Ch. 8 Similitude and Dimensional Analysis
Chapter 8 Similitude and Dimensional Analysis
8.0 Introduction
8.1 Similitude and Physical Models
8.2 Dimensional Analysis
8.3 Normalization of Equations
Objectives:
- learn how to begin to interpret fluid flows
- introduce concept of model study for the analysis of the flow phenomena that could not be solved by analytical (theoretical) methods
- introduce laws of similitude which provide a basis for interpretation of model results
- study dimensional analysis to derive equations expressing a physical relationship between quantities
8-1 Ch. 8 Similitude and Dimensional Analysis
8.0 Introduction
Why we need to model the real system?
Most real fluid flows are complex and can be solved only approximately by analytical methods.
Real System Set of assumptions
Conceptual Model Simplified version of real system
Field Study Physical Modeling Mathematical Model
(Scaled and simplified [Governing equations
version of prototype) + Boundary (initial) conditions]
Analytical solution Numerical solution (model)
(FDM/FEM/FVM)
8-2 Ch. 8 Similitude and Dimensional Analysis
▪ Three dilemmas in planning a set of physical experiments or numerical experiments
1) Number of possible and relevant variables or physical parameters in real system is huge and
so the potential number of experiments is beyond our resources.
2) Many real flow situations are either too large or far too small for convenient experiment at
their true size.
→ When testing the real thing (prototype) is not feasible, a physical model (scaled version of
the prototype) can be constructed and the performance of the prototype simulated in the
physical model.
3) The solutions produced by numerical simulations must be verified or numerical models
calibrated by use of physical models or measurements in the prototype.
▪ Model study
Physical models have been used for over a hundred years.
Models began to be used to study flow phenomena that could not be solved by analytical
(theoretical) methods.
[Ex]
Civil and environmental engineering: models of hydraulic structures, river sections, estuaries and coastal bays and seas
Mechanical engineering: models of pumps and turbine, automobiles
Naval architect: ship models
Aeronautical engineering: model test in wind tunnels
8-3 Ch. 8 Similitude and Dimensional Analysis
▪ Justification for models
1) Economics: A model, being small compared to the prototype, costs little.
2) Practicability: In a model, environmental and flow conditions can be rigorously
controlled.
▪ Laws of similitude
- provide a basis for interpretation of physical and numerical model results and crafting both physical and numerical experiments
▪ Dimensional analysis
- derive equations expressing a physical relationship between quantities
8-4 Ch. 8 Similitude and Dimensional Analysis
8.1 Similitude and Physical Models
Similitude of flow phenomena not only occurs between a prototype and it modelbut also may exist between various natural phenomena.
There are three basic types of similitude; all three must be obtained if complete similarity is to exist between fluid phenomena.
Geometrical similarity
Kinematic similarity
Dynamic similarity
1) Geometrical similarity
8-5 Ch. 8 Similitude and Dimensional Analysis
~ Flow field and boundary geometry of model and of the prototype have the same shape.
→ The ratios between corresponding lengths in model and prototype are the same.
[Cf] Distorted model
~ not geometrically similar > ()ldrr
~ The flows are not similar and the models have to be calibrated and adjusted to make them
perform properly.
~ used models of rivers, harbor, estuary
~ Numerical models are usually used in their place.
For the characteristic lengths we have
dlpp dlrr= = = dlmm
• Area
22 Adpp l p = = Admm l m
• Volume
33 Volpp d l p = = Volmm d l m
8-6 Ch. 8 Similitude and Dimensional Analysis
In addition to the flowfields having the same shape, the ratios of corresponding velocities and
accelerations must be the same through the flow.
→ Flows with geometrically similar streamlines are kinematically similar. VV 12pp= VV12mm aa34pp = aa34mm
3) Dynamic similarity
In order to maintain the geometric and kinematic similarity between flowfields, the forces acting on corresponding fluid masses must be related by ratios similar to those for kinematic similarity.
Consider gravity, viscous and pressure forces, and apply Newton’s 2nd law F123p F p F p Ma pp 4 = = = (8.2) F123m F m F m Ma mm 4
Define inertia force as the product of the mass and the acceleration FI = Ma
4) Complete similarity
~ requires simultaneous satisfaction of geometric, kinematic, and dynamic similarity.
8-7 Ch. 8 Similitude and Dimensional Analysis
→ Kinematically similar flows must be geometrically similar.
→ If the mass distributions in flows are similar, then kinematic similarity (density ratio for
the corresponding fluid mass are the same) guarantees complete similarity from Eq. (8.2).
From Fig. 8.1, it is apparent that
F123p++= F p F p Ma pp 4 (a)
F123m++= F m F m Ma mm 4 (b)
If the ratios between three of the four corresponding terms in Eq.(a) and Eq.(b) are the same,
the ratio between the corresponding fourth terms be the same as that the other three. Thus,
one of the ratio of Eq.(8.2) is redundant. If the first force ratio is eliminated,
Mapp4 Mamm4 = (8.3) FF22pm
Mapp4 Mamm4 = (8.4) FF33pm
▪ The scalar magnitude of forces affecting a flow field
2 Pressure force: Fp =∆=∆( p) A pl
2 3V 22 Inertia force: FI = Ma =ρρ l = V l l
3 Gravity force: FG = Mg = ρ lg
8-8 Ch. 8 Similitude and Dimensional Analysis
dv V 2 Viscosity force: FV =µ A = µµ l = Vl dy l
2 Elasticity force: FE = EA = El
Surface tension: FlT = σ
Here l and V are characteristic length and velocity for the system.
[Re] Other forces
Coriolis force of rotating system → Rossby number
Buoyant forces in stratified flow → Richardson number
Forces in an oscillating flow → Strouhal number
▪ Dynamic similarity
To obtain dynamic similarity between two flowfields when all these forces act, all corresponding force ratios must be the same in model and prototype.
8-9 Ch. 8 Similitude and Dimensional Analysis
(i) FFρρ V22 V II= = = (8.5) FF∆∆ p p pppmpm
ρ Define Euler number, Eu= V 2∆p
Eupm= Eu
FII Fρρ Vl Vl (ii) = = = FF µµ VVpmpm (8.6)
Vl Define Reynolds number, Re = ν
→ Repm= Re Reynolds law
F FV22 V (iii) II= = = F F gl gl GGpmpm (8.7)
V Define Froude number, Fr = gl
→ Frpm= Fr Froude law
FFρρ V22 V (iv) II= = = FF E E EEpmpm (8.8)
ρV 2 Define Cauchy number, Ca = E
8-10 Ch. 8 Similitude and Dimensional Analysis
Capm= Ca
V [Cf] Define Mach number, Ma= Ca = E ρ
Mapm= Ma
F Fρρ lV22 lV (v) II= = = FF σσ TTpmpm (8.9)
ρ lV 2 Define Weber number, We = σ
Wepm= We
Only four of these equations are independent. → One equation is redundant according to the
argument leading to Eq. (8.3) & (8.4). → If four equations are simultaneously satisfied, then
dynamic similarity will be ensured and fifth equation will be satisfied.
In most engineering problems (real world), some of the forces above (1) may not act, (2) may
be of negligible magnitude, or (3) may oppose other forces in such a way that the effect of
both is reduced.
→ In the problem of similitude a good understanding of fluid phenomena is necessary to
determine how the problem may be simplified by the elimination of the irrelevant, negligible,
or compensating forces.
8-11 Ch. 8 Similitude and Dimensional Analysis
1. Reynolds similarity
~ Flows in pipe, viscosity-dominant flow
For low-speed submerged body problem, there are no surface tension phenomena, negligible compressibility effects, and gravity does not affect the flowfield.
→ Three of four equations are not relevant to the problem.
→ Dynamic similarity is obtained between model and prototype when the Reynolds numbers
(ratio of inertia to viscous forces) are the same.
Vl Vl =Repm = Re = (8.6) νν pm
[Re] Reynolds similarity
Ratio of any corresponding forces will be the same.
