Lecture (part 1):

1. Dimensional analysis.
* Dimensional homogeneity.
* The power product method.
* Buckingham Pi Theorem.
* Dimensionless equations.
* Dimensionless groups.

Dimensional analysis.
Many fluid mechanics problems cannot be solved
satisfactorily by the integral and differential (analytical)
methods discussed in Chapters 3 and 4. Usually this is
because the flow becomes turbulent or partially turbulent,
because the geometry is too complex for analysis, or some
combination of these and unknown physical or chemical
effects. Under such circumstances (and even if a theoretical
solution exists) it is necessary to plan experiments and make
measurements. Planning the experiments, and organizing
results and their interpretation, requires frequent use of
dimensional analysis methods.
The basis of dimensional analysis is the principle of
dimensional homogeneity. Physical laws such as the
conservation of mass, momentum, and energy that we have
been applying to fluid mechanics may have several additive
terms and contain many different dimensional quantities in
different combinations, but each of the terms in a correct
physical equation must have the same units. If they do not,
the equation is wrong, or at least the dimensionally
inconsistent term is wrong.
For example, Newton might have felt intuitively that
the acceleration of objects should depend on their mass and
the forces applied, but equations like
a = f m ; a = f^2 / m ; a = m^2 / f
can be rejected immediately because they are dimensionally
inhomogeneous. With the idea that the acceleration is only a
function of mass and acceleration, Newton could have
applied the power product method of dimensional analysis;
that is,
a = const. f^a m^b
where a and b are constants necessary to make the equation
dimensionally homogeneous knowing the units of
acceleration, force and mass. Thus
L/t^2 [=] (ML/t^2)^a M^b
so from L: 1 = a ; from t: -2 = -2a (redundant) ; and from M:
0 = a + b ; so b = -1. Hence,
a = const. f / m
is the only dimensionally correct combination.
Unfortunately, Newton may not have known the dimensions
of force at first.
He may have used the other requirement for a
physical law to be true: correct physical laws must be
reasonable. Forces increase with mass, and also increase
with the acceleration of the mass. It may not have been
obvious that anything as simple as the relationship f = ma
would work. This is a subject requiring experiment (see the
AMES 170 Manual, Experiment 6, p 96). If it had turned
out that f = m^2 a^3 (which is not obviously unreasonable),
then the units of force would have been M^2 L^3 t^-6.

The power product method.
Lots of powerful results can be obtained for physical
problems involving only a few dimensional variables using
the power product method.
Consider the time T required to dissolve crystals of salt with
diffusivity D and surface to volume ratio R. If these are the
only relevant physical parameters of the problem, then set
T = const. D^a R^b
and solve for a and b. From T [=] t ; D [=] L^2 t^-1 ; R =
L^-1, we find
t: 1 = -a ; L: 0 = 2a - b ; so b = -2 ,
which gives
T = const. / D R^2
as the relationship indicated by the assumptions. We know
it is dimensionally homogeneous from our selection of a and
b. Is it physically reasonable? Large diffusivity should
reduce the dissolution time, so the D^-1 dependence is
reasonable. Small particles have large surface to volume
ratios, and it is reasonable that they should dissolve more
rapidly than large particles. Thus we have deduced a
reasonable, dimensionally consistent physical law.
Consider the velocity difference V between two points
separated by a distance L in a turbulent fluid. Kolmogorov
suggested in 1941 that if the separation is larger than the
viscous scale L_K = ( nu^3 eps^-1)^1/4 then the velocity
difference should depend only on L and eps. Thus, by
dimensional analysis using the power product method
V (L,eps) = const. L^a eps^b
where V [=] L^1 t^-1 and eps [=] L^2 t^-3 . For L: 1 = a +
2b ; and for t: -1 = - 3b . Hence, b = 1/3 = a , so
V = const. (L eps)^1/3
according to these assumptions. This relation has been
tested in the laboratory, the atmosphere, the ocean, and in
the interstellar medium, and has been found to be correct. The
constant is universal (like pi in the expression C = pi D for
cylinders that we derived last time).

