I know 3 phase is 120v line-neutral, and 208v line-line, but what is it when
all three lines are used? Or is it just not referred to at all, since
voltage between three lines doesn't mean anything?

Can 208v be called 2 phase? I know 2 phase is properly an obsolete system
that involved 2 120v line, 180 degrees out of phase; but is the term now
ever used for 208v?

"Toller" wrote in message
...
I know 3 phase is 120v line-neutral, and 208v line-line, but what is it
when
all three lines are used? Or is it just not referred to at all, since
voltage between three lines doesn't mean anything?

Can 208v be called 2 phase? I know 2 phase is properly an obsolete system
that involved 2 120v line, 180 degrees out of phase; but is the term now
ever used for 208v?


Rather than try to go into detail I took a look on the web. There are many
sites that give good explanations:


http://www.du.edu/~jcalvert/tech/threeph.htm

Toller wrote:

I know 3 phase is 120v line-neutral, and 208v line-line, but what is it when
all three lines are used? Or is it just not referred to at all, since
voltage between three lines doesn't mean anything?

Can 208v be called 2 phase? I know 2 phase is properly an obsolete system
that involved 2 120v line, 180 degrees out of phase; but is the term now
ever used for 208v?



Hi,
Sounds like you are confused between real 3 phase(star ot Y) and Edison
circuit. 2 phase? Maybe bi-phase?
Draw them out on a piece of paper then you'll wee what is what.
Tony

Tony Hwang wrote:

Toller wrote:

I know 3 phase is 120v line-neutral, and 208v line-line, but what is
it when all three lines are used? Or is it just not referred to at
all, since voltage between three lines doesn't mean anything?

Can 208v be called 2 phase? I know 2 phase is properly an obsolete
system that involved 2 120v line, 180 degrees out of phase; but is the
term now ever used for 208v?



Hi,
Sounds like you are confused between real 3 phase(star ot Y) and Edison
circuit. 2 phase? Maybe bi-phase?
Draw them out on a piece of paper then you'll wee what is what.
Tony
Hi,
Oops, Star = Y or delta.
Tony

3 phase is used for things like large electric motors.

These motors use a 3 wire connection. Like A, B, and C.

And these motors will run more "efficiently" on three phase.

It all starts at the electrical generation plant. An electrical generator is
turning around in circles. So you take 3 wires off of the generator like
"pieces of a pie" cut into 3 pieces, then send these 3 wires out on the
electric lines (notice high up electric lines have 3 wires - they are 3
phase).

Then usually businesses will have a 3 phase service, almost never a home.
The 3 phase services will have 4 wires coming in. The 4th is a neutral.

Then the business may have a large electric motor, and 3 wires are
connected. Then the 3 wires power or push the electric motor from the 3
different "pie pieces", or 3 different points on a clock, just as it was
received from the generator.

Sort of like a 3 cylinder engine instead of a one cylinder engine!

Why three phase. Remember power is proportional to voltage squared. If you
look at the sum of the power for each of the three phases you will find that
it is constant. So motors that use three phase power are much smoother than
2 phase motors and smaller for the same horsepower I believe. Also large DC
power sources use three phase power as it requires much less filtering.

Notice high-voltage transmission lines are always in groups of three for the
same reason. The voltage of three phase power supplied in moderate settings
is 208 so that 120 can be obtained using a "Y" type configuration.


"Toller" wrote in message
...
I know 3 phase is 120v line-neutral, and 208v line-line, but what is it
when all three lines are used? Or is it just not referred to at all, since
voltage between three lines doesn't mean anything?

Can 208v be called 2 phase? I know 2 phase is properly an obsolete system
that involved 2 120v line, 180 degrees out of phase; but is the term now
ever used for 208v?

When all three lines are used,it is three phase: A-B, B-C, and A-C



"Toller" wrote in message
...
I know 3 phase is 120v line-neutral, and 208v line-line, but what is it
when all three lines are used? Or is it just not referred to at all, since
voltage between three lines doesn't mean anything?

Can 208v be called 2 phase? I know 2 phase is properly an obsolete system
that involved 2 120v line, 180 degrees out of phase; but is the term now
ever used for 208v?

"RBM" rbm2(remove wrote in message
...
When all three lines are used,it is three phase: A-B, B-C, and A-C

So, if I wanted to refer to three phase, where L-N was 120v, I would call it
208v 3 Phase since a-b, b-c, and a-c are all 208v?

On Sun, 20 Nov 2005 04:50:31 GMT, "Toller" wrote:


"RBM" rbm2(remove wrote in message
...
When all three lines are used,it is three phase: A-B, B-C, and A-C

So, if I wanted to refer to three phase, where L-N was 120v, I would call it
208v 3 Phase since a-b, b-c, and a-c are all 208v?


Correct. The voltage naming convention refers to the phase-to-phase
voltage, not the phase-to-ground voltage. If you know the phase to
phase voltage, you may obtain the phase-to-ground voltage by dividing
the phase-to-phase voltage by the square root of 3 (which is 1.73).

208/1.73 = 120
480/1.73 = 277
etc.

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.

AC Single Phase Circuits actually deliver varying sinusoidoil pulses
of power that are a zero 120 times per second (the zero crossings of
the sinusoid).

AC Three phase circuits supplying a balance load such as a motor are
delivering the same level of power continuously, (but rotating in
intensity among the 3 conductors). It is interesting that DC
circuits also deliver power constantly without the zero crossing
breaks.

In practice, 3-phase motors can be cheaper, smaller, quieter, easier
to start, and run cooler and more efficiently for a given HP.

And just as he begins to understand the three phase wye, we'll throw in a
delta with a wild leg!!



"Beachcomber" wrote in message
...
On Sun, 20 Nov 2005 04:50:31 GMT, "Toller" wrote:


"RBM" rbm2(remove wrote in message
...
When all three lines are used,it is three phase: A-B, B-C, and A-C

So, if I wanted to refer to three phase, where L-N was 120v, I would call
it
208v 3 Phase since a-b, b-c, and a-c are all 208v?


Correct. The voltage naming convention refers to the phase-to-phase
voltage, not the phase-to-ground voltage. If you know the phase to
phase voltage, you may obtain the phase-to-ground voltage by dividing
the phase-to-phase voltage by the square root of 3 (which is 1.73).

208/1.73 = 120
480/1.73 = 277
etc.

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.

AC Single Phase Circuits actually deliver varying sinusoidoil pulses
of power that are a zero 120 times per second (the zero crossings of
the sinusoid).

AC Three phase circuits supplying a balance load such as a motor are
delivering the same level of power continuously, (but rotating in
intensity among the 3 conductors). It is interesting that DC
circuits also deliver power constantly without the zero crossing
breaks.

In practice, 3-phase motors can be cheaper, smaller, quieter, easier
to start, and run cooler and more efficiently for a given HP.

Also when you have a building which has a 3 phase service, a lot of the
stuff is not 3 phase - lighting, 120 V outlets, etc. So you want to balance
the load in the panel. Some circuits will be on one phase, others on the
next, and the rest on the third.

So this is why you might see three big huge wires and a little teeny tiny
4th wire for a 3 phase business electrical service. Most of the load is
balanced between the 3 phases.

Same as with a house where you want half on one leg and the other half on
the other leg.

FYI - Connecting the wires on an electric motor for 3 phase can be complex
to say the least, as there are different types of 3 phase service. Scroll
down to "Three Phase Motors-Single Speed" on the following link...
http://www.patchn.com/motor_connections.htm

"Bill" wrote in message
...
Also when you have a building which has a 3 phase service, a lot of the
stuff is not 3 phase - lighting, 120 V outlets, etc. So you want to
balance
the load in the panel. Some circuits will be on one phase, others on the
next, and the rest on the third.

So this is why you might see three big huge wires and a little teeny tiny
4th wire for a 3 phase business electrical service. Most of the load is
balanced between the 3 phases.

Same as with a house where you want half on one leg and the other half on
the other leg.

I doubt that you would see a building that has 120 V outlets and a tiny
return. Too much of a chance of loading one phase.

But when the loads are hard wired to 3 phase the return can be small or not
in existance. I remember one military application where the power was a
grounded delta.



FYI - Connecting the wires on an electric motor for 3 phase can be complex
to say the least, as there are different types of 3 phase service. Scroll
down to "Three Phase Motors-Single Speed" on the following link...
http://www.patchn.com/motor_connections.htm

This is Turtle.

Hummmmmm, i see Pie are round and Corn Pond Are square.

TURTLE

On 20 Nov 2005 09:46:21 -0800, "TURTLE" wrote:

This is Turtle.

Hummmmmm, i see Pie are round and Corn Pond Are square.

TURTLE

Reminds me of a boy downhome in the Missouri Ozarks, his folks sent
him off to college to get him educated. Since they had spent so much
money on him, his dad wanted him to impress the locals, so when he
came home, his dad said "Say something smart, Junior!" Junior thought
a while and said "Pie are square" Dad was totally humiliated, "Damn
fool is stupider now than when left, everyone knows pie are round,
cornbread are square".

Dan

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."

Synergy is the word for 3 phase. The three wires working in concert is what
gives you a gain in efficiency.

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."

To be real specific, sometimes references are to "120/208 volts", "277/480
volts", etc.. Reference is more commonly to the phase-to-phase voltages such
as "208", "240", "480", etc..
Don Young
"Toller" wrote in message
...

"RBM" rbm2(remove wrote in message
...
When all three lines are used,it is three phase: A-B, B-C, and A-C

So, if I wanted to refer to three phase, where L-N was 120v, I would call
it 208v 3 Phase since a-b, b-c, and a-c are all 208v?

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.

bud--

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

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.

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


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.

Jeff


--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.

bud--


--
Jeffry Wisnia

(W1BSV + Brass Rat '57 EE)

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

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. Thus the
percent losses are lower for 3 phase making the efficiency per wire
higher. (Actually losses should have been divided by the power supplied
to the wire.)

If neutrals are included in calculating loss per wire, the 3 phase
efficiency is improved.

bud--

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.


Thus the percent losses are lower for 3 phase making the efficiency per wire
higher. (Actually losses should have been divided by the power supplied
to the wire.)

If neutrals are included in calculating loss per wire, the 3 phase
efficiency is improved.

Agreed, three phase will take less pounds of copper per unit of
delivered power, at the same voltage of course.

Peace,

Jeff

--
Jeffry Wisnia

(W1BSV + Brass Rat '57 EE)

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

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--

"Bud--" wrote in message

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--

It would be great if the world used 3 phase power. All the wires could then
be the same color. Out system now has black, white, green, sometimes red.
It just gets so confusing.