of the following examples, the non dimensional heat transfer equation is
considered as an example for Laplace’s equation
dealing with a rectangular domain as depicted in Figure 1. There are eight
equal elements used implying eight nodes for the temperature and sixteen
for the interpolation of the flux according to the double nodes concept.
The nodes are represented as blue circles, whereas the dots represent
Gauss point coordinates for the numerical integration.
Figure 1: Meshing of the
rectangular domain. Geometrical Nodes (o) and Gauss points.
First, the boundary conditions are
They correspond to the example
As result, we get the temperature
field from Figure 2.
2: Temperature field for given BC.
Due to symmetry of both domain and
BC, the (analytical) solution is not dependent on spatial coordinate
and is just a linear function
y . Figure 3 shows the same
information as a contour plot. Besides, the flux vectors indicate the
direction and the quantity of the heat transfer. The calculation of the
flux is done by finite differences, therefore the results get inaccurate
on the boundary.
3: Temperature and flux field
next plot in Figures 4 shows the solution of the temperature field for the
boundary conditions from the course project as a contour plot.
applied, leading again to a linear solution as a function of
. In the interior of the domain, the temperature is evaluated using the
integral equation whereas values on the boundary are interpolated from the
nodal solution. The flux is determined numerically by the gradient of the
temperature field, depicted by blue arrows. Again, the boundary element
error gets obvious.
4: Temperature and flux field.
This boundary element error is
reduced by introducing smaller element sizes. The boundary of the
rectangle is now divided in sixteen equal elements (Figure 5).
Figure 5: Finer mesh of the
Again, the same boundary
conditions are imposed, and if Figures 4 and 6 are compared, it gets
obvious that the boundary element error has been reduced a lot.
6: Temperature field with a finer mesh size, compare with Figure 4.
problems that have been presented so far both possess linear solution.
Finally, a problem is presented where we cannot expect the solution to be
a linear function of spatial coordinates
The following BC are considered:
This means that on the right
boundary on the bottom part, some strong heat flux is entering the domain.
Besides, flux can only enter at the top and bottom boundary, since the
rest of the boundary is fully isolated. The evaluation is performed for
the interior of the domain, giving the temperature profile in Figure 7 and
the contour plot in Figure 8.
Figure 7: Temperature field for given BC
Figure 8: temperature and
fluxes for given BC.
If Laplace’s equation is
extended by a inhomogeneous part , one obtains Poissons equation
In the following consideration,
is considered to be a constant. If
is positive, it is representing a heat sink, whereas a negative value
stands for a heat source. If the boundary conditions as in the first
are applied again, and
one obtains the temperature field
in Figure 9. The solution is not linear as a function of y , but behaves
in a parabolic way. It gets clear from this solution that a positive
takes out energy of the system.
9: Interior temperature field for
, heat is introduced into the domain of the system, and therefore the
temperature level increases. (Figure 10)
10: Interior temperature field for
finds one of its major application in fluid acoustics, where time harmonic
behavior is described by Helmholtz equation
is the acoustic (excess) pressure and
the acoustic wave number. In general, there are two classes of acoustic
problems referring to the domain. One of them has a domain which is
enclosed completely by its boundary, called an Interior Problem. In the
other class, the domain is unbounded, thus going to infinity. In this
presentation, examples of both classes are shown in 2-D. A 2-D application
stands also for a domain with a constant thickness and corresponding
boundary conditions, such that the 3-D physics can be reduced to two
The so called Helmholtz resonator
just represents a tube with rigid walls and certain boundary conditions at
the two ends. Even though it can be threatened as a one dimensional
problem, it is also possible to calculate it in two dimensions. The domain
is shown in Figure 1 as well as nodal points. Between nodes 1-57, a rigid
wall is assumed, leading to Neumann boundary conditions which are equal to
zero. Between nodes 57-1, the system is excited by applying Dirichlet BC
with the amplitude equal to one. Note that this amplitude is normalized.
The excitation frequency is
Figure 1: Domain of the Helmholtz
The medium in the domain is
considered to be air. As a result, a sinusoidal
standing wave builds up in the domain with maxima and minima (Figure
2). The gradient at the rigid wall (
) is equal to zero. The peaks have an amplitude of about 1. In the contour
plot (Figure 3), the pressure gradient is shown as well. Maxima (red and
blue) and nodes (yellow) are a result of interference of incident wave and
its reflection at the rigid wall. When the frequency is lowered to 100Hz,
one obtains the solution from Figures 4 and 5. One of the resonance
frequencies is at 99Hz, thus we are operating in the close vicinity of
that, and therefore the amplitude increases from about 1 to nearly 15. All
of these numerical results in BEM have been verified by comparing its
solution to the analytical one. However, a discussion is beyond the
purpose of this presentation.
Figure 2: Acoustic pressure in
the Helmholtz resonator at 200Hz
Figure 3: Acoustic pressure in the Helmholtz
resonator at 200Hz
100Hz (Resonance at 99Hz)
Figure 4: Acoustic pressure in
the Helmholtz resonator at 100Hz
Figure 5: Acoustic pressure in the Helmholtz
resonator at 100Hz
Now a quadratic domain is
considered (Figure 6) as well as element nodes (o).
Figure 6: Domain for the
exterior acoustic field and BEM discretisation.
BC are applied on the whole boundary as follows: On the upper, lower and
right side, a rigid wall is assumed thus leading to zero boundary
conditions. The left wall vibrates with a normalized amplitude of one with
different frequencies. Therefore, an outgoing wave is generated and
propagating mainly to the left. However, the wave starts diffracting
around the corners and continues finally in all directions, even though
only a very low wave is propagating to the right. The blank rectangle is
at the position of the excluded domain.
following pictures, the acoustic pressure field and the intensity field is
depicted in the figures for different frequencies. It can be seen easily
that the near field distance is increased by increasing the frequency. The
near field characteristics are underlined by the numerous side lobes
occurring at high frequencies (
). Above 300Hz, the number of elements has to be increased in such a way
that sufficient results are obtained.
Figure 7: Acoustic pressure field
Figure 8: Intensity field at 100Hz
Figure 10: Acoustic pressure
field at 200Hz
Figure11: Intensity field at
Figure 12: Acoustic pressure
field at 300Hz
Figure 13: Intensity field at
Figure 14: Acoustic pressure
field at 520Hz with 16 elements per side length.
Figure 15: Intensity field at
520Hz with 16 elements per side length
Figure 16: Intensity field at
900Hz with 32 elements per side length.
Figure 17: Intensity field at
1800Hz with a 0.05 spatial field point resolution and 64 elements per side