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Ok here is a quick question. My well pump is 1/2 hp. It's 115 volt and says Max amps 12.4. If the max amps is 12.4 would this include the start up draw? If it does then it would only take 1426 watts to run it. Is this right or am I missing something?
Bill
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Your Right:
On some motors startup can double the run. When motor has to start under load it can realy draw the power. This will last till the motor reaches its run speed.
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I was just looking at the pump again and I missed something the first time. It says Max amps 12.4 / 6.2 . Does this make any difference?
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The 1/2 HP motor under full load may go as high as 12.4 Amps, unless the specifications declare that is the maximum starting current.
Normally, a motor, to start, needs from 5 to 7 times the running current for several seconds that could be as high as 5 to 7 seconds.
The power factor at starting is low, around 0.4, so at the starting current is just a bit lower than the locked rotor current, which is defined mostly by the winding inductance of the motor.
One can use capacitance to correct the power factor during start up, also one can correct the running power factor with capacitance.
Sometimes, the efficiency of the motor as generator can be as high as 80 % if properly done, though the voltage should go to the low end of the desired output for highest efficiency.
For the starting Power Factor correction, an automatic arrangement can be made using a current transformer and a small circuit and a Solid State Relay, which is energized at start then dropping once the current drops at around running current.
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Soapman
If the grid powers this pump i would let it continue to do so and do other things.
Like maybe feed an element in the water heater with the turbine.
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First let me say that I'm not extremely knowledgable on these things, but I seem to remember my brother (a licensed journeyman electrician) telling me that the "rule of thumb is 1.5x the rated current for startup.
This much I do know: if there are 2 amp ratings, there will be 2 volts ratings as well.
Look closely at the data plate, it may say something like this:
VOLTS 240/120
AMPS 6.2/12.4
The 6.2 amp rating is for 240v, the 12.4 amp rating is for 120v.
Personally, I would figure 1.5 x 6.2a = 9.3 amps starting current for 240v OR
1.5 x 12.4a = 18.6 amps for 120v, and size my conductors according to the startup amps.
These numbers are just an example, and may or may not correspond to your particular motor. The only way to know for sure how much your motor draws is to slap an amp-clamp on it while it's running. Starting current may indeed be 2x running current, but since it usually takes less than half a second to get an electric motor up to speed, 1.5x the nominal current is usually sufficient for sizing your components, i.e. wire, ckt brkrs, etc. Hope this helps!
BrianH
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The grid does power this pump but I was wondering about when the power goes out. It would be nice to get water for the washroom. Even if I could get it to run for a few minutes every once and a while it would help. How much power would it take to start and run this pump? We have 2 kids of our own and my wife runs a daycare during the week. No water for toilets can be a very messy thing. I'm trying to avoid storing a large amout of water for these times. Last summer it seemed that every little storm knocked the power out for 1/2 hr or more and we're coming up to that time of year again. Any ideas?
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Just wanted to add that the actual required amps will also be dependent on how deep the water is. If the pump is rated to lift water 200 Feet, then that may be when you will see the max amps draw. The best method to figure the draw may be to read it with a meter. However, if you read it during the rainy season, it may still be higher during drought, when the pump lifts the water higher.
Rod
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i would be very carefull trying to run this off a inverter. they dont generally like inductive loads. i am sure someone will corect me on the details but i think the inverter sees the motor as a dead short on start up. and being a pump it is going to be a full load start up. you can get round it in a couple of ways. what i did with a tv that the inverter didnt like was to put a light bulb in series with it to take up the surge. not sure if this would work with a motor. the other option is to get hold of a rotary inverter. they are not very effiecient but for a low duty cycle like this they should be fine. just wire it up to the pump with its own battery and a small charger if you have grid power and that will get you out of trouble. put a relay in circuit on the grid side of the float switch so it will switch on when the grid goes down and there you are. dont know where you find them but they are around. i have one but the postage would be horredious from the UK.
bob golding
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I'd read this -- http://www.otherpower.com/otherpower_waterpumping.html -- if you haven't already...
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Hi
Build yourself a cistern high up. On your roof or in the attic if you're in an area with cold winters. Size it for the amount of water you expect to use in a week. Let the pump feed the cistern with a float switch. If your power goes out, you will have water for a week longer if you suspect the power will stay out and start conserving.
That's the way everyone (who can afford the cistern) does it in third world countries where both water- and electric utilities can be unreliable.
Regards,
Owen Morgan
Yacht Magic
Admirality Bay, Bequia