240V from a 3 phase main ?

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This much shouldn't be a problem for either of the 3 phase setups that have been discussed here - the voltage across any two legs will be the same. Of course if the equipment needs a neutral as well and you have delta power then it will matter which legs you use.

For 240V delta power, two of the legs will give you 120V compared to neutral. For Y 208V power, any of the legs will be 120V.
Since you haven't had any problems you probably have a 208V Y configuration (or you just happened to use the right legs). For what it's worth, before this thread I had no idea that anyone provided delta power. I knew what it was (although not the name) because that's what you get from a phase converter where you're using two legs as they are and generating the third. But every industrial building I've worked in around here (near San Jose, CA) had 208V Y power. Sometimes they'll have some higher voltage as well, but since that tends to just go to the HVAC systems I've never paid any attention to its configuration.
allan
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snip---.
For what

That's an interesting observation. I've had three phase power in three different structures, all of them in residential areas, and it's been no problem at all to get three phase delta. The one problem, however, is getting them to run it in underground. I have been told by more than one EE that there are fire problems associated with service of that nature. Not sure I understand it, but I've learned to live with the three transformers overhead.
Harold
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Harold & Susan Vordos wrote:

Fire nature is based on the IR drop. You draw the current through the smaller than needed current and drop some voltage. I^2R or E^2/R that is power. The heated wire doesn't dissapage heat except down the wire. Some place gets to hot.
There is times when water leaks in and there is more IR drop - mostly to water.
Such is life.
Martin
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snipped-for-privacy@iwvisp.com (Everett M. Greene) wrote in message wrote:

It isn't that the motor won't run, but that a) the efficiency and life of the motor will likely be reduced because b) the current draw will go up at the lower voltage.
The speed of induction and synchronous motors (which most 240V tool motors will be, single or three phase) is determined primarily by the line frequency. The power needed by the motor is determined by the load-the motor doesn't care how much current it draws... it draws what it need to to meet the power demand of the load at the run speed. At the lower voltage (about 12%) the current will be higher (again, about 12%), leading to greater heat production in the motor and greater I^2R losses in the motor and supply wiring. If the load is near the 240V rating of the motor(compressor motor, large power tool, etc) then at 208V, the motor will like lose some or all of it's magic smoke and cease to function, especially if that 208V supply is really only 200V (5% either way is very common with system load variation, 10% not unusual, especially in the summer when lots of AC units are on).
Motors are generally not conservatively rated-it isn't economical to overrate. You get the 10% or so maximum supply variation built in, and that's it. Go below that, and you need to begin derating the motor rapidly, go above that, and the likelyhood of insulation failure goes way up.
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enl snipped-for-privacy@yahoo.com (e) writes:

I agree that power = I * V * cos(pf), so the current will be greater for a lower voltage at a constant power draw. I also agree that a fixed load device such as an air compressor could have trouble with a lower supply voltage, but variable load devices will simply have less power available -- if your table saw is a little weaker due to the voltage being somewhat low, just don't push the work through so fast.
Although motors are sized to meet their (presumed) load, motors are built in discrete sizes -- 1/2, 3/4, 1, 2, 5,...HP (or the equivalent in watts). If a load is 7/16 HP, the manufacturer is going to use a 1/2 HP motor which is then going to have >10% reserve. Motors are also designed to work in a particular voltage /class/, so they'll produce their rated power output at the minimum supply voltage without damage (or at least without catching fire). If the voltage is higher than the minimum, a motor can produce more than its rated output without damage. It would be a violation of various standards (UL, etc.) to manufacture a device that relies on a more than minimum supply voltage to safely produce the necessary power.
Industrial installations are another matter, but this discussion started as question about a "home" woodworking shop.
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Everett M. Greene wrote:

I think that would be hard for the user to judge. You would need a motor starter with properly sized heaters to ensure you're not drawing too much current.
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The thing to remember here is that with induction motors (as opposed to smaller permanent magnet rotor motors) the torque is proportional to the cross product of the stator and rotor currents. To a first approximation, if the supply voltage is 86.6% (208/240) then the torque will be 75% of that at rated voltage.
RB
Everett M. Greene wrote:

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Your math escapes me. If the supply voltage is lower, the current will be higher for the same power output, thus the torque would be higher, not lower, by your statement.
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It doesn't work that way. The impedance of the motor's winding doesn't change just because you have a lower voltage. To get the same torque you need a higher current but you won't get it.
If as the voltage is lowered the current rises then when the voltage is zero (as in a short across the input to the motor) the current will be how much?
If the voltage is 86% the torque developed will be 0.86 x 0.86 or 75% of that at rated voltage (240 volts.)
RB
Everett M. Greene wrote:

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torque != power
We're talking about two different things
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Everett M. Greene wrote:

I understand that, but for constant speed torque and power are directly proportional. The motors used for most tools are synchronous so the useful rotation speed will be close to constant. Torque=k*2*pi*r*(rotation speed)*Power.
Unlike older motors, what is being produced today has little design margin. If you try to raise the current above the nameplate levels as the NI product goes up flux saturtion is likely to occur. Not a good, or useful point to be operating at.
RB
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Are you sure about that? I would have thought most machines would have induction motors. Synchronous machines keep absolutely constant RPM as load increases, then suddenly stall (sometimes with disasterous results) when the phase angle hits 90 degrees. Induction machines gradually slow down as they get overloaded and have rather more benign stall behavior.
For example,
http://www.1-home-improvement.com/specialty-tools/Baldor-WWL3606-3-B00002 23WG.html
is almost certainly a 2-pole induction motor. The 3450 RPM rating is the giveaway, and it only runs at that speed at its rated load. Under no load, it'll run at close to (but not quite) 3600 RPM, and under overload, it'll run slower. The motors you commonly see that are rated at 1725 (or thereabouts) RPM are 4-pole induction motors (twice the torque but half the speed for a given size/weight/power).
A 2-pole 60 Hz syncronous motor would have a nameplate speed of 3600 RPM.
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On Sat, 17 Jan 2004 18:14:27 PST, snipped-for-privacy@iwvisp.com (Everett M. Greene) wrote:

Power is a rate, so it requires a time element. Power is equal to the product of torque and speed (time function). Since speed in an induction motor is a function of the power line frequency, it doesn't change as voltage changes. So power is proportional to torque.
However, I'd like to take exception to one thing that RB said. The impedance of the motor windings is a function of load, speed, and slip. As long as the speed, load, and slip are constant, impedance is constant. So lowering the applied voltage does lower the current. *But* an induction electric motor tries to compensate for this when available power falls below load demand by increasing slip.
As slip increases, the winding impedance falls, and current can increase, even at a lower supply voltage. Increased current yields increased torque, and at the same speed, increased power.
The fly in this ointment is that winding *resistance* doesn't change. Energy dissipated in the windings is a function of the square of current and the winding resistance (P=I^2 * R). So the windings heat more rapidly at a lower line voltage, but speed is constant, so the amount of cooling air remains constant. That causes winding temperature to rise, leading ultimately to insulation failure, and all the magic smoke is let out of the motor.
Gary
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But, again, we're talking about shop tools, most of which are going to be hand fed their work and whose motors are going to be lightly loaded most of the time. If the voltage is less, the available power will be less so you have to feed the work to the tool a little bit more slowly. In other words, you won't be pushing the motor load to the limit and causing the smoke to start rising. Up to a point, you can even overload a motor for a brief period without causing any harm.
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On Mon, 19 Jan 2004 19:08:24 PST, snipped-for-privacy@mojaveg.iwvisp.com (Everett M. Greene) wrote:

The available power will not be less. Remember, increasing slip decreases effective reactance in the motor, so at a given load, current will automatically increase to satisfy load demand when voltage decreases. Since P = I * E, power can remain the same when voltage is decreased. It is how electric motors work. So you will not have any sensible feedback telling you to slow down.
Now for some power tools, such as a table saw, you can consciously and deliberately reduce the load by decreasing feed rate so that power demanded decreases to match decreased voltage, and then current will not increase. But you can't do that by noting the motor is bogging, it won't. You'd have to continuously monitor motor current to stay within the safe area. The tool itself won't give you any feedback telling you to slow down, until you note the smoke coming from the motor.
For some motor operated loads, such as an air compressor, you have to change pulley ratios to reduce the load. While this will reduce running current, it may make the compressor hard to start, and possibly damage motor, contactors, or capacitors anyway.
In short, you can run a lightly loaded motor on reduced voltage, but you need some way to monitor current to ensure you really are loading it lightly enough to keep it from overheating. And you have to be wary of other factors which may come into play, such as excessive current draws required to come up to speed on reduced voltage.
Gary
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P != I * E
You're saying the utilities cannot reduce their demand load by reducing voltage because the motor loads will just draw increased current?

I would think you'd reduce the size of the driving pulley so reduce the motor load, thus making it easier to start.

And, again, we're talking about voltages that are within the motors' voltage rating class.
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On Mon, 26 Jan 2004 10:52:00 PST, snipped-for-privacy@mojaveg.iwvisp.com (Everett M. Greene) wrote:

Ok, to be a pedant, P = I * E * cos(theta)
For a non-zero theta, there will be circulating reactive currents as well as load currents. The former won't contribute to motor power, but will contribute to I^2 * R winding heating. So they make the situation worse than a simple P=I*E calculation would imply.

That's true for motors, it isn't true for resistive loads. Since the load on the grid is a mixture of motor and resistive loads, the utilities can decrease demand by decreasing voltage, but only the resistive loads will have reduced demand. The motors will just draw more current until they overheat and fail.
This is an unfortunate fact of life in some parts of the world where brownouts are common. Motors fail by overheating under low voltage conditions.
Gary
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Everett M. Greene wrote:

Power does = E * I for D.C. Power does = E * I * Cos(theta) for A.C. theta is the phase angle between E and I. Sometimes Cosine equals 1. :-) Those are the facts.

The swinging transformers move taps raising and lowing voltage. These are massive horz. transformers in substations.
They often drop voltages to shed load. Many motors and compressors don't start. Those that do will draw more but with the drop off's and the resistive load loss it is a win.
They often run this valley on between 68 and 93 volt not the normal 120-140v.
This is when storms take out a substation - they back-strap this valley from another.
I caught them one Sunday a.m. - TV worked and most things - computer UPS didn't like it one bit. I called it in - got service on the line - Naw that just can't be. I asked for a service person to verify the substation as I have checked my house from stem to stern. I gave the service person my number.
A super nice response engineer called from the substation. He was about to unlock, but though to call first. I told him I used my Beckman and my Tektronix true RMS voltmeters.
Once he heard the true RMS - he knew I knew something. We talked as he into the station - and found the main line and the back strap installed. That is when the fun came.
He had to undo a double hot backstop line voltage and heat. He asked me to stand by and call in for him if he didn't come back. I did and he did. I verified 120V was on the lines. We chatted as he locked up and off we both went. Him home, me computering.

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This reminds me of a class we had in engineer school. The object was to calculate the horsepower required to move a quantity of dirt in a scraper over a certain soil at a certain grade and speed. You did the math and came up with the required horsepower. However, There you are out in the field, you load your scraper and it either goes up the hill or it doesn't. If it doesn't, you take off some of the load, or go a different way. You don't go to your Company Commander and ask for a bigger motor because your theta ain't cosigned with the delta max. Paul
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Quite true. The subject line says 240V, 3-phase, so we have to presume AC is being discussed.

If they do this very often, you have grounds for some loud complaints to your state utility commission. Electric utilities don't guarantee much of anything, but they do have to stay within shouting distance of the nominal voltage of the service class. 68V isn't even close.
[snip]
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