Consider Drag force, D= Cρ Vl22
DD = FFIIpm
DD = ρρ22 22 Vlpm Vl
8-12 Ch. 8 Similitude and Dimensional Analysis
For flow of incompressible fluids through closed passages
Geometric similarity:
= (dd21) pm( dd 21)
ll = dd11pm
Assume roughness pattern is similar, surface tension and elastic effect are nonexistent.
Gravity does not affect the flow fields
Accordingly dynamic similarity results when Reynolds similarity, Eq. (8.6) is satisfied.
Repm= Re
Eq. (8.5) is satisfied automatically.
F F pp−− pp II=→=12 12 (8.5) ρρ22 FFPPpm Vpm V
8-13 Ch. 8 Similitude and Dimensional Analysis
◈ Reynolds similarity
① Velocity:
Rep Repm= Re =1,Rer = 1 Rem
Vd Vd V νν1 d p = →=mm =⋅ m ννpmVdp ν pdm ν pm
d p
−1 Vd If νν=→=mm mp Vdpp
② Discharge: Q= VA
22 Q dV dνν1 d mmmmm= = = mm ννd Qppppp dV d m ppd d p
③ Time:
l m 2 TVl11 l dν p l m= mm = = m m = m lVν ν Tp p lp mm ld p p mp l ν Vp Vpp
④ Force:
2 2ρ 32 F (mlTmm m) ( mmm llT m) µ ρ p mm= = = 2 32µρ Fp(mlTpp p) (ρ ppp llT p) pm
⑤ Pressure:
22 P µρm pp l m = µ Pp p ρmm l
8-14 Ch. 8 Similitude and Dimensional Analysis
[IP 8.1] Water flows in a 75 mm horizontal pipeline at a mean velocity of 3 m/s.
−3 Prototype: Water 0 C µ p =×⋅1.781 10 Pa s
3 ρ p = 99.8 kg/m
−3 1.781× 10 − v = =1.78 × 1062 m /s p 998.8
dpp=75mm, V = 3m/s, ∆= pl 14 kPa,p = 10m
−4 Model: Gasoline 20 C µm =×⋅2.9 10 Pa s (Table A 2.1)
3 ρm =×=0.68 998.8 679.2kg/m
− = × 72 vm 4.27 10 m /s
dm = 25mm
[Sol] Use Reynolds similarity; Repm= Re
−1 −7 Vm vd mm 4.27× 10 25 = = − /= 0.753 × 6 Vp vd pp 1.78 10 75
∴=Vm 0.753(3) = 2.26m/s
Eupm= Eu
∆∆pp 22= eVpm eV
14 ∆p = m [998.8×× (3)22 ] [679.2 (2.26) ]
∴∆pm =5.4 kPa
8-15 Ch. 8 Similitude and Dimensional Analysis
2. Froude similarity
~ Open channel flow, free surface flow, gravity-dominant flow.
For flow field about an object moving on the surface of a liquid such as ship model (William
Froude, 1870)
~ Compressibility and surface tension may be ignored.
~ Frictional effects are assumed to be ignored.
VV Frpm= =Fr = (8.7) gl gl pm
V gl m= mm Vp gl pp
8-16 Ch. 8 Similitude and Dimensional Analysis
[Re] Combined action of gravity and viscosity
For ship hulls, the contribution of wave pattern and frictional action to the drag are the same
order.
→ Frictional effects cannot be ignored.
→ This problem requires both Froude similarity and Reynolds similarity.
vv Vm gl mm Frpm== Fr = →= gl gl (a) pmVp gl pp
Vl Vl Vmmν lp Repm== Re = →= (b) ννpmVlpν pm
Combine (a) and (b)
0.5 1.5 gl ν lp ν glmm mm= m → m = glp pν pm l ν p glpp
This requires
(a) A liquid of appropriate viscosity must be found for the model test.
(b) If same liquid is used, then model is as large as prototype.
