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
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
• The ball is moving with out spin
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| Fig.10.41 |
♦ The streamlines are symmetric above and below the ball
3. Fig.10.41(b) shows a ball moving with spin
✰ That means, velocity of air is the same above and below the ball
■ So there is no pressure difference above and below the ball3. 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
• 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
♦ 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:
• 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 shows an aerofoil moving through air
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| Fig.10.42 |
♦ The streamlines are crowded above the aerofoil
♦ The streamlines are rarefied below the aerofoil
• 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
♦ 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
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 × 105 kg. Its total wing area is 500 m2 . 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
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
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
♦ 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 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: $\mathbf\small{\rm{P_a+P_B+\frac{1}{2} \rho v_B^2+ \rho g \times 0}}$
• 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 vA
♦ Air in the lower portion is moving with a velocity of vB
• 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
♦ Air in the lower portion is moving with a velocity of vB
• 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
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 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: $\mathbf\small{\rm{P_a+P_B+\frac{1}{2} \rho v_B^2+ \rho g \times 0}}$
• 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:
♦ vA = 234 km/h = 65 m s-1
♦ vB = 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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