Dimensional Analysis

Dimensional Analysis

Section-II UNIT 4

•Dimensions of physical quantities•Dimensional homogeneity•Buckingham pi Theorem•Important dimensionless numbers (Reynolds, Froude and Mach)•Model analysis

Fluid Mechanics

2

Dimensional AnalysisDimensional Analysis

• This technique is useful in all branches of science and engineering

• Basic premises are1. Physical quantities have dimensions

2. All equations developed from basic laws of physics are dimensionally homogeneous

• We can develop a useful theory from this information

3

• Consider viscosity, μ, which has SI units kg/(m•sec)

• We write

• [μ] means “dimensions of μ” while M, L and T denote dimensions of mass, length and time, respectively

• If temperature is relevant, we denote its dimension by Θ

• M, L, T and Θ are called independent dimensions

4

• All dimensions can be expressed in terms of the independent dimensions

• For example, consider pressure, p, which has dimensions of force per unit area…appealing to Newton’s law, we have

• Thus, for dimensional analysis purposes, pressure has SI units of kg/(m•sec2)

• The 2/3 of all fluid problems are too complex,– both geometrically and– physically, to be solved analytically

• They must be tested by experiment.– Their behavior is reported as experimental data

• Such data are much more useful if they are expressed in compact, economic form


• Graphs are especially useful, since tabulated data cannot be absorbed, nor can the trends and rates of change be observed, by most engineering eyes

• These are the motivations for dimensional analysis.

• The technique is traditional in fluid mechanics and is useful in all engineering and physical sciences,– with notable uses also seen in the biological and social

sciences


• The method of dimensional analysis is used in every field of engineering,– especially in such fields as fluid dynamics and – thermodynamics

where problems with many variables are handled.

• By constructing non-dimensional quantities expressing the relationship among the variables, it is possible to– summarise the experimental results and – to determine their functional relationship.

Dimensional analysis

• The scaling laws can convert data from a

cheap, small model to design information for an expensive, large prototype.

• We do not build a million-dollar airplane and see whether it has enough lift force.

• We measure the lift on a small model and use a scaling law to predict the lift on the full-scale prototype airplane.

• There are rules we shall explain for finding scaling laws.

• When the scaling law is valid, we say that a condition of similarity exists between the model and the prototype

Similarity

• In order to determine the characteristics of a full-scale

device through model tests,– besides geometrical similarity, – similarity of dynamical conditions between the two

is also necessary. • When the above dimensional analysis is employed, if

the appropriate non-dimensional quantities such as– Reynolds number and – Froude number

• are the same for both devices then the results of the model device tests are applicable to the full-scale device.

Similarity

• When the dimensions of all terms of an equation are equal the equation is dimensionally correct.

• In this case, whatever unit system is used, that equation holds its physical meaning.

• Utilising this principle that the terms of physically meaningful equations have equal dimensions,

• The method of obtaining dimensionless groups of which the physical phenomenon is a function is called dimensional analysis.


• If a phenomenon is too complicated to derive a formula describing it, dimensional analysis can be employed to identify groups of variables which would appear in such a formula

• By supplementing this knowledge with experimental data, an analytic relationship between the groups can be constructed allowing numerical calculations to be conducted

• In order to perform the dimensional analysis, it is convenient to use the π theorem


Dimensional Analysis Theorems

• Dimensional Homogeneity Theorem: Any physical quantity is dimensionally a power law monomial - [Any Physical Quantity] = MaLbTcQd

• Buckingham Pi Theorem: If a system has – n physical quantities that depend on – m independent dimensions(M,L,T,θ), – then there are a total of n-m independent dimensionless

terms. – Each term is called a term. 1, 2, …, k-r.

• The behavior of the system is describable by a dimensionless equation

F(1, 2, …, n-m)=0

•Application of fluid mechanics in design

makes use of experiments results.•Results often difficult to interpret.•Dimensional analysis provides a strategy

for choosing relevant data.•Used to help analyse fluid flow•Especially when fluid flow is too complex for mathematical analysis.•Uses principle of dimensional homogeneity

•Specific Need

• help design experiments• Informs which measurements are important• Allows most to be obtained from experiment: e.g. What runs to do. How to interpret.

