XYLENE POWER LTD.

SPHEROMAK WALL

By Charles Rhodes, P.Eng., Ph.D.

INTRODUCTION:
This web page develops a function Z(R) which specifies the position of the spheromak wall. A spheromak's wall geometry can be identified by its nominal linear dimension parameter Ro and by its shape parameter:
Kr = (Rs / Rc)
This web page develops expresions for the spheromak wall parameters: poloidal turn length Lp, toroidal turn length Lt, minimum inside diameter Rc, Maximum outside diametrer Rs, surface charge density:
S = So (Ro / R)^2,
and surface area A.

A numerical method is outlined for finding the spheromak height: 2 Ho.
 

SPHEROMAK ISSUES:
In a spheromak the electric field inside the spheromak wall is zero.
Spheromak theory indicates that the spheromak charge Q is uniformly distributed along the length Lh of the charge filament.

spheromak theory further indicates that on the spheromak wall the surface charge density takes the form:
S(R) = So (Ro / R)^2
Where So is the surface charge density at R = Ro

Assume Z(R) is a symetrical quasi-toroid where the detailed shape is unknown./P> The function Z(R) must comply with all of the following requirements.

A cross section of the spheromak wall can be described by a function:
Z(R) where Z is in the range:
- Ho < Z < Ho
 

DEFINITIONS:
R = radius from the Z axis (axis of cylindrical symmetry);
Ro = value of R at (dZ / dR) = 0
Rs = maximum value o fR;
Rc = minimum value of R;
Ho = maximum value of |Z|;
Ka = [2 Ho / (Rs - Rc)];
Kb = [(Rs - Rc) / 2 Ro];
 

SPHEROMAK SYMMETRY:
Due to spheromak symmetry, in the range:
Rc < R < Rs:
Z(R) = - Z(R)
Z(Ro) = Ho, - Ho
Z (Rc) = 0
Z(Rs) = 0

Alternatively:
R(z) has two positive real solutions.
R(z) = R (-Z).
R(Ho) = R(-Ho) = Ro

At Z = 0:
R(Z) = Rc, Rs
 

SPHEROMAK SCALING:
There is a scaling requirement that:
Ka = 2 Ho / (Rs - Rc)
and
Kb = [(Rs - Rc) / 2 Ro ]
and
Kr = (Rs / Rc)
where Ro = value of R at which (dZ / dR) = 0

We anticipate showing that Ka Kb is a function of physical constants.
 

SPHEROMAK SHAPE:
(dZ /dR) = 0 at R = Ro, -Ro.

|(dZ / dR)| goes to +/- infinity at R = Rc, Z = 0 and at R = Rs, Z = 0.
 

SPHEROMAK Lt VALUE:
Lt = Integral from R = Rc to R = Rs of:
2 dLt
or
Lt = Integral from R = Rc to R = Rs of:
2 [(dZ)^2 + (dR)^2]^0.5
or
Lt = Integral from R = Rc to R = Rs of:
2 [(dZ / dR)^2 + 1]^0.5 dR
 

SPHEROMAK SURFACE AREA:
A = Integral from R = Rc to R = Rs of:
2 Pi R 2 dLt
or
A = Lp Lt Ca =Integral from R = Rc to R = Rs of:
4 Pi R [(dZ / dR)^2 + 1]^0.5 dR
 

The spheromak surface area is:
A = Lp Lt
Where:
Lp = one poloidal turn length
Lt = one toroidal turn length
Lp and Lt each scale with Ro.
Ro = value of R where (dZ / dR) = 0.
 

SPHEROMAK SURFACE CHARGE:
S(R) = So (Ro / R)^2
where:
So = charge / unit area on the spheromak wall at R = Ro
 

SPHEROMAK CHARGE:
Q = Integral from R = Rc to R = Rs of: 2 S(R) dA
or
Q = Integral from R = Rc to R = Rs of:
So (Ro / R)^2 4 Pi R [(dZ / dR)^2 + 1]^0.5 dR
or
Q = Integral from R =Rc to R = Rs of:
So (Ro^2 / R) 4 Pi [(dZ / dR)^2 + 1]^0.5 dR
where:
So = charge / unit area at R = Ro, Z = Ho,and at Z = - Ho, R = Ro
and
S(R) = So (Ro^2 / R^2)
 

FIND So:
Rearrange above expression to get:
So = Q / {Integral from R = Rc to R = Rs of:
(Ro^2 / R) 4 Pi [(dZ / dR)^2 + 1]^0.5 dR}
 

SPHEROMAK CONSTRAINT:
There is a further constraint that the spheromak charge Q must be independent of Ro.
 

The importance of So is that it allows determination of S at R = Rs, Z = 0. At this point the electric and magnetic fields can be matched, potentially allowing determination of Nt, Then knowldege of (Np / Nt) allows determination of Np. These should point toward the Planck Constant.
 

