[BC] Lightning and grounded masts

[JYRussell@academicplanet.com]

Thank you Clive for all the work on this.  It'll take me a while to follow
along, but that I shall... I've never seen all this before...
Jason
----- Original Message -----
From: "Clive Warner" <clive at citiria.com>
To: <broadcast at radiolists.net>
Sent: Wednesday, May 11, 2005 7:59 PM
Subject: [BC] Lightning and grounded masts


> > Actually, Grounded towers don't do much better. It seems that when the
> charge
> > travels down the tower to ground, it induces voltage in the outrigger
> wires.
> >
> > -D
>
> Well, let's see if I can put some figures to that assertion.
> The magnetic flux density B, from a strike, is mu*H
> or 4*PI*(10 to the -7) * I/2*PI*R where R is the distance from the current
> path in metres.
>
> The shunt-fed is of course connected at the top (= 0 metres) and at the
> base, probably about 2 metres from the mast. So let's (conveniently) call
> the average distance 1 metre.
>
> A strike has current characteristics varying from 10 KA to over 200 KA.
> Here I will adopt a fairly high value of 100 KA.
>
> The worst case (lightning strikes contain a full spectrum) is at DC -
17KHz,
> so I will choose the worst case.
> At 3m from the stroke path, the nearest my tables will go, I see a flux
> density of 36 dBGauss.
> Graphing with Excel and using regression gives me an estimated value of 45
> dBG.
> 6dBG = 2 Gauss.    45dBG = approx. 160 Gauss.
> (See "EMI Control Methodology and Procedures", Don White Assoc., for refs.
> above)
>
> Assuming a wavelength of 600 KHz then lamda = 500 metres and one-tenth
> lambda = 50 metres.
> Therefore we can say the situation is an induced field of 160 Gauss across
a
> 'cable' of 50 metres.
> The mast end is earthed, we can calculate the induced voltage using
> Faraday's Law.
>
> This states that E = dW/dT where dW is the change in magnetic flux, in
> Webers/sq.m, of the event, and dT is the time over which the change takes
> place, in seconds.
> Unfortunately our field units are in Gauss not Webers so first I must
change
> that to 0.016 Webers/sq.m
> (How I HATE American obsolete units. My heavens I was in 6th grade when we
> changed from that crap (cgs) to proper MKS units!)
> Now I need to know how many sq.m cutting field lines that I have on the
> adjacent conductor. Let's say the downlines are 2m across at the base, 50m
> high, and taper to a point at the top. Using the formula for the area of a
> triangle gives us half base x height, = 1x50 or 50 sq. metres.
> I looked up the approximate pulse duration for lightning and discovered a
> big variation in pulse length, but for a high current strike we can work
> with 20 microseconds.
> Substituting, this gives V = 0.016/0.000002 = 800 volts.
> ======================================
> This looks like a lot less than the equivalent lightning strike on a
> base-insulated mast. DWC give 300 MV as the voltage of the same 'sample'
> strike. Obviously the LPD discharge horns will limit this, but I suggest
> that the breakdown voltage of the discharge horns is likely to be many
times
> the 800V I calculated above.
>
> Anyone care to challenge my figures?
>
> Empirically, I have worked on many shunt-fed, and never ever saw any
damage
> caused by lightning.
>
> Best wishes to all
> Clive Warner (staggering back upstairs with arms full of giant reference
> tomes)
> www.citiria.com
>
>
>
>
>
>
>
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