• It depends on the correct identification of variables

• Relates these variables together• Doesn’t give the complete answer• Experiments necessary to complete solution

•Specific uses:

•Any physical situation can be described by familiar properties.

e.g. length, velocity, area, volume, acceleration etc.•These are all known as dimensions.•Dimensions are of no use without a magnitude. i.e. a standardised unit

e.g metre, kilometre, Kilogram, a yard etc.•Dimensions can be measured.•Units used to quantify these dimensions.•In dimensional analysis we are concerned with the natureof the dimension i.e. its quality not its quantity

Significant Dimensionless Groups in Fluid Mechanics

• Reynolds Number (inertial to viscous forces)

Mach Number (inertial to elastic forces)

Important in all fluid flow problems

Important in problems with compressibility effects


• Froude Number (inertial to gravitational forces)

Important in problems with a free surface


• Euler Number (pressure to inertial forces)

Important in problems with pressure differences

Weber Number (inertial to surface tension forces)

Important in problems with surface tension effects


Dimensional Analysis 9.22

Whenever an engineering problem cannot be solved by using the

equations and analytical procedures described previously, we must

resort to experimentation. The tool used to plan, design and interpret the

experiments is called Dimensional Analysis.


Consider the steady flow of an incompressible Newtonian fluid through a

circular pipe. The quantity of interest is the pressure drop that develops

along the pipe as a result of friction. For turbulent flow this problem

cannot be solved analytically.


Pressure drop is expected to depend on

1.pipe length, ℓ,

2.pipe diameter, D,

3.velocity, V,

4.fluid density, ,

5.fluid viscosity, and

6.pipe roughness, :

P = (ℓ, D, V, , , )

Our goal is to form dimensionless groups from these variables.

Dimensional AnalysisBuckingham Pi Theorem:

If an equation involving n variables is dimensionally homogeneous, it can

be reduced to a relationship among n-m independent dimensionless

products, where m is the minimum number of reference dimensions

required to describe the variables.

The dimensionless products are frequently referred to as “pi terms”. A

relationship of the form shown below is thus obtained:

,....),,( 4321

Method of selecting Repeating variable

•The number of repeating variables = The number of fundamental dimensions of the problem.•The choice of repeating variables is governed by the following considerations:1.The dependant variable should not be selected as repeating variable.2.The repeating variable should be selected in such a way that

i. One variable contains geometric variable ( length, diameter, Height etc..,)

ii. Second variable contains flow property ( velocity, acceleration, etc..,)

iii. Third variable contains fluid property ( viscosity, density, specific weight, etc..,)

Method of selecting Repeating variable

3. The repeating variable selected should not form a dimensionless group

4. The repeating variable together must have the same number of fundamental dimensions.

5. No two repeating variables should have the same dimension

Thumb rule:The choice of repeating variables can be1) d,v,ρ2) l,v,ρ3) l, v, μ4) d , v, μ

Dimensional Analysis: Pipe flow example

• Step 1: List all the variables that are involved in the problem

• Step 2: Express each of the variables in terms of basic dimensions

• Step 3: Determine the required number of pi terms.

• Step 4: Select a number of repeating variables, where the number required is equal to the number of fundamental dimensions.

Dimensional Analysis: Pipe flow example• Step 5: Form a pi term by multiplying one of the non-repeating variables

by the product of the repeating variables, each raised to an exponent that will make the combination dimensionless.

• Step 6: Repeat step 5 for each of the remaining non-repeating

variables.

• Step 7: Check all the resulting pi terms to make sure they are

dimensionless

Dimensional Analysis: Pipe flow example• Step 8: Express the final form as a relationship among the pi

terms.

D ,

DV

D

V

P

D ,

D,

DVV

P

2

2

) D

(Re, f

• The drag force on a smooth sphere is found to be affected by the velocity of flow, V, the diameter D of the sphere and the fluid properties density ρ and viscosity μ.

• Using dimensional analysis obtain the dimensionless groups to correlate the parameters.

