# CE 121

Document Sample

```					ENGR-320 Fluid Mechanics – Chapter 8: Dimensional Analysis & Similitude

Fluid Mechanics
Summer 2005

Prof. Mesut Pervizpour
Office: KH #349B
Ph: x4046

1
Modified from original notes of Dr. Lennon, CE Dept. Lehigh University
Dimensional Analysis
• Assume quantities are related by
multiplication and division
• Dimensions and Units
 P Theorem
• Assemblage of Dimensionless Parameters
• Dimensionless Parameters in Fluids
• Model Studies and Similitude

2
Pendulum Example
What is the expression for period of the pendulum?
•  Important parameters L, W, g
•  One way  plot Tp-Lp, Tp-W, Tp-g
•  Relationship: Tp = f (L, W, g)                Lp
•  Dimensions: [Tp] = T
[Lp] = L
W
[g] = L/T  2

[W] = F or ML/T2
Note W has only F dimensions (can’t divide W by anything
to eliminate F, so Tp is not a function of W)
 Tp = f (L, g)
By observation What is the only way to combine L & g to have same
dimensions as Tp?     L                      L 
T              
Tp  f    
g                    g              3
Pendulum Example (cont.)
L
•   Run one experiment and plot Tp vs. g
•   Match the slope: Tp = a + mx (where x =   L
g   )
•   a = 0, and m = 2
•   The full relationship:
L
Tp  2
g
Tp

L
g
4
The Buckingham P Theorem
• “in a physical problem including n quantities in
which there are m dimensions, the quantities
can be arranged into n-m independent
dimensionless parameters”
No. Dimensionless Groups = No. Variables Involved – No. Basis Dimensions (3 max)

• We reduce the number of parameters we need
to vary to characterize the problem!

5
Example – Drag on a falling body
Drag on a model blimp  1530 N
FD
What is the force on prototype?
FD = fnc ( , , V, D)             W
Dimensions:               [FD] = ML/T2
[] = FT/L2 = M/LT          V
[] = M/L3 or FT2/L4
[V] = L/T
[D] = L
Our variables: FD, , , V, D                n=5
Our dimensions: M, L, T                      m=3
Number of  groups = n – m = 2

Choose 3 repeating variables (usually one fluid property, one
flow property, and one scale): , V, D
6
Example – (cont.)
FD = fcn (, , V, D)
Two equations for  groups:
1 = a Vb Dc FD         Eqn-1
2 = A VB DC           Eqn-2
Recognize that i = L0 T0 F0
L0 T0 F0 = (FT2L-4)a (LT-1)b (L)c F
L0 T0 F0 = L(-4a+b+c) T(2a-b) F(a+1)
F0 : 0 = a+1                  a = -1
T0 : 0 = 2a – b               b = 2a = -2
L0 : 0 = -4a + b + c          c = -2
FD
  1   V D FD 
1  2  2
(Normalized drag   force)
Similarly for 2:
V D2 2

        1
 2   V D  
1 1 1

VD Re                             7
Example (cont.)
1 = f (2)
 f Re 
FD
Where f(Re) = CRe : Drag coefficient
V D
2 2

FD
 C Re
V D
2 2

FD  CRe V 2 D2 For sphere projected Area Ap = D2 /4 = const. D2

1  2
FD  C D Ap  V           Formula for drag of an object in air
2   

8
Dimensional Analysis
• Want to study pressure drop as function of
p1
velocity (V1) and diameter (do )                         p0
• Carry out numerous experiments with
different values of V1 and do and plot the         V1     V0
data                                                     A0
A1
V12          V02
p1        p0  
2            2
p 

2
V02  V12 
   d 4 
p  V12   1   1
5 parameters:
2   d0 
              p,  , V1, d1, do
          
4
p      d1 
   1
 2  d0               2 dimensionless parameters:
V1                       p/(V2/2), (d1/do)
2
d                 Much easier to
Cp  f  1 
d 
 0                establish functional
relations with 2                       9
parameters, than 5
Exponent Method - revisit
1.    List all n variables involved in the problem
• Typically: all variables required to describe the problem
geometry (D) or define fluid properties (,  ) and to indicate
external effects (dp/dx)
2.    Express each variables in terms of MLT dimensions (j)
3.    Determine the required number of dimensionless parameters (n – j)
4.    Select a number of repeating variables = number of dimensions
• All reference dimensions must be included in this set and each
must be dimensionalls independent of the others
5.    Form a dimensionless parameter by multiplying one of the
nonrepeating variables by the product of the repeating variables,
each raised to an unknown exponent
6.    Repeat for each nonrepeating variable
7.    Express result as a relationship among the dimensionless parameters
10
Example (8.7)
• Find: Drag force on rough sphere
is function of D, ,  , V and k.
Express in form:
 3  f ( 1 ,  2 )

FD     D                      V      k      1   ( D aV b  c )

MLT-2    L    ML-3    ML-1T-1   LT-1   L      M 0 L0T 0  ( ML1T 1 )( L) a ( LT 1 ) b ( ML3 ) c
M:        0  1 c       c  1
L : 0  1  a  b  3c          a  1
T : 0  1  b            b  1
n=6       No. of dimensional parameters
j=3       No. of dimensions
                        VD
k = n - j = 3 No. of dimensionless parameters    1              or  1   
DV                        

Select “repeating” variables: D, V, and 
Combine these with nonrepeating variables: F,  & k

11
Example (8.7)
FD     D                         V     k     3  FD ( D aV b  c )
MLT-2   L     ML-3       ML-1T-1   LT-1   L    M 0 L0T 0  ( MLT 2 )( L) a ( LT 1 ) b ( ML3 ) c
M:        0  1 c       c  1
Select “repeating” variables: D, V, and 
Combine these with nonrepeating                L : 0  1  a  b  3c  a  2
variables: F,  & k                            T : 0  2  b  b  2

 2  k ( D aV b  c )
FD
3 
M 0 L0T 0  ( L)( L) a ( LT 1 )b ( ML3 ) c          V 2 D 2
M:        0c         c0
L : 0  1  a  b  3c  a  1
T : 0  b  b  0
FD     VD k
 f(    , )
2 
k                                         V D
2 2        D
D

12
Assemblage of Dimensionless
Parameters
• Several forces potentially act on a fluid
• Sum of the forces = ma (the inertial force)
• Inertial force is always present in fluids
problems (all fluids have mass)
• Nondimensionalize by creating a ratio with
the inertial force
• The magnitudes of the force ratios for a
given problem indicate which forces
govern
13
Forces on Fluids
•   Force         parameterdimensionless
•                      
Mass (inertia) ______
•   Viscosity          
______     ______
R Reynolds#
•   Gravitational ______g     ______ #
F Froude
•   Pressure           p
______     ______Pressure Coeff.
Cp
•                      s
Surface Tension______     ______ #
W Weber
•   Elastic        ______
K      ______ #
M Mach

Dependent variable
14
Inertia as our Reference Force
F
• F=ma       F  a          f  a

• Fluids problems always (except for statics)
include a velocity (V), a dimension of flow (l),
and a density ()

L=l       l          M=  l3
f      M                             T=V
L2T 2                                                   2
V
f i= 
l
15
Viscous Force
• What do I need to multiply viscosity by to
obtain dimensions of force/volume?
f                  l
f   C      C              Ll    T     M  l 3
                   V
 M                                              V2
 L2T 2    C 
1                        fi  
C         
LT                                l
M 
 LT 
                  V2
      f i Vl
V           fi                        Vl
C  2              l                R
l           fμ    V
 2    fμ             
l              Reynolds number
16
Gravitational Force
 M                  l
fg           L2T 2     Ll   T        M  l     3

Cg           Cg                      V
g             L                                   2
T 2 
                        fi  
V
l
M            Cg  
Cg  3
L

V2
fi  l
                fi V 2

V
fg   g                      F
f g gl       gl
Froude number
17
Pressure Force
 M                   l
fp           L2T 2    Ll     T       M  l 3
Cp           Cp                       V
p            M 
 LT 2                             V2
                           fi  
l
1
Cp            Cp 
1
L              l
V2
                                2p
fi       l      f i V 2        Cp 
                                 V 2
fp     p        fp   p
l
Pressure Coefficient
18
Dimensionless parameters
Vl
R
• Reynolds Number                           
V
F
• Froude Number                             gl
V 2 l
• Weber Number                          W
s
• Mach Number                           M
V
c
2Drag
 2p C d 
•   Pressure Coefficient             Cp 
V  2         V 2 A
• (the dependent variable that we measure experimentally)

19
Common Dimensionless No’s.
• Reynolds Number (inertial to viscous forces)                  Vd

• Important in all fluid flow problems                       

• Froude Number (inertial to gravitational forces)         F
V
• Important in problems with a free surface                   gh

• Euler Number (pressure to inertial forces)                      p
• Important in problems with pressure differences      Cp 
V 2
• Mach Number (inertial to elastic forces)                        V    V
• Important in problems with compressibility effects   M        
E/ c

• Weber Number (inertial to surface tension forces)           LV 2
• Important in problems with surface tension effects
W
s

20
Application of Dimensionless
Parameters
• Pipe Flow
• Pump characterization
• Model Studies and Similitude
•   dams: spillways, turbines, tunnels
•   harbors
•   rivers
•   ships
•   ...

21
Example: Pipe Flow
• What are the important forces?
______, ______. Therefore _________
Inertial viscous                Reynolds
number.
• What are the important geometric
parameters?
diameter, length, roughness height
_________________________
• Create dimensionless geometric groups
l/D      e/D
______, ______
• Write the functional relationship            l e
C p  f R , , 
 D D
22
 l e
Example: Pipe Flow                               C p  f R , , 
 D D
• How will the results of dimensional
analysis guide our experiments to
determine the relationships that govern
pipe flow?
• If we hold the other two dimensionless
parameters constant and increase the length
to diameter ratio, how will Cp change?
D       e                2p
Cp proportional to l      Cp     f  ,R        Cp 
l       D                V 2
 D      e   
f  Cp   f  , R      f is friction factor
  l     D 
23
Frictional Losses in Straight Pipes
0.1

0.05
0.04
0.03

0.02

e
0.015
friction factor

0.01

f                                                                                0.008
0.006
0.004
D
laminar
0.002

0.001
0.0008

0.0004
0.0002
0.0001
0.00005
0.01                                                       smooth

1E+03       1E+04   1E+05       1E+06   1E+07   1E+08
R
24
What did we gain by using
Dimensional Analysis?
• Any consistent set of units will work
• We don’t have to conduct an experiment
on every single size and type of pipe at
every velocity
• Our results will even work for different
fluids
• Our results are universally applicable
• We understand the influence of
temperature
25
Model Studies and Similitude:
Scaling Requirements
• dynamic similitude (forces)
• geometric similitude
• all linear dimensions must be scaled identically (LR,
AR, ..)
• roughness must scale
• kinematic similitude
• constant ratio of dynamic pressures at corresponding
points (VR, QR)
• streamlines must be geometrically similar
• _______, __________, _________, and _________
Mach Reynolds Froude                    Weber
numbers must be the same
Cp = f (M, R, F, W, geometry)
26
Relaxed Similitude
Requirements
• Impossible to have all force ratios the same
same ____
unless the model is the _____ size as the
prototype
• Need to determine which forces are
important and attempt to keep those force
ratios the same

27
Similitude Examples
•   Open hydraulic structures
•   Ship’s resistance
•   Closed conduit
•   Hydraulic machinery

28
Example
• Consider predicting the drag on a thin
rectangular plate (w*h) placed normal               FD  f ( w, h,  ,  ,V )
to the flow.
