XYLENE POWER LTD.

SPHEROMAK ENERGY:

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

This web page relies on results derived on the web page titled THEORETICAL SPHEROMAK. The various components of the total spheromak field energy are found in terms of the spheromak geometry and the field energy densities:
 

RELEVANT RESULTS FROM THE WEB PAGE TITLED: Theoretical Spheromak

FIELD ENERGY DENSITY OUTSIDE THE SPHEROMAK WALL:
Assume that in the region outside the spheromak wall the field energy density is given by:
Up = Uo {[Ro^2 / [(K Ro - R)^2 + Z^2]}^2.

Note that K is a constant slightly more than (1 / 2) which causes the peak value of Up:
Up = Upo
at R = 0, Z = 0 to exceed Uo.
 

FIELD ENERGY DENSITY INSIDE THE SPHEROMAK WALL:
In the quasi-toroidal region inside the spheromak wall where:
the total field energy density is given by:
Ut = Uto (K Ro / R)^2
 

SPHEROMAK WALL GEOMETRY REVIEW:
At the spheromak wall:
Ut = Up
which gives the equation of the spheromak wall as:
[Uto / Uo]^0.5 = {[Ro Rw / K] / [(K Ro - Rw)^2 + Zw^2]}

Choose Ut = Uto at Rw = K Ro

At Rw = K Ro this spheromak wall equation gives:
[Uo / Uto]^0.5 = [+ Zw|(Rw = K Ro)]^2 / Ro^2 = [Ho^2 / Ro^2]
where:
Ho = Z|(Rw = K Ro)

Thus the equation for the spheromak wall becomes:
[Ho / Ro]^2 = [K /(Ro Rw)][(K Ro - Rw)^2 + Zw^2]
or
Ho^2 = [K Ro / Rw][(K Ro - Rw)^2 + Zw^2]
or
Zw^2 = [Rw / K Ro] [Ho^2] - [(K Ro - Rw)^2]

The maximum overall length of the spheromak occurs at dZw / dRw = 0 where:
2 Zw dZw = 0 = [1 / K Ro][Ho^2] + 2 (K Ro - Rw)
or
Rw = Rm = [1 / (2 K Ro)][Ho^2] + K Ro
or
[Rm / Ro] = [1 / 2 K](Ho^2 / Ro^2) + K
= [1 / 2 K][1 + (1 / Nt)] + K
CHECK

EXPRESSIONS FOR Rc and Rs:
The spheromak wall crosses the equatorial plane at Zw = 0 or at:
Ho^2 - [K Ro / Rw][(K Ro - Rw)^2] = 0
which is a quadratic equation in Rw:
Rw Ho^2 = K Ro (K^2 Ro^2 -2 K Rw Ro + Rw^2)
or
K Ro Rw^2 + (-2 K^2 Ro^2 - Ho^2) Rw + K^3 Ro^3 = 0
which has real solutions:
Rc = K Ro + (Ho^2 / 2 K Ro) - (Ho^2 / 2 K Ro)[(4 K^2 Ro^2 / Ho^2) + 1]^0.5
and at
Rs = K Ro + (Ho^2 / 2 K Ro) + (Ho^2 / 2 K Ro)[(4 K^2 Ro^2 / Ho^2) + 1]^0.5
 

SPHEROMAK SHAPE FACTOR:
Define the spheromak shape factor [Ho / Ro]^2 by:
[Ho / Ro]^2 = [Uo / Uto]^0.5 = 4 K^2 = Np / Nt

We anticipate finding that a quantum spheromak adopts a shape factor defined by:
[Ho^2 / Ro^2] = 1 + (1 / Nt) = (Nt + 1) / Nt = Np / Nt
 

SPHEROMAK PROFILE:
Recall that:
Zw^2 = [Rw / K Ro] [Ho^2] - [(K Ro - Rw)^2]
or
Zw = +/- {[Rw / K Ro] [Ho^2] - [(K Ro - Rw)^2]}^0.5
or
Zw / Ro = +/- {[Rw / K Ro][Ho^2 / Ro^2] - [K - (Rw / Ro)]^2}^0.5

SUMMARY OF RESULTS FROM THE WEB PAGE TITLED: Theoretical Spheromak

FIELD ENERGY DENSITY INSIDE THE SPHEROMAK WALL:
In the toroidal region inside the spheromak wall where:
the field energy density is given by:
Ut = Uto (Ro / R)^2

