Why is Three Phase More Efficient?

An electrician was telling me the other day that three phase equipment is
more efficient than single phase. If you have a single phase 208V device,
and it only draws energy on a single phase of a three phase circuit, why
should that be any less efficient than a three phase implementation of the
same device, which draws lower amps on three separate phases?
Reply to
CHANGE USERNAME TO westes
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On Fri, 8 Oct 2004 12:14:03 -0700, CHANGE USERNAME TO westes put forth the notion that...
Not everything that is three phase is more efficient. Motors tend to be, and are also easier and cheaper to build. Resistive loads such as heaters use the same amount of power whether they're three phase or single phase, although you can get more power for the amount of wire that you need to run with three phase. For example, it takes two power conductors to run a single phase 240 volt circuit, but it only takes one more conductor to run a three phase circuit, which can power three single phase loads at the other end. Most of the efficiency of a three phase system lies in the distribution.
Reply to
Checkmate
"CHANGE USERNAME TO westes"
equipment is
phase 208V device,
circuit, why
implementation of the
phases?
...single phase power uses a displacement capacitor or shunt (with shade pole motors) to achieve the phase or magnetic displacement necessary to get rotation...it is not quite as efficient as three phase power which has a natural configuration that provides a rotating field directly...
Phil Scott
Reply to
Phil Scott
And this is only an advantage for applications with rotative mechanical properties, like a motor?
Reply to
CHANGE USERNAME TO westes
"CHANGE USERNAME TO westes"
message
shunt
as
mechanical
As for efficient energy use yes...but for lighing etc there are logistics and systems balance and wire size advantages with various other voltage and transformer configurations...but not three phase lighting. (lighting is a straight resistance load across a single element, so no rotation needed, no three phase needed per element at least)
Large electric heaters may use three phase, but its still only a single phase across each element... that would allow one to get more KW in heat with smaller wire than if only a single phase power were supplied. Heavier wire costs money to run.
Phil Scott
Reply to
Phil Scott
Even if you neglect the decreased amount of material required to manufacture three-phase motors and their power distribution systems, three-phase will still retain an efficiency advantage over single-phase.
A single phase motor is often described in terms of two counter rotating fields. At startup, these have slips of one and minus one. If you do manage to start the motor with an initial spin, the the slip for the wrong rotation will be almost -2. This is a parasitic drage that cannot be removed.
Bill
Reply to
Repeating Rifle
3 phase is not inherently more efficient at all. This is strictly a myth.
What is true is that higher voltage loads lose less energy to I2R losses because the current is lower for a given load then with a lower voltage supply.
"CHANGE USERNAME TO westes" wrote in message news:TvGdnYfO-960e snipped-for-privacy@giganews.com...
Reply to
Bob Peterson
From another viewpoint...
A single phase motor does not receive constant power at every instant in time. During the zero crossings of the voltage and current waveforms, the power delivered from source to load is zero for an instant in time. The mechanical rotational inertia of the motor keeps it spinning smoothly (like a flywheel) during these zero crossings.
For a three phase motor (or any three phase circuit) the reverse is true. The instantaneous power delivery rate is constant because when the power in one leg goes to zero for an instant in time, the other two legs will deliver the vector total sum of the power being delivered. Obviously this cycles equally among all three legs.
An analogous situation would be to supply power to two different incandescent lamps, one with AC and one with DC. The AC lamp appears to stay lit when the voltage waveform crosses zero 120 times a second because the filament cannot cool fast enough and appears at constant brightness.
The DC lamp stays at constant brightness because it receives constant power, just like our 3-phase motor.
Both lamps could be of equal wattage, or the single-phase motor could be consuming an equal amount of power as the 3-phase motor.
Thus it could be said that the Three-Phase system is more efficient as an energy delivery system in the same sense that Edison said that DC is a more efficient energy delivery system than AC. He was talking about distribution and using the least amount of conductive copper for the energy delivered.
Beachcomber
Reply to
Beachcomber
(Big snip)
Actually, if you want to argue about it (which I don't), the old 2 phase 3 wire system (90 degree phase displacement) is even more efficient than 3 phase in terms of pounds of copper per kW delivered. However, to achieve this efficiency requires that the conductors be of unequal sizes, and this poses both mechanical and electrical complications due to the non-symmetrical nature of the resulting circuits and devices. Many very early transmission systems used 2 phase which was converted to 3 phase and single phase at the distribution end. The simplicity of the balanced 3 phase system won out in the early 1930's however.
