Re: Calculating streamer breakout of top-loads

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Original poster: "Jim Lux" <jimlux-at-earthlink-dot-net> 


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From: "Tesla list" <tesla-at-pupman-dot-com>
To: <tesla-at-pupman-dot-com>
Sent: Saturday, August 30, 2003 11:46 AM
Subject: Calculating streamer breakout of top-loads


 > Original poster: "Gavin Dingley" <gdingley-at-ukf-dot-net>
 >
 > 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?

Precisely, although surface roughness has a very large effect on corona
formation...

 >
 > 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
 >
<snip>
 >
 > 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?

Yes, but....
see previous comment about surface finish, and also, bear in mind that it's
unlikely your sphere is perfectly isotropic, field wise.  You've probably
got it supported somehow (by something other than electrostatic forces or
blowing air.. i.e. the support has epsilon not equal epsilon0) and there are
probably things in the vicinity (particularly conductors) that make the
field not perfectly uniform...

However, the 70 kV/inch or 30 kV/cm is good starting point (at least it
bounds it!... if you have a AWG10 wire with a radius of 0.05 inches, you
know, for sure, it will corona if you put 20 kV on it...)

 >
 >
 > Thanks in advance,
 >
 > Gavin
 >
 >
 >
 >