microsoft power point - bmcf 2223 - ch 9 dimensional analysis e2

8/3/2019 Microsoft Power Point - BMCF 2223 - CH 9 Dimensional Analysis e2

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DIMENSIONAL ANALYSISDIMENSIONAL ANALYSIS

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DimensionDimension

5 basic dimensions?5 basic dimensions?

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Most physicalMost physical

quantities can bequantities can beexpressed in terms ofexpressed in terms ofthe combinations ofthe combinations of

MassMass MM

LengthLength LL

AbbreviationAbbreviationDimensionsDimensionsDimensionDimension

3

i.e. speed is always ai.e. speed is always alength divided a timelength divided a time

So dimensions ofSo dimensions of

speed are lengthspeed are lengthdivided by time or L/Tdivided by time or L/T

meme

ElectricElectric

currentcurrent

II

TemperatureTemperature


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DimensionDimension

What is theWhat is the

dimensions of thedimensions of thefollowing quantities?following quantities?

VolumeVolume

AreaArea

DensityDensity

LL33

LL22

M/LM/L33

22

4

ForceForce ML/TML/T22


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Dimensional HomogeneityDimensional Homogeneity

Any equation describing a physical situationAny equation describing a physical situationwill only be true if both sides have the samewill only be true if both sides have the same

dimensions, i.e.dimensions, i.e.1 meter + 3 kilograms = 4 days (not true)1 meter + 3 kilograms = 4 days (not true)

1 meters + 3 meters = 4 meters (true)1 meters + 3 meters = 4 meters (true)

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In other words, the dimensions must beIn other words, the dimensions must beconsistentconsistent

The property of dimensional homogeneityThe property of dimensional homogeneitycan be useful for:can be useful for: --

Checking units of equationsChecking units of equations

Converting between two sets of unitsConverting between two sets of units

Defining dimensionless relationshipsDefining dimensionless relationships


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Dimension and UnitsDimension and Units

Dimensions must have aDimensions must have astandardised unit i.e. meter, foot,standardised unit i.e. meter, foot,

yardyardThere are 2 systems of unit beingThere are 2 systems of unit beingwidely used:widely used: --

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(International system) units(International system) units

US (United States) unitsUS (United States) unitsHere we are using SI unitsHere we are using SI units


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Some basic SI unitsSome basic SI units

LengthLength metermeter

metremetre

mm

SymbolSymbolNameNameQuantityQuantity

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MassMass kilogramkilogram kgkg

TimeTime secondsecond ss

TemperatureTemperature kelvinkelvin KK


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Between dimensions and units:Between dimensions and units: Dimensions are properties which can beDimensions are properties which can be

measuredmeasured Unit are the standard elements we use toUnit are the standard elements we use to

quantify these dimensionsquantify these dimensions

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Dimensional AnalysisDimensional Analysis

Useful technique in allUseful technique in allexperimentally based areas ofexperimentally based areas ofengineeringengineering

If you know the factors involved inIf you know the factors involved in

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,,analysis can form a relationshipanalysis can form a relationshipbetween thembetween them


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In dimensional analysis we areIn dimensional analysis we areonly concerned with the natureonly concerned with the natureof the dimension, i.e. theof the dimension, i.e. the qualityqualitynotnot thethe quantityquantity

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The man weights 50 kg,standing at 60 m height

and 9.81 m/s2 gravity

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In fluid mechanics, all the physical propertiesIn fluid mechanics, all the physical propertiescan be presented with L, T, M and Fcan be presented with L, T, M and F

Force can also be presented in LTMForce can also be presented in LTM

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QuantityQuantity SI unitSI unit DimensionDimension

VelocityVelocity m/sm/s msms--11 LTLT--11

AccelerationAcceleration m/sm/s22 msms--22 LTLT--22

ForceForce N or kgm/sN or kgm/s22 kg mskg ms--22 MLTMLT--22

EnergyEnergy J or kgmJ or kgm22/s/s22 kg mkg m22 ss--22 MLML22TT--22

-- --

Dimensions of Some Common Physical Quantities:Dimensions of Some Common Physical Quantities:

