Abstract:
This patent discloses methods for increasing the power output from earlier disclosures regarding a heat engine based on temporary electrical and magnetic remanence by R. O. Cornwall, for example WO00/0064038 entitled “Thermodynamic Cycles and Method for Generating Electricity” and filed Apr. 19, 2000. Presented is a means of cancelling the power limitation caused by re-magnetisation field and methods for varying the turns-ratio of the output coils as the flux decays.

Description:
FIELD 
       [0001]    The disclosed technology relates to thermodynamic cycles, and in particular relates to improved power collection schemes from the collapse of magnetic fluxes in a sample for certain types of thermodynamic cycles. 
       BACKGROUND 
       [0002]    It has previously been proposed to use a sample having temporary magnetic remanence to perform a thermodynamic cycle in which a varying magnetic field is used to produce a magnetic flux in the sample during a first part of the cycle, with the field being removed during a second part of the cycle, leading to collapse of the magnetic domains in the sample and the creation of an independent magnetic flux. It has been proposed to use this principle to convert heat energy into electricity, and also for refrigeration. 
         [0003]    An example of a system of this type is described in International Publication No. WO 00/64038 entitled “Thermodynamic Cycles and Method for Generating Electricity” and filed Apr. 19, 2000. Described in WO 00/64038, energy is recovered from the sample once per cycle of the excitation field. In a first step of the cycle, the sample is magnetised, with very little corresponding temperature rise, as the sample is far from the ferromagnetic phase transition point of the sample. The sample is then demagenetised by the removal of the field from the sample, after which the temperature of the sample falls as thermal energy within the sample is expended in working to re-randomise the domains in the sample against the field arising from the remnant magnetism of the sample. After a short time, heat from the surroundings warms the sample, and the magnetic domains within the sample become randomly oriented, and this leads to the generation of an independent magnetic flux as the magnetic field arising from the alignment of the magnetic domains collapses. This independent magnetic flux delivers power to the field generation apparatus. 
       SUMMARY 
       [0004]    Disclosed below are representative embodiments of methods, apparatus, and systems that should not be construed as limiting in any way. Instead, the present disclosure is directed toward all novel and non-obvious features and aspects of the various disclosed methods, apparatus, systems, and equivalents thereof, alone and in various combinations and sub-combinations with one another. The present disclosure is not limited to any specific aspect or feature, or combination thereof, nor do the disclosed methods, apparatus, and systems require that any one or more specific advantages be present or problems be solved. 
         [0005]    In one exemplary embodiment of the disclosed technology, an apparatus for performing a thermodynamic cycle is disclosed. The exemplary apparatus comprises: a sample that exhibits temporary magnetic remanence; and a magnetisation apparatus for magnetising the sample, wherein the magnetisation apparatus is operable, during a first part of the thermodynamic cycle, to produce a cyclically-varying magnetising field comprising a wavetrain of a single or plurality of consecutive cycles, and to remove the magnetising field from the sample during a second part of the thermodynamic cycle, wherein demagnetisation of the sample during the second part of the thermodynamic cycle causes the generation of an independent magnetic flux. 
         [0006]    In another exemplary embodiment of the disclosed technology, an apparatus for performing a thermodynamic cycle is disclosed. The exemplary apparatus comprises: a sample that exhibits temporary electrical remanence; and a polarizing apparatus for polarizing the sample, wherein the polarizing apparatus is operable, during a first part of the thermodynamic cycle, to produce a cyclically-varying electrical field comprising a wavetrain of a single or plurality of consecutive cycles, and to remove the electrical field from the sample during a second part of the thermodynamic cycle, wherein depolarisation of the sample during the second part of the thermodynamic cycle causes the generation of an independent electric flux and a transient independent magnetic flux associated with the changing independent electric flux. 
         [0007]    In certain desirable embodiments, the periodically-varying magnetising field is a substantially unipolar or bipolar:—“cross-over” sinusoidal or half sawtoothed or pulse varying magnetising field. Preferably, the magnetising field produces a substantially exponentially-decaying remnant magnetic flux in the sample. The sample can be, for example, a ferrofluid, a ferroset slurry or a ferroset with heat conduction channels therethrough. In some embodiments, the wavetrain comprises at least five consecutive cycles of the magnetising field (e.g., at least ten consecutive cycles of the magnetising field). In certain embodiments, the magnetisation apparatus also comprises a power collection apparatus, in which power is generated during the second part of the thermodynamic cycle by the independent magnetic flux. 
         [0008]    In other certain desirable embodiments, the periodically-varying electric field is a substantially unipolar or bipolar:—“cross-over” sinusoidal or half sawtoothed or pulse varying polarising field. Preferably, the polarising field produces a substantially exponentially-decaying remnant electric flux in the sample. The sample can be, for example, an electret with heat conduction channels therethrough. In some embodiments, the wavetrain comprises at least five consecutive cycles of the polarising field (e.g., at least ten consecutive cycles of the polarising field). In certain embodiments, the polarizing apparatus also comprises a power collection apparatus, in which power is generated during the second part of the thermodynamic cycle by the independent electric flux or a transient magnetic field resulting from the independent electric flux. 
         [0009]    In another exemplary embodiment, a method of converting energy is disclosed. The exemplary comprises: providing a sample that exhibits temporary magnetic remanence; magnetising the sample, during a first part of a thermodynamic cycle, by producing a cyclically-varying magnetising field comprising a wavetrain of a plurality of consecutive cycles; and removing the magnetising field from the sample during a second part of the thermodynamic cycle, thereby allowing the sample to demagnetise, the demagnetisation of the sample causing the generation of an independent magnetic flux. In certain embodiments, the method further comprises the step of converting at least some of the independent magnetic flux into an electric current. 
         [0010]    In another exemplary embodiment, a method of converting energy is disclosed. The exemplary comprises: providing a sample that exhibits temporary electrical remanence; polarizing the sample, during a first part of a thermodynamic cycle, by producing a cyclically-varying polarizing field comprising a wavetrain of a plurality of consecutive cycles; and removing the polarizing field from the sample during a second part of the thermodynamic cycle, thereby allowing the sample to depolarize, the depolarizing of the sample causing the generation of an independent electric field. In certain embodiments, the method further comprises the step of converting at least some of the independent electric flux into an electric current. 
         [0011]    In another exemplary embodiment, a method of converting energy is disclosed. The exemplary comprises: providing a sample that exhibits temporary electrical remanence; polarizing the sample, during a first part of a thermodynamic cycle, by producing a cyclically-varying polarizing field comprising a wavetrain of a plurality of consecutive cycles; and removing the polarizing field from the sample during a second part of the thermodynamic cycle, thereby allowing the sample to depolarize, the depolarizing of the sample causing the generation of an independent electric field and by Maxwell&#39;s equations, a magnetic field. In certain embodiments, the method further comprises the step of converting at least some of the transient independent magnetic flux into an electric current. 
         [0012]    In another exemplary embodiment, an apparatus for performing a thermodynamic cycle is disclosed comprising: a sample that exhibits temporary magnetic remanence; and a magnetisation apparatus for magnetising the sample, wherein the magnetisation apparatus is operable, during a first part of the thermodynamic cycle, to produce a substantially periodically-varying magnetising field that is substantially unipolar or bipolar:—“cross-over” sinusoidal or half sawtoothed or pulse varying, thus producing a substantially unipolar or bipolar:—“cross-over” sinusoidal or half sawtoothed magnetic flux in the sample with a decaying exponential element at the discontinuities, and to remove the magnetising field from the sample during a second part of the thermodynamic cycle, wherein demagnetisation of the sample during the second part of the thermodynamic cycle causes the generation of an independent magnetic flux. The sample can be, for example, a ferrofluid or a ferroset with heat conduction channels therethrough. 
         [0013]    In a further exemplary embodiment, another method for converting energy is disclosed. The exemplary method comprises: providing a sample that exhibits temporary magnetic remanence; magnetising the sample, during a first part of a thermodynamic cycle, by producing a substantially unipolar or bipolar:—“cross-over” sinusoidal or half sawtoothed or pulse varying field, thus producing a substantially unipolar or bipolar:—“cross-over” sinusoidal or half sawtoothed magnetic flux in the sample with a decaying exponential element at the discontinuities; and removing the magnetising field from the sample during a second part of the thermodynamic cycle, thereby allowing the sample to demagnetise, the demagnetisation of the sample causing the generation of an independent magnetic flux. In certain embodiments, the method further comprises the step of converting at least some of the independent magnetic flux into an electric current. 
         [0014]    In another exemplary embodiment, an apparatus for performing a thermodynamic cycle is disclosed comprising: a sample that exhibits temporary electric remanence; and a polarising apparatus for polarising the sample, wherein the polarisation apparatus is operable, during a first part of the thermodynamic cycle, to produce a substantially periodically-varying polarising field that is substantially unipolar or bipolar:—“cross-over” sinusoidal or half sawtoothed or pulse varying field, thus producing a substantially unipolar or bipolar:—“cross-over” sinusoidal or half sawtoothed electric flux in the sample with a decaying exponential element at the discontinuities, and to remove the polarising field from the sample during a second part of the thermodynamic cycle, wherein depolarisation of the sample during the second part of the thermodynamic cycle causes the generation of an independent electric flux. The sample can be, for example, a ferroset with heat conduction channels therethrough or a plurality of ferrosets suspended in a non-conductive fluid. 
         [0015]    In a further exemplary embodiment, another method for converting energy is disclosed. The exemplary method comprises: providing a sample that exhibits temporary electric remanence; polarising the sample, during a first part of a thermodynamic cycle, by producing a substantially unipolar or bipolar:—“cross-over” sinusoidal or half sawtoothed or pulse varying field, thus producing a substantially unipolar or bipolar:—“cross-over” sinusoidal or half sawtoothed electric flux in the sample with a decaying exponential element at the discontinuities; and removing the polarizing field from the sample during a second part of the thermodynamic cycle, thereby allowing the sample to depolarise, the depolarisation of the sample causing the generation of an independent electric flux or transient magnetic flux resulting from the changing electric flux. In certain embodiments, the method further comprises the step of converting at least some of the independent electric flux or transient magnetic flux resulting from the changing electric flux into an electric current. 
         [0016]    In a further exemplary embodiment, another method for converting energy is disclosed. The exemplary method comprises: providing a sample that exhibits temporary magnetic remanence; magnetising the sample, during a first part of a thermodynamic cycle, by producing a substantially unipolar or bipolar:—“cross-over” sinusoidal or half sawtoothed or pulse varying field, thus producing a substantially unipolar or bipolar:—“cross-over” sinusoidal or half sawtoothed magnetic flux in the sample with a decaying exponential element at the discontinuities; and removing the magnetising field from the sample during a second part of the thermodynamic cycle, thereby allowing the sample to demagnetise, the demagnetisation of the sample causing the generation of an independent magnetic flux. In certain embodiments, the method further comprises the step of converting at least some of the independent magnetic flux into an electric current by superimposing a chopped, inverted copy of the induced current, at a frequency higher than the relaxation rate of the flux remanence, to cancel the re-magnetising field from the induced current to achieve higher output power. 
         [0017]    In a further exemplary embodiment, another method for converting energy is disclosed. The exemplary method comprises: providing a sample that exhibits temporary magnetic remanence; magnetising the sample, during a first part of a thermodynamic cycle, by producing a substantially unipolar or bipolar:—“cross-over” sinusoidal or half sawtoothed or pulse varying field, thus producing a substantially unipolar or bipolar:—“cross-over” sinusoidal or half sawtoothed magnetic flux in the sample with a decaying exponential element at the discontinuities; and removing the magnetising field from the sample during a second part of the thermodynamic cycle, thereby allowing the sample to demagnetise, the demagnetisation of the sample causing the generation of an independent magnetic flux. In certain embodiments, the method further comprises the step of converting at least some of the independent magnetic flux into an electric current by varying the turns-ratio of the output coil and/or the load resistance as the flux decays. 
         [0018]    In a further exemplary embodiment, another method for converting energy is disclosed. The exemplary method comprises: providing a sample that exhibits temporary electric remanence; polarising the sample, during a first part of a thermodynamic cycle, by producing a substantially unipolar or bipolar:—“cross-over” sinusoidal or half sawtoothed or pulse varying field, thus producing a substantially unipolar or bipolar:—“cross-over” sinusoidal or half sawtoothed electric flux in the sample with a decaying exponential element at the discontinuities; and removing the polarising field from the sample during a second part of the thermodynamic cycle, thereby allowing the sample to depolarise, the depolarisation of the sample causing the generation of an independent electric flux. In certain embodiments, the method further comprises the step of converting at least some of the independent electric flux into an electric current by causing the generation of a transitory independent magnetic flux from the change in the electric flux. In certain embodiments, the method further comprises the step of converting at least some of the independent magnetic flux into an electric current by superimposing a chopped, inverted copy of the induced current, at a frequency higher than the relaxation rate of the flux remanence, to cancel the re-magnetising field from the induced current to achieve higher output power. 
         [0019]    In a further exemplary embodiment, another method for converting energy is disclosed. The exemplary method comprises: providing a sample that exhibits temporary electric remanence; polarising the sample, during a first part of a thermodynamic cycle, by producing a substantially unipolar or bipolar:—“cross-over” sinusoidal or half sawtoothed or pulse varying field, thus producing a substantially unipolar or bipolar:—“cross-over” sinusoidal or half sawtoothed electric flux in the sample with a decaying exponential element at the discontinuities; and removing the polarising field from the sample during a second part of the thermodynamic cycle, thereby allowing the sample to depolarise, the depolarisation of the sample causing the generation of an independent electric flux. In certain embodiments, the method further comprises the step of converting at least some of the independent electric flux into an electric current by varying the load resistance as the flux decays. 
         [0020]    The foregoing and other objects, features, and advantages of the invention will become more apparent from the following detailed description, which proceeds with reference to the accompanying figures. 
     
    
     
       DESCRIPTION OF THE DRAWINGS AND APPENDIX 
         [0021]      FIG. 1  is a plant diagram [ 1 ] with heat exchanger  2 , working substance  3  (ferrofluid for example), the pump  4 , the power extraction area  5 . 
           [0022]      FIG. 2  shows a typical configuration of the temporary magnetic remanence device  1 , the working substance  3 , the co-material  6  and heat conduction holes  7  should the working substance be solid and not able to be pumped and the magnetising coils  8  and/or the power output coils  9 . 
           [0023]      FIG. 3  shows a bipolar sinusoidal “cross-over” excitation waveform  10  (or plurality of as a wavetrain) and the response of the temporary remanence working substance (typically ferrofluid)  11 , with the temporary remanence effect  12  after zero exciter waveform input. 
           [0024]      FIG. 4  is a typical circuit for the bipolar sinusoidal “cross-over” excitation waveform with SCR/triac  13 , trigger waveform  14 , storage capacitor  15 , electrical output load  16  and a tap from the output  17  to recharge the storage capacitor  15  with the energy and current  25  generated from the working substance  3 ,  6 . The switch  18  is only closed when the exciter waveform  10  is off to avoid needless transformer action and power diversion of the exciter circuit into the load, so that only the dipole work from the decaying remnant flux  12  goes to the load  16 . 
           [0025]      FIG. 5  is a unipolar half-sawtooth or pulse excitation waveform  10  (or the H field) and the response of the temporary remanence working substance  12  (the magnetisation M). The graph shows their combined magnetic fields summed as by B=μ 0  (H+M). 
           [0026]      FIG. 6  is a typical circuit for the unipolar half-sawtooth or pulse excitation waveform. The ramp-on is controlled by the transistor (typically)  19  which is switched by the control signal  14  (shown bipolar to switch the transistor off “hard”) and the magnetising coil&#39;s  8  inductance. A “flyback” circuit composed of a small coil around the arrangement  3 ,  6 ,  8 ,  9  and wound in anti-sense in series with the diode  20 , captures the field energy as the magnetising field collapses and returns it to the storage capacitor  15 . 
           [0027]      FIG. 7  shows a graph of energy into a resistive load  21 ,  22  per cycle vs. 1/R along with graphs of i(t) vs. t and M(t) vs. t on the lower half of the page. The upper graph shows the linear “Faraday Law” portion  21  and the plateau phase  22  and the energy input as magnetisation energy  23 . 
           [0028]      FIG. 8  shows a graph of energy into a resistive load per cycle vs. 1/R with the Field Cancellation Method along with graphs of i(t) vs. t and M(t) vs. t on the lower half of the page. 
           [0029]      FIG. 9  shows a series of sub-figures illustrating the method of field cancellation in the time domain: 
           [0030]      FIG. 9A  shows the current  25  in the power output circuit ( 9 ,  16 ,  18 ) being copied, inverted and chopped (second sub-figure on right) to form the field cancellation current  26 . This ( FIG. 9B ) is then added on to a copy of the current to leave a chopped positive waveform of the current (second sub-figure on right). The magnetic field from this current impinges on the working substance  3 ,  6  where it is lowpass filtered and due to the high frequency content from the chopping, the magnitude of the re-magnetising field (see description) is much reduced. 
           [0031]      FIG. 9C  shows the same scheme as  FIGS. 9A and 9B  (clockwise from the top left) but with a more negative copy of the current  25  and much high frequency to leave virtually no response to the re-magnetising field. 
           [0032]      FIG. 10A  shows the method of field cancellation in the frequency domain. The topmost figure shows the frequency spectrum of the current  25 . The middle figure shows the spectrum of the inverted chopped signal. The lower figure shows the summation of the field from the current  25  and the field cancellation method current  26 .  FIG. 10B  how the signals are filtered by the field cancelling and power output circuits. 
           [0033]      FIGS. 11A , B and C explain the field cancellation method with equivalent circuits. 
           [0034]      FIG. 12  shows a graph of the variation of parameters and energy output per cycle with the Field Cancellation Method at the plateau portions  22 ,  23  and  24  of  FIG. 8 . 
           [0035]      FIG. 13A  shows the Field Cancellation Circuit. A high voltage source (powered ultimately by the energy obtained from the working substance) is periodically switched  28  into storage capacitor  29 . A current mirror  32  copies current  25  and chops it with high frequency signal  33  to make field cancelling current  26  flow through the magnetising coil  8  or another coil if the same coil is not desired to be used from the magnetising part of the cycle (as opposed to the demagnetising power step of the cycle). A flyback circuit  30 ,  31  makes sure that the chopping is done efficiently, i.e. re-generatively. The sum of the fields from coil  8  (now the H-field cancelling field) and coil  9  (which produces the re-magnetising field) acts on the working substance  3 ,  6 . A choke/inductor  34  acts to stop power being wasted from the H field cancellation circuit to the Power output circuit. 
           [0036]      FIG. 13B  an example of a current mirror made from bipolar transistors producing one third of the copied current which can be advantageous in the design of the circuit. 
           [0037]      FIGS. 14A , B, C and D show the Non Linear Method and energy into a resistive load per cycle with the resistance varying over the cycle and also the turns ratio varying.  FIG. 14D  shows a further advantage of the method in speeding up the decay waveforms at low resistance/high inductance. 
           [0038]      FIG. 15  shows the circuit for the Non Linear Method. 
           [0039]      FIG. 16  shows an embodiment of the electrostatic temporary remanence device as a schematic showing the plates  36 , the electrostatic working substance  3 , the electrostatic co-material  35  and the non-conductive co-magnetic  6  embedded as a plurality of preferably small entities with preferably high anisotropy in the direction of the transient magnetic field. 
           [0040]      FIG. 17  shows the an embodiment of the electrostatic temporary remanence device as a capacitor with power take off by magnetic power coils  37 , the magnetic co-material  6 , the electrostatic co-material  35  and heat conduction channels  7 . 
           [0041]      FIG. 18  shows the electric and magnetic fields associated with the electrostatic temporary remanence device as a capacitor. The device has heat conduction holes  7  and power coil windings  37 . 
       
