Physics 6 Tutorial 10 - The Bernoulli Effect
This tutorial is for students of the Welsh Board and Eduqas who are doing Physics in Sport Option.
This is quite a long tutorial.
In 1732, the Swiss physicist and mathematician, Daniel Bernoulli (1700 - 1783), noticed that an increase in speed of a fluid resulted in a decrease in pressure. The equation we will look at is called the Bernoulli Equation, but was actually the work of Leonhard Euler (1707 - 1783). At the time, it was a physical observation. Now we know that it is important in the way that the wing of an aeroplane lifts the machine off the ground, and enables it to fly. This is summed up in the picture below:
The wing section is often called an aerofoil. The air is split and travels faster around the top of the wing than the bottom. There is a region of high pressure underneath the wing as a result of the slower moving air. There is a region of low pressure above the wing due to the faster passage of the air. The pressure difference (multiplied by the area) results in an upwards force called lift. If the lift is greater than the weight, the aeroplane will take off. The effect can be increased by lowering flaps, and the pilot lowers the flaps to take off at a lower speed. Flaps a lowered more as the aeroplane comes in to land.
The downside for this is that the aerofoil causes drag. Some aeroplanes have a laminar flow aerofoil that is symmetrical. This will support the aeroplane with little drag, but needs a high speed for it to work. Flaps are particularly important for flying at low speed, taking off and landing. If the flaps fail, a flapless landing can be decidedly scary!
In motor sport, racing cars have an inverted aerofoil that works in the opposite sense to the wing on an aeroplane. As the car goes faster, the aerofoil puts an extra pressure on the rear tyres to increase grip as the car travels at speeds of between 50 m s-1 and 100 m s-1.
Image by Brian Snelson - Wikimedia Commons
You can see the aerofoils on the front and back of this (rather dated) racing car. They should not be confused with the "go-faster" bits of plastic that some manufacturers add to their sporty models. They were popular in the 1980s and 1990s, as shown in this picture:
Image by Rupert Kent - Wikimedia Commons
At road legal speeds (30 m s-1) a spoiler like this is almost ineffective, impairs the view from the rear-view mirror, causes more drag, and could present a serious hazard to a cyclist or biker who runs into the back of the car. (OK, they shouldn't, but they do.)
Pressure as Energy Density
Energy density is defined as energy per unit volume. The units are joules per cubic metre (J m-3). An obvious example is the energy from a cubic metre of fuel. In Materials Tutorial 3, we saw that the area under the stress-strain graph was energy per unit volume. A stretched longbow in an archery competition has an energy per unit volume. Another example would be the stretched spring in a clay-pigeon trap, which is a device that hurls out clay pigeons (ceramic discs) for shooters to fire at in shooting competitions.
We know that pressure is defined as force per unit area.
We know that if we push against a force for a distance, s, in the direction of the force, we do a job of work:
W = Fs
So we can multiply the pressure equation on both sides by s:
We can rearrange this to give:
We know that area × distance = volume, and that force × distance moved (in the direction of the force) = work, so we can say:
This is energy per unit volume, in other words, energy density.
By converting to base units, show that the units for pressure are consistent with those for energy density.
Energy density from other sources can be described as pressure. Consider a star. There is a lot of energy in a large volume. Therefore the star has an energy density, better known as radiation pressure, which prevents the star from collapsing under the influence of gravity.
Kinetic Energy as Energy Density
Any moving body has kinetic energy, regardless whether it's solid, liquid, or gas. Of course, we know the equation:
We are in the game of turning things into energy density, so we divide the kinetic energy of the fluid by the volume:
We have the term mass ÷ volume in the equation. That is density, r. So we can write:
Remember that energy density is pressure. This is the kinetic energy per unit volume, often called the kinetic pressure, or the dynamic pressure.
A stream of water has a density of 1.00 × 103 kg m-3. It is moving at 3.5 m s-2.
What is the pressure it exerts due to its movement?
Potential Energy and Energy Density
We know that the formula for potential energy is:
We can convert the above equation into energy per unit volume:
Just as we did above, we can say that the mass ÷ volume term is the density:
This equation is used as the equation that describes how pressure increases with depth. It is more commonly written as:
p = hrg
A diver dives to 20 m off the coast to explore the seabed. The density of seawater is 1025 kg m-3.
What is the pressure acting on her?
The Bernoulli Equation
We saw above that pressure is equivalent to energy density. So the Bernoulli Equation can be considered in terms of the Law of Conservation of Energy:
Total energy density = Energy density + kinetic energy density + potential energy density
Reference pressure = pressure + kinetic pressure + potential pressure
The reference pressure is often taken as atmospheric pressure, 1.035 × 105 Pa.
In the air, the potential pressure (energy per unit volume) is very small compared to the atmospheric pressure. So we can simplify the equation to:
In the syllabus, the equation has been rearranged to:
We are interested in the pressure differential, rather than the absolute value for the pressure. So we can write:
A Frisbee is a toy that looks like the lid on a biscuit tin. It is thrown by one person to another, and it floats on the air. A cross-section of the toy is like this:
The Frisbee is 30 cm across and is flying at 5.5 m s-1.
The atmospheric pressure is 1.01 × 105 Pa.
The density of air is 1.23 kg m-3
(a) Assuming that the floating is a result of the Bernoulli Effect, work out the magnitude of the drop in air pressure acting on the top surface of the toy.
(b) Work out the magnitude of the force acting on the toy.
(c) State and explain the direction of the force acting on the toy.
(d) Which data item did you not use?
