Chapter 7 Dimensional Analysis Modeling, and Similitude

FUNDAMENTALS OF FUNDAMENTALS OF FLUID MECHANICS FLUID MECHANICS Chapter 7 Dimensional Analysis

Chapter 7 Dimensional Analysis Modeling, and Similitude

Modeling, and Similitude

Jyh Jyh - - Cherng Cherng Shieh Shieh

Department of Bio

Department of Bio--Industrial Industrial MechatronicsMechatronics Engineering Engineering National Taiwan University National Taiwan University

MAIN TOPICS MAIN TOPICS

™™Dimensional AnalysisDimensional Analysis

™™Buckingham Pi TheoremBuckingham Pi Theorem

™™Determination of Pi TermsDetermination of Pi Terms

™™Comments about Dimensional AnalysisComments about Dimensional Analysis

™™Common Dimensionless Groups in Fluid MechanicsCommon Dimensionless Groups in Fluid Mechanics

™™Correlation of Experimental DataCorrelation of Experimental Data

™™Modeling and SimilitudeModeling and Similitude

™™Typical Model StudiesTypical Model Studies

™™Similitude Based on Governing Differential EquationSimilitude Based on Governing Differential Equation

Dimensional Analysis

Dimensional Analysis

™™A typical fluid mechanics problem in which A typical fluid mechanics problem in which

experimentation is required, consider the steady flow of an experimentation is required, consider the steady flow of an

incompressible Newtonian fluid through a long, smooth incompressible Newtonian fluid through a long, smooth--

walled, horizontal, circular pipe.

walled, horizontal, circular pipe.

™™An important characteristic of this system, which would An important characteristic of this system, which would be interest to an engineer designing a pipeline, is the

be interest to an engineer designing a pipeline, is the

pressure drop per unit length that develops along the pipe pressure drop per unit length that develops along the pipe

as a result of friction.

as a result of friction.

Dimensional Analysis

Dimensional Analysis

™™The first step in the planning of an experiment to study The first step in the planning of an experiment to study this problem would be to decide on the factors, or

this problem would be to decide on the factors, or

variables, that will have an effect on the pressure drop.

variables, that will have an effect on the pressure drop.

™™Pressure drop per unit lengthPressure drop per unit length

) V , , , D ( f

p = ρ µ

∆

99 Pressure drop per unit length depends on FOUR variables:Pressure drop per unit length depends on FOUR variables:

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

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

Dimensional Analysis

Dimensional Analysis

™™To perform the experiments in a meaningful and To perform the experiments in a meaningful and

systematic manner, it would be necessary to change on of systematic manner, it would be necessary to change on of the variable, such as the velocity, which holding all other the variable, such as the velocity, which holding all other

constant, and measure the corresponding pressure drop.

constant, and measure the corresponding pressure drop.

™™Difficulty to determine the functional relationship between Difficulty to determine the functional relationship between the pressure drop and the various facts that influence it.

the pressure drop and the various facts that influence it.

Series of Tests

Series of Tests

Dimensional Analysis

Dimensional Analysis

™™Fortunately, there is a much simpler approach to the Fortunately, there is a much simpler approach to the problem that will eliminate the difficulties described problem that will eliminate the difficulties described

above.

above.

DDCollecting these variables into two Collecting these variables into two nondimensionalnondimensional combinations of the variables (called dimensionless combinations of the variables (called dimensionless

product or dimensionless groups) product or dimensionless groups)

99 Only one dependent and one Only one dependent and one independent variable

independent variable

99 Easy to set up experiments to Easy to set up experiments to determine dependency

determine dependency

⎟⎟ ⎠

⎜⎜ ⎞

⎝

⎛

µ φ ρ

ρ =

∆ VD

V p D

2l

Plot of Pressure Drop Data Using

Plot of Pressure Drop Data Using … …

0 0 0 2

1 2

4

3

2 F L T

) FT )(

T FL

(

) L / F ( L V

p

D = =

ρ

∆

−

− &

l

0 0 0 2

1 2

4

T L ) F

T FL

(

) L )(

LT )(

T FL (

VD = =

µ ρ

−

−

−

&

dimensionless product or dimensionless product or dimensionless groups

dimensionless groups

Buckingham Pi Theorem

Buckingham Pi Theorem

™™ A fundamental question we must answer is how many A fundamental question we must answer is how many

dimensionless products are required to replace the original list dimensionless products are required to replace the original list of of

variables ? variables ?

