ch 5 dimensional analysis

Fluid Mechanics I (MDB2013)

Chapter 5Dimensional Analysis, Similitude, and Modeling

Lecturer: Dr Shiferaw R. Jufar

Universiti Teknologi PETRONAS, Bandar Seri Iskandar, 32610 Tronoh, Perak, Malaysia | Tel: +605 368 7045 | Fax: +605 365 5670

e-mail : [email protected]

mailto:[email protected]

1. Understand concepts and behavior of fluids in static

and flowing condition.

2. Understand the concept and applications of control

volume.

3. Apply the knowledge of dimensional analysis.

4. Apply the concepts to the design of simple system

involving fluid.

Course Objectives

Learning objectives

o Apply the knowledge of dimensional analysis in

solving complex engineering problems.

o After completing this chapter, you should be able to:

apply the Buckingham pi theorem.

develop a set of dimensionless variables for a given

flow situation.

discuss the use of dimensionless variables in data

analysis.

apply the concepts of modeling and similitude to

develop prediction equations.

Contents

o Dimensional Analysis

o Units and Dimensions

o Synthesis of Experimental Data

o Buckingham Pi Theorem

o Determination of Pi Terms

o Correlation of Experimental Data

Dimensional Analysis

o Dimensional analysis is a powerful means in the design, the ordering, the performance and the analysis of experiment and also the synthesis of the resulting data.

o The great majority of experiment requires methods of measurement that use numerical scales from both defined units and dimensions.

o Rare exceptions to this are, for example, botany and anatomy where classification can be in terms of graphical descriptions of shape and colour though even here some measure of size is commonly used.

o Measurement is used as a basis of science and engineering and hence of dimensional analysis.

Units and Dimensions, cont’d

“In physical science a first essential step in the direction of

learning any subject, is to find principles of numerical reckoning,

and methods for practicably measuring, some quality connected

with it. I often say that when you can measure what you are

speaking about, and express it in numbers, you know something

about it; but when you cannot measure it, when you cannot

express it in numbers, your knowledge is of a meagre and

unsatisfactory kind: it may be the beginning of knowledge, but you

have scarcely, in your thought, advanced to the stage of science,

whatever the matter may be.”

Kelvin, 1883

Units and Dimensions, cont’d

o Addition of physical quantities is only meaningful when both the dimensions and the units are identical. There is no useful meaning in adding a length to a force; equally, nor is there in adding acres directly to hectares.

o It follows that an equality is under the same restrictions. This principle, though simple, is the foundation of the development of dimensional analysis.

o It is the first stage in the logic of this subject: it is the primary statement as being an acceptable affirmation from it being self evident. Thus it forms the basic premiss for the present work.

Dimensional System

Table 1: Symbols of Dimensions

Dimensional System, cont’d

Table 2: Dimensions of physical quantities

Dimensional System, cont’d

Table 3: Dimensions in the calculus

Where the symbol means “dimensionally equal to”

Buckingham Pi Theorem

o 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.

Determination of the Pi Terms

o Step 1:

o List all variables that are involved in the problem

Expermienter’s knowledge of the problem

The physical laws that govern the phenomenon

Typically the variables will include those that are

necessary to describe the:

• Geometry of the system (D, l)

• Fluid (material) properties (μ, ρ)

• External effects that influence the system (∆Pl)

Determination of the Pi Terms cont’d

o Step 2:

o Express each of the variables in terms of basic

dimensions


o Step 3:

o Determine the required number of pi terms

Buckingham Pi Theorem

Determine number of pi terms is equal to k – r where

k=5 is the number of variables in the problem and r=3

is the number of basic dimensions required to

describe these variables then according to the pi

theorem (5 – 3 = 2) there will be or two pi terms

required


o Step 4:

o Select a number of repeating variables, where the

number required is equal to the number of

reference dimensions

Select from the original list of variables several of

which can be combined with each of the remaining

variables to form a Pi term

The dependent variable should appear in only one Pi

term.

Thus do not choose the dependent variable as one of

the repeating variables, since the repeating variables

will generally appear in more than one pi term.


o Step 5:

o Form a pi term by multiplying one of the

nonrepeating variables by the product of the

repeating variables, each raised to an exponent

that will make the combination dimensionless

The Pi term will be of the form:

(ui )(u1)a(u2)

b(u3)c

Where: ui is one of the nonrepeating variables

u1, u2, and u3 are the repeating variables


o Step 6:

o Repeat Step 5 for each of the remaining

nonrepeating variables

o Step 7:

o Check all the resulting pi terms to make sure they

are dimensionless.

o Step 7:

o Express the final form as a relationship among

the pi terms, and think about what it means


o Example 1:

The steady flow of an incompressible Newtonian fluid

through a long, smooth-walled, horizontal circular pipe.

