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3rd year uids and engineering analysis

Tristan RobinsonDepartment of Civil, Environmental and Geomatic Engineering

University College London

October 19, 2012
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Table of contents

1 Dimensional analysis and similarityDimensionsNon-dimensional equationsSimilarityBuckingham - theorem
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1 Dimensional analysis and similarity

Dimensions Non-dimensional equations Similarity Buckingham - theorem
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Dimensions and units

Dimension: Measure of a physical quantity (length (L), time(T), mass (M)).

Units: Assignment of a number to a dimension ((m), (s),(kg)).

7 Primary dimensions (SI Units)1. Length Metre, m L2. Mass Kilogram, kg M3. Time Second, s T

4. Temperature Kelvin, K Q5. Current Ampere, A I 6. Luminosity Candela C d 7. Matter Mole N

Secondary units are made up from combinations ofprimary units.
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Dimensional analysis

Generate non-dimensional parameters that help in thedesign of experiments (physical and/or numerical) and inreporting of results

Obtain scaling laws so that prototype performance can bepredicted from model performance.
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The process of non-dimensionalisingequations

Law of dimensional homogeneity: every additive term in anequation must have the same dimensions If we divide each term in the equation by a collection of

variables that have the same dimensions, the equation isrendered non-dimensional

In the process of non-dimensionalising an equation,non-dimensional parameters ofter appear (ie Re and Fr)
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Non-dimensional Bernoulli equation

Divide by 0U 20 and set = 1 (incompressible condition)

p + 12

U 2 + g z = C

Since g = 1Fr 2 where Fr =

U 0 gL

p + 12 U

2

+ 1Fr 2 z = C

This is the non-dimensional Bernoulli equation.

Scaled according to the Froude number.
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Different types of similarity models

Geometric similarity - each dimension must be scaled bythe same factor and hence the angles in the model are

equal to the prototype. Kinematic similarity - velocity as any point in the model

must be proportional

Dynamic similarity - all forces in the model ow scale by aconstant factor to corresponding forces in the prototypeow.

Complete similarity is achieved only if all 3 conditions aremet.
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Complete similarity

This would require that both Fr p , m and Re p , m of theprototype and model match

Re p = U p Lp

p = Re m =

U m Lm m

Lm Lp

= m U p p U m

Fr p = U p

gLp = Fr m =

U m

gLm Lm Lp

=U m U p

2

Hence m p

= Lm Lp

3

2

The ratio of the viscosity is a function of the length scale.

More common is partial similarity when only similarity ofthe most important effects are ensure d.
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Buckingham - theorem

A fundamental question is how many dimensionlessproducts are required to replace the list of variables.

A system has k variables it can be described by k r indepedent products, where r is the minimumn number ofreference dimensions required to describe the variables

The dimensionless producst are referred to as pi terms(1 , 2 . . . k r )

This method does not predict the exact mathematical formof the equation.
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Apply the - theorem

Dene the number of independent dimensional parameters(n ) governing the behaviour of the system.

Dene the primary dimensions ( k ). (usually length [L], time[T ] and mass [M ]).

Number of non-dimensional parameters: m = n k . Construct the non-dimensional groups 1 , . . . , m as

products of unknown powers of dimensional parameters. All groups are non-dimensional, that is the total power of

each dimension should be 0.
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Example: Pipe ow

For the case of a fully developed pipe ow.

Generate a nondimensional relationship between shearstress and other parameters.


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Pipe ow: Dimensions

Quantity Units Dimension

Dependent Shear stress w [ML 1T 2 ]parameterIndependent Flow velocity U [LT 1]parameters Denisty [ML 3]

Viscosity [ML 1T 1 ]Roughness height [L]

Diameter D [L]Primary Length D [L]dimensions Time t [T ]

Mass m [M ]
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Pipe ow: Repeated variables

The expected number of nondimensional functions ( ) is

6

no . of parameters 3

no . of primary dimensions= 3

1 , 2 , 3.

Choose 3 repeating parameters1 Cannot pick dependent variable.2 Cannot pick parameters with the same dimensions.3 Pick parameters that you can measure.

Let the repeating variables be U , D and Generate the 1 , 2 and 3 functions
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Pipe ow: Generate 1

Non-dimensional shear stress1 = w U a D b c

Dimensions[1] = M 1L 1T 2 L1T 1 a [L]b M 1L 3 c

Power of each primary dimension should be 0L : 1 + a + b 3c = 0T : 2 a = 0M : 1 + c = 0

We have a = 2, b = 0 and c = 1.1 =

w

U 2
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Pipe ow: Generate 2

Non-dimensional viscosity2 = U a D b c

Dimensions[1] = M 1L 1T 1 L1T 1 a [L]b M 1L 3 c


We have a = 1, b = 1 and c = 1.2 =

UD


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Pipe ow: Generate 3

Non-dimensional roughness3 = U a D b c

Dimensions[1] = L1 L1T 1 a [L]b M 1L 3 c


We have a = 0, b = 1 and c = 0.3 =

D
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Variation of friction factor
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Exercise: theorem

Find an expression of the elevation of a ball in a vacuum.The elevation z of the ball must be a function of time t ,initial vertical speed w 0 , initial elevation z 0 and gravitationalacceleration g

Find the height as a function of independent variables andconstants

z = f (t , w 0 , z 0 , g )
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Ball in a vacuum: Repeated variables

The expected number of nondimensional functions ( ) is

5

no . of parameters 2

no . of primary dimensions= 3

1 , 2 , 3

.

Choose 2 repeating parameters1 Cannot pick dependent variable.2 Cannot pick parameters with the same dimensions.3 Pick parameters that you can measure.

Let the repeating variables be w 0 and z 0 Generate the 1 , 2 and 3 functions
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Ball in a vacuum: Generate 1

Non-dimensional elevation1 = zw a 0 z

b 0

Dimensions

[1] = [L] LT 1 a [L]b

Power of each primary dimension should be 0L : 1 + a + b = 0T : 0 a = 0

We have a = 0 and b = 1.1 =

z z

0
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Non-dimensional time2 = tw a 0 z

b 0

Dimensions

[1] = [T ] L1T 1 a [L]b

Power of each primary dimension should be 0L : a + b = 0

T : 1 a = 0 We have a = 1 and b = 1.

2 = tw 0

z 0


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Non-dimensional gravity3 = gw a 0 z

b 0

Dimensions

[1] = LT 2 LT 1 a [L]b

Power of each primary dimension should be 0L : 1 + a + b = 0

T : 2 a = 0 We have a = 2 and b = 1.

3 = gz 0w 2

0
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Mathematical derivation

Object falling by gravity through a vacuumd2z d t 2

= g

with initial height z 0 and velocity w 0 in the z -direction. Solve

z = z 0 + w 0 t 12

gt 2

Non-dimensional

z = 1 + t 1

2Fr 2t 2
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Advanced problem: Pumps

Quantity Units Dimension

Parameter Flow rate Q [L3T 1 ]Pressure head gH [L2T 2 ]Power P [ML2T 3]Speed of rotation N [T 1]Rotor diameter D [L]Density [ML 3 ]

Viscosity [ML 1

T 1

]Primary Length D [L]Time t [T]Mass m [M]
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Pump characteristic: Generate 1

Non-dimensional ow rate1 = QD a N b c

Dimensions[1] = L3T 1 [L]a T 1 b ML

3 c

Power of each primary dimension should be 0L : 3 + a + 3c = 0T : 1 b = 0M : 0 + c = 0

We have a = 3 b = 1 and c = 0.1 =

Q

D 3N
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Non-dimensional pressure head2 = gHD a N b c

Dimensions[1] = L2T 2 [L]a T 1 b ML

3 c

Power of each primary dimension should be 0L : 2 + a + 3c = 0T : 2 b = 0M : 0 + c = 0

We have a = 2, b = 2 and c = 0.2 =

gH

D 2N

2
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Non-dimensional power3 = PD a N b c

Dimensions

[1] = ML2T 3 [L]a T 1b

ML 3

c

Power of each primary dimension should be 0L : 2 + a 3c = 0T : 3 b = 0M : 1 + c = 0

We have a = 5, b = 3 and c = 1.3 =

P

D 5N

3
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Non-dimensional viscosity4 = D a N b c

Dimensions[1] = ML 1T 1 [L]a T 1 b ML

3 c

Power of each primary dimension should be 0L : 1 + a 3c = 0T : 1 b = 0M : 1 + c = 0

We have a = 2, b = 1 and c = 1.4 =

D 2

N
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Exercise

1 = Q D 2

/ ND uid velocity bladevelocity

Kinematic similarity of two

geomtrically similar machines is achieved if Q / ND 3

is thesame

3 = 3 1 = Q D 2

/D

uid velocity Reynolds number

4 =

412

= P / QgH

power transferred

turbine power This

represents the hydraulic efciency
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Exercise

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