™™ The answer to this question is supplied by the basic theorem of The answer to this question is supplied by the basic theorem of dimensional analysis that states

dimensional analysis that states

™If an equation involving k variables is dimensionally

homogeneous, it can be reduced to a relationship among k-r independent dimensionless products, where r is the minimum number of reference dimensions required to describe the variables.

Buckingham Pi Theorem

Buckingham Pi Theorem

™™Given a physical problem in which the dependent variable Given a physical problem in which the dependent variable is a function of k

is a function of k--1 independent variables.1 independent variables.

DDMathematically, we can express the functional relationship Mathematically, we can express the functional relationship in the equivalent form

in the equivalent form

) u

,..., u

, u ( f

u

=

0 )

u ,..., u

, u , u (

g

=

Buckingham Pi Theorem

Buckingham Pi Theorem

™™The Buckingham Pi theorem states that: Given a relation The Buckingham Pi theorem states that: Given a relation among k variables of the form

among k variables of the form

™™The k variables may be grouped into kThe k variables may be grouped into k--r independent r independent dimensionless products, or

dimensionless products, or ΠΠ terms, expressible in terms, expressible in functional form by

functional form by

0 )

u ,..., u

, u , u (

g

=

0 )

, , , , ,

, (

or

) ,

, , , ,

(

r k 3

2 1

r k 3

2 1

= Π

Π Π

Π φ

Π Π

Π φ

= Π

−

−

r ??

r ?? Π Π ?? ??

Buckingham Pi Theorem

Buckingham Pi Theorem

™™The number r is usually, but not always, equal to the The number r is usually, but not always, equal to the

minimum number of independent dimensions required to minimum number of independent dimensions required to specify the dimensions of all the parameters. Usually the specify the dimensions of all the parameters. Usually the

reference dimensions required to describe the variables reference dimensions required to describe the variables will be the basic dimensions M, L, and T or F, L, and T.

will be the basic dimensions M, L, and T or F, L, and T.

™™The theorem does not predict the functional form of The theorem does not predict the functional form of φφ or or ϕϕ . The functional relation among the independent . The functional relation among the independent

dimensionless products

dimensionless products ΠΠ must be determined must be determined experimentally.

experimentally.

™™The kThe k--r dimensionless products r dimensionless products ΠΠ terms obtained terms obtained from the procedure are independent.

from the procedure are independent.

Buckingham Pi Theorem

Buckingham Pi Theorem

™™A A ΠΠ term is not independent if it can be obtained from a term is not independent if it can be obtained from a product or quotient of the other dimensionless products of product or quotient of the other dimensionless products of

the problem. For example, if the problem. For example, if

then neither

then neither ΠΠ₅₅ nor nor ΠΠ₆₆ is independent of the other is independent of the other dimensionless products or

dimensionless products or dimensionless groupsdimensionless groups..

32 4 / 13 3 6

2

5 2 1 or

Π

= Π Π Π

Π

= Π Π

Determination of Pi Terms

Determination of Pi Terms

™™Several methods can be used to form the dimensionless Several methods can be used to form the dimensionless products, or pi term, that arise in a dimensional analysis.

products, or pi term, that arise in a dimensional analysis.

™™The method we will describe in detail is called the The method we will describe in detail is called the METHOD of repeating variables.

METHOD of repeating variables.

™™Regardless of the method to be used to determine the Regardless of the method to be used to determine the dimensionless products,

dimensionless products, one begins by listingone begins by listing all all dimensional variables

dimensional variables that are known (or believed) to that are known (or believed) to affect the given flow phenomenon.

affect the given flow phenomenon.

™™Eight steps listedEight steps listed below outline a recommended below outline a recommended procedure for determining the

procedure for determining the ΠΠ terms.terms.

Determination of Pi Terms

Determination of Pi Terms

™ ™ Step 1 List all the variables. Step 1 List all the variables.

>>List all the dimensional variables involved.List all the dimensional variables involved.

>>Keep the number of variables to a minimum, so that we Keep the number of variables to a minimum, so that we can minimize the amount of laboratory work.

can minimize the amount of laboratory work.

>>All variables must be independent. For example, if the All variables must be independent. For example, if the cross

cross--sectional area of a pipe is an important variable, sectional area of a pipe is an important variable, either the area or the pipe diameter could be used, but either the area or the pipe diameter could be used, but not both, since they are obviously not independent.

not both, since they are obviously not independent.

