Owen

The fusing/melting current for 1/2" copper pipe is probably well above
20kA, even for 100ms pulses. A more interesting potential failure mode
might be from the mechanical forces due to the magnetic field. (see,
e.g., quarter shrinking or can-crushing)



Hi Jim,

I note the "probably" in your comment, and the "dunno" in N5MK's
response.
That's because I was lazy and didn't want to actually compute it. I've
put multi tens of kA pulses through 1/4" copper pipe, but they're not
100ms long.



The uncertainty in my statement is over the exact lightning scenario,
they vary, and the circuit response (ie current waveshape, amplitude,
duration, ringing etc) depend on the specific excitation and circuit
elements (parameters of the down conductor, nature of the earth system,
ground, environment etc).

One could certainly use the standard double exponential approximations..
either a 2/50 waveform for a strike or the longer surge impulse (I can't
remember the exact rise/fall times for the surge..)



As far as supposition as to the fusing current for conductors, that is
determinable for a given scenario. I have at hand the Protective Earthing
Code of Practice published by the Electricity Authority of NSW June 1975
and it shows that a 35mm^2 copper conductor has a fault current withstand
of 20kA for 100ms. (I have considered implementing the underlying
formulas in an online calculator.)

Preece or Onderdonk?
(http://home.earthlink.net/~jimlux/hv/fuses.htm

N5MK stated "A #10 wire can handle that job". If he is talking copper, I
understand that #10 means 2.5mm diameter, or ~5mm^2, or less than 15% of
the recommended conductor csa for the stated scenario.


Preece equation gives fusing current for AWG10 (2.5mm diameter, as you
say) as 316 amps, but that's sort of for a steady state.

Onderdonk's equation, plugging in 100 ms for the melt time, gives 4.7kA,
which I can believe. I've blown up a lot of AWG10 wire with those sorts
of currents in a quarter shrinker. Partly melting, partly mechanical
stresses in that application.

The purpose of the National Electrical Code (National, here, referring
chauvinistically to the U.S.) required AWG 6 (diam 0.15 inches, 3.8 mm)
bonding wire for grounds is NOT to carry the lightning current (which it
wouldn't, in most cases) but to carry fault currents from things like
shorts from line to grounding conductor, which are usually in the
hundreds of amps range. Say an energized power line falls down and hits
the antenna. You want the antenna's grounding conductor to carry the
likely fault current and not go open, and carry enough current to trip
any overcurrent protective devices.

Lightning protection is usually things like 2/0 (0.364 inch diameter,
9.25 mm), which has a fusing current (viz Onderdonk) of 65kA.

I am not familiar
with your water pipe sizes. If it were, say, a half inch diameter #19, it
has a CSA of around 35mm^2, so the #10 wire should melt before the pipe
electrode, thus protecting the pipe electrode from failure. Yes,
mechanical forces are also relevant to lightning conductors, but my
comment was about the fusing current.

35 mm^2 would have a fusing current of around 30-35 kA.

1/2" Copper pipe is 0.625" od and 0.545" id (very close to 1mm wall)
so, has about 47 mm^2 area.


In this part of the world there is an Australian Standard (AS1768)
relating to lightning protection, there may be a similar standard or
"code" in other jurisdictions, and they would not be a bad place to start
in understanding lightning protection and designing a protection scheme.

Another source of information is to walk around a mobile phone base
station and look at the earthing system from the outside. It is even more
enlightning (no pun) to look inside. These things withstand lightning
events quite well. Are they over engineered? Probably not, they do suffer
damage from time to time.

It is my view that there is a significant risk that an inadequate
lightning protection scheme may be much worse than doing nothing.

I would agree..



Owen

....

Jim, a lot of interesting stuff with which I generally agree.

The approach that my reference took to rating the conductor for a
lightning discharge includes a safety factor (as you might expect), and
so will rate the conductor at lower I^2*t than finding the conditions to
melt the wire. In real life, you would want the conductor to withstand a
second strike or fault soon after, and you would want to allow some
tolerance for other variables, hence the safety factor. The approach is
to find the I^2*t that raises the conductor one third of the way from
ambient (323K) to melting point. The calculator you used might assume
resistivity is at 0°C , ambient is 0°C, and the material is raised to
melting point with no heat loss, and that would give a fusing current
close to double of the approach that I used.

BTW, we have half inch copper water pipe over here (we still do but it
has a nominal metric size) and it is half in od... whereas half inch
galvanised steel pipe is half inch nominal bore... actually about 5/8"
id. Don't you like consistency in the same field!

Some years ago I did extensive modelling of a double exponential
excitation of structures and facilities (not lightning, faster than
lightning) and it was interesting how much the circuit configuration
affected the transformation of the excitation waveform to structure
current, including ringing. The same software could run a lightning
scenario, but that wasn't the main goal of the analysis so my experience
with the lightning scenario is more limited. So, as I said, the nature of
the current waveform is the big uncertainty and so measures are usually
quite conservative to cover that uncertainty.


Owen

Owen Duffy wrote:
...

Jim, a lot of interesting stuff with which I generally agree.

The approach that my reference took to rating the conductor for a
lightning discharge includes a safety factor (as you might expect), and
so will rate the conductor at lower I^2*t than finding the conditions to
melt the wire. In real life, you would want the conductor to withstand a
second strike or fault soon after, and you would want to allow some
tolerance for other variables, hence the safety factor. The approach is
to find the I^2*t that raises the conductor one third of the way from
ambient (323K) to melting point. The calculator you used might assume
resistivity is at 0°C , ambient is 0°C, and the material is raised to
melting point with no heat loss, and that would give a fusing current
close to double of the approach that I used.

BTW, we have half inch copper water pipe over here (we still do but it
has a nominal metric size) and it is half in od... whereas half inch
galvanised steel pipe is half inch nominal bore... actually about 5/8"
id. Don't you like consistency in the same field!
But they're not the same field.. the stuff made of copper is actually
"tubing" and the stuff made of steel is "pipe", and historically,
they've been measured differently.

Tubing is usually soldered/sweated/brazed into fittings with a
receptacle, so the OD is important, because even with different wall
thicknesses, the fittings are all the same.

Pipe is based on something else (King John's toe diameter or something)

Some years ago I did extensive modelling of a double exponential
excitation of structures and facilities (not lightning, faster than
lightning) and it was interesting how much the circuit configuration
affected the transformation of the excitation waveform to structure
current, including ringing. The same software could run a lightning
scenario, but that wasn't the main goal of the analysis so my experience
with the lightning scenario is more limited. So, as I said, the nature of
the current waveform is the big uncertainty and so measures are usually
quite conservative to cover that uncertainty.

There's some fascinating papers out there that use NEC to model response
to a nearby lightning stroke (a much more common occurance than a direct
hit). It's actually quite involved, since they model the traveling
impulse of the stroke.


Owen

We had a near by lightning strike last night... All my radios and
antennas survived... My son's Dish Network receiver did not... He's
quite bummed out...

denny

Jim Lux wrote in
:

The purpose of the National Electrical Code (National, here, referring
chauvinistically to the U.S.) required AWG 6 (diam 0.15 inches, 3.8
mm) bonding wire for grounds is NOT to carry the lightning current
(which it wouldn't, in most cases) but to carry fault currents from
things like shorts from line to grounding conductor, which are usually
in the hundreds of amps range. Say an energized power line falls down
and hits the antenna. You want the antenna's grounding conductor to
carry the likely fault current and not go open, and carry enough
current to trip any overcurrent protective devices.

I meant to comment:

I think that it is common in electricity distribution level networks,
that they are designed to hold fault current to about 20 times the
maximum working current.

In this part of the world, a single phase 240 home probably has a 80A
rated service, and fault current would usually be not worse than than
about 1600A, so the specified 6mm^2 earthing conductor and 4mm^2 bonding
conductor should withstand that current for 100ms until the protective
device operates. (4mm^2 withstands 3200A for 0.1s with a safety factor of
3.)

You probably know the numbers for your own distribution network practice,
they may be of interest to readers.

So you raise a good point, that if your tower falls onto power lines, it
would be good if your earth system could withstand the likely fault
current to take out the protection on the power lines and leave the tower
un-energised. In this part of the world with LV distribution, the
protection is probably a 500A HRC fuse with a fault current level of 10
+kA. It may be much lower for you if power lines are HV where the fault
current level should be a good deal lower.

Owen

Owen Duffy wrote in
:
protective device operates. (4mm^2 withstands 3200A for 0.1s with a
safety factor of 3.)

That should read:


protective device operates. (4mm^2 withstands 2400A for 0.1s with a
safety factor of 3.)