The Buckingham Pi method.
If there are several dimensional parameters relevant
in a problem, a more elegant method is needed, due to
Buckingham in 1914. A dimensional matrix (n,k) is
constructed, where n is the number of relevant dimensional
variables and k is the number of fundamental dimensions
included in these variables. For example, if we are
interested in the drag force F on an object of size L moving
at constant velocity V through a fluid of density rho and
viscosity mu, and we take these to be the only relevant
variables, then n = 5 (F,L,V, rho , mu) and k = 3 (M,L,t).
Usually the rank of the matrix r is equal to k, where r is the
size of the largest non-zero determinant that can be formed
from the matrix by reordering the rows and columns. After r
has been determined, then a core group of r variables is
selected, and the remaining variables are made dimensionless
with the core group variables, forming the Pi groups. Since
k = r = 3 for our example there will only be two Pi groups
whose relationship must be determined experimentally,
rather than 5. This vastly reduces the cost of the
experiments required.
The core group of variables must contain all the
fundamental dimensions between them (otherwise the
remaining variables containing the missing dimensions can't
be made dimensionless with such a core group), and each
must have different dimensions that the others. The most
important variables should not be included in the core group,
but we see this selection may be a matter of taste. Therefore,
dimensional analysis solutions may not be unique.
For our example, the matrix (FLV rho mu , MLt) is
(1 0 0 1 1 , 1 1 1 -3 -1 , -2 0 -1 0 -1). The rank is 3. If we
choose LV rho as the core group we include all the
dimensions, the dimensions of each variable is different, and
we can use them to make F and mu dimensionless.
Making F dimensionless gives the Euler number Eu
= F / rho V^2 L^2 . Making mu dimensionless gives the
inverse of the Reynolds number; that is mu / rho V L = Re^-
1. Thus, we now must measure Eu versus Re as the
solution to the problem. If we choose L,V,mu as the
core group instead, we get the Scott number F/LVmu
and the Reynolds number rho LV/mu as the Pi groups.
The Scott number is CdRe and is usually constant at small Re.
The Euler number, or Cd, is usually constant for large Re.

Dimensionless equations.
The reason dimensional analysis works can be
understood by making the physical equations describing two
physical situations dimensionless. If the dimensionless
equations are identical, the dimensionless solutions must
also be identical. Dimensionless "Pi" groups emerge in the
process of "normalizing" the describing equations. For
example, many flows are described by the Navier Stokes
equations of momentum conservation. One form of these
equations we discussed was
d,t v = v cross w - grad B + nu del^2 v
where v is the velocity, t is the time, w is the vorticity, B is
the Bernoulli group, and nu is the kinematic viscosity. Each
of the terms in the equation has the units of velocity/time, as
expected from dimensional homogeneity. We can make the
equation dimensionless by multiplying each term by L/V^2,
which has the dimensions of time/velocity. The new
equation is
d,t* v* = v* cross w* - grad* B* + del*^2 v* / Re
where dimensionless time t* = tV/L, dimensionless v* =
v/V, w* = wL/V, grad* = grad L, B* = B/V^2, del* = del
L, Re = LV/nu, and L and V are characteristic length and
velocity scales, respectively.
Notice that the Reynolds number emerges naturally
from this dimensionless equation. It represents the ratio of
each of the other terms to the viscous forces per unit mass nu
del^2 v. For example the inertial vortex forces per unit mass
v cross w divided by nu del^2 v is approximately the
Reynolds number VL/nu. We can see this by approximating
the inertial vortex forces by V times V/L, or V^2/L, and the
viscous forces by nu V/L^2. The ratio (V^2/L)/ (nu V/L^2)
= VL/nu = Re. Consequently, all flows described by these
equations with the same boundary conditions, initial
conditions, and Re values must be identical. Note that the
dimensionless continuity equations and energy equations
must also be identical, but this is true for our example.
Many other forms of the momentum equations may
be needed to describe different flows with other relevant
forces, and when these are made dimensionless other
dimensionless groups emerge. Table 5.2 gives a list of 15,
but there are hundreds of others.

