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CIVE3077 - Fluid Mechanics

Chapter 7: Dimensional Analysis

Instructor: Assoc. Prof. Dr. Ho Viet Hung

Homepage: http://hungtlu.wordpress.com

Learning Objectives

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.

Fluid Mechanics 2

7.1 Dimensional Homogeneity

◼ All theoretically derived equations are dimensionally homogeneous—that

is, the dimensions of the left side of the equation must be the same as those

on the right side.

◼ The three basic dimensions: L, T, and M are required. Alternatively, L, T, and

F could be used, where F is the basic dimensions of force.

◼ We accept that all equations describing physical phenomena must be

dimensionally homogeneous.

Fluid Mechanics 3

2

( )

( )

Mass ( )

( ),

Length L m

Time T s

M kg

Force F N F MLT −

Dimensions Associated with Common Physical Quantities

Fluid Mechanics 4

7.2 Buckingham Pi Theorem

◼ Buckingham pi theorem (Basic theorem of dimensional analysis): 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.

◼ Equation involving k variables:

◼ We can rearrange the equation into a set of dimensionless products (pi

terms): Π1 = Φ(Π2, Π3,…, Πk-r).

◼ Usually, the reference dimensions required to describe the variables will be

the basic dimensions M, L, T or F, L, and T.

◼ in some instances perhaps only two dimensions, such as L and T, are

required, or maybe just one, such as L.

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1 2 3( , ,..., )ku f u u u=

7.3 Determination of Pi Terms

Method of repeating variables:

◼ Step 1: List all the variables that are involved in the problem.

The variables are necessary to describe the geometry of the system, to define

any fluid properties, and to indicate external effects that influence the system.

◼ Step 2: Express each of the variables in terms of basic dimensions (M, L, T

or F, L, T).

◼ Step 3: Determine the required number of pi terms (using Buckingham Pi

theorem).

◼ Step 4: Select a number of repeating variables, where the number required

is equal to the number of reference dimensions (usually the same as the

number of basic dimensions).

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Method of repeating variables

◼ Step 5: Form a pi term by multiplying one of the nonrepeating variables by

the product of repeating variables, each raised to an exponent that will make

the combination dimensionless.

Each pi term will be of the form 𝑢𝑖𝑢1𝑎𝑖𝑢2

𝑏𝑖𝑢3𝑐𝑖 where 𝑢𝑖 is one of the

nonrepeating variables; 𝑢1, 𝑢2, 𝑢3 are the repeating variables.

◼ Step 6: Repeat Step 5 for each of the remaining nonrepeating variables.

◼ Step 7: Check all the resulting pi terms to make sure they are dimensionless

and independent.

◼ Step 8: Express the final form as a relationship among the pi terms and think

about what it means.

Fluid Mechanics 7

Example 1 (7.1)

◼ 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 and h, the fluid viscosity

μ and density ρ, and the velocity V of the fluid approaching the plate.

◼ From the statement of the problem we can write the relation

◼ The dimensions of these 6 variables are:

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• We will select 3 repeating variables such as w, V, and ρ, which are dimensionally

independent.

• The first pi term can be formed by combining D with the repeating variables

→

to define the six variables so that three pi terms will be needed (k – r = 6 – 3 = 3).• All three basic dimensions are required

◼ For Π1 to be dimensionless it follows that

◼ For M: 1 + c = 0

◼ For L: 1 + a + b - 3c = 0

◼ For T: -2 – b = 0

◼ Therefore a = -2; b = -2; c = -1

◼ The last nonrepeating variable is μ so that

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• The second pi term Π2 for nonrepeating

variable, h:

• We should check to make sure they are dimensionless.

• We can express the results of the dimensional

analysis in the form:

Example 2

◼ At low velocities (laminar flow), the volume flowrate Q through a small-bore tube

is a function only of the tube radius R, the fluid viscosity μ, and the pressure drop

per unit tube length dp/dx. Using the pi theorem, find an appropriate

dimensionless relationship.

