Hello, welcome to my talk. This talk is on the physical modeling technology, Reynolds
similitude, Part 2: wind tunnel test and examples.
In this talk, the focus is on how we can achieve the similitude laws in the wind
tunnel tests, especially for the model tests in the high speed flows. And in
this talk, I will use some examples to illustrate how we can solve the problems.
In reality, take an example of an aerofoil uniformly flying in the still air at a
speed U0. If we watch the flying aerofoil from a fixed point on the
ground, the reference coordinate, we will see the problem like this: the
flying aerofoil is moving in the air, and for an air particle at a certain
position, as seen in here. At the time t0, the flying aerofoil is far away
from the air particle, hence the air particle is still; and at the next
time step t0 + delta_t, the airfoil is flying close to the particle, and
push the air particle away; and at the time step t0+2 delta_t, the air
particle is just about the flying airfoil, and it would move slightly to the airfoil
surface as this. In such a reference coordinate, the problem is unsteady.
We can look at the problem in a different way: suppose we take the
reference point fixed on the flying aerofoil, see here the reference coordinate.
and in such a reference coordinate, the airfoil is fixed in the space and
the air is coming to the airfoil at a speed U0, and for an air particle in the
same position related to the fixed aerofoil, its motion in such a coordinate is
the same for different time t0 + delta_t and t0 + 2* delta_t,
so in this regard the problem would be steady. Luckily, we have the Galilean
transformation for these two different systems because of the uniform moving
speed. so the Galilean transformation would guarantee these two dynamic
systems are same. Therefore this transformation would simplify the
problem very much, and this is a same simplification as we have seen in
the numerical modeling.
Having said the Galilean transformation for the flying structure
in the air and the models in the wind tunnel, see the example, a MD-11 model in
the wind tunnel. Principally there should be same, due to the Galileann transformation
principle. In reality, there will be differences between the flying
structure in the air and the model test in the wind tunnel, including: the
structure flying in the still air, there is no turbulence in the still air, but in
wind tunnel the incoming flow is always with some turbulence. The question is: is
there an effect of the turbulence in the freestream in the wind tunnel? the answer is
'YES', so for the wind tunnel test, we needed to deduce the turbulence in the freestream;
for the wind tunnel test, the model is normally smaller than the flying
structure, therefore, we may have the problem of scaling effect.
Ideally we need to match the Reynolds number and the Mach number in the wind
tunnel test; the flying structure is in the open air, but the model test in
the wind tunnel is in the confined working section, therefore, there may be
blocking effect in the wind tunnel test. The solution would be a small model must
be used for the relevant working section;
the structure flying in the space is free from any other objects, but the
model test in the wind tunnel must be supported use some supporting systems,
therefore, we may have the influence from the supporting system in the wind tunnel
test. The solution is use the small or better supporting system.
if we have two geometrically similar systems, here S1 and S2, with
a scale factor EPSILON, given as this, so in the conventional wind (tunnel) test, the
air density and viscosity would be same for both systems, S1 and S2,
that is, RHO_1 = RHO_2 and MU_1 =MU_2. The Reynolds
similitude requires this equation, and we can cancel out the density and the
viscosity, so we have the expression as this: U20 for the scale model would
be the scale factor times U10, so, that is, for the scale model, since
EPSILON is always larger than 1, thus the Reynolds similitude would require
a higher air speed than that in the prototype. If we look at the Mach number
similitude, it requires the equation as this, and since the speed of the sound
in the conventional air is the same: a1 = a2, therefore, we have the
requirement for the freestream velocity U20 = U10, this means the
Mach number similitude would require a same air speed as that in the prototype,
regardless what the scale factor is used. So from this analysis we can see for a
scaled model, where EPSILON would be larger than 1, in the conventional
wind tunnel test, the Reynolds similitude and Mach number similitude cannot be
satisfied at same time. This is quite similar as we have seen in the flowude
similitude, but here we don't have a simple solution for such a modeling
problem as we have seen in the Froude similitude.
