From: Paul
Date: Sat, 23 Dec 2000 16:03:33 +0000
Subject: Re: [TSSP] Final solution...great chance for it
I wrote: > > Yes, that's the time taken to solve the network for > > AC steady state conditions at a single frequency. boris petkovic wrote: > > > ----- > I guessed you would say that. > What about (computing time), simulating 100 turn coil > response when step function applied ,say at pentium > processor? > ----- Funny you should say that. I have a time domain simulator which just happens to have 100 elements...its a while since I've used it, and it doesn't do the longitudinal coupling stuff. I don't have any timings for it, but its *very* slow, I seem to recall an hour or more to get a solution for a transient. boris petkovic wrote: > The purpose of his technical paper was not just to > predict current and voltage waveforms over winding but > to emphasize difference between standard (our),well > known approach with the chain of coupled qudruples > and his method. > Physical background of the process is better explained > in his case. Understood. I'd like to see more - will look forward to the post arriving. > Once,you re intersted in predicting Q of secondaries > ,right? > Now,I will ask you quite a diffucult question. > What do you think could be the highest possible Q of > the structure like Tesla's 100 turn 100"*100'' extra > coil from CS? > Imagine there is perfect ,loseless isolation,super > conducting winding and enviroment,and the only > losess comes becouse of radiating. > I don't expect concrete number ,but try to guess the > ballpark. > ------ OK, the only loss is due to radiation. I worked this out once, a long time ago. Can't find my figures so I'll do it again... Consider the lossless TC resonating in free space with angular freq w and stored energy E. The Q factor is then Q = wE/P where P is the radiated power. From a sufficient distance away it resembles a Herztian dipole with a dipole moment qh where q is the displaced charge and h the coil length. Since V = sqrt(2E/Cee) then q = CesV = Ces sqrt(2E/Cee). The total power radiated from a Herztian dipole is obtained by integrating the Poynting power density over the surface of a large bounding sphere. I'll spare the details and say P = (1/12pi) * z * w^4 * q^2 * h^2 / c^2 = (1/6pi) * z * w^4 * Ces^2 * h^2 * E/Cee/c^2 Watts where c = 300 * 10^6 metre/sec, z = 377 ohms. Thus Q = wE/P = 6pi * c^2 * Cee / (z * w^3 * Ces^2 * h^2) Say Cee = Ces = 30pF; w = 2.pi.10^5 rad/sec; h = 1 metre. Then Q is approx 6 * 10^8, pretty high. Now is that in the right ballpark? Are we safe to ignore radiation? I get the feeling you're going to tell me I'm missing something :) Cheers, -- Paul Nicholson, Manchester, UK. --
Maintainer Paul Nicholson, paul@abelian.demon.co.uk.