FUNDAMENTALS OF FLUID MECHANICS Chapter 7 Dimensional … classes... · 2015-04-28 · 1 1 FUNDAMENTALS OF FLUID MECHANICS Chapter 7 Dimensional Analysis Modeling, and Similitude

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FUNDAMENTALS OFFLUID MECHANICS

Chapter 7 Dimensional Analysis

Modeling, and Similitude

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MAIN TOPICS

Dimensional AnalysisBuckingham Pi TheoremDetermination of Pi TermsComments about Dimensional AnalysisCommon Dimensionless Groups in Fluid MechanicsCorrelation of Experimental DataModeling and SimilitudeTypical Model StudiesSimilitude Based on Governing Differential Equation

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Dimensional Analysis1/2

Example: Pressure drop per unit length

Pressure drop per unit length depends on FOUR parameters:sphere size (D); speed (V); fluid density (ρ); fluid viscosity (m)

Difficult to know how to set up experiments to determine dependencies

Difficult to know how to present results (four graphs?)

)V,,,D(fp

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Plot of Pressure Drop Data Using …

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Dimensional Analysis2/2

Example: Pressure drop per unit length

Only one dependent and one independent variable Easy to set up experiments to determine dependency Easy to present results (one graph)

VD

V

pD2

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Buckingham Pi Theorem 1/5

A fundamental question we must answer is how many dimensionless products are required to replace the original list of variables ?

The answer to this question is supplied by the basic theorem of 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 Pi terms

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Given a physical problem in which the dependent parameter is a function of k-1 independent parameters.

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

)u,.....,u,u(fu k321

0)u,.....,u,u,u(g k321

Where g is an unspecified function, different from f.

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

The k parameters may be grouped into k-r independent dimensionless ratios, or Π parameters, expressible in functional form by

0)u,.....,u,u,u(g k321

0),,,,,,(or

0),,,,,(

rk321

rk321

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Where the number r is usually, but not always, equal to the minimum number of independent dimensions required to specify the dimensions of all the parameters. Usually the reference dimensions required to describe the variables will be the basic dimensions M, L, and T or F, L, and T.

The theorem does not predict the functional form of or . The functional relation among the independent dimensionless Π parameters must be determined experimentally.

The k-r dimensionless Π parameters obtained from the procedure are independent.

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A Π parameter is not independent if it can be obtained from a product or quotient of the other parameters of the problem. For example, if

then neither Π5 nor Π6 is independent of the other dimensionless parameters.

2

3

4/3

1

6

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1

5or

2

6

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Determination of Pi Terms1/11

Regardless of the method to be used to determine the dimensionless parameters, one begins by listing all dimensional parameters that are known (or believed) to affect the given flow phenomenon.

Do not be afraid to include all the parameters that you feel are important. If the parameter is extraneous, an extra Π parameter may result, but experiments will later show that it may be eliminated from consideration.

Six steps listed below outline a recommended procedure for determining the Π parameter.

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Determination of Pi Terms 2/11

Step 1

List all the dimensional parameters involved.

Let k be the number of parameters.

Example: For pressure drop per unit length pl,D,,,V and k=5

)V,,,D(fp

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Step 2

Select a set of fundamental (primary) dimensions.

For example: MLT, or FLT.

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

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Step 3

List the dimensions of all parameters in terms of primary dimensions.

Let r be the number of primary dimensions.

Example: For pressure drop per unit length r = 3.

1

224

3

LTV

TFLTFL

LDFLp

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Step 4Select a set of r dimensional parameters that includes

all the primary dimensions ( repeating parameters).These repeating parameters will all be combined with

each of the remaining parameters. No repeating parameters should have dimensions that are power of the dimensions of another repeating parameter.

Example: For pressure drop per unit length ( r = 3) select ρ , V, D.

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Step 5

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 k – r equations.