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PressurePressure Pa or kg/m/sPa or kg/m/s22 kg mkg m--11 ss--22 MLML--11TT--22

DensityDensity kg/mkg/m33 kg mkg m--33 MLML--33

Specific weightSpecific weight N/mN/m33 or kg/mor kg/m22/s/s22 Kg mKg m--22ss--22 MLML--22TT--22

viscosityviscosity N s/mN s/m22 or kg/msor kg/ms Kg mKg m--11 ss--11 MLML--11TT--11


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Result of Dimensional AnalysisResult of Dimensional Analysis

The results of performingThe results of performingdimensional analysis (on adimensional analysis (on a

physical problem) isphysical problem) is a singlea singleequationequation

This equationThis equation relatesrelates all of theall of the

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p ys ca ac orsp ys ca ac ors var a es, .e.var a es, .e.velocity, force, time) involved tovelocity, force, time) involved toone anotherone another


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Result of Dimensional AnalysisResult of Dimensional Analysis

For example:For example: --

If we want to find theIf we want to find the forceforce on aon a

propeller blade we must first decidepropeller blade we must first decidewhat might influence this forcewhat might influence this force

It would be reasonable to assume thatIt would be reasonable to assume thatthe force F de ends on the followinthe force F de ends on the followin

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physical propertiesphysical properties diameter, ddiameter, d

forward velocity of the propeller, uforward velocity of the propeller, u

fluid density,fluid density,

revolutions per second, Nrevolutions per second, N

fluid viscosity,fluid viscosity,


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Before we do any analysis we write thisBefore we do any analysis we write thisequation:equation:

F =F = (d, u,(d, u, , N,, N, ))

oror0 =0 = 11(F, d, u,(F, d, u, , N,, N, ))

WhereWhere andand 11 are unknown functionsare unknown functions

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These can be expanded into an infiniteThese can be expanded into an infiniteseries which can itself be reduced toseries which can itself be reduced to

F = K dF = K dmm uupp qq NNrr ss

Where K is a numerical constantWhere K is a numerical constantAnd m, p, q, r, s are unknown constant powerAnd m, p, q, r, s are unknown constant power


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From dimensional analysis weFrom dimensional analysis we

obtain these powerobtain these power

form the variable into severalform the variable into severaldimensionless groupsdimensionless groups

The value of K and the functionThe value of K and the function andand 11 must be determinedmust be determinedfrom experimentfrom experiment

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The knowledge of theThe knowledge of thedimensionless group oftendimensionless group oftenhelps in deciding whathelps in deciding what

experimental measurementsexperimental measurementsshould be takenshould be taken


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BuckinghamsBuckinghams TheoremsTheorems

One of the methods used to performOne of the methods used to performdimensional analysisdimensional analysis

Provides better generalised strategy inProvides better generalised strategy inobtaining a solution (compared to indicialobtaining a solution (compared to indicialmethod)method)

11stst theorem:theorem: --

A relationship between n variables (physicalA relationship between n variables (physical

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properties such as velocity, density etc.) canproperties such as velocity, density etc.) canbe expressed as a relationship betweenbe expressed as a relationship betweennn m nonm non--dimensional groups of variablesdimensional groups of variables(called(called group), where m is the number ofgroup), where m is the number ofthe fundamental dimensions (such as mass,the fundamental dimensions (such as mass,length and time) required to express thelength and time) required to express thevariablesvariables


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BuckinghamsBuckinghams TheoremsTheorems

So if a physical problem can beSo if a physical problem can beexpressed as:expressed as: --

