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 sa=
ws. I can wire it either for 220v or 110v. The only difference I see is t=
hat 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 obvi=
ously run the heavier wire with a 30 amp breaker for future loads.
Question: Does the motor care? Is there a preference?
Thanks for all the in put.
Ivan Vegvary
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
"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)]
Run 3, (L1,L2,N), #10AWG with 2P-30a c'kbr and make life
easy on yourself.
Lew
Ivan asks:
>> Question: Does the motor care? Is there a preference?
>> --------------------------------------------
I answered:
>> 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)]
---------------------------------------------------------------
"Dan Coby" wrote:
>
> 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.
--------------------------------------------------------
I write:
YES, it definitely makes a difference.
It took you several sentences to address the heart of the matter.
Total power consumed equals "Work" of the saw plus line losses
of the distribution system to deliver the power to the "Work".
The WORK is internal to the motor and as you suggest either
120V or v240V gets the job done; however, the distribution losses
equal I^2*R and are in addition to the work performed.
Reducing the line current by 50% reduces the line losses by 75%.
There is a reason utilities distribute at high voltages, it reduces
the distribution line losses.
Lew
.
On Sunday, September 23, 2012 7:49:52 PM UTC-7, 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 22=
0v.=20
Higher voltage is preferred, if it's available. Partly because it works wi=
th smaller wires,
but also because it makes the starter switches/power switches a little more=
resistant
to dust and dirt. 220V will burn through a bigger bit of sawdust than 120V=
will.
Switches and wires both take smaller run-time currents, and (all else equal=
) stay cooler
with the higher voltage.
On Sun, 23 Sep 2012 19:49:52 -0700 (PDT), Ivan Vegvary
<[email protected]> 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?
240v is the way to go. Quicker startup, better torque. If you move to
a 5hp beastie in the future, you'll thank yourself for using the
larger gauge wiring now, but I doubt a 3hp motor would feel it.
Using the 10ga/30a is a no brainer.
--
Never trouble another for what you can do for yourself.
-- Thomas Jefferson
On Sun, 23 Sep 2012 20:55:23 -0700, Dan Coby <[email protected]>
wrote:
>
>No.
>
Yes, the 240 will start faster.
in 1538967 20120924 110406 Ed Pawlowski <[email protected]> wrote:
>On Sun, 23 Sep 2012 20:55:23 -0700, Dan Coby <[email protected]>
>wrote:
>
>
>
>>
>>No.
>>
>
>Yes, the 240 will start faster.
Context??
On 9/25/2012 1:21 AM, Bob Martin wrote:
> in 1538967 20120924 110406 Ed Pawlowski <[email protected]> wrote:
>> On Sun, 23 Sep 2012 20:55:23 -0700, Dan Coby <[email protected]>
>> wrote:
>>
>>
>>
>>>
>>> No.
>>>
>>
>> Yes, the 240 will start faster.
>
> Context??
>
?