Chapter 10.14 - Stokes Law and Reynolds Number

In the previous section, we saw viscosity. In this section we will see Stokes law
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:

Fig.10.47

• In the fig.10.47, the sphere will try to pull layer-1 and layer-1’ towards the bottom
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

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Let us see a practical applications of the law. Here we try to explain the 'falling of a rain drop'. It can be written in 12 steps:
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 (v_t)
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 × 10^₃ 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

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• 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:
    ♦ R_e 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 cm^₃ = (1000 × 10^_-6) m^₃ = 10^_-3 m^₃
    ♦ 1 min = 60 seconds
(ii) So 0.48 L/min = $\mathbf\small{\rm{\frac{0.48\times 10^{-3}}{60}}}$ = 8 × 10-6 m^₃ 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 m^₃ 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. R_e is greater than 2000. So the flow will be turbulent

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In the next section, we will see Surface tension



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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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Chapter 10.12 - Magnus Effect

In the previous section, we saw speed of efflux. In this section, we will see another practical application of the Bernoulli's equation, which is: The spray gun. Later in this section, we will see Magnus effect

The working of a spray gun can be explained in steps:
1. Consider the spray gun shown in fig.10.40(a) below:

• The piston is shown in the retracted position. So the cylinder is filled with air
2. Now, the piston is moved with a great speed towards the right
• The air inside flows out. This is shown in fig.b
3. At the outlet, the area of cross section is very low
• So, based on the equation of continuity, the velocity of outflow will be very large
4. Based on Bernoulli’s equation, we know that, when velocity increases, pressure decreases
• So the fluid inside the container will rise up
• The air jet will separate the rising fluid into a large number of minute droplets
• Thus we get the spray

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Magnus effect can be explained in 6 steps:

1. Fig.10.41(a) below, shows a ball moving through the air
• The ball is moving with out spin

Fig.10.41

2. Arrangement of streamlines in the fig. can be described as:

    ♦ The streamlines are symmetric above and below the ball

          ✰ That means, velocity of air is the same above and below the ball

■ So there is no pressure difference above and below the ball
3. Fig.10.41(b) shows a ball moving with spin

• Arrangement of streamlines in the fig. can be described as:

    ♦ The streamlines are crowded above the ball
    ♦ The streamlines are rarefied below the ball

          ✰ That means, velocity of air is greater above the ball

4. We know that, when velocity increases, pressure decreases
• So the air pressure above the ball is lesser than the air pressure below
5. So the air below the ball exerts a net upward force on the ball
■ This dynamic lift due to spinning is called Magnus effect
6. Dynamic lift is different from the 'other type of lift (buoyancy)' which a floating body experiences

    ♦ Buoyancy is experienced even when the body is stationary

    ♦ But dynamic lift can be obtained only when the body is in motion in a fluid
 

Aerofoil
• Details about aerofoil can be written in 5 steps:

1. Aerofoil is a special shape designed in such a way that, when it moves through a fluid, a dynamic lift is obtained
• Fig.10.42 shows an aerofoil moving through air

Fig.10.42

 2. Arrangement of streamlines in the fig. can be described as:

    ♦ The streamlines are crowded above the aerofoil
    ♦ The streamlines are rarefied below the aerofoil

          ✰ That means, velocity of air is greater above the aerofoil

3. We know that, when velocity increases, pressure decreases
• So the air pressure above the aerofoil is lesser than the air pressure below
4. So the air below the aerofoil exerts a net upward force

■ That means, aerofoil experiences magnus effect
5. Aircraft wings have shape similar to the aerofoil
• So the air exerts a net upward force on the wings
• Thus the wings are able to float
• The wings support the whole weight of the aircraft

 

Let us see some solved examples

 

Solved example 10.26

A fully loaded Boeing aircraft has a mass of 3.3 × 10_⁵ kg. Its total wing area is 500 m_² . It is in level flight with a speed of 960 km/h. (a) Estimate the pressure difference between the lower and upper surfaces of the wings (b) Estimate the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface. [The density of air is ρ = 1.2 kg m_^-3]

Solution:

Part (a):
1. Mass of the aircraft = 3.3 × 10⁵ kg

Weight of the aircraft = (3.3 × 10⁵ × 9.8) N 

2. Pressure on the wings

= $\mathbf\small{\rm{\frac{Weight}{Area\,of\,wings}=\frac{3.3 \times 10^5 \times9.8}{500}}}$ = 6468 N m_^-2
3. The air must provide a net upward pressure of 6468 N m_^-2

4. That means:
    ♦ The difference in pressure between
    ♦ Upper portion of the wings
    ♦ And lower portion of the wings
    ♦ Must be equal to 6468 N m_^-2

Part (b):

1. Let us mark two points
    ♦ Point A is at the upper portion of the wing

    ♦ Point B is at the lower portion of the wing

• We will assume that, a horizontal line through B is the datum

• Let the level difference between A and B be h  

2. Let us apply Bernoulli’s equation at the two points
• At A we get: $\mathbf\small{\rm{P_a+P_A+\frac{1}{2} \rho v_A^2+ \rho g h}}$
• At B we get: $\mathbf\small{\rm{P_a+P_B+\frac{1}{2} \rho v_B^2+ \rho g \times 0}}$

3. Equating the two, we get: $\mathbf\small{\rm{P_a+P_A+\frac{1}{2} \rho v_A^2+ \rho g h=P_a+P_B+\frac{1}{2} \rho v_B^2}}$