Example: For pressure drop per unit length .

cba1 VDp

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Step 5 (Continued)


1c,2b,1a

0c2b:T

0c4ba3:L

0c1:F

TLF)TFL()LT()L)(FL( 000c24b1a3

21 V

Dp

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Step 5 (Continued)

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 k – r equations.


cba2 VD

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Step 5 (Continued)


1c,1b,1a

0c2b1:T

0c4ba2:L

0c1:F

TLF)TFL()LT()L)(TFL( 000c24b1a2

DV2

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Step 6

Check to see that each group obtained is dimensionless.


0000002

00000021

TLMTLFDV

TLMTLFV

Dp

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Step 7

Express the result of the dimensional analysis.


DVV

Dp2

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EXAMPLE-I 1/4

Drag force on a PLATE

Step 1:List all the dimensional parameters involved. D,w,h, ρ,μ,Vk=6 dimensional parameters.

Step 2:Select primary dimensions M,L, and t.

Step 3:List the dimensions of all parameters in terms of primary dimensions. r=3 primary dimensions

)V,,,h,w(fD

13

11

2

LTVML

TMLLh

LwMLTD

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EXAMPLE-I 2/4

Step 4:Select repeating parameters w,V,. r=3 repeating parameters.

Step 5:combining the repeating parameters with each of the other parameters in turn, to form dimensionless groups.

1c,2b,2a

0b2:T

0c3ba1:L

0c1:M

TLMMLLTLMLTVDw 000c3b1a2cba1

221 Vw

D

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EXAMPLE-I 3/4

1c,1b,1a

0b1:T

0c3ba1:L

0c1:M

TLMMLLTL)TML(Vw 000c3b1a11cba3

0c,0b,1a

0b:T

0c3ba1:L

0c:M

TLMMLLTLLVhw 000c3b1acba2

w

h2

wV3

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EXAMPLE-I 4/4

Step 6: Check to see that each group obtained is dimensionless

The functional relationship is

wV,

w

h

Vw

D

or),,(

22

321

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EXAMPLE-II 1/3

Drag force on a smooth sphere

Step 1:List all the dimensional parameters involved. F,V,D, ρ,μk=5 dimensional parameters.

Step 2:Select primary dimensions M,L, and t.

Step 3:List the dimensions of all parameters in terms of primary dimensions. r=3 primary dimensions

),,V,D(fF

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EXAMPLE-II 2/3

Step 4:Select repeating parameters ρ,V,D. r=3 repeating parameters.


2c2b1a02b:t01cba3:L01a:M

tLMt

MLL

t

L

L

MFDV 000

2

c

ba

3

cba

1

221 DV

F

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EXAMPLE-II 3/3

Step 6: Check to see that each group obtained is dimensionless

2f1e1d01e:t01fed3:L

01d:M

tLMLt

ML

t

L

L

MDV 000f

ed

3

fed

2

VD2


VDf

DV

F

or),(f

22

21

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Selection of Variables 1/3

One of the most important, and difficult, steps in applying dimensional analysis to any given problem is the selection of the variables that are involved.

There is no simple procedure whereby the variable can be easily identified. Generally, one must rely on a good understanding of the phenomenon involved and the governing physical laws.

If extraneous variables are included, then too many pi terms appear in the final solution, and it may be difficult, time consuming, and expensive to eliminate these experimentally. If important variables are omitted, then an incorrect result will be obtained; and again, this may prove to be costly and difficult to ascertain.

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

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

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

Variables can be classified into three general group:

Geometry: lengths and angles.

Material Properties: relate the external effects and the responses.

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

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Points should be considered in the selection of variables:

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

Consider the basic laws that govern the phenomenon.

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

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

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

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

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Determination of Reference Dimension 1/3

Typically, in fluid mechanics, the required number of reference dimensions is three, but in some problems only one or two are required.

EXAMPLEAn open, cylindrical tank having a diameter D is supported around its bottom circumference and is filled to a depth h with a liquid having a specific weight . The vertical deflection, , of the center of the bottom is a function of D, h, d, , and E

E,,d,f,Df

212

223

TMLE,FLETML,FL

LdLhLDL

For F,L,T. Pi terms=6-2=4For M,L,T Pi terms=6-3=3

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

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

D

E,

D

d,

D

h

DD

E,

D

d,

D

h,

D 4321

A closer look at the dimensions of the variables listed reveal that only two reference dimensions, L and MT-2 are required.

