A question about motors and capacitors

Eric R Snow wrote:


Only with a DC motor, not with an AC motor.
To start an AC well pump motor more easily, a flywheel and motor could run continuously from the generator right over at the well head. When the well pump started, the flywheel would drive the motor as a generator to provide a small synchronous starting surge. I am not exactly sure how you wire one of these or if accidents can happen this way. It's like a phase converter, sort of. More like a momentary kinetic UPS.
Doug
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I think all posters have good valid points. I just want to add that a power factor correcting capacitor will definitely help start an air compressor from a generator. I think the starting surge is highly reactive (inductive) and the capacitor can greatly reduce its effect on the generator. I think it would also reduce the line loss. I would be tempted to try a capacitor current of 10 amps or so during starting. Don Young
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wrote:

What value of capacitance? And is the cap just connected across the incoming power? My well pump uses a starting cap that's located in a box at the well head. It's called a "three wire pump" in the local vernacular. Probably the same name everywhere else too. Since I haven't looked at my neighbor's setup yet I don't know what he is using. And if he has a three wire pump which wires get the cap? I don't remember the HP of my pump but the instructions said that if 10 gauge wire is used then the control box, which contains the cap and some other stuff I don't remember, can be up to 200 feet away from the pump. The pump is a 240 volt model. ERS
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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:

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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

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Eric R Snow wrote:

You state: 200' well + 150` to garage, (single phase - my assumtion) 350' x 2 = 700' total run. Motor load:13.4A.@120V. or 6.7A. @240V. NFPA-70 Gives: "#10 AWG. Cu. (NonMetallic conduit or Cable)- AC. Z:,(Impedance),: 1.1 OHMS/1000' " I get: 10.3V. drop @ 120V.- (8.6%), 5.2V. drop @ 240V. - (2.2%) Its probably wired at 120V. and the starting current, (2-2.5 x full load), is doubling the drop, keeping the motor from starting.
The 'expensive & correct` solution, #6 wire for the whole run.
The 'quick & dirty` solution would be a bucking autotransformer to boost the voltage by about 10%. Use a 120 to 12V. transformer, (you only need about 2.5 KVA.rating), ( probably about $!00.00 retail), but have an electrician wire it up for you. - It's simple, but a little tricky to explain, and getting it wrong might be expensive or messy. MadDog
"Please don't hesitate to let me know if there is anything else with which I can be of absolutely no help at all."
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On 6 Jul 2006 03:41:58 -0700, snipped-for-privacy@yahoo.com wrote:

I'm betting his pump is 240 volts. Everybody around here uses 240 volt pumps for installations where the pump resides at the bottom. What I don't understand is why 350' turns into 700'. I understand that the voltage has to travel 700' to return back to the source, but the pump is 350' along that run so wouldn't the voltage measured at the pump only be what counts? And wouldn't it only be subject to the resistance of 350' of wire? ERS
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There is no "return" for AC (and the total length of wire is needed for DC as well).
Think of it as current doing "work". Some of that work is done by the pump (useful work) and some is wasted "work" such as heating the wire. It does not matter if the wire is heated on one side of the pump or another, it is still wasted and not given to the pump/
i
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On Thu, 06 Jul 2006 14:29:50 GMT, Ignoramus2158

Iggy, I understand that part about the wasted power. What I don't understand is why the voltage drop calculation should be based on 700' instead of 350'. Unless you are saying that MadDog is wrong when stating that 700' is to be used when figuring the voltage drop in this case. ERS
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The entire length of wire should be used to calculate losses.
Losses = Current * resistance
Resistance = Resistance per foot * number of total feet of conductor
i
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A picture is worth ...
+ --------- (Motor) --------- -
Here, it's a DC motor. You can move the motor to the left of the cable or to the right. In your setup, it's in the middle and the + and - are at the same point (by chance). Without doubt, if the motor would be to the left or right you would use the full lenghth. So why not if it is in the middl?
HTH, Nick
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On Thu, 6 Jul 2006 19:22:21 +0200, snipped-for-privacy@gmx.de (Nick Mller) wrote:

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
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"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.
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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
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On Thu, 6 Jul 2006 22:26:42 +0200, snipped-for-privacy@gmx.de (Nick Mller) wrote:

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
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Eric R Snow wrote:

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.
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On 6 Jul 2006 15:33:20 -0700, snipped-for-privacy@yahoo.com wrote:

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
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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.
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