Discussion:
Thermonuclear Fusion (generating algorithm of print out)
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Christopher Strevens
2013-01-26 15:07:19 UTC
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'This computer simulation shows the power likely to be generated by a
fusion engine like mine using different parameters.

'There are several conflicting factors so the best solution is
discovered by human intelligence.





'Program to find the inductance of a multilayer coil by summing the
inductance of each layer and the effect of mutual inductance

'Inductance = 4.piE-7.A.n^2/l



'to find the final diameter of a multilayer coil and the length of wire

'each layer adds thickness to layer so first work out turns per layer

let Pi=3.14215926

'for room temp

Let ResistivityOfCopper=1.68E-8

Let DensityOfCopper=8.94E3

'Kg per m^3

'for superconductor

Let ResistovityOfCopper=0

'for LN temp

'Let ResistivityOfCopper=2.647E-9

Let mu=4E-7*Pi

Let e0=8.85E-12

[InductanceLoop]



input "diameter of wire in mm " ; DiameterOfWire

Input "Diameter Of Lumen mm "; DiameterOfLumen

input "Thickness of Insulation mm ";ThicknessOfInsulation

input "diameter of former in mm ";DiameterOfFormer

input "length of former in mm " ;LengthOfFormer

'Input "Resonant Frequency Hz ";ResonantFrequency



let DiameterOfWire=DiameterOfWire/1000

Let DiameterOfLumen=DiameterOfLumen/1000

Let ThicknessOfInsulation=ThicknessOfInsulation/1000

Let DiameterOfFormer=DiameterOfFormer/1000

Let LengthOfFormer=LengthOfFormer/1000

input "number of turns "; NumberOfTurns

let
TurnsPerLayer=LengthOfFormer/(DiameterOfWire+2*ThicknessOfInsulation)

let NumberOfLayers=NumberOfTurns/TurnsPerLayer

Let
ThicknessOfWinding=NumberOfLayers*(DiameterOfWire+2*ThicknessOfInsulatio
n)

Let DiameterOfWinding=DiameterOfFormer+2*ThicknessOfWinding

'length of wire for each layer is the layer diameter multiplied by Pi

LayerN=0

LengthOfWire=0

Let TotalInductance=0

Let TotalCapacitance=0

[ForLoop]

LayerN=LayerN+1

Let
LayerDiameter=DiameterOfFormer+(LayerN-1)*(DiameterOfWire+2*ThicknessOfI
nsulation)

Let LengthOfWire=Pi*LayerDiameter*TurnsPerLayer+LengthOfWire

let
LayerInductance=mu*Pi*((LayerDiameter/2)^2)*TurnsPerLayer^2/LengthOfForm
er

Let
LayerCapacitance=e0*Pi*LayerDiameter*((DiameterOfWire+2*ThicknessOfInsul
ation)/(DiameterOfWire+2*ThicknessOfInsulation))*TurnsPerLayer

Let
InterLayerCapacitance=e0*Pi*LayerDiameter*LengthOfFormer/(DiameterOfWire
+2*ThicknessOfInsulation)

Let TotalInductance=LayerInductance+TotalInductance

Let
TotalCapacitance=LayerCapacitance+InterlayerCapacitance+TotalCapacitance

if LayerN<NumberOfLayers then goto [ForLoop]

Let
ResistanceOfCoil=ResistivityOfCopper*LengthOfWire/(((DiameterOfWire/2)^2
)-((DiameterOfLumen/2)^2))*Pi

Let TotalInductance=TotalInductance*NumberOfLayers^2



Print "Turns per layer ";TurnsPerLayer

Print "Number of Layers ";NumberOfLayers

Print "Diameter of coil ";DiameterOfWinding

Print "Length of Wire ";LengthOfWire; " Meters"

Print "Resistance of Coil "; ResistanceOfCoil ; " Ohms"

Print "Inductance of Multilayer Coil ";TotalInductance;" Henry"

Print "Capacitance of Mutilayer Coil ";TotalCapacitance*1E6; "
microfarads"

'input "1=exit 2=InductanceLoop ";OK

'if OK=2 then goto [InductanceLoop]

'to calculate dynamic impedance



'dynamic impedance is L/Cr where r is the resitance of the coil.









[ResonanceLoop]

'input "Resonant Frequency Hz "; ResonantFrequency











input "capacitance of external capacitor "; ExternalCapacitance

Let Capacitance=TotalCapacitance+ExternalCapacitance

'input "Capacitance (F) "; Capacitance

let impedance = TotalInductance/(Capacitance*ResistanceOfCoil)

print "dynamic impedance "; impedance

'print "dynamic impedance - Infinite"



'program to calculate resonant frequency

'f=1/(2.pi.sqr(l.c))









'input "Inductance of coil "; Inductance

let ResonantFrequency=1/(2*Pi*(sqr(TotalInductance*Capacitance)))

print "Resonant frequency is "; ResonantFrequency



input "1 for exit 2 for loop "; ok



if ok=2 then goto [InductanceLoop]