Reply to
BFoelsch
in article JOKdnS6fE4zZk snipped-for-privacy@giganews.com, BFoelsch at snipped-for-privacy@comcast.ditch.this.net wrote on 10/9/04 9:13 AM:
I find that hard to believe. Even if you do not wish to argue over your statement, it is intriguing enough for me to want to see a rationale for it.
Bill
Reply to
Repeating Rifle
| An electrician was telling me the other day that three phase equipment is | more efficient than single phase. If you have a single phase 208V device, | and it only draws energy on a single phase of a three phase circuit, why | should that be any less efficient than a three phase implementation of the | same device, which draws lower amps on three separate phases?
Certain equipment can be more efficient in three phase versions. Motors are the big thing. Efficiency is measured in total cost, too, so you do factor in the manufacturing and maintenance costs, as well as the energy and delivery costs.
Power supplies that convert AC to DC can also be more efficient in terms of simpler less costly design, and some reduction in heat loss, when they operate from three phase. This is generally only realized on largs scale designs.
In terms of total power with a given set of conductors (cost, weight, or whatever), three phase can come out ahead, depending on the way things are wired. Lets say you have 21600 watts of incandescent lighting. You could wire this up a number of ways:
1. A single pole 120 volt circuit totalling 180 amps. The single phase wire is 180 amps, and the neutral is 180 amps.
2. A single phase two pole (2 phase angles at 180 degrees are considered to be single phase) circuit with each pole being 120 volts to ground, with 90 amps on each pole. The neutral only needs to be 90 amps.
3. A three phase three pole circuit with each pole being 120 volts to ground, with 60 amps on each pole. The neutral only needs to be 60 amps.
So the total cost of wiring, expressed in the sum of current capacity, is:
1. 180x2 = 360
2. 90x3 = 270
3. 60x4 = 240
While light bulbs typically cannot be connected phase to phase, many other loads can be. That can eliminate the neutral being connected at all (the separate equipment ground would stay for the ground fault protection).
Suppose we have heating elements for a water heater which we can choose the voltage for. I'll use 21600 watts again.
4. 240 volt elements connected to opposite poles on single phase. The two poles have 90 amps.
5. Three separate 208 volt elements connected phase to phase over the three phases equally. The current in each element will be 34.641 amps. But the wires on each phase will be serving two elements each, with currents 120 degrees out of phase. The total current is NOT 69.282 amps. Since some of the current can run between elements across the point of connection due to the phase angle difference, the end result is 60 amps of current in each phase wire.
So now we have wiring costs of:
4. 90x2 = 180
5. 60x3 = 180
Some of these savings comes from reducing the size of the neutral, or by eliminating it entirely. But some also comes from the fact that currents are lower. Since the loss in the wire is proportional to the square of the current, reducing the current is a substantial savings if the size of the wire remains constant. But even if we cut the size of the wire in half, there is still some savings. The loss is also affected by the resistance, which goes up as we cut the size of the wire. But that only compensates for half the loss due to increased current.
Note that in example 5 I used only 208 volts instead of 240. This might not seem fair, but in both cases the voltage relative to ground is 120 volts. If you did this with a 240 volte delta load, the current would be reduced by 15.47 percent. But a power source providing that is going to have at least some conductors at a higher voltage relative to ground. To power such a load with a WYE source, the voltage to ground would be 139 volts.
When a comparison is made between say, 240 volts single phase and 240 volts three phase (or 480 or whatever), you're really dealing with a voltage increase, relative to ground (1.1547 times as much). That changes the picture.
Basically, if you double the voltage, halve the current, and halve the current capacity (approximately double the resistance), you still have cut the loss approximately in half. This works because while the wire resistance goes up by about 2 times, the total system impedance goes up 4 times for the same watts or volt-amps. The wire then becomes a smaller proportion of the circuit impedance. The cross section of the wire is half as much and the current is half as much. The heat produced per cross section is the same, but this is only half as much, and smaller wire has more surface for a given cross section. So it stays cooler, or can be shaved in size even more. Utility transmission lines can get very very hot (because they cut the size of the wire as much as they can for many reasons, such as its weight and supporting requirements) and not really be losing all that much power. The "magic" is that the overall system impedance, compared to the resistance in the wire, is very high at transmission voltages.