(Q(Q11, Q, Q22, Q, Q33, , Q, , Qnn) = 0) = 0

Then, according to the above 1Then, according to the above 1

stst

theorem, this can be expressed as:theorem, this can be expressed as: --

11((11,, 22,, 33, ,, , nn mm) = 0) = 0

In fluid we can normall taken m = 3In fluid we can normall taken m = 3

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(corresponding to M, L and T)(corresponding to M, L and T)


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22ndnd theorem:theorem: --

EachEach group is a function ofgroup is a function ofm repeating variables plusm repeating variables plusone of the remainingone of the remainingvariablesvariables

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== (Q(Q11, Q, Q22, , Q, , Qmm, Q, Qm+1m+1) = 0) = 0


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Choice of Repeating VariableChoice of Repeating Variable

From the 2From the 2ndnd theorem there cantheorem there canbe m (= 3) repeating variablebe m (= 3) repeating variable

When combined, these repeatingWhen combined, these repeatingvariables must contain all of thevariables must contain all of thedimensions (M, L, T)dimensions (M, L, T)A combination of the repeatingA combination of the repeatingvariables will not form avariables will not form a

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mens on ess groupmens on ess group

The repeating variables do notThe repeating variables do nothave to appear in allhave to appear in all groupsgroups


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Choice of Repeating VariableChoice of Repeating Variable

The repeating variables shouldThe repeating variables shouldbe chosen to be measurable inbe chosen to be measurable inan experimental investigation.an experimental investigation.

They should be the majorThey should be the majorinterest to the designerinterest to the designer

In fluid mechanics, it is usuallyIn fluid mechanics, it is usuallyossible to takeossible to take uu andand dd asas

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the three repeating variablesthe three repeating variablesThis freedom of choice resultsThis freedom of choice resultsin there being many differentin there being many different group which can be formed,group which can be formed,

and all are valid (there is noand all are valid (there is noreally a wrong choice)really a wrong choice)


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Example:Example:

Taking the example discussedTaking the example discussedpreviously, force F induced on apreviously, force F induced on apropeller blade, we have thepropeller blade, we have theequationequation

F =F = (d, u,(d, u, , N,, N, ))n = 6 and m = 3n = 6 and m = 3

There are nThere are n m = 3m = 3 groups, sogroups, so

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((11,, 22,, 33) = 0) = 0The choice ofThe choice of , u, d as the, u, d as therepeating variables satisfies therepeating variables satisfies thecriteria above:criteria above: --

They are measurableThey are measurable good design parameters andgood design parameters and

in combination contain all thein combination contain all thedimensions M, L, Tdimensions M, L, T


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Example:Example:

We can now form the three groupsWe can now form the three groupsaccording to the 2according to the 2ndnd theorem,theorem,

11 == aa11uubb11ddcc11FF

22 == aa22uubb22ddcc22NN

33 == aa33uubb33ddcc33

As theAs the groups are allgroups are alldimensionless i.e. they havedimensionless i.e. they have

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dimensions Mdimensions M

00

LL

00

TT

00

we use thewe use theprinciple of dimensionalprinciple of dimensionalhomogeneity to equate thehomogeneity to equate thedimensions for eachdimensions for each groupgroup


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For the firstFor the first group,group,

11 == aa11

uu

bb11

dd

cc11

FFIn terms of dimensionsIn terms of dimensions

MM00LL00TT00 = (ML= (ML--33))aa11(LT(LT--11))bb11(L)(L)cc11(MLT(MLT--22))

For each dimension the powers must be equalFor each dimension the powers must be equal

on both sides of the equation, soon both sides of the equation, soFor M:For M: 0 = a0 = a11 + 1+ 1

aa11 == 11

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For L:For L: 0 =0 = 3a3a11 + b+ b11 + c+ c11 + 1+ 10 = 4 + b0 = 4 + b11 + c+ c11For T:For T: 0 =0 = bb11 22

bb11 == 22

cc11 == 44 bb11 == 22

GivingGiving 11 asas

11 == --11uu--22dd--22FF

11 = F/= F/uu22

dd22


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or:or:

11 = F/= F/uu22dd22

FF == 11 uu22dd22

Where,Where, 11 is a force coefficientis a force coefficient

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Example:Example:Similar procedure is followed for the otherSimilar procedure is followed for the other

group, i.e.group, i.e.22 ==

aa22uubb22ddcc22NN

MM

00

LL

00

TT

00

= (ML= (ML

--33

))

aa22

(LT(LT

--11

))

bb22

(L)(L)

cc22

(T(T

--11

))For M:For M: 0 = a0 = a22For L:For L: 0 =0 = 3a3a22 + b+ b22 + c+ c22

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22 22

For T:For T: 0 =0 = bb22 11bb11 == 11

cc11 == 11

GivingGiving 22 asas22 ==

00uu--11dd11NN

22 = Nd/u= Nd/u


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Example:Example:For the third,For the third,


MM00LL00TT00 = (ML= (ML--33))aa33(LT(LT--11))bb33(L)(L)cc33(ML(ML--11TT--11))

For M:For M: 0 = a0 = a33 + 1+ 1aa33 == 11

For L:For L: 0 =0 = 3a3a33 + b+ b33 + c+ c33 11

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33 ++ 33==

For T:For T: 0 =0 = bb33 11

bb33 == 11

cc33 == 11


--11uu--11dd--11

33 == //udud


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Example:Example:Thus, the problem may be described by theThus, the problem may be described by the

following function of the three nonfollowing function of the three non--dimensionaldimensional groupgroup

((11,, 22,, 33) = 0) = 0

(F/(F/uu22dd22, Nd/u,, Nd/u, //ud) = 0ud) = 0

May also be written:May also be written:

F/F/ uu22dd22 == Nd/uNd/u // udud

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or:or:FF == uu22dd22 11(Nd/u,(Nd/u, //ud)ud)


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ADVANTAGE OF PERFORMING DIMENSIONAL ANALYSISADVANTAGE OF PERFORMING DIMENSIONAL ANALYSIS

To determine experimentally howTo determine experimentally how

F =F = (d, u,(d, u, , N,, N, ))

If number of level to be testedIf number of level to be testedfor each parameter is 5for each parameter is 5

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For uFor u22,,

22, N, N

22,,

22

dd11 F = ?F = ?

dd22 F = ?F = ?

dd33 F = ?F = ?

dd44 F = ?F = ?

dd55 F = ?F = ?

For uFor u33,, 33, N, N33,, 33

dd11 F = ?F = ?

dd22 F = ?F = ?

dd33

F = ?F = ?

dd44 F = ?F = ?

dd55 F = ?F = ?

For uFor u11,, 11, N, N11,, 11

dd11 F = ?F = ?

dd22 F = ?F = ?

dd33 F = ?F = ?

dd44 F = ?F = ?

dd55 F = ?F = ?


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ADVANTAGE OF PERFORMING DIMENSIONAL ANALYSISADVANTAGE OF PERFORMING DIMENSIONAL ANALYSIS

F =F = (d, u,(d, u, , N,, N, ))

Number of independent parameters, n = 5Number of independent parameters, n = 5

For number of level = 5,For number of level = 5,

Total runTotal run = 5n = 55 = 3125 experiments

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Through dimensional analysis:Through dimensional analysis:

number ofnumber of group = ngroup = n m = 3m = 3

11 == ((22,, 33))

Total run =Total run = 52 = 25 experiments!


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If 1 day you can perform 25 experimentsIf 1 day you can perform 25 experimentsFor 3125 experiments = 125 daysFor 3125 experiments = 125 days

if you work 5 days per week = 25 weeksif you work 5 days per week = 25 weeks

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= .= .

= YEAR +!= YEAR +!