γγ=ρ=ρ××g, that is, γg, that is, γ,ρ,ρ, and g are not independent., and g are not independent.

Determination of Pi Terms

Determination of Pi Terms

™ ™ Step 1 List all the variables. Step 1 List all the variables.

>>Let k be the number of variables.Let k be the number of variables.

>>Example: For pressure drop per unit length, k=5. (All Example: For pressure drop per unit length, k=5. (All variables are

variables are ∆∆pp_ll,, D,D,µµ,ρ,ρ, and , and V )V )

) V , ,

, D ( f

p = ρ µ

∆ _l

Determination of Pi Terms

Determination of Pi Terms

™ ™ Step 2 Express each of the variables in terms of Step 2 Express each of the variables in terms of basic dimensions. Find the number of reference basic dimensions. Find the number of reference

dimensions.

dimensions.

>>Select a set of fundamental (primary) dimensions.Select a set of fundamental (primary) dimensions.

>>For example: MLT, or FLT.For example: MLT, or FLT.

>>Example: For pressure drop per unit length , we choose Example: For pressure drop per unit length , we choose FLT.FLT.

2 4

3 D L FL T

FL p

−

−

−

− = ρ =

=

∆ _l & & &

r=3r=3

Determination of Pi Terms

Determination of Pi Terms

™ ™ Step 3 Determine the required number of pi terms. Step 3 Determine the required number of pi terms.

>>Let k be the number of variables in the problem.Let k be the number of variables in the problem.

>>Let r be the number of reference dimensions (primary Let r be the number of reference dimensions (primary dimensions) required to describe these variables.

dimensions) required to describe these variables.

>>The number of pi terms is The number of pi terms is kk--rr

>>Example: For pressure drop per unit length k=5, r = 3, Example: For pressure drop per unit length k=5, r = 3, the number of pi terms is

the number of pi terms is kk--rr=5=5--3=2.3=2.

Determination of Pi Terms

Determination of Pi Terms

™ ™ Step 4 Select a number of repeating variables, Step 4 Select a number of repeating variables,

where the number required is equal to the number where the number required is equal to the number

of reference dimensions.

of reference dimensions.

>>Select a set of r dimensional variables that includes all Select a set of r dimensional variables that includes all the primary dimensions ( repeating variables).

the primary dimensions ( repeating variables).

>>These repeating variables will all be combined with These repeating variables will all be combined with each of the remaining parameters. No repeating

each of the remaining parameters. No repeating

variables should have dimensions that are power of the variables should have dimensions that are power of the dimensions of another repeating variable.

dimensions of another repeating variable.

>>Example: For pressure drop per unit length ( r = 3) Example: For pressure drop per unit length ( r = 3)

Determination of Pi Terms

Determination of Pi Terms

™ ™ Step 5 Form a pi term by multiplying one of the Step 5 Form a pi term by multiplying one of the nonrepeating

nonrepeating variables by the product of the variables by the product of the

repeating variables, each raised to an exponent that repeating variables, each raised to an exponent that

will make the combination dimensionless.

will make the combination dimensionless.

>>Set up dimensional equations, combining the variables Set up dimensional equations, combining the variables selected in Step 4 with each of the other variables

selected in Step 4 with each of the other variables

(nonrepeating(nonrepeating variables) in turn, to form dimensionless variables) in turn, to form dimensionless groups or dimensionless product.

groups or dimensionless product.

>>There will be k There will be k –– r equations.r equations.

>>Example: For pressure drop per unit lengthExample: For pressure drop per unit length

Determination of Pi Terms

Determination of Pi Terms

™ ™ Step 5 (Continued) Step 5 (Continued)

c b

1 = ∆ p D a V ρ

Π _l

1 c

, 2 b

, 1 a

0 c

2 b

: T

0 c

4 b

a 3

: L

0 c

1 : F

T L F )

T FL

( ) LT

( ) L )(

FL

( ^{3 a 1 b 4 2 c 0 0 0}

−

=

−

=

=

⇒

= +

−

=

− +

+

−

= +

− =

−

− &

1 2

V D p

ρ

= ∆

Π

Determination of Pi Terms

Determination of Pi Terms

™ ™ Step 6 Repeat Step 5 for each of the remaining Step 6 Repeat Step 5 for each of the remaining nonrepeating

nonrepeating variables. variables.