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,,Dfh

222 T

M

TL

MLL

Dh

L

F

L

FLL

Dh

3

EXAMPLE

2200T

0211L

1100M

Dh

0000T

1311L

1100F

Dh

D

h1

222 DD

h

D

MLT system FLT system

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

Dimensional matrix

MLT system FLT system Rank=2Pi term=4-2=2

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Uniqueness of Pi Terms1/2

The Pi terms obtained depend on the somewhat arbitrary selection of repeating variables. For example, in the problem of studying the pressure drop in a pipe.

VD

V

DP12

l

VD

V

DP2

2

l

)V,,,D(fp

Selecting D,V, and as repeating variables:

Selecting D,V, and as repeating variables:

Both are correct, and both would lead to the same final equation for the pressure drop.There is not a unique set of pi terms which arises from a dimensional analysis.

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Uniqueness of Pi Terms2/2

321

,

3

'

211,

EXAMPLE

b

3

a

2

'

2 Form a new pi term

V

DPVD

V

DP 2

l

2

l

VD

V

DP2

2

l

VD

V

DP12

l

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Common Dimensionless Groups

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Reynolds Number

Discovered by Osborne Reynolds, the British engineer, in the 1880s.

This parameter is a criterion by which the tube flow may be categorized as laminar or turbulent regime.

The Reynolds number is the ratio of inertia forces to viscous forces.

Flows with “large” Reynolds number generally are turbulent. Flows in which the inertia forces are “small” compared with the viscous forces are characteristically laminar flows.

VLVL

Re

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Euler number

This ratio is named after Leonhard Euler, the Swiss mathematician. Euler was the first to recognize the role of pressure in fluid mechanics.

Euler’s number is the ratio of pressure force to inertia forces. It is often called the pressure coefficient, Cp.

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Cavitation Number

Where p is the pressure in the liquid stream, and pv is the liquid vapor pressure at the test temperature.

The cavitation number is used in the study of cavitation phenomena.

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

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Froude Number

Discovered by Froude and this son, Robert Edmund Froude.

Froude number is the ratio of inertia forces to gravity forces.

Froude number is significant for flows with free surface effects.

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

3

222

2

gL

LV

gL

VFr

gL

VFr

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Weber Number

Weber number is the ratio of inertia forces to surface tension forces.

Weber number is indicative of the existence of, and frequency of, capillary waves at a free surface.

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Mach Number

Introduced by Ernst Mach, the Austrian physicist, in the 1870s.Mach number is the ratio of inertia forces to forces due to

compressibility, where V is the flow speed and c is the local sonic speed.

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

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

2

22

2

LE

LVMor

E

V

ddp

V

c

VM

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Correlation of Experimental Data

Problems with one Pi term.

Problems with two or more terms.c

1

)(21

321

,

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Modeling and Similitude 1/2

Model ? A model is a representation of a physical system that may be used to predict the behavior of the system in some desired respect. Mathematical or computer models may also conform to this definition, our interest will be in physical model.

Prototype? The physical system for which the prediction are to be made.

Models that resemble the prototype but are generally of a different size, may involve different fluid, and often operate under different conditions. Usually a model is smaller than the prototype.

A model test must yield data that can be scaled to obtain the forces, moments, and dynamic loads that would exist on the full-scale prototype.

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Modeling and Similitude 2/2

What conditions must be met to ensure the similarity of model and prototype flows?

Geometric SimilarityModel 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 SimilarityForces on model and prototype differ only by a constant scale

factor.

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Theory of Models 1/4

What are the conditions that ensure dynamic similarity between model and prototype flows ?To develop the procedures for designing models so that the model and prototype will behave in a similar fashion.

For any problem, only a knowledge of the general nature of the physical phenomenon and the variables involved is required, can be described in term of a set of Π parametersas

),,,,(n321

This equation applies to any system that is governed by the same variables.

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If this equation describes the behavior of a particular prototype, a similar relationship can be written for a model of this prototype; that is,

where the form of the function will be the same as long as the same phenomenon is involved in both the prototype and the model.

