Hello welcome to my talk, on physical
modeling, Froude similitude for physical modeling of marine structures.
This is the last part of three talks on this topic, and in this talk I will
introduce what are the criteria for physical model test and the examples
illustrate how these criteria can be met.
This is the video for a model test of two WaveScan models, which are in the
different scale ratios, 1:8 and 1:16, respectively. This wave tank
test was conducted in the large ocean tank at Ecole Centrale
de Nantes in France. This wave tank is 50m long, 30m wide and 5m deep.
These two buoys are very small models tested in such a large wave tank, but
the purpose of the test is to examine the wave buoy's performance in large waves,
especially in the extreme waves. So we may ask one question: what is the
smallest scale model suitable for a tank test on principle? here we are not
talking about the practical operations in the test model, for example,
in this test, the smaller WaveScan model has a diameter of 0.17 meter and
a weight about 0.7 kg. On the larger model we mounted 3 fluorescent balls,
but on the smaller model we have only two.
Suppose we have the geometrically similar systems, S1 and S2, here S2
is the target small-scale model for a tank testing,
the dynamic similarity would require the Froude similitude, for which the Froude
numbers for S1 and S2 must be same as this, and we can cancel out
the constant, the gravitational acceleration, g, so we have the relation
in this form. so if we take the system 1 as the full scale model, and the
system 2 as the scale model, and then we can define the scale factor EPSILON as this,
and based on Froude similitude, we can obtain the ratio of the
velocities for the full scale device and the model scale device, given in this
form, which is the square root of the model scale factor EPSILON.
For example if we have a scale model of a scale factor: EPSILON equals to 50, so we
can see the full scale system as S1 in this, and the scale model would be
like this, very very small, so we can enlarge this, we can see the details of
the system S2 or the scaled model S2, so we can calculate the velocity
requirement for the scaled model based on the formula as this.
so this means, based on the Froude similitude, the velocity of the scale
model is about 1/7 of that of the prototype. This is very in favour of
the scale model test in the test tanks, because we have a smaller
velocity. And based on this, we can calculate
the Reynolds number for the scale model, Rem, defined as this, and use the
scale ratio for the velocity and the length, we got the expression as this, so
we can see the Reynolds number for the model is 1/350 of the
full scale model. This means, based on the Froude similitude, the Reynolds number
for the scaled model would be much smaller than that of the prototype.
Now we return back to the USS aircraft carrier, the overall length of the carrier
is 333m and a displacement 100,000 long tons, and the maximum speed
over 56 km/h, that is, 30 knots, or 15.56 m/s.
so if we supposed a scale factor of 50 used for testing the carrier model
in the towing tank, and based on the Froude similitude, we can calculate
the model length, given by Lm, so it is 6.66m, this is
a suitable for many tank testing; and the model mass is calculate by this m_m, so
it is slightly more than 800 kg, again this is a mass of the model
suitable for many tank testings; and the model maximum speed Um is
accurated in this formula, it is 2.2 m/s. And this is very good for the tank
testing. So for such a model, the size, mass and maximum speed all very
suitable for tank testing, and in fact, it can be tested in many towing tanks in
the world. For a comparison the Reynolds number for the full scale model
is calcualted as about 5 billion, and the scale model is about 15 million.
Again the Reynolds number of the scale-model is much smaller than
that of the full scale ship.
Now we look at the Reynolds similitude. Based on the dynamic similarity, the Reynolds
similitude for these two geometrically similar systems S1 and S2
requires the Reynolds number for both systems S1 and S2 are equal,
expressed as this. Since we normally use same fluids for the full scale and the
scaled model. For example, the sea water and the fresh
water in the tank, then we can assume the densities are same and the fluid viscosities
are same, thus we have the relation based on the Reynolds similitude as this,
and we can calculate the model speed from the Reynolds similitude, we have
the expression as this, therefore, the Reynolds similitude would
lead (to) a larger model speed than that of the full scale model, hence this condition is
not easy to be satisfied for a smaller model.
Same as the previous example, take the scale factor
EPSILON as 50, based on the Reynolds similitude, the velocity for the
scale model is 50 times of that of the full scale ship.
and this relation could produce a practical problem on how we can achieve
such a velocity for the model. And for a comparison, we can calculate
the Froude number for the scale model given a this, based on the
Reynolds similitude, and from this calculation we can see
that Froude number for the scale model would be 350 times larger than
that of the full scale ship. This means based on the Reynolds
similitude, the Froude number for the scale model would be larger than that of
the prototype.