Lecture (part 2):

1. Modeling---flow similarity.
* Geometrical similarity.
* Kinematic similarity.
* Dynamical similarity.
2. Examples.
3. Design Project D6.1 hints.
4. Exam 2 review

Dimensional analysis and modeling.
The power of dimensional analysis is that it permits a
few judiciously selected experimental results on models to be
extrapolated to a wide range of actual, or prototype, flows of
interest. It is much cheaper to carry out a few dozen
measurements at different Reynolds numbers of the drag
coefficients on models of a new automobile or airplane
design in a wind or water tunnel, than to build the prototypes
and find out the bad news on the road or in the air. It is
better to go through a sequence of model tests that can
suggest changes in the prototype design (and the next
models to be tested), and finally build a carefully modeled
prototype to have full confidence that it will work.
More is needed in modeling than simply the correct
list of relevant dimensional parameters of the problem
arranged into Pi groups. The boundary conditions and initial
conditions of the prototype flow must also be represented by
the model: the flows must be "similar" in various ways,
even if the model and prototype flows are not identical.

Geometric similarity.
All linear dimensions of the model and prototype
must be proportional, with the same angles and shapes to
achieve geometrical similarity. Photographs from all angles
must be indistinguishable. Furthermore, the model must be
aligned to the flow at the same angles as the flow over the
prototype for the flows to be considered geometrically
similar. Geometrical similarity is a necessary but insufficient
condition for modeling of a prototype flow.

Kinematic similarity.
Kinematic similarity of flows requires not only the
same length scale ratios of geometric similarity, but also the
same time scale ratios so that identical velocity scale ratios
are achieved. "The motions of two systems are
kinematically similar if homologous particles lie at
homologous points at homologous times" according to
Langhaar in his book on "Dimensional Analysis and the
Theory of Models".
We can see this will work for ideal flows because
reducing the length scales must reduce the velocity scales by
the same factors from the expressions for velocity in terms
of the stream function psi and potential function phi. Given
del*^2 phi* = 0 and geometrically similar boundary
conditions will give the same solution for model and
prototype phi* (x*,y*) in dimensionless coordinates x* = x/L
and y* = y/L normalized with a characteristic dimension L and a
characteristic time T, where phi* = T phi /L^2.
The velocities grad* phi* = (u*,v*) must therefore be
identical, so both the length scale ratios and the time scale
ratios are identical, as required for kinematic similarity.
Frictionless flows with the same Froude number Fr
= V^2/gL will be kinematically similar because V_m/V_p =
t_m/t_p = alpha^1/2 = (L_m/L_p)^1/2. This follows
because (V^2/gL)_p = (V^2/gL)_m , so V_m / V_p =
(L_m/L_p)^1/2 = alpha^1/2, and because t_m/t_p =
(L/V)_m/(L/V)_p = alpha^1/2. Therefore the homologous
position displacement of a model particle moving at velocity
V_m for time t_m will be a distance L_m = V_m t_m = alpha
L_p as required. Equal Froude numbers is an example of
dynamical similarity, because the buoyancy to inertial force
ratios are identical. If other forces such as viscous, surface
tension, and so on are important, then these force ratios must
also be equal for kinematic similarity to be achieved.

Dynamical similarity.
Dynamical similarity includes not only geometrical
(length scale ratio) and kinematic (time scale ratio) similarity,
but also force scale similarity. This means that all the
relevant force ratios such as the Reynolds number
(inertial/viscous), Froude number (inertial/buoyancy),
Rossby (inertial/Coriolis), Euler number (pressure/inertial),
Weber number (inertial/surface tension), etc. that are relevant
must be equal.
Fortunately for the movie industry, which uses
models extensively to cut filming costs, it is not necessary
that dynamical similarity be achieved at all scales for the
flows to look alike. A burning model house on film looks
the same as a real burning house as long as the viewer can't
resolve the viscous scale of the flames in the movie. As long
as the Reynolds number at the smallest visible scale of the
burning model is above the critical value for turbulence to
exist, then turbulence looks like turbulence, independent of
scale (this is part of the universal similarity hypothesis of
high Reynolds number turbulence proposed by Kolmogorov
in 1941).


