Jeff Wisnia wrote:
Bud-- wrote:

Jeff Wisnia wrote:

Bud-- wrote:

Jeff Wisnia wrote:

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 3000W supplie




conductors, divided by power entering the line) should be
constant if
the voltage and pounds of copper used stay the same.

Comments?

Jeff

For single phase 120V line to neutral loads - common neutral, assume
1000W each load:
--Single phase supply 3wire (ABN) 2000W supplied
- 2000W/3 = 667 Watts per wire --3 phase supply (208/120V) 4
wire (ABCN) 3000W supplied
- 3000W/4 = 750 Watts per wire - 12% higher

For 3 phase 240V line to line loads - assume 10A per wi
--Single phase supply 2 wire (AB)- watts supplied = 2400W
- 2400W/2 = 1200 Watts per wire
--3 phase supply (240V delta) 3 wire (ABC) watts supplied = 10 X
240 X SQR(3) = 4157W mo
- 4157/3 = 1386 watts per wire - 15.5% higher

3 phase is also a significant advantage in all but small motors.
Also can be in power supplies.




Here's where I was coming from guys:

Say the 2000W load is composed of two 1000W loads in series,
connected across that single phase 240V supply.

So, there's zero current in the neutral.

And, each of the two wires (A&B) is carrying 2000/240 = 8.33 Amps.

Thus the line losses (in Watts) are equal to that current times the
resistance of each wire. Let's assume 1 ohm resistance in each wire,
so the total line losses for the two wires (AB) will be 16.66W while
powering a 2000W load.

True [will be the same loss if one 1000W load is turned off]

Now take the 3 phase supply (208/120V) 4 wire (ABCN), and let the
load be three 1000 watt loads connected in a star pattern,
dissipating a total of 3000 watts. (It's easier for me to do
visualize current flows with a star rather than a delta.)

Because everything is balanced there's zero current in the neutral in
this case too.

The current in each of the three wires (ABC) is 1000/120 = 8.33 Amps,
just like the single phase example. If they are the same 1 ohm wires,
the total line losses for the three wires (ABC) are 24.99 Watts

Also true [will probably also be the same loss with any total load of
2000W]


Now, 2000/3000 = 16.66/24.99, so the "power tranmission efficiency"
per wire is the same, IF the loads are balanced and you can get away
WITHOUT a neutral wire.

I agree that's not usually a code permitted case, so 'ya got me on
the "wire count efficiency" if the neutral has to be there, even
though it's not carrying any current and dissipating any losses.

If you can ignore the neutral also true. (But not counting the neutral
is cheating.)

Beachcomber's post was the power delivered per per pound of copper, in
which case 3 phase provides more power per wire (and per pound of
copper).

For efficiency, if we use your example ignoring the neutral, 3 phase
is still more efficient since the losses per wire are the same but the
losses are divided by a larger power delivered for 3 phase.


At the risk of making a total PIA of myself over this, bud, I was
showing that the losses per wire (In my example 8.33 Watts per wire.)
were the same for each wire, in both the single phase (2 wires) and
three phase (3 wires) examples.

The power in the three phase example (3000 Watts) was 1.5 times the
power of the single phase one (2000 watts) as were the losses in both
examples.

So, I still can't agree that the losses expressed as a percentage of the
delivered power (or the supplied power, which as you point out below is
"correcter".) would be different for the two examples I gave.


Yea that works, but only if you ignore the neutrals. It also requires
the loads to have the same I-V characteristic. In the real world that is
unlikely which would result in too high a voltage on the lower wattage
loads (which is why it is a code violation unless designed as part of a
listed apparatus).

Ignoring the code, the real world connection would likely be all 240V
loads supplied by single phase 240V 2 wire, or 3 phase 240V 3 wire
delta. Load matching would not be an issue then. In that case the 3
phase delivers more power per wire.

bud--