1Dr. Muhammad Ali Shamimali.shamim@uettaxila.edu.pk051-9047652; Internal 652051 9047652; Internal 652

2` Concept of Dimensional Analysis in fluidphenomenonp

` Solid understanding of fluid flow in pressureconduits

` Understanding of forces on immersed bodies

` Various types of hydraulic machinery andtheir working

By the end of the course, students will be ableto

` Perform Dimensional analysis for Hydraulicmodel studies

` Understand the basics of fluid flow throughpipes, differentiate between laminar andturbulent flow and velocity profiles for both.

3` Understand the formation of boundary layerin immersed bodies, surface drag and form, gdrag.

` Construction and working of turbo-machinesused in the field and how to achieve themaximum efficiency in each of them.

` Dimensional Analysis Rayleigh and Buckingham's Pi-theorem and their

applications.

4` Fluid flow in pipes

Reynold's number and its significance, Instability ofviscous flow.

Viscous flow through circular pipes. Turbulent flowthrough circular pipes.

Semi-empirical theories of turbulence. Velocityprofile in turbulent flow.p

Pipe roughness, Nukuradse's experiments. Moody'sdiagrams.

Pipe networks

` Forces on immersed bodies Development of boundary layer on immersed

Bodies. Elementary theory of surface drag and form drag. Simple lift and drag equations and their

applications to simple engineering problems. Separation of boundary layer.

5` Forces on vanes and turbomachinery Impulse momentum equation.Impulse momentum equation. Forces on moving flat and curved vanes. Impulse turbine, construction features and

operations, Specific speed. Reaction turbine, types, construction features and

operation, specific speed, cavitation, draft tube.

` Centrifugal pumps. Classification, construction features and operation,

specific speed, cavitation.` Reciprocating Pumps

Single acting and double acting pumps, acceleration head, maximum suction lift. Use of air vessels

6` Assessment

20% sessional (quizzes, assignments, presentations etc)

20% mid semester examination 20% laboratory work 40% final exam (end semester)

7what a man means by a term is to be found out by observing what he does with it, not what he says about it.

Percy W. Bridgman

1882-1961

The principal use of the dimensional analysis is todeduce from a study of dimensions of variables in

h l l dany physical system certain limitations and anypossible relationship between those variables(Bridgman).

It is a pure mathematical technique to establish arelationship between physical quantities involved ina fluid phenomenon by considering theirdimensions.

8` Consistent problem solvingapproach

` Reduces errors in algebra` Reinforces unit conversion` Simplifies computation` Improves understanding of math

li iapplications` Multiple ways to solve the same

problem

` Successful problem solvingstrategy for advanced orspecial needs students

` Vertically aligns withstrategies for Physics andCh iChemistry

` Improves Math scores` Easy to assess and grade

9` Identify what you are being asked.` Write down what is given or known.` Look for relationships between knowns

and unknowns (use charts, equations).` Rearrange the equation to solve for the

unknown.` Do the computations, cancel the units,

check for reasonable answers

z EX. How many quarts is 9.3 cups?

9.3 cups ? quarts=9.3 cups ? quarts

10

9.3 cups ? quarts=

9.3 cups x

* Use charts or tables to find relationships

9.3 cups xcups

quart

4

1

*units of known in denominator (bottom) first*** units of unknowns in numerator (top)

11

9.3 cups xcups

quart

4

1

9.3 cups xcups

quart

4

1

1 x 4=

9.3 x 1

4=

9.3= 2.325 quarts

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

ALL IS NOT LOST BECAUSE OF

Secret #2: Dimensional Analysis

Dimensional Analysis

Dimensional Analysis refers to the physicalnature 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

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` Heat Loss Surface Area (L2)

` Mass Volume (L3)

H L /M A /V l` 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)

Polar Bear (L = 2 m)1/L 1/(2 )1/L 1/(0.05 m)

= 20 m-11/L = 1/(2 m)

= 0.5 m-1

20 : 0.5 or 40 : 1

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` Gulliver was 12x theLilliputians

` How much should theyfeed him?12x their food ration?

` A persons food needs are` A persons food needs arerelated to their mass(volume) This dependson the cube of the lineardimension.

Let LG and VG denote Gullivers linear and volume dimensions.Let LL and VL denote the Lilliputians linear and volume dimensions.