),,,,,(nmm3m2m1

The prototype and the model must have the same phenomenon.

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Model design (the model is designed and operated) conditions, also called similarity requirements or modeling laws.

The form of Φ is the same for model and prototype, it follows that

nmnm33m22.....

m11

This is the desired prediction equation and indicates that the measured of Π1m obtained with the model will be equal to the corresponding Π1 for the prototype as long as the other Π parameters are equal.

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Theory of Models 4/4 - Summary


The model is designed and operated under the following conditions (called design conditions, also called similarity requirements or modeling laws)

The measured of Π1m obtained with the model will be equal to the corresponding Π1 for the prototype.

nmnm33m22.....

m11 Called prediction equation

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Theory of Models –EXAMPLE 1

Example: Considering the drag force on a sphere.


Design conditions.

Then …

m

mmm

12

m

2

mm

mDV

fDV

F

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Theory of Models –EXAMPLE 2

Example: Determining the drag force on a thin rectangular plate (wx h in size)


Design conditions.

Then …

V,,,h,wfD

Vw

,h

w

Vw

D222

m

mmm

m

m2

m

2

m

2

m

mwV

,h

w

Vw

D

VwwV

,h

w

h

w

m

mmm

m

m

m

2

mm

2

m

2

m

2

m

2

m

m

222D

V

V

w

wD

Vw

D

Vw

D

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Theory of Models –EXAMPLE 3 1/5

Example:A long structural component of a bridge has the cross section shown in Figure. It is known that when a steady wind blows past this type of bluff body, vortices may develop on the downwind side that are shed in a regular fashion at some definite frequency. Since these vortices can create harmful periodic forces acting on the structure, it is important to determine the shedding frequency. For the specific structure of interest, D=0.2m, H=0.4m, and a representative wind velocity 50km/hr. Standard air can be assumed. The shedding frequency is to be determined through the use of a small-scale model that is to be tested in a water tunnel. For the model Dm=20mm and the water temperature is 20 .

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Determine the model dimension, Hm, and the velocity at which the test should be performed.

If the shedding frequency ω for the model is found to be 100Hz, what is the corresponding frequency for the prototype?

For air at standard condition

For water at 20 ,

[ SOLUTION ]

35 m/kg23.1,sm/kg1079.1 3

water3

water m/kg998,sm/kg101

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Step 1:List all the dimensional parameters involved. ωD,H,V,ρ,μ n=6 dimensional parameters.

Step 2:Select primary dimensions F,L and t. Step 3:List the dimensions of all parameters in terms of

primary dimensions. r=3 primary dimensions

Step 4:Select repeating parameters D,V, μ, r=3 repeating parameters.

TMLTFLLTVLH

LDT

224

1

1

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VDVD

D

HVHD

V

DVD

333

222

111

cba

3

cba

2

cba

1


VD

,D

H

V

D

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The Design conditions.

Then….

m

mmm

m

m

m

mmDV

,D

H

V

D

m

m

D

H

D

H

m

mmm DVVD

...DD

HH m

m ...V

D

DV

mm

m

m

...D

D

V

Vm

m

m

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Model Scales

The ratio of a model variable to the corresponding prototype variable is called the scale for that variable.

2

m2

1

m1

m2

m1

2

1

m

V

Vm

V

m

Length Scale

Velocity Scale

Density Scale

Viscosity Scale

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Validation of Models Design 1/3

Most model studies involve simplifying assumptions with regard to the variables to be considered. Although the number of assumptions is frequently less stringent than that required for mathematical models, they nevertheless introduce some uncertainty in the model design.

It is, therefore, desirable to check the design experimentally whenever possible.Validation of model design ?

In some situation the purpose of model design is to predict the effects of certain proposed changes in a given prototype, and in this instance some actual prototype data may be available.

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The model can be designed, constructed, and tested, and the model prediction can be compared with these data. If the agreement is satisfactory, then the model can be changed in the desired manner, and the corresponding effect on the prototype can be predicted with increased confidence.

Another useful and informative procedure is to run tests with a series of models of different sizes, where one of the models can be thought of as the prototype and the others as “models” of this prototype.

