electrical efficiency

The higher the voltage, the lower the current for a given load. The lower the current, the lower the IsquaredR losses will be in the wiring, transformers, etc.

Charles Perry P.E.

Whats "considerable loss"? The windings are in parallel not series @ 208V so resistance/reactance is 1/4. I'm with you, Mike... It's a wash Chuck EE

what kind of equipment? what will maintenance / replacement cost be vs 208?

you could , with a very straight face, win the argument by saying: "it's OK, we will just spec equipment which uses oxygen free copper"

Better check the utility rates before you jump into the fray. Around here there is difference in the way that the rates are calculated. Of course if your over 800 kw demand it will make no sense not to have the higher voltages.

| I had a question and I thought that this would be the best place to | search for an answer. I got into a discussion with a fellow engineer | about voltages and efficiency. His claim was that operating equipment | at 480 volt would save considerable energy as compaired to 208 volts | or something simmilar. I understand the savings during the | construction and installation..... smaller wire size, fuses | disconnects and so on, but I told him that you pay for power in kW and | not amps, so there should be little difference. | | I also understand that there is power loss through transformers when | you do voltage reductions, however, I didnt think there was a | substantial savings by operating equipment at 480 volts as compaired | to 208 volts.

Just to make it work out simpler, I'll change the comparison to single phase 240 volt vs. 480 volt. Suppose you have a water heater that has

If you have them wired separate and switch to wired in parallel, you basically would parallel the wires, or double the cross section, to handle the capacity.

Now consider going from the wired separate to the wired in series. If the power is a split single phase 240/480 system (240-0-240), and if the two separate circuits are on opposite phase, you can connect the two neutrals together at the heaters. Now you have THE SAME CURRENT as before running in the hot wires, but now there is NO CURRENT in the two neutral wires. So the neutral wires are not longer heating up, and no longer contributing to the losses and voltage drop.

It basically comes down to the fact that losses and voltage drop are in proportion to the SQUARE of the current times the resistance. So if you go from 40 amps at 240 volts, to 20 amps at 480 volts, and use wire that has half the cross section (thus doubling its resistance) the losses are you have 1/4 the loss attributed to 1/2 the current, offset by 2 times the loss attributed to resistance, with a net change of 1/2 the loss. So the higher the voltage, the lower the loss, even though you are using smaller wire.

Back to 480 and 208 volts for three phase (277 and 120 between hot and ground). That ratio is even greater at 2.3094. So if your wire has only 1/2.3094 (0.433) the cross sectional area when you switch to 480, you still have 1/2.3094 of the loss, because 1/2.3094 of the current means 1/5.3333 (0.1875) of the loss on the same size wire (and back to

So are you ready to convert your home to 480Y/277, yet? I have been for a long time (but 277 volt incandescent light bulbs are expensive and there are code issues to deal with).

I understand that part, but electrical consumption is charged in kWH. Is there any significant reduction in consumption by using a higher voltage.

It's true, there will be a savings in reduced voltage drop for the service wiring to the motor at the higher voltage.

However, most of the additional I^2R losses will be in the motor itself for the lower (240 V.) voltage requiring 2 times the current for the same power output. A larger chunk of each input kilowatt is going to go toward heating the motor wiring and you will have that much less power delivered as mechanical torque.

Thus, for the same mechanical output, a motor wired for 480 V. is going to operate cooler then one wired for 240 V. and your savings are going to be in how much LESS you are going to be paying to heat up the motor coils at the 480 V. level

Basically just restating the rule that higher voltages are more efficient in electrical transmission, distribution, and consumption.

Beachcomber

| I understand that part, but electrical consumption is charged in | kWH. Is there any significant reduction in consumption by using a | higher voltage.

The "I squared R" losses are reduced by the higher voltage.

