Wind Tunnel testing basics

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.