• Drag is a function of: w, h, , , V                  1  f ( 2 ,  3 )
• Dimensional analysis shows:                       FD          w Vw
 f( ,           )
• And this applies BOTH to a model                 w V
2   2       h 
and a prototype
• We can design a model to predict the
drag on a prototype.                                1m  f ( 2m ,  3m )
• Model will have:                                FDm          w  V w
 f ( m , m m m)
wm  mVm
2      2      hm        m
• And the prototype will have:
 1 p  f ( 2 p ,  3 p )
FDp          w p  pV p w p
 f( ,             )
w p  pV p
2       2     hp     p
29
Example
•Similarity conditions
Geometric similarity
wm w p               hm
 2m   2 p                     wm     wp        Gives us the size of the model
hm h p               hp
Kinematic similarity
mVm wm  pV p w p                 m  p w p
 3m   3 p                                  Vm             Vp
m       p                       p m wm
Then                                     Gives us the velocity in the model

Dynamic similarity
2           2
FDp                wp   p Vp 
 FDp             F
FDm
 1m   1 p                2
wm  mVm
2     2
w p  pV p
2            w   m  V  Dm
 m        m

30
Example (8.28)
• Given: Submarine moving below surface in sea water
( =1015 kg/m3, n   /  = 1.4x10-6 m2/s).
Model is 1/20th scale in fresh water (20oC).
• Find: Speed of water in the test dynamic similarity and the ratio
of drag force on model to that on prototype.
• Solution: Reynolds number is significant parameter.
Re m  Re p                             Fm

Fp

V L     V L                            mVm lm  pV p l 2
2 2       2
p
p p
m m   
nm        np                             Fm  mVm lm
2 2

Lp n m                         F p  pV p l 2
2
p
Vm              Vp
Lm n p                                           2
1000  28.6   1 
2
              
20 1                                    1015  2   20 
        2m / s
1 1.4                            Fm
Vm  28.6 m / s                            0.504                      31
Fp
Scaling in Open Hydraulic
Structures
• Examples
• spillways
• channel transitions
• weirs
• Important Forces            NCHRP Request For Proposal on “Effects of Debris on Bridge-Pier Scour “
p://www4.trb.org/trb/crp.nsf/All+Projects/NCHRP+24-26
• inertial forces
• gravity: from changes in water surface elevation           V
F
• viscous forces (often small relative to gravity forces)     gl
• Minimum similitude requirements                                                       Vl
• geometric                                                                        R

• Froude number
32
V
F                                                  Fm  Fp
gl         Froude similarity
V  2
Vp2
• Froude number the same in model and                      
m
g mLm g pLp
prototype
Vm2
Vp2

• ________________________ L m L p
difficult to change g
Lp
• define length ratio (usually larger than 1)             Lr 
Lm
• velocity ratio    Vr  L r
Lr
• time ratio            tr        Lr
Vr
• discharge ratio   Qr  Vr Ar  L r L r L r  Lr  5/ 2

• force ratio                           3 Lr
Fr  M r a r   r L r 2  L3r
tr
33
Example: Spillway Model
• A 50 cm tall scale model of a proposed 50
m spillway is used to predict prototype
flow conditions. If the design flood
discharge over the spillway is 20,000 m3/s,
what water flow rate should be tested in the
model?
Fm  Fp         Lr  100

Qr  L
5/ 2
r       100,000
20,000 m3 s
Qm               0.2 m3 s
100,000
34
Ship’s Resistance
• Skin friction ______________
Viscosity, roughness
• Wave drag (free surface effect) ________
gravity
• Therefore we need ________ and Froude
Reynolds        ______
similarity

2Drag           e , R, F 
 Cd  f           
V A
2
l         

35
Reynolds and Froude Similarity?
Reynolds                               Froude
Vl                                    V
R                                     F
                                      gl
 mVmlm  pV p l p   Water is the only     Vr  L r

m     p         practical fluid
Vmlm  V pl p
Vp lm
                       1
Vm l p                          Lr   Lr = 1
1                  Lr
Vr 
Lr
36
Ship’s Resistance
• Can’t have both Reynolds      2D total       e       
 C d  f  , R, F 
        
and Froude similarity         V A
2
D       
• Froude hypothesis: the two
D total  D f  D w
forms of drag are
independent
V 2 A  e        
• Measure total drag on Ship     Df                f  ,R 
2         D 
• Use analytical methods to                          analytical
calculate the skin friction
V 2 A
• Remainder is wave drag           Dw                f F 
2
empirical
37
Closed Conduit Incompressible
Flow
• Forces
• __________
viscosity
• __________
inertia
• If same fluid is used for model and
prototype
• VD must be the same
velocity
• Results in high _________ in the model
• High Reynolds number (R)
• Often results are independent of R for very high
R
38
Example: Valve Coefficient
 2p
• The pressure coefficient, Cp         , for a
V 2
600-mm-diameter valve is to be
determined for 5 ºC water at a maximum
velocity of 2.5 m/s. The model is a 60-mm-
diameter valve operating with water at 5
ºC. What water velocity is needed?