In this slide we will see how the Mach number and Reynolds number have the
effect on the flows. Take the airflow past an aerofoil as an example, for the
airflow, we have very different ranges of Mach number: in the low-speed flows, where
the Mach number is smaller than 0.3, then the classic incompressible flow can be
taken and the Mach number has no influence on the flow;
and we have the subsonic flow, where the compressibility needs to be
considered, and for a higher flow velocity, from transonic, to supersonic, to
hypersonic flows, different shock waves and expansions would be
developed and the problems become more complicated.
For the effects of Reynolds number on the flows, if the Reynolds number is
very small, less than 100, the flow is a creep flow, where the viscous effect
dominates the flow; when the Reynolds number is larger than 100 and less
than 100,000, the vortex shedding and separation bubbles may
appear; when the Reynolds number increases more
to larger than 100,000 and less than 1,000,000, then we have the transitional
flow. In this transitionalflow, both laminar
and turbulent boundary layers exist in the flow; when the Reynolds number is
larger than 1,000,000, the flow would become fully turbulent, the inertia
effect would dominate in this fully turbulent flow, with the delayed
separation; delayed stall etc.
In the wind tunnel model test, ideally we need to achieve the correct Reynolds
number and Mach number, so the flow can be correctly modelled or
at least we need to model the flow in the same category. For instance, using a
laminar flow to model a turbulent flow, you will never obtain the correct answer;
or if use the low speed flow to model the transonic flow, it's
not a correct choice.
Next we will discuss the flow modeling for different scenarios. The 1st
scenario is the low speed flows, in which the Mach number has a very
small or no effect on the flows.
The first example is the flow in a horizontal pipe, here the Moody diagram
is given based on the published experiment data for the friction
coefficient of the viscous flow in the horizontal pipe Cf, against the Reynolds
number. so from this plot we can see, there are three different types of
flows: the laminar flow,
transitional flow and turbulent flows.
When the Reynolds number is less than 2300, the flow is generally
laminar, and the friction coefficient is calculated by this
formula, and the pipe roughess has no effect on the friction
coefficient. when the Reynolds number is medium, the
flow is transitional and the fluid friction coefficient is
most complicated.
when the Reynolds number is large enough, the flow becomes fully turbulent, and
the flow inertia effects would dominate, and the corresponding friction
coefficient would be a constant, see here, which would be independent of the Reynolds
number. However for different relative roughness,
the corresponding Reynolds number for the fully turbulent is
different, and the higher the relative roughness, the larger the friction
coefficient, can see here.
The second example is the flow past an aerofoil, NACA 23012, at
different Reynolds number between 1.7 to 4.5 (million), All these flows
should be fully turbulent.
At the low angle of attack, say ALPHA < 8 deg, the lift
coefficient cl would be independent of the Reynolds number, can be seen here. At the
medium angle of attack, between 8 degrees to 12 degrees, the lift
coefficient would be weakly dependent on Reynolds number.
in the high angle of attack, ALPHA > 12 degrees, up to the full stall,
the lift coefficient is strongly dependent on the Reynolds number:
basically the higher the Reynolds number, the higher the maximum cl. The reason for
this is the higher Reynolds number could delay the flow separation, thus
a higher maximum lift coefficient, see here.
The 2nd scenario is the Mach number similitude.
the scaling issues in the Reynolds number have been widely studied and correlated
using the wind tunnel test data and the flight data, the methods have been
developed and experience has been accumulated, but there's no perfect
answer for this issue yet, the Mach number can be a more difficult scaling
factor, especially in the high speed flows, with the evident compressibility,
even in the relatively low Mach numbers. When the Mach number is less than 0.8, we
can still see the flow passing the aerofil, the supersonic flow can be
generated on the airfoil surface and thus the shock wave. And in the higher
speed flow, strong shock waves and expansions may happen, and this
phenomenon can be modlled if the collect Mach number is attained.
In the next few slides, the GOAHEAD project would be discussed.
