371 lec01 bucking ham pi theorem(1)

8/3/2019 371 Lec01 Bucking Ham Pi Theorem(1)

http://slidepdf.com/reader/full/371-lec01-bucking-ham-pi-theorem1 1/15

MECH 371

Dimensional Analysis & Similarity

MECH 371

Dimensional Analysis and Similarity

Overview

The control volume and differential approach are the analytic tools used by engineers to solve

flow problems. Unfortunately, purely analytic methods are limited:

• lack of complete information (turbulence)

• difficulty of math and computations required.

100 years of research has yet to yield a complete theory for turbulent flow in a pipe.

• the variables are known but not their relationship



MECH 371

2 1f

Turbulent flow in a pipe. Investigate the head loss.

p pgh

ρ

−∼


where ~ is an order of ma

( )f

gnitude estimation

We can write L, D, V, , , where is a measure of the wall roughness.

Finding the function is the objective of the experiment. need not be an analytical function.

Ta

gh F

F F

ρ μ ε ε =

6

bles, charts, and curve fits are perfectly acceptable.

If we varied each parameter for 10 data points we require 10 experiments including experiments

with fluid of different and .

Also, anyone usin

ρ μ

g your data would have to make a 6 parameter interpolation. To reduce the

number of parameters, we take advantage of . Dimensional Analysis

MECH 371

:

• a packaging or compacting technique used to reduce the complexity of experimental

ro rams and increase the eneralit ofthe results

Dimensional Analysis


.

Turbulent p f

2

L VDipe flow , ,

1 D DV

2

• helps our way of interpreting and planning an experiment or theory. It suggests ways of

writing equations.

ghF

ρ ε

μ

⎛ ⎞⇒ = ⎜ ⎟

⎝ ⎠

• gives a great deal of insight into the form of physical relationships.

• provides scaling laws which can convert data from models to prototypes.



MECH 371

To use dimensional analysis, we must make use of the (PDH).

PDH: Any equation that completely describes a physical phenomenon must be valid regardless

principle of dimensional homogeneity


of the units of measurements.

• Dimensions of all additive terms must be the same.

• Not affected by derivatives and integrals.

Example

1 1 2 21 2

V VThe Bernoulli's equation is dimensiona

2 2

p pg z g z

ρ ρ + + = + + lly homogeneous.

MECH 371

( )

Example

However, the Manning Equation for open channel flow (English units)

is dimensionally homogeneous semi-empirical :not


2 1

3 2n n

1.49V = R S , R = radi

nus (ft)

S = slope

= number

n

V = velocity (ft/sec)

• The Manning Equation is valid only for English units.

Any dimensionally homogeneous equation or process can be written in nondimensional form.

Nondimensionalization of experimental data or equation requires the use of the

. Buckingham Pi Theorem



MECH 371

Buckingham Pi(Π) Theorem:


# of dimensionless parameters (Π) = # of important parameters involved - # of dimensions involved.

For dimensional analysis there are in general four dimensions used:

M , L , Ω, T or F , L , Ω, T where Ω = time

For most of this course we neglect heat transfer so we'll usually use only three:

(M , L , Ω) or (F , L , Ω)

MECH 371

Steps in Applying the Buckingham Pi Theorem

Pipe Flow Example: (more detail)


( )f

f

L, D, V, , ,

where represents

gh F

gh

ρ μ ε =

mechanical energy loss

f

. ta n mportant parameters o t e pro em

i) flow properties: V,

ii) geometry: L, D,

gh

ε

iii) fluid properties: , ρ μ



MECH 371


2. Determine the number of dimensionless parameters you need to construct = 7

*

N

=

group among themselves.

i M s usually equal to the number of fundamental dimensions involved in the

dimensional parameters, . It is never greater than . We have M , L , T so = 3.

To find initially, assume a

K K K

M M = K nd try to find a set of dimensional parameters

that can't form a group. If then stop.

M

M = K Π

If reduce by 1 and try again. M K M ≠

MECH 371


[ ]

[ ]

Hint: Try to find parameters whose dimensions are as close to "pure" as possible.

L = L

D = L pure L

-1

-3

-1

V = L closest to pure

= ML closest to pure M

= ML

ρ

μ

Ω Ω⎣ ⎦

⎡ ⎤⎣ ⎦

Ω

[ ]

-1

= L

We can choose D , V , and since we cannot form a with these.

ε

ρ

⎡ ⎤⎣ ⎦

Π

= 7 3 = 4 groups

Rare cases when are when certain dimensions (usually mass and time) oc

M = 3 N - M

M < K

∴ ⇒ − Π

2

cur only in

Mfixed combinations. For example, in stress analysis:

⎡ ⎤⎢ ⎥Ω⎣ ⎦



MECH 371


* 3. Nondimensionalization of the parameters.

Rules ranked in order of importance:

• Donot usethe de endent arameter hf .

• Do not use a parameter that may be important only part of the time ,μ ( )

• Use parameters whose dimensions are as close to "pure" as possible.

Important: Select as repeating variables of the dimensional parameters with all of the

fundamental dimensions. , V, D

Tak

M

ε

ρ

( )f e the remaining parameters and combine them with repeating ones , L, , ,gh μ ε

( )

a b c

1 f

to form the (4) groups.

