above
(ii) An example:
    ♦ Let the distance of the layer-3 from the fixed plate be y₃
    ♦ Layer-3 has a velocity v₃
• So we get two coordinates: (v₃,y₃)
(iii) Another example:
    ♦ Let the distance of the layer-7 from the fixed plate be y₇
    ♦ Layer-5 has a velocity v₇
• So we get two coordinates: (v₇,y₇)
• In this way, we get two coordinates for each layer
(iv) All those coordinates will lie on a straight line
    ♦ In our present case, the straight line is shown in red color
(v) The 'straight line' indicates that, the variation of velocity is uniform
• If the variation was not uniform, the graph would have been a curve
7. Forces between the layers:
(i) Consider any one layer, say layer-4
    ♦ The layer-3, which is just above, will pull layer-4 towards the right
    ♦ The layer-5, which is just below, will pull layer-4 towards the left
(ii) So there is a force acting between each layer
• For the layer-4:
    ♦ Rightward force
    ♦ is greater than
    ♦ Leftward force
• So layer-4 moves towards the right
(iii) But it is obvious that, layer-4 experiences a 'resistance to motion'
• This is due to the pull by layer-5
(iv) Layers exert forces on each other because of 'friction' existing between them
■ The resistance to fluid motion is called viscosity
8. Even if viscosity is large, the layers will indeed slide past the lower layers
■ The flow of fluids in layers is known as laminar flow
9. Now consider the flow through a pipe
• In the previous case of glass plates, upper boundary was movable. Only lower boundary was fixed
• But in the case of flow through pipes, the boundary all around is stationary
• Here, laminar flow can be explained in 4 steps:
(i) The flowing liquid can be considered to be made up of concentric cylinders
• This is shown in fig.10.44 (a) below:

Fig.10.44

(ii) Velocities of various cylinders:
    ♦ The cylinder which is in contact with the inner surface of the pipe will have zero velocity
    ♦ The cylinder along the axis of the pipe will have the maximum velocity
    ♦ The outer cylinders will have lesser velocities
    ♦ The inner cylinders will have greater velocities
(iii) A 2D representation is shown in fig.10.44(b)
In both figs.(a) and (b), the layer closest to the pipe is shown to be stationary
(iv) Here also, the ‘variation of velocity’ from the center of the pipe towards the outer surface is linear
• Consider any one cylinder in the fig.
    ♦ The velocity of that cylinder will be a constant

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• We have seen the basics about viscosity. Next we will derive an equation to calculate the viscosity of a fluid. It can be done in 9 steps:

1. In fig.10.45 below, ABCD indicates the original shape of the oil
• Due to the motion of the top glass, the oil deforms to the shape AEFD

Fig.10.45

2. We have seen this type of deformation in the case of solids. It is called shear deformation (Details here)

• We can apply a similar procedure here also:
    ♦ $\mathbf\small{\rm{\frac{F}{A}}}$ is the shear stress
    ♦ Δx is the shear strain

• We apply a force F and measure the resulting Δx

    ♦ Then $\mathbf\small{\rm{\frac{\Delta x}{l}}}$ gives the strain

3. In the case of solids, there is no flow. But in the case of fluids, there is flow

• So for liquids, 'time of flow' has to be considered
    ♦ Because, if the flow takes place for a greater duration of time, Δx will be greater
    ♦ Consequently, the strain will be greater
4. So we consider the strain taking place in unit time
• For that, we divide $\mathbf\small{\rm{\frac{\Delta x}{l}}}$ by Δt

    ♦ Where Δt is the duration in which the flow takes place

• When we divide strain by time, we get the strain rate
■ So $\mathbf\small{\rm{\frac{\Delta x / l}{\Delta t}}}$ is the strain rate
5. But $\mathbf\small{\rm{\frac{\Delta x}{\Delta t}}}$ is the velocity v

So we get: strain rate = $\mathbf\small{\rm{\frac{v}{l}}}$ 

6. Now we consider the proportionality:

    ♦ For solids, shear stress is proportional to shear strain

    ♦ For liquids, Shear stress is proportional to shear strain rate

• So for liquids, we can write: $\mathbf\small{\rm{\frac{F}{A} \propto \frac{v}{l}}}$

⇒ $\mathbf\small{\rm{\frac{F}{A} = a\; constant \; \times \frac{v}{l}}}$

⇒ $\mathbf\small{\rm{\frac{F}{A} = \eta \; \times \frac{v}{l}}}$

    ♦ Where 𝜂 (Greek small letter eta) is the constant of proportionality

    ♦ 𝜂 is called the coefficient of viscosity

    ♦ It is defined as the ratio of shearing stress to strain rate

7. So we can write: $\mathbf\small{\rm{\eta = \frac{\frac{F}{A}}{ \frac{v}{l}}}}$

Thus we get:

Eq.10.12: $\mathbf\small{\rm{\eta = \frac{F\,l}{v\,A}}}$

8. The unit for measuring 𝜂 can be calculated as:

$\mathbf\small{\rm{\frac{(N)\,(m)}{(m s^{-1})\,(m^2)}=N\,s\,m^{-2}}}$

• But N m_^-2 is Pa. So another unit for 𝜂 is: Pa s

9. The dimensions of 𝜂 can be calculated as:

$\mathbf\small{\rm{\frac{[MLT^{-2}]\,[L]}{[LT^{-1}]\,[L^2]}=ML^{-1}T^{-1}}}$

 

Now we will see a solved example:

Solved example 10.29

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. 10.46. 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.10.46

Solution:

1. Tension (T) in the string will be equal to the weight of the 0.010 kg mass

• So we get: T = 0.01 × 9.8 = 0.098 N

2. Shearing force is the force acting parallel to the top surface of the liquid film

• So the Weight of the metal block will not cause any shearing effect

• It is the tension T that causes shearing

• So we can write: F = T = 0.098 N

3. We have Eq.10.12: $\mathbf\small{\rm{\eta = \frac{F\,l}{v\,A}}}$

• Substituting the known values, we get: $\mathbf\small{\rm{\eta = \frac{0.098 \times 0.3 \times 10^{-3} }{0.085 \times 0.1}}}$ = 3.45 × 10_^-3 Pa s

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In the next section, we will see Stokes law


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