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

Now VG (LG)3 and VL (LL)3, soVG / VL = (LG)3 / (LL)3

= (12 LL)3 / (LL)3= 123= 1728

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

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` Development of an equation for fluidphenomenonp e o e o

` Conversion of one system of units to another

` Reducing the number of variables required inan experimental program

` Develop principles of hydraulic similitude formodel study

` MLT system` FLT systemsyste

These two systems are inter-relatedF = ma 2nd Law of motionF = MLT-2M = FL-1T-2M FL T

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[x], Length L[m], Mass M[t], Time T[v], Velocity LT-1[a], Acceleration LT-2

[], Mass Density ML-3[P], Pressure ML-1T-2[E], Energy ML2T-2[I], Electric Current QT-1[q], Electric Change Q[ ],

[F], Force MLT-2g

[E], Electric Field - MLQT-2

` Principle of Dimensional Homogeneity (1822-Baron Joseph Fourier) The fundamental dimensions and their respective

powers should be identical on either side of the sign of equality.

Q A V ti it ti (h )Q = A.V continuity equation (homogeneous)

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` Rayleighs Method` Buckinghams -method

Functional relationship between variables is expressed in the form of an exponential p prelation which must be dimensionally homogeneous

if y is a function of independent variables x1,x2,x3,..xn, then

),.......,,( 321 nxxxxfy =

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In exponential form as

]),.......()(,)(,)[( 321z

ncba xxxxy = ])()()()[( 321 ny

` Write fundamental relationship of the given data` Write the same equation in exponential form` Select suitable system of fundamental dimensions

S b i di i f h h i l i i` Substitute dimensions of the physical quantities` Apply dimensional homogeneity` Equate the powers and compute the values of the

exponents` Substitute the values of exponents` Simplify the expression` Ideal up to three independent variables, can be

d f fused for four.

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` For further understanding, lets explore theequation for the velocity (V) of a pressureq y pwave through a fluid.

` First it should be visualised what physicalfactors actually influence the velocity

` Compressibility density and kinematicviscosity are the physical factors influencingthe motion

E

` The dimensions of these quantities will be

MFL

==

=

=

=LM

LTM

LFE

TLV

2

3

22

,

,

TL3,

21

dbaCEV =C is the dimensionless constant. Substituting

the dimensionsdba

TL

LM

LTM

TL

=

2

32

` For dimensional homogeneity, the exponents of each dimension must be identical on both sides.

` For M` For L` For T

ba +=0dba 231 +=

da += 21

22

` Solving the above equations, we get2/1=a

` So finally,

02/1

==

db

ECV =

` Dimensional analysis was developed in suchway by Lord Rayleigh.y y y g

` Very serviceable method but has beensuperseded.

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` Developed by E. Buckingham (1915)-a more generalizedmethod of dimensional analysis.

` If n is the total number of variables in a dimensionallyhomogenous equation containing m fundamentaldimensions, then they may be grouped into (n-m) terms.

if y=f(x1, x2, xn)then the functional relationship will be written as

0).......,( 21 = mn` Suitable where` Not applicable if (n-m) = 0

4n

` List all physical variables and note n and m.n = total no. of variablesm = eq containing fundamental dimensionsm = eq. containing fundamental dimensions

` Compute number of -terms` Write the eq. in functional form` Write eq. in general form` Select repeating variables. Must have all of the m

fundamental dimensions and should not form a among themselves

` Solve each -term for the unknown exponents by p ydimensional homogeneity.

0.],.........,,[ 321 =

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` Consider the factors affecting the drag force on a sphere.p

` These include size of the sphere D, velocity of the sphere V, density and viscosity . So

0),,,,( =VDFf D` Here we see that n=5

` Choose a Dimensionless system (MLT or FLT)and determine the number of fundamentaldimensions involved in the system (m).

` We will choose MLT system so thecorresponding dimension will be

MMLML

` We can see that here m=3LTM

LM

TLL

TML ,,,, 32

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` Determine the number of -terms needed. Inthis case they will be n-m=5-3=2.y

` Select the primary or repeating variables suchthat they must contain all of the mfundamental dimensions and must not form a (dimensionless group) among themselves.

` Choose , D and V as the repeating variables.The -terms will then be

cba VD 1111 =

` Using the principle of dimensionalhomogeneity, we can solve for the exponentson each side of the equation.

Dcba FVD 2222 =

` Since -terms are dimensionless, they can bereplaced by M0L0T0.

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` So our expressions will look like

130:10:

111

1

3000

1

1

1

++=+=

=

cbaLaM

LTM

TLL

LMTLM

cb

a

10: 1 = cT

` Solving,

1111

1

111 1;1;1

===

===

DVDV

VD

thuscba

` As

/Re DV=

27

11 R=

222 VDF

and

D

=

( )( )0,

0,

221

21

=

=

VD

FR D

` This shows that the drag force depends upon theReynolds number which is the ratio of inertialf fforces to viscous forces.

` Dimensional analysis only provides a partialsolution to the fluid problems as it dependsentirely on the ability of the individual to perceivethe factors influencing a fluid phenomenon.

` So if an important variable is omitted, then theresults could be entirely different.

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