Basic details about the law can be written in 8 steps:
1. Consider a small sphere falling through a liquid
• There will be a small amount of friction between the sphere and the liquid
2. Because of this friction, the sphere will try to pull the nearby layers downwards
• This is shown in fig.10.47 below:
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| Fig.10.47 |
3. But layer-1 is in contact with layer-2
• So layer-2 will be trying to stop the downward motion of layer-1
♦ Layer-2 will exert an upward force to stop layer-1
• But since the sphere is moving downwards, the downward force wins
• Thus layer-1 moves downwards, pulling layer-2 along
4. Similarly, on the other side:
• Layer 2’ tries to stop layer-1’
• But layer-1’ moves downwards, pulling layer-2' along
5. We can write the net effect:
• On the left side:
♦ Layer-1 pulls down layer 2
• On the right side:
♦ Layer-1' pulls down layer 2'
6. Now, since layer-2 is pulled downwards, the layer-3 will also experience a downward force
• On the right side, layer-3’ will also experience a downward force
7. In this way, a ‘relative motion between different layers’ will begin
• All the layers will be trying to stop the sphere
■ As a result, the sphere will experience a retarding force
8. An English scientist Sir George G. Stokes discovered that, the retarding force will be equal to 6π𝜂av
• So we can write:
Eq.10.13: F = 6π𝜂av
♦ Where
✰ F = Retarding force due to viscosity
✰ 𝜂 = Coefficient of viscosity of the liquid
✰ a = Radius of the sphere
✰ v = velocity of the sphere
• This is called Stokes’ law
1. Consider a rain drop falling through the sky
• We know that, any falling object will be subjected to the acceleration due to gravity
• Due to the acceleration, the velocity of our rain drop goes on increasing
2. But from Eq.10.13, it is clear that, when velocity increases, the drag force also increases
♦ So we have two increasing quantities:
✰ One is velocity
✰ The other is drag force
3. We know that, the drag force acts upwards
• There is one more force which acts in the upward direction. It is the buoyant force
• Since the rain drop is fully immersed in the atmospheric air, that air will exert an upward force
• So the total upward force = Drag force + Buoyant force
♦ This is shown in fig.10.47(b) above
• Out of the two component forces, the drag force is continuously increasing
♦ So the total upward force is continuously increasing
4. Now we consider the downward forces
• There is only one downward force. It is the gravitational force mg
♦ Where m is the mass of the rain drop
• We know that, this force is a constant
5. Let us compare the forces:
♦ In (3) we have an increasing upward force
♦ In (4) we have a constant downward force
6. We will soon reach a point where the upward force becomes equal to the downward force
• At that point, the net force will be zero
7. If net force is zero, it means that acceleration is zero
• If acceleration is zero, it means that the rain drop is moving with constant velocity
8. When the velocity is constant, drag force also becomes constant
• This is obvious because, all other quantities on the right side of Eq.10.13 are constants
9. Thus, the point mentioned in (6) is an important point
• From that point on wards, the rain drop will begin to fall with a constant velocity
■ This constant velocity is called terminal velocity (vt)
10. Let us write the actual forces just when the rain drop passes the point mentioned in (6):
(i) Viscous drag force = $\mathbf\small{\rm{6 \pi \eta a v_t}}$
♦ Where a is the radius of the drop and 𝜎 is the density of air
(ii) Buoyant force = Weight of the air displaced = $\mathbf\small{\rm{\frac{4}{3} \pi a^3 \sigma g}}$
(iii) Weight of the drop = $\mathbf\small{\rm{\frac{4}{3} \pi a^3 \rho g}}$
Where ρ is the density of the rain drop
11. Applying the condition for equilibrium, we get:
$\mathbf\small{\rm{6 \pi \eta a v_t+\frac{4}{3} \pi a^3 \sigma g=\,\frac{4}{3} \pi a^3 \sigma g}}$
⇒ $\mathbf\small{\rm{6 \pi \eta a v_t=\frac{4}{3} \pi a^3 (\rho-\sigma) g}}$
⇒ $\mathbf\small{\rm{v_t=\frac{\frac{4}{3} \pi a^3 (\rho-\sigma) g}{6 \pi \eta a }}}$
Thus we get:
Eq.10.14: $\mathbf\small{\rm{v_t=\frac{2 a^2 (\rho-\sigma) g}{9 \eta}}}$
■ This equation can be used in the general case
♦ a will be the radius of the body which falls through a fluid with viscosity 𝜂
♦ 𝜎 is the density of the fluid and ρ is the density of the body
12. If we neglect the effect of buoyancy, the step (11) will become:
$\mathbf\small{\rm{6 \pi \eta a v_t=\,\frac{4}{3} \pi a^3 \rho g}}$
• Thus we get:
Eq.10.15: $\mathbf\small{\rm{v_t=\frac{2 a^2 \rho g}{9 \eta}}}$
Let us see a solved example:
Solved example 10.30
In Millikan's oil drop experiment, what is the terminal speed of the uncharged drop of radius 2.0 × 10-5 m and density 1.2 × 103 kg m-3. Take the viscosity of air at the temperature of the experiment to be 1.8 × 10-5 Pa s. How much is the viscous force on the drop at that speed ? Neglect buoyancy of the drop due to air
Solution:
Part (a). To find the terminal speed:
• Substituting all the known values in Eq.10.15, we get:
$\mathbf\small{\rm{v_t=\frac{2 (2\times 10^{-5})^2 (1.2\times 10^{3}) (9.8)}{9 (1.8 \times 10^{-5})}}}$ = 0.058 m
Part (b). To find the viscous force:
• We have Eq.10.13: F = 6π𝜂av
• Substituting the known values, we get:
F = (6)(3.14)(1.8 × 10-5)(2.0 × 10-5)(0.058) = 3.9 × 10-10 N
• Next, we will see the details about Reynold's number
• It can be written in 6 steps
1. We have seen that, when speed of increases beyond a certain value called critical speed, the flow becomes turbulent (Details here)
2. When the flow is turbulent, velocity at any point varies rapidly and randomly. This can be explained in two steps:
(i) Mark any point in the path of a turbulent flow
(ii) At any convenient time t1, measure the speed of flow v(t1) at that point
(iii) At any other convenient time t2, measure the speed of flow v(t2) at that same point
(iv) There is no guarantee that, v(t1) will be equal to v(t2)
(Recall that, in streamline flow, both the velocities will be the same)
(v) The change in velocity at a point will be rapid. That means, the point will experience different velocities in short intervals of time
(vi) The change in velocity at a point will be random. That means, the velocities experienced at the point will have random values.