c b

2

= µ D

V ρ Π

1 c

, 1 b

, 1 a

0 c

2 b

1 : T

0 c

4 b

a 2

: L

0 c

1 : F

T L F )

T FL

( ) LT

( ) L )(

T FL

( ^{2 a 1 b 4 2 c 0 0 0}

−

=

−

=

−

=

⇒

= +

−

=

− +

+

−

= +

− =

−

− &

ρ

= µ

Π ² DV

Determination of Pi Terms

Determination of Pi Terms

™ ™ Step 7 Check all the resulting pi terms to make Step 7 Check all the resulting pi terms to make sure they are dimensionless.

sure they are dimensionless.

>>Check to see that each group obtained is dimensionless.Check to see that each group obtained is dimensionless.

>>Example: For pressure drop per unit length .Example: For pressure drop per unit length .

0 0 0 0

0 2 0

0 0 0 0

0 0 1 2

T L M

T L DV F

T L M

T L V F

D p

= ρ =

= µ Π

= ρ =

= ∆ Π

&

&

&

&

l

Determination of Pi Terms

Determination of Pi Terms

™ ™ Step 8 Express the final form as a relationship Step 8 Express the final form as a relationship

among the pi terms, and think about what is means.

among the pi terms, and think about what is means.

>>Express the result of the dimensional analysis.Express the result of the dimensional analysis.

>>Example: For pressure drop per unit length .Example: For pressure drop per unit length .

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛

ρ φ µ

ρ =

∆

V DV D p

l2

) ,

, , , ,

( _{2 3 k r}

1 = φ Π Π Π −

Π

Dimensional analysis will not provide Dimensional analysis will not provide the form of the function. The function the form of the function. The function

can only be obtained from a suitable set can only be obtained from a suitable set

of experiments.

of experiments.

Determination of Pi Terms

Determination of Pi Terms

™™The pi terms can be rearranged. For example, The pi terms can be rearranged. For example, ΠΠ₂₂, could , could be expressed as

be expressed as

⎟⎟ ⎠

⎜⎜ ⎞

⎝

⎛

µ φ ρ

ρ =

∆ VD

V D p

l2 µ

= ρ

Π VD

2

Example 7.1 Method of Repeating Example 7.1 Method of Repeating

Variables Variables

zz A thin rectangular plate having a width w and a height h is A thin rectangular plate having a width w and a height h is located so that it is normal to a moving stream of fluid.

located so that it is normal to a moving stream of fluid.

Assume that the drag, D, that the fluid exerts on the plate Assume that the drag, D, that the fluid exerts on the plate

is a function of w and h, the fluid viscosity,

is a function of w and h, the fluid viscosity, µµ ,and ρ,and ρ, ,

respectively, and the velocity, V, of the fluid approaching respectively, and the velocity, V, of the fluid approaching the plate. Determine a suitable set of pi terms to study this the plate. Determine a suitable set of pi terms to study this

problem experimentally.

problem experimentally.

Example 7.1

Example 7.1 Solution Solution

™™Drag force on a PLATEDrag force on a PLATE

>>Step 1:List all the dimensional variables involved. Step 1:List all the dimensional variables involved.

D,w,h,

D,w,h, ρρ,μ,μ,V,V k=6 dimensional parameters.k=6 dimensional parameters.

>>Step 2:Select primary dimensions M,L, and T. Express Step 2:Select primary dimensions M,L, and T. Express each of the variables in terms of basic dimensions

each of the variables in terms of basic dimensions

) V , ,

, h , w ( f

D = ρ µ

1 3

1 1

2

LT V

ML T

ML

L h

L w

MLT D

−

−

−

−

−

=

= ρ

= µ

=

=

=

&

&

&

&

&

&

Example 7.1

Example 7.1 Solution Solution

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

kk-r-r=6-=6-3=33=3

>>Step 4:Select repeating variables w,V,ρStep 4:Select repeating variables w,V,ρ..

>>Step 5~6:combining the repeating variables with each Step 5~6:combining the repeating variables with each of the other variables in turn, to form dimensionless of the other variables in turn, to form dimensionless groups or dimensionless products.

groups or dimensionless products.