$\mathbf\small{\rm{P_A+\frac{1}{2} \rho v_A^2+ \rho g h=P_B+\frac{1}{2} \rho v_B^2}}$

• Compared to the size of the aircraft, h will be small. So we can ignore it 

• So the equation becomes:

$\mathbf\small{\rm{P_A+\frac{1}{2} \rho v_A^2=P_B+\frac{1}{2} \rho v_B^2}}$

⇒ $\mathbf\small{\rm{P_B-P_A=\Delta P=\frac{1}{2} \rho \, (v_A^2 - v_B^2)}}$

⇒ $\mathbf\small{\rm{\Delta P=\frac{1}{2} \rho \, (v_A + v_B)(v_A - v_B)}}$

$\mathbf\small{\rm{(v_A - v_B)=\frac{2\Delta P}{\rho(v_A + v_B)}}}$

4. So we have the 'increase in speed'

• Now, $\mathbf\small{\rm{\frac{Increase \;in\;speed}{Original \;speed}}}$ will give the fractional increase in speed  

• Original speed can be taken as the average speed

• So we get:

Fractional increase in speed = $\mathbf\small{\rm{(v_A - v_B)\div \left[\frac{(v_A + v_B)}{2}\right ]}}$

5. So we divide both sides of (3) by $\mathbf\small{\rm{\frac{(v_A + v_B)}{2}}}$

• We get:

Fractional increase in speed = $\mathbf\small{\rm{\frac{2\Delta P}{\rho(v_A + v_B)} \div \frac{(v_A + v_B)}{2}}}$

⇒ Fractional increase in speed = $\mathbf\small{\rm{\frac{2\Delta P}{\rho(v_A + v_B)} \times \frac{2}{(v_A + v_B)}}}$

⇒ Fractional increase in speed = $\mathbf\small{\rm{\frac{\Delta P}{\rho} \times \left[\frac{2}{(v_A + v_B)}\right ]^2}}$  

6. Finding $\mathbf\small{\rm{\frac{(v_A + v_B)}{2}}}$:

    ♦ Air in the upper portion is moving with a velocity of v_A
    ♦ Air in the lower portion is moving with a velocity of v_B
• We can assume that, the average velocity of the above two, it the velocity of the aircraft
• So we get: $\mathbf\small{\rm{\frac{(v_A + v_B)}{2}}}$ = 960 km/h = 267 m s^_-1

7. Substituting this in (5), we get:

Fractional increase in speed

= $\mathbf\small{\rm{\frac{6468}{1.2} \times \left[\frac{1}{267}\right ]^2}}$ = 0.075

8. In percentage form, 0.075 is equal to (0.075 × 100) = 7.5%

• So we can write:
The speed of air in the upper portion needs to be 8% higher than the speed in the lower portion

Solved example 10.27

In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s^-1 and 63 m s^-1 respectively. What is the lift on the wing if its area is 2.5 m² ? Take the density of air to be 1.3 kg m^-3 .

Solution:

1. Let us mark two points
    ♦ Point A is at the upper portion of the wing

    ♦ Point B is at the lower portion of the wing

• We will assume that, a horizontal line through B is the datum

• Let the level difference between A and B be h   

2. Let us apply Bernoulli’s equation at the two points
• At A we get: $\mathbf\small{\rm{P_a+P_A+\frac{1}{2} \rho v_A^2+ \rho g h}}$
• At B we get: $\mathbf\small{\rm{P_a+P_B+\frac{1}{2} \rho v_B^2+ \rho g \times 0}}$

3. Equating the two, we get: $\mathbf\small{\rm{P_a+P_A+\frac{1}{2} \rho v_A^2+ \rho g h=P_a+P_B+\frac{1}{2} \rho v_B^2}}$

$\mathbf\small{\rm{P_A+\frac{1}{2} \rho v_A^2+ \rho g h=P_B+\frac{1}{2} \rho v_B^2}}$

• Compared to the size of the aircraft, h will be small. So we can ignore it 

• So the equation becomes:

$\mathbf\small{\rm{P_A+\frac{1}{2} \rho v_A^2=P_B+\frac{1}{2} \rho v_B^2}}$

⇒ $\mathbf\small{\rm{P_B-P_A=\Delta P=\frac{1}{2} \rho \, (v_A^2 - v_B^2)}}$

4. Substituting the known values, we get: ΔP = 605.15 N m^-2

• This is the net upward pressure

• So the lift = (Pressure × Area) = (605.15 × 2.5) = 1.5 × 10³ N 

  

Solved example 10.28

A plane is in level flight at constant speed and each of its two wings has an area of 25 m 2 . If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be 1 kg m –3 ).

Solution:

1. Given that:

    ♦ v_A = 234 km/h = 65 m s^-1 

    ♦ v_B = 180 km/h = 50 m s^-1 

2. From the previous example, we have:

$\mathbf\small{\rm{P_B-P_A=\Delta P=\frac{1}{2} \rho \, (v_A^2 - v_B^2)}}$

3. Substituting the known values, we get: ΔP = 862.5 N m^-2

• This is the net upward pressure

• So the lift = (Pressure × Area) = (862.5 × 2 × 25) N

4. This lift is equal to the weight of the plane

So the mass = $\mathbf\small{\rm{\frac{862.5\times 2 \times 25}{9.8}}}$ = 4400 kg

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In the next section, we will see viscosity



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