# 220v conversion question

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• posted on September 2, 2009, 7:33 pm
On 09/02/2009 01:08 PM, snipped-for-privacy@gmail.com wrote:

Consider a motor that draws 15A at 120V. Let's call it a 90A inrush current, with a wiring length of 30 feet (20 feet from the panel, 10-foot cord) and #12 wiring.
Power loss in the wiring is:
Pc = I^2 * R
For 30' of #12 copper, R=0.048 ohm
At 120V, assuming 90 A inrush: Pc = 389W
At 240V the inrush should be half, or 45A: Pc = 97W
This makes sense, we double the voltage and cut power loss by a factor of 2^2.
So for a total inrush power draw of 10800W (90*120 or 45*240), at 120V we lose 3.6% of the power to supply losses, while at 240V we lose 0.9%.
We get 2.7% more power delivered to the motor by switching to 240V. I wouldn't be able to notice the difference, so I call that "small".
Chris

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• posted on September 2, 2009, 7:50 pm
"Chris Friesen" wrote:

The inrush current can easily be in the order of 8-10 times the FLA.
Using your example above, a 15 FLA motor could easily have 120A-150A inrush.
Lew

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• posted on September 2, 2009, 8:14 pm

In fact, since the motor is an inductor, the inrush is only limited by the resistance. The lower the source impedance the higher the inrush, for a shorter time. Shorter time => faster start. QED.

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• posted on September 2, 2009, 9:40 pm
Chris Friesen wrote:

I believe the resistance of the motor's windings figures into the formula as well as it's part of the circuit.
--
Jack Novak
Buffalo, NY - USA

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• posted on September 2, 2009, 9:49 pm

The line losses of pushing 1000 watts through a given conductor at a higher voltage creates less resistance than at a lower voltage. Hence 500KV power lines. Current is your enemy, voltage is your friend.
Speaking of Watts... Charlie did NOT, I repeat, DID NOT quit he Rolling Stones.

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• posted on September 3, 2009, 12:10 am
On Wed, 02 Sep 2009 13:33:20 -0600, Chris Friesen

It's not 30' of copper. Go *all* the way back, both legs.

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• posted on September 3, 2009, 10:40 am

I agree, small. But noticable to me.

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• posted on September 2, 2009, 7:20 pm
"Chris Friesen" wrote:

You would be surprised.
Don't forget to include the cost of the male plug in your cost comparisons.
Lew

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• posted on September 2, 2009, 3:50 am
On Tue, 01 Sep 2009 16:40:47 -0600, Chris Friesen

The resistance of the wire is the same. ...unless you're wiring your house with 16AWG.

No, you're not. The house wiring resistance is the same. Any drop in the house wiring will shop up in the motor power (squared).

Wire resistance.

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• posted on September 2, 2009, 2:25 pm
On 09/01/2009 09:50 PM, krw wrote:

The resistance of the motor changes when rewired for 240V.
Chris

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• posted on September 2, 2009, 3:33 pm

The resistance of the wire does not.

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• posted on September 3, 2009, 4:19 am
On Wed, 02 Sep 2009 08:25:07 -0600, Chris Friesen

Yes, generally it drops to roughly quarter.. Windings go from parallel to series. 2 windings of 2 ohms each in parallel is 1 ohm, in seies it is 4

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• posted on September 1, 2009, 8:29 pm
On Tue, 01 Sep 2009 12:02:53 -0600, Chris Friesen

At 240v there's slightly more power available to the motor because of less power loss in the circuit wiring due to the higher current at 120v. This is especially evident under high current conditions like startup when the motor is temporarily drawing several times it's normal current draw.
If the 120v circuit had half the resistance of the 240v circuit, there would likely be little or no difference in the behavior of the motor. But that's typically not the case. A 20 amp 120v circuit would typically be wired with 12ga wire. A 20 amp 240v circuit would also typically be wired with 12ga wire. Therefore, all other things being the same, the 120v and the 240v circuits would have roughly the same resistance. I'd expect the motor running on 240 under those conditions to have faster startup characteristics as well as being slightly harder to stall.
Tom Veatch Wichita, KS USA

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• posted on September 1, 2009, 11:01 pm
On 09/01/2009 02:29 PM, Tom Veatch wrote:

Technically true. But on a properly designed circuit the NEC requires at most 5% voltage loss. Even assuming that we got perfect efficiency at 240V I defy you to notice a tablesaw coming up to speed 5% faster.
This is not a "wow, start up is almost instant" sort of difference.
Chris

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• posted on September 2, 2009, 3:50 am
On Tue, 01 Sep 2009 17:01:46 -0600, Chris Friesen

Try it. You *will* notice the difference.

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• posted on September 2, 2009, 12:52 pm

I (OP) noticed a difference.

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• posted on September 2, 2009, 2:29 pm
On 09/02/2009 06:52 AM, DLB wrote:

As I mentioned in my first email, if it makes that much difference then it's likely the motor itself was not operating optimally at 120V.
Aside from supply losses which have been discussed extensively :) in this thread, there's no inherent reason why a motor would operate more efficiently at higher voltage.
Chris

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• posted on September 2, 2009, 3:34 pm

...and it is exactly the supply losses that make the difference, so...

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• posted on September 2, 2009, 4:28 pm
On 09/02/2009 09:34 AM, snipped-for-privacy@gmail.com wrote:

As Dan Coby pointed out, with a 314ft run of 12awg you'd have 6.8% more power available due to reduced supply losses if drawing 10A at 120V. With a typical shop wiring configuration the difference would be less.
There's no way that the reduction in supply losses results in a "wow, start up is almost instant" sort of difference when switching to 240V unless the 120V circuit was way undersized.
Chris

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• posted on September 2, 2009, 5:29 pm
On Wed, 02 Sep 2009 10:28:36 -0600, Chris Friesen

Now square that.

Try it and you'll change your tune. Your numbers are only good for steady state current draw. There is a large inrush to start the motor, which will be limited by the supply impedance. At 240V that impedance is the same, though the potential is doubled.