1. Dimensional analysis.

* Dimensional homogeneity.

* The power product method.

* Buckingham Pi Theorem.

* Dimensionless equations.

* Dimensionless groups.

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Dimensional analysis.

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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.

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The power product method.

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Lots of powerful results can be obtained for physical

problems involving only a few dimensional variables using

the power product method.

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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.

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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).

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The Buckingham Pi method.

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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.

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Dimensionless equations.

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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.

1. Modeling---flow similarity.

* Geometrical similarity.

* Kinematic similarity.

* Dynamical similarity.

2. Examples.

3. Design Project D6.1 hints.

4. Exam 2 review

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Dimensional analysis and modeling.

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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.

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Geometric similarity.

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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.

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Kinematic similarity.

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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.

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Dynamical similarity.

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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).

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Examples

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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.

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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.

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Problem 5.12.

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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.

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Design project hints.

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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.

are negligible.

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).

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Exam 2 review.

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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

momentum?

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?

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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 .

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You might be asked to solve dimensional analysis

problems. Consider the following example:

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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.

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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.