Buck Rogers is shown streaking across the sky in a rocket
ship. The wake of the rocket has no eddies and rises
perceptibly due to buoyancy. The motion of the rocket is
jerky, and sparks fly off at speeds many times that of the
rocket. Why are we not convinced this is a real rocket ship?
<img src="Buck.Rogers.GIF">

We know from television pictures of the space
shuttle that wakes of real rockets have huge Reynolds
numbers so they are turbulent at all visible
scales. Turbulence has eddies (turbulence is an
eddy-like state of fluid motion where the inertial-vortex
forces of the eddies are larger than any of the other forces
that tend to damp out the eddies). Therefore the wake is a
phony because the model Reynolds number is too small.
The ratio of the rocket velocity to the buoyancy velocity
(squared) is the Froude number which is very small for real
rockets. The wake is also a phony because we can notice that it
is rising. Any jerks in a real rocket's motion imply
enormous accelerations that would be hazardous to the health
of real astronauts. Sparks moving with enormous mach
numbers are not emitted by real rockets.

A large scale example of universal similarity between
wingtip vortices is shown by a radio telescope image
of a galaxy in the Perseus cluster with an active nucleus
(probably a large black hole). The spinning magnetic
field squirts out powerful jets of material that become
turbulent and interact with the strong intergalactic wind
of the dense galaxy cluster, much like wing tip vortices
generated by an airplane wing interact with the wind
over the wing.


Problem 5.12.
The velocity of sound a of a gas varies with pressure
p and density rho . Show by dimensional reasoning that the
proper form must be a = const. (p/ rho )^1/2 .
Solution: Use the power product method. Set a =
const p^c rho^d and equate dimensions. Thus for L: 1 = -c
-3d ; t: -1 = -2c ; M: 0 = c + d . Hence, c = 1/2 = -d, so a =
const (p/ rho )^1/2: QED.

Design project hints.
The hydroponic garden perforated pipe watering
system should be designed to have nearly uniform flow
through the holes along the pipe. The flow rate is not
specified, so an obvious easy solution is to make the holes
so small that there will be only a small change in the pressure
75 kPa along the pipe due to pipe frictional losses and
Bernoulli effects. Please don't do this. You are free to
decide what drill bit sizes are available to the "typical"
machine shop. From the sketch on p 386 the author has in
mind that the watering flow is high enough for the pressure
for the first holes near the pump to be significantly reduced,
so these holes must be larger to keep their watering rate the
same (nearly) as the later holes. Keep in mind that time is
money, and the profit margin for hydroponic gardens is very
low...don't spend much time on this problem trying to
perfectly optimize the solution.
It is best to assume that gravity and frictional effects
are negligible.
Consider control volumes around segments
containing individual holes and apply the Reynolds Transport
Theorems for mass and momentum, and Bernoulli's equation
to determine the pressure variation along the pipe and the
velocity of the jets emerging from the pipe holes. The mass
balance gives a linear decrease in the velocity along the pipe
since Q/50 volume per unit time is removed by each of the
50 holes. The momentum balance does not give the pressure,
but does give the longitudinal force on the pipe walls, in case
you needed this information. Bernoulli's equation shows
that the pressure increases downstream because the velocity
decreases. The velocity through each jet therefore increases
along the pipe (vj = (2 pj / rho )^1/2 , where vj is the jet
velocity, rho is the fluid density, and pj is the pressure for
the jth jet). The volume flow rate for each jet should be Q/50
which is vj x pi Dj^2 / 4 . Thus the jet hole diameter Dj may
be calculated from the pressure pj along the pipe, and your
assumed value of Q. Make sample calculations as you proceed
to be sure your computer or calculator program is working.
The new PowerMacs have Excel 5, which can easily do the
calculations and plots you need (339 EBII).