For ggmp=
1.5 1.5 ν lm lm m = →=νν mpl ν p lp p
lm 1 ν If =→=ν m lp 10 31.6
− Water: µ =×⋅1.0 103 Pa s
8-17 Ch. 8 Similitude and Dimensional Analysis
− Hydrogen: µ =×⋅0.21 104 Pa s
~ Choose only one equation → Reynolds or Froude
~ Correction (correcting for scale effect) is necessary.
8-18 Ch. 8 Similitude and Dimensional Analysis
◈ Froude law
① Velocity
V gl m= mm Vp gl pp
l ② Time T = v
TlVp l glpp g pl mm= = m = m Tp lV p m l p g mm l g m l p
③ Discharge Q= vA
2 2 0.5 2.5 Q Vlm gl lm gl mm m= m = mm = Qp V plp glpp lp gl pp
④ Force
3 F ρmm l m = ρ Fp pp l
⑤ Pressure
P ρmm l m = ρ Pp pp l
8-19 Ch. 8 Similitude and Dimensional Analysis
[IP 8.2] Ship model (Free surface flow)
l =120 m , l = 3m , V = 56 km h D = 9N p m p m
Find model velocity and prototype drag.
[Sol]
Use Froude similarity
VV = (8.7) gl gl pm
lm 31 lr = = = lp 120 40
3 12 ( gl)m 56× 10 3 VVmp= = = 2.46 m s ( gl) p 3600 120
• Drag force ratio
DD = (8.15) ρρ22 22 Vlpm Vl
22 ρ 3 2 2 ( Vl) p 56× 10 3600 120 DDpm= =×9 ×=575.8 kN ρ 22 ( Vl)m 2.46 3
8-20 Ch. 8 Similitude and Dimensional Analysis
3. Mach similarity
Similitude in compressible fluid flow
~ gas, air
~ Gravity and surface tension are ignored.
~ Combined action of resistance and elasticity (compressibility)
Vppν l Re=→= Re m pmVlν (a) m mp
VV Mapm= Ma = = (8.8) aa pm
E where a = sonic velocity = ρ
Va pp= Va (b) mm
Combine (a) and (b)
lp ν pm a = lm ν m a p
→ gases of appropriate viscosity are available for the model test.
8-21 Ch. 8 Similitude and Dimensional Analysis
• Velocity
Va Eρ p mm= = m Vap p E pmρ
• Time
TlVEppρ l m= m = mm Tp lV pm E mρ p l p
• Discharge
22 Q llmmVEppρ mm= = Qp llpp VEm mpρ
8-22 Ch. 8 Similitude and Dimensional Analysis
4. Euler Similarity
~ Modeling of prototype cavitation
~ For cavitation problem, vapor pressure must be included.
[Ex.1] Cavitating hydrofoil model in a water tunnel
Fig. 8.6 (a)
Here gravity, compressibility, and surface tension are neglected.
Dynamic similitude needs Reynolds similarity and Euler similarity.
Vl Vl Repm= Re = = ννpm
pp00−−vv pp σσpm= =22 = (8.5) ρρVV00 pm pp− σ = 0 v = cavitation number ρV 2
= absolute pressure p0
= vapor pressure pv
~ Virtually impossible to satisfy both equation.
~ Cavitation number must be the same in model and prototype. 8-23 Ch. 8 Similitude and Dimensional Analysis
[I.P.8.3] Model of hydraulic overflow structure → spillway model
3 Qp = 600 m s
lm 1 lr = = lp 15
[Sol]
Since gravity is dominant, use Froude similarity.
0.5 2.5 Q glmm m = Q glpp p
2.5 2.5 lm 1 QQmp= = 600 lp 15
=0.69 m3 s = 690 l/s
8-24 Ch. 8 Similitude and Dimensional Analysis
8.2 Dimensional Analysis
Dimensional analysis
~ mathematics of the dimensions of quantities
~ is closely related to laws of similitude
~ based on Fourier′s principle of dimensional homogeneity (1882)
→ An equation expressing a physical relationship between quantities must be dimensionally
homogeneous.
→ The dimensions of each side of equation must be the same.
~ cannot produce analytical solutions to physical problems.
~ powerful tool in formulating problems which defy analytical solution and must be solved
experimentally.
