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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Chapter 10.10 - The Venturimeter

In the previous section, we saw the basics of Bernoulli's equation. In this section, we will see some practical applications of the equation. First we will see venturimeter. It can be explained in 12 steps:

1. Fig.10.37(a) below shows a simple water supply system

• Water is taken out from the yellow tank

    ♦ First the water flows through the red pipe

    ♦ Then it flows through the magenta pipe

    ♦ A horizontal magenta pipe with a tap is taken out from the vertical magenta pipe

Fig.10.37

2. A special device is inserted in the horizontal magenta pipe

• This special device is called venturimeter. It is shown in yellow color

• It's main features can be written in 4 steps:

(i) Water enters the venturimeter from the left end

    ♦ At the point of entrance, the venturimeter has the same diameter as the magenta pipe

    ♦ This uniform diameter continues upto a short distance to the right

(ii) But as we proceed further to the right, the diameter gradually reduces to a minimum

(iii) As we proceed further, the diameter again increases gradually and become equal to the diameter of the magenta pipe

(iv) This can be clearly seen in fig.b

    ♦ Fig.b is the enlarged view of the venturimeter in fig.a

3. Position of the venturimeter:

• The venturimeter

    ♦ is at a vertical distance of z₁ from the top surface of water

    ♦ is at a height of h₁ from the datum

4. Two points A and B are marked inside the venturimeter

    ♦ A is at the wider section

    ♦ B is at the narrow section

5. Let us apply Bernoulli’s equation at the two points

• At A we get: $\mathbf\small{\rm{P_a+P_A+\frac{1}{2} \rho_w v_A^2+ \rho_w g h_1}}$

• At B we get: $\mathbf\small{\rm{P_a+P_B+\frac{1}{2} \rho_w v_B^2+ \rho_w g h_1}}$

    ♦ Where ρ_w is the density of water

6. Equating the two, we get: $\mathbf\small{\rm{P_A+\frac{1}{2} \rho_w v_A^2+ \rho_w g h_1=P_B+\frac{1}{2} \rho_w v_B^2+ \rho_w g h_1}}$

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

(Note that, since the venturimeter is horizontal, h₁ is same for both points)

■ It is clear that:

    ♦ If velocity increase at any one of the points, the pressure at that point will decrease
    ♦ If velocity decrease at any one of the points, the pressure at that point will increase
7. Rearranging the equation in (6), we get:

$\mathbf\small{\rm{P_A-P_B=\frac{1}{2} \rho_w \left (v_B^2-v_A^2 \right)}}$
8. So we want the pressure difference (P_A - P_B) between the two points A and B
• For that, a U-tube manometer as shown in fig.b is used. It can be explained in 4 steps:
(i) In fig.b, we see that:
    ♦ One end of the manometer is connected to the wide section of the venturimeter
    ♦ The other end is connected to the narrow section of the venturimeter
(ii) Now we apply the equation of continuity: A_Av_A = A_Bv_B
    ♦ It is clear that, when area decreases, velocity increases
    ♦ Velocity at B will be greater than the velocity at A
(iii) Consequently, the pressure at B will be lesser than the pressure at A
(iv) The greater pressure at A will push the mercury upwards
• As a result, there will be a level difference between the mercury surfaces in the two limbs of the manometer
• This level difference is denoted as ‘h’ in fig.b
• The pressure difference between A and B can be calculated using ‘h’

• Once we know the pressure difference, we can calculate the rate of flow

• The following steps from (9) to (12) show the derivation of the equation for rate of flow:

9. The pressure difference can be calculated in 3 steps:

(i) In fig.b, let the dashed line be the datum. Then we get:

• Pressure at A = P_a + P_A + Pressure corresponding to 'x meters of water'

• Pressure at B = P_a + P_B + Pressure corresponding to '(x-h) meters of water' + Pressure corresponding to 'h meters of mercury'

(ii) Equating the two pressures, we get:

[P_a + P_A + Pressure corresponding to 'x meters of water']

= [P_a + P_B + Pressure corresponding to '(x-h) meters of water' + Pressure corresponding to 'h meters of mercury']

= [P_a + P_B + Pressure corresponding to 'x meters of water' - Pressure corresponding to h meters of water' + Pressure corresponding to 'h meters of mercury']

(iii) Thus we get:

[P_A] = [P_B - Pressure corresponding to 'h meters of water' + Pressure corresponding to 'h meters of mercury']

⇒ [P_A - P_B] = [Pressure corresponding to 'h meters of mercury' - Pressure corresponding to 'h meters of water']

⇒ [P_A - P_B] = [ρ_mgh - ρ_wgh] = [ρ_m - ρ_w]gh N m^-2

    ♦ Where ρ_m and ρ_w are the densities of mercury and water respectively   

10. Next we write v_A in terms of v_B

• From (8.ii), we have: $\mathbf\small{\rm{v_A=\frac{A_B V_B}{A_A}}}$

11. substituting (9) and (10) in (7), we get:

$\mathbf\small{\rm{[\rho_m - \rho_w] gh=\frac{1}{2} \rho_w \left (v_B^2-(\frac{A_B v_B}{A_A})^2 \right)}}$

⇒ $\mathbf\small{\rm{v_B^2=2\left (\frac{\rho_m - \rho_w}{\rho_w} \right )gh \left(\frac{A_A^2}{A_A^2-A_B^2} \right )}}$

