The  Static  Field  Converter:  An   Engineering  Analysis

    In the drawing of the Static field Converter (hereinafter SFC), the dewar is a conventional glass dewar with it's reflective finish consisting of silver, or possibly chromium or titanium of a thickness < 1 micron, eliminating eddy currents induced by it's exposure to changing magnetic fields while retaining it's property of reflecting most radiant energy.

    The axles are hollow through which supports B support the magnet/steel energy source (hereinafter referred to as the primary). The rotor (as described in the patent) is supported by the axles which are supported by the superconducting bearings in such manner that the axle/rotor combination freely revolves around the primary. The superconducting bearings are supported by supports A. The rotor is magnetically coupled to the electric motor outside the dewar. The motor is of conventional design well known to the art as is the magnetic coupling.

    Surrounding the dewar and outside of the dewar are the sensing coils (hereinafter referred to as the secondary). They may be strategically designed and isolated from the dewar to minimize convection heat transfer due to I2R losses.
 
    Both the superconducting hemisphere and the superconducting bearings may be composed of quasi-crystals of of yttrium 1,2,3 of the type heavily researched by Intermagnetics General Corp. and Praxair Corp.

    The low-temperature refrigeration apparatus may be located at the top of the dewar and it's design is also well known to the art.

    The magnet will be one of the current neodymium rare-earth permanent magnets with an energy product of up to 45 MGOe and a residual flux in excess of 11,000 gauss. Common cold-rolled steel will serve as the flux guide because eddy currents will never complete their path in that metal.

    Let us examine a machine with a given set of parameters to get a glimpse of it's possibilities.

    Given: The diameter of the primary is 6 inches. Let us assume that the neodymium compound is a sphere of 4 inch diameter with one inch of cold rolled steel surrounding it. A sphere is not the ideal shape for this magnetic material, long cylindrical configurations are. However, even if the magnet was designed in it's worst possible configuration, it would lose less than 25% of it's energy product. Conservatively, the worst we could do is an energy product of 30 MGOe. Let's plug in some numbers.

          4 inches X 2.58 cm/inch = 10.32 cm                                     d = 10 cm          r = 5 cm
                                                                                                      d = diameter        r = radius

          The volume of a sphere = V = 4/3  X   pi  X  r3
                                              = 4   X  53 = 500 cm3

              30 MGOe = 3.0 X 107 ergs/cm3
              1 joule = 107  ergs                     (3.0 X 107 ergs/cm3)    X   (10-7 joules/erg ) = 3 joules/cm3

                                                                               (3 joules/cm3)  X  500 cm3 = 1500J

    1 J/s = 1 watt. Our primary will induce 1500 watts/revolution/s. How much of it is useable will depend upon the design of the secondary, because it will produce a fundamental wave form, with higher order harmonics, there will be an rms factor as well as inductance and I2R losses to be considered. Once again, let us consider the worst possible scenario. We will assume 50% efficiency. We will assume 750 watts/revolution/s are useable.

    Is that possible? Won't the magnet de-magnetize after one revolution? If not, when will the magnet de-magnetize?

    The conditions to which the magnet are subjected are not unique. They are well known and well understood. And once again they are a function of secondary design.

    The only de-magnetizing energy the primary will be subjected to is the de-magnetizing energy due to the secondary. How will the permanent magnet react to that de-magnetizing energy?

    Once again, let us assume the worst possible scenario. Let us assume the secondary generates a de-magnetizing field of equal magnitude to it's inducing field. 100% efficiency. It will still not de-magnetize a neodymium compound magnet. The magnet will operate on a minor hysteresis loop, generating milliwatts of power loss, but it will still return to it's original condition after the de-magnetizing energy has been removed, regardless of the frequency of the de-magnetizing field. These magnets also improve at colder temperatures1.

    As a given, let us assume the machine will operate at 3600 rpm.

                        60 rev/s  X  750 W/rev/s = 45 kilowatts

    We now have 45 kilowatts to spend. Let's look at our losses.

    The superconducting bearings inside the dewar produce virtually no heat. Even if the field of the secondary was equal to that of the primary (in reality it is nowhere near the magnitude of the field of the primary, particularly if that is by design), the hysteresis losses would be measured in milliwatts. The dewar has no direct link to the outside, so the great heat producers, such as the motor and secondary, can be isolated from the system.

    Accurate figures for the efficiency of a glass dewar were not available, however let's once again think the worst.

    Let's assume that 50 watts get transferred to the dewar by radiation and convection (don't forget the dewar is surrounded by an evacuated jacket). At 1% efficiency (once again assume the worst) we would have to spend 5 of our 45 kilowatts to maintain cryogenic temperatures.

    We are down to 40 kilowatts. There is also the electric motor. If the rotor velocity is 3600 rpm and the dewar is evacuated (extremely high vacuums are possible in glass dewars), once the rotor is brought up to it's rated velocity, once the necessary kinetic energy has been transferred to the rotor, the only energy necessary to keep it rotating at 3600 rpm is the energy it loses due to friction from both the bearings and whatever atmosphere remains in the vessel. Again: worst case scenario: 2 horsepower. We must spend 1.5 kilowatts more. We are left with 38.5 kilowatts to do with as we please.

Power Density Calculations (worst case scenario):

Dewar I.D. = 10"
Dewar O.D. = 12"
Dewar Height (including refrigerator) = 6'
Coil  d = 18"
Motor Volume = 200 in3

The volume of a cylinder = V2 = pi X rh + 200 in3         h = height

                                    = (3.14)(9)(72) + 200 in= 2,200 in3

Power Density = 40,000/2,200 = 18 W/in3
Energy Density = Very High Numbers/unit3

There is no reason not to increase the rotor velocity.
 
 
 
 

 

1R.J. Parker, Advances in Permanent Magnetism, (John Wiley, New York, 1990) pp 83-97, 328.


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