On Thu, 02 Sep 2004 20:29:38 GMT Beachcomber wrote: | |>Now consider going from the wired separate to the wired in series. If |>the power is a split single phase 240/480 system (240-0-240), and if the |>two separate circuits are on opposite phase, you can connect the two |>neutrals together at the heaters. Now you have THE SAME CURRENT as |>before running in the hot wires, but now there is NO CURRENT in the |>two neutral wires. So the neutral wires are not longer heating up, and |>no longer contributing to the losses and voltage drop. |>

|>It basically comes down to the fact that losses and voltage drop are in |>proportion to the SQUARE of the current times the resistance. So if you |>go from 40 amps at 240 volts, to 20 amps at 480 volts, and use wire that |>has half the cross section (thus doubling its resistance) the losses are |>you have 1/4 the loss attributed to 1/2 the current, offset by 2 times |>the loss attributed to resistance, with a net change of 1/2 the loss. |>So the higher the voltage, the lower the loss, even though you are using |>smaller wire. |>

| | It's true, there will be a savings in reduced voltage drop for the | service wiring to the motor at the higher voltage. | | However, most of the additional I^2R losses will be in the motor | itself for the lower (240 V.) voltage requiring 2 times the current | for the same power output. A larger chunk of each input kilowatt is | going to go toward heating the motor wiring and you will have that | much less power delivered as mechanical torque. | | Thus, for the same mechanical output, a motor wired for 480 V. is | going to operate cooler then one wired for 240 V. and your savings are | going to be in how much LESS you are going to be paying to heat up the | motor coils at the 480 V. level | | Basically just restating the rule that higher voltages are more | efficient in electrical transmission, distribution, and consumption.

Now, at what point is it worthwhile to choose a higher voltage motor when your supply voltage is fixed, requiring you to use a transformer to drive the motor. Can the difference in losses in the motor ever overcome the loss in the transformer being inserted to do this?

I disagree with the statement that the motor has increased I^2R losses at the lower voltages.

The resistance of a three phase motor varies by the ratio of voltages squared so the resistive losses come out the same assuming the motor design has the same cross section of copper per amp for the different voltages.

Ed

-------------- Not true!

- consider a motor rated at 240 or 480 V. At 240V the windings are parallel and each has a current I when the motor is loaded. Total current 2I , total resistance half that of a single winding- copper loss =(R/2)(2I)^2 = 2R(I^2) At 480V the windings are in series and the winding current is still I and the total current is I. Copper losses =2R(I^2) same as at 240V. Think on it.

It's a wash as far as the motor is concerned. As far as the supply external to the motor - then the current at 240V is twice that at 480V and then the economics of higher voltages may kick in. (sometimes the higher voltage is less economical than a lower voltage- depends on load)

- -- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer

As people have said, I^2R losses are lower with the higher voltage. So, there is *some* savings. But what do you consider 'significant'? If a motor is 88% efficient to begin with, 12% are losses. But of all the losses in a motor, I^2R are only a portion. Let's take a guess and say that I^2R losses are about 1/3 of all the losses (mechanical, hysteresis and eddy currents being others). That would be about 4% of the total. So, *if* you could just double the voltage and cut the current by 2, you would have 1/4 of the losses you had before (1% instead of 4%). So you save 3% of the total energy drawn by the motor. Is that 'significant' to you?

But it doesn't work out that well. As others have pointed out, the higher voltage motor probably has smaller wire (or is wired in series vs. parallel). So, for example if the higher voltage motor has 1/2 the current, but twice the resistance, it would mean you would only save 1/2 the I^2R losses (saving you only 2% of the total energy drawn in the previous example).

Now, some people will contend that there is additional savings in the I^2R losses in the feeder supply cables etc... But I would argue that is only true if you use the same size wiring. If you pull new wire, want to bet whether it is sized for the expected current draw and is thus smaller with a higher resistance?? There is some savings, but it is a lot smaller than you might first guess.

The irony is that the cable is sized for its ability to dissipate a certain amount of heat without raising the insulation temperature (really long runs are sized for voltage drop, but that's another matter). Same insulation temperature limit, same amount of heat per linear foot. (actually, it varies a little because of surface area and other details, but to the first approximation, it's the same amount of energy per foot). Similar with motor design. Lower current (because of higher voltage) means the designer can use less copper and stay under the temperature rise limitations of the winding. Ends up with about the same amount of energy losses.

daestrom