3. An English scientist Osborne Reynolds discovered that, two conditions are helpful in preventing turbulent flow:
(i) The fluid must be viscous
(ii) The rate of flow must be low
4. He found out that, the various properties of a fluid can be used to calculate a certain number
♦ If this number is below 1000, the flow would be streamline or laminar
♦ If this number is above 2000, the flow would be turbulent
♦ If this number is between 1000 and 2000, the flow would be unsteady
(Unsteady flow is that flow in which the properties like speed pressure etc., changes with time. But those changes would not be rapid or random. Those changes will obey specific rules. We will see unsteady flow in higher classes)
5. This number is called Reynolds number
It can be calculated using Eq.10.16: $\mathbf\small{\rm{R_e=\frac{\rho v d}{\eta}}}$
Where:
♦ Re is the Reynolds number
♦ ρ is the density of the fluid
♦ d is the diameter of the pipe
♦ 𝜂 is the viscosity of the fluid
6. Let us do a dimensional analysis of the right side of Eq.10.16. We get:
$\mathbf\small{\rm{\frac{[ML^{-3}][LT^{-1}][L]}{[ML^{-1}T^{-1}]}=[M^0L^0T^0]}}$
■ So Reynolds number is just a number. It has no units or dimensions
Let us see a solved example:
Solved example 10.31
The flow rate of water from a tap of diameter 1.25 cm is 0.48 L/min. The coefficient of viscosity of water is 10 Pa s. After some time, the flow rate is increased to 3 L/min. Characterize the flow for both the flow rates
Solution:
Case 1. Flow rate = 0.48 L/min
1. First we calculate the speed of flow as follows:
(i) Given that. rate of flow = Av = 0.48 L/min
♦ 1 L = 1000 cm3 = (1000 × 10-6) m3 = 10-3 m3
♦ 1 min = 60 seconds
(ii) So 0.48 L/min = $\mathbf\small{\rm{\frac{0.48\times 10^{-3}}{60}}}$ = 8 × 10-6 m3 s-1
(iii) Thus from (i), we get: $\mathbf\small{\rm{\frac{\pi d^2 v}{4}=8\times 10^{-6}m^3 s^{-1}}}$
⇒ $\mathbf\small{\rm{\frac{\pi (1.25\times 10^{-2})^2 v}{4}=8\times 10^{-6}m^3 s^{-1}}}$
⇒ v = 0.065 m s-1
2. Now we calculate Reynolds number using Eq.10.16: $\mathbf\small{\rm{R_e=\frac{\rho v d}{\eta}}}$
Substituting the known values, we get:
Re = $\mathbf\small{\rm{R_e=\frac{(1000)(0.065)(0.0125)}{(10^{-3})}}}$ = 812.5
3. Re is less than 1000. So the flow will be steady
Case 2. Flow rate = 3 L/min
1. First we calculate the speed of flow as follows:
(i) Given that. rate of flow = Av = 3 L/min
(ii) 3 L/min = $\mathbf\small{\rm{\frac{3\times 10^{-3}}{60}}}$ = 5 × 10-5 m3 s-1
(iii) Thus from (i), we get: $\mathbf\small{\rm{\frac{\pi d^2 v}{4}=5\times 10^{-5}m^3 s^{-1}}}$
⇒ $\mathbf\small{\rm{\frac{\pi (1.25\times 10^{-2})^2 v}{4}=5\times 10^{-5}m^3 s^{-1}}}$
⇒ v = 0.4076 m s-1
2. Now we calculate Reynolds number using Eq.10.16: $\mathbf\small{\rm{R_e=\frac{\rho v d}{\eta}}}$
Substituting the known values, we get:$\mathbf\small{\rm{R_e=\frac{(1000)(0.4076)(0.0125)}{(10^{-3})}}}$ = 5095
3. Re is greater than 2000. So the flow will be turbulent
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