Upload
others
View
1
Download
0
Embed Size (px)
Citation preview
1
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.
Fluid Mechanics 5
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).
Fluid Mechanics 6
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:
Fluid Mechanics 8
• 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
Fluid Mechanics 9
• 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.
Fluid Mechanics 10
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.
Fluid Mechanics 12
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.
Fluid Mechanics 13
7.6 Common Dimensionless Groups in Fluid Mechanics
Fluid Mechanics 14
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.
Fluid Mechanics 15
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.
Fluid Mechanics 16
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.
Fluid Mechanics 17
A plot of these two pi terms can be
made with the results shown in figure
below
Assignment
◼ Homework Assignment #7
Fluid Mechanics 18