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),
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 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:
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 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 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 Plnnck Constant.

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

Recall that:
Lp = 2 Pi Ro Ca

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

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 = ((Rs / Ro) - 1)(1 - (Rc / Ro))

Hence:

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

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

Now express Ro in terms of Kc:
At R = Ro:
[dZ / dR]^2 = 0
which implies that:
{(Kc^2 Ro^2) / (Rs - Ro) (Ro - Rc)]} = 1
or
Kc^2 = [(Rs / Ro) - 1)(1 - (Rc / Ro)]

Alternatively:
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)
= {(Rs + Rc) +/- [(Rs - Rc)^2 - (4 Kc^2 Rs Rc)]^0.5} / 2 (1 + Kc^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.

For real solutions Kc^2 can take values in the range:
0 < Kc^2 < (Rs - Rc)^2 / 4 Rs Rc

When Kc takes its maximum value of:
(Rs - Rc)^2 / 4 Rs Rc
then
Ro = (Rs + Rc) / 2 (1 + Kc^2)
= [(Rs + Rc) 4 Rs Rc] / [2 (4 Rs Rc + (Rs - Rc)^2]
= [(Rs + Rc) 4 Rs Rc] / [2 (Rs + Rc)^2]
= 2 Rs Rc / (Rs + Rc)

Spheromak scaling implies that:
Rs / Rc = Kr= constant.

Then in general:
Ro = {(Rs + Rc) +/- [(Rs - Rc)^2 - (4 Kc^2 Rs Rc)]^0.5} / 2 (1 + Kc^2)
= {(Kr + 1) Rc +/- [ (Kr - 1)^2 Rc^2 -(4 Kc^2 Kr Rc^2]^0.5 / 2 (1 + Kc^2)
= Rc {(Kr + 1) +/- [ (Kr - 1)^2 - (4 Kc^2 Kr)]^0.5 / 2 (1 + Kc^2)

Then when Kc^2 takes its maximum value:
Ro = 2 Rs Rc / (Rs + Rc)
= 2 Kr Rc / (Kr + 1)
or
[Ro / Rc] =2 Kr / (Kr + 1)
or
[Ro / Rc}^2 = 4 Kr^2 / (Kr + 1)^2
 

On the web page title:
spheromak Approximation we find that the operating value of [Ro / Rc]^2 is given by:
[Ro / Rc]^2 = 2 Kr^2 / (2 Kr + 1)