SIMILARITY
• In order to obtain the same flow
behavior over prototype in wind
tunnel as that of model at real
operating condition, it is necessary to
establish the similarities between
them in three aspects.
TYPES OF SIMILARITIES TO BE ESTABLISHED
• Geometrical similarity
• Kinematic similarity
• Dynamic similarity
GEOMETRICAL SIMILARITY
• When the ratios of the dimensions of model
to corresponding dimensions of prototype are
same then the geometrical similarity is said to
be established between model and prototype.
• Let ‘L’ , ‘T’, ‘A’ and ‘V’ be the characteristic length, maximum thickness, surface area and volume of an aerofoil (model) & let ‘l’ , ‘t’, ‘a’ and ‘v’ be the corresponding dimensions of an aerofoil (prototype) then
L/l = T/t = r & A/a = r2 & V/v = r3
r – scaling factor
KINEMATIC SIMILARITY
• When the ratios of the flow parameters
( velocity, acceleration etc ) of an model to
corresponding flow parameters of aerofoil are
same along with their vector points
(directions), then kinematic similarity is said to
be established between them.
• Let (V1)m & (V2) m be the velocities of a fluid flow over a model and (A1) m and (A2) m be the acceleration of fluid over a model at point 1 and 2 respectively
‘AND’ (v1)p ,(v2) p, (a1) p and (a2) p be the corresponding
flow parameters over a prototype at their corresponding points 1 & 2 respectively, then
• (V1)m / (v1)p = (V2) m /(v2) p = R V
R V - Velocity ratio
(A1) m / (a1) p = (A2) m / (a2) p = R A
R A – Acceleration ratio
R A and R v need not to be same as they represent
different flow characteristics.
DYNAMIC SIMILARITY
• When the ratios of the forces acting on the
model to corresponding forces acting on the
prototype along with their vectors (directions)
are same, then dynamic stability is said to be
established between them.
• Let Fi, Fv and Fg be the inertia force, viscous force and gravity force at any point ‘A’ on model and fi , fv and fg be the corresponding forces acting on the prototype at corresponding point ‘A’, then
Fi / fi = Fv / fv = Fg / fg = Rf
Rf - Force ratio
FORCES ACTING ON A FLUID FLOW
• Inertia force• Viscous force• Gravity force• Pressure force• Surface tension force• Elastic force
INERTIA FORCE
• It is equal to the product of mass (m) and acceleration (a) .
• It acts in the direction opposite to direction of acceleration of fluid
Fi = m x a = ρ x volume x V/T = ρ x volume/T x V
= ρ x A x V X V = ρ x A x V2 = ρ x L2 X V2
VISCOUS FORCE
• It is present in the fluid flow problems when viscous is considered.
• It is the product of shear stress and flow surface area.
Fv = τ x A = μ x du/dy x A
= μ x (V/L) x A = μ x V x L
GRAVITY FORCE
• Gravity force exists on every matter on earth hence it is associated with fluid flow on earth.
• It is product of mass and acceleration due to gravity.
Fg = m x g = ρ x volume x g = ρ x L3 x g
= ρ x A x L x g
PRESSURE FORCE
• It is well considered in case of internal flow.• It is equal to the product of pressure intensity
and cross – sectional area.
Fp = P x A = P x L2
SURFACE TENSION FORCE
• It is equal to the product of surface tension
and fluid flow length.
Fs = σ x L
ELASTIC FORCE
• It is equal to the product of elastic stress and
fluid flow area.
Fs = k x A = k x L2
NON- DIMENSIONLESS NUMBERS
• Since only non dimensional numbers remains
unaffected by scaling of model, it is of great
importance and helps to obtain same flow
behavior as that of real operating conditions.
• These are obtained as the ratio inertia force to
remaining other five forces.
IMP. NON-DIMENSIONLESS NUMBERS
• REYNOLD’S NUMBER
• FROUDE’S NUMBER
• EULER’S NUMBER
• WEBER’S NUMBER
• MACH NUMBER
REYNOLDS NUMBER (Re)• It is defined as ratio of inertia force to viscous force.• It is of great importance for low speed incompressible flow
regimes.