GOAHEAD project is an EU FP6 project: GOAHEAD
short for Generation Of an Advanced Helicopter Experimental Aerodynamic
Database for CFD code validation, and I participated in the project when I was
with University of Glasgow, in the tunnel test and the database establishment. From
the title of the project, we can see there are two main research activities: the
wind tunnel tests of a 1/4 helicopter model and the data collection;
as well as the CFD code validation, use the collect experimental data. In this talk
I will focus only on the experimental campaign. Here I would like to thank
Dr. Schwarz of DLR, who sent me the files and gave me the permission to use the
GOAHEAD photos and plots.
The GOAHEAD helicopter model has been tested in the DNW wind tunnel of a working
section: 8m*6m. The 1/4 helicopter model is an assembled
model, the actual scale ratio is 1:3.88:
the fuselage is the slightly modified NH90 helicopter, with 422 pressure
sensors (130 unsteady and 292 steady sensors) and other measurement sensors;
4 instrumented 7AD blades for the main motor, with 122 pressure sensors
plus other measurement sensors; 2 instrumented BO 105 blades for the
tail rotor, with 38 pressure sensors and other measurement sensors.
In the experimental campaign, seven partners were involved in the wind
tunnel test. The main reason for this it would be very difficult for one partner to
provide so many sensors and different data acquisition systems.
In total, there are more than 710 sensors on the test model+ 3 PIV systems.
Different data acquisitions from different partners; the test operation is (was)
made by DLR, including: model assembly, instrumentation, test signal
transmissions and the data acquisition; one important issue is the data
synchronization to ensure the test data have been synchronized for the different
acquisition systems.
Basically in the experiment campaign, 4 typical flight condition have been
tested. In all wind tunnel tests, the tip Mach numbers for the main rotor and tail
rotor, are all same, as 0.617 and 0.563,
respectively. the flight conditions include the low-speed pitch up
condition, in which the helicopter forward speed is Mach number of 0.059.
the test includes the different angles of the fuselage; and the test includes
the cruise and tail shake condition, with the Mach number of 0.204,
in which the different fuselage angles were tested; for the dynamic stall
test, at Mach number 0.259, the different fuselage angles were
tested; and in the high-speed condition, with Mach number 0.28,
only one fuselage angle is tested. The database was established by the
University of Glasgow, with the total test data more than 400GB;
and the data postprocessor was provided for the database, together with
the comprehensive documentation for the database. More details can be found in
the GOAHEAD publications, the link here.
The 3rd scenario is the high Reynolds number wind tunnel test, in which we will
see how we can achieve the full Reynolds numbers or the large Reynolds numbers in
the wind tunnel test.
In the scale model test in wind tunnel, we may need to increase the Reynolds number
of the model to a large number or even equalling to the full Reynolds number. For
achieving the goal, based on the definition of Reynolds number, given in
here, we can increase the Reynolds number in different ways,
such as: increase the size of the model, so we can increase the Reynolds number, but
the question is: the model size would be limited by the working section size of
the wind tunnel; we can increase the free stream velocity U0, but this would
be limited by wind tunnel maximum velocity or the Mach number; we can increase the
density of the flow in the wind tunnel, for example, we can apply the high
pressure in the wind tunnel; or we can reduce the temperature of the fluid in the
wind tunnel. Based on this gas equation, we can achieve a large Reynolds number by
pressured wind tunnel or the cryogenic wind tunnel, and we can also reduce the
fluid viscosity, this can be done by reducing the fluid temperature in the
wind tunnel, such as, the cryogenic wind tunnel. This is different for the liquid
like a water, temperature would be increased for reducing viscosity, as shown
in my talk, 'Froude Similitude', the 'sunna tank'.
For achieving Mach number, we can use the Mach number definition: the velocity of
freestream U0, divided by the speed of sound, and the speed of
sound is given by this formula, here R and GAMMA can be regarded as
constants. As such, we can increase the Mach number of the model to a large
number, even equalling to the Mach number of the prototype, there are two different
ways: to increase the velocity of the free stream, the problem is we may
have a limit of the wind tunnel maximum velocity; another way for increasing the
Mach number is reduce the temperature of the wind tunnel, here by this equation, we can
see reduces the temperature we can reduce the speed of the sound, so this
is the special issue in the cryogenic wind tunnel.