Example 7.1

Example 7.1 Solution Solution

1 c

, 2 b

, 2 a

0 b

2 : T

0 c

3 b a

1 : L

0 c

1 : M

T L M )

ML (

) LT ( ) L )(

MLT (

V

Dw ^{a b c 2 a 1 b 3 c 0 0 0}

1

−

=

−

=

−

=

>>

=

−

−

=

− +

+

= +

=

= ρ

=

Π ^{− − −}

= ρ

Π

V w

D

0 b

: T

0 c

3 b

a 1

: L

0 c

: M

T L M )

ML (

) LT

( ) L ( L V

hw ^{a b c a 1 b 3 c 0 0 0}

2

=

=

− +

+

=

=

= ρ

=

Π ^{− −}

w

2 = h

Π

Example 7.1

Example 7.1 Solution Solution

1 c

, 1 b

, 1 a

0 b

1 : T

0 c

3 b

a 1

: L

0 c

1 : M

T L M )

ML (

) LT

( ) L )(

T ML

( V

w^{a b c 1 1 a 1 b 3 c 0 0 0}

3

−

=

−

=

−

=

>>

=

−

−

=

− +

+

−

= +

=

= ρ

µ

=

Π ^{− − − −}

ρ

= µ

Π

wV

Example 7.1

Example 7.1 Solution Solution

>>Step 7: Check all the resulting pi terms to make sure Step 7: Check all the resulting pi terms to make sure they are dimensionless.

they are dimensionless.

>>Step 8: Express the final form as a relationship among Step 8: Express the final form as a relationship among the pi terms.

the pi terms.

The functional relationship is The functional relationship is

⎟⎟ ⎞

⎜⎜ ⎛

ρ φ µ

=

Π Π

φ

= Π

h , D

or ), ,

(

2 2

3 2

1

Selection of Variables

Selection of Variables

™™One of the most important, and difficult, steps in applying One of the most important, and difficult, steps in applying dimensional analysis to any given problem is the selection dimensional analysis to any given problem is the selection

of the variables that are involved.

of the variables that are involved.

™™There is no simple procedure whereby the variable can be There is no simple procedure whereby the variable can be easily identified. Generally, one must rely on a good

easily identified. Generally, one must rely on a good understanding of the phenomenon involved and the understanding of the phenomenon involved and the

governing physical laws.

governing physical laws.

™™If extraneous variables are included, then too many pi If extraneous variables are included, then too many pi

terms appear in the final solution, and it may be difficult, terms appear in the final solution, and it may be difficult,

time consuming, and expensive to eliminate these time consuming, and expensive to eliminate these

experimentally.

experimentally.

Selection of Variables

Selection of Variables

™™If important variables are omitted, then an incorrect result If important variables are omitted, then an incorrect result will be obtained; and again, this may prove to be costly will be obtained; and again, this may prove to be costly

and difficult to ascertain.

and difficult to ascertain.

™™Most engineering problems involve certain simplifying Most engineering problems involve certain simplifying assumptions that have an influence on the variables to be assumptions that have an influence on the variables to be

considered.

considered.

™™Usually we wish to keep the problems as simple as Usually we wish to keep the problems as simple as possible, perhaps even if some accuracy is sacrificed possible, perhaps even if some accuracy is sacrificed

Selection of Variables

Selection of Variables

™™A suitable balance between simplicity and accuracy is an A suitable balance between simplicity and accuracy is an desirable goal.~~~~~

desirable goal.~~~~~

™™Variables can be classified into three general group:Variables can be classified into three general group:

BBGeometry: lengths and angles.Geometry: lengths and angles.

BBMaterial Properties: relate the external effects and the Material Properties: relate the external effects and the responses.

responses.

BBExternal Effects: produce, or tend to produce, a change External Effects: produce, or tend to produce, a change in the system. Such as force, pressure, velocity, or

in the system. Such as force, pressure, velocity, or gravity.

gravity.

Selection of Variables

Selection of Variables

™™Points should be considered in the selection of variables:Points should be considered in the selection of variables:

BBClearly define the problem. WhatClearly define the problem. What’’s the main variable s the main variable of interest?

of interest?

BBConsider the basic laws that govern the phenomenon.Consider the basic laws that govern the phenomenon.

BBStart the variable selection process by grouping the Start the variable selection process by grouping the variables into three broad classes.

variables into three broad classes.

Continued Continued

Selection of Variables

Selection of Variables

™™Points should be considered in the selection of variables:Points should be considered in the selection of variables:

BBConsider other variables that may not fall into one the Consider other variables that may not fall into one the three categories. For example, time and time dependent three categories. For example, time and time dependent variables.

variables.