Re = Fi / Fv = (ρ x L2 X V2 ) / (μ x V x L)
= (ρ x V x L) / μ = (V x L) / ν μ – Dynamic viscosity ν – kinematic viscosity
FROUDE’S NUMBER (Fr)
• It is defined as the square root of ratio of the inertia force to gravity force.
Fr = (Fi / Fg )1/2 = ((ρ x A x V2 )/(ρ x A x L x g)) 1/2
= V/(L x g ) 1/2
EULER’S NUMBER (Eu)
• It is defined as the square root of the ratio of inertia force to pressure force.
Eu = (Fi / Fp)1/2 = ( (ρ x A x V2 )/(P x A) )1/2
= V/(P/ρ)1/2
WEBER NUMBER (We)
• It is defined as the square root of the ratio of inertia force to surface tension force.
We = (Fi / Fs )1/2 = ( (ρ x L2 X V2 ) / (σ x L)) 1/2
= V / (ρ x L) 1/2
MACH NUMBER (M)
• It is defined as the square root of the ratio of inertia force to elastic force.
• It is of great for compressible fluid flow regimes.
M = (Fi / Fe )1/2 = ((ρ x L2 X V2)/(k x L2 )) 1/2
= V / (k /ρ )1/2 = V / a
Where a = (k /ρ )1/2
DIMENSIONAL ANALYSIS
• It is a technique of establishing a relationship between different physical quantities.
• So obtained physical quantities can be dimensional or non-dimensional.
Application in fluid flow• To derive a rational formulae for fluid flow problems.
• To check the dimensional homogeneity of fluid flow equations.
• To derive an equations in terms of non-dimensional parameters so that it can be applied to any model even after scaling.
• To plan the design of model for scaling.
DIMENSIONAL ANALYSIS TECHNIQUES
• RAYLEIGH’S METHOD
• BUCKINKHAM’S ‘π’ METHOD
BUCKINGHAM’S ‘π’ METHOD• It is more convenient method than Rayleigh’s
method as it can deal with large number of dependent variables.
• If there are ‘n’ no. of variables (dependent and independent) in a dimensionally homogeneous equation and if these variables contain ‘m’ fundamental dimensions (M,L,T etc.) then the variables can be arranged into (n-m) dimensionless terms which are called as ‘π’ terms.
PROBLEM 1
• Derive an expression for lift force (FL) and lift
coefficient (CL) generated by an aerofoil. Assume that
FL = f (ρ,V, l, μ,α)
ρ- density of air (kg/m3)
V- velocity of flow of air (m/s)l- chord length of an aerofoil (m)μ- dynamic viscosity (kg/m/s)α- geometrical angle of attack (degrees)
Step 1
• Form a function F1 such that
F1 = f1 (FL, ρ,V, l, μ,α) = 0
FL = f (ρ,V, l, μ,α)
FL - FL = f(ρ,V, l, μ,α) - FL = 0= F1
f1 (FL, ρ,V, l, μ,α) = 0 = F1
STEP 2
• Identify the number of variables (n) and its dimensions in standard form (M,L,T), no. of dimensions m=3
f1 (FL, ρ,V, l, μ,α) = 0
n=6
• FL - M L T-2
• ρ - M L-3 • V - L T-1
• l - L• μ - M L-1 T-1 • α - M0 L0 T0
STEP 3
• Generate a function ‘f’ in terms of (n-m) non dimensional parameters ‘π’.(n-m=6-3=3)
f1 (FL, ρ, V, l, μ,α) = 0 = F1 = f1 ( π1, π2, π3 )
• Identify the no. of repeating variables. Generally in case fluid flow problems repeating variables are ρ, V, D or l.
STEP 4
• Assign the value for π1, π2, π3 consisting of (m+1) variables and introduce the constants a1,b1,c1,…..c3 to repeating variables in π1, π2, π3 terms.
π1 = la1. Vb1 . ρ c1 .FL
π2 = la2. Vb2 . ρ c2 . μ
π3 = la3. Vb3 . ρ c3 . α
STEP 5
• Determine the values of the constants a1,b1,c1,…..c3.
