Here is a graph of the coefficient of drag as a function of Re number for fluids flowing
over a sphere.
We have 2 different cases.
One the upper case with a small Re number and the lower case with a large Re number.
These experimental results are independent in to the diameter of the sphere, the viscosity
of the fluid, ect., For example the case on the left might the upper case might be some
small Re number.
In the lower case it might be some large Re number and if we know the Re number we immediately
now the coefficient of drag.
It is powerful because we have combined all of these different parameters the diameter
of the sphere, the viscosity, and the density of the fluid.
On to one single curve, and it does work.
The question is how can you figure out before hand to organize the data on terms of the
Re number and the coefficient of drag.
So to figure out these dimensionless parameters we will use what is known as the Buckingham
Pi Theorem.
In the first step, which is often the hardest step.
Is to choose variables or parameters that may affect the experimental results.
This is challenging because it is hard to see it at the beginning.
So what we ultimately want to measure is the force of drag, and the parameters that might
affect it.
I would say the velocity of the fluid would affect it.
The viscosity of the fluid, the density of the fluid, and I would also expect that the
diameter of the sphere would affect the experimental results or affect the force.
So then you count up the number of variables that you would believe would affect the experimental
results, and this number of variables we call it, we give it the variable name k, and in
this case k is equal to 5, because we have 5 variables.
The second step is to choose what is known as the repeating variables for the analysis.
When you select these variables do not select what we call the dependent variable.
So in this case our dependent variable would be the force of drag.
So we can choose whatever we want.
Lets choose the diameter of the sphere (D), the velocity of the fluid (V), and the fluid
density (rho) as our repeating variables.
Third step is to count the number of dimensions that we need to consider.
So we will have dimensions of mass for example in the density we have mass in the numerator.
kilogram per cubic meter.
We got length of course because it describes the diameter, and we also got the dimension
of time in our analysis.
Velocity is length per unit time.
We got three dimensions.
In this case r, the number of dimensions that we need is 3.
In step four is to calculate the number of pi terms that these relationships could be
reduced too.
So in this case the Buckingham Pi Theorem.
It says the number of Pi terms is k-r, and in this case 5-3, and that equals 2.
In this case what we will find is the first pi term is the Re number, and the second pi
term is the coefficient of drag.
The fifth step is to actually form the different pi terms.
To do that we are going to say that pi 1 is equal, lets use one of our none repeating
variables the viscosity.
We will say that the viscosity times diameter to the a power, velocity to the b power, and
the density to the c power.
Our goal is the rearrange these to figure out exponents a, b, and c.
Such that there are no dimensions for this pi term.
These three parameters are what is known as our repeating variables.
So we will repeat these 3 when we calculate the second pi term.
Say that the dimensions of viscosity are units of mass (M) per length (L)* time (T).
The dimensions of the diameter is length (L), and we are going to raise this the the a power.
The dimensions of velocity are m/s or length (L) per time (T) raised to the b power, and
the dimensions of density are mass (M) per length (L) cubed and that is raised to c power.
So for each of these dimensions we can write an equation.
Lets look at mass for example.
We want in our pi term to have dimensions of mass to the 0 power.
So we have mass in the first term.
We have mass in the numerator of 1, and there is nothing in this term or this term, but
in the density we have mass raised to the c power.
Lets do the same thing for the length.
Again we want no dimensions of length.
In the first term the viscosity we have length the the -1 power, and in the diameter scaled
to the the a power plus b for velocity and we have length to the negative 3 c power.
We will do the same thing for time.
-1-b.
Since we have a series of 3 equations and 3 unknowns.
What we find is c equals -1, a is equal to -1, and b is equal to -1.
So if we write this out pi 1 is equal to the viscosity divided by the product of the diameter,
the velocity, and the density.
So to make sure we are going it right.
This better be a dimensionless term.
So we got a numerator of kg/ms for viscosity.
one over meters for diameters in the denominator.
We have seconds per meter the inverse of the velocity, and finally m^3 per kg/ Kilograms
cancel out, second cancel out, we have meters cubed in the denominator, and meters cubed
in the numerator and we have indeed done this right.
We get a dimensionless number for pi 1.
It turns out, because it is dimensionless pi 1 prime for example, and we can say pi
1 prime can simply be equal to one over pi 1.
It is equal to rho, V, D over mu.
This actually has the name as the Re number.
To find the second pi term you can do something similar.
Instead of using the viscosity I am now using the force.
Here is my other three repeating variables V, D, rho raised to an exponent.
Here are dimensions for force ML/T^2, the diameter is L, and velocity is L/T and so
on for the density.
I will do the same thing, write out three algebraic equations according to the dimensions.
What I come up with is that c is equal to -1, a is equal to -2, and D is equal to -2.
So using these exponents I can write pi 2 is equal to the force divided by the diameter
squared, velocity squared, multiplied by the density.
It terms out we can say that the area is simply pi over four times D^2 or we can rearrange
that and say that D^2 is equal 4 over pi times the A.
We do not care about this 4 over pi.
All we care about are the dimensionless groups, and because this constant does not affect
the scaling we will just say that pi 2 prime, and what you will see is that it is characterized
a force divided by 1/2 times a projected area, times the velocity squared times the density.
When it is written this way.
This pi 2 prime is now known as the coefficient of drag.
So again if we have done it right force of drag would be kg*/s^2.
we have 1 over a or 1 over m^2 and we have velocity squared (s^2/m^2) and the density
m^3/kg, and the kg again fall out.
Seconds squared will fall out.
We have m^4 in both the numerator and the denominator and indeed the coefficient of
drag is dimensionless.
So if we made a graph of this.
This the original graph you have seen.
We have pi 2 prime is the coefficient of drag, pi 1 is the Re number, and what that means
is that the coefficient of drag is a function of the Re number and the Re number only.
So for all those different geometries are all those fluids.
All you need to know is the Re number some experimental data and that will give you the
coefficient of drag.