BBBe sure to include all quantities that may be held Be sure to include all quantities that may be held constant (e.g., g).

constant (e.g., g).

BBMake sure that all variables are independent. Look for Make sure that all variables are independent. Look for relationships among subsets of the variables.

relationships among subsets of the variables.

Determination of Reference Dimension Determination of Reference Dimension

™™When to determine the number of pi terms, it is important When to determine the number of pi terms, it is important to know how many reference dimensions are required to to know how many reference dimensions are required to

describe the variables.

describe the variables.

™™In fluid mechanics, the required number of reference In fluid mechanics, the required number of reference

dimensions is three, but in some problems only one or two dimensions is three, but in some problems only one or two

are required.

are required.

™™In some problems, we occasionally find the number of In some problems, we occasionally find the number of reference dimensions

reference dimensions needed to describe all variables is needed to describe all variables is smaller than the number of basic dimensions

smaller than the number of basic dimensions. Illustrated in . Illustrated in Example 7.2.

Example 7.2.

Example 7.2 Determination of Pi Terms Example 7.2 Determination of Pi Terms

zz An open, cylindrical tank having a diameter D is An open, cylindrical tank having a diameter D is

supported around its bottom circumference and is filled to supported around its bottom circumference and is filled to

a depth h with a liquid having a specific weight

a depth h with a liquid having a specific weight γγ. The . The vertical deflection,

vertical deflection, δδ , of the center of the bottom is a , of the center of the bottom is a function of D, h, d,

function of D, h, d, γγ, and E, where d is the thickness of , and E, where d is the thickness of

the bottom and E is the modulus of elasticity of the bottom the bottom and E is the modulus of elasticity of the bottom

material. Perform a dimensional analysis of this problem.

material. Perform a dimensional analysis of this problem.

Example 7.2

Example 7.2 Solution Solution

™™The vertical deflectionThe vertical deflection

( ^{D h d E} )

f , , , γ ,

δ =

For F,L,T. Pi terms=6

For F,L,T. Pi terms=6--2=42=4 For M,L,T Pi terms=6

For M,L,T Pi terms=6--3=33=3

2 1

2

2 2

3, ML T FL

L d

L h

L D

L

−

−

−

−

−

− γ =

= γ

=

=

=

= δ

Example 7.2

Example 7.2 Solution Solution

™™For F,L,T system, Pi terms=6For F,L,T system, Pi terms=6--2=42=4

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛ Φ γ δ =

γ ⇒

= Π

= Π

= δ Π

=

Π D

, E D , d D

h D

D , E

D , d

D , h

D ^{2 3 4}

1

D and

D and γγ are selected as repeating variablesare selected as repeating variables

4 4

3 3

2 2

1 1

b 4 a

b 3 a

b 2 a

b 1 a

ED dD

hD D

γ

= Π

γ

= Π

γ

= Π

γ δ

= Π

Example 7.2

Example 7.2 Solution Solution

™™For M,L,T system, Pi terms=6For M,L,T system, Pi terms=6--3=3 ?3=3 ?

A closer look at the dimensions of the variables listed A closer look at the dimensions of the variables listed

reveal that only two reference dimensions,

reveal that only two reference dimensions, L and MTL and MT^-2-2 are required.

are required.

Determination of Reference Dimension Determination of Reference Dimension

( ^{γ σ} )

=

∆ h f D , ,

2 2

2 T

M T

L L M

L

D

h γ σ

∆

L F L

L F L

D h

3

σ γ

∆ EXAMPLE

EXAMPLE

MLT SYSTEM

MLT SYSTEM FLT SYSTEMFLT SYSTEM

Pi term=4-3=1 Pi term=4-2=2

Determination of Reference Dimension Determination of Reference Dimension

2 2

0 0

T

0 2

1 1

L

1 1

0 0

M

D h

−

−

−

σ γ

∆

0 0

0 0

T

1 3

1 1

L

1 1

0 0

F

D h

−

−

σ γ

∆

D

1 = ∆h Π

⎟⎞

⎜⎛ σ Φ

∆ = σ ⇒

=

Π h

Set Dimensional Matrix Set Dimensional Matrix

MLT SYSTEM

MLT SYSTEM FLT SYSTEMFLT SYSTEM

Rank=2 Pi term=4

Rank=2 Pi term=4--2=22=2

Uniqueness of Pi Terms

Uniqueness of Pi Terms

™™The Pi terms obtained depend on the somewhat arbitrary The Pi terms obtained depend on the somewhat arbitrary selection of repeating variables. For example, in the

selection of repeating variables. For example, in the problem of studying the pressure drop in a pipe.

problem of studying the pressure drop in a pipe.

