240volt vs. 120volt

[Robert Galloway]

120 volt light circuit maybe 14-3/G Most of us have shop circuits wired with 12-3/G and an occasional 10-3/G. 12-3/G goes most places in my shop 120 or 240.

rhg

J. Clarke wrote:

[Alan Sadler]

[J. Clarke]

You're supposed to size the wire for the current and the length of the run so that the voltage drop at max load is allowable. If you do that then saw should not be "bogging" due to voltage drop in the wiring under heavy load.

[Richard Cline]

Keep in mind that the light bulb example is totally different than the saw. In the light bulb, the watts go into heating the filament. A 60 watt bulb is just that regardless of the voltage/current relationship to create the power.

In the case of the saw, most of the energy goes into cutting wood. Only a small amount goes into heating the motor. The motor heating goes as I^2*R. As the lower voltage requires a higher current, there is more heating in the lower voltage situation.

Dick

[Richard Cline]

All incandescent lamps use tungsten filaments. There are some trace elements added that tend to prolong life but that material is not altered for different voltages.

Halogen lampsd have one of the halogen gasses in the envelope. This creates a "halogen cycle" where the evaporated tungsten combines with a halogen gas molecule and then breaks down when the combined molecule strikes the filament again. Thus the lamp can operate at a higher temperature before failure.

Dick

[BadgerDog]

Here's my rational for proposing that the 240V configuration gives the motor slightly more power.

First consider the 120V configuration: Lets suppose that the motor draws 16A from the 120V source. The motor windings will be connected in parallel, so for two windings in parallel, each winding will have 8A flowing through it and each will have close to

120V accross it (it is a little less than 120V because of the voltage drop in the wiring). If the wiring has a resistance of 0.3 ohms (not unreasonable for a run of 12 AWG wire), the voltage drop due to the wiring would be 0.3 ohm * 16 A = 4.8 V. So, the each motor winding would see 120V - 4.8 V = 115.2V. Note, at this point we can calculate the resistance due to the motor windings: 115.2 V / 8 A = 14.4 ohm. To double check the calculations: For a single winding of 14.4 ohm, the resistance of two winding in parallel will be 1/2 that of one (i.e. 7.2 ohm). The 120 V circuit "sees" a total resistance of 0.3 ohm + 7.2 ohm = 7.5 ohm; 120 V / 7.5 ohm = 16A (as we originally supposed). In terms of power, the wiring will use 16A * 4.8V = 76.8 Watts. The motor will use 16A * 115.2 V = 1843.2 Watts. Total power is 76.8 W + 1843.2 W = 1920 W, in agrrement with 16 A * 120 V = 1920 W for the total circuit.

Now consider the same motor configured for 240V operation: Now the two winding are in series with each other and the wiring, so doubling the voltage does not exactly halve the current (since the total circuit resistance changes). The total circuit resistance will be 0.3 ohm (assuming same run of wiring is used) + 28.8 ohm (two 14.4 ohm windings in series) = 29.1 ohm. The current will be 240 V / 29.1 ohm = 8.25 A. Now we can calculate the voltage drops: for the wiring we have 0.3 ohm * 8.25 A =

2.48V (close to 1/2 the 4.8 V drop for the 120 V configuration); 14.4 ohm * 8.25 A = 118.8 V for EACH winding. The winding together in series give a drop of 2 * 118.8 V = 237.6 V; total circuit voltage drop of 2.475V + 237.6V = 240.075V (slightly off from 240V due to rounding of current to 8.25A). So, in the 240 V configuration, each winding sees a little more voltage AND a little more current, so therefore the motor gets a little more power. We can also calculate the power involved: the wiring uses 8.25A * 2.475 V = 20.4 W and the motor will use 8.25 A * 237.6 V = 1960.2 W, for a combine total of 20.4 W + 1960.2 W = 1980.6 W.

To summarize:

120 V configuration: 1920 Watts total, 76.8 Watts for the wiring and 1843.2 Watts for the motor 240 V configuration: 1980.6 Watts total, 20.4 W for the wiring and 1960.2 Watts for the motor

Compared to the 120V configuration, when wired for 240 V the motor gets

1960.2 W - 1843.2 W = 117 W more power, or a (117/1843.2)*100 = 6.3% increase.

Granted, I've simplified the analysis by only considering the resistive loads (i.e. I've implicitly assumed that the reactance due to the motor's inductance has been balances out by the reactance of the capacitor). Also, I realize that motors are more complicated than I have described them; for example the amount of current they draw depends on the level of mechnical resistance they are working against (e.g. freely spinning saw blade versus cutting 2 inch thick maple). However, I hope I've shed some light on the an aspect of the 120V versus 240V debate in order to show that it is not as simple as "double the voltage, halve the current" response that is given so often.

Disclaimer: I am not an electrician, but I do have a background with some things electrical.

BadgerDog

[Doug Miller]

Point taken; just the same, a dual-voltage motor *does* run cooler at

240V than it does at 120V.

[John Sellers]

Whoops, got the motor connections backwards. And that was after I checked and swapped them - twice.

I agree about the increased power draw when the motor load increases; and so do the parasitic losses. But, that's why I said it's really a moot issue. There's no such thing as constant load with a woodworking machine. So, it doesn't make much sense to debate the issue of small current changes. Bottom line - 240V makes for easier use of the power, with less wasted energy.

Sorry for the confusion. John Sellers

[Nova]

I wasn't referring to the resistance of the filament but rather the conductors feeding a motor being operated at 110V verses 220V. The amperage being double when the equipment is operated at 110V would cause the conductors to generate more heat decreasing the overall efficiency. Granted, with a short run the difference would be minimal.

-- Jack Novak Buffalo, NY - USA (Remove "SPAM" from email address to reply)

[Robert Galloway]

Correct. As I mentioned in a different post, in an ideal world, every application of a power tool would have copper coming to it of such size that the maximum load of the tool wouldn't produce a discernible voltage drop. This would include the cord provided by the manufacturer of the tool. In the real world, I've been happier with tools running on 220. I've spend a lot of time with most of them both ways. You pay's your money and takes your choice. My advice is, if the tool stays in one place all the time and 220 is handy, use it. YMMV.

rhg

J. Clarke wrote:

[Robert Galloway]

This would be true if you didn't rewire the motor from a parallel connection to a series connection when going from 110 to 220. Current, therefore I^2R losses, should be the same. Most small motors, however don't change the connection to the start winding so it DOES carry twice the current during the brief period it's active.

rhg

Richard Cl> In article ,

[Edwin Pawlowski]

Interesting. I'm not sure about the "simplified" part though ;) Ed

[Robert Galloway]

More consumption in the 240 volt mode. Makes sense. Doesn't work out that way. The rotating motor functions as a generator. Its generated emf opposes the supplied emf. Its ability to counter the supplied voltage is what limits the net draw. If there's voltage drop in the line, (which is more pronounced in the 120v mode) then the motor is less effective as a generator, less effective at opposing the applied emf and generates more temp rise. Motors run best at full rated voltage. However you want to supply this, the result is the same. On

120 volt, use the largest gauge wire. For a given gauge, 240v will do a better job of keeping your motor spun up to the full rated RPM and therefore do the best job of generating counter emf.

bob g.

BadgerDog wrote: