This preview shows page 1. Sign up to view the full content.
Unformatted text preview: V. Modeling, Similarity, and Dimensional Analysis
To this point, we have concentrated on analytical methods of solution for fluids
problems.
However, analytical methods are not always satisfactory due to:
(1) limitations due to simplifications required in the analysis,
(2) complexity and/or expense of a detailed analysis.
The most common alternative is to:
Use experimental test & verification procedures.
However, without planning and organization, experimental procedures can :
(a) be time consuming,
(b) lack direction,
(c) be expensive.
This is particularly true when the test program necessitates testing at one set of
conditions, geometry, and fluid with the objective to represent a different but
similar set of conditions, geometry, and fluid.
Dimensional analysis provides a procedure that will typically reduce both the time
and expense of experimental work necessary to experimentally represent a desired
set of conditions and geometry.
It also provides a means of "normalizing" the final results for a range of test
conditions. A normalized (nondimensional) set of results for one test condition
can be used to predict the performance at different but fluid dynamically similar
conditions ( including even a different fluid).
The basic procedure for dimensional analysis can be summarized as follows:  v v V1  eText Main Menu  Textbook Table of Contents  Study Guide 1. Compile a list of relevant variables (dependent & independent) for the
problem being considered,
2. Use an appropriate procedure to identify both the number and form of
the resulting nondimensional parameters.
This procedure is outlined as follows for the Buckingham Pi Theorem
Definitions:
n = the number of independent variables relevant to the problem
j’ = the number of independent dimensions found in the n variables
j = the reduction possible in the number of variables necessary to be
considered simultaneously
k = the number of independent pi terms that can be identified to describe the
problem, k = n  j
Summary of Steps:
1. List and count the n variables involved in the problem.
2. List the dimensions of each variable using {MLTΘ} or {FLTΘ}. Count the
number of basic dimensions ( j’) for the list of variables being considered.
3. Find j by initially assuming j = j’ and look for j repeating variables which do
not form a pi product. If not successful, reduce j by 1 and repeat the process.
4. Select j scaling, repeating variables which do not form a pi product.
5. Form a pi term by adding one additional variable and form a power product.
Algebraically find the values of the exponents which make the product
dimensionless. Repeat the process with each of the remaining variables.
6. Write the combination of dimensionless pi terms in functional form: Πk = f( Π1, Π2, …Πi) Consider the following example for viscous pipe flow. The relevant variables for
this problem are summarized as follows: ∆P = pressure drop
µ = viscosity ρ = density
ε = roughness V = velocity
L = length {∆ P Seven pipe flow variables: dependent D = diameter ρ, V, D, µ, ε, L }
independent  v v V2  eText Main Menu  Textbook Table of Contents  Study Guide Use of the Buckingham Pi Theorem proceeds as follows:
1. Number of independent variables: n = 7
2. List the dimensions of each variable ( use m L t Θ ): ∆P variables ρ 1 2 V
3 dimensions mL t mL Lt 1 D µ L ε mL t 1 1 L L L The number of basic dimensions is j’ = 3.
3. Choose j = 3 with the repeating variables being ρ, V, and D. They do
not form a dimensionless pi term. No combination of the 3 variables will
eliminate the mass dimension in density or the time dimension in velocity.
4. This step described in the above step. The repeating variables again are ρ,
V, and D and j = 3. Therefore, k = n – j = 7 – 3 = 4 independent Π
terms.
5. Form the Π terms: Π1 = ρa Vb Dc µ−1 = (mL3)a ( Lt1)b Lc ( mL1t1 )−1 In order for the Π term to have no net dimensions, the sum of the
exponents for each dimension must be zero. Therefore, we have:
mass:
time:
length: a  1 =0, a= 1
 b + 1 = 0, b = 1
3a + b + c + 1 = 0, c = 3 – 1 – 1 = 1 We therefore have Π1 = ρ V D /µ = Re = Reynolds number Repeating the process by adding the roughness ε  v v V3  eText Main Menu  Textbook Table of Contents  Study Guide Π2 = ρa Vb Dc ε1 = (mL3)a ( Lt1)b Lc ( L )1
Solving:
mass:
time:
Length: Π2 = ε / D a =0, a= 0
 b = 0, b = 0
3a + b + c + 1 = 0, c = – 1
Roughness ratio Repeat the process by adding the length L. Π3 = ρa Vb Dc L1 = (mL3)a ( Lt1)b Lc ( L )1
Solving:
mass:
time:
length: a =0, a= 0
 b = 0, b = 0
3a + b + c + 1 = 0, c = – 1 Π3 = L / D lengthtodiameter ratio These three are the independent Π terms.
