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Dimensional Analysis (Partial Analysis) DA is a mathematical method of considerable value to problems in

science and engineering especially physics and fluid mechanics. All physical quantities can usually be expressed in terms of certain

primary quantities which in mechanics are: Length (L), Mass (M),

and Time (T).

e.g. Force = Mass x Acceleration

= Mass x Length/Time2 = M L T-2

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The principle of homogeneity of dimensions can be used to:

1. Check whether an equation has been correctly formed

(apples = apples) ;

2. Establish the form of an equation relating a number of

variables (dimensional analysis); and

3. Assist in the analysis of experimental results (experimental

design).

Checking Equations:

Eg. 1: Show by D. A. that the equation:

p + ½ ρ v2 + ρ g z = H

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is a possible relationship between pressure p, velocity v, and height z for a frictionless flow along a streamline of a fluid of density ρ, and determine the dimension of H.

Solution:

p = Force/Area = m a/Area = M L T-2 L-2 = M L-1 T-2

ρ = mass/volume = M L-3

v = length/time = L T-1

g = L T-2, Z = L

p = M L-1 T-2

½ ρ v2 = M L-3 x L2 T-2 = M L-1 T-2

ρ g z = M L-3 x L T-2 x L = M L-1 T-2

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All terms on the l.h.s have the same dimensions and the equation is

physically possible if H also has dimension of M L-1 T-2. ½ is a

pure number hence not amenable to D. A. Constant could have

been another number and the equation is still valid.

Therefore, the analysis is not complete. We need additional

information to confirm that the constant is indeed ½.

Eg. 2. Check that Einstein’s famous equation is indeed

dimensionally correct: E = mc2

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Methods of dimensional analysis: grouping of variables.

Many methods are available but we will consider the main ones:

1. Rayleigh indicial method

2. Buckingham π

3. Hunsaker and Rightmire (Whittington’s)

4. Matrix Method

The methods above are given roughly in chronological order. Also

from most tedious to use to almost automatic (if matrix inversion

routine available).

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Rayleigh’s method:

Example: The velocity of propagation of a pressure wave through a liquid can be expected to depend on the elasticity of the liquid represented by the bulk modulus K, and its mass density ρ. Establish by D. A. the form of the possible relationship. Assume: u = C Ka ρb

U = velocity = L T-1, ρ = M L-3, K = M L-1 T-2

L T-1 = Ma L-a T-2a x Mb L-3b

M: 0 = a + b

L: 1 = -a – 3b

T: -1 = - 2a

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Therefore: a = ½ , b = -½, and a possible equation is: ρKCu =

Rayleigh’s method is not always so straightforward. Consider the situation of flow over a U-notched weir.

Q = f(ρ, µ, H, g)

[Q] = [C ρa µb Hc gd] [ ] => dimensions of

Using the M, L, T system,

[L3 T-1] = [ML-3]a [M L-1 T-1]b [L]c [L T-2]d

M: 0 = a + b (1)

L: 3 = -3a –b +c + d (2)

T: -1 = - b – 2d (3)

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We have only 3 equations, but there are 4 unknowns. Need to

express a, b, c, in terms of d.

b = 1 – 2d

a = -b = 2d -1

c = 3 + 3a + b – d = 1 + 3d

Q = C ρ(2d-1) µ(1-2d) H(1+3d) g(d)

d

gHHC

= 2

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µρ

ρµ

= 2

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µρφ

µρ gHHQ 2 dimensionless groups. Please check.

This method can be very tedious if there are more variables.

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Buckingham’s π theorem

This is the most well-known method and is the basis for the matrix

and other methods.

“If there are n variables in a problem and these variables contain m

primary dimensions (e.g. M, L, T), the equation relating the

variables will contain n-m dimensionless groups”. Buckingham

referred to these dimensionless groups as π1, π2, …, πn-m, and the

final equation obtained is: π1 = φ (π2, π3,…, πn-m).

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E.g. consider the previous problem with 5 variables, hence there

will be 5 – 3 = 2 dimensionless groups.

The dimensionless groups can be formed as follows:

1) Select 3 repeating variables from the list of variables which

together must contain M, L,T. eg. ρ (ML−3), H (L), g (LT-2)

2) Combine repeaters with one other variable in turn,

3) Never pick the dependent variable as a repeater.

Consider: Q = f(ρ, µ, H, g) 5 variables

Pick repeaters: ρ, H, g

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Combine with one other variable, say, Q.

