Chapter 11-Dimensional Analysis2013 (1)

CHE 493 Fluid Mechanics

Chapter 11:

Dimensional analysis

Faculty of Chemical Engineering

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Learning Outcomes: Develop a set of dimensionless variables for a given

flow situation using the Buckingham Pi theorem

Evaluate the results of the dimensional analysis in

the required form

Discuss the primary purposes of dimensional

analysis

State examples of common dimensionless numbers

e.g. Re, Fr, f

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Learning Outcome Faculty of Chemical Engineering


Learning Outcomes: mathematical technique which makes use of the

study of the dimensions for solving several

engineering problems.

helps in determining a systematic arrangement of

the variables in the physical relationship, combining

dimensional variables to form non-dimensional

parameters

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Dimensional Analysis Faculty of Chemical Engineering


Introduction

• Dimension – a measure of a physical quantity (without numerical value)

• Unit – a way to assign a number to that dimension

• Eg: Length is a dimension, measured in units such as microns (µm), feet (ft), centimeters (cm, meters (m), kilometers (km)

Faculty of Chemical Engineering

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7 basic dimensions


• To generate non-dimensional parameters that help in

designing of experiments and in the reporting of

experimental results

• To test the dimensional homogeneity of any equation of

fluid motion.

• To derive rational formulae for a flow phenomenon.

• To derive equations expressed in terms of non-

dimensional parameters to show the relative

significance of each parameter.

• To obtain scaling laws so that prototype performance

can be predicted from model performance

• To predict trends in the relationship between

parameters 6

Purposes of dimensional analysis Faculty of Chemical Engineering


• Rayleigh’s method

• Buckingham’s π-method

• Bridgman’s method

• Matrix-tensor method

• By visual inspection of the variables involved

• Rearrangement of differential equations

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Methods of dimensional analysis Faculty of Chemical Engineering


A quantity that has no units is known as a non-dimensional (or dimensionless) quantity

A dimensionless proportion has the same value regardless of the measurement units used to calculate it.

Examples: relative density, strain and angle measured in radians, Reynolds number, Froude number, Euler number.

[Strain]=[Extension]/[original length]

= [L]/[L]

= [1] 8

Dimensionless quantities Faculty of Chemical Engineering


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Purposes of dimensional analysis Faculty of Chemical Engineering

Dimensionless Group

Name Interpretation

𝜌𝑣𝑙

𝜇

Reynolds number, Re 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑓𝑜𝑟𝑐𝑒

𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑓𝑜𝑟𝑐𝑒

𝑣

𝑔𝑙 Froude number, Fr

𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑓𝑜𝑟𝑐𝑒

𝑔𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑓𝑜𝑟𝑐𝑒

𝑝

𝜌𝑣2 Euler number, Eu 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑓𝑜𝑟𝑐𝑒

𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑓𝑜𝑟𝑐𝑒

𝜌𝑣2

𝐸𝑣

Cauchy number, Ca 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑓𝑜𝑟𝑐𝑒

𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑏𝑖𝑙𝑡𝑦 𝑓𝑜𝑟𝑐𝑒

𝑣

𝑐 Mach number , Ma 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑓𝑜𝑟𝑐𝑒

𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑦 𝑓𝑜𝑟𝑐𝑒

𝜔𝑙

𝑣

Strouhal number, St 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑙𝑜𝑐𝑎𝑙 𝑓𝑜𝑟𝑐𝑒

𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑣𝑒 𝑓𝑜𝑟𝑐𝑒

𝜌𝑣2𝑙

𝜎

Weber number, We 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑓𝑜𝑟𝑐𝑒

𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒


By using primary dimensions, verify that the Archimedes and Grashof number are indeed dimensionless

Ar=(ρsgL3/μ2)(ρs-ρ)

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Examples Faculty of Chemical Engineering


Ar=(ρsgL3/μ2)(ρs-ρ)

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Examples Faculty of Chemical Engineering


• Buckingham π-method is a key theorem in

dimensional analysis

• Provides a way of generating sets of

dimensionless parameters

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Buckingham π-method Faculty of Chemical Engineering


• Theorem states that if we have an equation

involving a certain number, n, of variables, and

these variables are expressible in terms of k,

independent fundamental physical quantities

• To determine number of p (π)

p = n − k

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Determination of π term Faculty of Chemical Engineering


1.List all the variables, n that are

involved in the problem

2. Express each of the variables in terms of basic

dimensions

3. Determine the required number of

pi terms 4. Select a number of

repeating variables. The number required is equal to

the number of reference dimensions

5. Form a pi term by multiplying one of the non repeating variables by the product of the repeating

variables.

