| I work for an architectural firm and I am curious as to how electrical
| engineers compute the amp load on electric panels. I have even asked a
| couple of their employees about this and all they can tell me is that
| they enter the loads into a program and it tells them what the amp load
| is. Can someone explain it to me?

The software would likely be following the calculation methods described
by the National Electrical Code. A good engineer should at least understand
how to do that by hand, though certainly letting a computer do the work is
a lot easier. It is rather complicated.

| If a 120v single phase panel has, oh lets say, 5000 watts total load
| per leg, what would be the amp load on this panel?

For a specific wattage and voltage, it's easy. But coming up with that
wattage actually needed for a given building is where all the "magic" is.
It's complicated by having to consider things line concurrent usage and
non-concurrent usage. For example, if a home has multiple stoves, they
are allowed to factor in certain probability that not all of the stoves
would be used at the same time, or if they are, other applicances won't
be in use.

| How would the calculations differ on a three phase circuit?

If you convert all line-to-line loads to their equivalent line-to-neutral
loads, the calculations go easier. On three phase, the ratio is the square
root of 3 (1.732 is good enough for this purpose). If you have 3 100 amp
loads connected line-to-line and evenly distributed among the phases, then
each phase conductor will have 173.2 amps of current. If you have total
building volt-amps (watts divided by the power factor), and know that it is
evenly distributed on all three phases, just divide the VA by the L-N volts
and you get the total amps. Divide that by 3 to spread those amps over the
three phase conductors where evenly distributed.

5000 watts, divided by 120, then divided by 3, is 13.888 amps on each wire.
That assumes even distribution. For small load amounts, that's less likely
to be true. For a larger example, assume 1440 kVA in a large commercial
building. That's 1440000/120, then divided by 3. That's 4000 amps. That's
a huge and difficult to manage current. It's more likely to be provided as
480Y/277 at 1732.05 amps or 600Y/346 at 1385.64 amps. That still big but
the benefit in lower current at that level overcomes the higher voltage.
Some buildings may have even higher service voltages. Then a variety of
step-down transformers, such as one per floor, can provide 208Y/120 in a
more practical and safer way.

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| Phil Howard KA9WGN (ka9wgn.ham.org) / Do not send to the address below |

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