• Taking π1 term
π1 = la1. Vb1 . ρc1 .FL
M0 L0 T0 = L a1 . ( L T-1 ) b1. ( M L-3 ) c1 . M L T-2
M0 L0 T0 = L a1+b1-3c1 . M c1+1 . T -b1-2
Solving for M, 0 = c1+1Solving for L , 0 = a1+b1-3c1+1Solving for T, 0 = -b1-2
c1=-1, b1=-2, a1=-2
• Taking π2 term
π2 = l a2. Vb2 . ρc2 . μ
M0 L0 T0 = L a2 . ( L T-1 ) b2. ( M L-3 ) c2 . M L-1 T-1 M0 L0 T0 = L a2+b2-3c2-1 . M c2+1 . T –b2-1
Solving for M, 0 = c2+1Solving for L , 0 = a2+b2-3c2-1Solving for T, 0 = -b2-1 c2=-1, b2=-1, a2=-1
• Taking π3 term
π3 = l a3. Vb3 . ρc3 . α
M0 L0 T0 = L a3 . ( L T-1 ) b3. ( M L-3 ) c3 . M0 L0 T0
M0 L0 T0 = L a3+b3-3c3+0 . M c3+0 . T –b3+0
Solving for M, 0 = c3+0Solving for L , 0 = a3+b3-3c3+0Solving for T, 0 = -b3+0 c3 =0, b3=0, a3=0
STEP 6
• Form the function F1 by substituting the values of constants a1,b1,c1,…..c3.
π1 = FL / ρ. V2 . l2
π2 = μ / ρ. V . l
π3 = ρ0. V0 . l0 . α
F1 = f1 (FL / ρV2l2 , μ / ρVl , α )=0
STEP 7• Determination of lift force FL .
F1 = f1 (FL / ρV2l2 , μ / ρVl , α )=0
F1 = f2 ( μ / ρVl , α ) - FL / ρV2l2 =0
FL / ρV2l2 = f2 ( μ / ρVl , α )
FL = ρV2l2 f2 ( μ / ρVl , α )
f2 ( μ / ρV l , α ) corresponds to lift coefficient (CL)
CL = f2 ( μ / ρV l , α ) = FL / ρV2l2 = FL / ρV2S
PROBLEM 2• Derive an expression for thrust (T) developed by a
propeller. Assume that T= f(ρ,V,D, μ,a,ω)
ρ- density of air (kg/m3)
V- velocity of flow of air (m/s)D- Diameter of a Propeller (m)μ- dynamic viscosity (kg/m/s)a- velocity of sound (m/s)ω- angular velocity (rad./s)
STEP 1
• Form a function F1 such that
F1 = f1 (T, ρ,V,D, μ,a, ω) = 0
T - T = f(ρ,V,D, μ,a, ω) – T = 0= F1
f1 (T, ρ,V,D, μ,a, ω) = 0 = F1
STEP 2
• Identify the number of variables (n) and its dimensions in standard form (M,L,T),no. of dimensions m=3
f1 (T, ρ,V,D, μ,a, ω) = 0
n=7
• T - M L T-2 • ρ - M L-3 • V - L T-1
• D - L• μ - M L-1 T-1 • a - L T-1
• ω - T-1
STEP 3
• Generate a function ‘f’ in terms of (n-m) non dimensional parameters ‘π’.(n-m=7-3=4)
f1 (T, ρ,V,D, μ,a, ω) = 0 = F1 = f1 ( π1, π2, π3 , π4)
• Identify the no. of repeating variables. Generally in case fluid flow problems repeating variables are ρ, V, D or l.
STEP 4• Assign the value for π1, π2, π3 consisting of (m+1) variables and
introduce the constants a1,b1,c1,…..c3 to repeating variables in π1, π2, π3 terms.
π1 = Da1. Vb1 . ρc1 .T π2 = Da2. Vb2 . ρc2 . μ π3 = Da3. Vb3 . ρc3 . a
π4 = Da4. Vb4 . ρc4 . ω
STEP 5
• Determine the values of the constants a1,b1,c1,…..c3.
