chapter 11-dimensional analysis2013 (1)
DESCRIPTION
fluid mechanicTRANSCRIPT
CHE 493 Fluid Mechanics
Chapter 11:
Dimensional analysis
Faculty of Chemical Engineering
1
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
2
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
3
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
4
7 basic dimensions
5
• 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
7
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
9
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)
10
Examples Faculty of Chemical Engineering
Ar=(ρsgL3/μ2)(ρs-ρ)
11
Examples Faculty of Chemical Engineering
• Buckingham π-method is a key theorem in
dimensional analysis
• Provides a way of generating sets of
dimensionless parameters
12
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
13
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
14
Steps Faculty of Chemical Engineering
1
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
15
Buckingham π-method Faculty of Chemical Engineering
16
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.
17
Buckingham π-method Faculty of Chemical Engineering
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
18
p = n − k
Buckingham π-method Faculty of Chemical Engineering
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
19
Buckingham π-method Faculty of Chemical Engineering
• Do not select dependent variable as one of repeating variable
Thus, we will select ρ, V and d
20
Buckingham π-method Faculty of Chemical Engineering
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.
21
Buckingham π-method Faculty of Chemical Engineering
22
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
2
Buckingham π-method Faculty of Chemical Engineering
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
23
Buckingham π-method Faculty of Chemical Engineering
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
Buckingham π-method Faculty of Chemical Engineering
EXAMPLE 2
25
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.