'to calculate Q from Q= 2.Pi.f.L/R

Let Q=2*Pi*ResonantFrequency*TotalInductance/ResistanceOfCoil



Print "Q " ;Q



'Print "Q infinite"

'to calculate the total mass of copper in the winding Area times lengh
times density



let
TotalWindingMass=Pi*(((DiameterOfWire/2)^2)-((DiameterOfLumen/2)^2))*Len
gthOfWire*DensityOfCopper

Print "Winding Mass ";TotalWindingMass;" Kg"





'to calculate slow wave



Print "Half Wavelength "; LengthOfFormer/2

Print "Velocity "; LengthOfFormer*ResonantFrequency

Print "Peak Potential "; 1.4*6000

Let
PeakCurrent=1.4*6000/(TotalInductance*2*Pi*ResonantFrequency+ResistanceO
fCoil)

Print "Peak Current ";PeakCurrent

Let PeakAmpereTurns=PeakCurrent*NumberOfTurns

Let PeakAmpereTurnsPerMeter=PeakAmpereTurns/(LengthOfFormer/3)

Print "Peak Ampere Turns "; PeakAmpereTurns

Print "Plasma Current "; PeakAmpereTurns

Let PlasmaCurrentDensity=PeakAmpereTurns/((1.5E-3)*LengthOfFormer/3)

Print "Plasma Current Density ";3*PlasmaCurrentDensity; " Amp per
meter^2, near centre"

Let PressureOnPlasma=
((3*PlasmaCurrentDensity*PeakAmpereTurnsPerMeter)*mu)/((DiameterOfFormer
-(6E-3))/2)

Print "Additional Pressure of Plasma ";PressureOnPlasma; " Pa"

Print "Power lost through resistance ";
ResistanceOfCoil*((PeakCurrent^2)/1.4); " Watt"

'Power = 0.5 watt/m^3/kPa^2 10torr=133 Pa

Let
PowerGenerated=0.5*(LengthOfFormer/3)*3^2*2*Pi*((133+PressureOnPlasma)/1
000)^2

Print "Power Generated ";PowerGenerated;" Watt"

input "1 exit 2 loop "; ok

if ok=2 then goto [InductanceLoop]





End





Dr Chris

http://www.cs003o327.webspace.virginmedia.com/
Christopher Strevens
2013-01-27 07:56:32 UTC
Permalink
Raw Message
This theory depends on the existence of a force towards the axis of a
coil with alternating current in the winding. This means the force on a
conducting loop inside the coil being higher near the wire. I showed
this in "reactor theory" using Math Cad for which I have a non
commercial license. These gif images were deleted from my web page by
the police who did other adjustments to my web page without my knowledge
or permission.



Dr Chris

http://www.cs003o327.webspace.virginmedia.com/



From: Christopher Strevens
[mailto:***@hotmail.co_1.uk_1]
Posted At: 26 January 2013 15:07
Posted To: sci.physics.fusion
Conversation: Thermonuclear Fusion (generating algorithm of print out)
Subject: Thermonuclear Fusion (generating algorithm of print out)



'This computer simulation shows the power likely to be generated by a
fusion engine like mine using different parameters.

'There are several conflicting factors so the best solution is
discovered by human intelligence.





'Program to find the inductance of a multilayer coil by summing the
inductance of each layer and the effect of mutual inductance

'Inductance = 4.piE-7.A.n^2/l



'to find the final diameter of a multilayer coil and the length of wire

'each layer adds thickness to layer so first work out turns per layer

let Pi=3.14215926

'for room temp

Let ResistivityOfCopper=1.68E-8

Let DensityOfCopper=8.94E3

'Kg per m^3

'for superconductor

Let ResistovityOfCopper=0

'for LN temp

'Let ResistivityOfCopper=2.647E-9

Let mu=4E-7*Pi

Let e0=8.85E-12

[InductanceLoop]



input "diameter of wire in mm " ; DiameterOfWire

Input "Diameter Of Lumen mm "; DiameterOfLumen

input "Thickness of Insulation mm ";ThicknessOfInsulation

input "diameter of former in mm ";DiameterOfFormer

input "length of former in mm " ;LengthOfFormer

'Input "Resonant Frequency Hz ";ResonantFrequency



let DiameterOfWire=DiameterOfWire/1000

Let DiameterOfLumen=DiameterOfLumen/1000

Let ThicknessOfInsulation=ThicknessOfInsulation/1000

Let DiameterOfFormer=DiameterOfFormer/1000

Let LengthOfFormer=LengthOfFormer/1000

input "number of turns "; NumberOfTurns

let
TurnsPerLayer=LengthOfFormer/(DiameterOfWire+2*ThicknessOfInsulation)

let NumberOfLayers=NumberOfTurns/TurnsPerLayer

Let
ThicknessOfWinding=NumberOfLayers*(DiameterOfWire+2*ThicknessOfInsulatio
n)

Let DiameterOfWinding=DiameterOfFormer+2*ThicknessOfWinding

'length of wire for each layer is the layer diameter multiplied by Pi

LayerN=0

LengthOfWire=0

Let TotalInductance=0

Let TotalCapacitance=0

[ForLoop]