Let's start with = V D .gh ρ

Π

Π

MECH 371


1Find a , b , and c so is dimensionless:Π

[ ] [ ]a b c2 -2 -3 -1

1 L ML L L

⎡ ⎤ ⎡ ⎤ ⎡ ⎤Π = Ω Ω⎣ ⎦ ⎣ ⎦ ⎣ ⎦

[ ] [ ]

[ ] [ ] [ ]

2-3a+b+c -2-ba

o o o

= M L

= M L

=

⎡ ⎤ Ω⎣ ⎦

Ω

2 + b = 0 a

2 3a + b + c = 0

⎪⎬⎪− ⎭

= 0 , b = 2 , c = 0−



MECH 371


f 1 2

3. continued

gh∴ Π =

( ) [ ] [ ] [ ]1-3a+b+c -ba b c a

2 L V D = M L

a = 0

b = 0

1 3a + b + c + 0 a = 0 , b = 0 , c = 1

ρ Π = Ω

− ∴ −

2

L

DΠ =

MECH 371


( ) a b c

3 V D 1 + a = 0

1 3a + b + c = 0

1 b = 0

μ ρ Π =

− −

− −

( )

3

a b c

4

a = 1 , b = 1 , c = 1

VD

V DD

So we can write

μ

ρ

ε ε ρ

− − −

∴ Π =

Π = =

( )1 2 3 4

L

2

, ,

orV

F

ghF

Π = Π Π Π

=L

, ,D VD D

μ ε

ρ

⎛ ⎞⎜ ⎟⎝ ⎠



MECH 371


4. Rearrange the groups to correspond with customary usage. Use traditional types of Π

.

Rearrangement: We are allowed to

• multiply s by a constant

• raise the

Π

Π

1

s to any positive or negative power

• multiply any power of a with any power of another .

VD L VDh h h ε −

Π Π

⎛ ⎞ ⎛ ⎞2

2 2

2 , so we can write , ,1 1V VD D D

V V2 2

G ρ μ μ

= = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

MECH 371


Physical Significance of Dimensionless Parameters

Standard rou s can be inter reted as a ratio of t ical values of two h sical entitiesΠ

(force or energy).

Imagine a moving fluid particle under the influence of an inertia force ( ) along a streamline

Estimate the inertia force

ma

3dV L

~⎛ ⎞ ⎡ ⎤

2 d

where ~ is an order of magnitude estimation

t Ω⎝ ⎠ ⎣ ⎦



MECH 371


2

since V, L, are characteristic velocity, length, and time, we can write

V V V

Ω

23 2 2

aL L

V

Vma L V L

L

Estima

ρ ρ

=Ω ⎛ ⎞

⎜ ⎟⎝ ⎠

∴ =

∼ ∼

∼

te the viscous shear force

viscous shear force shear stress area

ushear stress = =

y

τ μ

×

⎛ ⎞∂⎜ ⎟

∂⎝ ⎠

∼

MECH 371


Assuming the velocity changes from 0 to the characteristic value V over the length L

u u V 0 V∂ Δ −

y y L L

Vthen the viscous force is L

Lμ

=∂ Δ

⎛ ⎞⎜ ⎟⎝ ⎠

∼

∼2

2 2

= VL

Note then that the ratio of

inertia force V L

μ

ρ ∼

v scous orce

VL= Re the Reynolds number

ρ

μ



MECH 371


2

2

2 2 2

Pressure force pressure difference area = L

pressure force L= = coefficient of pressure

1 1inertia force p

p

p pC

= × Δ

Δ Δ=∼

3

2 2

gravity force L g ρ ∼

2 2 2

3

inertia force V L V= Froude number squared

gravity force L Lg g

ρ

ρ ∴ =∼

L

L VStrouhal number = =

1V

ω

ω

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠

char. flow time

char. oscillation time

∼

MECH 371


Standard Dimensionless Parameters

Two generic problems:

Internal flow External flow

The functional relation for these problems:

Dimensional arameter of interest = size sha e fluid velocit fluid ro ertiesF dimensional constants)

In general, the type of parameters on the right hand side changes very little from

problem to problem.



MECH 371


On the left hand side we are usually interested in

External flow: force on the body by the fluid (total force, lift force or drag) or

osc at on ).

Internal flow: wall fo

ω

f

rce (shear stress), pressure drop, mechanical energy loss,

or oscillations ( ).

ˆFor our example we will use drag (D) and mechanical energy loss ( ), respectively.

ˆ

gh

ω

= V w

f w

w

, , , , , , , , , ,

= (L, V, , , a, , , , , , T )

a speed of sound, surface tension, T wall temperatu

p v

p vgh F c c g ρ μ σ ε

σ ≡ ≡ ≡( )re

MECH 371


Using the drag equation we put it in fundamental dimensional form:

ML L M M L= L , , , , ,F

⎛ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

2

2

L L

M L,

Ω Ω Ω Ω⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎝

⎡ ⎤⎢ ⎥Ω Ω⎣ ⎦

[ ] [ ]2

2 2 2

L L, , , L , T

T T

The drag equation has N = 12 dimensional paramenters and = 4 fundamental

dimensions M, L, , T .

K

⎞⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥Ω Ω⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎠

Ω

nce , , , an canno orm a , = an pc we use em as e

repeating variables.

# of Pi groups = = 12 4 = 8

N - K −



MECH 371


2 2

w

Perform dimensional analysis on the original equation to get

D̂ VL V V L V T=

pc ρ ρ ε ⎛ ⎞

2 2 0

, , , , , ,1 a L L T

V L2

Each of the groups on the right hand side is a "stand

vg cμ σ ρ ⎝ ⎠

ard" dimensionless parameter in

fluid mechanics.

The standard dimensionless parameter are used as tools to classify flow as laminar or

turbulent, compressible or incompressible, etc.

Dimensionless parameters and their significance are given by the following table.

MECH 371




MECH 371 Notes

Questions?

MECH 371

Notes

See you next time.

371 lec01 bucking ham pi theorem(1)

Documents