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Wrong Choice of Physical PropertiesWrong Choice of Physical Properties

If important / influentialIf important / influential

variable was missed, thenvariable was missed, thenthe importantthe important group will begroup will bemissingmissing

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on these results may misson these results may missthe significant behaviouralthe significant behaviouralchangeschanges

Therefore, the choice ofTherefore, the choice ofvariables is very importantvariables is very important


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Wrong Choice of Physical PropertiesWrong Choice of Physical Properties

If extra/unimportant variableIf extra/unimportant variable

are introduced, then theare introduced, then theextra/unimportantextra/unimportant groupsgroupswill be formwill be form

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They will play very little roleThey will play very little roleinfluencing the physicalinfluencing the physicalbehaviour of the problembehaviour of the problem

And should be identify duringAnd should be identify duringthe experimental workthe experimental work


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Manipulation of TheManipulation of The GroupGroupManipulation of theManipulation of the group isgroup ispermittedpermitted

These manipulation did notThese manipulation did notchange the number of groupschange the number of groupsinvolved, butinvolved, but

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May change their appearanceMay change their appearance


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Manipulation of TheManipulation of The GroupGroup

Any number of groups can beAny number of groups can becombined by multiplication orcombined by multiplication or

((11,, 22,, 33, ,, , nn mm) = 0) = 0

Taking the definingTaking the definingequation as:equation as:

Then,Then,

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which replace one of thewhich replace one of theexisting, i.e.existing, i.e.

11 andand 22 is combined to formis combined to form

1a1a == 11//22 so thatso that

((1a1a,, 22,, 33, ,, , nn mm) = 0) = 0


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The reciprocal of any group isThe reciprocal of any group isvalid, i.e.valid, i.e.

((11,, 22,, 33, ,, , nn mm) = 0) = 0


Then,Then,

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((11, 1/, 1/22,, 33, , 1/, , 1/nn mm) = 0) = 0


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Any group may be raised toAny group may be raised toany power, i.e.any power, i.e.

((11,, 22,, 33, ,, , nn mm) = 0) = 0


Then,Then,

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((1122

,, 221 21 2

,, 3344

, ,, , nn mm) = 0) = 0

M i l i f ThM i l i f Th GG


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Any group may be multipliedAny group may be multipliedby a constant, i.e.by a constant, i.e.

((11,, 22,, 33, ,, , nn mm) = 0) = 0


Then,Then,

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(k(k1111,, 22, k, k2233, ,, , nn mm) = 0) = 0

M i l ti f ThM i l ti f Th GG


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Any group may be expressedAny group may be expressedas a function of the otheras a function of the other

((11,, 22,, 33, ,, , nn mm) = 0) = 0


Then,Then,

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, . ., . .

22 == 11((11,, 33, ,, , nn mm))

M i l ti f ThM i l ti f Th GG


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((11, 1/, 1/22,, 33ii, , k, , knn mm) = 0) = 0

In general, the definingIn general, the definingequation may look like:equation may look like:

42


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CommonCommon GroupsGroups

During dimensional analysis,During dimensional analysis,several groups will appearseveral groups will appear

again and again for differentagain and again for differentproblemproblem

43

their namestheir names


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CommonCommon GroupsGroups

Reynolds numberReynolds number Inertial, viscous forceInertial, viscous forceratioratio

Euler numberEuler number Pressure, inertial forcePressure, inertial force

ud=Re

2En

p=

Some commonSome commondimensionless groups are:dimensionless groups are:

44

Froude numberFroude number Inertial, gravitationalInertial, gravitationalforce ratioforce ratio

Weber numberWeber number Inertial, surface tensionInertial, surface tension

force ratioforce ratioMach numberMach number Local velocity, localLocal velocity, local

velocity of sound ratiovelocity of sound ratio

gd

u2Fn =

ud=We

c

u=Mn

Example:Example:


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Example:Example:

Determine the drag forceDetermine the drag forceexerted on a submergedexerted on a submergedsphere as it moves through asphere as it moves through aviscous fluidviscous fluid

Step 1: Consider which physicalStep 1: Consider which physicalfactors influence the drag forcefactors influence the drag force

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Diameter of the sphere, dDiameter of the sphere, d