To increase the Reynolds number in the wind tunnel test,
one way is to increase the pressure in the wind tunnel, such as those pressurized
wind tunnels. the conventional pressurised wind tunnel
might be operated under an appled pressure and the normal temperature, for
instance, 15 degrees (Celsius). Principally, the gas equation can be used
for this situation, here the gas constant and the temperature both are
constants, thus the air density would be proportional to the applied pressure, see
this figure. So for increasing the Reynolds number in
such a wind tunnel, for instance, the air viscosity would be taken as a constant,
since the change would be very small, thus the Reynolds number would be
increased proportionally to the density, thus to the pressure applied.
A typical range of Reynolds number in such a pressurized wind tunnel can be seen in
this figure, from the Reference [2]. For instance, if we keep the Mach number 0.2
unchanged, so in the conventional pressure, we may have
the Reynolds number 1.75 million. If we apply the pressure 3.85
atmospheric pressure, the Reynolds number would be increased to
6.8 million, 3.85 times large in the Reynolds number.
The more complicated wind tunnel test would be in the cryogenic wind tunnel, in
which the temperature can be lowered to -150 degrees Celsius or lower,
it should be noted in such a low temperature, Nitrogen would be used for
the fluid, and the pressure can be increased to be more than 4 bars.
Suppose the prototype structure in the conventional atmosphere in the air, so
the properties of the air would be: density 1.227 kg/m^3;
the viscosity is about 1.8*10^(-5) Pascal second.
the speed of sound is about 340 m/s.
so in the low temperature and the applied pressure, the Nitrogen would
have the properties as these: density would be 11.5 kg/m^3,
so the density is increased about 10 times; the viscosity here is reduced when
compared to the viscosity at the conventional atmosphere, so it's about
half of that viscosity; and the speed of the sound is reduced to 220 m/s,
it's about 0.65 times of the speed of
the sound in the conventional atmosphere.
The Reynolds similitude requires the Reynolds numbers for both prototype and
the scale model are equal, Re1 = Re2. Use the definition of for
Reynolds number, we have the equation as this. Here subscript '0' means the
properties in the conventional atmosphere; this side of the parameters would be
the parameters in the cryogenic wind tunnel. Based on this equation, we can
calculate the scale factor, EPSILON, given in this form, so we can calculate this
scale factor, based on the Reynolds similitude as this.
the Mach number similitude requires the Mach numbers for two systems are same: M1 = M2,
and use the Mach number definition, we have the
expression as this. thus, we can calculate the velocity ratio,
U0/U equals to 1.53. So we put these Reynolds and
Mach numbers similitudes together, we can obtain the desired scale factor,
EPSILON, as this: 13.11. This means in such a cryogenic wind tunnel,
we can apply the scale ratio of 13.11 to guarantee
both Reynolds and Mach number similitudes.
In this slide the European FP6 project, FLIRET, is introduced. FLIRET is short
for Flight Reynolds number testing. This project was led by the European transonic
wind tunnel, a cryogenic wind tunnel.
this plot is re-drawn, based on the data in the reference. Suppose we have an
aerofoil of a chorl length 0.22 meter. In the conventional wind tunnel, at the
test condition, 1 atmospheric pressure, at 12.5 degrees Celsius,
the Reynolds number would be 8 million.
if we apply the pressure in the wind tunnel only, up to 3.5 atmospheric pressure,
the Reynolds number can be increased to 28 million.
if we lower the temperature in the wind tunnel to -172 degrees Celsius at
the normal atmospherice pressure, the Reynolds number can be increased to 50
million. if we combine both high pressure and
low temperature in the cryogenic wind tunnel (this is the cryogenic
wind tunnel used for), the pressure is 4.5 atmospheric pressure, the
temperature is -172 degrees Celsius, the Reynolds number can be increased to
136 million, this is 56 times when compared to the
conventional test condition, and this Reynolds number is in the range of
Reynolds number for the full aeroplanes. For a given Reynolds number, we can choose
different the combinations of the pressure and the temperature for achieving the
Reynolds number. It should be noted that this type of
wind tunnel test would come at a cost: it would be very expensive for a test in
such a wind tunnel in both preparation and operation.
