air compressor Power

I have no idea why the motor would smooth out unless you had so much voltage drop in the 110 volt circuit you were using that the motor was drawing excess current trying to carry the load. In theory, there should be no difference, as you have the same amount of current flowing through each winding regardless of whether the internal windings are connected in parallel for 110 or series for 220.

Keith

Each half of the windings in the motor only run on 120 volts, anyhow.

For the lower voltage, the sets of windings are connected in parallel, and fed 120 Volts.

For the higher voltage, the windings are connected in series, and the 240 volts divides across them, so each winding is still only running on 120 volts.

Aren't you glad you didn't ask a political question!

If you aren't popping breakers, then it's unlikely your compressor is drawing 15 amps. If it is popping breakers, then switching to 240 will solve that problem.

Yes, it draws 1/2 the Amps, but at double the Voltage so power used, WATTS remains the same.

Watts, or actually Kilowatt Hours, are what you are buying to run it, and it will draw the same total watts at 120 Volts or 240 Volts.

One advantage to running it at 240 Volts is that the wiring from the breaker box to the compressor, or an extension cord of a given length, can be half the size on 240 Volts as on 120 Volts.

The advantage of using 220 is that since only half as much current is drawn, you can use smaller wire. Additionally, there is less loss in the wiring because of the lower current. For example, a compressor which draws 15 amps requires a 20 amp circuit (they're only supposed to be loaded to 80% of breaker capacity) run with #12 wire. On 220, it would draw 7.5 amps, and could be easily run on a 15 amp circuit wired with #14. At higher powers, it becomes impractical to run a load on 120 volts due both to the size of the wire required to carry the current and the load imbalance resulting from drawing significantly more current from one leg than the other.

I'm not that familiar with how power company meters respond if the load isn't balanced, so there may be a (probably) small difference in measured comsumption when you're drawing current in both hot legs instead of only one. Breaker panels are supposed to be wired so that, on average, the same amount of current is drawn on both hot legs and the load remains balanced. In practice that rarely actually happens, but I'd be really concerned if the power company meters had a large error when the load isn't balanced. If I get a few minutes, maybe I'll look up how pole meters work & find out for sure.

One other thing you'll notice if you re-wire it for 220 is a reduction in lights flickering when it starts. Starting current is divided between both hot legs instead of only one, with a corresponding reducting in the voltage dip resulting from starting current draw.

Keith

PS...I'm only a practicing electical engineer with 30+ years of professional experience !:)

Just stating the facts.

The only real source of error I could find is that they don't correct for power factor. If you have a load with a poor power factor, it'll draw reactive current which is added to the useful current in the meter. This has nothing to do with whether the load is connect hot-to-hot or hot-to-neutral. Of course, the error is in the power company's favor!!

Old, not to keep harping on this, but if you get a chance ask your power company to explain to you why they said the same load will cost less to operate on 220 than on 110. I'd really like to know if I'm missing something.

Thanks,

Keith

Actually, you were using shorthand to say 220 is slightly more efficient at delivering power, & so running at double the voltage but 1/2 the watts will give a slight use savings. That is correct.

Somehow, reading your first posting & the way it was worded, it seemed you were saying the savings would be 50%. Odd how the mind plays tricks.

What you meant to say is accurate. The way it was reading, sounds odd. :)

--->Paul

Take an example of using 50 feet of #12 wire to run the compressor.
50 feet of #12 has a resistance of .0811 ohms.

At 120 volts the compressor draws 15 amps. 120 volts times 15 amps= 1800 watts. 15 amps through .0811 ohms will drop the voltage 1.22 volts. That is about 1% voltage drop.

1.22 volts times 15 amps = 18.3 watts lost in heat in the wire.

At 240 volts the compressor will draw 7.5 amps. 240 times 7.5 = 1800 watts.

7.5 amps through .0811 ohms will drop the voltage .6 volts. That is about .25% voltage drop.

.6 volts times 7.5 amps = 4.5 watts lost in heat in the wire.

So you can see that the wasted POWER in the wire is 4 times worse with the 120 volt circuit. However it is a small amount when compared to the power drawn by the compressor. This can become significant with long runs of wire though.

These figures will be slightly different because the compressor will not see the full voltage by the amount dropped in the wire and the compressor current may be slightly different because of it but this gets you close to what happens.

Regards
Gary

In my above example of 18.3 watts lost in the wire on 120 and 4.5 watts lost in the wire on 240 volts, that�s a difference of 13.8 watts.

At 8 cents per kilowatt hour for the cost of electricity, that 13.8 watts difference would require the compressor to run for about 72 hours to cost 8 cents more. (1000/13.8 = 72.)

The motor itself will run as efficiently on 120 as it will on 240. As others have said, the motor windings are in parallel for 120 and series for 240. Same amount of loss in windings either way.

It can be left to readers to determine if 72 hours of run time is worth worrying about 8 cents or not.

Regards
Gary

--->Paul

Yes Paul that may be correct in extreme cases. In the amount of power lost in the wire in my example, 13.8 watts, some additional amount would probably be expended in heat in the motor also. But that amount would be barley detectable in the motor. As far as lasting longer, I doubt it would have much effect at the levels discussed here.

The reason that I say the lost power in the wire would also be lost in the motor is because when the motor has a constant load, that is delivering a certain horse power to a load, it requires a certain voltage and current to deliver that power. So If the voltage is lowered slightly then the motor must draw a little more current to compensate in order to deliver the same amount of power. That additional current in the motor windings will result in additional heat in the motor.

This also raises the power lost in the wire slightly above those given as the current in the supply wire will be greater too.

Regards
Gary