The motor is a 120/240V, 19.5/9.5A table saw in a commercial wood shop. Is there any reason to convert the current 120V setup to 240V? Watts=watts, but what about efficiency, longevity, strength under load, etc?
Dual votage motor
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if you had 14AWG conductors, you would want to wire it for 240. A lower voltage motor tends to run warmer than a higher voltage motor of same HP and RPM.
If you are installing a new motor, you can save some money running smaller conductors and higher voltage.
If you are installing a new motor, you can save some money running smaller conductors and higher voltage.
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Why? The current though the windings should be the same no matter what voltage you are using, and if the current is the same than the I²R losses are too.
This makes absolutely no sense whatsoever, but if the motor is connected for 240, it will have more pull-up and breakdown torque.
In other words, the guys who use it will very likely notice more power.
Even if the wire for 240 is smaller, there's still a noticeable difference.
Yes, I realize this violated the laws of physics, but the actual fact is.......
So yes; I would absolutely connect it for 240.
In other words, the guys who use it will very likely notice more power.
Even if the wire for 240 is smaller, there's still a noticeable difference.
Yes, I realize this violated the laws of physics, but the actual fact is.......
So yes; I would absolutely connect it for 240.
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The motors torque doesn't ever change. Who cares what voltage it is. The motors rated for a set duty. If you wire that motor with the correct conductors the demand for electrons will always be their.
In theory, yes. In actual reality, no; it'll have more torque all the way around when connected for the higher voltage. Ask any old-time carpenter if his saw runs better on '110' or '220' and every one of them will say the same thing. 220.
Years ago, when I was a carpenter, my portable tablesaw had a basic 1.5 HP single phase motor. I installed a dual-voltage switch on it so I could run it on either 120 or 240.
It had considerably more power being run on 240 with a #16 cord than it did on 120 with #12s.
Yes, I know what theory says; I also know that my actual experience is different.
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If I'm not mistaken it should also draw a lower inrush current spike at startup when wired for 240v.
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And there is is...
Percentage? No.
Power? No.
Current? Yes, it is lower in terms of Amperes. And therein lies the typical reason for the PERCEPTION that the motor provides mor "power" at the higher voltage. If, when starting, the high starting CURRENT causes a voltage drop in the wires, that relates DIRECTLY to the starting TORQUE that the motor can deliver.
Starting torque varies by the SQUARE of the voltage change. So if, at 115V, you experience a VD of just 10% at the very instant you hit the button because the current is high, that motor is putting out only 81% of its maximum starting torque. If you use the SAME SIZE WIRE and connect it for 230V, your wire is now twice as big relative to the circuit current, PLUS your starting current is 1/2 as high. So you likely experience NO voltage drop in that case, and effectively, you have almost 20% more starting torque to the shaft, which means in accelerates faster and RECOVERS faster from a step change in load.
That said, IF you start with conductors rated for twice the motor current rating at 115V, you may not likely see any perceivable different in performance. The problem is, nobody does that. 19.5A, people will likely run #10 and call it good, and the NEC says that's acceptable. But that motor will pull upward of 117A at startup. If you use 230V at 9.5 A, the starting current only gets to 57A. If you look at the VD from 117A on 50Ft circuit length of #10 wire, it doesn't look as good as the VD of 57A on the same wire, maybe not even as good as 57A on #12. But if, at 115V, you ran #4, no problem. Yet who is going to run #4 for a 19.5A load?
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I picked that avatar because it's a half way real representation of what I look like, although I have a beard.
I also love that show. :thumbup:
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GREAT Post! Thanks! :thumbup:
How did you arrive at this inrush current? Roughly 6x the continuous load?
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Yes, 600% is the safe bet. A few motors will pull more, many will be less, but 600 is a good place to land and be safe in your assumptions.
Side note / pet peeve:
"Inrush" current is officially only supposed to mean peak current that happens when magnetizing the windings, and on motors can be anywhere from 800-2000%, but only for about 1 to 1-1/2 cycles. You also have Inrush Current on transformers, even solenoids; anything with a coil.
The 600% in a motor is technically "Starting" current, aka "Locked Rotor Current" and will stay that high until the motor gets to about 90% speed.
I know virtually everyone uses "Inrush" to mean "Starting" current, it's been that way for all of my career and I get that I'm beeing picky. It's just that nobody ever says anything about it being wrong, until they experience the REAL inrush current causing breakers to trip, yet they can't see it happen because their meters are too slow to catch it. They see the Starting current and it doesn't exceed the breaker trip settings, so they can't figure out what's causing the breaker to trip. I tell them it's the "Inrush current" and they proceed to explain to me again that they already know that the 600% is not causing it... <sigh>
I've always called the initial current magnetizing current. This is followed by inrush (aka locked rotor), then running current.
I guess this is not technically correct, since inrush would be more akin to magnetizing.
But yes, the first cycle or two is basically a bolted fault. The current is limited only by the impedance of the winding. No counter EMF at all.
I guess this is not technically correct, since inrush would be more akin to magnetizing.
But yes, the first cycle or two is basically a bolted fault. The current is limited only by the impedance of the winding. No counter EMF at all.
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