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛

µ Φ ρ

ρ =

∆ VD

V D p

2 1 l

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛

µ Φ ρ

µ =

∆ VD

V D p

2 l 2

) V , , , D ( f

p = ρ µ

∆

Selecting D,V, and ρ as repeating variables:

Selecting D,V, and µ as repeating variables:

Uniqueness of Pi Terms

Uniqueness of Pi Terms

Both are correct, and both would Both are correct, and both would

lead to the same final equation for lead to the same final equation for

the pressure drop.

the pressure drop. There is not a There is not a unique set of pi terms

unique set of pi terms which which

arises from a dimensional analysis arises from a dimensional analysis. .

The functions

The functions ΦΦ₁₁ and and ΦΦ₂₂ are will are will be different because the dependent be different because the dependent

pi terms are different for the two pi terms are different for the two

relationships.

relationships.

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛

µ Φ ρ

ρ =

∆ VD

V D

pl2 1

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛

µ Φ ρ

µ =

∆ VD

V D p

2 l 2

Uniqueness of Pi Terms

Uniqueness of Pi Terms

(

)

1 = Φ Π ,Π Π

(

) (

)

2 1

1

= Φ Π , Π = Φ Π , Π Π

EXAMPLE EXAMPLE

3b a2

'2

= Π Π Π

Form a new pi term Form a new pi term

All are correct

Uniqueness of Pi Terms

Uniqueness of Pi Terms

µ

= ∆ µ

× ρ ρ

∆

V D p

VD V

D

p ²

2 l

l

⎟⎟ ⎠

⎜⎜ ⎞

⎝

⎛

µ Φ ρ

µ =

∆ VD

V D p

2 l 2

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛

µ Φ ρ

ρ =

∆ VD

V D p

2 1

l Selecting D,V, and ρ as

repeating variables:

Common Dimensionless Groups

Common Dimensionless Groups

™ A list of variables that commonly arise in fluid mechanical problems.

™ Possible to provide a

physical interpretation to the dimensionless groups which can be helpful in assessing their influence in a particular application.

Froude Number

Froude Number

™™ In honor of William Froude (1810~1879), a British civil engineer, In honor of William Froude (1810~1879), a British civil engineer, mathematician, and naval architect who pioneered the use of towi mathematician, and naval architect who pioneered the use of towing ng

tanks for the study of ship design.

tanks for the study of ship design.

™™ Froude number is the ratio of the forces due to the acceleration of a Froude number is the ratio of the forces due to the acceleration of a fluid particles (inertial force) to the force due to gravity (gr

fluid particles (inertial force) to the force due to gravity (gravity avity forces).

forces).

™™ Froude number is significant for flows with free surface effectsFroude number is significant for flows with free surface effects..

™™ Froude number less than unity indicate subcritical flow and valuFroude number less than unity indicate subcritical flow and values es greater than unity indicate supercritical flow.

greater than unity indicate supercritical flow.

3 2 2 2 2

gL L V gL

Fr V gL

Fr V

ρ

= ρ

=

>>

=

Froude Number

Froude Number

*

*

* s s 2 s

s s

s ds

V dV V

ds V dV dt

a dV

= l

=

=

l s s

V

V_s^* = V^{s *} =

ds m V dV

F V _*

*

* s s 2

I = l

l l

l g

V g

V ds

V dV g

V F

r F F

2

*

*

* s s 2

G

I = ≡ ≡

=

Reynolds Number

Reynolds Number

™™ In honor of Osborne Reynolds (1842~1912), the British engineer In honor of Osborne Reynolds (1842~1912), the British engineer who first demonstrated that this combination of variables could who first demonstrated that this combination of variables could be be

used as a criterion to distinguish between laminar and turbulent

used as a criterion to distinguish between laminar and turbulent flow.flow.

™™ The Reynolds number is a measure of the ration of the inertia foThe Reynolds number is a measure of the ration of the inertia forces rces to viscous forces.

to viscous forces.

