This thread peaked my curiosity...
Neglecting the RF characteristics of a strike, impedance, & skin effects...
Calculate how long it would take a copper rod, 2" dia and 1 meter long to
reach melting point (from ambient temp) under a 20,000 amp load.
No wonder even small wires can pass huge currents for short durations....
KM5VI
-----Original Message-----
From: TowerTalk [mailto:towertalk-bounces@contesting.com] On Behalf Of Jim
Lux
Sent: Saturday, November 17, 2012 11:14 PM
To: towertalk@contesting.com
Subject: Re: [TowerTalk] Grounding
On 11/17/12 6:32 PM, K8RI wrote:
> On 11/17/2012 10:51 AM, Jim Lux wrote:
>> On 11/17/12 4:40 AM, K1TTT wrote:
>>>> Dale - WD4IFR
>>>> In any lightning strike, even a 2" solid piece of copper takes on the
>>>> characteristics of a light bulb filament.
>
> I've never seen a conductor melt that has take a lightning strike here.
>
>>>
>>> It is simple statements like this that make the rest of what you say
>>> questionable. The fact is that MOST lightning strokes hitting even
>>> small
>>> conductors will not significantly heat them. To get any significant
>>> warmth
>>> in 2" copper is probably even beyond even the largest measured mega
>>> strokes.
>>
> I'm not sure I'd go quite that far as I've seen holes blown right
> through the wings of large aircraft from mega strokes.
>
Aluminum melts at a MUCH lower temperature than copper (660 vs 1085) AND
a lower density (so temperature rise for a given amount of energy, in
the same volume, is higher).
When doing exploding wires, I did a lot of aluminum wires for this reason.
>> I agree... take a look at the Preece or Onderdonk equations for fusing
>> current. Onderdonk takes into account the pulse length.
>
> Many lightning strikes are a repetition of strokes that add up to
> substantial time (relative to the length of a single stroke). Yet I have
>
> The current is a summation of a rather complex wave form with relatively
> steep rise and fall times. Most strikes with thousands of peak amps
20-50kA peak. 2 us rise (10%-90%), 50 us fall to 50%
> probably have less than a hundred amps average for less than one second.
> (I'd guess at less than a 1/4 second for the event) If you are
> calculating fuse time, how can you not take into account duration which
> would be a summation of the current wave form for the duration of the
> event. IE you need to know how many jouls the fuse can stand for a
> specific interval.
>
It's actually called the "action", which is basically the integral of
I^2*t (combined with the resistance of the conductor, that turns into
joules, which is what melts the metal). So, it's really the peak
current that's important, because that's the squared term, more than the
time.
But in any case, there are well known relationships for lightning that
cover the multiple pulses over time issue.
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