Re: Calculating streamer breakout of top-loads

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Dr. Resonance" <resonance-at-jvlnet-dot-com> 


Gavin:

This would apply to a single pulse operation when the pulse rate would be
perhaps 1 shot per minute or greater.  In the case of Tesla coils it would
only apply when the coil is charged perhaps on DC through a very high
resistive value so the single shot is only one every minute.

If fired more often, especially in the TC case of perhaps 120 shots/sec.,
this would be totally invalid because of all the residuals ions and
molecules in the air around the terminal.

As an example, we have a 10 inch dia. sec. that produces sparks up to 6.8
feet long but the actual potential is only 148 kV, ie, in a single shot mode
the spark length is only approx 8 inches to a ground sphere.  The Tesla coil
spark is a "growing" process that has no connection to the potential but is
linked to the power level.

Your work would apply directly to Van de Graaff spheres, Marx impulse
generators, and other "single-shot" devices.

Dr. Resonance

Resonance Research Corporation
E11870 Shadylane Rd.
Baraboo   WI   53913
 > Hi all,
 > I was playing around with some math to figure out the voltage at which
 > corona will form around a sphere. First, am I right in thinking that in
 > general air breaks down at an electric field strength of 3MV/m?
 >
 > Here is that math I was playing around with:
 >
 > Electric field strength (E) in free-space is given by    E = D / e0    equ
1
 >
 > where D is the electric flux density in C/m^2, and e0 is the permitivity
of
 > free space equal too 8.85e-12 F/m
 >
 > But electric flux density is given by    D = Q / A    equ 2
 >
 > where Q is the charge on the top-load, and A is the surface area.
 >
 > For a sphere the surface area is given by    A = 4 pi r^2
 >
 > so equ 2 becomes    D = Q / ( 4 pi r^2 )    equ 3
 >
 > The charge on a top-load (Q) is given by    Q = C V    equ 4
 >
 > where C is the isotropic capacity of the top-load, and V is the voltage
 > applied to it (the potential between the top-load and ground)
 >
 > For a sphere the isotropic capacity is given by    C = 4 pi r e0
 >
 > So equ 4 becomes    Q = 4 pi r e0 V
 >
 > and in turn equ 3 becomes    D = ( 4 pi r e0 V ) / (4 pi r^2)
 >
 > which reduces to    D = ( eo V ) / r
 >
 > Substituting this into equ 1 becomes
 >
 >      E = D / e0    equ 1
 >
 >      but
 >
 >      D = ( eo V ) / r
 >
 >      therefor
 >
 >      E = ( e0 V ) / r e0
 >
 >      E = V / r
 >
 >      and so
 >
 >      V = E r
 >
 >
 > For air this becomes    V = 3e6 r
 >
 > if V is in volts and r is in metres
 >
 > or    V = 75 r
 >
 > if V is in kV and r is in inches.
 >
 > That is a 2 inch radius sphere will break out at V = 75 * 2 = 150 kV
 >
 > Is this correct?
 >
 >
 > Thanks in advance,
 >
 > Gavin
 >
 >
 >
 >