So again we look at the USS carrier, and based on the Reynolds similitude, we
take the scale factor as 50, so based on the Reynolds similitudes,
we can calculate the model length as this, so this is the same length as that for
the Froude similitude, because this is geometrical scaled model, and the
model mass given as this, this weight is still good for the tank test,
and the model maximal speed is calculated as this, so we can see the model speed the
requirement would be 777 m/s, this is a huge speed for a
model. So it can be seen the maximal speed requires for the scale model
based on the Reynolds similitude would be 777 m/s, which is
so high, no testing facility can satisfy this condition. And based on the Reynolds
similitude, we can calculate the Froude number for the full scale model,
we got 0.272, but the scale model we got the Froude
number 96.25, so we can see the Froude numbers for the
full scale and the scale-model are very different.
From the previous two examples, we can see, it is a reality, for a scale model, we
cannot achieve the Reynolds similitude and the Froude similitude at same time.
However the similitude laws require both similitudes. In this slide, an
interesting example would show: how we can achieve the Reynolds and the Froude
similitudes at same time. This is the proposed Soding's 'Sauna Tank', for which
the water is supposed to be heated up to 90 degrees Celsius for the model test.
For the full scale ship, the water is supposed at 20 degrees Celsius, while the
scale model at 90 degrees, so we can see the density would reduce by about 3%
from 999 to 966 kg/m^3, but the viscosity would reduce
from MU_1 to MU_2: MU_2 is less than 1/3 of MU_1. To maintain the
Reynolds number, we have this relation, and we can work out this relation, using the
different densities and viscosities. To maintain Froude similitude, we have the
relation as this, and this. So if we put these together, we can
obtain the scale ratio EPSILON as 2.11.
This means at 90 degrees Celsius if the model scale factor is taken as 2.11,
which is roughly a 1/2 model, the Reynolds and the Froude similitude can
be both satisfied. But in reality there's no such 'Sauna Tank', because it would
be expensive to heat the tank up to 90 degrees Celsius, and in many cases
the scale factor 2.11 is too small for practical model test.
In this slide we will examine the practical similitude: on how we can
achieve this in practical applications. Look at the non-dimensional
navier-stokes equation, we can see, we have 2 non-dimensional parameters in
the equation, that is, Froude number and the Reynolds number, defined in these
formulae. However, for most turbulent flows,
Reynold number Re would be large, due to the small viscosity for both
water and air. So based on the recommendation in the
reference here, we can assume when Reynolds number is larger than 100,000, the fluid
viscosity effect, the terms with the Reynolds number would be very small when
compared to other terms, therefore, the viscous terms can be neglected.
In the case of a high Reynolds number, the non-dimensional navier-stokes
equation can be simplified as this, so this non-dimensional navier-stokes
equation for the transformed standard system can be governed by the Froude
number only. This means in the scale model, the dynamic similarities can be
guaranteed by the Froude similitude alone. As we have shown in the previous
examples, the requirement of the Froude similitude can be easily
satisfied for the scale models, this is why for most of the model tests of
marine structures, Froude similitude is the sole similitude,
and the Froude similitude alone could guarantee the dynamic similarities for
the scale-model test.
In this slide, we are looking at what could be the minimum model size. Take
the USS aircraft carrier as an example, from the recommendation for the
model Reynolds number Rem= 100,000, and we also have the full
scale Reynolds number as 5.2*10^9.
so to achieve the required Reynolds number for the scale model, the scale factor
can be calcualted as this, so we can obtain the scale factor, would be about 1400.
so based on the scale factor, we can calculate the model length Lm, given by
this, 0.239m; and the model mass calculated
by this, would be 0.0376kg; and based on
the Froude similitude, the model's maximum speed would be given by
this, it is 0.417 m/s. So this extreme model
would be very small: it is smaller and lighter than a 10-inch tablet. This example
is to illustrate the critical Reynolds number for the scale model can be easily met.
Based on the Froude similitude, we can easily work out the scaling
factors for different physical parameters, see in the table. Some of
these parameters would be used to scale the parameters to set up the test
environment. Take the example: EPSILON equals to 50,
for a cylindrical oscillating water column wave energy conveter. In the test, we need
to scale the wave parameters: the wave height needs to be scaled down by 50
here, so if the wave height in the sea is 3m, in the tank, it would be 60mm,
the wave period is scaled down by the square root of the scale
factor, thus the ocean wave period of 12s
would be about 1.7s in the tank. The other use of these scale factors is
scaling the parameters from the model test up to the full scale, for instance
the force obtained from the scale model should be enlarged by 125,000 times
for the full scale structure, which is given by the scale factor,
power of 3.
This is the cylindrical oscillating water column model test in regular
waves, with the scale factor EPSILON, equalling to 50, the regular wave
height is 60 mm and the wave period is 1.7s, but the period of
pitch motion is double of the wave period, and this is the motion of
coupled motion of heave and roll motions.