QgH zyx 1111 ρπ =

( ) ( ) ( ) ( )1323 111 −−−= TLLTLMLzyx

M: x1 = 0

L: -3x1 + y1 + z1 + 3 = 0

T: -2z1 – 1 = 0

Therefore: z1 = -½ , y1 = -5/2

QgH 2/12/501

−−= ρπ

2/52/11 HgQ

=π

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µρπ 2222

zyx gH=

( ) ( ) ( ) ( )1123 222 −−−−= TMLLTLMLzyx

M: x2 + 1 = 0

L: -3x2 + y2 + z2 -1 = 0

T: -2z2 – 1 = 0

x2 = -1, z2 = -1/2, y2 = 3/2

µρπ 2/12/312

−−= gH

2/12/32 gHρµπ =

That is:

= 2/12/32/52/1 gHHg

Qρ

µφ

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Ex. The head loss per unit length (∆h/L) of pipe in turbulent flow through a smooth pipe depends on v, D, ρ, g, and µ. Use Buckingham’s π theorem to determine the general form of the equation. F(∆h/L, v, D, ρ, µ, g) = 0; 6 variables, 3 dimensions, 3 π terms. That is: π1 = φ(π2, π3). Choose 3 repeating variables: v, D, and ρ

µρπ 1111

xxx Dv=

( ) ( ) ( ) 1131 111 −−−−= TMLMLLLTxxx

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M: z1 + 1 = 0

L: x1 + y1 – 3z1 – 1 = 0

T: -x1 – 1 = 0

Therefore: x1 = -1, z1 = -1, and y1 = -1.

µρπ 1111

−−−= Dv or vDρµ or Re=

µρvD

Final solution: { }2Re,FrLh φ=

∆

Where Re = Reynold’s number, and Fr = Froude number = gDv

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Hunsaker and Rightmire’s Method (1947)

This method is easier to use and quicker than Buckingham’s

method. The method is similar the Buckingham’s method by use of

repeating variables, but express them in term of the variables

themselves.

e.g. D to represent Length, [L] = D

v to represent Time, [T] = L/v = D/v

r to represent Mass, [M] = ρL3 = ρD3

π1: g = L T-2 = v2D D-2 = v2 D-1, therefore π1 = gDv2

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π2: µ = M L-1 T-1 = vDDvD ρρ

=2

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Therefore: π2 = Re=µ

ρvD

π3 = Lh∆ , as before.

Example: A spherical drop of liquid of diameter D oscillates under the influence of its surface tension. Investigate the frequency of oscillation f. F(f,σ, D, ρ) = 0 Ans: 3D

Kfρσ

=

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Matrix Method Best for problems with many variables. Can be solved using a matrix inversion routine. Consider a problem with 7 variables in 4 dimensions (M, L, T, θ) First form the dimensional matrix: a b c d e f g M L T θ

A B

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We need to transform the above matrix to: a b c d e f g

a b c d

1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1

I = A-1A D = A-1B A-1 = adj A/ |A|

Whatever operation was done to get the unit matrix on the left must be also done on the right to get D.

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Eg. Consider the earlier problem of the head loss per unit length. F(∆h/L, v, g, D, ρ, µ) = 0 ρ v D g µ ∆h/L M L T

1 0 0 -3 1 1 0 -1 0

0 1 0 1 -1 0 -2 -1 0

A B Inverse of A can be obtained using Minitab, Matlab, or by hand.

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−=−

113100

0011A D =

−−−

−

012011010

113100

001

D =

− 011012010

The three π terms are:

∆Lh

vDvgD ,,2 ρ

µ

The matrix method can obtain the π terms all in one go instead of

one term at a time.

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Comments about Dimensional Analysis

• Most important but most difficult problem in applying DA to any problem is the selection of the variables involved. There is no easy way of identifying the correct variables without specialized knowledge about the phenomenon being investigated.

• If you select too many variables, you get too many π terms and

may require much additional experimentation to eliminate them.

• If important variables are omitted, then an incorrect result will

be obtained, which may prove to be costly and difficult to ascertain.

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Type of variables: 1. Geometry – length, angles, diameter, or area.

2. Material properties – ρ, µ, elasticity, etc.

3. External effects – any variable that tends to produce a

change in the system e.g. forces, pressures, velocities,

gravity, etc.

You need to keep the number of variables to a minimum, and that

they are independent. E.g. D and Area need not be included

together because one is derived from the other. Therefore, heavy

thinking is required in variable selection similar to DOE.

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Points to consider in the selection of variables:

1. Clearly define the problem. What is the main response variable of interest? That is, what is Y?

2. Consider the basic laws that govern the phenomenon. Even a

crude theory may be useful.

3. Start the selection process by grouping the variables in the 3 broad classes: geometry, material properties, and external effects.

4. Consider other variables that may not fall into one of the

three categories, e.g. time, temperature, colour, equipment, etc.

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5. Be sure to include all quantities that enter the problem even though some of them may be held constant e.g. g. For D.A. it is the dimensions of the quantities that are important – not specific values.