Your electrician as basically correct, but it's not so much that three phase is somehow better (it is for certain uses), but rather the advantage is in how we use three phase, or the voltages that three phase comes in, such as 400 (Europe), 480 (USA), 600 (Canada), 690 (Europe), 1000 (mining services where electrical equipment is run over cables extended sometimes many miles underground), and even higher for distribution and transmission. Even single phase loads like lights are often run at voltages like 277 (the voltage to ground of 480Y/277) and 346 (of 600Y/346).
Reply to
phil-news-nospam
Well, if you will accept some hand-waving for the moment, here goes.
Let's assume a 20 kW, 100% PF load on a 3 phase 3 wire 100 volt circuit.
Current drawn is 20,000/1.732 X 100 = 115.5 amperes, equal in all phases.
(hand-waving starts here)
Therefore, line loss will be calculated on the basis of three conductors carrying a total of 346.5 amperes.
In the 2 phase, 3 wire case however, to deliver the same 20 kW load we end up with two conductors carrying 100 amperes, and the common conductor carrying 141.4 amperes, for a total in all 3 wires of 341.4 amperes. Obviously, less copper is required to provide the same line loss with 341.4 amperes than with 346.5 amperes as in the 3 phase case.
Subtle, only about 1.5%, but real.
Reply to
BFoelsch
circuit.
conductors
Think your maths is astray.
American 2 phase is not really it is just one phase with the centre earthed and called the neutral
To get 20,000watts you need 200 amps in each side if you stick with 100 volts across the load or If you have 200 volts across the load then there will be 100 amps in each wire and ZERO in the centre tap.
There is no 1.414 or 1.732 involved in a calculation where the 2 ends are 180degrees apart or just the opposite ends of a 200 volt supply.
Reply to
John G
I am discussing the legacy 2 phase system with two phases displaced by 90 degrees, not the standard 120/240 center-tapped system.
Reply to
BFoelsch
On its own "efficient" is a word of dubious interpretation.
Efficient in its own use of energy (ie, lowest heat losses)?
Efficient in the use of energy from source to comsumption (lowest heat loss at the generator, distribution system, and end end user equipment?
Efficient in the use of material, or mass, or volume, to make the equipment for a given performance?
Efficient in some other way?
Sylvia.
Reply to
Sylvia Else
in article tsOdnXa57aOe6 snipped-for-privacy@giganews.com, BFoelsch at snipped-for-privacy@comcast.ditch.this.net wrote on 10/9/04 4:36 PM:
I think this shows some false logic. It looks like there is a mixing of phase current and line current concepts. If the line to neutral voltage is 100 volts then the current required for each line is 20000kW/((three-phases)*100V)=67 A/phase. This compares to a two phase system delivering with currents in the three conductors as described. Please explain I^2*R losses from two 100 amp conductors and a 141 amp conductor being smaller than from three 67 amp conductors.
Bill
Reply to
Repeating Rifle
in article gJ%9d.291$ snipped-for-privacy@nnrp1.ozemail.com.au, John G at snipped-for-privacy@ozemail.com.au wrote on 10/9/04 5:52 PM:
You are describing the Edison three-wire system. Two phase as used here really refers to a four phase system with neutral where only two phases 90 degrees apart are used.
Bill
Reply to
Repeating Rifle
-------------- Actually- the math is a bit out. The neutral conductor will be carrying (root(2))*115.5 =. 163.3A. In addition, summing the currents is not valid as losses are dependent on the current squared so the loss in the neutral- to be the same as in the other conductors- requires half the resistance and twice the copper cross section. This leads to the copper needed being 4/3 that of the 3 phase case. 3 phase 3 wire leads 3 conductors carrying 115.5 A and the line loss will be 3(115.5^2)R =40020R watts Now with the 2 phase 90 degree apart situation there is a loss in two of the wires of 26,680R watts. If the neutral is sized for the same loss as the phase leads then the loss in the neutral is 13,340R watts and the current is 163.34 A. At this current then R must be half that of the phase leads so the conductor cross section must be twice that of the phase conductor so the neutral conductor weighs twice as much as the phase conductors. This leads to a relative weight/kW of copper needed which is twice that for the 3 phase case. You get a better deal if you increase the conductor area by root(2) -the loss/power ratio goes up by 4/3 but the required conductor weight/kW drops to 1.71 times that of the 3 phase case.