6. Check all the resulting pi terms to make sure they are

dimensionless

7. Express the final form as a

relationship among the pi terms

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Steps Faculty of Chemical Engineering

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Buckingham Pi Theorem

Assume the pipe is a smooth pipe. Using Buckingham π (Pi) theorem, determine the dimensionless Pi parameters involved in the problem of determining pressure drop along a straight, horizontal circular pipe. We are

interested in the pressure drop per unit length, Dpl, along the pipe

l

D V r, m

Dpl = (p1-p2)/l

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1.List all the variables, n that are involved in

the problem Relevant flow parameters

Dpl pressure drop, ρ density, V average

velocity, μ viscosity, D pipe diameter.

Therefore, the pressure drop is a function,

Buckingham π-method

Dpl= f(ρ , V, μ, D)

Dimensional Analysis

2.Express each of the variables in terms of basic dimensions

Primary dimensions There are a total of three (3) primary dimensions involved: M, L, and T.

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Dpl MT-2L-3

ρ ML-3

V LT-1

μ ML-1T-1

d L


3.Determine the required number, p of pi terms

Since there are five (n=5) variables, and three required

reference dimensions (k=3:M,L,T), then according to the pi theorem (5-3), two (2) pi terms required

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p = n − k



4.Select a number of repeating variables, m. The number required is equal to the number of reference dimensions

• Must contain jointly all the fundamental dimension.

• Must not form the non-dimensional parameters among themselves.

• Geometric properties (l,d,h), flow property (v,a) and fluid property (ρ,μ).

• Pick simple parameters. Now, select a set of dimensional parameters that

collectively they includes all the primary dimensions. Select m = three since we have three

primary dimensions involved in the problem

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• Do not select dependent variable as one of repeating variable

Thus, we will select ρ, V and d

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P Groups

5.Form a pi term by multiplying one of the nonrepeating variables by the product of the repeating variables.

Set up dimensionless π groups by combining the parameters selected

previously with as Dpl or μ.

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The first group: π1= Dpl r

a Vb Dc, a, b & c exponents are needed to non-dimensionalize the group. In order to be dimensionless:

π1= Dpl Da Vb ρc

(MT-2L-2)(L)a(LT-1)b(ML-3)c = M0L0T0

1+c = 0 (For M) -2+a+b-3c = 0 (For L) -2-b = 0 (For T)

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

Thus, π1 = Dpl D / ρ V

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The process is repeated for the remaining nonrepeating variable, m.

π2= μ Da Vb ρc

(ML-1T-1)(L)a(LT-1)b(ML-3)c = M0L0T0

1 + c = 0 (For M)

-1 + a + b -3c = 0 (For L)

-1-b = 0 (For T)

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

π2= μ/ DVρ

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6. Check all the resulting pi terms to make sure they are dimensionless

7. Express the final form as a relationship among the pi terms

Dpl D/ρV2 = 𝜱(μ/ DVρ)

Dimensional analysis will not provide form of the function 𝜱. Thus, the pi terms can be rearranged, that is, the reciprocal of

μ/DVρ could be used. Thus

π2= DVr/m

The relationship between π1 and π2

Dpl D/ρV2 = 𝜱(μ/ DVρ) 24



EXAMPLE 2

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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. Assume the drag (D) that the fluid exerts on the plate is a function of w,h, the fluid viscosity (μ), density(ρ) and velocity (v) of the fluid approaching the plate. Determine a suitable set of π-terms to study this problem experimentally.

Chapter 11-Dimensional Analysis2013 (1)

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