™™ If the Reynolds number is small (Re<<1), If the Reynolds number is small (Re<<1), thithi sis an indication that sis an indication that the viscous forces

the viscous forces arar dominant in the problem, and it may be dominant in the problem, and it may be possible to neglect the inertial effects; that is, the density o

possible to neglect the inertial effects; that is, the density of the fluid f the fluid will no be an important variable.

will no be an important variable.

= υ µ

= ρ V l V l

Re

Reynolds Number

Reynolds Number

™™ Flows with very small Reynolds numbers are commonly referred to Flows with very small Reynolds numbers are commonly referred to as as ““creeping flowscreeping flows””..

™™ For large Reynolds number flow, the viscous effects are small For large Reynolds number flow, the viscous effects are small relative to inertial effects and for these cases it may be possi

relative to inertial effects and for these cases it may be possible to ble to neglect the effect of viscosity and consider the problem as one neglect the effect of viscosity and consider the problem as one

involving a

involving a ““nonviscousnonviscous”” fluid.fluid.

™™ Flows with Flows with ““largelarge”” Reynolds number generally are turbulent. Flows Reynolds number generally are turbulent. Flows in which the inertia forces are

in which the inertia forces are “small“small”” compared with the viscous compared with the viscous forces are characteristically laminar flows

forces are characteristically laminar flows..

Euler number Euler number

™™ In honor of Leonhard Euler (1707~1783), a famous Swiss In honor of Leonhard Euler (1707~1783), a famous Swiss

mathematician who pioneered work on the relationship between mathematician who pioneered work on the relationship between

pressure and flow.

pressure and flow.

™™ EulerEuler’’s number is the ratio of pressure force to inertia forces. It iss number is the ratio of pressure force to inertia forces. It is often called the

often called the pressure coefficient, Cp.pressure coefficient, Cp.

2

2 V

p V

Eu p

ρ

≡ ∆

= ρ

Cavitation

Cavitation Number Number

™™ For problems in which For problems in which cavitationcavitation is of concern, the dimensionless is of concern, the dimensionless group is commonly used, where

group is commonly used, where pp_vv is the vapor is the vapor pressure and p

pressure and p_rr is some reference pressure.is some reference pressure.

™™ The The cavitationcavitation number is used in the study of number is used in the study of cavitationcavitation phenomena.phenomena.

™™ The smaller the cavitationThe smaller the cavitation number, the more likely cavitationnumber, the more likely cavitation is to is to occur.

occur.

2 v r

2 V 1

p Ca p

ρ

= −

2 21

v

r p ) / V

p

( − ρ

Cauchy Number & Mach Number

Cauchy Number & Mach Number

™™ The Cauchy number is named in honor of AugustinThe Cauchy number is named in honor of Augustin Louis de Louis de Cauchy (1789~1857), a French engineer, mathematician, and Cauchy (1789~1857), a French engineer, mathematician, and

hydrodynamicist hydrodynamicist..

™™ The Mach number is named in honor of Ernst Mach (1838~1916), The Mach number is named in honor of Ernst Mach (1838~1916), an Austrian physicist and philosopher.

an Austrian physicist and philosopher.

™™ Either number may be used in problems in which fluid Either number may be used in problems in which fluid compressibility is important.

compressibility is important.

E Ca Ma V

V E E

V c

Ma V E

a V C

2 2

2 ρ =

ρ =

= ρ

= ρ =

=

υ υ υ

υ

Cauchy Number & Mach Number

Cauchy Number & Mach Number

™™ Both numbers can be interpreted as representing an index of the Both numbers can be interpreted as representing an index of the ration of inertial force to compressibility force, where V is th

ration of inertial force to compressibility force, where V is the flow e flow speed and c is the local sonic speed.

speed and c is the local sonic speed.

™™ Mach number is a key parameter that characterizes compressibilitMach number is a key parameter that characterizes compressibility y effects in a flow.

effects in a flow.

™™ When the Mach number is relatively small (say, less than 0.3), tWhen the Mach number is relatively small (say, less than 0.3), the he inertial forces induced by the fluid motion are not sufficiently

inertial forces induced by the fluid motion are not sufficiently large large to cause a significant change in the fluid density, and in this

to cause a significant change in the fluid density, and in this case the case the compressibility of the fluid can be neglected.

compressibility of the fluid can be neglected.

™™ For truly incompressible flow, c=For truly incompressible flow, c=∞∞ so that M=0.so that M=0.