Hi, Gang;
I scribbled a few calculations concerning the temperature rise if a
copper pipe filled with water conducts a lightning strike.
I assumed a 0.5" OD pipe with 0.050" wall thickness and I assumed a
30,000 A current for a duration of 1 millisecond. That makes a worse
than average strike.
The resistance of that pipe at 1 MHz is 0.002 Ohm/ft. I used that value
for the entire bandwidth of a strike.
The power dissipated is 30,000 A squared times 0.002 Ohm, which is 1.8
MegaWatts.
The energy is 1.8 MW times 0.001 seconds, which is 1800 Watt-seconds or
0.5 W-Hr. That is 0.43 kilocalories, which is 1.7 BTU.
Assume all that energy goes into heating the water, not the pipe. That
pipe holds 1.5 cu in per foot (0.2" radius times pi times 12" per foot).
A pound of water is about 29 cu. in.. 1.5 cu in is about 0.05 pounds.
The temperature rise will be 1.7 BTU divided by 0.05 pounds which is 33
degrees F. (A BTU is the energy that will raise the temperature of 1
pound of water 1 degree F.)
There are several simplifying assumptions in this calculation, but most
are conservative. While 30,000 A is only an average value, for the
first stroke, first strokes last for only tens of microseconds, not
milleseconds and the current rises and falls in that time rather than
remaining at 30,000 A. There are subsequent strokes, typically of less
than half the peak current of the first one. Calculating the energy
dissipated by 30,000 A for one millisecond yields a higher value than
the total energy dissipated in a foot of conductor by a typical strike.
The pipe is assumed to be 0.5" OD, which is smaller than 1/2" water
pipe. The volume of water in the assumed pipe is similarly smaller than
in a 1/2" water pipe. The calculation assumes all the energy goes to
heat the water, not the pipe, which is conservative.
Anyway, I conclude that the water in a pipe conducting a strike will not
be heated to steam in many cases. There may be some mechanical stress
on the pipe that I haven't addressed.
Larger pipe will offer less resistance; hence, less energy will be
dissipated in the pipe. A larger pipe will hold more water; hence, the
temperature rise will be less by both the effect of more water and of
less energy dissipated.
Someone care to calculate all this more precisely? I hope I didn't make
any gross mistake, like moving a decimal point or inverting an
operation. Guess I'll take my chances and read the corrections as they
appear.
73 de Red
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