6. Make sure that all variables are independent – look for

relationships among subsets of the variables (same as DOE).

Remember that after a dimensional analysis, you still need to carry out the experiment to relate the dimensionless groups. Hence DOE may be needed unless you have only 1 or 2 π terms. More on this later.

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More on π terms Specific π terms obtained depend on the somewhat arbitrary

selection of repeating variables. For example, if we choose:

µ, D, g instead of ρ, D, v, we would end up with a different set of

π terms. Both results would be correct, and both would lead to

the same final equation for ∆h/L, however, the function relating

the different π terms would be different.

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Hence, there is not a unique set of π terms which arises from a

dimensional analysis. However the required number of π terms is

fixed, and once a correct set is determined all other possible sets

can be developed from this set by combination of products of the

original set.

e.g. if we have problem involving 3 π terms,

π1 = φ(π2, π3)

we can combine the π terms to give a new π term: ba32

'2 πππ =

Then the relationship could be expressed as: ( )3'21 ,ππφπ =

Or even: ( )3'21 ,ππφπ =

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All these would be correct; however, the required number of π

terms cannot be reduced by this manipulation; only the form of

the π terms is altered.

e.g. µ

ρπρ

π vDvpD

=∆

= 221 ;

vpDvD

vpD

µµρ

ρπππ

2

22111

∆=

∆==

Which form of p terms is best? There is no simple answer. Best to keep it simple. Some p terms are well-known dimensionless numbers like Reynolds, Froude, Mach, Weber, Cauchy, Euler, etc.

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Correlation of Experimental Data

One of the most important uses of dimensional analysis is an aid

in the efficient handling, interpretation, and correlation of

experimental data.

As noted earlier, DA cannot provide a complete answer and a

specific relationship among the π terms cannot be determined

without experimentation. The degree of difficulty obviously

depends on the number of π terms.

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Problems with:

1 π term π = C where C = a constant

2 π terms π1 = f(π2) simple regression problem

> 2 π terms π1 = f(π2, π3) multiple regression problem

With more and more π terms, a DOE approach may be needed as

in Problem 4 of assignment 2 and may require the use of RSM if

relationship is nonlinear.

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Eg 1: The pressure drop per unit length, ∆p/L for the flow of blood through a horizontal small diameter tube is a function of flow rate Q, diameter D, and the blood viscosity µ. For a series of test with D = 2 mm and µ = 0.004 Ns/m2, the following data were obtained for ∆p measured over a length of 300 mm.

Q (m3/s) 3.6 x 10-6 4.9 x 10-6 6.3 x 10-6 7.9 x 10-6 9.8 x 10-6

∆p (N/m2) 1.1 x 104 1.5 x 104 1.9 x 104 2.4 x 104 3.0 x 104

Perform a DA for this problem and make use of the data to determine a general relationship between ∆p and Q, one that is valid for other values of D, L and µ.

Solution: 4 variables, F (∆p/L, D, Q, µ) = 0, i.e. 4 – 3 = 1 π term

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From D.A. (try this yourself), we get: QDLp

µ

4)/(∆ = constant

Substituting the values used in the experiment,

QDLp

µ

4)/(∆ = Qp

Qp ∆

×=×

∆ −84

1033.1004.0

)002.0(3.0/ . Using the data obtained

from the experiment,

QDLp

µ

4)/(∆ = [ 40.6, 40.7, 40.1 40.4 40.7]

Average for constant C = 40.5, hence: 45.40DQ

Lp µ

=∆

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Eg 2: A liquid flows with a velocity v through a hole in the side of a tank. Assume that v = f(h, g, ρ, σ). Where h is the height of water above the hole, ρ is the density of the fluid, and σ the surface tension. The density ρ is 1000 kg/m3, and s = 0.074 N/m. The data obtained by changing h and measuring v are:

V (m/s) 3.13 4.43 5.42 6.25 7.00 h (m) 0.50 1.00 1.50 2.00 2.50

Plot the data using appropriate dimensional variables. Could any of the original variables have been omitted? Solution: 5 variables, F(v, h, g, ρ, σ) = 0, 2 π terms.

From dimensional analysis,

=

σρφ

2ghghv

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ρgh2/σ 3.31 x 104 13.3 x 104 28.8 x 104 53.0 x 104 82.9 x 104

v/(gh)1/2 1.41 1.41 1.41 1.41 1.41

Plotting the data will show that ghv is independent of

σρ 2gh , which

means that ρ and σ can be omitted. Of course this is well-known if one were to apply the Bernoulli

equation to solve the problem.

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References:

Chapter 2, 3, 4 notes from course website.

Thomas Szirtes (1998): Applied Dimensional Analysis and

Modeling, McGraw Hill, 790 pages.

Most books in physics and fluid mechanics.