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An in-depth analysis of the pressure drop in fluid dynamics using dimensional analysis. It explains the concept of Buckingham Pi Theorem, determination of Pi terms, and the uniqueness of Pi terms. The document also includes examples and solutions for different systems.
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Di
i
l A
l
i
Di
i
l A
l
i
Di
mensional Analysis
Dimensional Analysis
Buckingham Pi TheoremBuckingham Pi Theorem
Determination of Pi TermsDetermination of Pi Terms
Comments about Dimensional AnalysisComments about Dimensional Analysis
Common Dimensionless Groups in Fluid MechanicsCommon Dimensionless Groups in Fluid Mechanics
Correlation of Experimental DataCorrelation of Experimental Data
pp
Modeling and SimilitudeModeling and Similitude
Typical Model StudiesTypical Model Studies
Typical Model StudiesTypical Model Studies
Similitude Based on Governing Differential EquationSimilitude Based on Governing Differential Equation
2/42/
2/42/
TheThe first step in the planning of an experiment
first step in the planning of an experiment to study
to study
this problem would be tothis problem would be to decide the factors, or variables
decide the factors, or variables,
that will have an effect on thethat will have an effect on the pressure drop
pressure drop.
Pressure drop per unit lengthPressure drop per unit length
p p
g
p p
Pressure drop per unit length depends on FOUR variables:Pressure drop per unit length depends on FOUR variables:sphere size (D); speed (V); fluid density (
ρ
); fluid viscosity
sphere size (D); speed (V); fluid density (
ρ
); fluid viscosity
sphere size (D); speed (V); fluid density (
ρ
); fluid viscosity
sphere size (D); speed (V); fluid density (
ρ
); fluid viscosity
(m)(m)
3/43/
3/43/
To perform the experiments in a meaningful andTo perform the experiments in a meaningful andsystematic manner, it would be necessary to change thesystematic manner, it would be necessary to change thevariable, such as the velocity, which holding all othervariable, such as the velocity, which holding all otherconstant, and measure the corresponding pressure drop.constant, and measure the corresponding pressure drop.
Difficulty toDifficulty to determine the functional relationship between
determine the functional relationship between
the pressure drop and the various factsthe pressure drop and the various facts that influence it.
that influence it.
4/44/
4/44/
Fortunately, there is aFortunately, there is a much simpler approach
much simpler approach to the
to the
problem that will eliminate the difficulties describedproblem that will eliminate the difficulties describedabove.above.
Collecting these variables into two nonCollecting these variables into two non-
-dimensional
dimensional
gg
combinations of the variablescombinations of the variables (called dimensionless
(called dimensionless
product orproduct or dimensionless groups
dimensionless groups)
Only one dependent and oneOnly one dependent and oneindependent variableindependent variable
Easy to set up experiments toEasy to set up experiments todetermine dependencydetermine dependency
Easy to present results (one graph)Easy to present results (one graph)
2
Easy
to present results (one graph)
Easy to present results (one graph)
0
0
0
3
T L F ) L / F ( L p D
2
1
2
4
2
T
L
F
)
FT
)(
T
FL
(
V
0 0 0 1 2 4
)
L
)(
LT
)(
T
FL
(
VD
0
0
0
2
T
L
F
)
T
FL
(
)
L
)(
LT
)(
T
FL
(
VD
dimensionless product ordimensionless product ordimensionless groupsdimensionless groups
2/52/
2/52/
Given a physical problem in which theGiven a physical problem in which the dependent variable
dependent variable
is a function of kis a function of k-
-1 independent variables
1 independent variables.
)
u
,.....,
u , u ( f u
Mathematically, we can express the functional relationshipMathematically, we can express the functional relationshipin the equivalent formin the equivalent formin the equivalent formin the equivalent form
0
)
u
,.....,
u , u , u ( g
k
)
,
,
,
,
(
g
k
wwhere g is an unspecified function, different from f.
here g is an unspecified function, different from f.
3/53/
3/53/
The Buckingham Pi theorem states that: Given a relationThe Buckingham Pi theorem states that: Given a relationamong k variables of the formamong k variables of the form
ThTh
k
i bl
k
i bl
b
d i t
b
d i t
kk
i d
d
t
i d
d
t
0
)
u
,.....,
u , u , u ( g
k
3
2
1
TheThe k variables
k variables may be grouped into
may be grouped into k
k-
-r independent
r independent
dimensionless productsdimensionless products, or
terms, expressible in
, or
terms, expressible in
f
ti
l f
b
f
ti
l f
b
functional form byfunctional form by
r
k
3
2
1
r
k
3
2
1
5/55/
5/55/
term is not independent if it can be obtained from a
term is not independent if it can be obtained from a
product or quotient of the other dimensionless products ofproduct or quotient of the other dimensionless products ofthe problem. For example, ifthe problem. For example, if
4
/
3
2
3
4 / 3 1 6 3 2
1
5
then neither
then neither
55
nor
nor
66
is independent of the otheris independent of the other
3
3
2
then
neither
then neither
55
nor
nor
66
is independent of the otheris independent of the other
dimensionless products ordimensionless products or dimensionless groups
dimensionless groups.
1/121/
1/121/
Several methods can be used to form the dimensionlessSeveral methods can be used to form the dimensionlessproducts, or pi term, that arise in a dimensional analysis.products, or pi term, that arise in a dimensional analysis.
Regardless of the method to be used to determine theRegardless of the method to be used to determine thedimensionless products,dimensionless products, one begins by listing
one begins by listing all
all
pp
g
y
g
g
y
g
((dimensional
dimensional)
) variables that are known (or believed) to
variables that are known (or believed) to
affect the given flow phenomenonaffect the given flow phenomenon.
Eight steps listedEight steps listed
below outline a recommendedbelow outline a recommended
procedure for determining the
terms.
procedure for determining the
terms.
procedure for determining the
terms.
procedure for determining the
terms.
3/123/
3/123/
22
Let k beLet k be the number of variables
the number of variables.
Example: For pressure drop per unit length, k=5. (AllExample: For pressure drop per unit length, k=5. (All variables arevariables are
pp
andand V )
variables arevariables are
pp
, and, and V )
4/124/
4/124/
Select a set of fundamental (primary) dimensionsSelect a set of fundamental (primary) dimensions
Select a set of fundamental (primary) dimensions.Select a set of fundamental (primary) dimensions.
For example: MLT, or FLT.For example: MLT, or FLT.
Example: For pressure drop per unit length , we chooseExample: For pressure drop per unit length , we choose FLT.FLT.
1
2
2
4
3
p
r=3r=
6/126/
6/126/
Select a set ofSelect a set of r dimensional variables
r dimensional variables that
that includes all
includes all
the primary dimensionsthe primary dimensions (repeating
(repeating variables).
variables).
These repeating variables will all be combined withThese repeating variables will all be combined with each of the remaining parameters.each of the remaining parameters.
Example: For pressure drop per unit length ( r = 3)Example: For pressure drop per unit length ( r = 3) selectselect
ρρ
1
2
2
4
3
T
FL
L
D
FL
p
1
2
LT
V
T
FL
7/127/
7/127/
11
Set up dimensional equations,Set up dimensional equations, combining the variables
combining the variables
selected in Step 4selected in Step 4 with
with each of the other variables
each of the other variables
(nonrepeating variables) in turn, to form dimensionless(nonrepeating variables) in turn, to form dimensionlessgroups or dimensionless product.groups or dimensionless product.
There will be kThere will be k –
r equations.
Example: For pressure drop per unit lengthExample: For pressure drop per unit length
