3. Computer simulation of the Brown generator
Our model bases on one magnetic circuit, see fig.4 + 5.
The source of the magnetism is a permanent magnet of constant magnetic
. The magnetic flux due to his voltage is limited by the network of magnetic
resistances in the circuit consisting of the inner magnetic resistance
of the permanent magnet ,
of the magnetic resistance of the stray fields ,
and of the resistance
of the airgap plus the rotor which consists of a rotating piece of iron
in the air gap controling the magnetic flux .
We describe it as variable resistance
from position of the angle
during rotation. A coil in the magnetic circuit connected with a impedance
of constant resistance R gives off the magnetic field energy excited by
the rotating iron in the form of a time dependent electric current I(t).
By this current I the coil with n windings has a magnetic voltage =-n.I
. (The sign of
has to be minus because - as we will see later- the differential equation
of the generator reduces to an usual LR-circuit under stationary conditions
with =0 only
if this is the case.) In order to derive the differential equation of this
generator we have to find the flux
at the rotor and through the coil. Therefore we analyse the network at
the black dot in fig.5. At this node Kirchhoff's law holds
For the three magnetic resistors involved Ohms law applies
From the last two equations we obtain the magnetic voltage
whereby we used the abbreviations
and . As it will
be shown in the appendix
can be identified with the flux through the frontal surface of the permanent
magnet and b is a constant whose value depends from geometry and distance
l of the airgap, if the magnetic circuit is closed by the rotor. Therefore,
using (2c) and (3)
Using , U=R.I, =-n.I
and the definition
the differential equation for the generator is derived to
If we add a capacitor C parallel to the resistor R (see dotted capacitor
in fig.5) we would have to insert I=U/R + C.dU/dt in =-n.I
and in (4) and we would obtain an differential equation of second order
for the generator
where we abbreviated .
Before we solve these equations we have to specify and all other model parameters.
contains two components which are added on in series: is the magnetic resistance of the rotor which is neglected because -the time dependent magnetic resistance of the airgap- is much bigger. We split the last resistance again in in two parts: the constant resistance of the remaining airgap if the rotor closes the magnetic cycle nearby at in fig.4 and resistance of rest of airgap left open additionally at other angels. If we use the formula for the magnetic resistance of a volume of length l and area A which is filled with matter of the ferromagnetic constant , then we can estimate all values numerically. The values used in our calculation are compiled in tab.3.
it is neglected and we get at frequency f
(In order to facilitate iteration in our calculation the edges at
of the |cos|-function were smoothed in the intervals
by a biquadratic spline .
The coefficient c and d can be found assuming the function and its first
derivative to be continuous.)
In order to estimate the efficiency we have to calculate the electrical and the mechanical work done after half a cycle. The electrical work output at half a cycle of frequency f=1/T can be calculated to
The mechanical work input is
whereby the torque for a ferromagnetic
stick has been calculated to
In the third line we used the approximation ,
for ferromagnetic materials, where :=magnetic
permeability. Deriving the last line we used
and =B.A, i.e. the definitions
for the magnetic resistance and magnetic flux.
Starting condition of our cycle was t=0, I=0 at 50 Hz normally. A standard Runge Kutta library routine has been used which can be found in .
Our numeric results are periodic.The initial phase starting the generator plays a role only for low load, see upper trace in fig.6. The lower trace there shows current I versus time t for a rotation frequency of 50Hz and R= 240 Ohm. The diagram is periodic for half a cycle. We see that the generator produces a pair of current spikes of opposite polarity especially at angles where the rotor closes the magnetic circuit.The first spike is higher than the second. Fig. 7 show the electric and mechanic work versus time t at every moment measured at the resistor R at 50Hz and 240 Ohm. After half a cycle (0.01 sec) ~0.5 J are consumed by the resistor. Therefrom, we can calculate the mean electric output power of 50 W at 50 Hz delivered by the generator estimated by our calculation. The mechanic work done at this time is 2.3D-05 J. Therefrom we can estimate an efficiency 2.17D4. Fig. 8 shows versus . It is the magnetic work diagram of the coil for half a cycle and is identical to the electrical work delivered after half a cycle acc. to equ.8. It has an area of ~0.5 J. Fig.9 shows versus resistance R, fig. 10 shows versus frequency f.