The pressure drop per unit length, Δpl along the pipe

as illustrated in the figure. Determine a suitable set of

Pi terms to study this problem experimentally.


o Step 1:

List all of the variables

Δpl= pressure drop per unit length

D= pipe diameter

ρ = fluid density

µ = viscosity

V =velocity

o Step 2:

Express all the variables in terms of basic dimensions

Using F, L, and T or M, L, and T as basic dimensions


o Step 3:

Determine the number of Pi terms required which are equal to k – r where k = 5 is the number of variables in the problem and r = 3 is the number of basic dimensions required to describe these variables then according to the pi theorem (5 – 3 = 2) there will be or two Pi terms required.

o Step 4

Select a number of repeating variables, equal to the number of basic dimensions

repeating variables need to be selected from the list Δpl, D, ρ, µ, V. Those are D, V, ρ, because these are dimensionally independent

• D is a length, L

• V involves both length and time, and L and T

• ρ involves force, length, and time L, T and F.


o Step 5:

Form the Pi terms by combining the dependent

variable with the repeating variables.

Since this combination is dimensionless, it follows that:

Since the resulting combination is dimensionless, we

can write:


The solution to the above system of algebraic

equations gives the desired values of a, b and c.

a = 1, b = -2, c = -1

and, therefore:

o Step 6:

The process is now repeated for the remaining

nonrepeating variables. In this example there is only

one additional variable (μ) so that:


and, therefore:

Solving these equations simultaneously it follows that:

a = -1, b = -1, c = -1

So that,


o Step 7:

Check to make sure the Pi terms are actually

dimensionless

or alternatively,

FLT

MLT


o Note that dimensional analysis will not provide the form of

the functional relation between the Pi terms. This can only

be obtained from a suitable set of experiments.

o If desired, the Pi terms can be rearranged; that is,

reciprocal of μ/DVρ could be used and, of course, the

order in which we write the variables can be changed.

Thus, for example, π2 could be expressed as:


o Step 8:

Express the relationship between the Pi terms, i.e. π1 and

π2 as:

o The dimensionless product DVρ/ μ is a very famous

one in fluid mechanics – the Reynolds number.


o Example 2

Flow past a flat plate. See Example 7.1 pp354

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 as shown in Fig. Assume the drag,D, that the fluid

exerts on the plate is a function of

w and h, the fluid viscosity and

density, and , respectively, and the

velocity V of the fluid approaching

the plate. Determine a suitable set

of pi terms to study this problem

experimentally

Number of Pi terms

6 - 3 = 3

Three repeating variables selected are w, V,

and ρ

it would be incorrect to use both w and h as

repeating variables since they have the same

dimensions.

The 1st pi term can be formed by combining D

with the repeating variables such that


Some comments about Dimensional Analysis

o There are also other methods in dimensional

analysis but the method of repeating variables is the

easiest.

o There is not a unique set of Pi terms which arises

from a dimensional analysis. However, the required

number of pi terms is fixed.

o Typically, in fluid mechanics, the required number of

reference dimensions is three, but in some problems

only one or two are required.

Common Dimensionless Groups in Fluid Mechanics

Common Dimensionless Groups in Fluid Mechanics

o Re no. can only be neglected in flow regions away from high-velocity gradients, e.g. away from the solid surface, jets, or wakes.

o Eu no. is only important when the pressure drops low enough to cause vapor formation (cavitation) in a liquid.

o Fr no. is totally unimportant if there is no free surface.

o We no. is important only if it is of order of unity or less, which typically occurs when the surface curvature is comparable in size to the liquid depth, e.g. in droplets, capillary flows, ripple waves, and very small hydraulic models.

Correlation of Experimental Data

o As noted previously, a dimensional analysis cannot

provide a complete answer to any given problem, since

the analysis only provides the dimensionless groups

describing the phenomenon, and not the specific

relationship among the groups.

o To determine this relationship, suitable experimental data

must be obtained.

o As the number of required pi terms increases, it becomes

more difficult to display the results in a convenient

graphical form and to determine a specific empirical

equation that describes the phenomenon.


o Make use of the data given below to obtain a general

relationship between the pressure drop per unit

length and the other variables.


o For problems involving more than two or three Pi

terms, it is often necessary to use a model to predict

specific characteristics

The graphical presentation of data for problems involving three pi terms.


o It may also be possible to determine a suitable empirical

equation relating the three pi terms.

o However, as the number of pi terms continues to

increase, corresponding to an increase in the general

complexity of the problem of interest, both the graphical

presentation and the determination of a suitable empirical

equation become intractable.

o For these more complicated problems, it is often more

feasible to use models to predict specific characteristics

of the system rather than to try to develop general

correlations.

