I seem to recall a discussion that if you assume equal voltages, currents,
and resistances, thus equal loss in each conductor, then adding the third
conductor increases losses by 50 percent (1.5 times the two conductor loss)
for a power delivery increase of 73 percent (1.73 times the two conductor
power). I think in a balanced three phase system that Power equals RMS
Current times RMS Voltage times the Square Root of Three (1.73). Right now I
don't have the math handy to show it but I think it is correct.
Don Young
"Jeff Wisnia" wrote in message
...
Beachcomber wrote:


snipped


In general terms, a three phase circuit is more powerful than a single
phase circuit because it delivers more power from point A to B per
unit of copper conductor and hence, is much more efficient.


I'm having difficulty following that one point, Beachcomber.

If by "unit of copper conductor" you mean the pounds of copper needed to
make the conductors going between points A and B, then;

It seems to me that for each indivdual conductor running between points A
and B, the power loss in that conductor is just going to be equal to the
rms current squared times the total resistance of the conductor, and the
power delivered to the load by that conductor is going to be equal to the
rms current times the rms voltage at the load, assuming that the load has
a unity power factor of course.

Those two powers (power loss and power delivered) remain the same whether
that conductor happens to be part of a single phase or a multiphase
transmission system, so the efficiency of the transmission system (power
delivered to the load less power lost in heating the conductors, divided
by power entering the line) should be constant if the voltage and pounds
of copper used stay the same.

Comments?

Jeff
--
Jeffry Wisnia

(W1BSV + Brass Rat '57 EE)

"Truth exists; only falsehood has to be invented."