"Edwin Pawlowski" wrote in message
...

"BadgerDog" wrote in message

To first order, won't the saw consume the same amount of power when
using
120V or 240V (assuming of course the motor is wired correctly for the
appropriate voltage)?

Yes.

That said, in reality I think that saw will consume a
little more power (maybe 5-10% more for a typical installation) in the
240V
configuration than it would in the 120V configuration.

How do you figure that? It is not logical.



Here's my rational for proposing that the 240V configuration gives the motor
slightly more power.

First consider the 120V configuration:
Lets suppose that the motor draws 16A from the 120V source. The motor
windings will be connected in parallel, so for two windings in parallel,
each winding will have 8A flowing through it and each will have close to
120V accross it (it is a little less than 120V because of the voltage drop
in the wiring). If the wiring has a resistance of 0.3 ohms (not
unreasonable for a run of 12 AWG wire), the voltage drop due to the wiring
would be 0.3 ohm * 16 A = 4.8 V. So, the each motor winding would see 120V -
4.8 V = 115.2V. Note, at this point we can calculate the resistance due to
the motor windings: 115.2 V / 8 A = 14.4 ohm. To double check the
calculations: For a single winding of 14.4 ohm, the resistance of two
winding in parallel will be 1/2 that of one (i.e. 7.2 ohm). The 120 V
circuit "sees" a total resistance of 0.3 ohm + 7.2 ohm = 7.5 ohm; 120 V /
7.5 ohm = 16A (as we originally supposed). In terms of power, the wiring
will use 16A * 4.8V = 76.8 Watts. The motor will use 16A * 115.2 V = 1843.2
Watts. Total power is 76.8 W + 1843.2 W = 1920 W, in agrrement with 16 A *
120 V = 1920 W for the total circuit.

Now consider the same motor configured for 240V operation:
Now the two winding are in series with each other and the wiring, so
doubling the voltage does not exactly halve the current (since the total
circuit resistance changes). The total circuit resistance will be 0.3 ohm
(assuming same run of wiring is used) + 28.8 ohm (two 14.4 ohm windings in
series) = 29.1 ohm. The current will be 240 V / 29.1 ohm = 8.25 A. Now we
can calculate the voltage drops: for the wiring we have 0.3 ohm * 8.25 A =
2.48V (close to 1/2 the 4.8 V drop for the 120 V configuration); 14.4 ohm *
8.25 A = 118.8 V for EACH winding. The winding together in series give a
drop of 2 * 118.8 V = 237.6 V; total circuit voltage drop of 2.475V + 237.6V
= 240.075V (slightly off from 240V due to rounding of current to 8.25A).
So, in the 240 V configuration, each winding sees a little more voltage AND
a little more current, so therefore the motor gets a little more power. We
can also calculate the power involved: the wiring uses 8.25A * 2.475 V =
20.4 W and the motor will use 8.25 A * 237.6 V = 1960.2 W, for a combine
total of 20.4 W + 1960.2 W = 1980.6 W.


To summarize:
120 V configuration: 1920 Watts total, 76.8 Watts for the wiring and 1843.2
Watts for the motor
240 V configuration: 1980.6 Watts total, 20.4 W for the wiring and 1960.2
Watts for the motor

Compared to the 120V configuration, when wired for 240 V the motor gets
1960.2 W - 1843.2 W = 117 W more power, or a (117/1843.2)*100 = 6.3%
increase.

Granted, I've simplified the analysis by only considering the resistive
loads (i.e. I've implicitly assumed that the reactance due to the motor's
inductance has been balances out by the reactance of the capacitor). Also,
I realize that motors are more complicated than I have described them; for
example the amount of current they draw depends on the level of mechnical
resistance they are working against (e.g. freely spinning saw blade versus
cutting 2 inch thick maple). However, I hope I've shed some light on the an
aspect of the 120V versus 240V debate in order to show that it is not as
simple as "double the voltage, halve the current" response that is given so
often.

Disclaimer: I am not an electrician, but I do have a background with some
things electrical.

BadgerDog