Nature of Dimensional Analysis

There are 5 dimensional quantities

There are 3 independent dimensions, i.e., M, L and T The Buckingham π Theorem tells us there are 5 – 3 = 2

dimensionless groupings Drag force depends on FOUR parameters:

sphere size (D); speed (V); fluid density (); fluid viscosity ()

Buckingham Pi Theorem

• Step 1:List all the dimensional parameters involved

Let n be the number of parameters

Example: For drag on a sphere, F, V, D, , , and n = 5

Buckingham Pi Theorem• Step 2

Select a set of fundamental (primary) dimensions

Example: For drag on a sphere choose MLt


List the dimensions of all parameters in terms of primary dimensions

Let n be the number of primary dimensions

Example: For drag on a sphere n = 3


• Step 4Select a set of r dimensional parameters that includes all the primary dimensions

Example: For drag on a sphere (m - n = 3)

select , V, D


Set up dimensional equations, combining the parameters selected in Step 4 with each of the other parameters in turn, to form dimensionless groups

There will be n – m equations

Example: For drag on a sphere

Buckingham Pi Theorem• Step 5 (Continued)


M a+1=0L -3a+b+c+1=0T -b-2=0


a = -1 b = -2 c = -2


Check to see that each group obtained is dimensionless


• π2 = μ ρa Vb Dc

Solving we get,

• π2 = μ/ρVD

• A rectangular plate of height, a and width, b is held perpendicular to the flow of a fluid. The drag force on the plate is influenced by the dimensions a and b, the velocity u, and the fluid properties, density ρ and viscosity μ. Obtain a correlation for the drag force in terms of dimensionless parameters.

• There are 6 parameters and three dimensions. Hence three π terms can be obtained.

• Selecting b, u and ρ as repeating variables

1 + c = 0, 1 + a + b – 3c = 0, – 2 – b = 0

c = – 1, b = – 2, a = – 2

c = 0, 1 + a + b – 3c = 0, – b = 0,

a = – 1

1 + c = 0, – 1 + a + b – 3c = 0, – 1 – b = 0.

b = – 1, c = – 1, b = – 1

Convective heat transfer coefficient in free convection over a surface is found to be influenced by the density, viscosity, thermal conductivity, coefficient of cubical expansion, temperature difference, gravitational acceleration, specific heat, the height of surface and the flow velocity. Using dimensional analysis, determine the dimensionless parameters that will correlate the phenomenon.

The variables with dimensions in the MLT θ set is tabulated below.

• There are nine variables• There are four dimensions• Hence five π terms can be identified• ρ, μ, x and k are chosen as repeating

variables• π1 = ΔT ρa μbxckd

• π2 = β ρa μbxckd

• π3 = g ρa μbxckd

• π4 = c ρa μbxckd

• π5 = h ρa μbxckd

• π1 = ΔT ρa μbxckd

a + b + d = 0, – 3a – b + c + d = 0, b – 3d = 0, 1 – d = 0,

a = 2, b = – 3, c = 2, d = 1

• π2 = β ρa μbxckd

a + b + d = 0, – 3a – b + c + d = 0, – b – 3d = 0,– 1 – d = 0,a = – 2, b = 3, c = – 2, d = – 1

• π3 = g ρa μbxckd

a + b + d = 0, 1 – 3a – b + c + d = 0, – 2 – b – 3d = 0, d = 0,

a = 2 b = – 2 c = 3 d = 0

• π4 = c ρa μbxckd

a + b + d = 0, 2 – 3a – b + c + d = 0, – 2 – b – 3d = 0, – 1 – d = 0,

a = 0, b = 1,c = 0, d = – 1

• π5 = h ρa μbxckd

1 + a + b + d = 0, – 3a – b + c + d = 0, – 3 – b – 3d = 0, – 1 – d = 0,

a = 0, b = 0, c = 1, d = – 1

Wrong choice of physical properties

• If, extra, unimportant variables are chosen :• * Extra π groups will be formed• * Will have little effect on physical performance• * Should be identified during experiments

• If an important variable is missed:• A π group would be missing.• Experimental analysis may miss significant behavioural

changes.

Initial choice of variables should be done with great care.

(The Secrets )

OutlineA. What’s the secret of being a Scientist or an

Engineer?