    
    
       [0042]    Appendix 1 contains the MatLab computer simulation code for the magnetic temporary remanence device state equations. 
       DETAILED DESCRIPTION 
       [0043]    This document concerns improvements to a method of generating electricity given by the plant diagram  1  in  FIG. 1 . Heat energy from the environment via an heat exchanger  2  (between C and D) and heats a working substance  3  which is pumped  4  around a circuit to the power extraction area  5  (between A and B). In region  5 , the working substance performs dipole work, generates electricity and cools. The cycle then continues via the pump  4  to the heat exchanger  2 . 
         [0044]    The power extraction area has the working substance  3  in a typical configuration given in  FIG. 2  of the working substance and magnetic co-material  6 , which boosts the susceptibility. If it is desired to have a solid working substance, heat conduction channels  7  are drilled or otherwise formed through it to allow heat transfer fluid to execute the circuit ABCD given in  FIG. 1 , instead of the working fluid itself. Around the power extraction area  5  are magnetising coil  8  and power extraction coil  9 . 
         [0045]    With reference to  FIG. 13A , we shall see later that additional coils such as the flyback coil  30  for the Field Cancellation Method are placed around the same former and discussion of this is left until later. 
         [0046]    With reference to  FIG. 3 , the basic cycle consists of a magnetisation phase where the coil  8  is energised to provide a substantially unipolar or bipolar:—“cross-over” sinusoidal or half sawtoothed or pulse varying field  10 , thus producing a substantially unipolar or bipolar:—“cross-over” sinusoidal or half sawtoothed electric flux  11  in the sample  3 . The form of the field  10  can be singular pulses or a pulsetrain of the said pulses. 
         [0047]    With reference to  FIG. 4 , the second half of the cycle results in the excitation field being substantially off. A temporary remnant flux  12  results in the sample which is amenable to dipole-work that will directly convert heat energy into electricity via the power output circuit elements  3 ,  6 ,  9 ,  16 ,  18  and  25 . The switch  18  is only closed when the excitation waveform  10  is zero to avoid transformer action and wasting of the power in the excitation circuit, which runs in an efficient, recuperative manner. 
         [0048]    In this section we explain a temporal phenomenon that seems to limit the dipole-work to being under the supplied magnetisation energy—if the load resistance is only linear; the implication being that a non-linear impedance will allow the dipole-work to exceed the magnetisation energy input or by some other method (the Field Cancellation Method). 
       The Electrical State Equations for the Magnetic Temporary Remanence Cycle 
       [0049]    A mathematical model can be constructed for the working substance and electrical output circuit. Let us first consider the ferrofluid flux decaying into a linear resistor. 
         [0050]    Reference  FIG. 4  (and the description text earlier) elements  3 ,  6 ,  9 ,  16 ,  18  and  25 : 
         [0051]    The flux linkage is given by (μ r  is the relative permeability): 
         [0000]      λ=NAB           NAμ   0 μ r ( H+M )   eqn. 1
 
         [0052]    The magnetic field is given by: 
         [0000]    
       
         
           
             
               
                 
                   H 
                   = 
                   
                     
                       N 
                       D 
                     
                      
                     i 
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
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                   2 
                 
               
             
           
         
       
     
         [0053]    (reference is also made  FIGS. 3 and 5 ) 
         [0054]    Where i is the current through the coil, N is the number of turns and D is the length. The ferrofluid or super-paramagnetic material in general obeys a 1 st  order equation and implicit in this is the convolution of the H field  10  with magnetisation M  11 ,  12 . Advantageously, the H field  10  is switched on slowly relative to the relaxation rate of the working substance  3  so that dissipative processes are minimised.: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                        
                       M 
                     
                     
                        
                       t 
                     
                   
                   = 
                   
                     
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                         1 
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                       ( 
                       
                         M 
                         - 
                         
                           χ 
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         [0055]    That is, the rate of change of the magnetisation is negatively proportional to the existing magnetisation minus the driving contribution of the magnetic field (boosted by the susceptibility χ and permeability of the co-material 6 μ r l ), thus when H is substituted, the following is obtained: 
         [0000]    
       
         
           
             
               
                 
                   
                     
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         [0056]    The LR circuit  9 ,  16 ,  18 , on analysis considering the voltages yields the following, another state space equation: 
         [0000]    
       
         
           
             
               
                 
                   
                     
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         [0057]    Or substituting 
         [0000]    
       
         
           
             
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         [0000]    from eqn. 4, 
         [0000]    
       
         
           
             
               
                 
                   
                     
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                    
                   6 
                 
               
             
           
         
       
     
         [0058]    With reference to  FIG. 7 , the two equations eqn. 4 and eqn. 6 are suitable for coding on a digital computer and  FIG. 7  shows typical output from Matlab code (reference appendix 1, the parameters are at the start of the simulation code). The energy delivered to the load  16  per cycle by the dipole work  21 ,  22  is shown versus 1/R. We can see that there is a linear Faraday Law portion  21  where the energy is proportional to: 
         [0000]    
       
         
           
             
               
                 
                   
                     W 
                     dw 
                   
                   = 
                   
                     
                       ∫ 
                       0 
                       ∞ 
                     
                      
                     
                       
                         
                           
                             ( 
                             
                               
                                  
                                 λ 
                               
                               
                                  
                                 t 
                               
                             
                             ) 
                           
                           2 
                         
                         R 
                       
                        
                       
                           
                       
                        
                       
                          
                         t 
                       
                     
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   7 
                 
               
             
           
         
       
     
         [0059]    A simple electrical load only returns part of the magnetisation energy  23  and that dipole work plateaus  22 . We shall investigate this with the following theory: 
         [0060]    With reference to  FIG. 6 , simulation and experiment have found that a simple resistive load always returns less energy than the magnetisation energy input  23 . The work done magnetising is given by: ∫HdB·dV of which the “H” field energy is discarded, as this can be returned with total efficiency if done by a mechanical magnetisation process or very nearly so with an electronic process ( FIGS. 6 )  8 ,  14 ,  15 ,  19 ,  20  and  20   a,  leaving: 
         [0000]    
       
         
           
             
               
                 ∫ 
                 
                   M 
                   · 
                   V 
                 
               
                
               
                 
                   μ 
                   0 
                 
                  
                 
                   μ 
                   r 
                 
                  
                 H 
                  
                 
                     
                 
                  
                 
                   
                      
                     M 
                   
                   · 
                   
                      
                     V 
                   
                 
               
             
             = 
             
               
                 μ 
                 0 
               
                
               
                 HM 
                 ′ 
               
                
               V 
             
           
         
       
     
         [0061]    The integrand has been resolved with the relative permeability of the co-material  6  in close proximity to the working substance subsumed into M′. We can further write the integrand by M′=μ r χH as (dropping the primes): 
         [0000]    
       
         
           
             
               
                 
                   
                     E 
                     mag 
                   
                   = 
                   
                     
                       
                         μ 
                         0 
                       
                       
                         χμ 
                         r 
                       
                     
                      
                     
                       M 
                       2 
                     
                      
                     V 
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   8 
                 
               
             
           
         
       
     
         [0062]    With reference to the lower figures in  FIG. 7 , we shall now show that the lower returned dipole-work is due to the phenomenon of the slowing of the current waveform  25 : Taking the Laplace Transform of eqn. 4 and eqn. 6 and the set solved for I(s) in the s-domain to yields 
         [0000]    
       
         
           
             
               
                 
                   
                     ( 
                     
                       
                         with 
                          
                         
                             
                         
                          
                         L 
                       
                       = 
                       
                         
                           
                             μ 
                             0 
                           
                            
                           
                             AN 
                             2 
                           
                         
                         D 
                       
                     
                     ) 
                   
                   , 
                   
                     
 
                   
                    
                   
                     
                       I 
                        
                       
                         ( 
                         s 
                         ) 
                       
                     
                     = 
                     
                       
                         
                           DM 
                           0 
                         
                         N 
                       
                       
                         
                           
                             s 
                             2 
                           
                            
                           
                             τ 
                             ferro 
                           
                         
                         + 
                         
                           s 
                            
                           
                             ( 
                             
                               
                                 
                                   R 
                                   L 
                                 
                                  
                                 
                                   τ 
                                   ferro 
                                 
                               
                               + 
                               
                                 ( 
                                 
                                   1 
                                   + 
                                   
                                     
                                       μ 
                                       r 
                                     
                                      
                                     χ 
                                   
                                 
                                 ) 
                               
                             
                             ) 
                           
                         
                         + 
                         
                           R 
                           L 
                         
                       
                     
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   9 
                 
               
             
           
         
       
     
         [0063]    The dominant pole near the origin sets the dynamics, and a binomial series expansion of the roots of the denominator gives: 
         [0000]    
       
         
           
             
               
                 as 
                 2 
               
               + 
               bs 
               + 
               c 
             
             = 
             
               0 
               ⇒ 
               
                 s 
                 ≅ 
                 
                   
                     - 
                     
                       b 
                       
                         2 
                          
                         a 
                       
                     
                   
                   ± 
                   
                     
                       b 
                       
                         2 
                          
                         a 
                       
                     
                      
                     
                       ( 
                       
                         1 
                         + 
                         
                           
                             1 
                             2 
                           
                            
                           
                             ( 
                             
                               
                                 
                                   - 
                                   4 
                                 
                                  
                                 ac 
                               
                               
                                 b 
                                 2 
                               
                             
                             ) 
                           
                         
                         + 
                         
                           
                             1 
                             2 
                           
                           · 
                           
                             - 
                             
                               1 
                               2 
                             
                           
                           · 
                           
                             1 
                             
                               2 
                               ! 
                             
                           
                           · 
                           
                             
                               ( 
                               
                                 
                                   
                                     - 
                                     4 
                                   
                                    
                                   ac 
                                 
                                 
                                   b 
                                   2 
                                 
                               
                               ) 
                             
                             2 
                           
                         
                         + 
                         
                           O 
                            
                           
                             ( 
                             
                               n 
                               3 
                             
                             ) 
                           
                         
                       
                       ) 
                     
                   
                 
               
             
           
         
       
     
         [0064]    The dominant pole gives the response: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       s 
                       ≅ 
                       
                         c 
                         b 
                       
                     
                     ⇒ 
                     
                       - 
                       
                         1 
                         
                           τ 
                           ferro 
                           ′ 
                         
                       
                     
                   
                   = 
                   
                     - 
                     
                       1 
                       
                         
                           τ 
                           ferro 
                         
                         + 
                         
                           
                             L 
                              
                             
                               ( 
                               
                                 1 
                                 + 
                                 
                                   
                                     μ 
                                     r 
                                   
                                    
                                   χ 
                                 
                               
                               ) 
                             
                           
                           R 
                         
                       
                     
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   10 
                 
               
             
           
         
       
     
         [0065]    Thus 
         [0000]    
       
         
           
             
               τ 
               ferro 
               ′ 
             
             = 
             
               
                 
                   
                     τ 
                     ferro 
                   
                   + 
                   
                     
                       L 
                        
                       
                         ( 
                         
                           1 
                           + 
                           
                             
                               μ 
                               r 
                             
                              
                             χ 
                           
                         
                         ) 
                       
                     
                     R 
                   
                 
                 ⇒ 
                 
                   τ 
                   ferro 
                 
               
                
               
                 | 
                 
                   R 
                   → 
                   0 
                 
               
                
               
                 → 
                 
                   τ 
                   
                     elec 
                      
                     
                         
                     
                   
                 
               
             
           
         
       
     
         [0000]    the 2 nd  term is a purely electrical circuit effect (inductor-resistor circuit) which dominates at high loading (R→0). The current  25  induced into the power coil  9  is then: 
         [0000]    
       
         
           
             
               
                 
                   
                     i 
                      
                     
                       ( 
                       t 
                       ) 
                     
                   
                   = 
                   
                     
                       
                         
                           DM 
                           0 
                         
                         N 
                       
                        
                       
                          
                         
                           
                             - 
                             t 
                           
                           / 
                           
                             τ 
                             ferro 
                             ′ 
                           
                         
                       
                     
                     = 
                     
                       
                         
                           DM 
                           0 
                         
                         N 
                       
                        
                       
                          
                         
                           
                             - 
                             t 
                           
                            
                           
                               
                           
                            
                           
                             R 
                             / 
                             
                               L 
                                
                               
                                 ( 
                                 
                                   1 
                                   + 
                                   
                                     
                                       μ 
                                       r 
                                     
                                      
                                     χ 
                                   
                                 
                                 ) 
                               
                             
                           
                         
                       
                     
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   11 
                 
               
             
           
         
       
     
         [0066]    The electrical work delivered to the load is: 
         [0000]    
       
         
           
             
               ∫ 
               0 
               ∞ 
             
              
             
               
                 
                   i 
                   2 
                 
                  
                 
                   ( 
                   t 
                   ) 
                 
               
                
               R 
                
               
                   
               
                
               
                  
                 t 
               
             
           
         
       
     
         [0000]    by which we can calculate the work as the time constant stretches to infinity (the plateau  22  of the dipole-work on  FIG. 7 ): 
         [0000]    
       
         
           
             
               
                 
                   
                     W 
                     
                       
                         dw 
                         · 
                         
                           L 
                           / 
                           R 
                         
                       
                       → 
                       ∞ 
                     
                   
                   = 
                   
                     
                       1 
                       2 
                     
                      
                     
                       
                         μ 
                         0 
                       
                       
                         ( 
                         
                           1 
                           + 
                           
                             χ 
                              
                             
                                 
                             
                              
                             
                               μ 
                               r 
                             
                           
                         
                         ) 
                       
                     
                      
                     
                       M 
                       2 
                     
                      
                     V 
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   12 
                 
               
             
           
         
       
     
         [0067]    This expression for the ultimate simple dipole-work, eqn. 12 is seen to be less than the magnetisation energy eqn. 8. 
       The “H-Field Cancellation” Method 
       [0068]    Reference  FIGS. 8, 9A, 9A, 9C, 10A, 10B, 11A, 11B, 11C, 12, 13A and 13B  and the description texts earlier. 
         [0069]    In the previous section it was shown that a resistive electrical load  16  on its own only returned part of the input magnetisation work. What was manifest was a slowing in the time constant ( FIG. 7 , lower figures) of the induced current  25  waveform. Heuristically this effect arises due to the re-magnetisation term χμ r H in eqn. 3, as without it the ferrofluid would relax at its native rate. This is shown underlined below: 
         [0000]    
       
         
           
             
               
                  
                 M 
               
               
                  
                 t 
               
             
             = 
             
               
                 - 
                 
                   1 
                   τ 
                 
               
                
               
                 ( 
                 
                   M 
                   - 
                   
                     
                       
                         χμ 
                         r 
                       
                        
                       H 
                     
                     _ 
                   
                 
                 ) 
               
             
           
         
       