39
Example: Valve Coefficient
• Note: roughness height should scale!
• Reynolds similarity
Vl           VD            ν = 1.52 x 10-6 m2/s
R          R
              n
Vm Dm       Vp Dp          Vp Dp
           Vm 
nm          np                Dm

(2.5m / s)0.6m 
Vm                               Vm = 25 m/s
0.06m
40
Example: Valve Coefficient
(Reduce Vm?)
• What could we do to reduce the velocity in
the model and still get the same high    Vl
R
Reynolds number?                          
VD
Decrease kinematic viscosity          R
n
Use a different fluid
Use water at a higher temperature

41
Example: Valve Coefficient
• Change model fluid to water at 80 ºC
VD
R                                         νm = 0.367 x 10-6 m2/s
_____________
n
νp = ______________
1.52 x 10-6 m2/s
Vm Dm         Vp Dp                n mV p D p
                   Vm 
nm         np                   n p Dm


0.367 x10 m / s (2.5m / s)0.6m
6   2
Vm
1.52 x10 m / s 0.06m
6    2
Vm = 6 m/s
42
Approximate Similitude at High
Reynolds Numbers
• High Reynolds number means inertial
______
forces are much greater than _______
viscous
forces
• Pressure coefficient becomes independent
of R for high R

43
Pressure Coefficient for a
Venturi Meter
10
Cp

 2p
Cp 
V 2

1
1E+00 1E+01 1E+02 1E+03 1E+04 1E+05 1E+06
R            Vl
Similar to rough pipes in                 R

Moody diagram!                                      44
Hydraulic Machinery: Pumps
• Rotational speed of pump or turbine is an
• additional dimensionless parameter is the ratio
of the rotational speed to the velocity of the
water _________________________________
streamlines must be geometrically similar
• homologous units: velocity vectors scale _____Vr  lr
• Now we can’t get same Reynolds Number!
• Reynolds similarity requires           1
Vr 
lr
• Scale effects

45
Dimensional Analysis Summary
Dimensional analysis:
• enables us to identify the important
parameters in a problem
• simplifies our experimental protocol
• does not tell us the coefficients or powers
of the dimensionless groups (need to be
determined from theory or experiments)
• guides experimental work using small
models to study large prototypes
46
Port Model
• A working scale model was used to eliminated danger to boaters
from the "keeper roller" downstream from the diversion structure

http://ogee.hydlab.do.usbr.gov/hs/hs.html                  47
Hoover Dam Spillway
A 1:60 scale
hydraulic model
of the tunnel
spillway at
Hoover Dam for
investigation of
cavitation damage
preventing air
slots.

http://ogee.hydlab.do.usbr.gov/hs/hs.html
48
Irrigation Canal Controls
http://elib.cs.berkeley.edu/cypress.html

49
Spillways
Frenchman Dam and spillway (in use).
Lahontan Region (6)

50
Dams
Dec 01, 1974
Cedar Springs Dam, spillway & Reservoir
Santa Ana Region (8)

51
Spillway
Mar 01, 1971
Cedar Springs Spillway construction.
Santa Ana Region (8)

52
Kinematic Viscosity
1.00E-03
kinematic viscosity 20C (m2/s)

1.00E-04

1.00E-05

1.00E-06

1.00E-07

53
Kinematic Viscosity (m /s)
Kinematic Viscosity of Water
2.0E-06
2

1.5E-06

1.0E-06

5.0E-07

0.0E+00
0   20     40      60      80   100
Temperature (C)

54

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