Exam 2 review.
The exam next time will focus on the material of Chapters 3,
Integral Relations for a Control Volume, Chapter 4,
Differential relations for a fluid particle, and Chapter 5,
Dimensional analysis and similarity. Since some of the early
lectures provided derivations of the differential equations that
appear in Chapters 3 and 4, and since you may need to demonstrate
that you can do these derivations, you should review all
lectures that cover material in Chapters 3, 4 and 5. You should
know how to derive differential equations for the
conservation of mass, chemical species, momentum,
vorticity, and energy, and how to prove vector identities.
Why is curl grad phi = 0? What is the velocity field in the
vicinity of a fluid particle? Can you prove (v dot del)v =
grad (v^2 /2) - v cross w? Can you derive the Bernoulli equation?
Can you derive the Reynolds Transport Theorems for mass and
Review the conditions and assumptions of ideal flow problems
and be prepared to solve and superpose such flows.
You might be asked to solve a viscous flow problem.
For example, what is the flow between two parallel plates
with various boundary conditions of surface velocity and
pressure gradient?

Example: What is the velocity for laminar flow
between two parallel plates where the upper plate moves at
velocity -U in the x direction, the lower plate is fixed, and
there is a negative pressure gradient dp/dx = -K in the
streamwise, or x, direction, where K is a positive constant?
Neglect gravity and other body forces. Assume the flow is
steady and the fluid is Newtonian with constant viscosity mu
and constant density rho.
Solution: From the continuity equation div dot v = 0
for rho = const. The velocities v and w are zero at steady
state because there are no forces to sustain constant velocities
in these directions. Therefore du/dx = 0, so u is a function
only of y and possibly z. There are no variations in the z
direction given, so u is only a function of y, the "vertical"
direction. The only non-zero momentum is in the x
direction, with conservation equation
d,t u + u d,x u + v d,y u + w d,z u = - d,x (p/rho) +
g_x + nu del^2 u
which reduces to
0 = - d(p/rho)/dx + nu d^2,x u = K/rho + nu d^2,y u
which gives after two y integrals
u = -(K/mu)y^2 /2 + C_1 y + C_2
where C_1 and C_2 are constants of integration. These are
evaluated from the boundary conditions that u = -U at y = h
(BC1) and u = 0 at y = 0 (BC2). From BC2 we find that
C_2 = 0. From BC1 it follows that C_1 = (K/mu)h /2 -
U/h. Substituting into the equation for u
u = -(K/mu)y^2 /2 + [(K/mu)h /2 - U/h] y
which is the solution to the problem. Note that the density
of the fluid does not enter in the solution because the inertial
forces are negligible. It is interesting to solve for the
position where the velocity gradient du/dy = 0 = -Ky/mu
-[U/h - Kh/2mu], at y = h/2 - muU/Kh. For small muU/Kh
values the velocity is nearly parabolic with maximum
velocity near the centerline. As muU/Kh increases the
maximum approaches y = 0 and then vanishes when U =
h^2K/2mu. With further increases the velocity
approaches a linear profile u = -Uy/h .

You might be asked to solve dimensional analysis
problems. Consider the following example:

Example: You are a bioengineering student, and
want to design a new animal to be used on a planet the
AMES Department is planning to help colonize. The
question arises as to how thick to make the legs of the
animal. After some discussion you conclude that the leg
diameter D should depend only on the animal's mass M, the
strength of the leg material gamma, and the gravitational
acceleration g_p of the planet.
Solution: Use the power product method to find D =
f(g_p , M , gamma ) = const. g_p^a M^b gamma ^c .
Equate units to find L [=] (L/t^2)^a (M)^b (M/Lt^2)^c .
Solving gives
D = const. [g_pM/gamma]^1/2 .
This equation is dimensionally homogeneous, and is
reasonable with respect to all three dimensional variable.
The constant of proportionality can be found by examining
the legs of any other animal with geometrically similar legs,
known mass, known gravity, and known strength.

Be sure you know how to do Buckingham Pi
analysis and power product analysis. The "power product" method of
dimensional analysis is a subset of of Buckingham Pi analysis,
corresponding to the case where there is only one Pi group. For example,
if the rank of the dimensional matrix is three, you can use the power
product method if there are only four relevant dimensional variables
needed to describe the problem.