     
         [0070]    The technique is to provide a cancelling magnetic field that has no effect on the ferrofluid or the power extraction circuit and this is depicted in  FIGS. 8 and 9 . If the re-magnetisation term is removed (or cancelled) then the energy per cycle curve depicted in  FIG. 8  results. It can be seen that the curve  24  by this method exceeds the magnetisation energy  23  and the energy returned into a simple resistive load  22 . 
         [0071]    Essentially what occurs is that the current  25  in the output power coil  9  re-magnetises the working substance  3 ,  6  and slows the current waveform ( FIG. 7 , lower figures). By generating an opposing field, this term can be neutralised by another coil  8  ( FIG. 13A  or use of the same magnetisation coil in step  1  of the cycle) around the power extraction area  5 . 
         [0072]    It is not a simple matter of just supplying an opposite field as a null-transformer will result. We shall discuss this shortly in the next section looking at the electrical analysis. Essentially, a superimposed field around the working substance needs to be created, that is of such high frequency, that it is averaged to substantially zero by the slow relaxation rate of the working substance  3 . 
         [0073]      FIG. 9A  shows how the current  25  is copied, inverted and chopped by a high frequency signal. This copy current produces its own field and this sums with the field produced by the current  25  to result in  FIG. 9B . The working substance largely filters the resulting chopped re-magnetising field to a much lower magnitude (last legend in  FIG. 9B ). The chopping process results in the cancelling field includes periods corresponding to the zero regions in the inverted current, which are separated from each other by a time period which is less than the relaxation time of the sample, and preferably by a time period which is one third or less than the relaxation time of the sample. 
         [0074]      FIG. 9C  shows an even better method that uses an asymmetric principle of copying a multiple (x1.5) of the current  25  (or more turns on the former to produce a bigger field) and a higher chopping frequency. The final figure in the legend shows that the re-magnetising field is virtually eliminated by the technique. 
         [0075]    Comparing the current and magnetisation vs. time traces in the lower figures of  FIGS. 7 and 8 , we can see that both the current and magnetisation traces are speeded up by the method ( FIG. 8 ), compared to the dipole-work into a simple electrical load ( FIG. 7 ). 
         [0076]    With reference to  FIGS. 10A  and B, another way of viewing the technique can be seen in the frequency domain. The re-magnetising field from the current is shown in the first legend  FIG. 10  in the frequency domain. This current is copied and inverted in the next legend and the chopping produces harmonics, as per the modulation theorem: 
         [0000]      sin  A· sin  B= ½cos( A−B )−½cos( A+B )   eqn. 13
 
         [0077]    The lowest legend represents the superposition of the field from the current  25  and the field from the cancellation current  26 . It can be seen that the low frequency component, that the working substance  3  would respond to, is obliterated. In  FIG. 10B , we can see how the frequency response of the field cancellation circuit ( FIG. 13A ) and the working substance  3  and power output circuit respond to the low frequency power signal and high frequency components of the cancellation signal: in the first case, the power signal does not interfere with the cancellation circuit and in the second case, the cancellation signal is filtered out. 
         [0078]    The Electrical Circuit and Electrical analysis of the work required by the H-field cancellation circuit 
         [0079]    With reference to  FIGS. 11A , B, C and  13 A and B, the cancellation circuit can be implemented by a current mirrorb  32  which is switched on and off rapidly by an high frequency clock signal  33 . The current  26  resulting can impinge on the working substance  3 ,  6  by the same excitation field coils  8  or another coil. A flyback circuit  30 ,  31  recoups the field energy used in the chopping process and returns it to storage capacitor  29 . The chopping process will have some losses, so periodically the capacitor  29  is charged via switch  28  from a high voltage source  27 . Ultimately all the power needs of this circuit comes from the power generated by the working substance  3 ,  6  and the power output circuit. 
         [0080]    We proceed to analyse the energetics of the scheme by the equivalent circuit of a null transformer ( FIG. 11A ), that is, a transformer with two windings in intimate contact with equal and opposite currents flowing through the windings. The result is that there is obviously no effect in this case and by deviating from the arrangement, we show the validity of the cancellation method. 
         [0081]    The sense of the currents and voltages from the self and mutual inductances and the decaying ferrofluid (working substance  3 ,  6 ) flux, 
         [0000]    
       
         
           
             
               V 
               ff 
             
             = 
             
               - 
               
                 
                    
                   
                     λ 
                     ff 
                   
                 
                  
               
             
           
         
       
     
         [0000]    is shown. It is quite clear that the LHS current mirror does work against the decaying ferrofluid flux and this is of course at least equal to the work that is supposed to be delivered onto the RHS into the load. It is obvious that no power is delivered to the load. Another way of putting this is, of course, that it is a null transformer with changes in magnetic field excluded from the coils&#39; interior. Another way, still, is to note that the current in the LHS circuit is equal and opposite to the RHS and that this is induced into the RHS circuit nullifying all current. 
         [0082]    Next we note the addition of the filtering circuit elements, the high pass (and storage capacitor) on the LHS and the high frequency inductor (hf choke) on the RHS in  FIG. 11B . Now the situation is different: Firstly the high frequency series inductor (or “choke”) blocks the high frequency chopped current from the LHS being induced into the RHS and the “high pass” capacitor blocks the low frequency current from the RHS being induced in LHS. 
         [0083]    This can be understood by a simple potential divider effect ( FIG. 11C ): In the left-hand figure, the dynamic impedance of the current source is represented by R CS . It is a simple matter to find the voltage across the current source induced by the changing ferrofluid flux and deduce that with a relatively low rate of change of ferrofluid flux, with a small capacitance  29  ( FIG. 13A ), very little voltage and hence work is done against the current source. 
         [0000]    
       
         
           
             
               
                 
                   
                     V 
                     
                       R 
                       CS 
                     
                   
                   = 
                   
                     
                       V 
                       ff 
                     
                     ( 
                     
                       
                         R 
                         CS 
                       
                       
                         
                           1 
                           
                             j 
                              
                             
                                 
                             
                              
                             ω 
                              
                             
                                 
                             
                              
                             
                               C 
                               HP 
                             
                           
                         
                         + 
                         
                           R 
                           CS 
                         
                       
                     
                     ) 
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   14 
                 
               
             
           
         
       
     
         [0084]    The current  26  in the LHS circuit is set-up by the current mirror  32  ( FIG. 13A ) and this can be a fraction of the current on the RHS ( FIG. 13B ), all that matters is that the turns-ratio of the coil is sufficient to cancel the H-field from the RHS circuit. In the limit of small current, it is obvious that the electrical work performed by the changing ferrofluid flux is less than that performed on the RHS circuit. 
         [0085]    Further to the argument, the current source mainly performs electrical work establishing the cancellation magnetic field on the LHS. This can be recouped with high efficiency by a “flyback” circuit  30 ,  31  ( FIG. 13A ). The current source performs net work against the electrical resistance of the left-hand circuit and this can be made arbitrarily small, in fact to labour the point, a fractional current mirror ( FIG. 13B ) can be used (since all that matters is the current multiplied by the left-hand turns to establish the cancelling field) and this makes it even more obvious. 
         [0086]    Considering now the work of the chopping circuit on the right-hand power output circuit ( FIG. 13A ), in  FIG. 11C  we can see a similar potential divider effect: 
         [0000]    
       
         
           
             
               V 
               R 
             
             = 
             
               
                 V 
                 chop 
               
               ( 
               
                 R 
                 
                   
                     j 
                      
                     
                         
                     
                      
                     ω 
                      
                     
                         
                     
                      
                     
                       L 
                       choke 
                     
                   
                   + 
                   R 
                 
               
               ) 
             
           
         
       
     
         [0087]    This time we note that, the high frequency chopping field results in an high impedance from the choke  34 ; very little electrical work is thus expended by the chopping circuit on the power output circuit. 
       1.1.1. Dynamic Analysis of the H-Field Cancellation Method and the Utimate Electrical Work 
       [0088]    We now follow the same procedure with the state equations of eqn. 3, eqn. 4 and eqn. 6 but with the re-magnetising H-field removed from equation 3, to yield the transform of the induced current  25 : 
         [0000]    
       
         
           
             
               
                 
                   
                     I 
                      
                     
                       ( 
                       s 
                       ) 
                     
                   
                   = 
                   
                     
                       
                         DM 
                         0 
                       
                       N 
                     
                     
                       
                         
                           s 
                           2 
                         
                          
                         
                           τ 
                           ferro 
                         
                       
                       + 
                       
                         s 
                          
                         
                           R 
                           L 
                         
                          
                         
                           τ 
                           ferro 
                         
                       
                       + 
                       
                         R 
                         L 
                       
                     
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   15 
                 
               
             
           
         
       
     
         [0089]    Whereupon the current in the time domain by the dominant pole is: 
         [0000]    
       
         
           
             
               
                 
                   
                     i 
                      
                     
                       ( 
                       t 
                       ) 
                     
