Being the avid soccer and science enthusiast that I am, it was only right for me to do my Extended Essay on a topic that could encompass both of my passions. In soccer, one of the most prized abilities and techniques for a player to master is that of the curved free kick. In order to understand this, one must understand both Bernoulli’s Principle and the Magnus Effect.
Bernoulli’s Theorem is composed of Bernoulli’s Principle and Bernoulli’s Equation, both are rather similar however with some differences which will be explained later. Bernoulli's Theorem was first derived in 1738 by Swiss mathematician Daniel Bernoulli in 1738. Bernoulli’s Theorem pertains to fluid dynamics, the subdiscipline of fluid mechanics with regards to the flow of liquids and gases. It is defined as the “relation among the pressure, velocity, and elevation in a moving fluid (liquid or gas), the compressibility and viscosity (internal friction) of which are negligible and the flow of which is steady, or laminar.”
In other words, Bernoulli’s Theorem is generally used in hydrodynamics when examining the relation between speed, pressure, and width of a pipe, or in aerodynamics when calculating lift or dealing with projectile motion.
Bernoulli’s Principle:
Bernoulli’s Principle is a highly counterintuitive statement relating to the speed of a flow in a horizontal pipe and the pressure at the given point. The Principle states, “At points along a horizontal streamline, higher pressure regions have lower fluid speed and lower pressure regions have higher fluid speed.” The reason the principle states the pipe must remain horizontal is due to the complications which arise with large variations in gravitational potential energy. This caveat is later addresses in Bernoulli’s Equation. This Principle can be used to estimate speeds and pressures in pipes with non-constant widths. An incompressible liquid must speed up in narrower parts of the pipe (i.e. areas with less pressure) due to the higher pressure imposed on the liquid in wider areas of the pipe. In other words, volume flow rate varies directly with (is a function of) pressure. In narrower areas of a pipe less liquid can flow throw if it is moving at a constant speed. This violates the law of conservation of energy as the transfer of energy would no longer equal the amount of work being done seeing as volume flow rate would decrease. In order to remain within the barriers of the law of conservation of energy, the pressure must decrease as speed increases. This allows for the same quantity of liquid per second to pass through all sections of the pipe and therefore of work to equal change in energy. When a fluid flows from a wide to narrow area, the force directed toward the constricted part of the pipe is greater than the force directed from the constricted section to the wide section. This causes the fluid to accelerate. This increase in kinetic energy is the direct result of the work being done by the constricting walls of the pipe.
Deriving Bernoulli’s Equation
Much like in Bernoulli’s principle, the conditions for Bernoulli’s Equation address the relationship between flow rate, speed, and width of pipe however with one key difference: changes in gravitational potential energy.
As demonstrated by the figure to the right, not only are points 1 and 2 different widths, but at different heights as well. As the water moves upward it gains gravitational potential energy Ugand kinetic energy K. Before continuing it is imperative for understanding that all variables be defined. The variables used will be as follows:
U= Gravitational Potential Energy
W= Work
K= Kinetic Energy
A= Area
P= Pressure
F= Force
V=Volume
v= Velocity
m= Mass
g= Acceleration due to gravity
h=Height
ρ= Density
We must start with the simple outline of our equation which can be derived by setting the work done equal to the change in the total energy of the system. This gives us: Wnet=(K+U)systemFor convenience's sake, it will be assumed that the liquid is streamlined (does not contain any crossing streams) and non viscous. Thus there will be no negative forces opposing the flow of the liquid. Additionally, it will be assumed the force of gravity is work that is internal to the system and not external, thus no changes in gravitational force will interfere with the law of conservation of energy. This means that Wexternal=(K+U)system. The only areas of the pipe displayed above which can do work are those outside of the system: P1 and P2. Next, W=Fd and the pressure formula for force is F=PA, if this is plugged in to the original formula we get F=PAd. The water to the left of P1is doing positive work and can therefore be modeled with W1=P1A1d1and the water to the right of point P2is pushing with force W2=-P2A2d2.The work being done near P2is negative because it is being done in the direction opposite that of P1.
We then take the original equation, Wnet=(K+U)system, and plug in the values for W:
(P1A1d1)+(-P2A2d2)=(K+U)system.Because it is assumed that the liquid being used is incompressible and therefore must have a constant volume flow rate throughout the pipe, A1d1=A2d2(note that Ad=volume). The equation can then be arranged as follows:
(P1A1d1)+(-P2A2d2)=(K+U)system
(P1V)+(-P2V)=(K+U)system
The left-hand side of the equation is now complete however we must take into account the difference energy when comparing the liquid in the above diagram located at region 1 and 2. The liquid in region above has received all the transformations and increases in energy necessary to flow through the pipe at a constant volume flow rate, while region 1 has not. This can be accounted for as follows:
(P1V)+(-P2V)=(K+U)system
(P1V)+(-P2V)=(K2+U2)-(K1+U1)
Next, we can replace K and U with the equations for kinetic and potential energy: K=12mv2and U=mgh.By doing so we get:
(P1V)+(-P2V)=(K2+U2)-(K1+U1)
(P1V)+(-P2V)=(12mv22+mgh2)-(12mv21+mgh1)
The reason the ‘m’s do not have a subscript is because volume flow rate must stay constant, the mass of water moving through the pipe is constant as well. Therefore, the mass at region 1 is equal to the mass at region 2 and we can use the same value for ‘m’.
Next, simple algebra is all that is necessary to derive the final version of the equation. We can divide both sides by ‘V’ as follows:
(P1V)V+(-P2V)V=12mv22V+mgh2V-12mv21V+mgh1V
We can cancel out ‘V’s on the left hand side and remove the parentheses. Since ρ=mvwe can substitute ρ in on the right hand side and get the final form of the equation:
P1-P2=(12ρv22)+(ρgh2)-(12ρv12)-(ρgh1)
P1+(12ρv12)+(ρgh1)=P2+(12ρv22)+(ρgh2)
P+(12ρv2)+(ρgh)=constant
This is the final form of Bernoulli’s Equation; note that the three equations above are interchangeable and look different due to algebraic manipulation.