A question about motors and capacitors

Nope, just a rough sort of guess. The conductors are probably doing all kinds of twisting, there could be kinks, etc.. its all conjecture until somebody does some tests.

Gregm

Reply to
Greg Menke
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The overheated coiled extension cord could as easily be plain old heating. The popcorn popper sounds like a different phenomenon.

Gregm

Reply to
Greg Menke

Eric - looks like the discussion got off on a tangent - the short answer is "no", capacitance won't help because this is an AC motor - use 8 gauge wire instead.

Bill

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Reply to
William B Noble (don't reply t

Thanks Don and everyone else who answered my original question. From the answers I got it looks like using a capacitor to help with the starting current won't work. Especially with AC. Is this because it just takes much more energy than is feasible to store in capacitors? Thanks, Eric P.S. My neighbor is going to move the generator next to the well head. But we are both curious about the feasibility of using capacitors as a way to store and release energy fast, similar to a cap in a flash unit. Not as fast though (I suspect).

Reply to
Eric R Snow

You cannot use a capacitor for storing energy to supplement the generator for starting, because capacitors store DC and your motor needs AC.

You can use a capacitor to correct (improve) power factor and reduce current, thus reducing voltage drop, likely with very marginal improvement.

Bigger wires could help. You can just double 10 gauge wires, and halve your voltage drop this way.

i

Reply to
Ignoramus1869

Reactance in this case is identical to inductance, and does not cause heating. (It is possible that the large magnetic field from a coiled SINGLE conductor can cause self-heating due to the magnetic field and eddy current effects, but not so for the paired cable.)

Jon

Reply to
Jon Elson

A capacitor, on AC, only stores energy from one half-cycle of the line frequency to the next. It gets completely discharged and recharged 120 times a second, independent of load. You need to store energy for a second or so in this application. So, a capacitor across the line would be useless in this application. Putting a pair of transformers on the line, one at the generator to step up to 480 V, then back down at the well head to 240 would reduce voltage drop by a factor of 4 (voltage drop at 480 /2 is half, and the current would be halved at the higher voltage, so halve it again). That's why substations have those huge transformers, and they run 45,000 up to 750,000 volts to send power long distances.

Jon

Reply to
Jon Elson

Sounds to me like the solution is not a capacitor but rather larger wires. The capacitor is going to do nothing when the pump is running and if the voltage has dropped so low the motor will not start then you are only going to bugger up the motor by running it much under the required voltage.

Reply to
Roger Shoaf

How could resistance or inductance not cause heating, please? Where does the energy go then?

I wonder if we're not seeing a "theoretical world" vs "real world" situation here.

Reply to
Dave Hinz

Measure the voltage at the pumphead and see how much drop you have. My bet is that by the time that the voltage gets down to the pump, it will be under

85V (for a 120V motor). This is way too low and you will need to either reset the generator to produce a higher voltage or put a booster transformer on the line to get the voltage up. You may have to go and provide two voltages to the head if there is electronics there or do the boosting out there at the wellhead to get things running right. Please note that if you increase the voltage with a transformer, you will be drawing more current as you need so many VA and with the low source voltage at the wellhead, the current necessarily goes up. Also note that the voltage on a power line can be anywhere between 108V and 135V and still be good.

-- Why do penguins walk so far to get to their nesting grounds?

Reply to
Bob May

Thanks Jon, I kept thinking about the start winding on a capacitor start motor and how it is provided with power that's out of phase with the power that goes to the run winding, because of the capacitor. Of course the capacitor does get charged and discharged and charged as the AC changes. But that's OK becuase the cap is used for phase shifting and not for a one-shot charge to the start windings. Now I see it. Thanks. ERS

Reply to
Eric R Snow

The resistance will cause heating, but the inductance will not. When looked at on an instantaneous basis, with the inductance the energy gets stored and released 120 times a second (for 60 Hz), but is never consumed.

jk

Reply to
jk

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

Reply to
Doug Goncz

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

Reply to
Don Young

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."

Reply to
MadDogR75

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

Reply to
Eric R Snow

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

Reply to
Eric R Snow

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
Reply to
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
Reply to
Eric R Snow

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
Reply to
Ignoramus2158

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