                   
                   = 
                   
                     
                       
                         
                           DM 
                           0 
                         
                         N 
                       
                        
                       
                          
                         
                           
                             - 
                             t 
                           
                           / 
                           
                             τ 
                             ferro 
                             ′ 
                           
                         
                       
                     
                     = 
                     
                       
                         
                           D 
                            
                           
                               
                           
                            
                           
                             M 
                             0 
                           
                         
                         N 
                       
                        
                       
                          
                         
                           
                             - 
                             tR 
                           
                           / 
                           L 
                         
                       
                     
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   16 
                 
               
             
           
         
       
     
         [0090]    The dipole-work by the cancellation method in the limit is obtained, once again, by 
         [0000]              ∫   0   ∞              i   2          (   t   )                     R           t          :               W   dw.cancel.L/R→∞ =½μ 0   M   2   V    eqn. 17
 
         [0091]    This is seen to be the magnetic field energy of the ferrofluid flux (the plateau  24 ,  FIG. 8 ). 
         [0092]    The cancellation method has been proven in experiment and simulation (appendix 1) in the first instance by the simple expedient of zeroing the re-magnetisation term: 
         [0000]    
       
         
               
             
           
               
                   
               
             
             
               
                 %########################################################## 
               
               
                 dMdt = (1./tor) .* fac .* ( M − 0.*H.*Xchi.*CONST_perm ); 
               
               
                   
               
             
          
         
       
     
         [0093]    The results are displayed in  FIG. 8 . Straightaway, logically, one can see the effect of the ferrofluid relaxing at its native rate in  FIG. 8 , lower traces, compared to  FIG. 7 , lower traces. 
         [0094]    A more physical simulation, other than the “trick” of zeroing the H-field is implemented at the end of appendix 1 by a high frequency cancellation H-field: 
         [0000]    
       
         
               
             
           
               
                   
               
             
             
               
                   %###################################################### 
               
               
                   dMdt = −(1./tor) .* fac .* ( M − H.*(1−Cancel(t)) ... 
               
               
                    .*Xchi.*CONST_perm); 
               
               
                 ... 
               
               
                  function field = Cancel(t) 
               
               
                   % Generate square waveform 0-1.5 at frequency “freq” 
               
               
                   freq = 20000; % freq = 100/tor; 
               
               
                   field = 1.5.*(0.5+0.5.*sign(0.5+freq.*t-round(freq.*t+0.5))); 
               
               
                 end 
               
               
                   
               
             
          
         
       
     
         [0095]    Though this code is much slower to run due to the fine time-scale needed to simulate the cancellation field and the potentially long time scale of the electrical circuit. 
         [0096]    With reference to  FIG. 12 , comparison is now made to the plateau limit ratios ( FIG. 8 ) of equations eqn. 8, eqn. 12 and eqn. 17, that is: 
         [0000]    
       
         
           
             
               1 
               
                 χμ 
                 r 
               
             
              
             
               : 
             
              
             
                 
             
              
             
               1 
               
                 2 
                  
                 
                   ( 
                   
                     1 
                     + 
                     
                       χμ 
                       r 
                     
                   
                   ) 
                 
               
             
              
             
               : 
             
              
             
                 
             
              
             
               1 
               2 
             
           
         
       
     
         [0000]    with variation of the paramenter χμ r  which is the effective susceptibility of the ferrofluid/working substance  3  with the high permeability co-material  6  present and this is plotted in  FIG. 12 . For all variation of parameters, the magnetisation energy is always greater than the dipole-work without the cancellation method. However if χμ r &gt;2 the dipole-work, with the cancellation method, will exceed the magnetisation energy input. 
         [0097]    The power produced by the device is then: 
         [0000]        P= ( W   dw.cancel   −E   mag   −W   losses ) F   cycle    eqn. 18
 
       Method of Excess Power by Non-linear Impedances 
       [0098]    Reference is made to  FIGS. 7, 14A , B, C and D and  15 . 
         [0099]      FIG. 4  shows an improvement to the basic setup of  FIG. 2 . The main feature is a variable turns ratio for the output coil and variable resistance for the output resistance. These impedances are functions of current and time and are under computer control for the duration of the power cycle. 
         [0100]    The first problem we shall address is the electrical time constant dominating and swamping the quicker time constant of the ferrofluid. As has been seen in the linear case, we aim for low output impedances (which can always be matched to a load or the load is just used as a heating element for a conventional Carnot cycle) as these obtain the most energy on each cycle, however they are the slowest ( FIG. 7  upper and lower legends). 
         [0101]    The first method is, under computer control for the computer to vary the turns ratio of the output coil and the output resistance. This method is able to achieve the highest energy returned from the decaying ferrofluid in a finite time scale.  FIG. 7 , lower legend shows the linear case and the lowest resistances give the highest returned energy but the leftmost traces show incredibly long time constants; we cannot start another cycle until this cycle complete and this implies that the power output from the device is low. Immediately this suggests a constrain condition: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       
                         L 
                          
                         
                           ( 
                           
                             i 
                             , 
                             t 
                           
                           ) 
                         
                       
                       
                         R 
                          
                         
                           ( 
                           
                             i 
                             , 
                             t 
                           
                           ) 
                         
                       
                     
                     ≤ 
                     
                       
                         K 
                         1 
                       
                        
                       
                         τ 
                         ferro 
                       
                     
                   
                    
                   
                     
 
                   
                   ⇒ 
                   
                     
                       R 
                        
                       
                         ( 
                         
                           i 
                           , 
                           t 
                         
                         ) 
                       
                     
                     ≥ 
                     
                       
                         
                           μ 
                           0 
                         
                          
                         
                           μ 
                           r 
                         
                          
                         
                           
                             
                               AN 
                               2 
                             
                              
                             
                               ( 
                               
                                 i 
                                 , 
                                 t 
                               
                               ) 
                             
                           
                           D 
                         
                       
                       
                         
                           K 
                           1 
                         
                          
                         
                           τ 
                           ferro 
                         
                       
                     
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   19 
                 
               
             
           
         
       
     
         [0102]    The inductance has been expanded into a well-known form for the inductance of a long solenoid. 
         [0103]    Another constraint can be found from adjusting the rate of non-linear power output (just EMF 2 /R) to be greater than the linear case magnetising the ferrofluid as a baseline. 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       μ 
                       0 
                       2 
                     
                      
                     
                       A 
                       2 
                     
                      
                     
                       
                         
                           N 
                           2 
                         
                          
                         
                           ( 
                           
                             i 
                             , 
                             t 
                           
                           ) 
                         
                       
                       
                         R 
                          
                         
                           ( 
                           
                             i 
                             , 
                             t 
                           
                           ) 
                         
                       
                     
                      
                     
                       
                         ( 
                         
                           
                              
                             M 
                           
                           
                              
                             t 
                           
                         
                         ) 
                       
                       2 
                     
                   
                   = 
                   
                     
                       K 
                       2 
                     
                      
                     
                       E 
                       mag 
                     
                      
                     
                       t 
                       
                         
                           K 
                           3 
                         
                          
                         τ 
                       
                     
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   20 
                 
               
             
           
         
       
     
         [0104]    Where the constants K 2  is a multiple of the magnetising energy and K 3  is the time scale of linear magnetisation (the ferrofluid is switched on slower than its 3-dB point so that needless dissipation doesn&#39;t occur. K 3  works out about 3). 
         [0105]    Into eqn. 20 is substituted eqn. 3 and we solve this for R(i,t) which is then substituted into inequality eqn. 19 leading to eventually: 
         [0000]    
       
         
           
             
               
                 
                   
                     N 
                      
                     
                       ( 
                       
                         i 
                         , 
                         t 
                       
                       ) 
                     
                   
                   ≤ 
                   
                     
                       [ 
                       
                         
                           M 
                            
                           
                             ( 
                             t 
                             ) 
                           
                         
                         ± 
                         
                           
                             
                               
                                 K 
                                 2 
                               
                                
                               
                                 E 
                                 mag 
                               
                                
                               t 
                             
                             
                               
                                 μ 
                                 0 
                               
                                
                               
                                 μ 
                                 r 
                               
                                
                               
                                 K 
                                 1 
                               
                                
                               
                                 K 
                                 3 
                               
                                
                               
                                 τ 
                                 ferro 
                                 2 
                               
                             
                           
                         
                       
                       ] 
                     
                      
                     
                       D 
                       
                         
                           χμ 
                           r 
                         
                          
                         
                           i 
                            
                           
                             ( 
                             t 
                             ) 
                           
                         
                       
                     
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   21 
                 
               
             
           
         
       
     
         [0106]    Overall the solutions for N(i,t) and R(i,t) are constrained (physically) as: 
         [0000]    
       
         
           
             i 
              
             
                 
             
              
             ε 
              
             
                 
             
              