Our model shows that overunity efficiency is possible principally from theory. We think that the tendency of our numerical results are correct surely, because a first glance at equ.6 for shows that the mechanical input work can be reduced arbitrarily choosing as high as possible without influencing the electrical output (equ.4) at all. This indicates the possibility of overunity efficiency without any numeric calculation. Furthermore, our calculation reproduces a curious behaviour described by Kromrey: The generator protects itself against being short circuited and produces less power at very low load, see fig.9. We believe that the output values are calculated in right order of magnitude. A calculation with 4 x 4 cm^2 cross section iron core resulted in 70 W output. If we realize that Brown used electromagnets for excitation whose iron cores are probably in the magnetic saturated stated to reduce back emf then it is very probable that these magnets have more than 1 T field strenght at their surface. And because output power goes with ~ I^2 or~^2 then it seem possible that his (probably RMS-) values can be correct. However, Brown's experimental input values can not be calculated correctly. This could have following causes:
Any friction is neglected in our model. For a real machine a input power loss P due to friction has to be added to the input. Its dependence is quadratic proportional to the rotation frequency, i.e. . The proportional constant has to be measured or estimated according experience.
2) saturation of iron:
Magnetic material shows a saturation. This can be accounted for in the equations by making dependent from in our equations according to the material chosen.
3) hysteretic losses and finite transmission time of magnetism
This problem is inherent and can be reduced only by appropriated choice of magnetic material.
4) eddy currents:
Although Brown took care to avoid this problem using laminated transformer cores his setup had metallic ground plates which were electrically excited by the rotation of the machine as he proved himself. Therefore, the use of non- metallic material in the neighborhood of the machine is recommended. If possible the permanent magnets should laminated as well - or should be made out of non conducting magnetic ceramic.
5) finite time of transmission of the magnetic flux
Browns generator used laminated transformer cores, which are probably to slow for the faster harmonics of the spikes. The problem can be overcame using appropriate iron core materials.
6) bad description of the magnetic resistance
We made the approximative assumption that the field across the surface of the magnet is constant. The assumption for the dependence of the magnetic resistance from angle is simple but arbitrary.
The sense of our model was to show qualitativ dependences with present data. With complete data the calculations can be improved by exact field calculation of the magnetic resistance at every angle -or- by using empirically measured values of the magnetic resistance for every . The measurement can be done by determing L at each angle and calculating the magnetic resistance acc. to the equation which can be found by comparing coefficients of the generator equation under stationary conditions and =0 with the ordinary L/R-circuit.
Appendix: Derivation of
and b from magnet data
1.) According to  the the magnetic field along the axis of an permanent
magnet obeys the formula
2.) If two identic cylindric magnets face one another with poles opposed, Br must be doubled at the point P.
3.) Using the same geometry like 2.) but connecting the magnets with soft iron on the back L has to be replaced by 2L and Br has to be doubled.
see Part 3 this article
Fig.4: Our model of the Brown-Ecklin generator
Fig.5: The Brown-Ecklin generator represented as closed magnetic circuit of magnetic flux current
Fig.6: Current I versus time of the Brown generator at 50 Hz and
lower trace 2 Ohm. upper trace 240 Ohm
Fig7: Work W vs. time t of the Brown generator at 50Hz and 240 Ohm load
Fig.8: The magnetic work diagram versus of half a cycle of 50 Hz at 240 Ohm load
Fig.9: Power P versus load R of the Brown generator at 50 Hz
Fig.10: Power P versus frequency f of the Brown generator at 240 Ohm load