In this space, visitors are invited to post any comments, questions, or skeptical observations about Philo T. Farnsworth's contributions to the field of Nuclear Fusion research.Subject: Proof of conductor diameters
Date: Nov 21, 00:24 am
Poster: Joshua ResnickOn Nov 21, 00:24 am, Joshua Resnick wrote:
>>I have started building a HV feed for my new chamber. One of my professors used to design HV power feed through devices and was a wealth of information. My design is based on the fact that the electric field between the conductors in a coaxial cable can be minimized if the ratio of outer diameter to inner diameter is e. So Do/Di = e. This formula is not trivial to derive but you don't have to be a rocket scientist either.
>
>Joshua,
>
>Thanks for sharing this!!!! This is sage advice >from a real designer of real world stuff and >counts in the grand scheme of things. As a >grizzled old engineer myself, info like this is >appreciated and cherished.
All,
While working on some theoretical Fusor stuff under one of the HV profs. at school, I was given an assignment to prove that the outer the above formula Do/Di = e as the ideal radius. I think I did the proof right, I don't know if its the same that is in the books, but here it is anyway, if there is something wrong with it let me know:
Solve for the optimal feed through diameter for a high voltage feed:
Term for the electric field some distance between two coaxial conductors:
> E(r):=phi/(epsilon*2*pi*r);
phi
E(r) := 1/2 ------------
epsilon pi r
Substituting the term for electric flux with a term relative to the electric field associated with the outer conductor:
> E(r)=(epsilon*Eout*2*Pi*ro)/(epsilon*2*pi*r);
Eout Pi ro
E(r) := ----------
pi r
Pi cancels:
> E(r)=(Eout*ro)/(r);
Eout ro
E(r) := -------
r
We are interested in the voltage at some radius between the cylinders in terms of the electric field between the two, so the first step is to integrate our term for E from the inner radius to the outer radius:
> V=Int(Eout*ro/r,r=ri..ro);
ro
/
| Eout ro
V = | ------- dr
| r
/
ri
Substituting the term for Eout in terms of q, and replacing ro with a unit value based on the symmetry of the problem (L is the length of the conductor):
> v=q/(2*pi*epsilon*L)*Int(1/r,r=ri..ro);
ro
/
|
q | 1/r dr
|
/
ri
v = 1/2 --------------
pi epsilon L
Integrating:
> v=q/(2*pi*epsilon*L)*int(1/r,r=ri..ro);
q (ln(ro) - ln(ri))
v = 1/2 -------------------
pi epsilon L
Solving for q:
> v*2*pi*epsilon*L=q*(ln(ro)-ln(ri));
2 v pi epsilon L = q (ln(ro) - ln(ri))
> q=(v*2*pi*epsilon*L)/(ln(ro)-ln(ri));
v pi epsilon L
q = 2 ---------------
ln(ro) - ln(ri)
Using the relationship (r = inner radius):
> E=q/(2*Pi*epsilon*L*r);
q
E = 1/2 --------------
Pi epsilon L r
> E:=(v,ro,r)->(v*2*epsilon*L)/(ln(ro)-ln(r))/(2*epsilon*L*r);
v
E := (v, ro, r) -> ------------------
(ln(ro) - ln(r)) r
Finding the minumum yields the point of lowest electrical stress:
> E1:=simplify(diff(E(v,ro,r),r));
v (-1 + ln(ro) - ln(r))
E1 := - -----------------------
2 2
(ln(ro) - ln(r)) r
> solve(E1=0,r); Solve for E1=0 algebraically
exp(-1) ro
Therefore the optimal value for r is:
> r=exp(-1)*ro;
r = exp(-1) ro
The the optimal ratio for ro/r is:
> ro/r=exp(1);
ro
---- = exp(1)
r