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Old 06-30-2009, 02:47 PM   #16
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Do you think a tool running at 240v works better than at 120v?


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Originally Posted by hayewe farm View Post
[The reason that high voltage is used for transmission lines is because the higher the voltage the lower the amperage the smaller the wire needed.]
The voltage drop and consequently the efficiency factor only come into play with long distance transmission lines where distance and size of wires are concerned. In a hand-held, plug-in tool, there is no discernible difference. Whether it runs on 240v. and 6.5 Amps or 120v. and 13Amps. the power output is the same. P=ExI...! Additional proof is the fact that there is no compensation (for losses) at 120v.!!!Don't Drink and Drive!!!

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Old 06-30-2009, 02:56 PM   #17
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Do you think a tool running at 240v works better than at 120v?


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Is there a way to ban wiki and google links?
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Old 06-30-2009, 03:07 PM   #18
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Do you think a tool running at 240v works better than at 120v?


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no discernible difference
Then I'll go with a theoretical difference.
Higher current = higher operating temp. = shorter service lifetime.

13A into 2 ohms is 2x the heat and 2x the temp. rise above amb. of 6.5A into 4 ohms.
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Old 06-30-2009, 03:23 PM   #19
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Do you think a tool running at 240v works better than at 120v?


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Nah, it's pretty clear that, like many others,he thinks of the grounded neutral as simply a sink for current, instead of the source that it really is on half of the cycle. For some reason, people think that because you ground one wire and equalize its potential relative to grounded objects, that it magically becomes a dead wire.

The fact is, a neutral is a hot wire, and current flows to and from it, depending on which half of the cycle it's on.
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Old 06-30-2009, 04:56 PM   #20
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Do you think a tool running at 240v works better than at 120v?


Large commercial buildings use 347 volts for the flourescent lighting, for the simple reason that more fixtures can be run from a #14 conductor fed from a 15 amp breaker, than those powered by 120 volts. This means major savings in wire and conduit sizes.
Horse power has a direct relationship with wattage. So if you can achieve the same wattage on small gauge wire by using a higher voltage, less power is lost in the supply conductors and more is available to the load.
This would be of greater advantage on a long feeder run vs. a short run.
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Old 06-30-2009, 05:37 PM   #21
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Do you think a tool running at 240v works better than at 120v?


The heat loss in a motor (part of which is called the copper loss) represents a loss of energy available to perform useful work. The copper loss is due to the resistance of the windings, and is computed as IxIxR, where I is the current flow through the windings, and R is the resistance of the windings. At 120 volts, the current flow through the windings is twice as much as the current flow at 240 volts. The resistance of the windings is essentially independent of the voltage, at least at relatively low voltage like 120 volts. Therefore, the copper loss is 4 times as great at 120 volts as at 240 volts, representing less energy available to drive the tool, meaning lower performance and higher cost to run the tool when comparing 240 volts versus 120 volts.

The main reason small motors are built to run at 120 volts is because it costs more to insulate the motor windings at higher voltage, and 240 volts are not always available for small tools like a drill or a saw. The same tool would work better at higher voltage, however only in commercial establishments are you likely to encounter 480 volts and higher. The tool will run better using three phase power than single phase, for different reasons than the copper loss, but three phase power is rarely available in residential settings, hence you don't see three phase motors very often outside of commercial settings.
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Old 06-30-2009, 05:45 PM   #22
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Do you think a tool running at 240v works better than at 120v?


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Therefore, the copper loss is 4 times as great at 120 volts as at 240 volts
That was my first mistake; the winding resistance at 240v is 2x that of 120v, so you end up with 2x the loss instead of 4x.
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Old 06-30-2009, 07:08 PM   #23
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Do you think a tool running at 240v works better than at 120v?


Y'all think about ohms law for a minute.
The nameplate says 6.5 amps at 240 volts and 13 amps at 120 volts. Did the copper resistance of the windings change by a factor of 4?

Of course not. You're paralleling the windings at 120 volts and seriesing (however you spell that) them at 240 volts. Each winding sees the same voltage and current in either case. Therefore the losses in the motor itself are the same in either case. From an infinite buss, there is no difference.

In the real world, I prefer 240 volts. At startup, the inrush will be less at 240 than 120 causing less voltage drop in the circuit which in turn (as noted previously) gets you a quicker start with less flicker in the rest of the house.
Secondly, you get the same advantage when you hit that huge knothole that bogs the saw down. Obviously the advantages of this vary with the size and length of the circuit.

As for line losses while it's running, the more current, the more line losses....but at this level it's fairly negligible.

As to the transmission line losses, all the above arguments are true (unless I missed a weird one). For a given size conductor, doubling the voltage halves the current and causes line losses to fall by a factor of four (not 2). Sure, you can put up bigger wire, but bigger wire means stronger towers, more towers (closer spacing due to weight), etc. It is all cost driven.
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Old 06-30-2009, 08:08 PM   #24
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Do you think a tool running at 240v works better than at 120v?


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Originally Posted by WFO View Post
Y'all think about ohms law for a minute.
The nameplate says 6.5 amps at 240 volts and 13 amps at 120 volts. Did the copper resistance of the windings change by a factor of 4?

While this is true, there must be some other factor involved, because I KNOW my saw works better at 240 than at 120. I'm thinking it must be some self inductance. Or it may just be the voltage sag on the line is less at start up or under heavy load, thereby allowing enough current in to compensate for the load.