             ℝ 
           
         
       
       
         
           
             t 
              
             
                 
             
              
             ε 
              
             
                 
             
              
             
               ℝ 
               + 
             
              
             
                 
             
              
             t 
              
             
                 
             
              
             is 
              
             
                 
             
              
             monotonically 
              
             
                 
             
              
             increasing 
           
         
       
       
         
           
             
               R 
                
               
                 ( 
                 
                   i 
                   , 
                   t 
                 
                 ) 
               
             
              
             ε 
              
             
                 
             
              
             
               ℝ 
               + 
             
           
         
       
       
         
           
             
               N 
                
               
                 ( 
                 
                   i 
                   , 
                   t 
                 
                 ) 
               
             
              
             ε 
              
             
                 
             
              
             
               ℕ 
               + 
             
           
         
       
       
         
           
             K 
             &gt; 
             1 
           
         
       
     
         [0107]    With the conditions: 
         [0108]      FIGS. 14A , B, C and D show the results of a simulation and specifically the parameters K 1 =5, K 2 =3 and K 3 =3 and is able to achieve non-linear dipole work tending to the linear case in finite time; observe  FIG. 14D  and we see that the current waveform is only about ten times slower than the ferrofluid magnetisation waveform and this, itself, is hardly changed from the no load value (when R co).  FIG. 15  gives a circuit schematic. 
         [0109]    Device based on Electrical Remanence 
         [0110]    Reference is made to  FIGS. 1, 15, 16, 17 and 18  (and the description text earlier). 
         [0111]    In constructing the electrostatic dual of the temporary remanence cycle, there are subtle similarities and differences. Both involve a charging and discharging phase: one with magnetic flux and an energy cost of the magnetising energy, the other electric or polarisation flux and the energy cost of polarisation energy. Both too would seem to have a “lossy” tank where this input energy is converted to internal energy (“heat”) at the rate a function of: 
         [0000]    
       
         
           
             
               
                  
                 φ 
               
               
                  
                 t 
               
             
             = 
             
               - 
               
                 
                   φ 
                   τ 
                 
                 . 
               
             
           
         
       
     
         [0000]    However, as we saw with the magnetic system: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       
                          
                         M 
                       
                       
                          
                         t 
                       
                     
                     = 
                     
                       
                         
                           
                             - 
                             
                               1 
                               τ 
                             
                           
                            
                           
                             ( 
                             
                               M 
                               - 
                               
                                 
                                   χμ 
                                   r 
                                 
                                  
                                 H 
                               
                             
                             ) 
                           
                         
                          
                         
                           
 
                         
                         ⇒ 
                         
                           
                              
                             M 
                           
                           
                              
                             t 
                           
                         
                       
                       = 
                       
                         
                           - 
                           
                             1 
                             τ 
                           
                         
                          
                         
                           ( 
                           
                             M 
                             - 
                             
                               
                                 χμ 
                                 r 
                               
                                
                               
                                 N 
                                 D 
                               
                                
                               i 
                             
                           
                           ) 
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   And 
                 
               
               
                 
                   ( 
                   
                     eqn 
                     . 
                     
                         
                     
                      
                     4 
                   
                   ) 
                 
               
             
             
               
                 
                   
                     
                       - 
                       
                         
                            
                           λ 
                         
                         
                            
                           t 
                         
                       
                     
                     - 
                     
                        
                        
                       
                           
                       
                        
                       R 
                     
                   
                   = 
                   0 
                 
               
               
                 
                   ( 
                   
                     used 
                      
                     
                         
                     
                      
                     to 
                      
                     
                         
                     
                      
                     derive 
                      
                     
                         
                     
                      
                     
                       eqn 
                       . 
                       
                           
                       
                        
                       6 
                     
                   
                   ) 
                 
               
             
           
         
       
     
         [0112]    The re-magnetising field is in the same sense (if one can imagine the entry and exit wires of the solenoid as parallel to the axial field) as the current and original magnetising field; furthermore, this re-magnetising field can be cancelled by the field cancellation method to leave, via eqn. 5, a means of getting dipole work that exceeds the input energy cost; the difference in the two is the thermal energy converted (see Cornwall&#39;s thesis). 
         [0113]    Eqn. 5 comes directly from the 2 nd  Maxwell/Faraday&#39;s Law equation in integral form, which is then equated to the potential drop across the resistor. No such law exists in the electrostatic case regarding the flux and discharge and this always leads to the return of the electrostatic field energy ½ ε 0 EdE·dV and polarisation energy EdP·dV (from EdD·dV): 
         [0114]    From the 1 st  Maxwell equation/Gauss&#39; Law: 
         [0000]    
       
         
           
             
               
                 
                   
                     ∇ 
                     
                       · 
                       
                         ( 
                         
                           E 
                           + 
                           
                             P 
                             
                               ɛ 
                               0 
                             
                           
                         
                         ) 
                       
                     
                   
                   = 
                   
                     
                       ρ 
                       free 
                     
                     
                       ɛ 
                       0 
                     
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   22 
                 
               
             
           
         
       
     
         [0115]    And hence, 
         [0000]    
       
         
           
             
               
                 
                   
                     ∯ 
                     
                       
                         ( 
                         
                           E 
                           + 
                           
                             P 
                             
                               ɛ 
                               0 
                             
                           
                         
                         ) 
                       
                       · 
                       
                          
                         A 
                       
                     
                   
                   = 
                   
                     
                       Q 
                       free 
                     
                     
                       ɛ 
                       0 
                     
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   23 
                 
               
             
           
         
       
     
         [0116]    This represents a combination of the electric field at the plates of the capacitor and the electric field from the polarisation. The movement of the free charges is the circuit current, thus: 
         [0000]    
       
         
           
             
               
                 
                   
                     A 
                      
                     
                       ( 
                       
                         
                           
                              
                             E 
                           
                           
                              
                             t 
                           
                         
                         + 
                         
                           
                             1 
                             
                               ɛ 
                               0 
                             
                           
                            
                           
                             
                                
                               P 
                             
                             
                                
                               t 
                             
                           
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       1 
                       
                         ɛ 
                         0 
                       
                     
                      
                     
                       
                          
                         i 
                       
                       
                          
                         t 
                       
                     
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   24 
                 
               
             
           
         
       
     
         [0117]    Multiplying both sides the voltage across the plates, i.e. 
         [0000]    
       
         
           
             v 
             = 
             
               
                 ∫ 
                 d 
               
                
               
                 E 
                 · 
                 
                     
                 
                  
                 
                    
                   l 
                 
               
             
           
         
       
     
         [0000]    yields, 
         [0000]    
       
         
           
             
               Ed 
               · 
               
                 A 
                  
                 
                   ( 
                   
                     
                       
                          
                         E 
                       
                       
                          
                         t 
                       
                     
                     + 
                     
                       
                         1 
                         
                           ɛ 
                           0 
                         
                       
                        
                       
                         
                            
                           P 
                         
                         
                            
                           t 
                         
                       
                     
                   
                   ) 
                 
               
             
             = 
             
               
                 1 
                 
                   ɛ 
                   0 
                 
               
                
               v 
                
               
                 
                    
                   i 
                 
                 
                    
                   t 
                 
               
             
           
         
       
     
         [0118]    Which upon integration w·r·t. time, 
         [0000]                (ε 0 EdE+EdP) V=vi    eqn. 25
 
         [0119]    This is just seen to be the differential electrostatic work EdD and the instantaneous electrical power. 
         [0120]    We also note in this case too, that the first state equation becomes: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                        
                       P 
                     
                     
                        
                       t 
                     
                   
                   = 
                   
                     
                       - 
                       
                         1 
                         τ 
                       
                     
                      
                     
                       ( 
                       
                         P 
                         - 
                         
                           
                             χɛ 
                             0 
                           
                            
                           
                             ɛ 
                             r 
                           
                            
                           E 
                         
                       
                       ) 
                     
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   26 
                 
               
             
           
         
       
     
         [0121]    No de-polarisation cancelling method can be made to strike out the term χε 0 ε r E, when we realise that the potential across the load resistor is negative and acts to increase the rate of decay further. This only reflects energy leaving the capacitor “tank” (in competition to that being converted to heat), as it should. 
         [0122]    The power extraction area  5  can be implemented with the electrical dual of magnetism, electrostatics. By the 4 th  Maxwell equation, the changing electrical field from the temporary polarisation creates a temporary magnetic field. This then amounts immediately to an analogous situation with the magnetic device and a further embodiment ( FIG. 18 ). The input electrostatic and polarisation energy then becomes magnetisation energy and the field cancellation method then applies, to net the difference between magneto-static energy and the magnetisation energy as thermal work converted. Let us explore this: 
         [0123]    Where P is the polarisation and E is the electric field strength. The electrically polarisable working substance  3  can have its electrical susceptibility x increased with electrically polarisable co-material  35  ( FIGS. 16, 17 ) whose relaxation rate is substantially faster than the relaxation rate of the working substance. We shall see shortly that it is advantageous to have an electrically non-conducting magnetic co-material  6  too. 
         [0124]    From the definition of electrical polarisation by the first Maxwell equation as being the electric field (E-field) produced by bound charges, we can write: 
         [0000]    
       