SOLUTION REQUIREMENTS:
1) Lp = A / Lt

2) Lt = 2 Integral from R= Rc to R = Rs of:
dLt = Integral from R = Rc to R = Rs of:
2 [(dZ / dR)^2 + 1]^0.5 dR
where:
Lt must be proportional to Ro.

3) THE TOTAL AREA A OF THE SPHEROMAK WALL MUST BE PROPORTIONAL TO Ro^2.
A = Integral from R = Rc to R = Rs of:
4 Pi R [(dZ / dR)^2 + 1]^0.5 dR

4) Q must be independent of Ro, where:
Q = Integral from R = Rc to R = Rs of:
4 Pi R [So / R^2] [(dZ / dR)^2 + 1]^0.5 dR

5) The aforementioned integrals must converge.

6)(dZ / dR) = 0 at R = Ro.
 

SPHEROMAK SOLUTION:
Invoke Achem's Razor that the simplest function that meets all the constraint conditions is likely the right one. The required function must produce to quasi-toroid shape and must meet the spheromak performance conditions on Lt, A, and Q.

Attempt solution:
[dZ / dR]^2 = {[Rx^2 / (Rs - R) (R - Rc)]^0.5 - 1}

This function has hidden complexity.

Region of validity is:
Rc <= R <= Rs

At R = Ro:
[dZ / dR]^2 = 0
or
[Rx^2 / (Rs - Ro) (Ro - Rc)] = 1

Note that (dZ / dR) changes sign at R = Ro at which point (dZ / dR) = 0.

Then:
Rx^2 = (Rs - Ro)(Ro - Rc)

[dZ / dR] = +/- {[Rx^2 / (Rs - R) (R - Rc)]^0.5 - 1}^0.5

For Z > 0 and Rc < R < Ro choose (dZ / dR) > 0
For Z > 0 and Ro < R < Rs choose (dZ / dR) < 0
For Z < 0 and Rc < R < Ro choose (dZ / dR) < 0
For Z < 0 and Ro < R < Rs choose (dZ / dR) > 0

Conclusion:
Integration of (dZ / dR) to find the function Z(R) can only be done via a sequence of line integrals, Where the sign of(dZ / dR) is dependent on the sign of Z and the sign of (R - Ro).
Each of the integrations required to find Lt, A and Q must take into account this (dZ / dR) sign change.
 

FIND Ro in terms of Rx:
(dZ / dR) = 0 at R = Ro implies that:
Rx^2 = (Rs - Ro)(Ro - Rc)
or
Rx^2 = -Ro^2 + Ro (Rs + Rc) - Rs Rc
or
Ro^2 - (Rs + Rc) Ro + (Rs Rc + Rx^2) = 0

Hence:
Ro = {(Rs + Rc) +/- [(Rs + Rc)^2 - 4 (1)(Rs Rc + Rx^2)]^0.5} / 2
or
Ro = {(Rs + Rc) +/- [(Rs - Rc)^2 - (4 Rx^2)]^0.5} / 2
Note that for real values of Ro:
Rx < [(Rs - Rc) / 2]

This equation indicates that Ro has two distinct real solutions, possibly indicating for a positive charge particle that one solution corresponds to a proton and one solution corresponds to a positron. Alternatively for a negative charged particle one solution corresponds to an anti- proton and one solution correponds to an electron.

Recall that:
[dZ / dR]^2 = {[(Rx^2) / (Rs - R) (R - Rc)]^0.5 - 1}

Note that at R = Ro:
[dZ / dR]^2 = 0

Thus:
[dZ / dR] = +/- {[(Rx^2) / (Rs - R) (R - Rc)]^0.5 - 1}^0.5

For real solutions Rx can take values in the range:
0 < Rx < [(Rs - Rc) / 2]

Note that when Rx = 0 the Ro values are Ro = Rc and Ro = Rs

Recall that:
[dZ / dR]^2 = {[(Rx^2) / (Rs - R) (R - Rc)]^0.5 - 1}

Note that (dZ / dR) changes sign at R = Ro where:
(dZ / dR) = 0

The range of validity is: Rc <= R <= Rs

Z(Rs)= Z|R = Rc + Integral from R = Rc to R = Ro of:
(dZ / dR) dR
+ Integral from R = Ro to R = Rs of:
(dZ / dR) dR
where Z(Rc) = Z(Rs) = 0
and the sign of (dZ / dR) switches at R = Ro
 

SUBSTITIONS:

Let X = R / Ro
Xc = Rc/Ro
Xs = Rs / Ro

Hence:
Integral from X = Xc to X = Xs of:
{dX / [-X^2 + X (Xs + Xc) - (Xs Xc)]^0.5}

b^2 = (Xs + Xc)^2 = Xs^2 + 2 Xs Xc + Xc^^2
4 a c = 4 (-1)(- Xs Xc) = 4 Xs Xc
Generally b^2 > 4 a c

Dwight #380.001 gives:
Integral from Xc to Xs of:
dX/ [(Xs - X) (X - Xc)]^0.5
for Xc < X < Xs
= [-1 / (- a)^0.5][arc sin{(2 a X + b) / (b^2 - 4 a c)^0.5} evaluated from Xc to Xs

= -1 arc sin{(-2 X + Xs + Xc) / ((Xs + Xc)^2 - (4 Xs Xc) )^0.5 evaluated from Xc to Xs

=-1 arc sin{(-2 Xs + Xs + Xc) / ((Xs + Xc)^2 - (4 Xs Xc) )^0.5}
+ 1 arc sin{(-2 Xc + Xs + Xc) / ((Xs + Xc)^2 - (4 Xs Xc) )^0.5}

= - 1 arc sin{( -Xs + Xc) / ((Xs - Xc)^2)^0.5}
+ 1 arc sin{(+ Xs - Xc) / ((Xs - Xc)^2}^0.5}

= - arc sin(-1) + arc sin(1)
= -(- Pi / 2) + (Pi / 2)
=Pi

Thus we have the important intermediate result that:
for Xc < X < Xs:
Integral from Xc to Xs of:
dX / [(Xs - X) (X - Xc)]^0.5
= Pi
 

b^2 = (Xs + Xc)^2 = Xs^2 + 2 Xs Xc + Xc^^2
4 a c = 4 (-1)(- Xs Xc) = 4 Xs Xc
Generally b^2 > 4 a c
 

FIND Lt:
Recall that:
Lt = Integral from R = Rc to R = Rs of:
2 [(dZ / dR)^2 + 1]^0.5 dR

Recall that we already have the important interim result that for Xc < X < Xs:
Integral from Xc to Xs of:
dx / [(Xs - X) (X - Xc)]^0.5
= Pi

Lt = Integral from R = Rc to R = Rs of:
2 ( Rx) / [(Rs - R) (R - Rc)]^0.5 dR

FIND AREA A:
A =Integral from R = Rc to R = Rs of:
4 Pi R [(dZ / dR)^2 + 1]^0.5 dR

= Integral from R = Rc to R = Rs of:
4 Pi R [{[( Rx^2) / (Rs - R) (R - Rc)] - 1} + 1]^0.5 dR

= Integral from R = Rc to R = Rs of:
4 Pi Rx {R dR / [[Rs - R) (R - Rc)]^0.5}

= Integral from X = Xc to X = Xs of:
4 Pi Rx Ro {(X dX /[Xs - X) (X - Xc)]^0.5}

X = R / Ro
dX = dR / Ro
a = -1
b = (Xc + Xs)
c = - Xs Xc

Recall that we have the important interim result that:
Integral from Xc to Xs of:
dx / [(Xs - X) (X - Xc)]^0.5
for Xc < X < Xs
= Pi

Dwight 380.011 gives:
= Integral from X = Xc to X = Xs of:
{X dX /[Xs - X) (X - Xc)]^0.5}
= [- b / 2a] Pi
= (Xs + Xc) Pi / 2

Hence Spheromak surface area A is given by:
A = Integral from X = Xc to X = Xs of:
4 Pi Rx Ro {(X dX /[Xs - X) (X - Xc)]^0.5}
=[4 Pi Rx Ro](Xs + Xc) Pi / 2
= 4 Pi Rx (Rs + Rc) Pi / 2
= 2 Pi^2 Rx (Rs + Rc)

FIND LP:
Lp = A / Lt
= 2 Pi^2 Rx (Rs + Rc) / (2 Rx Pi) = Pi (Rs + Rc) = [2 Pi] [(Rs + Rc) / 2]
Note that Rs and Rc scale with Ro so that Lt and Lp scale with Ro.

The values of Lt, Lp and so are used on the web page titled Spheromak Approximation
 

FIND So:
Q = Integral from R = Rc to R = Rs of:
4 Pi R dLt [So Ro^2 / R^2]
= Integral from R = Rc to R = Rs of:
4 Pi [So Ro^2 / R] [{[(Rx^2) / (Rs - R) (R - Rc)] - 1} + 1]^0.5 dR
= Integral from R = Rc to R = Rs of:
4 Pi [So Ro^2 Rx] (1 / R) / [(Rs - R) (R - Rc)]^0.5 dR
= Integral from R = Rc to R = Rs of:
4 Pi [So Ro^2 Rx / Ro] (Ro / R) / [(Rs - R) (R - Rc)]^0.5 dR