Velocity of the sphere, uVelocity of the sphere, u

Density of the fluid,Density of the fluid,

Viscosity of the fluid,Viscosity of the fluid,

Thus we can write:Thus we can write:Drag force, F =Drag force, F = (d, u,(d, u, ,, ))

n = 5n = 5

Example:Example:


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Example:Example:

Step 2: List the dimensions of eachStep 2: List the dimensions of eachvariablevariable

dd LL

uu LTLT--11

MLML--33

--11 --11

46

FF MLTMLT--22So,So,

m = 3m = 3 (L, M, T)(L, M, T)

Example:Example:


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Example:Example:

Step 3: Determine the number ofStep 3: Determine the number ofdimensionless group (dimensionless group ( group)group)

nn m = 5m = 5 3 = 23 = 2

So we write:So we write:

1st p theorem:1st p theorem: --A relationship between nA relationship between nvariables can be expressed asvariables can be expressed as

47

((11,, 22) = 0) = 0 a re at ons p etween na re at ons p etween n mmdimensionless groups, where mdimensionless groups, where mis the number of theis the number of thefundamental dimensionsfundamental dimensionsrequired to express therequired to express the

variablesvariables

Example:Example:


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Example:Example:

Step 4: Select m (= 3) repeating variablesStep 4: Select m (= 3) repeating variables

Generally it helps to choose variables thatGenerally it helps to choose variables thatrelates to:relates to: --

GeometryGeometry dd

Fluid PropertyFluid Property

?

?

48

Example:Example:


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Example:Example:

Step 5: Write theStep 5: Write the groups according to 2groups according to 2ststtheorem:theorem: --

EachEach group is a function of m (= 3)group is a function of m (= 3)repeating variables plus one of therepeating variables plus one of the

remaining variablesremaining variablesremaining variablesremaining variables

49

From the problemFrom the problemSo we write,So we write,



F =F = (d, u,(d, u, ,, ))repeating variablesrepeating variables

?

?

Example:Example:


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Example:Example:

Step 6: Solve for the exponents and theStep 6: Solve for the exponents and theforms of the dimensionless groupsforms of the dimensionless groups

50

Example:Example:


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For the firstFor the first group,group,


MM00LL00TT00 = (ML= (ML--33))aa11(LT(LT--11))bb11(L)(L)cc11(ML(ML--11TT--11))

For M:For M: 0 = a0 = a11+ 1+ 1

aa11 == 11

For L:For L: 0 =0 = 3a3a11 + b+ b11 + c+ c11 11

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11 ++ 11==

For T:For T: 0 =0 = bb11 11bb11 == 11

cc11 == 11


--11uu--11dd--11

11 == //ud orud or ud/ud/ = Re= Re

For the secondFor the second group,group,


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For the secondFor the second group,group,


In terms of dimensionsIn terms of dimensions

MM00LL00TT00 = (ML= (ML--33))aa22(LT(LT--11))bb22(L)(L)cc22(MLT(MLT--22))

For each dimension the powers must be equalFor each dimension the powers must be equal

on both sides of the equation, soon both sides of the equation, soFor M:For M: 0 = a0 = a22 + 1+ 1

aa22 == 11

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For L:For L: 0 =0 = 3a3a22 + b+ b22 + c+ c22 + 1+ 1

0 = 4 + b0 = 4 + b22 + c+ c22For T:For T: 0 =0 = bb22 22

bb22 == 22

cc22 == 44 bb22 == 22GivingGiving 22 asas

22 == --11uu--22dd--22FF

22 = F/= F/uu22

dd22

Example:Example:


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pp

Step 7: Rearrange theStep 7: Rearrange the group as desiredgroup as desired

Thus, the problem may beThus, the problem may bedescribed by the following functiondescribed by the following functionof the 2 dimensionlessof the 2 dimensionless groupgroup

((11,, 22) = 0) = 0

(Re, F/(Re, F/uu22dd22) = 0) = 0

53

That is:That is:

F/F/uu22dd22 == 11(Re)(Re)

or:or:

FF == uu22dd22 11(Re)(Re)

Example:Example:


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pp

FF == uu22dd22 11(Re)(Re)

Thus dimensional analysisThus dimensional analysis

indicates that the drag force isindicates that the drag force isproportional toproportional to ,, uu22 and dand d22, and, andalso depends on the Realso depends on the Re

54

WHAT IFWHAT IF


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WHAT IFWHAT IF

nn m = 0 ?m = 0 ?