Modeling and Similitude

o A model is a representation of a physical system that

may be used to predict the behavior of the system in

some desired respect.

o The physical system for which the predictions are to be

made is called the prototype.

Modeling and Similitude (Cont’d)

o Model Design Conditions (Similarity Requirements or

Modeling Law)

To achieve similarity between model and prototype

behavior, all the corresponding pi terms must be

equated between model and prototype

3p3m

2p2m

3m2m1m ,,

Geometric Similarity

Dynamic Similarity

Kinematic Similarity


o Example:


Geometric Similarity

A model and prototype are geometrically similar if an only if all body

dimensions in all three coordinates have the same linear-scale ratio.

All angles are preserved in geometric similarity.

All flow directions are preserved.

The orientations of model and prototype with respect to the

surroundings must be identical.


Kinematic Similarity

Velocities are related to the full scale by a constant scale factor. They

also have the same directions as in the full scale.


Dynamic Similarity

Forces are related to full scale by a constant factor. Also requires

geometric and kinematic similarity.


A long structural component of a bridge has an elliptical 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= 0.1 m, H = 0.3 m, and a representative wind velocity is 50 km/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 and the water temperature is Dm = 20 mm

and the water temperature is 20 °C.

Determine the model dimension, Hm,and the velocity at which the test

should be performed. If the shedding frequency for the model is found to

be 49.9 Hz, what is the corresponding frequency for the prototype?

Strouhal number = f (geometric parameter D/H and Reynolds number)

To maintain similarity between model and prototype

From the first similarity requirement

First Similarity Requirement

Second Similarity Requirement

This is a reasonable velocity that could be readily achieved in

a water tunnel. With the two similarity requirements satisfied,

it follows that the Strouhal numbers for prototype and model

will be the same so that

The drag, D, on a sphere located in a pipe through which a fluid is

flowing is to be determined experimentally. Assume that the drag is a

function of the sphere diameter, d, the pipe diameter, D, the fluid velocity,

V, and the fluid density,ρ

(a) What dimensionless parameters would you use for this problem?

(b) Some experiments using water indicate that for d=0.5cm, D=1.3

cm, and V= 0.6m/s, the drag 0.0067N. If possible,

estimate the drag on a sphere located in a 0.6m diameter pipe

through which water is flowing with a velocity of 1.8 m/s. The sphere

diameter is such that geometric similarity is maintained. If it is not

possible, explain why not.

Example 5

Solution in the class room

In terms of F L T

D = F

d = L

D = L

V = LT-1

ρ = F T2L-4

D = f (d, D, V, ρ)

Number of Pi (π) terms

5-3=2 pi terms

Number of repeating variables

d, V, ρ

1st Pi term

π1= D da Vb ρc

(F) (L)a (LT-1)b (FT2L-4)c = Fo LoTo

For (F) 1+c = 0

For (L) a + b - 4c = 0

For (T) -b + 2c = 0

c = -1

-b + 2 (-1) = 0

b = -2

a + b - 4c = 0

a + (-2) – 4(-1) = 0

a = -2

Substitute a b c values in 1st pi term

There for π1= d-2V-2ρ-1

Similarly for 2nd pi term

π2=DdaVbρc

(L) (L)a (LT-1)b (FT2L-4)c = FoLoTo

For (F) c = 0

For (L) 1 + a + b - 4c = 0

For (T) -b + 2c = 0

c = 0

-b + 2 (0) = 0

b = 0

1 + a + b - 4c = 0

1 + a + (0) -4(0) = 0

a = -1

π2=Dd-1V0ρ0

π1 = f (π2)

Continued in the class

Modeling and Similitude (cont.)

Distorted Modelso Models for which one or more of the similarity requirements are not

satisfied are called distorted models.

o Distorted models are rather commonplace, and they can arise for a

variety of reasons, i.e. perhaps a suitable fluid cannot be found for

the model

o Distorted models can be successfully used but the interpretation of

the results obtained with this type of model is obviously more difficult

than the interpretation of results obtained with true models for which

all the requirements are met.

o Models involving high-speed flows are often distorted w.r.t Re number

similarity but Ma number similarity is maintained.

End of Chapter 5

Thank you!Q and A

ch 5 dimensional analysis

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