A question about motors and capacitors

So in other words the voltage seen by the motor will be the same no matter where on the wire it is located. And if I were to measure the voltage across the motor leads this voltage would be the same no matter where the motor is located. Thanks Nick and Iggy ERS

"Eric R Snow" wrote: (clip) And if I were to measure the voltage across the motor leads this voltage would be the same no matter where the motor is located. (clip) ^^^^^^^^^^^^^^^^^ I think the explanations given so far have been correct, but another way of expressing it can never hurt. There is a voltage drop in the wire from the source to the motor. There is an identical voltage drop in the return lead, from the motor back to the generator. Let's say that with 120 volts across the generator terminals, the drop is 10v in each leg. This means that the voltage the motor sees will be 100v. It's not running between "hot" and generator "ground." It's running between whatever appears on the motor terminals.

Yes. But with a normal power cord, you don't have that much choice. Both ends will have the same length. :-) I showed you a "mental experiment" that will (did it?) help to understand.

Yes.

Nick

Yes Nick. your illustration did indeed help. I just kept thinking of the cord as being one ended, which, of course, is not the case. Thanks, Eric

Eric, You may be right. I didn't think the drop at full load for 240V. would stall the pump, but on further consideration the start current just might cause enough drop to do so. the booster transformer will still work at 240V single phase.

The autotransformer connection looks like this:

Neutral (120V.) Line (120V.) - For 120V. Feed

Line (240V.) Line (240V.) - For 240V. feed | | | | o---------Primary coil-------o---Secondary Coil----------o | | | | Motor Motor

Since the primary only carries the 'difference` load, it need not be sized for the motor load. A .25 KVA. 240/12V. transformer would have a secondary sized for 21 A. and would be sufficient, wired for a 5% boost.

A lot has been said here about inductive resistance, but unless the wire is coiled a lot it shouldn't apply to this case. I assume your runs are relatively straight. I don't think the transformer would 'choke` the start current but it might be wise to go up a size. The electrician can advise you.

Again, I've tried to explain it, but the tenor of your questions lead me to advise you to have a pro check the installation and wire it up for you. these things come with multiple leads for different purposes and there could be some

expensive confusion.

Best of luck, MadDog

"When in the course of human events it becomes neccessary..." - to actually DO something, it's a good time to be elsewhere.

Perhaps equally important is the proximity effect. An extension cord coiled up will have higher resistance because of the proximity effect.

Dan

Thanks for the replies. I was just confusing myself. As to the actual wiring, if I help him with any wiring it will all be by the book. I wired my shop with an electrician's help and guidance. The only fault the inspector found was where I assumed something and didn't check code. Even though the bathroom wiring was in conduit running down the middle of the studs nail plates were still required. I had assumed they would not be because of the conduit. ERS

The power factor correction capacitor should be connected as close to the motor as possible. You can use a simple graphical method to determine the amount of capacitance. I have posted how a number of times, but I don't think anyone other than myself has ever tried it. So assuming that you won't try using the graphic method.......... The way most people do it is to use a meter that can measure the AC current. First measure the amount of current the pump draws with no PF cap. Then add some capacitance across the line and measure the current going to it and the motor. See how much reduction there is in the current. If you have some more capacitors, add another one and measure the current again. For the same amount of capacitance, the current reduction will be less for the second capacitor. Don't try to get the absolute minimum, just keep adding capacitors until the amount of current drop does not seem worthwhile. You really should NOT try to get the absolute minimum. 1. It is not the economic solution. 2. When you have the absolute minimum, the circuit is a resonant circuit tuned to the line frequency. Probably not a problem, but not really a good thing.

Dan Eric R Snow wrote:

Although you will not have as much inductance in closely coiled parallel conductors as in conductors that are not parallel, you will have some. This will drop the voltage to the motor, and the motor will therefore draw more current. In addition to that, you will get proximity effect ( somewhat similar to skin effect ) which will increase the resistance ( not inductance ) of the wire. This will also drop the voltage to the motor and cause it to draw more current.

I have not got a good feel for how much these two things will effect the current. My guess in not too much, but still a measureable amount. If it is a nice day tomorrow, I might see what the effect is in coiling a 100 entension cord. One experiment is worth more than as lot of conjecture.

Jerry Martes wrote:

But inductive voltage drop in a line to a motor means the motor is operating at a lower voltage and therefore draws more current, which produces heat.

Dan Leo Lichtman wrote:

Greetings Dan, I remember you showing me how to graphically find the proper cap for power factor correction. In fact, I still have the drawing you made and gave to me at Grant's house. If my neighbor wants me to try to correct the power factor I'll be using your method. Thanks, Eric R Snow

wrote: But inductive voltage drop in a line to a motor means the motor is operating at a lower voltage and therefore draws more current, which produces heat. ^^^^^^^^^^^^^^^^^^ Running a motor on lower voltage will cause it to lose torque, so it will slow down. This will reduce the air cooling, and cause a temperature rise. But I don't see how lowering the voltage will increase the current, unless the motor slows down so much that the starting windings kick in--I don't think we're talking about that.

It doesn't "go", that's the trick. Current flowing through a large inductance will always be at 90 degrees to the voltage waveform. It is mathematically impossible for this current to deliver any power (or heat). It is called imaginary power for that reason.

No, the theoretical basis is quite "real".

Jon

The capacitor in there now is obviously not supplying enough current to create a magnetic field strong enough to begin turning the motor. You could use a bigger capacitor.

If the motor is 1/8 horse power, try a capacitor used in a 1/6 horse power motor.

The weather today was not bad and I was curious so did a little experiment. I measured the voltage at a 1.5 hp motor with the power going through a coiled 100 foot extension cord and again with the same extension cord uncoiled. Without the motor connected the power measured 120.6 volts. In both cases the voltage with the motor connected was about 111.5. Apparently the motor fluctuates a bit in how much current it draws, as the voltage varied by a few tenths in both cases when the motor was running.

Dan

snipped-for-privacy@krl.org wrote:

Greetings Leo, An induction motor will try to draw more current in order to speed up in a low voltage situation. This is why brownouts are bad for refrigerators. ERS

Maybe - but then there is current in one wire going this way and the current in the other going the other way so any field is canceled. The wires are even twisted...

Martin

Martin H. Eastburn @ home at Lions' Lair with our computer lionslair at consolidated dot net NRA LOH & Endowment Member NRA Second Amendment Task Force Charter Founder IHMSA and NRA Metallic Silhouette maker & member

Greg Menke wrote: