Road Test: Strong Technobeam, page 40 - PLSN.com

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FOCUS ON FUNDaMENTaLS
Me So Stupid
A
Neutral
Fig.1
120V
208V
1500W
I1
o
C
120V
I2
120V
1500W
208V
B
=
A
208V
1500W
I1
C
I2
B
1500W
208V
“
It is our responsibility as scientists,
knowing the great
progress which comes from
a satisfactory philosophy of
ignorance, the great progress
which is the fruit of freedom of
thought, to proclaim the value
of this freedom; to teach how
doubt is not to be feared but
welcomed and discussed; and
to demand this freedom as our
duty to all coming generations...”
— Richard Feynman, “What Do You
Care What Other People Think?”
PROJECTION LIGHTS & STAGING NEWS
Let me put this right up front: I’m not the
brightest lamp on the truss. At times, I can
make Jessica Simpson look like Marilyn vos
Savant, who is listed in the Guinness Book of
World Records for five years under “Highest IQ”
for both childhood and adult scores. I’m not
particularly proud or embarrassed about that,
but admitting it has served me well. An empty
head can be like being an empty vessel looking
for understanding with which to be filled.
Last month I was in the Washington D.C.
area leading a seminar on electricity for the
entertainment electrician. There were a lot of
very experienced and very good electricians
By Richardcadena
The Tao of electrical load calculations
fig.1
in attendance. When we started talking about
how to calculate the feeder cable current in a
three-phase system, the Jessica Simpson in me
made her appearance.
The classic formula for power in a threephase
system is: watts = volts × amps × power
factor × 1.73. By manipulating that formula we
can come up with the formula for three-phase
current, which is: I = W ÷ (V × PF × 1.73).
One of the electricians in the class said
that he uses a different formula for figuring
out the current in a three-phase system: I =
(W × 2) ÷ V. Even I could figure out that these
two formulas are nothing alike.
At this point, I could have invoked the
teacher/student clause, which is, “I am the
teacher, you are the student; now shut up and
listen to me.” But I realized that dismissing his
approach would be tantamount to dismissing
his 30 years of experience. Admitting you don’t
know something, particularly in front of your
colleagues, is kind of like when someone is
waving to a person behind you, but you think
they’re waving at you. So you wave back and
then you realize...and you feel…so…stupid.
After fumbling for an answer to this conundrum,
I punted. I called a break to think about
it and when the break was over, I not-so-gracefully
moved on, promising myself to address it
once and for all in the next day’s class.
That night, I couldn’t sleep. I went to bed
and for two solid hours, I stared at the ceiling
and contemplated the question of why
the two approaches yielded different results.
My mind went blank. Eventually, the words
of Lao-tsu came to my mind: “The usefulness
of what is depends on what is not.” In other
words, the usefulness of your mind depends
not on what you already know, but what you
might gain by not knowing, contemplating,
and then understanding. The answer suddenly
popped into my head. I fell asleep.
The next day, I went to the class and here’s
what I tried to convey. Suppose we have, for
example, a single 1500-watt 208V automated
light connected across phase A and phase B
of a three-phase wye-connected transformer.
With nothing else is connected, the line current
will be 1500W ÷ 208V = 7.2 amps (assuming
unity power factor). In this case we can use the
formula for a single-phase system because we
only have one single-phase load connected.
If we then connect another automated
light across phase B and phase C, then we
have a more complex situation. Phase B is
now feeding current to both loads. How do
we figure out the resulting line current?
To simplify it and make it easier to understand,
we can redraw it as a delta-configured
secondary with the same two loads, as shown
in the right half of the illustration in Fig. 1. If
we pay close attention to the two drawings
we’ll see that they are exactly the same even
though they are illustrated differently. Now
we can see that the line current in B is being
drawn from two different places; phase A and
phase B. Does that mean that we can simply
add the magnitude of the two currents, I1
and I2, to get the resulting current?
Not exactly.
Because these two currents are 120 degrees
out of phase with each other, the resulting
current is something less than twice the
current in one of the phases. The mathematical
solution to the addition of these two out
of phase sinusoidal currents is a little bit complex,
but the simple answer is that they add
up to 1.73 times the magnitude of any of the
phase currents.
continued on page 34
60 PLSN March 2008
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