Dimensional Analysis 2nd Mech April 8 2015 2

ALEXANDRIA UNIVERSITY

FACULTY OF ENGINEERING

Mechanical Engineering Department

Hydraulic Machines and Fluid Mechanics Branch

اإلسكندریةجامعة كلیة الھندسة

ھندسة المیكانیكیةالقسم اآلالت الھیدرولیكیة ومیكانیكا الموائع شعبة

------------------------------------------------------------------------------------------------------------------------------------------------------- Dr. Ihab G. Adam

1

Similarity and Dimensional Analysis

Review: Dimension: Measure of a physical quantity, e.g., length (L), time (t), mass (M) Units: Assignment of a number to a dimension, e.g., (m), (sec), (kg) 7 Primary Dimensions: Mass M (kg) Length L (m) Time t “T” (sec) Temperature T (K) Current I (A) Amount of Light C (cd) Amount of matter N (mol) All non-primary dimensions can be formed by a combination of the 7 primary dimensions Examples

{Velocity} = {Length/Time} = {L/T} {Force} = {Mass Length/Time2} = {ML/T2}

Law of dimensional homogeneity (DH):

Every additive term in an equation must have the same dimensions Example: Bernoulli equation:

{p} = {force/area}={mass x length/time x 1/length2} = {M/(T2L)} {1/2 V2} = {mass/length3 x (length/time)2} = {M/(T2L)} {gz} = {mass/length3 x length/time2 x length} ={M/(T2L)}







------------------------------------------------------------------------------------------------------------------------------------------------------- Dr. Ihab G. Adam

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Non-dimensionalization of Equations: Given the law of DH, if we divide each term in the equation by a collection of variables and constants that have the same dimensions, the equation is rendered non-dimensional In the process of non-dimensionalizing an equation, non-dimensional parameters often appear, e.g., Reynolds number and Froude number Example: Bernoulli equation: To nondimensionalize, for example, the Bernoulli equation, the first step is to list primary dimensions of all dimensional variables and constants {p} = {M/(T2L)} {} = {M/L3} {V} = {L/T} {g} = {L/T2} {z} = {L} Next, we need to select Scaling Parameters. For this example, select L, U0, 0 By dividing each term in above B.E. by 0U0

2 which has the same dimensions as the pressure, dimensionless B.E. can be expressed as:

Where:







------------------------------------------------------------------------------------------------------------------------------------------------------- Dr. Ihab G. Adam

3

The laboratory model and the full-size Prototype must be completely similar.

Types of Similarity: 1. Geometric Similarity 2. Kinematic Similarity 3. Dynamic Similarity

Geometric Similarity: It is a similarity of shape which includes the solid (flow field) boundary and model dimensions

rm

P

m

P

m

P lb

b

d

d

l

l ........

where lr is the scale ratio or factor.

Also, 2r

m

P lArea

Area and 3

rm

P lVolume

Volume

Example: Flow in a pipeline

rm

P

m

P ld

d

l

l

m

P

should be equal to lr , but is not easy to make it equal lr

Kinematic Similarity: It is a Similarity of motion To have a kinematic similarity, we must have geometric similarity and dynamic similarity.

rm

P vv

v at all corresponding points ( similarly located points)

where vr is the velocity ratio.

Also, r

rr

m

P lt

t

t

v

r

rr

m

P

la

a

a v2

2v rrrm

P lQQ

Q

Dynamic Similarity: It is a Similarity of forces To have a dynamic similarity, we must have geometric similarity and kinematic similarity.

rm

P fF

F or

2

1

2

1

m

m

P

P

F

F

F

F , where fr is the force ratio.







------------------------------------------------------------------------------------------------------------------------------------------------------- Dr. Ihab G. Adam

4

Type of Forces:

1. Inertia force v22lmaFI

2. Gravity (Body) force gmgF lG3

3. Pressure force lPPAFP2

4. Viscous force ll

Ady

duF lV v

v 2

5. Surface Tension lFT (Liquid-gas interface)

6. Elastic force lEEAFE2

rFF

F

F

F

G

G

I

I

m

P

m

P ..........

V

I

V

I

m

m

P

P

F

F

F

F ,

G

I

G

I

m

m

P

P

F

F

F

F ,

P

I

P

I

m

m

P

P

F

F

F

F ,

E

I

E

I

m

m

P

P

F

F

F

F ,…...

Dimensionless Numbers:

1. Reynolds Number l

F

FR

V

IN

v

2. Froude’s Number glF

FFr

G

IN

v2

gl

FrN

v

3. Euler’s Number 2

22v

pl

l

F

FEu

P

IN

p

EuN

2

v

4. Mach’s Number 2

22v

El

l

F

FM

E

I

a

vv

E

M

5. Weber’s Number l

l

F

FWe

T

IN

v22

l

M

v







------------------------------------------------------------------------------------------------------------------------------------------------------- Dr. Ihab G. Adam

5

Dimensional Analysis

1. It reduces the number of variables by grouping these variables in dimensionless groups (Combinations).

2. By putting the equations in dimensionless form, it eliminates any dimensional mistakes.

3. It is helpful in experimental work, by making the results as widely applicable as possible (generalization of the experimental data).

4. It helps to find the relationship between the prototype and the model. In this section, we will consider the turbulent flow through pipe as an example:

),,v,,( dfl

p

gd

lf

w

phl 2

v2

, where f is f (ReN and ε/d)

π-Theorem:

1. List all variables involved in the problem

Dependent variable: l

p and ,,v,, d

2. Express each variable in terms of the basic dimensions (M, L, T)

2233 , , , , v,

TL

M

L

F

l

p

LT

MLLd

T

L

L

M

3. Select a number of repeating variables equals to the number of reference dimensions (number of basic dimensions appeared in step (2) ≤ 3 (M, L, T))

a. Do not choose the dependent variable. b. All reference dimensions should be represented c. Each repeating variable must be dimensionally independent of the others (the

dimensions of any repeating variable can not be reproduced by any combinations of the dimensions of the others)

d. Each repeating variables should represents one of the following: i. The geometry (d ,l , ……) ii. The fluid property ( ,, ……) iii. The fluid motion ( v, Q, F, p, Power,……)

d, v, ρ







------------------------------------------------------------------------------------------------------------------------------------------------------- Dr. Ihab G. Adam

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4. Determine the required number of π-groups = Total number of variables – number of reference dimensions (repeating variables) = number of the non-repeating variables)

=6 -3 = 3

5. Form the dimensionless π-groups for the non-repeating variables By multiplying each non-repeating variables by the product of the repeating variables, each raised to an exponent that will make the group dimensionless.

21

000223

1

v

2 ,1 ,1

02

01

023

v

l

pd

cba

cT

bM

cbaL

TLMTL

M

T

L

L

ML

l

pd

cba

cba

N

cba

cba

d

cba

cT

bM

cbaL

TLMLT

M

T

L

L

ML

d

Re

1

v

1 ,1 ,1

01

01

013

v

2

0003

2







------------------------------------------------------------------------------------------------------------------------------------------------------- Dr. Ihab G. Adam

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d

cba

cT

bM

cbaL

TLMLT

L

L

ML

dcb

a

cba

3

0003

3

0 ,0 ,1

0

0

013

v

6. Express the dependent group as a function of the other groups

dff

gd

lfh

gd

l

dfh

df

dl

p

df

lpd

f

N

LNL

N

N

,Re hence;

2

v similar to is which ,

v,Re

,Rev

,Rev

/

,

22

2

2

321

Dimensional Analysis 2nd Mech April 8 2015 2

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