Either way, I have done the test to see, and it seems pretty clear to me that my saw works better at 240.
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Old 06-30-2009, 09:16 PM   #25
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Do you think a tool running at 240v works better than at 120v?


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While this is true, there must be some other factor involved, because I KNOW my saw works better at 240 than at 120.

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In the real world, I prefer 240 volts.
I think we have an accord!
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Old 07-01-2009, 11:43 AM   #26
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Do you think a tool running at 240v works better than at 120v?


Well, the answer is that the tool will run more efficiently at a higher voltage - and dual voltage motors will perform better at higher voltages.

However, you'd be hard pushed on a lot of tools to notice any real difference in real-world performance.

Here in the UK, a lot of professionals use an inverter to switch down the voltage from 240v to 110v and run their power tools at 110v. The benefit is the additional safety from running electrical systems at 110v rather than the higher voltage.
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Old 07-01-2009, 11:06 PM   #27
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Do you think a tool running at 240v works better than at 120v?


While the resistance of the motor is changed by a factor of 4
R=E/I 120/13= 9.23 ohms 240/6.5= 36.92
The copper loss or IIR is the same
IIR 13 X13 X 9.23=1559.87 6.5 X 6.5 X 36.92 = 1559.87
The wattage of the motor is the same so the same amount of heat is created and the same amount of work is produced.

14 ga wire has a resistance of 2.525 per 1000 feet and 12 ga wire has a resistance of 1.588 per 1000 feet the difference in resistance will equalize the cable loss for the 2 voltages. So there is virtually no difference in operation. A 1/2 HP motor produces 1/2 Hp on either 120 v or 240 volt.
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Old 07-02-2009, 11:46 AM   #28
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Do you think a tool running at 240v works better than at 120v?


To the OP: the jury is still out.
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Old 07-02-2009, 12:04 PM   #29
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Do you think a tool running at 240v works better than at 120v?


Not to beat a dead horse, but there appears to be some misunderstanding about losses in a motor. The effective resistance of a motor may be computed as V/I = R, where V is the RMS voltage, and I is the current. However, that is NOT the resistance of the windings, it is the total effective resistance of the motor including windings, stator, eddy currents, and back emf of the motor. You do not compute IxIxR losses based on the total current, since most of the current actually performs work (i.e. driving the tool).

The copper losses are computed based on the resistance of the windings themselves, which are always much less than the total effective resistance of the motor. The resistance of the windings is due to the resistance of the copper wire in the windings, which can be minimized through a variety of techniques, such as using larger diameter wire, using hollow wire for AC motors (check out skin effect for an explanation of why hollow wire is useful), and proper insulation of the wire. However, for a GIVEN MOTOR, the resistance of the windings (NOT THE RESISTANCE OF THE MOTOR) is essentially fixed, and is independent of the voltage applied.

So, the IxIxR copper losses in the winding are only a function of the current flow I. If you double the current, as you do if you run at 120 volts versus 240 volts, you multiply the winding losses by 4 times, and you decrease the power output of the tool. The confusion seems to be in using the effective resistance of the motor INSTEAD of the winding resistance to compute losses in the winding.

The total available power to the motor is VxI, so the total theoretical power available to the tool is the same whether you run at 120 volts or 240 volts. However, the tool only utilizes a percentage of the available power to produce work. Typically, the efficiency of the tool is somewhere around 80 to 85 percent. You measure efficiency by taking the actual shaft power produced by the tool (measure with dynamometer) and dividing by the power available at the input wires (VxI). Tool losses include motor efficiency loss (copper losses, eddy current) and mechanical losses in the tool (friction).

If you do the analysis carefully, you will find that the electrical losses are greater at 120 volts than at 240 volts, hence the tool will have more power at the shaft at 240 volts than at 120 volts. The exact difference depends on a variety of factors, ONE OF WHICH is the resistance of the windings.
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Old 07-02-2009, 11:20 PM   #30
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Do you think a tool running at 240v works better than at 120v?


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So, the IxIxR copper losses in the winding are only a function of the current flow I. If you double the current, as you do if you run at 120 volts versus 240 volts, you multiply the winding losses by 4 times, and you decrease the power output of the tool. The confusion seems to be in using the effective resistance of the motor INSTEAD of the winding resistance to compute losses in the winding.
I disagree....not in your statement about current vs. losses, but in the fact that the current through the windings is not changing.
There are two windings in the motor. Each is designed for a specific current range and voltage, in this case 120 volts.
When you change the taps at the junction box, you are simply either connecting these windings in series (i.e., 240 volt input, but 120 volts per winding as described in Kerchoffs Law), or in parallel for 120 volts(each winding still sees 120 volts and the same current per winding.) Sure, the line current feeding the motor changes, but the windings of the motor see the same voltage and the same current PER WINDING in either configuration. If that stays the same, the losses stay the same.

The only practical thing you are impacting is how the circuit feeding the motor reacts to additional voltage drop, and how the motor reacts to that voltage drop. With that said, the 240 volt hookup minimizes the voltage drop through lower line current. Ergo (it's not often I get to use "ergo") the 240 volt hookup is better, IMHO. Marginally, perhaps, but still better.

If going 240 volts means re-wiring the entire circuit, go 120. It's probably not going to make or break your saw either way unless you have an unusually long circuit.

Hook it up 240 and enjoy your saw.
Hook it up 120 and enjoy your saw.

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