Force on a spinning ball
If a ball is given spin, the we get the following:
The force makes the ball deviate from a straight line.
State what shape the path will be. Explain your answer.
What assumption have you made?
The force acting on the ball is sometimes called the Magnus Effect.
A tennis ball has a diameter of 6.75 cm and a mass of 58.0 g. It is spinning at 500 rpm and has a forward speed of 20 m s-1.
The atmospheric pressure is 1.01 × 105 Pa.
The density of air is 1.23 kg m-3
(a) Show that the rotational speed of the circumference of the ball is about 1.8 m s-1.
(b) Calculate the value of the total difference in pressure acting on the ball.
(c) Calculate the value of the force acting on the ball.
(d) Work out the radius of the path of the ball, assuming its forward speed remains constant.
Footballers do a similar trick to make the ball curve through the air (hence bend it like Beckham). It requires considerable intuitive skill, which I never had. I am hopeless at all ball games and cannot kick, catch, hit, or throw a ball to save my life. I became a rower and a runner.
Extension - Fluids in a constriction
Consider a fluid of density r passing at a speed v through a pipe that has a constriction in it, like this:
The potential energy per unit volume should be considered as well as the kinetic energy per unit volume. So we write:
At the point p1, we can therefore write:
Similarly at point p2:
So we can equate the two expressions through p0:
This assumes that there is laminar (or smooth) flow that is in steady state, meaning that the flow is constant. It is a simple model. At university level you will find turbulence and other changes factored in. The mathematics is more complex.
Examples of this include:
The Venturi in a carburettor;
The water injectors on a high pressure steam boiler.
Drag is a result of collisions of air molecules on the body of a cyclist. The faster cyclist, the more frequent the collisions are, and the greater the change of momentum. We know that change in momentum results in force. You can feel drag for yourself by pedalling fast on a bicycle. When you are travelling slowly, there is little air resistance. The faster you go, the greater the air resistance becomes. And that means you have to work harder.
Drag on a racing bicycle can be reduced by having a carefully designed frame with streamlined tube cross-sections rather than round. The wheels have a narrow profile and fewer spokes. The rider wears close-fitting clothing made of Lycra ® and streamlined helmets. Some male cyclists shave the body hair on their legs and arms.
Image from Wikimedia Commons
The best of these machines, built with hi-tech materials, can cost more than a car (and are ridden by MAMILs and OMILs - Middle-aged Men In Lycra and Old Men In Lycra).
Drag depends on these factors:
The cross-sectional area facing the direction of movement (A);
The density of the fluid (r);
The square of the speed (v2);
The drag coefficient (CD).
The formula for the drag force is this:
The term ½ rv2 is the kinetic energy per unit volume, or dynamic pressure. This can be given the physics code q, so we can write:
What are the units for the drag coefficient? Explain your answer.
We can rearrange the equation to write:
Therefore the drag coefficient is the ratio of the drag force to the product of the dynamic pressure and area.
The drag coefficient is determined experimentally and is affected by a large number of different factors, some of which can be quite complex. In a sporting context the drag coefficient explains why fast moving projectiles like a javelin are long and thin. However there is a fly in the ointment - skin friction. The idea is shown in the table below:
Shape and Form
Form Drag / %
Skin Friction / %
You can see that a long thin shape has very low drag, but high skin friction. However the skin friction is rather less significant in causing drag than the cross-sectional area.
A javelin of mass 600 g has a diameter of 30 mm. It is thrown through the air at a speed of 20 m s-1. The athlete accelerates the javelin to its flight speed in 0.10 s.
(a) The acceleration of the javelin;
(b) The force required to accelerate the javelin;
(c) The drag force acting on the javelin if the drag coefficient is 0.01;
(d) The range of the javelin if it's thrown at an angle of 45o; (Look at Mechanics Tutorial 9)
(e) The kinetic energy of the javelin.
(f) The flight path is about 56 m. Work out the lost energy due to the drag force.
Density of air = 1.23 kg m-3
Acceleration due to gravity = -9.81 m s-2 (i.e. downwards).
The flight characteristics of a javelin are often studied by third year undergraduate students for their dissertations. The Question 8 is a simplification.
Skin Friction (Extension only)
This eight-oared racing shell is long and thin. It has a low form drag, but a high skin friction.
The skin friction on this eight-oared racing shell arises from turbulence between the surface of the shell and the water. The skin friction is the ratio of the skin shear stress (from the turbulence) and the dynamic pressure.
In our example of the javelin, the skin friction was very small. In the case of this racing shell, it is significant. The equation is:
Cf - skin friction (no units);
tw - skin shear stress (N m-2);
q - dynamic pressure (N m-2).
The dynamic pressure is given by:
r - density (kg m-3);
v - speed (m s-1).
A racing shell has a total mass of 750 kg. It is travelling at 5.7 m s-1. The crew stop rowing and allow the boat to run. It comes to a stop in a distance of 40 m.
Assume that all the kinetic energy is lost as a result of the drag force and skin force from the water.
(a) Show that the kinetic energy of the racing shell is about 12 kJ.
(b) Calculate the magnitude of the acceleration of the racing shell using an appropriate equation of motion.
(c) Calculate the value of the total skin friction and drag force.
(d) The area perpendicular to the direction the movement of the boat is about 0.047 m2. What is the drag force if the drag coefficient is 10 %?
(e) Calculate the skin friction. Give your answer to an appropriate number of significant figures.
Relating skin friction to drag force involves several different factors, which are beyond the scope of our discussion.