Now obtain the dependent Π term by adding ∆P Π4 = ρa Vb Dc ∆P1 = (mL3)a ( Lt1)b Lc ( mL1t2 )1
Solving:
mass:
time:
length: Π4 = ∆P a + 1 = 0 , a = 1
 b  2 = 0, b = 2
3a + b + c  1 = 0, c = 0 / ρ V2 Pressure coefficient  v v V4  eText Main Menu  Textbook Table of Contents  Study Guide Application of the Buckingham Pi Theorem to the previous list of variables
yields the following nondimensional combinations: ∆P ρ VD L ε = f
,,
ρ V2
µ D D or { Cp = f Re,L, ε } Thus, a nondimensional pressure loss coefficient for viscous pipe flow
would be expected to be a function of (1) the Reynolds number, (2) a nondimensional pipe length, and (3) a nondimensional pipe roughness. This
will be shown to be exactly the case in Ch. VI, Viscous Internal Flow.
A list of typical dimensionless groups important in fluid mechanics is given in
the accompanying table.
From these results, we would now use a planned experiment with data analysis
techniques to get the exact form of the relationship among these non dimensional parameters.
The next major step is concerned with the design and organization of the
experimental test program
Two key elements in the test program are:
* design of the model
* specification of the test conditions, particularly when the test must be
performed at conditions similar, but not the same as the conditions of
interest.
Similarity and nondimensional scaling
The basic requirement is in this process to achieve 'similarity' between the
'experimental model and its test conditions' and the 'prototype and its test
conditions' in the experiment.  v v V5  eText Main Menu  Textbook Table of Contents  Study Guide Table 5.2 Dimensional Analysis and Similarity
Parameter Definition
RE = Reynolds number Qualitative ratio
of effects
Inertia
Viscosity
Flow speed
Sound speed
Inertia
Gravity
Inertia
Surface tension
Pressure
Inertia ρ UL
µ U
A
U2
Fr =
gL MA = Mach number
Froude number
Weber number We = ρ U2 L
γ Cavitation number
(Euler number) Ca = p  pv
ρU 2 C pµ
k
U2
Ec =
c p To
c
γ= p
cv
ωL
St =
U Eckert number
Specificheat ratio
Strouhal number L Gr = Grashof number
Temperature ratio
Pressure coefficient Compressible flow
Freesurface flow
Freesurface flow
Cavitation
Heat convection
Dissipation Enthalpy
Internal energy
Oscillation
Mean speed
Wall roughness
Body length ε Roughness ratio Always Dissipation
Conduction
Kinetic energy
Enthalpy Pr = Prandtl number Importance β ∆TgL3 ρ2
µ2 Tw
To
p − p∞
Cp =
1/ 2 ρ U2 Compressible flow
Oscillating flow
Turbulent,rough
walls Buoyancy
Viscosity
Wall temperature
Stream temperature
Static pressure
Dynamic pressure Lift coefficient CL = L
1/ 2 ρ U2 A CD = D
1/ 2ρ U2 A Heat transfer
Aerodynamics,
hydrodynamics Lift force
Dynamic force Drag coefficient Natural convection Lift force
Dynamic force Aerodynamics
hydrodynamics
Aerodynamics,
hydrodynamics  v v V6  eText Main Menu  Textbook Table of Contents  Study Guide In this context, “similarity” is defined as
Similarity: All relevant dimensionless parameters have the same values
for the model & the prototype.
Similarity generally includes three basic classifications in fluid mechanics:
(1) Geometric similarity
(2) Kinematic similarity
(3) Dynamic similarity Geometric similarity
In fluid mechanics, geometric similarity is defined as follows: Geometric Similarity All linear dimensions of the model are related
to the corresponding dimensions of the
prototype by a constant scale factor SFG Consider the following airfoil section (Fig. 5.4): Fig. 5.4 Geometric Similarity in Model Testing
For this case, geometric similarity requires the following:  v v V7  eText Main Menu  Textbook Table of Contents  Study Guide rm L m Wm
=
=
=⋅ ⋅ ⋅
rp
L p Wp SFG = In addition, in geometric similarity,
All angles are preserved.