At the spheromak walls:
Ut = Up

Thus at R = Rc, Z = 0:
Uo [Ro^2 / (K Ro - Rc)^2)]^2 = Uto [Ro / Rc]^2
or
Uto = Uo [Rc / Ro]^2 [Ro^2 / (K Ro- Rc)^2]^2
= Uo [Rc Ro / (K Ro - Rc)^2]^2

Hence inside the spheromak wall:
Ut = Uto (K Ro / R)^2
= Uo {[Rc Ro / (K Ro - Rc)^2]^2} (K Ro / R)^2
 

SPHEROMAK WALL POSITION:
Recall that the locus of points defining the spheromak wall is given by:
Zw^2 = [Rw / K Ro] [Ho^2] - [(K Ro - Rw)^2]

FIELD ENERGY DENSITY SUMMARY:
Thus we have expressions for the field energy density everywhere around and within a spheromak and we have a closed form expression for the spheromak wall position. Via suitable integration we can use these expressions to find the total field energy of a theoretical spheromak in free space.
 

COMPONENTS OF FIELD ENERGY OF A SPHEROMAK:
The total field energy Ett associated with a spheromak can be expressed in terms of the spheromak nominal energy density Uo, the spheromak nominal radius Ro and its shape parameter [Ho / Ro] where:
[Ho / Ro]^2 = K [Uo / Uto]^0.5

The total spheromak field energy Ett is:
Ett = Ef - Efp + Eft
where:
 
Ef = maximum field energy if no spheromak wall was present so the energy density in all of space was given by:
U = Uo {Ro^2 / [(K Ro - R)^2 + Z^2]}^2
 
Efp = field energy that would be inside the spheromak wall if the energy density function inside the spheromak wall was the same as the energy density function outside the spheromak wall.
 
Eft = toroidal field energy that actually exists inside the spheromak wall

Each of these spheromak field energy components is evaluated below.

An important fraction is:
[Ett / Ef] = [Ef - Efp + Eft] / Ef
= [1 - (Efp / Ef) + (Eft / Ef)]
Each of these fractions is calculated herein.
Then:
Ett = [Ett / Ef] Ef

The value of Ro is dependent on the total amount of energy present. As shown herein integration over all space shows that the total energy present Ett is typically a few percent less than:
Ef = [Uo Ro^3 Pi^2 /(2 K)]

Thus a spheromak can potentially model a real particle or a real plasma with a total energy Ett and a characteristic radius Ro.
 

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FIELD ENERGY Ef:
The upper limit on the total field energy Ett is the field energy Ef which can be obtained by integrating the outside field energy density function over all space.

Use cylindrical elements of volume with energy density that changes along the cylinder length. The energy contained in one element of volume is given by:
Integral from Z = - infinity to Z = infinity of:
Uo {Ro^2 / [(K Ro - R)^2 + Z^2)]}^2 2 Pi R dR dZ

= Uo Ro^4 2 Pi R dR Integral from Z = - infinity to Z = infinity of:
dZ / [(K Ro - R)^2 + Z^2)]}^2

which Dwight 120.2 gives as:
= Uo Ro^4 2 Pi R dR from Z = - infinity to Z = infinity of:
{Z / {(2 (K Ro - R)^2 [(K Ro - R)^2 + Z^2]}}| + {1 / [2 (K Ro - R)^3]} arc tan(Z / (K Ro - R)}|

= Uo Ro^4 2 Pi R dR {1 / [2 (K Ro - R)^3]} arc tan(Z / (K Ro - R)}|(Z = infinity)
- Uo Ro^4 2 Pi R dR {1 / [2 (K Ro - R)^3]} arc tan(Z / (K Ro - R)}|(Z = - infinity)

= {Uo Ro^4 2 Pi R dR / [2 (K Ro - R)^3]}[Pi / 2]
- {Uo Ro^4 2 Pi R dR / [2 (K Ro - R)^3]}[- Pi / 2]

= {Uo Ro^4 2 Pi R dR / [2 (K Ro - R)^3]}[Pi]