B. What are Units and Dimensions anyway?

C. What is Dimensional Analysis and why should I care?

D. Why aren’t there any mice in the Polar Regions?

E. Why was Gulliver driven out of Lillipute?

F. But what do I really need to know about Dimensional Analysis so that I can pass the test?

G. But can it be used in the Lab ?

How to be a Scientist or Engineer

The steps in understanding and/or control any physical phenomena is to

1. Identify the relevant physical variables.

2. Relate these variables using the known physical laws.

3. Solve the resulting equations.

Secret #1: Usually not all of these are possible. Sometimes none are.

ALL IS NOT LOST BECAUSE OF

• Physical laws must be independent of arbitrarily chosen units of measure. Nature does not care if we measure lengths in centimeters or inches or light-years or …

• Check your units! All natural/physical relations must be dimensionally correct.

Secret #2: Dimensional Analysis

Rationale

Dimensional AnalysisDimensional Analysis refers to the physical nature of the quantity and the type of unit (Dimension) used to specify it.

•Distance has dimension L.

•Area has dimension L2.

•Volume has dimension L3.

•Time has dimension T.

•Speed has dimension L/T

Why are there no small animals in the polar regions?

• Heat Loss Surface Area (L2)

• Mass Volume (L3)

• Heat Loss/Mass Area/Volume = L2/ L3

= L-1

Heat Loss/Mass Area/Volume = L2/ L3

= L-1

Mouse (L = 5 cm) 1/L = 1/(0.05 m) = 20 m-1

Polar Bear (L = 2 m) 1/L = 1/(2 m) = 0.5 m-1

20 : 0.5 or 40 : 1

Gulliver’s Travels: Dimensional Analysis

• Gulliver was 12x the Lilliputians

• How much should they feed him?12x their food ration?

• A persons food needs arerelated to their mass (volume) – This depends on the cube of the linear dimension.

Gulliver is 12x taller than the Lilliputians, LG =12 LL

Now VG (LG)3 and VL (LL)3, so

VG / VL = (LG)3 / (LL)3 = (12 LL)3 / (LL)3 = 123

= 1728Gulliver needs to be fed 1728 times the amount of food each day as the Lilliputians.

Let LG and VG denote Gulliver’s linear and volume dimensions.Let LL and VL denote the Lilliputian’s linear and volume dimensions.

This problem has direct relevance to drug dosages in humans

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General ObservationsGeneral Observations

Dimensional analysis has at least four key uses…

1. Checking dimensional consistency of algebraic results

2. Deducing appropriate physical scales

3. Correlating experimental data

4. Extrapolating data from small to large models

Modeling

• An engineering (or physical) model is a representation of a physical system, that may be used to predict its behaviour.

• The physical system for which the predictions are to be made is called the prototype.

• Models resemble the prototype, but are generally of a different size (usually smaller), may involve different fluids and often operate under different conditions.

Theory of Models• Using the principles of dimensional analysis, any given

problem concerning the prototype can be described in terms of a set of pi terms as:

),....,( n321 • A similar relationship can be writted for a model of this prototype:

),....,( nmm3m2m1 • In order to ensure that our model behaves in a similar manner as the

prototype the following similarity requirements, or modeling laws must be met:

1m1

nnm

3m3

2m2

....

Flow Similarity and Model Studies• Geometric Similarity

– Model and prototype have same shape– Linear dimensions on model and prototype

correspond within constant scale factor

• Kinematic Similarity– Velocities at corresponding points on model and

prototype differ only by a constant scale factor

• Dynamic Similarity– Forces on model and prototype differ only by a

constant scale factor

Geometric similarity:The ratio of all corresponding dimensions in the model and prototype are equal.

Kinematic similarityThe similarity of time as well as geometry. It exists if:i. the paths of particles are geometrically similarii. the ratios of the velocities of are similar

Dynamic similarityIf geometrically and kinematically similar and the ratios of all forces are the same.

This occurs when the controlling π group is the same for model and prototype. The controlling π group is usually Re. So Re is the same for model and prototype:

Modelling and Scaling Laws:Measurements taken from a model needs a scaling law applied to predict the values in the prototype.An example:For resistance R, of a body moving through a fluid.R, is dependent on the following:

Flow Similarity and Model Studies• Example: Drag on a Sphere

Flow Similarity and Model Studies• Example: Drag on a Sphere

For dynamic similarity …

… then …

Flow Similarity and Model Studies

• Incomplete Similarity

Sometimes (e.g., in aerodynamics) complete similarity cannot be obtained, but phenomena may still be successfully modelled

MODEL AND PROTOTYPE

• In the engineering point of view model can be defined as the representation of physical system that may be used to predict the behavior of the system in the desired aspect.

• The system whose behavior is to be predicted by the model is called the prototype.

• The discussion in this lecture is about physical models that resemble the prototype but are generally smaller in size.

• These may also operate with different fluids, at different pressures, velocities etc.

• As models are generally smaller than the prototype, these are cheaper to build and test.

• Model testing is also used for evaluating proposed modifications to existing systems.

• The effect of the changes on the performance of the system can be predicted by model testing before attempting the modifications.

• Models should be carefully designed for reliable prediction of the prototype performance.

Nondimensional Parameters and Dynamic Similarity

• Arranging the variables in terms of dimensionless products is especially useful in presenting experimental data.

• Consider the case of drag on a sphere of diameter d moving at a speed U through a fluid of density ρ and viscosity μ. The drag force can be written as

D = f(d,U, ρ, μ).

• If we do not form dimensionless groups, we would have to conduct an experiment to determine D vs d, keeping U, ρ, and μ fixed.

• We would then have to conduct an experiment to determine D as a function of U, keeping d, ρ, and μ fixed, and so on.

• However, such a duplication of effort is unnecessary if we write the above equation in terms of dimensionless groups. A dimensional analysis of equation gives

• A dimensional analysis of equation reduces the number of variables from five to two, and consequently a single experimental curve

• Not only is the presentation of data united and simplified,the cost of experimentation is drastically reduced.

• It is clear that we need not vary the fluid viscosity or density at all;

• we could obtain all the data of Figure in one wind tunnel experiment in which we determine D for various values of U

Similitude• Similitude

– Predict prototype behavior from model results

– Models resemble prototype, but are • Different size (usually smaller) and may

operate in • Different fluid and under• Different conditions

– Problem described in terms of dimensionless parameters which may apply to the model or the prototype

– Suppose it describes the prototype

– A similar relationship can be written for a model of the prototype

)321 ,...,,( nf

)321 ,...,,( npppp f

)321 ,...,,( nmmmm f

Similitude

• If the model is designed & operated under conditions that

• then

npnm

pm

pm

...

33

22

pm 11

Similarity requirements or modeling laws

Dependent variable for prototype will be the same as in the model

Example• Consider predicting the drag on a

thin rectangular plate (w*h) placed normal to the flow.

• Drag is a function of: w, h, , , V

),,,,( VhwfFD

),(

),(

22

321

Vw

hw

fVw

F

f

D

),(

),(

22

321

m

mmm

m

m

mmm

Dm

mmm

wV

h

wf

Vw

F

f

),(

),(

22

321

p

ppp

p

p

ppp

Dp

ppp

wV

h

wf

Vw

F

f

• Dimensional analysis shows:• And this applies BOTH to a model

and a prototype

• We can design a model to predict the drag on a prototype.

• Model will have:

• And the prototype will have:

Example

pm

p

m

p

p

mm

p

ppp

m

mmmpm V

w

wV

wVwV

33

Dmm

p

m

p

m

pDp

ppp

Dp

mmm

Dmpm F

V

V

w

wF

Vw

F

Vw

F22

222211

•Similarity conditionsGeometric similarity

Dynamic similarity

Then

pp

mm

p

p

m

mpm w

h

hw

h

w

h

w 22 Gives us the size of the model

Gives us the velocity in the model

Exponent Method1. List all n variables involved in the problem2. Express each variables in terms of [M] [L] [T ]

dimensions (m)3. Determine the required number of dimensionless

parameters (n – m)4. Select a number of repeating variables = m

(All dimensions must be included in this set and each repeating variable must be independent of the others.)

5. Form a dimensionless parameter by multiplying one of the non-repeating variables by the product of the repeating variables, each raised to an unknown exponent.

6. Solved for the unknown exponents.7. Repeat this process for each non-repeating variable8. Express result as a relationship among the

dimensionless parameters – F(1, 2, 3, …) = 0.

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Dimensional Analysis