All flow directions are preserved.
Orientation with respect to the surroundings must be same for the model
and the prototype, ie.,
Angle of attack )m = angle of attack )p
Kinematic Similarity
In fluid mechanics, kinematic similarity is defined as follows:
Kinematic Similarity The velocities at 'corresponding' points on the
model & prototype are in the same direction
and differ by a constant scale factor SFk. Therefore, the flows must have similar streamline pattterns
Flow regimes must be the same.
These conditions are demonstrated for two flow conditions, as shown in the
following kinematically similar flows (Fig. 5.6). Fig. 5.6a Kinematically Similar Low Speed Flows  v v V8  eText Main Menu  Textbook Table of Contents  Study Guide Fig. 5.6b Kinematically Similar Free Surface Flows
The conditions of kinematic similarity are generally met automatically when
geometric and dynamic similarity conditions are satisfied. Dynamic Similarity
In fluid mechanics, dynamic similarity is typically defined as follows:
Dynamic Similarity This is basically met if model and prototype
forces differ by a constant scale factor at
similar points. This is illustrated in the following figure for flow through a sluice gate (Fig. 5.7). Fig. 5.7 Dynamic Similarity for Flow through a Sluice Gate  v v V9  eText Main Menu  Textbook Table of Contents  Study Guide This is generally met for the following conditions:
1. Compressible flows: model & prototype Re, Ma, are equal Rem = Rep, Mam= Map , γm = γp
2. Incompressible flows
a. No free surface
Rem = Rep
b. Flow with a free surface
Rem = Rep , Frm = Frp Note: The parameters being considered, e.g., velocity, density, viscosity,
diameter, length, etc., generally relate to the flow, geometry, and fluid
characteristics of the problem and are considered to be independent
variables for the subject problem.
The result of achieving similarity by the above means is that relevant non dimensional dependent variables, e.g., CD, Cp, Cf, or Nu, etc., are then equal for
both the model and prototype.
This result would then indicate how the relevant dependent results, e.g. drag force,
pressure forces, viscous forces, are to be scaled for the model to the prototype.
Equality of the relevant nondimensional independent variables Re, Ma, x/L, etc.,
indicates how the various independent variables of importance should be scaled.  v v V10  eText Main Menu  Textbook Table of Contents  Study Guide An example of this scaling is shown as follows:
Example: The drag on a sonar transducer prototype is to be predicted based on the
following wind tunnel model data and prototype data requirements. Determine
the model test velocity Vm necessary to achieve similarity and the expected
prototype force Fp based on the model wind tunnel test results.
Given: Prototype Model sphere sphere D 1 ft 6 in V 5 knots unknown? F ? 5.58 lbf ρ 1.98 ν slugs
ft 2
5 ft
1.4 *10 0.00238 3 slugs
ft 1.56 * 104 ft s 3 2 s From dimensional analysis: C D = f{ R e } F D 2 = f VD υ ρ V2 For the prototype, the actual operating velocity and Reynolds number are:
Prototype: na ⋅ mi
ft
1hr
ft
na ⋅ mi
ft
1hr
ft
6080
6080
= 8.44 Vp = 5
= 8.44
hr
na ⋅ mi 3600 s
s
hr
na ⋅ mi 3600 s
s V11  v v Vp = 5  eText Main Menu  Textbook Table of Contents  Study Guide Re =
p 8. 44 ft * 1ft
5
s
=
= 6. 03 * 10
2
1. 4 *10 −5 ft s VD υ p Equality of Reynolds number then yields the required model test velocity of Re = Re =
m VD p υ m = > Vm = 188 ft s Based on actual test results for the model, i.e. measured Fm, equality of model and
prototype drag coefficients yields ∴ C D = C D = > Fp = Fm
p m ρp 2
Vp ρm 2
Vm D 2
m D2
p = 37. 4 lb f Note: All fluid dynamic flows and resulting flow characteristics are not Re
dependent.
Example:
The drag coefficient for bluff bodies with a fixed point of separation; e.g.,
radar antennae, generally have a constant, fixed number for CD which is not
a function of Re.
CD = const ≠ f (R )
.
e  v v V12  eText Main Menu  Textbook Table of Contents  Study Guide ...
View
Full
Document
This note was uploaded on 10/27/2009 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.
 Spring '08
 Sakar

Click to edit the document details