⇒ $\mathbf\small{\rm{v_B=\left [\sqrt{\left (\frac{\rho_m}{\rho_w}-1 \right )\left (\frac{2gh}{A_A^2-A_B^2} \right ) }\;\right ]\times A_A}}$

12. Multiplying v_B by A_B, we get the rate of flow. So we get:

Eq.10.8: Rate of flow through the venturimeter =

$\mathbf\small{\rm{\left [\sqrt{\left (\frac{\rho_m}{\rho_w} -1 \right )\left (\frac{2gh}{A_A^2-A_B^2} \right ) }\;\right ]\times A_A A_B}}$

 

Solved example 10.20

A venturimeter has 30 cm diameter at the wide section and 15 cm diameter at the narrow section. If the level difference in the manometer is 20 cm mercury, what is the rate of flow of water through the venturimeter?
Solution:

• Area at the wide section A_A = 0.25 π d_A² = 0.07065 m^₂
• Area at the narrow section A_B = 0.25 π d_B² = 0.01766 m^₂
• Density of water ρ_w = 1000 kg m^_-3
• Density of mercury ρ_m = 13600 kg m^_-3

• h = 20 cm = 0.20 m

• g = 9.81 m s_^-2
• Substituting the known values in Eq.10.8, we get:

Rate of flow through the venturimeter = 0.12570 m_³ s_^-1

Solved example 10.21

Figures 10.38(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is in correct? Why?

Solution:

1. Bernoulli's equation tells us that:

$\mathbf\small{\rm{P+\frac{1}{2} \rho v^2+ \rho_w g h}}$ = A constant

2. In our present case, the venturimeter is horizontal. So we need not consider the third term. We can write:

$\mathbf\small{\rm{P+\frac{1}{2} \rho v^2}}$ = A constant 

3. So it is clear that,
    ♦ When velocity increases, pressure decreases
    ♦ When velocity decreases, pressure increases

Fig.10.38

4. From the equation of continuity, we know that:
At B, the velocity will be higher than the velocity at A
5. So the pressure at B will be lower than at A
So the liquid level at B will be lower than that at A
6. Thus we get:
The fig.a is incorrect

 

Solved example 10.22

The flow of blood in a large artery of an anesthetised dog is diverted through a venturimeter. The wider part of the meter has a cross-sectional area equal to that of the artery. A = 8 mm_². The narrower part has an area a = 4 mm_². The pressure drop in the artery is 24 Pa. What is the speed of the blood in the artery?

Solution:

• In this problem, we are given the pressure difference P directly
• So we will derive a new equation like the one we derived in Eq.10.8

1. Two points A and B are marked inside the venturimeter (see fig.10.38.b)

    ♦ A is at the wider section

    ♦ B is at the narrow section

2. Let us apply Bernoulli’s equation at the two points

• At A we get: $\mathbf\small{\rm{P_A+\frac{1}{2} \rho v_A^2+ \rho g h_1}}$

• At B we get: $\mathbf\small{\rm{P_B+\frac{1}{2} \rho v_B^2+ \rho g h_1}}$

3. Equating the two, we get: $\mathbf\small{\rm{P_A+\frac{1}{2} \rho v_A^2+ \rho g h_1=P_B+\frac{1}{2} \rho v_B^2+ \rho g h_1}}$

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

■ It is clear that:

    ♦ If velocity increase at any one of the points, the pressure at that point will decrease
    ♦ If velocity decrease at any one of the points, the pressure at that point will increase
7. Rearranging the equation in (6), we get:

$\mathbf\small{\rm{P_A-P_B=\frac{1}{2} \rho \left (v_B^2-v_A^2 \right)}}$
8. So we want the pressure difference (P_A - P_B) between the two points A and B

    ♦ But it is directly given in the question as 24 Pa

    ♦ We will denote it as ΔP

9. Next we write v_A in terms of v_B

• From equation of continuity, we have: $\mathbf\small{\rm{v_A=\frac{A_B V_B}{A_A}}}$

10. substituting (8) and (9) in (7), we get:

$\mathbf\small{\rm{\Delta P=\frac{1}{2} \rho \left (v_B^2-(\frac{A_B v_B}{A_A})^2 \right)}}$

⇒ $\mathbf\small{\rm{v_B^2=\frac{2\, \Delta P\,A_A^2}{\rho(A_A^2-A_B^2)}}}$

⇒ $\mathbf\small{\rm{v_B=\left [\sqrt{\frac{2\, \Delta P}{\rho(A_A^2-A_B^2)}}\;\right ]\times A_A}}$

11. Multiplying v_B by A_B, we get the rate of flow. So we get:

Eq.10.9: Rate of flow through the venturimeter =

$\mathbf\small{\rm{\left [\sqrt{\frac{2\, \Delta P}{\rho(A_A^2-A_B^2)}}\;\right ]\times A_A \times A_B}}$

12. Substituting the known values in Eq.10.9, we get:

Rate of flow of blood through the venturimeter =

$\mathbf\small{\rm{\left [\sqrt{\frac{2\, (24)}{(1060)(8^2-4^2)\times 10^{-12}}}\;\right ]\times 8 \times 4\times 10^{-12}}}$ = 9.83 ×10_^-7 m_³ s_^-1

13. So we have: 

• Rate of flow = A_Av_A= 9.83 ×10^_-7 m^₃ s^_-1

• So v_A = $\mathbf\small{\rm{\frac{9.83 \times 10^{-7}}{8 \times 10^{-6}}}}$ = 0.122 m s^_-1

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In the next section, we will see speed of efflux

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