• Taking π1 term
π1 = D a1. Vb1 . ρc1 .T
M0 L0 T0 = L a1 . ( L T-1 ) b1. ( M L-3 ) c1 . M L T-2
M0 L0 T0 = L a1+b1-3c1 . M c1+1 . T -b1-2
Solving for M, 0 = c1+1Solving for L , 0 = a1+b1-3c1+1Solving for T, 0 = -b1-2
c1=-1, b1=-2, a1=-2
• Taking π2 term
π2 = Da2. Vb2 . ρc2 . μ
M0 L0 T0 = L a2 . ( L T-1 ) b2. ( M L-3 ) c2 . M L-1 T-1 M0 L0 T0 = L a2+b2-3c2-1 . M c2+1 . T –b2-1
Solving for M, 0=c2+1Solving for L , 0=a2+b2-3c2-1Solving for T, 0=-b2-1 c2 =-1, b2 = -1, a2 = -1
• Taking π3 term π3 = Da3. Vb3 . ρc3 . a M0 L0 T0 = L a3 . ( L T-1 ) b3. ( M L-3 ) c3 . L T-1
M0 L0 T0 = L a3+b3-3c3+1 . M c3+0 . T –b3-1
Solving for M, 0 = c3+0Solving for L , 0 = a3+b3-3c3+1Solving for T, 0 = -b3-1 c3 = 0, b3 = -1, a3 = 0
• Taking π4 term π4 = Da4. Vb4 . ρc4 . ω M0 L0 T0 = L a4 . ( L T-1 ) b4. ( M L-3 ) c4 . T-1
M0 L0 T0 = L a4+b4-3c4 . M c4 . T –b4-1
Solving for M, 0 = c4+0Solving for L , 0 = a4+b4-3c4+1Solving for T, 0 = -b4-1 c4 = 0, b4 = -1, a4 = 1
STEP 5
• Form the function F1 by substituting the values of constants a1,b1,c1,…..c3.
π1 = T / ρ. V2 . D2
π2 = μ / ρ. V . D
π3 = a / V π4 = D.ω / V F1 = f1 ( T / ρV2D2 , μ / ρVD , a /V, D.ω /V ) = 0
STEP 6
• Determination of thrust force T .
F1 = f1 ( T / ρV2D2 , μ / ρVD , a /V, D.ω /V ) = 0
F1 = f2 (μ / ρVD , a /V, D.ω /V ) - T / ρV2D2 =0
T / ρV2D2 = f2 (μ / ρVD , a /V, D.ω /V )
T = ρV2D2 f2 (μ / ρVD , a /V, D.ω /V )
LIMITATION OF DIMESIONAL ANALYSIS
• Since we have to ourself consider the dependent variable for the physical quantities, any physical quantity/’s wrongly considered can lead to error in results.
• It doesn’t gives the complete information. It just gives relationship between the selected parameters and considered physical quantity.
PROBLEM 3
The thrust developed by the propeller prototype of
600 mm dia at 480 rpm is 300 N with a torque of 30
Nm and generates about 3 m/s forward speed. What
would be the forward speed, thrust and torque for a
model of 4.8 m dia at 120 rpm at sea level condition.
FORWARD SPEED
T = ρV2D2 f2 (μ / ρVD , a /V, D.ω /V )
Dp.ωp / Vp= Dm.ωm / Vm
V = R ω = π D N / 60
ω = π N / 30
Dp.Np / Vp= Dm.Nm / Vm
Substituting the values
(0.6 x 480) / 3 = ( 4.8 x 120 ) / Vm
Vm = 6 m/s
THRUST
Tp / ρpV2pD2
p = Tm / ρmV2
mD2m
Substituting the values
300 / (1.225 x 32 x0.62 ) = Tm / (1.225 x 62 x4.82 )
Tm = 76800 N
TORQUE
Efficiency = output / input = (thrust x velocity) / (torque x angular velocity) (Tp x Vp) / (τp x ωp) = (Tm x Vm) / (τm x ωm)
ω = π N / 30 Substituting the values
(300 x 3) / (30 x 480) = (76800 x 6) / (τm x 120)
τm = 61422 Nm
PROBLEM 3 If the drag (D) of a body is assumed as D=ρV2l2 f2 ( V l / ν) where ρ is density, ν is kinematic viscosity of fluid, l length of body, V velocity of flow, then determine the drag force experienced by the model in air
when its 1:8 scaled model tested in water gives drag of 220 N at 12 m/s flow speed.
