ce16 01 - dimensional analysis and similitude
TRANSCRIPT
Fluid Mechanics for Civil Engineering II
CE 16 Lecture Notes
Cornelio Q. Dizon
Institute of Civil Engineering
College of Engineering
University of the Philippines
Dimensional Analysis and Similitude 1
Dimensional
Analysis
& Similitude
Dimensional Analysis
• Definition : Dimensional analysis is a process of formulating fluid mechanics problems in
terms of dimensionless parameters or groups by studying the dimensions
of the quantities or variables
•Why is it used :
• Reduction in variables ( If F(A1, A2, … , An) = 0, then f(ΠΠΠΠ1, ΠΠΠΠ2, … ΠΠΠΠr < n) = 0,
where, F = functional form, Ai = dimensional variables, ΠΠΠΠj = dimensionless parameters, m = number of
important dimensions, n = number of dimensional variables, r= n – m ). Thereby the number of
experiments required to determine f vs. F is reduced.
• Based on the principle that an equation expressing a physical relationship between variables must be
dimensionally homogeneous, the dimensions on each side of the equation must be the same
• Provides qualitative rather than quantitative relationships
Dimensional Analysis and Similitude 2
• Provides qualitative rather than quantitative relationships
•When combined with experimental methods, results in accurate prediction equation
Variables in Fluid Mechanics
Boundary Geometry Fluid Flow Fluid Propertylength (L) time (T) velocity (LT-1) mass density (ML-3)area (L2) acceleration (LT-2) discharge (L3T-1) specific weight (ML-2T-2)volume (L3) force (MLT-2) pressure (ML-1T-2) absolute viscosity (ML-1T-1)
work (ML2T-2) energy (ML2T-2) elasticity (ML-1T-2)power (ML2T-3)
Applications Drag = f(V, L, r, m, c, t, e, T, etc.)
From dimensional analysis,
Dimensional Analysis and Similitude 3
Vortex formed behind a cylinder
Buckingham π Theorem
1. A prediction equation involving n variables in m dimensions may be reduced to a set of (n-m) dimensionless groups
– Typically: all variables required to describe the problem geometry (D) or define fluid properties (ρ, µ) and fluid flow (velocity, acceleration)
2. Express each variables in terms of MLT dimensions (m)
3. Determine the required number of dimensionless parameters (n – m)
Dimensional Analysis and Similitude 4
3. Determine the required number of dimensionless parameters (n – m)
4. Select a number of repeating variables = number of dimensions
– Each should represent boundary geometry, fluid flow and fluid property
– The most common repeating variables are Velocity, Mass Density, and a Length Parameter
5. The repeating variables are combined with each of the other nonrepeating variables and by a proper choice of exponents arrive at the dimensionless products
6. Repeat for each nonrepeating variable
7. Express result as a relationship among the dimensionless parameters
Manipulation of π GroupsOnce identified manipulation of the π groups is permitted. These manipulations do not change the number
of groups involved, but may change their appearance drastically.
Taking the defining equation as: φ ( π1 , π2 , π3 ……… πn-m ) = 0
Then the following manipulations are permitted:
•Any number of groups can be combined by multiplication or division to form a new group which
replaces one of the existing. E.g. π1 and π2 may be combined to form π1a = π1 / π2 so the defining
equation becomes
Dimensional Analysis and Similitude 5
φ ( π1a , π2 , π3 ……… πn-m) = 0
•The reciprocal of any dimensionless group is valid. So φ ( π1 ,1/ π2 , π3 ……… 1/πn-m ) = 0 is valid.
•Any dimensionless group may be raised to any power. So φ ( (π1 )2, (π2 )
1/2, (π3 )3……… πn-m ) = 0 is
valid.
•Any dimensionless group may be multiplied by a constant.
•Any group may be expressed as a function of the other groups, e.g.