= {Uo Ro^4 Pi^2 R dR / [(K Ro - R)^3]}

The energy potentially contained in the entire space is:
Ef = Uo Ro^4 Pi^2 Integral from R = 0 to R = infinity of:
R dR / [(K Ro - R)^3]
which from Dwight 91.3 is:
Ef = Uo Ro^4 Pi^2 {[- 1 / (K Ro - R)] + [K Ro / 2 (K Ro - R)^2]}|(R = infinity)
- Uo Ro^4 Pi^2 {[- 1 / (K Ro - R)] + [K Ro / 2 (K Ro - R)^2]}|(R = 0)

= Uo Ro^4 Pi^2 {[ 1 / (K Ro)] - [K Ro / 2 (K Ro)^2]}

= Uo Ro^4 Pi^2 {[1 / (K Ro)] - [1 / 2 (K Ro)]}

= Uo Ro^4 Pi^2 (+ 1 / 2 K Ro)
= Ef = (Uo / 2 K) Ro^3 Pi^2
This result has been double checked.

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TOROIDAL FIELD ENERGY Eft CONTENT OF SPHEROMAK:
The toroidal field energy content of the spheromak is found by integrating over elemental disks. Each disk has inside radius Rc(Zw), outside radius Rs(Zw) and thickness dZ.

The toroidal energy density function is:
Ut = Uto [K Ro / R]^2

The toroidal energy Eft is:
Eft = Integral from Zw = - Hm to Zw = + Hm, Integral from R = Rc(Zw) to R = Rs(Zw) of:
Ut 2 Pi R dR dZ

= Integral from Z = 0 to Z = + Hm, Integral from R = Rc(Zw) to R = Rs(Zw) of:
Ut 4 Pi R dR dZ

The web page titled: THEORETICAL SPHEROMAK gives:
Ut = Uto (K Ro / R)^2

Hence:
Eft = Integral from Z = 0 to Z = + Hm, Integral from R = Rc(Zw) to R = Rs(Zw) of:
Uto (K Ro / R)^2 4 Pi R dR dZ

= Integral from Z = 0 to Z = + Hm, Integral from R = Rc(Zw) to R = Rs(Zw) of:
Uto (K Ro)^2 4 Pi dZ (dR / R)
= Integral from Z = 0 to Z = + Hm of:
Uto (K Ro)^2 4 Pi Ln[Rs(Zw) / Rc(Zw)] dZ
= Integral from Z = 0 to Z = + Hm of:
Uto K^2 Ro^3 4 Pi Ln[Rs(Zw) / Rc(Zw)] (dZ / Ro)
= Integral from Z = 0 to Z = + Hm of:
[Uto / Uo] Uo K^2 Ro^3 4 Pi Ln[Rs(Zw) / Rc(Zw)] (dZ / Ro)
= Integral from Z = 0 to Z = + Hm of:
[1 / 4 K^2] Uo K^2 Ro^3 4 Pi Ln[Rs(Zw) / Rc(Zw)] (dZ / Ro)
= Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro) of:
Uo Ro^3 Pi Ln[Rs(Zw) / Rc(Zw)] (dZ / Ro)

For the solution:
Ho^2 / Ro^2 = 4 K^2
the web page titled: Theoretical Spheromak gives:
Rc(Zw) = 3 (K Ro) - [8 (K Ro)^2 - Zw^2]^0.5
and
Rs(Zw) = 3 (K Ro) + [8 (K Ro)^2 - Zw^2]^0.5

Rs(Zw) / Rc(Zw) = {3 + [8 - (Zw / K Ro)^2]^0.5} / {3 - [8 - (Zw / K Ro)^2]^0.5}

This integration needs numerical evaluation. Increment through:
0 < (Zw / Ro) < (Hm / Ro)
where:
(Hm / Ro) = 2 2^0.5 K

At each increment calculate the corresponding value of [Rs(Zw) / Rc(Zw)].
Then use that value to calculate dEft. Sum the dEft values.

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POLOIDAL FIELD ENERGY Efp LOST FROM REGION INSIDE THE SPHEROMAK WALL:
The energy related to the outside field that could potentially be inside the spheromak wall is found by dividing the inside region into elemental disks. Within each disk the Z value is constant. The inside radius of each elemental disk is Rc(Z). The outside radius of each elemental dsik is Rs(Z). The thickness of each dislk is dZ. This calculation is analogous to the calculation of Eft except that the energy density functionis different.