55

Do the following:Do the following: Check your list of parametersCheck your list of parameters

Check your algebraCheck your algebra

Reduce m by oneReduce m by one

Cengel pg 290Cengel pg 290


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Cengel pg. 290Cengel pg. 290

You are curious as to theYou are curious as to the

relationship between soap bubblerelationship between soap bubble

radius and the pressure inside theradius and the pressure inside the

soap bubblesoap bubble

You reason that pressure inside theYou reason that pressure inside the

56

bubble must be greater thanbubble must be greater than

atmospheric pressure, andatmospheric pressure, and

Shell of the bubble is underShell of the bubble is under

tension, alsotension, also

Surface tension must be importantSurface tension must be important

Not knowing other physic, you decide toNot knowing other physic, you decide to


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g p y yg p y y

approach the problem usingapproach the problem using

dimensional analysisdimensional analysis

Establish a relationship between:Establish a relationship between: --

Pressure different,Pressure different, p = pp = pinsideinside ppatmatm Soap bubble radius, RSoap bubble radius, R

57

ss

Assumptions:Assumptions:

Bubble is neutrally buoyant, and soBubble is neutrally buoyant, and so

gravity is not importantgravity is not importantNo other variables are importantNo other variables are important

List variables and its dimensionsList variables and its dimensions


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pp (ML(ML--11

TT--22

)) RR (L)(L)

ss (MT(MT--22))

n = 3 m = 3n = 3 m = 3nn m = 0m = 0

58

List variables and its dimensionsList variables and its dimensions


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pp (ML(ML--11

TT--22

)) RR (L)(L)

ss (MT(MT--22))

n = 3 m = 3n = 3 m = 3nn m = 0m = 0

Do the following:Do the following: Check your list of parametersCheck your list of parameters

Check our al ebraCheck our al ebra

59

SoSo

mm 1 = 31 = 3 1 = 21 = 2

Reduce m by oneReduce m by one


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NowNow

n = 3 m = 2n = 3 m = 2

nn m = 1m = 1

--

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11 groupgroup

2 repeating variables2 repeating variables

Now write:Now write:


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11 = R= Raa

ss

bb

ppIn terms of dimensionsIn terms of dimensions

MM00LL00TT00 = (L)= (L)aa(MT(MT--22))bb(ML(ML--11TT--22))

Solve for the exponentsSolve for the exponentsFor M:For M: 0 = b + 10 = b + 1

b =b = 11

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For L:For L: 0 = a0 = a 11a = 1a = 1

For T:For T: 0 =0 = 2b2b 22

b =b = 11GivingGiving 11 asas

11 = R= Rss--11p =p = RR p/p/ss

Rearrange theRearrange the group as desiredgroup as desired


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((11) = 0) = 0((RR p/p/ss) = 0) = 0

That is:That is:

RRp/p/ss == 11(Nothing)(Nothing)

This is possible only whenThis is possible only when

62

RRp/p/

ss == constantconstantThereforeTherefore

pp == constantconstant ss/R/R


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pp == constantconstant ss/R/R

Note:Note:

63

mens ona ana ys s canno pre cmens ona ana ys s canno pre c

the value of the constantthe value of the constantFurther experimental works reveals thatFurther experimental works reveals that

the constant is equal to 4the constant is equal to 4

microsoft power point - bmcf 2223 - ch 9 dimensional analysis e2

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