◼ Write the given relation and count variables: , 4 variables (k=4)

◼ Make a list of the dimensions of these variables

◼ By trial and error we determine that R, μ, and dp/dx cannot be combined into a pi

term. Then r = 3, and k – r = 4 - 3 = 1. There is only one pi term.

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There are three basic

dimensions (M, L, T)

Equate exponents:

◼ Mass: b + c = 0

◼ Length: a - b - 2c + 3 = 0

◼ Time: -b - 2c - 1 = 0

◼ Solving simultaneously, we obtain a = -4, b = 1, c = -1. Then

or

◼ Since there is only one pi group, it must equal a dimensionless constant. This is

as far as dimensional analysis can take us.

Fluid Mechanics 11

4dp RQ C

dx =

Selection of Variables

◼ Clearly define the problem. What is the main variable of interest (the

dependent variable).

◼ Consider the basic laws that govern the phenomenon. Even a crude theory

that describes the essential aspects of the system may be helpful.

◼ Start the variable selection process by grouping the variables into three

broad classes: geometry, material properties, and external effects.

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Selection of Variables

◼ Consider other variables that may not fall into one of the above categories.

For example, time will be an important variable if any of the variables are

time dependent.

◼ Be sure to include all quantities that enter the problem even though some of

them may be held constant. For a dimensional analysis it is the dimensions

of the quantities that are important—not specific values!

◼ Make sure that all variables are independent. Look for relationships among

subsets of the variables.

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7.6 Common Dimensionless Groups in Fluid Mechanics

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Example 3

◼ The relationship between the pressure drop per unit

length, Δpl, along a smooth horizontal pipe and the

variables (the pipe diameter, D, fluid density, ρ, fluid

viscosity, μ, and the velocity, V) is to be determined

experimentally. Find a functional relationship between the

pressure drop per unit length and the other variables.

◼ We will assume that the pressure drop per unit length,

Δpl, is a function of the pipe diameter, D, fluid density, ρ,

fluid viscosity, μ, and the velocity, V.

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Example 3

◼ The relationship between the pressure drop per unit length, Δpl, along a smooth

horizontal pipe and the variables (the pipe diameter, D, fluid density, ρ, fluid viscosity, μ,

and the velocity, V) is to be determined experimentally. Find a functional relationship

between the pressure drop per unit length and the other variables.

◼ We will assume that the pressure drop per unit length, Δpl, is a function of the pipe

diameter, D, fluid density, ρ, fluid viscosity, μ, and the velocity, V.

◼ ∆𝑝 = 𝑓(𝐷, 𝜌, 𝑉, 𝜇) → k = 5

◼ List dimension of each variables: → r = 3.

◼ There are 2 pi terms (k – r = 2).

◼ We will select 3 repeating variables such as D, V, and ρ.

◼ The first pi term can be formed by combining Δp with the repeating variables

◼ Π1 = ∆𝑝𝐷𝑎𝑉𝑏𝜌𝑐 → a = 1; b = -2; c = -1. So that: Π1 = ∆𝑝𝐷/𝑉2𝜌

◼ The second pi term: Π2 = (1

μ)𝐷𝑎𝑉𝑏𝜌𝑐 → a = 1; b = 1; c = 1; and Π2 =

𝐷𝑉𝜌

μ= 𝑅𝑒.

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Example 3 (continued)

◼ Application of the pi theorem yields two pi

terms

◼ Π1 =𝐷∆𝑝

𝜌𝑉2 and Π2 =𝐷𝑉𝜌

μ= 𝑅𝑒

◼ Hence𝐷∆𝑝

𝜌𝑉2 = Φ(𝑅𝑒)

◼ To determine the form of the relationship,

we need to vary the Reynolds number, Re,

and to measure the corresponding values

of 𝐷∆𝑝

𝜌𝑉2 . These are dimensionless groups so

that their values are independent of the

system of units used.

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A plot of these two pi terms can be

made with the results shown in figure

below

Assignment

◼ Homework Assignment #7

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