LayerN=LayerN+1

Let
LayerDiameter=DiameterOfFormer+(LayerN-1)*(DiameterOfWire+2*ThicknessOfI
nsulation)

Let LengthOfWire=Pi*LayerDiameter*TurnsPerLayer+LengthOfWire

let
LayerInductance=mu*Pi*((LayerDiameter/2)^2)*TurnsPerLayer^2/LengthOfForm
er

Let
LayerCapacitance=e0*Pi*LayerDiameter*((DiameterOfWire+2*ThicknessOfInsul
ation)/(DiameterOfWire+2*ThicknessOfInsulation))*TurnsPerLayer

Let
InterLayerCapacitance=e0*Pi*LayerDiameter*LengthOfFormer/(DiameterOfWire
+2*ThicknessOfInsulation)

Let TotalInductance=LayerInductance+TotalInductance

Let
TotalCapacitance=LayerCapacitance+InterlayerCapacitance+TotalCapacitance

if LayerN<NumberOfLayers then goto [ForLoop]

Let
ResistanceOfCoil=ResistivityOfCopper*LengthOfWire/(((DiameterOfWire/2)^2
)-((DiameterOfLumen/2)^2))*Pi

Let TotalInductance=TotalInductance*NumberOfLayers^2



Print "Turns per layer ";TurnsPerLayer

Print "Number of Layers ";NumberOfLayers

Print "Diameter of coil ";DiameterOfWinding

Print "Length of Wire ";LengthOfWire; " Meters"

Print "Resistance of Coil "; ResistanceOfCoil ; " Ohms"

Print "Inductance of Multilayer Coil ";TotalInductance;" Henry"

Print "Capacitance of Mutilayer Coil ";TotalCapacitance*1E6; "
microfarads"

'input "1=exit 2=InductanceLoop ";OK

'if OK=2 then goto [InductanceLoop]

'to calculate dynamic impedance



'dynamic impedance is L/Cr where r is the resitance of the coil.









[ResonanceLoop]

'input "Resonant Frequency Hz "; ResonantFrequency











input "capacitance of external capacitor "; ExternalCapacitance

Let Capacitance=TotalCapacitance+ExternalCapacitance

'input "Capacitance (F) "; Capacitance

let impedance = TotalInductance/(Capacitance*ResistanceOfCoil)

print "dynamic impedance "; impedance

'print "dynamic impedance - Infinite"



'program to calculate resonant frequency

'f=1/(2.pi.sqr(l.c))









'input "Inductance of coil "; Inductance

let ResonantFrequency=1/(2*Pi*(sqr(TotalInductance*Capacitance)))

print "Resonant frequency is "; ResonantFrequency



input "1 for exit 2 for loop "; ok



if ok=2 then goto [InductanceLoop]





'to calculate Q from Q= 2.Pi.f.L/R

Let Q=2*Pi*ResonantFrequency*TotalInductance/ResistanceOfCoil



Print "Q " ;Q



'Print "Q infinite"

'to calculate the total mass of copper in the winding Area times lengh
times density



let
TotalWindingMass=Pi*(((DiameterOfWire/2)^2)-((DiameterOfLumen/2)^2))*Len
gthOfWire*DensityOfCopper

Print "Winding Mass ";TotalWindingMass;" Kg"





'to calculate slow wave



Print "Half Wavelength "; LengthOfFormer/2

Print "Velocity "; LengthOfFormer*ResonantFrequency

Print "Peak Potential "; 1.4*6000

Let
PeakCurrent=1.4*6000/(TotalInductance*2*Pi*ResonantFrequency+ResistanceO
fCoil)

Print "Peak Current ";PeakCurrent

Let PeakAmpereTurns=PeakCurrent*NumberOfTurns

Let PeakAmpereTurnsPerMeter=PeakAmpereTurns/(LengthOfFormer/3)

Print "Peak Ampere Turns "; PeakAmpereTurns

Print "Plasma Current "; PeakAmpereTurns

Let PlasmaCurrentDensity=PeakAmpereTurns/((1.5E-3)*LengthOfFormer/3)

Print "Plasma Current Density ";3*PlasmaCurrentDensity; " Amp per
meter^2, near centre"

Let PressureOnPlasma=
((3*PlasmaCurrentDensity*PeakAmpereTurnsPerMeter)*mu)/((DiameterOfFormer
-(6E-3))/2)

Print "Additional Pressure of Plasma ";PressureOnPlasma; " Pa"

Print "Power lost through resistance ";
ResistanceOfCoil*((PeakCurrent^2)/1.4); " Watt"

'Power = 0.5 watt/m^3/kPa^2 10torr=133 Pa

Let
PowerGenerated=0.5*(LengthOfFormer/3)*3^2*2*Pi*((133+PressureOnPlasma)/1
000)^2

Print "Power Generated ";PowerGenerated;" Watt"

input "1 exit 2 loop "; ok

if ok=2 then goto [InductanceLoop]





End





Dr Chris

http://www.cs003o327.webspace.virginmedia.com/

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