         
           
             
               
                 
                   
                     ∇ 
                     
                       · 
                       
                         ( 
                         
                           
                             
                               ρ 
                               bound 
                             
                             
                               ɛ 
                               0 
                             
                           
                           + 
                           
                             E 
                             temp 
                           
                         
                         ) 
                       
                     
                   
                   = 
                   
                     
                       ρ 
                       free 
                     
                     
                       ɛ 
                       0 
                     
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   27 
                 
               
             
           
         
       
     
         [0125]    Note that the electric field has two components: The E-field from the polarisation and the E-field which results from the displacement current, E temp . 
         [0126]    Let us consider E temp  first. The fourth Maxwell equation includes the displacement current term. We are considering a dielectric so the current density term is left out: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       c 
                       2 
                     
                      
                     
                       ∇ 
                       
                         × 
                         B 
                       
                     
                   
                   = 
                   
                     
                       ∂ 
                       E 
                     
                     
                       ∂ 
                       t 
                     
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   28 
                 
               
             
           
         
       
     
         [0127]    The electric field results from the polarisation along the x-axis, which is the axis of the capacitor, so we can write: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       c 
                       2 
                     
                      
                     
                       ∇ 
                       
                         × 
                         B 
                       
                     
                   
                   = 
                   
                     
                       ɛ 
                       0 
                     
                      
                     
                       ɛ 
                       r 
                     
                      
                     χ 
                      
                     
                       
                         ∂ 
                         
                           P 
                           x 
                         
                       
                       
                         ∂ 
                         t 
                       
                     
                      
                     i 
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   29 
                 
               
             
           
         
       
     
         [0128]    Thus the curl operator can only have components in the yz plane: 
         [0000]    
       
         
           
             
               
                 
                   
                     ∇ 
                     
                       × 
                       B 
                     
                   
                   = 
                   
                      
                     
                       
                         
                           i 
                         
                         
                           j 
                         
                         
                           k 
                         
                       
                       
                         
                           
                             ∂ 
                             
                               ∂ 
                               x 
                             
                           
                         
                         
                           
                             ∂ 
                             
                               ∂ 
                               y 
                             
                           
                         
                         
                           
                             ∂ 
                             
                               ∂ 
                               z 
                             
                           
                         
                       
                       
                         
                           
                             B 
                             x 
                           
                         
                         
                           
                             B 
                             y 
                           
                         
                         
                           
                             B 
                             z 
                           
                         
                       
                     
                      
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   30 
                 
               
             
           
         
       
     
         [0129]    For simplicity we shall consider cylindrical symmetry and we know that the B-field will circulate around the changing P x  vector. Using Stoke&#39;s Identity to relate the line integral of the curl of B to the surface integral of the flux from P, we find: 
         [0000]    
       
         
           
             
               
                 c 
                 2 
               
                
               
                 
                   ∫ 
                   0 
                   r 
                 
                  
                 
                   B 
                   · 
                   
                       
                   
                    
                   
                      
                     
                       ( 
                       
                         2 
                          
                         π 
                          
                         
                             
                         
                          
                         l 
                       
                       ) 
                     
                   
                 
               
             
             = 
             
               
                 ∫ 
                 0 
                 r 
               
                
               
                 
                   ɛ 
                   0 
                 
                  
                 
                   ɛ 
                   r 
                 
                  
                 χ 
                  
                 
                   
                     
                       ∂ 
                       
                         P 
                         x 
                       
                     
                     
                       ∂ 
                       t 
                     
                   
                   · 
                   
                       
                   
                    
                   
                      
                     
                       ( 
                       
                         π 
                          
                         
                             
                         
                          
                         
                           l 
                           2 
                         
                       
                       ) 
                     
                   
                 
               
             
           
         
       
     
         [0130]    Thus the temporary independent magnetic flux is: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       B 
                       yz 
                     
                      
                     
                       ( 
                       t 
                       ) 
                     
                   
                   = 
                   
                     
                       ɛ 
                       0 
                     
                      
                     
                       ɛ 
                       r 
                     
                      
                     χ 
                      
                     
                       r 
                       
                         c 
                         2 
                       
                     
                      
                     
                       
                         ∂ 
                         
                           
                             P 
                             x 
                           
                            
                           
                             ( 
                             t 
                             ) 
                           
                         
                       
                       
                         ∂ 
                         t 
                       
                     
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   31 
                 
               
             
           
         
       
     
         [0131]    This B-field is itself changing and will lead to E temp  and so on, as a series in powers of 1/c 2 , so we safely truncate it to first order in 1/c 2 . The E-field is given by Maxwell&#39;s 2 nd  equation: 
         [0000]    
       
         
           
             
               
                 
                   
                     ∇ 
                     
                       × 
                       
                         E 
                         temp 
                       
                     
                   
                   = 
                   
                     - 
                     
                       
                         ∂ 
                         
                           
                             B 
                             yz 
                           
                            
                           
                             ( 
                             t 
                             ) 
                           
                         
                       
                       
                         ∂ 
                         t 
                       
                     
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   32 
                 
               
             
           
         
       
     
         [0132]    Which we know from Stoke&#39;s Identity will lead to an E-field perpendicular to the plane yz, that is, in the anti-x axis direction ( 
         [0000]    
       
         
           
             
                
               P 
             
             
                
               t 
             
           
         
       
     
         [0000]    is negative), increasing with magnitude with the radius (that is, our line integral path is an axially aligned loop through the centre of the capacitor, see  FIG. 18 ): 
         [0000]    
       
         
           
             
               
                 
                   
                     v 
                     temp 
                   
                   = 
                   
                     
                       
                         ∮ 
                         path 
                       
                        
                       
                         E 
                         · 
                         
                            
                           
                               
                           
                            
                           l 
                         
                       
                     
                     = 
                     
                       - 
                       
                         
                           ∂ 
                           λ 
                         
                         
                           ∂ 
                           t 
                         
                       
                     
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   33 
                 
               
             
           
         
       
     
         [0133]    The path at the centre contributes nothing, so we can write (V is the volume, n is the turns per unit length): 
         [0000]    
       
         
           
             
               
                 
                   
                     v 
                     temp 
                   
                   = 
                   
                     
                       - 
                       
                         ( 
                         
                           
                             μ 
                             0 
                           
                            
                           
                             μ 
                             r 
                           
                            
                           n 
                         
                         ) 
                       
                     
                      
                     V 
                      
                     
                       
                         
                           ɛ 
                           0 
                         
                          
                         
                           ɛ 
                           r 
                         
                          
                         χ 
                       
                       
                         c 
                         2 
                       
                     
                      
                     
                       
                         
                           ∂ 
                           2 
                         
                          
                         
                           
                             P 
                             x 
                           
                            
                           
                             ( 
                             t 
                             ) 
                           
                         
                       
                       
                         ∂ 
                         
                           t 
                           2 
                         
                       
                     
                   
                 
               
               
                 
                   eqn 
                   . 
                   
                       
                   
                    
                   34 
                 
               
             
           
         
       
     
         [0134]    A further embodiment of this device is thus apparent and shown in  FIG. 17 . The factor μ 0 μ r n has been included to allow for the possibility of magnetic co-material  6  (which must be non-conductive, e.g. ferrite) to boost this field and multiple turns around the capacitor. All discussion of the H-field cancellation technique and non-linear methods discussed earlier pertain to this technique too. 
         [0135]    Much the same argument as regards the magnetisation field ( FIG. 3 ) applies to the polarising field as regards the wavetrain, the off period and the slow switch-on to minimise dissipative losses. 
         [0136]    The non-linear approach can be applied too to this embodiment of the device by varying the load resistance ( FIG. 17 ) as the polarising flux decays or varying the capacitance with time by having plates  36  at further distances from the working substance. 
         [0137]    The following claims include dependent claims which are not repeated for all of the independent claims. However, unless wherein it would be inconsistent, it should be understood that the features of any of the dependent claims may be combined with any of the independent claims. 
         [0138]    When used in this specification and claims, the terms “comprises” and “comprising” and variations thereof mean that the specified features, steps or integers are included. The terms are not to be interpreted to exclude the presence of other features, steps or components. 
         [0139]    The features disclosed in the foregoing description, or the following claims, or the accompanying drawings, expressed in their specific forms or in terms of a means for performing the disclosed function, or a method or process for attaining the disclosed result, as appropriate, may, separately, or in any combination of such features, be utilised for realising the invention in diverse forms thereof.