Here we are going to use the idea of similitude,
and scaling using pi groups to determine ultimately
the force on an object.
So lets say that we are told that the force
on this object is a function of volumetric
flow rate, length, density, and gravity, and
from that we are asked to find our dimensionless
or pi groups.
So we used the Buckingham Pi Theorem.
We find that our pi terms 2 of them are F
divided by rho,g, L^2, which is a function
of Q divided by the square root of g*L^5.
Given that our volumetric flow rate on our
model is 0.01 m^3/s.
The flow rate for the prototype is 101.2 m^3/s,
and the length of the prototype is 1000 m.
We are asked what is the length scale.
Another words what is the length of the model
to that of the prototype.
So in order to determine that we need to find
the length of the model.
So we look at our 2 pi groups, and with the
information that we are given we realize that
we are going to have to use our 2nd pi group
in order to find the length of the model.
So Q divided by the square root of g*L^2 of
the model is going to equal Q divided by the
square root of g*L^5 of the prototype.
So we are going to assume that gravity stays
constant.
So Q over the square root of L^5 of the model
equals Q divided by the square root of L^5
of the prototype.
So lets simplify this and try to write it
in terms of L of the model.
So the Q of the model divided by the Q of
the prototype times the square root of Lp^5
equals the square root of the Lm^5.
If we square both sides we get (Qm/Qp)^2 times
Lp^5 equals the length of the model to the
5th.
Then we just raise them both the the 1/5 to
isolate the length of the model (Lm).
So this is (Qm/Qp)^(2/5) times the length
of the prototype is going to be then length
of the model.
When we put our numbers in from above, such
that (0.01/101.2 (both in m^3/s))^2/5 times
1000 m.
We end up with the length of the model equals
25 m.
So our scale which is Lm/Lp is 25/1000, which
equals 1/40.
So lets go look a little bit further in the
same situation, and lets say that we are asked
to find the force on the prototype equals.
If the force on the model is 3 N.
We are going to assume that this is being
done in the same fluid.
So now we are going to use our second pi group.
So Force divided by rho*g*L^3 of the model
is going to be equal to F over rho*g*L^3 of
the prototype.
Since it is the same fluid and we assume that
gravity is constant we can write this as the
force of the model (Fm) over L^3 of the model
(Lm) equals Force of the prototype (Fp) over
L of the prototype cubed (Lp), and this is
a little confusing.
So I am going to rewrite this as Lm^3.
So the Fp equals Fm times (Lp/Lm)^3 and now
you see why we had to use our first dimensionless
parameter because we would not have known
what the scale.
For the scaling of the prototype to the model
would be.
So lets take a look here.
Our force of the prototype equals the force
on the model.
Our scaling is 40 over 1.
That is the inverse of our 1/40 because it
is Lp/Lm and now we have to cube this.
So this equals our 3 N times (40/1)^3.
Therefore our force on the prototype is 192
kN.