Efp = Integral from Z = - Hm to Z = + Hm
Integral from R = Rc(Z) to R = Rs(Z)
Up (2 Pi R) dZ dR
 
= Integral from Z = 0 to Z = + Hm
Integral from R = Rc(Z) to R = Rs(Z)
2 Up (2 Pi R) dZ dR
 
OK
= Integral from Z = 0 to Z = + Hm
Integral from R = Rc(Z) to R = Rs(Z)
Uo {Ro^2 / [(K Ro - R)^2 + Z^2]}^2 (2 Pi R) 2 dZ dR
OK

Let:
X = (K Ro - R)
dX = - dR
R = K Ro - X

Then:
Efp = Integral from Z = 0 to Z = + Hm
Integral from R = Rc(Z) to R = Rs(Z)
Uo {Ro^2 / [(K Ro - R)^2 + Z^2]}^2 (2 Pi R) 2 dZ dR

OK
= Integral from Z = 0 to Z = + Hm
Integral from X(Z) = K Ro - Rc(z) to X(Z) = K Ro - Rs(Z)
Uo (4 Pi Ro^4) {1 / [X(Z)^2 + Z^2]}^2 ((+ K Ro - X(Z))) (-dX) dZ

OK
= Ro^-1 Uo 4 Pi Ro^4 Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
Integral from (X / Ro) = (K Ro - Rc) / Ro to (X / Ro) = (K Ro - Rs) / Ro
(X(Z) / Ro) d(X / Ro) d(Z / Ro) / [(X / Ro)^2 + (Z / Ro)^2]^2

- Ro^-2 Uo 4 Pi K Ro Ro^4 Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
Integral from (X(Z) / Ro) = (K Ro - Rc(Z)) / Ro to (X(Z) / Ro) = (K Ro - Rs(Z)) / Ro
d(Z / Ro) d(X / Ro) / [(X / Ro)^2 + (Z / Ro)^2]^2
OK

Define:
INT1 = Uo 4 Pi Ro^3 Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
Integral from (X / Ro) = (K Ro - Rc) / Ro to (X / Ro) = (K Ro - Rs) / Ro
(X / Ro) d(X / Ro) d(Z / Ro) / [(X / Ro)^2 + (Z / Ro)^2]^2

and
INT2 = Uo 4 Pi K Ro^3 Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
Integral from (X / Ro) = (K Ro - Rc) / Ro to (X / Ro) = (K Ro - Rs) / Ro
d(X / Ro) d(Z / Ro) / [(X / Ro)^2 + (Z / Ro)^2]^2

where:
Efp = INT1 - INT2

The above two integrations need to be focused on, one at a time.

Dwight 121.2 gives:
INT1 = Uo Ro^3 4 Pi Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
Integral from (X(Z) / Ro) = ((K Ro - Rc(Z)) / Ro) to (X(Z) / Ro) = ((K Ro - Rs(Z)) / Ro)
(X / Ro) d(X / Ro) d(Z / Ro) / [(X / Ro)^2 + (Z / Ro)^2]^2

= Uo Ro^3 4 Pi Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
{(- 1 / 2[(X(Z) / Ro)^2 + (Z / Ro)^2]|((X(Z) / Ro) = (K Ro - Rs(Z)) / Ro)
- (- 1 / 2[(X(Z) / Ro)^2 + (Z / Ro)^2]|((X(Z) / Ro) = (K Ro - Rc(Z)) / Ro)}
d(Z / Ro)

OK
= Uo Ro^3 4 Pi Integral from (Z / Ro) = 0 to (Z/ Ro) = + (Hm / Ro)
{(- 1 / 2 [(K Ro - Rs(Z)) / Ro)^2 + (Z / Ro)^2]
- (- 1 / 2(((K Ro - Rc(Z)) / Ro)^2 + (Z /Ro)^2)}
d(Z /Ro)

OK
= Uo Ro^3 4 Pi Integral from (Z / Ro) = 0 to (Z/ Ro) = + (Hm / Ro)
{(- 1 / 2 [(K Ro - Rs(Z)) / Ro]^2)
- (- 1 / 2[(K Ro - Rc(Z)) / Ro]^2)}
d(Z /Ro)


