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.
 

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 spheromakk 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:
2 Ho / (Rs - Rc) = Ka
and
(Rs - Rc) / 2 Ro = Kb
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 Ca
Where:
Lp = 2 Pi Ro
Lt = one toroidal turn length
Lp and Lt each scale with Ro.
Ca = shape dependent constant ~ 1
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 electic and magnetic fields can be matched, allowing determinationof Nt, Then knowldege of (Np / Nt) allows determination of Np. These should point toward the Plnnck Constant.

An important isssue is What is the value of Ca?
P>Recall that:
Lp = 2 Pi Ro

Hence:
Lt Ca = A / Lp
or
Ca = [A / ( Lp Lt)]
 

Recall that:
Lp = 2 Pi Ro

Hence:
Lt Cr = A / Lp
= A / 2 PiRo
= Integral from R = Rc to R = Rs of:
[2 R / Ro][(dZ / dR)^2 + 1]^0.5 dR
 

SOLUTION REQUIREMENTS:
1) Lp = 2 Pi Ro
where:
Ro = value of R where:
(dZ / dR) = 0

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, where Ro is unique.
 

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 = {[(Kc^2 R^2) / (Rs - R) (R - Rc)] - 1}

This function has hidden complexity.

Region of validity is:
Rc <= R <= Rs
and
Kc^2 R^2 => (Rs - R)(R - Rc)

Note that (dZ / dR) changes sign at R = Ro at which point (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
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 thesignof(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 intoaccount this (dZ / dR) sign change.
 

FIND Ro:
(dZ / dR) = 0 at R = Ro implies that:
Kc^2 Ro^2 = (Rs - Ro)(Ro - Rc)
or
Kc^2 Ro^2 = - Ro^2 + Ro (Rs + Rc) - Rs Rc
or
Ro^2 (1 + Kc^2) - (Rs + Rc) Ro + Rs Rc = 0
or
Ro = {(Rs + Rc) +/- [(Rs + Rc)^2 - 4 (1 + Kc^2) Rs Rc]^0.5} / 2 (1 + Kc^2)

Ro takes its unique value of (Rs + Rc) / 2 (1 + Kc^2)
at:
(Rs + Rc)^2 - 4 (1 + Kc^2) Rs Rc = 0
or at
(Rs - Rc)^2 - 4 Kc^2 Rs Rc = 0
or at
Kc^2 = (Rs - Rc)^2 / 4 Rs Rc

1 + Kc^2 = [(Rs - Rc)^2 + 4 Rs Rc] / [4 Rs Rc]
= (Rs + Rc)^2 / (4 Rs Rc)

Hence the unique value of Ro is:
Ro = (Rs + Rc) / 2 (1 + Kc^2)
= (Rs + Rc) / [2 (Rs + Rc)^2 / (4 Rs Rc)]
= 2 Rs Rc / (Rs + Rc)
 

At R = Ro:
Kc^2 Ro^2 = [(Rs - Rc)^2 / (4 Rs Rc)][2 Rs Rc / (Rs + Rc)]^2
= [(Rs - Rc) / (Rs + Rc)]^2 [Rs Rc]

At R = Ro:
(Rs - Ro) (Ro - Rc) = (Rs - [2 Rs Rc / (Rs + Rc)]) ([2 Rs Rc / (Rs + Rc)] - Rc)
= [(Rs^2 + Rs Rc - 2 Rs Rc)) / (Rs + Rc)] [(2 Rs Rc - Rc Rs -Rc^2 ) / (Rs + Rc)]
= [Rs (Rs - Rc) / (Rs + Rc)] [Rc (Rs - Rc) / (Rs + Rc)]
= [(Rs - Rc) / (Rs + Rc)]^2 [Rs Rc]

Thus the algebra seems consistent.

Hence:
[dZ / dR]^2 = {[Kc^2] [R^2) / (Rs - R) (R - Rc)] - 1}
or
[dZ / dR]^2 = {[(Rs - Rc)^2 / (4 Rs Rc)] [R^2 / (Rs - R) (R - Rc)] - 1}

[dZ / dR] = +/- {[(Rs - Rc)^2 / (4 Rs Rc)] [R^2 / (Rs - R) (R - Rc)] - 1}^0.5

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
 

DETERMINATION OF Q:
Q = Integral from R = Rc to R = Rs of:
4 Pi [So Ro^2 / R][(dZ / dR)^2 + 1]^0.5 dR
or
Q = Integral from R = Rc to R = Rs of:
4 Pi [So Ro^2 / R] [{[(Kc^2 R^2) / (Rs - R) (R - Rc)] - 1} + 1]^0.5 dR
or
Q = Integral from R = Rc to R = Rs of:
4 Pi [So Ro^2 / R] [[(Kc R) / [(Rs - R) (R - Rc)]^0.5 dR
or
Q = Integral from R = Rc to R = Rs of:
4 Pi [So Ro^2 Kc] / [(Rs - R) (R - Rc)]^0.5 dR
or
Q = Integral from R = Rc to R = Rs of:
[4 Pi So Ro^2 Kc] {dR / [-R^2 + R (Rs + Rc) - (Rs Rc)]^0.5}

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

Hence:
Q = Integral from X = Xc to X = Xs of:
[4 Pi So Ro^2 Kc] {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 interim result that:
Integral from Xc to Xs of:
dx / [(Xs - X) (X - Xc)]^0.5
for Xc < X < Xs
= Pi

Hence:
Q = [4 Pi So Ro^2 Kc] Pi

Rearranging gives:
So = Q / (4 Pi^2 Ro^2 Kc)

Recall that:
Ro = Rs Rc / (Rs + Rc)<
and
Kc^2 = (Rs - Rc)^2 / 4 Rs Rc

Hence:
So = {Q / [4 Pi^2 (Rs Rc)^2]}{[Rs + Rc]^2[4 Rs Rc]^0.5 / (Rs - Rc)
= Q [(Rs + Rc)^2 / (Rs - Rc)] [[1 / (2 Pi^2)] / [(Rs Rc)^1.5]
 

DETERMINATION OF Lt:

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

Xc = Rc / Ro
X = R / Ro
Xs = Rs / Ro
dX = dR / Ro

Lt = Integral from Xc to Xs of:
2 Kc Ro [X / [(Xs - X)(X - Xc)]^0.5 dX

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

From Dwight #380.011BR> Lt = 2 Kc Ro {[(a X^2 + b X + c)^0.5 / a]} evaluated from Xc to Xs

Lt = 2 Kc Ro { [(a Xs^2 + b Xs + c)^0.5 / a]
- (b / 2 a) Pi }
- 2 Kc Ro {[(a Xc^2 + b Xc + c)^0.5 / a] - (b / 2 a) Pi }

Lt = [2 Kc Ro][-b / 2 a][Pi]
= 2 Kc Ro (Xs + Xc) [(Pi / 2)]
Lt = Kc (Rs + Rc) Pi