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With the models designed and operated on the basis of the proposed design, a necessary conditions for the validity of the model design is that an accurate prediction be made between any pair of models, since one can always be considered as a model of the other.

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Distorted Models 1/6

In many model studies, to achieve dynamic similarity requires duplication of several dimensionless groups.

In some cases, complete dynamic similarity between model and prototype may not be attainable. If one or more of the similarity requirements are not met, for example, if , then it follows that the prediction equation is not true; that is,

MODELS for which one or more of the similarity requirements are not satisfied are called DISTORTED MODELS.

m22

m11

m11

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For example: To determine the drag force on a surface ship, complete dynamic similarity requires that both Reynolds and Froude numbers be duplicated between model and prototype.

2/1

p

p

p2/1

m

m

m gL

VFr

gL

VFr

p

pp

p

m

mm

m

LVRe

LVRe

Froude numbers

Reynolds numbers

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To match Froude numbers between model and prototype

To match Reynolds numbers between model and prototype

2/1

p

m

p

m

L

L

V

V

p

m

p

m

p

m

L

L

V

V

2/3

p

m

p

m

2/1

p

m

p

m

L

L

L

L

L

L

If Lm/Lp equals 1/100(a typical length scale for ship model tests) , then υm/υp must be 1/1000.>>> the kinematic viscosity ratio required to duplicate Reynolds numbers cannot be attained.

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It is impossible in practice for this model/prototype scale of 1/100 to satisfy both the Froude number and Reynolds number criteria; at best we will be able to satisfy only one of them.

If water is the only practical liquid for most model test of free-surface flows, a full-scale test is required to obtain complete dynamic similarity.

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For example: In the study of open channel or free-surface flows. Typically in these problems both the Reynolds number and Froude number are involved.

2/1

pp

p

p2/1

mm

m

m Lg

VFr

Lg

VFr Froude numbers

p

ppp

p

m

mmm

m

LVRe

LVRe

Reynolds numbers

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To match Froude numbers between model and prototype

To match Reynolds numbers between model and prototype

2/1

p

m

p

m

L

L

V

V

m

p

m

p

p

m

p

m

L

L

V

V

If Lm/Lp equals 1/100(a typical length scale for ship model tests) , then υm/υp must be 1/1000.>>> the kinematic viscosity ratio required to duplicate Reynolds numbers cannot be attained.

p

m

mp

pm2/3

m

p

mp

pm

p

m

p

m

/

/L

L

/

/

L

L

V

V

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Typical Model Studies

Flow through closed conduits.

Flow around immersed bodies.

Flow with a free surface.

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Flow Through Closed Conduits 1/2

This type of flow includes pipe flow and flow through valves, fittings, and metering devices. The conduits are often circular, they could have other shapes as well and may contain expansions or contractions.

For flow in closed conduits at low Mach numbers, and dependent pi term, such as pressure drop, can be expressed as

Dependent pi term=

V,,i

i321 ,.....,, Where and are series of length terms used to describe the geometric characteristics.

Particular length dimension for the system

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Flow Through Closed Conduits 2/2

To meet the requirement of geometric similarity

To meet the requirement of Reynolds number

mm

i

im

m

mi

m

im

mm

mm

m

mmm

V

VVV

If the same fluid is used, then

/VVV

Vm

m

m

The fluid velocity in the model will be larger than that in the prototype for any length scale less than 1.Reynolds number similarity may be difficult to achieve because of the large model velocities required.

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Flow Around Immersed Bodies 1/4

This type of flow includes flow around aircraft, automobiles, golf balls, and building.

A general formulation for these problems is

V,,i

i321 ,.....,, Where and are pertinent length terms used to describe the geometric characteristics.

Characteristic length of the system

Dependent pi term=

22D

V21

DC

The dependent pi terms would usually be expressed in the form of a drag coefficient

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To meet the requirement of geometric similarity

To meet the requirement of Reynolds number similarity

mm

i

im

m

mi

m

im

m

m

mm

mm

m

mmm

V

VVV

m

2

m

2

mm2m

2mm

m

22D

V

VD

V21

D

V21

D

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The same fluid is used, then

/VVV

Vm

m

m

The fluid velocity in the model will be larger than that in the prototype for any length scale less than 1.Reynolds number similarity may be difficult to achieve because of the large model velocities required.

