240V from a 3 phase main ?

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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
--
Martin Eastburn, Barbara Eastburn
@ home at Lion's Lair with our computer snipped-for-privacy@pacbell.net
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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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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.

--
Martin Eastburn, Barbara Eastburn
@ home at Lion's Lair with our computer snipped-for-privacy@pacbell.net
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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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There are several common 3phase hookups:
If the 3 phase is specified as 208V, then it is a 120/208Y setup. This is a four wire (wye-type) connection, with the threee phase legs and a neutral
B A \ / \ / \ / * <--neutral | | | C
measuring 208V from A to B, from B to C and from C to A. This will measure 120V from A, B, or C to the neutral. This is NOT split phase in any sense of the word, as, for example, A to neutral and B to neutral are 120 degrees out of phase, not 180 degrees. This phase angle diffence is why 120V+120V is 208V, not 240V.
Next is 277/480Y which is a very common (in the northeast US) hookup for commercial property. There are cheap 277V light fixtures (fluorescent), 440V motors, etc. A 240 tool would be at the limit on 277V, and might not take it.
Another common, though declining in popularity, hookup is the 240V delta. This HAS NO NEUTRAL.
A / \ / \ B--G--C
It has a Grounded conductor, usually a generating a 120-120/240 split phase. This is split phase like in a home, really 240V and 120V. Many power companies will no longer do these installations, and don't like to support them. Different transformer than the wye and it isn't as popular anymore as the wye hookups, so they don't wan't to pay to train their people or keep a stock of transformers, meters, etc.
the 240 delta is easy to work with. 240V 3 phase tools and 120 or 240V single phase plug right in. Watch the third leg from the grounded, tho, as it will be 208V and if you hook a 120V load to it, usually BOOM!
With 120/208, you may be in luck ,tho, as many single phase 240V tools will run on 208, either by rejumpering, or because they are really built for a compromise voltage, like 220 or 230V. A 220V tool will run on either 208V or 240V-- the spec is generally +-10% to match the requirements on thepower company, and both 208 and 240 are within 10% of 220 and 230V. Note that the power and efficiency will be a compromise as well. (230V motors seem more common, probably because line voltages are usually set to the high side of nominal)
For the 277/480, you can get a set of transformers, as tou can for 120/208 if you are reall concerned about the difference between 208 and 240V. They arn't that expensive new, and a dirt cheap second hand.

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E, I followed you, and thought I had it until you said the BOOM part. If you wrote what you wanted to say, how are you going to get 208V from that 120/240 volt setup? The way I see it, A to B is 240V and either A or B to G/C is 120V (each out of phase by 180 degrees). A or B to ground (G/C) is 120V and G/C to earth is 0 volts.
Al, you stated it right when you said "split phase" instead of the mis-named 2 phase. That subject caused a lot of noses to curl when I used it here for loss of a proper term. Gary C. explained many months back that 2 phase does or did exist but is rare. Just filter & follow the good advise here and with the Power Co. to be safe when you play with your landlord's power. New environments often bring surprises with them.
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There are four nodes in this setup-- The fourth point (G) is usually there in the US to give the 120/240 split phase (B to G is 120, C to G is 120, 180 out of phase) G is the grounded node-current carrying, but tied to ground at the service enterence, usually white in the US- for the load (user) side of the hookup, but does not have an equivelant on the power co side of the transformers. It is derived from a center tap on the user side solely to provide the two 120 phases-- this transformer is the same hookup as a normal split phase hookup at a residence. The third power node, A, is 120 degrees ahead of B and 120 degrees behind C.
This leads to a bunch of interesting behaviour, such as the voltage from A to G is 208V, 90 degrees out of phase with the voltages B-G, C-G and B-C. Node A is often called the 'wild leg', and is supposed to be colored orange under NEC, so as to differentiate it from the 120 legs. Another effect is that, though this hookup (in the standard three tranformer version, not one of the two transformer versions) is very stiff against load for 240V loads, the two 120's are not very stiff, and if not fairly well balanced, will drift. For example, at my shop, the wiring was done with one building WAY out of balance (5000W of lighting on one 120 leg, and nothing on the other) and this shows as several volts different between the phases at the transformer.
Another neat feature is the ability for the supplier to REALLY cheap out on the hookup (see my other post)
The delta is sometimes done a different way: one of the corners will be the grounded node. This will not provide 120V directly, but requiers a seperate transformer. Not real common other that for motor load only hookups. This setup won't provide 208.