I did work out a summary based on Po =Vp*Ip where Vp and Ip are the phase quantities and assuming unity pf as well as conductors sized for constant current density (i.e. area proportional to current and resistance inversely proportional to current.
Case power loss loss/kW conductor weight weight/kW 1phase 2 wire Po 2Lo =2(Ip^2)R 2K=2Lo/Po 2V 2M=2V/Po 1 ph 3 wire 2Po 2Lo K 2V+ M+ (+ ins due to neutral so 3V and 3M are more typical) 2ph 90 degrees 2Po 3.41Lo 1.71K 3.41V 1.71M (Losses lower and weight higher with larger neutral) 3ph 3wire 3Po 3Lo K 3V M 3ph 4 wire " " " 3V+ M+ (+ depends on neutral size) N phase N+1 wire as for 3 phase 4 wire
Considering losses in conductors single phase 3 wire falls into the class of N phase 360/N degrees apart with neutral (N+1) wires. Single phase, 3 wire and 3 phase come out far better than 2 phase 90 degrees. Three phase is better for use of material in machines.
As to the use of 3 phase only becoming dominant in the 1930's- I think that that is not so- the advantage of 3 phase was recognised at least by 1900. I note that Fortescue's classic paper on Symmetrical Components which was presented in the 1914 AIEE transactions dealt with 3 phase and I have seen more than one generator of the 1914 era and earlier which were 3 phase. Majot synchronous machine analysis papers (i.e. Park's equations) were prior to the 30's. I haven't seen an commercial 2 phase machine although I don't doubt their existence (once the Scott 2 to 3 phase transformer was a topic included in machines courses). I also recall (possibly wrongly) that Tesla's patent was based on a 3 phase unit and the first Niagara units were 3 phase.
Reply to
Don Kelly
The two phase systems with the 4 wires and the 90 deg. phase separation soundslike an interesting technical development that did not offer any real economic justification. Does anyone have information on these systems and when was it determined that they were obsolete? I've never seen a two-phase motor. I agree that the Scott Connection was once considered an essential big deal in Electrical Engineering Courses. There was also a Fortescue connection, was there not?
Beachcomber
Beachcomber
Reply to
Beachcomber
Why? The total power is the sum of the phase powers. Assuming for a moment a 4 wire 2 phase system, with two completely independent phases, it is obvious that each phase will deliver 10kW at 100V, and hence each conductor, will carry 100A not 115.5A. Combining the circuits will yield two conductors carrying 100A and one carrying 141.4A. Do we agree on this?
You are of course correct - I did warn about imminent hand-waving, however!!
Using the same conductors in the two phase case, with the corrected currents would we not see 2(100^2}R + (141.4^2)R = 40000R watts? Yes, I know, more hand waving. Read on.....
I think you may have used the wrong current values in these calcs.
Let's see. The two phase case has two conductors of 100A and one of 141.4A. Lets say that each of the conductors in the 3 phase case, carrying 115.5A, has an area of 1. To achieve the same loss, the two 100A conductors need an area proportional to the current squared, so their area should be (100/115.5)^2=.7496. So I need a total area of 1.499 to accomodate the 2 100A conductors.
The single 141.4 A conductor needs an area of (141.4/115.5)^2=1.499, for a total in the 2 phase case of 2.998 ( 2 conductors of .7496 plus one of 1.499), compared to a total in the 3 phase case of 3.000.
I need to go back and do the whole thing out to a few more significant figures.
I'll pick through this in the morning. It's late.............
Oops, I didn't say that, or mean that. I meant that 2 phase was still being installed into the early 1930's, and it wasn't until then that 3 phase completely displaced it.
Actually, the first Niagara Falls units were 2 phase, and C.F. Scott developed his famous connection so that the power could be transmitted to Buffalo as 3 phase. (Alternating Current Machinery, Bryant and Johnson, 1935, p268)
I of course realize that 3 phase dominated the industry since the early 1900's; my post was more or less a point of academic trivia. However, 2 phase was available as a new service until at least the early 1930's in some locations. Some "hot spots" for it were Utica/Rome/Albany, New York; Hartford/New Haven CT; and the Philadelphia area. We still, to this day, get 2 phase motors coming in to the shop (located in Connecticut) for rewind; they are primarily ventilation/HVAC motors installed in downtown buildings in the early 1900s. 2 phase (4 wire) safety switches and motor starters were standard catalog items until about 1980 to service the replacement market.
Reply to
BFoelsch

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