= Uo Ro^3 4 Pi Integral from (Z / Ro) = 0 to (Z/ Ro) = + (Hm / Ro)
{(+ 1 / 2 {[Rs(Z)/ Ro]^2 - [Rc(Z)/ Ro]^2)}
d(Z /Ro)
OK

The web page titled: Theoretical Spheromak gives:
Rc(Zw) = 3 (K Ro) - [8 (K Ro)^2 - Zw^2]^0.5
and
Rs(Zw) = 3 (K Ro) + [8 (K Ro)^2 - Zw^2]^0.5

Hence:
[Rc(Z) / Ro] = 3 K - [8 K^2 - (Zw / Ro)^2]^0.5
and
[Rs(Z) / Ro] = 3 K + [8 K^2 - (Zw / Ro)^2]^0.5

Thus:
INT1 = Uo Ro^3 4 Pi Integral from (Z / Ro) = 0 to (Z/ Ro) = + (Hm / Ro)
{(+ 1 / 2 {[Rs(Z)/ Ro]^2 - [Rc(Z)/ Ro]^2)}

= Uo Ro^3 4 Pi Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
6 K [8 K^2 - (Z / Ro)^2]^0.5 d(Z /Ro)

Thus INT1 can be numerically evaluated.

Recall that:
INT2 = Uo 4 Pi K Ro^3 Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
Integral from (X / Ro) = ((K Ro - Rc) / Ro) to (X / Ro) = ((K Ro - Rs) / Ro)
d(X / Ro) d(Z / Ro) / [(X / Ro)^2 + (Z / Ro)^2]^2

OK

Dwight 120.2 gives:
INT2 = Uo 4 Pi K Ro^3 Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
{[(X / Ro) / 2 (Z / Ro)^2 ((Z / Ro)^2 + (X / Ro)^2)] + [1 / 2 (Z / Ro)^3][arc tan(X / Z)]
|((X / Ro) = (K Ro - Rs) / Ro)
- [(X / Ro) / 2 (Z / Ro)^2 ((Z / Ro)^2 + (X / Ro)^2)] + [1 / 2 (Z / Ro)^3][arc tan(X / Z)]
|((X / Ro) = (K Ro - Rc)/ Ro)}
d(Z / Ro)

OK
= Uo 4 Pi K Ro^3 Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
{[((K Ro - Rs(Z)) / Ro) / 2 (Z / Ro)^2 ((Z / Ro)^2 + ((K Ro - Rs(Z)) / Ro)^2)] + [1 / 2 (Z / Ro)^3][arc tan((K Ro - Rs(Z)) / Z)]
- [((K Ro - Rc(Z))/ Ro) / 2 (Z / Ro)^2 ((Z / Ro)^2 + ((K Ro - Rc(Z)) / Ro)^2] + [1 / 2 (Z / Ro^3][arc tan((K Ro - Rc(Z)) / Z)]}
d(Z / Ro)

OK

Recall that:
Rc(Zw) = 3 (K Ro) - [8 (K Ro)^2 - Zw^2]^0.5
and
Rs(Zw) = 3 (K Ro) + [8 (K Ro)^2 - Zw^2]^0.5

or
[8 (K Ro)^2 - Zw^2]^0.5 = 3(K Ro) - Rc
and [8 (K Ro)^2 - Zw^2]^0.5 = Rs - 3 (K Ro)
or
[8 (K Ro)^2 - Zw^2] = 9 (K Ro)^2 - 6 (K Ro) Rc + Rc^2
and [8 (K Ro)^2 - Zw^2] = Rs^2 - 6 (K Ro) Rs + 9 (K Ro)^2
or
Zw^2 + (K Ro)^2 - 2 K Ro Rc + Rc^2 = 4 K Ro Rc
and Zw^2 + (K Ro)^2 - 2 K Ro Rs + Rs^2 = 4 K Ro Rs
or (K Ro - Rc(Z))^2 + Z^2 = 4 (K Ro) Rc(Z)
and
(K Ro - Rs(Z))^2 + Z^2 = 4 (K Ro) Rs(Z)
or
[(K Ro - Rc(Z))^2 + Z^2] / Ro^2 = 4 K Rc(Z) / Ro
and
[(K Ro - Rs(Z))^2 + Z^2] / Ro^2 = 4 K Rs(Z) / Ro

