- posted on September 24, 2012, 2:49 am

Question: Does the motor care? Is there a preference?

Thanks for all the in put.

Ivan Vegvary

- posted on September 24, 2012, 3:15 am

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

- posted on September 24, 2012, 3:55 am

On 9/23/2012 8:15 PM, Lew Hodgett 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.

Dan

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

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

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

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

- posted on September 24, 2012, 5:51 am

Ivan asks:

I answered:

---------------------------------------------------------------

"Dan Coby" wrote:

-------------------------------------------------------- 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 .

I answered:

---------------------------------------------------------------

"Dan Coby" wrote:

-------------------------------------------------------- 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 .

- posted on September 24, 2012, 10:04 am

wrote:

Yes, the 240 will start faster.

Yes, the 240 will start faster.

- posted on September 25, 2012, 6:21 am

Context??

- posted on September 25, 2012, 1:21 pm

On 9/25/2012 1:21 AM, Bob Martin wrote:

?

?

- posted on September 24, 2012, 4:40 am

On Sun, 23 Sep 2012 19:49:52 -0700 (PDT), Ivan Vegvary

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.

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

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.

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

- posted on September 26, 2012, 10:25 pm

On Sunday, September 23, 2012 7:49:52 PM UTC-7, Ivan Vegvary wrote:

Higher voltage is preferred, if it's available. Partly because it works with 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.

Higher voltage is preferred, if it's available. Partly because it works with 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.

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