It is given kinematic viscosity of air is 13 times that of water
and density of water is 810 times that of air.
D=ρV2l2 f2 ( V l / ν)
(V l / ν)m,a = (V l / ν)p,w
Vm, a = Vp, w x (lp, w / lm, a) x (νm, ax νp,w)
Substituting the values
= 12 x (1/8) x 13
Vm, a= 19.2 m/s
(D /ρV2l2 )m,a = (D /ρV2l2 )p,w
Dm,a = Dp,w x (lm,a / lp,w )2 x (Vm,a / Vp,w)2 x (ρm,a / ρp,w)
Substituting the values
Dm,a = 220 x 82 x (19.2 / 12 )2 x (1/810)
Dm,a= 45.9 N
PROBLEM 4
A scaled model of an aircraft with a scale ratio of 1:40 is tested in a water tunnel and gives the pressure drop of 7.5 kN/m2 . Determine the corresponding pressure drop of real model in air.
Take Density of air ρa = 1.24 kg/m3
Density of water ρw = 1000 kg/m3
Viscosity of air = 0.00018 poiseViscosity of water = 0.01 poise
Since Reynolds’ no. would be same for both model and prototype,
((ρ x V x L) / μ)m,a = ((ρ x V x L) / μ)p,w
Vp,w / Vm,a = (ρm.a / ρp,w) x (Lm,a / Lp,w) x (μm,a / μp,w)
Substituting the values
Vp,w / Vm,a = (1.24 / 1000) x 40 x (0.01 / 0.00018)
Vp,w / Vm,a = 2.755
Since Euler’s no. is the factor that remains same in both model and prototype and also consists pressure term it; can be used to determine the pressure drop.
(V / (P/ρ)1/2 )m,a = (V / (P/ρ)1/2 )p,w
Vm,a / (ΔPm,a /ρ m,a)1/2 = Vp,w / (ΔPp,w /ρp,w)1/2
(ΔPm,a /ρ m,a)1/2 = (Vp,w / Vm,a )x(ΔPp,w /ρp,w)1/2
Substituting the values,(ΔPm,a /1.24)1/2 = (2.755 )x(7500 /1000)1/2
ΔPm,a = 1.225 N/m2
PROBLEM 5
• Determine the scale factor (r) of a propeller which is capable of generating 80 kN thrust at 120 rpm if its scaled model generates 0.5 kN at 480 rpm in a wind tunnel test.
• Dp.Np / Vp= Dm.Nm / Vm
Vm / Vp = (Dm/Dp) x (Nm/Np)
Tp / ρpV2pD2
p = Tm / ρmV2
mD2m
Substituting the value of (Vm / Vp) and rearranging
Tm / Tp = (ρm/ρp ) x (Dm/Dp)4 x (Nm/Np)2
Substituting the values
80000/500 = (1 ) x (Dm/Dp)4 x (120/480)2
(Dm/Dp) = 7.11 = r
PROBLEM 6
The aerofoil of an aircraft is scaled to 1:40 and is tested in a wind tunnel. If the ratio of kinematic viscosity of model at real operating condition to that of prototype is 35 than determine the aircraft’s cruise altitude.
• Vm lm / νm = Vp lp / νp
Vm/Vp = (lp/l m) x (ν m / νp) = (1/40) x (35)
Vm/Vp = 0.875
Vm/am= Vp/ap
Vm /Vp = am/ap , am = 297.5 m/s
am = 297.5= ( γRT)1/2
T = 220.27 K
T = T0 –λh
T0 = 15°C = 288 K (sea level standard temperature)
λ = 6.50 K / km ( troposphere lapse rate)
h = 10.42km
SCALE EFFECTS
• The difference between the behavior of same physical quantity for different scale ratios is termed as ‘scale effect’. It can be positive or negative.
• Factors causing scale effect magnitude - type of problem - scale ratio - sometimes only predominant forces are considered
which may cause the considerable discrepancy.