On 9/23/2012 8:15 PM, Lew Hodgett wrote:
"Ivan Vegvary" wrote:

Still working on cobbling together the best of four Craftsmen table
saws. Biggest motor I have is an old (heavy) 3 hp outboard motor on
one of the saws. I can wire it either for 220v or 110v. The only
difference I see is that I could possibly put a smaller gauge wire
into my conduit if I run 220v. I have both wire (10 & 12 awg) and 20
or 30 amp breakers. I should obviously run the heavier wire with a 30
amp breaker for future loads.

Question: Does the motor care? Is there a preference?
--------------------------------------------
Yes.

240V allows the motor at 1/2 the current req'd for 120V which
translates into 1/2 the heat loss. [Watts = I^2(R)]

No.

A motor that can be wired for with 120 or 240 has two main
windings. (There is also a starting winding but lets ignore that
for this discussion.)

For 240 volt operation the two windings are in series. For 120
volt operation the two windings are in parallel. In both cases
the same amount of current is going through each separate winding
and the losses are the same.

An example: Assume each winding has 1 ohm of resistance and at
full load the winding will draw 10 amps of current.

With 240 volts, the windings are in series and motor will draw
10 amps. The 10 amp current flows through one winding and then
the same current flows through the other winding. The power loss
in one winding will be I^2*R = 10*10*1 = 100 watts. The total power
loss for the motor will be 200 watts.

With 120 volts, the windings are in parallel and the motor will
draw 20 amps (10 amps for each winding). The power loss in one
winding will once again be I^2*R = 10*10*1 = 100 watts. The total
power loss for the motor will be 200 watts which is the same as
the 240 volt example above.

Another way to look at the 120 volt case: The windings are in
parallel and a parallel combination of two 1 ohms resistors is
equivalent to a 1/2 ohm resistor. The total power lost in the
winding resistance is I^2*R = 20*20*(1/2) = 200 watts.




Things are different if one considers the losses in the wiring
going to the motor. The 120 volt motor will draw twice the current.
Unless the supply wiring is larger, there will be more voltage drop
in the supply wiring. This drop in the supply voltage to the motor
will mean the motor will need to draw even more current for a given
output power. This will increase the winding resistance losses.
(The output power = supply voltage * current - internal motor losses.)
(Yes, I am ignoring the power factor.) The external wiring losses
are more likely to be a problem with the higher currents of the 120
volt motor.


Dan