π2 = φ ( π1 , π3 ……… πn-m)
In general the defining equation could look like
φ ( π1 , 1/π2 ,( π3 )i……… 0.5πn-m ) = 0
Example 1
• Find: Drag force on rough sphere is function
of D, ρ, µ, V and l. Express in form:
),( 213 πππ f=
FD D ρ µ V l
Dimensional Analysis and Similitude 6
MLT-2 L ML-3 ML-1T-1 LT-1 L
µρ
πρ
µπ
ρµπ
VDor
DV
bbT
acbaL
ccM
MLLTLTMLTLM
VD
cba
cba
===
−=⇒−−=
−=⇒−++−=
−=⇒+=
=
=−−−−
Re
110:
1310:
110:
)()())((
)(
11
3111000
1
n = 6 No. of dimensional parameters
m = 3 No. of dimensions
r = n - m = 3 No. of dimensionless parameters
Select “repeating” variables: D, V, and ρCombine these with nonrepeating variables: F, µ & l
Example 1
FD D ρ µ V l
MLT-2 L ML-3 ML-1T-1 LT-1 L
VDl cba=2 )( ρπ
Select “repeating” variables: D, V, and ρCombine these with nonrepeating variables: F, µ & l
312000
3
110:
)()())((
)(
ccM
MLLTLMLTTLM
VDF
cba
cbaD ρπ
−=⇒+=
=
=−−−
Dimensional Analysis and Similitude 7
D
l
bbT
acbaL
ccM
MLLTLLTLM
VDl
cba
=
=⇒−=
−=⇒−++=
=⇒=
=
=−−
2
31000
2
00:
1310:
00:
)()())((
)(
π
ρπ
223
220:
2310:
110:
DV
F
bbT
acbaL
ccM
D
ρπ =
−=⇒−−=
−=⇒−++=
−=⇒+=
),(22 D
lVDf
DV
FD
µρ
ρ=
Example 2
0),,,,( =µρθ VLF
ρµπ
acbaL
ccM
MLLTLTMLTLM
VL
cba
cba
−=⇒−++−=
−=⇒+=
=
=−−−−
1310:
110:
)()())((
)(
3111000
1
Force F on a body immersed in a flowing fluid depends
on: L, V, ρ, and µ
n = 5 No. of dimensional parameters
m = 3 No. of dimensions
r = n - m = 2 No. of dimensionless parameters
F L V ρ µ
Dimensional Analysis and Similitude 8
µρ
πρ
µπ
VLor
LV
bbT
acbaL
===
−=⇒−−=
−=⇒−++−=
Re
110:
1310:
11Select “repeating” variables: L, V, and ρCombine these with the rest of the variables: F & µ
Reynolds number
F L V ρ µ
MLT-2 L LT-1 ML-3 ML-1T-1
110:
)()())((
)(
312000
2
ccM
MLLTLMLTTLM
VLF
cba
cba
−=⇒+=
=
=−−−
ρπ
F L V ρ µ
MLT-2 L LT-1 ML-3 ML-1T-1
Example 2
Dimensional Analysis and Similitude 9
(Re)
)(
220:
2310:
110:
22
12222
fLV
F
fandVL
F
bbT
acbaL
ccM
=
==
−=⇒−−=
−=⇒−++=
−=⇒+=
ρ
ππρ
π
Dimensionless force is a function
of the Reynolds number
Dimensional Analysis and Similitude 10
Dimensional Analysis and Similitude 11
Example 3
Dimensional Analysis and Similitude 12
Dimensional Analysis and Similitude 13
1 1 22
1 2 .5 12 lo g
3 .7 R e
k D
ff
= − +
Common Dimensionless Numbers
• Reynolds Number (inertial to viscous forces)
– Important in all fluid flow problems
• Froude Number (inertial to gravitational forces)
– Important in problems with a free surface
µρVd
=Re
gy
VFr =
Dimensional Analysis and Similitude 14
• Euler Number (pressure to inertial forces)
– Important in problems with pressure differences
• Mach Number (inertial to elastic forces)
– Important in problems with compressibility effects
• Weber Number (inertial to surface tension forces)
– Important in problems with surface tension effects
gy
c
V
K
VM ==
ρ/
σρ 2LV
W =
2V
pEu ρ
∆=
Similitude: Prototype and Model
Dimensional Analysis and Similitude 15
• Scales: model and prototype (full-scale)
• Selection of the model scale: governed by dimensional analysis and similitude
Physical Models
Dimensional Analysis and Similitude 16
Similitude• Definition : Flow conditions for a model test are completely similar if all significant
dimensionless parameters have the same corresponding values for model and prototype.