~ It points the way toward a maximum of information from a minimum of experiment by
the formation of dimensionless groups, some of which are identical with the force ratios
developed with the laws of similitude.
▪ Four basic dimension
~ directly relevant to fluid mechanics
~ independent fundamental dimensions
length, L
mass, M or force, F
time, t
thermodynamic temperature T
8-25 Ch. 8 Similitude and Dimensional Analysis
Newton′s 2nd law
ML F= Ma = t2
~ There are only independent fundamental dimensions.
(1) Rayleigh method
Suppose that power, P , derived from hydraulic turbine is dependent on QE,,γ T
Suppose that the relation between these four variables is unknown but it is known that these
are the only variables involved in the problem.
Pf= QE,,γ (a) ( T )
Q = flow rate
γ = specific weight of the fluid
ET = unit mechanical energy by unit weight of fluid (Fluid system → turbine)
Principle of dimensional homogeneity
→ Quantities involved cannot be added or subtracted since their dimensions are different.
Eq. (a) should be a combination of products of power of the quantities.
P= CQabγ E c (b) T
where C = dimensionless constant ~ cannot be obtained by dimensional methods
abc,,= unknown exponents
8-26 Ch. 8 Similitude and Dimensional Analysis
Eq. (b) can be written dimensionally as
ab c (Dimensions ofPQ) = ( Dimensions of) ( Dimensions of γ ) (Dimensions of ET )
2 3 a b ML L M c = (L) (c) t3 t Lt22
Using the principle of dimensional homogeneity, the exponent of each of the fundamental dimensions is the same on each side of the equation.
Mb:1=
L:2=−+ 3 a 2 bc
−=−− t:3 ab 2
Solving for a, b, and c yields
= = = abc1, 1, 1
Resubstituting these values Eq. (b) gives
P= CQγ E (d) T
C = dimensionless constant that can be obtained from
① a physical analysis of the problem
② an experimental measurement of PQ, ,,γ ET
Rayleigh method ~ early development of a dimensional analysis
8-27 Ch. 8 Similitude and Dimensional Analysis
(2) Buckingham theorem
~ generalized method to find useful dimensionless groups of variables to describe process (E.
Buckingham, 1915)
▪ Buckingham′s Π - theorem
1. n variables are functions of each other
→ Then k equations of their exponents (abc,,,)can be written.
k = largest number of variables among n variables which cannot be combined into a
dimensionless group
[Ex]
Drag force D~ f( lV ,,,,ρµ g) on ship
2. In most cases, k is equal to the number m of independent dimensions (M, L, t) ≤ km
3. Application of dimensional analysis allows expression of the functional relationship in terms of (nk− )distinct dimensionless groups.
[Ex] n=6, km = = 3 → nk −= 3 groups
D π = 1 ρ lV22
ρVl π =R = 2 e µ
V π 3 =Fr = gl
8-28 Ch. 8 Similitude and Dimensional Analysis
[Ex] Drag on a ship
f( Dl,,,,,ρµ V g) = 0
Three basic variables = repeating variables
Vl,,ρ t,, LM
Other variables Dg,,µ appear only in the unique group describing the ratio of inertia force to force related to the variable.
• Procedure:
1. Find the largest number of variables which do not form a dimensionless Π - group.
For drag problem, No. of independent dimensions is m= 3 and Vl ,ρ and cannot be formed into a Π - group, so km= = 3
2. Determine the number of Π - groups to be formed: n=6, km = = 3
∴ No. of Π - group = nk−=3
3. Combine sequentially the variables that cannot be formed into a dimensionless group, with each of the remaining variables to form the requisite Π - groups.
Π=11f( D,,,ρ Vl)
Π=22f(µρ,,, Vl)
Π=33f( g,,,ρ Vl)
8-29 Ch. 8 Similitude and Dimensional Analysis
4. Determine the detailed form of the dimensionless groups using principle of
dimensional homogeneity.
i) Π1
Π=Daρ b Vl cd (a) 1
Since Π1 is dimensionless, writing Eq. (a) dimensionally
a bc 0 00 ML M L d M Lt = (L) (b) t23 Lt
The following equations in the exponents of the dimensions are obtained
M:0= ab +
L:0=− a 3 bcd ++
=−− t:0 2 ac
Solving these equations in terms of a gives
b=−=−=− ac, 2, a d 2 a
a −−− D aaρ 22 aa Π=1 D Vl =22 ρ lV
The exponent may be taken as any convenient number other than zero.
If a = 1, then
8-30 Ch. 8 Similitude and Dimensional Analysis
D Π=1 (c) ρ lV22
ii) Π2
a b cd Π=2 µρVl
a bc 000 MML d M Lt = (L) Lt L3 t
M:0= ab +
L:0=−− a 3 bcd ++
t:0=−− ac
Solving these equations in terms of a gives
b=−=−=− ac,, ad a
a a− a −− aa µ Π=2 µρ Vl = ρ lV
If a = -1, then
Vlρ Π= =Re (d) 2 µ
iii) Π3
ab c d Π=3 glρ V
8-31 Ch. 8 Similitude and Dimensional Analysis
a cd 000 L b ML M Lt= 23 L t Lt
Mc:0=
L:0=+− ab 3 cd +
t:0=−− 2 ad
Solving these equations in terms of a gives
b= ac, = 0, d = − 2 a
a aa−2 a gl Π=3 glV = (e) V 2
If a = -1/2, then
V Π=3 =Fr gl
Combining these three equations gives
D ′ = f 22,Re,Fr 0 ρ lV
D = f ′′ ( Re, Fr) ρ lV22
Dimensional analysis
8-32 Ch. 8 Similitude and Dimensional Analysis
~ no clue to the functional relationship among Dρ lV22, Re and Fr
~ arrange the numerous original variables into a relation between a smaller number of
dimensionless groups of variables.
~ indicate how test results should be processed for concise presentation
[Problem 8.48] Head loss in a pipe flow
f( hL ,,,,,, Dlρµ V g) = 0
Pipe diameter Repeating variables: lV,,ρ
Π=11fhl( L ,,ρ , V)
Π=21f( Dl,,ρ , V)
Π=33f(µρ,, lV , )
Π=44f( gl,,ρ , V)
ab c d (i) Π=1 hlL ρ V
cd 000 abML M Lt= LL Lt3
Mc:0=
L:0=+− ab 3 cd +
td:0= − ba= −
a hL ∴ Π=1 l
8-33 Ch. 8 Similitude and Dimensional Analysis
hL If a =1: Π=1 l
ab c d (ii) Π=2 Dlρ V
cd 000 abML M Lt= LL Lt3
Mc:0= ①
L:0=+− ab 3 cd + ②
td:0= − ③
② : 0 =ab + ba= −
a D ∴ Π=2 l
D If a =1 : Π=2 l
ab c d (iii) Π=3 µρlV
a cd 000 M b ML M Lt= L LT L3 t
= + ① → = →=− M:0 ac cd c a =−+− + ② L:0 ab 3 cd
t:0=−− ad→ da= − ③
② db+−30 dd + =bd=→=− b a
aa−− a − a ∴ Π=3 µρlV 8-34 Ch. 8 Similitude and Dimensional Analysis
lVρ If a = −1 ∴ Π= =Re 3 µ
ab c d (iv) Π=4 glρ V
a cd 000 L b ML M Lt= L t23 Lt
Mc:0= ①
L:0=+− ab 3 cd + ②
t:0=−− 2 ad ③
③ da= −2
② 0=+−−ab 02 a→=ba
l a aa−2 a g Π=4 glV =2 V
1 V If a = − : Π=4 =Fr 2 gl
hlL f , ,Re,Fr= 0 lD
hlL = f ′,Re,Fr lD
8-35 Ch. 8 Similitude and Dimensional Analysis
Homework Assignment # 8
Due: 1 week from today
Prob. 8.6
Prob. 8.10
Prob. 8.14
Prob. 8.20
Prob. 8.24
Prob. 8.30
Prob. 8.56
Prob. 8.59
8-36