Download FM.4

Dimensional Analysis & Similarity
Uses:
Verify if eqn is always usable
Predict nature of relationship between quantities (like
friction, diameter etc)
Minimize number of experiments. Concept of DOE
Buckingham PI theorem
Scale up / down
Scale factors
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Dimensional Analysis
Basic Dimensions:
M,L,T (or F,L,T for convenience)
Temp, Electric Charge... (for other problems)
E  MC 2
E  Energy  Force  Length
 Mass  Acceleration  Length
 M LT 2 L  M 1L2T 2
pH   log( C )
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
C in gram  mole per litre
Dimensional Analysis
Ideal Gases
G  RT ln( P)  C0
Not dimensionally consistent
Can be used only after defining a standard state
P
G  Gs  RT ln( )
Ps
Empirical Correlations: Watch out for units
Write in dimensionally consistent form, if possible
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Dimensional Analysis
Is there a possibility that the equation exists?
Effect of parameters on drag on a cylinder
Choose important parameters
viscosity of medium, size of cylinder (dia, length?), density
velocity of fluid?
Choose monitoring parameter
drag (force)
Are these parameters sufficient?
How many experiments are needed?
  Pa  s  M L T
D  L1
  M 1L3 F  M 1L1T 2
1 1
1
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
V  LT
1
1
Is a particular variable important?
Need more parameters with temp
Activation energy & Boltzmann constant
Does Gravity play a role?
Density of the particle or medium?
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Design of Experiments (DoE)
How many experiments are needed?
DOE:
piece wise linear
Full factorial and Half factorial
Neglect interaction terms
(or quadratic ) models
Corner, center models
Levels of experiments (example 5)
Change density (and keep everything else constant) and measure
velocity. (5 different density levels)
Change viscosity to another value
Repeat density experiments again
change viscosity once more and so on...
5 levels, 4 parameters
5  625 experiment s
4
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Limited physical insight
Pi Theorem
Can we reduce the number of experiments and still get the
exact same information?
Dimensional analysis / Buckingham Pi Theorm
Simple & “rough” statement
If there are N number of variables in “J” dimensions, then
there are “N-J” dimensionless parameters
Accurate statement:
If there are N number of variables in “J” dimensions, then
the number of dimensionless parameters is given by (Nrank of dimensional exponents matrix)
Normally the rank is = J. Sometimes, it is less
 ,V , , D, Force, 
M , L, T
Min of 6-3 = 3 dimensionless groups
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Pi Theorem
Premise: We can write the equation relating these parameters in
dimensionless form
 (1 ,  2 ,  3 ....,  n )  0
 i is dimensionl ess
“n” is less than the number of dimensional variables (i.e.
Original variables, which have dimensions)
==> We can write the drag force relation in a similar way if we
know the Pi numbers
Method (Thumb rules) for finding Pi numbers
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Method for finding Pi numbers
1.Decide which factors are important (eg viscosity, density,
etc..).
Done
2.Minimum number of dimensions needed for the variables (eg
M,L,T)
Done
3.Write the dimensional exponent matrix
  M 1L1T 1
  M 1L3T 0
D  M 0 L1T 0
  M 0 L1T 0
F  M 1L1T 2
V  M 0 L1T 1
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
M L T
1 1 1
1 3 0
0 1
0
0 1
0
1 1 2
0 1 1
Method for finding Pi numbers
4.Find the rank of the matrix
=3
To find the dimensionless groups
Simple examination of the variables
D

5.Choose J variables (ie 3 variables here) as “common” variables
They should have all the basic dimensions (M,L,T)
They should not (on their own) form a dimensionless number
(eg do not choose both D and length)
They should not have the dependent variable
Normally a length, a velocity and a force variables are
included
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Method for finding pi numbers
D, V , 
Combine the remaining variables, one by one with the following
constraint
 i  D a V b  c ( variable )1  M 0 L0T 0
Solve for a,b,c etc (If you have J basic dimensions, you will get J
equations with J unknowns)
Note: “common” variables form dimensionless groups
among themselves ==> inconsistent equations
dependent variable (Drag Force) is in the common variable,
==> an implicit equation
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Pi numbers: Example
D, V , 
Length
Consider viscosity
1 
DV
Drag Force

2 
D

3 
3 
F
 V 2 D2
12 V D
What if you chose length instead of density? Or velocity?


 DV  
2
1
 3   1 ,  2 
F

V D   
, 
2
  D
Similarly, pressure drop in a pipe


 DV  
P  1 V 2  
, 
2
  D
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
F
2
Physical Meaning
Ratio of similar quantities
Many dimensionless numbers in Momentum Transfer are
force ratios
Re 
DV

V

D 2  Inertial Force

DV  Viscous Force
2

D




 2 D 4 Centrifugal Force
 D 
Strouhal 


 
2 2
V
V D
Inertial Force
 V 
2

V
V2
V 2 D 2 Inertial Force
Fr 



3
gD
Gravity Force
gL gL


V D   Inertial Force
Eu 

PD  Pressure Force
P
V D   Inertial Force
DV
We 

V 2
2
2
2
2

Ca 
V 2
Es
2
2
D 
Surface Tension Force
V D   Inertial Force  Inertial Force   V 

E D  Elastic Force Compressibility Force  C 
2
2
2
s
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
2
 Ma 2
DV
2

 P   V  g
Dt
N-S equation
Use some characteristic length, velocity and pressure to obtain
dimensionless groups
L,U , U 2
x
x 
L
*
V
V 
U
*
Ut
t 
L
*
P* 
P
U 2
 *  L

V *
1 *2 * 1 g
*
* *
* *
 V . V   P 
 V 
t *
Re
Fr g
Reynolds and Froude numbers in equation
Boundary conditions may yield other numbers, like Weber
number, depending on the problem
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Scaling (Similarity/Similitude)
Scale up/down
Practical reasons (cost, lack of availability of tools with
high resolution)
Geometric, Kinematic and Dynamic
Geometric - length scale
Kinematic - velocity scale (length, time)
Dynamic - force scale (length, time, mass)
Concept of scale factors
KL = L FULL SCALE/ L MODEL
KV = (Velocity) FULL SCALE / (Velocity) MODEL
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Examples
No baffles
Impeller
Turbine
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Baffles
Sketch from
Treybal
Examples
From “Sharpe Mixers” website
IIT-Madras, Momentum Transfer: July 2005-Dec 2005