Hence:
INT2 = Uo 4 Pi K Ro^3 Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
{[((K Ro - Rs(Z)) / Ro) / 2 (Z / Ro)^2 ((Z / Ro)^2 + ((K Ro - Rs(Z)) / Ro)^2)] + [1 / 2 (Z / Ro)^3][arc tan((K Ro - Rs(Z)) / Z)]
- [((K Ro - Rc(Z)) / Ro) / 2 (Z / Ro)^2 ((Z / Ro)^2 + ((K Ro - Rc(Z)/ Ro)^2)] + [1 / 2 (Z / Ro)^3][arc tan((K Ro - Rc(Z)) / Z)]}
d(Z/ Ro)


= Uo 4 Pi K Ro^3 Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
{[((K Ro - Rs(Z))/ Ro) / 2 (Z / Ro)^2 (4 K Rs(Z) / Ro)] + [1 / 2 (Z / Ro)^3][arc tan((K Ro - Rs(Z)) / Z)]
- [((K Ro - Rc(Z))/Ro) / 2 (Z /Ro)^2 (4 K Rc(Z) / Ro)] + [1 / 2 (Z /Ro)^3][arc tan((K Ro - Rc(Z)) / Z)]}
d(Z / Ro)


= Uo 4 Pi K Ro^3 Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
{[((K Ro - Rs(Z))/ Ro) / (Z / Ro)^2 (4 K Rs(Z))/Ro)] + [1 / (Z / Ro)^3][arc tan((K Ro - Rs(Z)) / Z)]
- [((K Ro - Rc(Z))/ Ro) / (Z / Ro)^2 (4 K (Rc(Z)/ Ro))] + [1 / (Z / Ro)^3][arc tan((K Ro - Rc(Z)) / Z)]}
d(Z / Ro)


= Uo 4 Pi K Ro^3 Integral from (Z /Ro) = 0 to (Z / Ro) = + (Hm / Ro)
{[(K) / (Z / Ro)^2 (4 K (Rs(Z)/ Ro))] + [1 / (Z / Ro)^3][arc tan((K Ro - Rs(Z)) / Z)]
- [(K) / (Z / Ro)^2 (4 K (Rc(Z) / Ro))] + [1 / (Z / Ro)^3][arc tan((K Ro - Rc(Z)) / Z)]}
d(Z / Ro)

= [Uo 2 Pi K Ro^3] Integral from (Z / Ro) = 0 to (Z / Ro) = + (Hm / Ro)
{[1 / [(Z / Ro)^2 (4 Rs(Z) / Ro)]] + [(Ro / Z)^3][arc tan((K Ro - Rs(Z)) / Z)]
- [1 / [(Z / Ro)^2 (4 Rc(Z) / Ro)]] + [(Ro / Z)^3][arc tan((K Ro - Rc(Z)) / Z)]}
d(Z / Ro)

INT2 needs numerical evaluation.

STUDY THESE TERMS AND ENSURE THAT THEY CONVERGE. THERE MIGHT BE A PROBLEM NEAR:
(Z / Ro) = 0

Recall that Efp = INT1 - INT2

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SPHEROMAK TOTAL STATIC FIELD ENERGY: The total spheromak static field energy is given by:
Ett = Ef - Efp + Eft
= Ef [1 - (Efp / Ef) + (Eft / Ef)]
 

Recall that:
Ef = [Uo Ro^3 Pi^2 / 2 K]

Note that for a theoretical spheromak total energy is proportional to Ro^3.

(Efp - Eft) / Efs
= _____________________________________

Note that everywhere inside the spheromak wall:
|Efp| > |Eft|
which implies that spheromaks can exist.
 

NUMERICAL EXAMPLE OF RELATIVE SIZES OF Ef, Eft AND Efp:
For:
Ef = [Uo Ro^3 Pi^2 / (2 K)]
Efp = [Uo Ro^3 Pi^2 / (2 k)] (0.20)
Eft = [Uo Ro^3 Pi^2 / (2 K)] (0.16)
Ett = [Uo Ro^3 Pi^2 / (2 K) (1 - 0.20 + 0.16)
= [Uo Ro^3 Pi^2 / 2 K] (1 - .04)
= 0.96 Ef

Thus the replacement of the outside energy distribution by the inside energy distribution causes a drop in total spheromak energy Ett of about 4%.

Note that Ett is the total field energy in a spheromak.
 

This web page last updated October 24, 2022.