� ΠΠΠΠi model = ΠΠΠΠi prototype
� Enables conversion of measured quantities from model to prototype
values
� However, complete similarity usually is not possible. Therefore, often it is
necessary to use Re, or Fr, or M scaling, i.e., select most important ΠΠΠΠ and
accommodate others as best possible
Models in Fluid Mechanics
Dimensional Analysis and Similitude 17
� Mathematical Models
� Physical Models – undistorted, distorted
• Similitude between Model and Prototype
� Geometric Similarity : all body dimensions in all three coordinates have the
same linear-scale ratios. Ratio of linear dimensions
� Kinematic Similarity : homologous (same relative position) particles lie at
homologous points at homologous times. Ratio of velocity components
� Dynamic Similarity : in addition to the requirements for kinematic similarity
the model and prototype forces ratio must be in a constant. Ratio of
homologous forces
Similitude
• Similitude
– Predict prototype behavior from
model results
– Models resemble prototype, but are
• Different size (usually smaller)
and may operate in
Dimensional Analysis and Similitude 18
and may operate in
• Different fluid and under
• Different conditions
– Problem described in terms of
dimensionless parameters which
may apply to the model or the
prototype
– Suppose it describes the prototype
– A similar relationship can be written
for a model of the prototype
)321 ,...,,( nf ππππ =
)321 ,...,,( npppp f ππππ =
)321 ,...,,( nmmmm f ππππ =
Similitude
• If the model is
designed & operated
under conditions that
npnm
pm
pm
ππ
ππ
ππ
=
=
=
...
33
22
Similarity
requirements or
modeling laws
Dimensional Analysis and Similitude 19
• then
npnm ππ =
pm 11 ππ = Dependent variable
for prototype will be
the same as in the
model
Dimensional Analysis and Similitude 20
µρVd
=Re
gy
VFr =
ρ/K
VM =
σρ 2LV
W =
Dimensional Analysis and Similitude 21
2V
pEu ρ
∆=
σ
Dimensional Analysis and Similitude 22
Similitude Ratios for Flow Variables
� Having decided on the Model Law, similitude ratios for
the various flow variables are determined in terms of the
Length Ratio LR
� Velocity Ratio corresponds to the Model Law
� With pressure computations, the Euler criteria is satisfied
Dimensional Analysis and Similitude 23
� With pressure computations, the Euler criteria is satisfied
p
p
m
m
p
m
p
pp
m
mm
pm
VL
LV
LVLV
νν
νν
=
=
= ReReFor example,
using Reynold’s Law
Similitude Ratios Using Froude’s Law
Dimensional Analysis and Similitude 24
Example• Consider predicting the drag on a
thin rectangular plate (L*B) placed
normal to the flow.
• Drag is a function of: B, L, µ, ρ, V
),,,,( BLVfFD µρ=
),(
),(
22
321
µρ
ρ
πππ
VL
B
Lf
VL
F
f
D =
=• Dimensional analysis shows:
• And this applies BOTH to a model
and a prototype
Dimensional Analysis and Similitude 25
),(
),(
22
321
m
mmm
m
m
mmm
Dm
mmm
LV
B
Lf
VL
F
f
µρ
ρ
πππ
=
=
),(
),(
22
321
p
ppp
p
p
ppp
Dp
ppp
LV
B
Lf
VL
F
f
µ
ρ
ρ
πππ
=
=
and a prototype
• We can design a model to predict the
drag on a prototype.
• Model will have:
• And the prototype will have:
Example•Similarity conditions
Geometric similarity
Dynamic similarity
p
p
mm
p
p
m
mpm L
B
BL
B
L
B
L=⇒== 22 ππ Gives us the size of the model
Dimensional Analysis and Similitude 26
p
m
p
m
p
p
mm
p
ppp
m
mmmpm V
L
LV
LVLV
ρ
ρ
µµ
µ
ρ
µρ
ππ =⇒== 33
Dm
m
p
m
p
m
p
Dp
ppp
Dp
mmm
Dmpm F
V
V
L
LF
VL
F
VL
F22
222211
=⇒==
ρ
ρ
ρρππ
Dynamic similarity
Then Gives us the velocity in the model
Example
• Given: Submarine moving below surface in
sea water
(ρ=1015 kg/m3, ν=µ/ρ=1.4x10-6 m2/s).
Model is 1/20-th scale in fresh water (20oC).
• Find: Speed of water in the test for dynamic
similarity and the ratio of drag force on 22
2222=
mmmm
ppp
p
mmm
m
lVF
lV
F
lV
F
ρ
ρρ
Dimensional Analysis and Similitude 27
similarity and the ratio of drag force on
model to that on prototype.
• Solution: Reynolds number is significant
parameter.
smV
sm
VL
LV
LVLV
m
pp
m
m
pm
p
pp
m
mm
pm
/6.28
/24.1
1
1
20
ReRe
=
=
=
=
=
νν
νν504.0
20
1
2
6.28
1015
100022
22
22
=
=
=
p
m
ppp
mmm
p
m
F
F
lV
lV
F
F
ρ
ρ
Sample Problems
Dimensional Analysis and Similitude 28