A different fluid is used in the model such that 1/m For example, the ratio of the kinematic viscosity of water to that of air is approximately 1/10, so that if the prototype fluid were air, test might be run on the model using water.

How to reduce the fluid velocity in the model ?

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This would reduce the required model velocity, but it still may be difficult to achieve the necessary velocity in a suitable test facility.

Other way to reduce the fluid velocity in the model.Same fluid with different density.. m>An alternative way to reduce Vm is to increase the air pressure in the tunnel so that m>. The pressurized tunnels are obviously complicated and expensive.

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Flow with a Free Surface 1/4

This type of flow includes flow in canals, rivers, spillways, and stilling basics, as well as flow around ship.

For this class of problems, gravitational ,inertial forces, and surface tension are important and, therefore, the Froude number and Weber number become important similarity parameters needed to be considered along with the Reynolds number.

A general formulation for these problems is

2

iV

,g

V,

V,,

i321 ,.....,, Where and are pertinent length terms used to describe the geometric characteristics.

Characteristic length of the systemDependent pi term=

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To meet the requirement of Froude number similarity

To meet the requirement of Reynolds number and Froude number similarity

m

mm

m

mm

m

V

Vgg

g

V

g

V

2/3m

m

m

mm

mm

m

mmm

V

VVV

The working fluid for the prototype is normally either freshwater or seawater and the length scale is small. It is virtually impossible to satisfy , so models involving free-surface flows are usually distorted.

2/3

m/

39

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To meet the requirement of Weber number similarity (surface tension effects)

2

2

m

2

mmm

2

m

m

2

mm

V

V

/

/VV

Same fluid cannot be used in model and prototype if we are to have similitude with respect to surface tension effects for .1

Fortunately, in many problems involving free-surface flows, both surface tension and viscous effect are small and consequently strict adherence to Weber and Reynolds number similarity is not required.

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Concern would occur if in a model the depths were reduced to the point where surface tension becomes an important factor.

Different horizontal and vertical length scales, which introduce geometric distortion, are often used to eliminate surface tension effects in the model.

How to overcome this problem ?

40

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79

Similitude Based on Governing Differential Equations 1/4

For a steady incompressible two-dimensional flow of a Newtonian fluid with constant viscosity.

The mass conservation equation is

The Navier-Stokes equations are

These equations are nonlinear differential equations for u,v, and ρ, and are difficult to solve for most flow.

Has dimensions of 1/time.

Has dimensions of force/volume

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How to non-dimensionalize these equations ?

t

tp

pp

V

vv

V

uu

yy

xx *

0

*****

2*

*2

22

2

*

*

x

uV

x

u

xx

u

x

uV

x

uu

22

22

y

v

x

v

VV

g

y

p

V

p

y

vv

x

vu

t

v

V

y

u

x

u

Vx

p

V

p

y

uv

x

uu

t

u

V

0y

v

x

u

22

220

22

20

Strouhal number

Euler number

41

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From these equations it follows that if two systems are governed by these equations, then the solutions (in terms of u*,v*,p*,x*,y*, and t*) will be the same if the four parameters are equal for the two systems.

The two systems will be dynamically similar. Of course, boundary and initial conditions expressed in dimensionless form must also be equal for the two systems, and this will require complete geometric similarity.

These are the same similarity requirements that would be determined by a dimensional analysis if the same variables were considered. These variables appear naturally in the equations.

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All the common dimensionless groups that we previously developed by using dimensional analysis appear in the governing equations that describe fluid motion when these equations are expressed in term of dimensionless variables.

The use of governing equations to obtain similarity laws provides an alternative to conventional dimensional analysis.

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FUNDAMENTALS OF FLUID MECHANICS Chapter 7 Dimensional … classes... · 2015-04-28 · 1 1 FUNDAMENTALS OF FLUID MECHANICS Chapter 7 Dimensional Analysis Modeling, and Similitude