| firstname.lastname@example.org wrote in wrote:
|>| I am not getting the correct results for horse power.
|>| An online catalog for a 3-phase motor states:
|>| 20/10 A
|>| Nonimal Effic: 89.5%
|>| If I use the formula: hp = (V*I*Effic) / 746
|>| I get 5.5hp.
|> Is this power 230 volt delta (L-L) 230 volt wye (L-N)? Is the current
|> that for the individual winding leads, or for the supply circuit?
| I'm not sure. It's a very old wiring system in an old factory. I heard it's
| something called a Bastard Leg but can't find much about that wiring
| Have you ever heard of it?
Yes. It's 240V delta, or a Scott-T derived equivalent. One could also make
a system equivalent enough for the purpose of this motor with 230/133 wye/star,
but such systems are generally non-existant. It's definitely NOT a 400/230 or
416/240 wye/star system. That was what my question was trying to figure out.
Given that, the steps *I*
use to figure this out is to convert the electrical
system to its wye/star equivalent, which in this case is 230/133. I multiply
the voltage of each leg relative to a (virtual) neutral (e.g. 133) times the
current (20) times the phases (3) times efficiency in units (0.895) times the
typical power factor (0.8), and _divide_
by 746 and get 7.66 HP.
Electrical power engineers generally do the arithmetic a little differently,
multiplying just the delta or line-to-line voltage by the square root of 3
(1.732). They skip multiplying by number of phases because by multiplying
by the square root of 3, that automatically includes all three phases. It's
still effectively the same thing. I just worked out my steps from the science
of three phase power as I learned it before I ever encountered the practiced
formula (I did not study this in formal education). Engineers like shortcuts
and their method is slightly shorter (less work, unless remembering the square
root of 3 is considered work, which I have found is not the case) than mine.
The current ratings given for a motor is not an indication of power used, but
rather of current drawn in the course of motor operation. Since the motor is
partially inductive, it will store some energy in the field and "generate" it
back towards the supply during part of the next voltage half-cycle. This is
seen in the fact that the current phase is shifted relative to the voltage
phase. The power factor is the ratio of the volts applied times current drawn
divided by power used and wasted. The power factor on _this_
motor may be a
tad bit lower than 0.8, or the actual current drawn may be a tad bit lower
than 20A. And the usual utilization voltage listed for motors is 230/460V
even though standard electrical supply in North America is 240/480V. Except
in extreme cases, the accuracy of these numbers is good enough for engineering
practice. It's good enough to select the proper motor and correctly rate the
circuit needed to power it.
BTW, I sometimes use extreme precision of arithmetic when exploring scientific
or mathematical theory related to electrical systems (or anything else I am
studying). It drives the engineers nuts :-)
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