# wire sizing issues

Shawn-H
Solar Expert Posts:

**107**✭✭
my well is 350 feet from my inverters. the well is another 300 feet deep the wire leading to the well pump from the cap is 8 gauge. what size wire do I need to run from my inverters to my well. the well pump is a two horse 240v 2000w run 7500w spike.

100% Off-grid in the White Mountains of Arizona. 36 Kyocera 265w mounted on four DPW 9 module pole top mounts, midnite solar combiners, breakers, & lightning arresters, 1 midnight solar classic 150, & 3 classic 150 lights, 3 x 1574 AH GB Industries forklift batteries total of 4722 AH @24v. Feeding a Outback power systems FW500 with 2 x VFX 3624, with the x240. 2 Honda EU3000is gensets with the 240 combiner and control box running LP, and 1 Honda EU2000is Gas.

System #2 is a off grid water system @ 1590w (6 Kyocera 265's) on a 6 module DPW top of pole mount. Feeding a Granfas deep well pump and pump controller at 580 feet. 2 x 2800 gallon above ground poly storage tanks, and 1 x 1200 gallon underground cistern and a Granfas 24v booster pump feeding a 90 gallon carbon fiber pressure tank.

Vag woodstove for heat.

Follow our journey at

https://www.facebook.com/ShawnpHarvey

System #2 is a off grid water system @ 1590w (6 Kyocera 265's) on a 6 module DPW top of pole mount. Feeding a Granfas deep well pump and pump controller at 580 feet. 2 x 2800 gallon above ground poly storage tanks, and 1 x 1200 gallon underground cistern and a Granfas 24v booster pump feeding a 90 gallon carbon fiber pressure tank.

Vag woodstove for heat.

Follow our journey at

https://www.facebook.com/ShawnpHarvey

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## Comments

30,633adminFirst, the needs of the motor. AC power is actually quite complex in its application to induction motors (vector math, inductors/capacitors, etc.). But the basics work out like this.

- 2 HP * ~746 Watts per HP = 1,492 Watts of power

You have said that it actually takes ~2,000 Watts (at full power) to run the pump. Or, its electrical efficiency is:- 1,492 Watts / 2,000 Watts = 0.746 ~ 75% efficiency (seems reasonable).

So, you need more power/current to run the pump due to motor losses.Next, is another set of "losses". With AC power, the Voltage Sine Wave is not always "in phase" with the Current Sine Wave... For induction motors, the current wave "lags" the voltage sine wave... The end result is that the motor "Takes more current" to run than you would expect from the "simple" power equation:

- 2,000 Watts / 240 VAC = 8.33 Amps (@ 240 VAC).

In reality, because the current lags the voltage wave form, the motor needs more current to operated at rated output power (this is sort of like pulling a car with a rope--You stand in front, all of your pull moves the car forward. Stand 45 degress to the side, and only ~71% of your "pulling force" is moving the car forward, the rest is trying to pull the car sideways).For induction motors, this can be talked about as VA (Volt Amps) -- The amount of current and voltage needed to make the motor run at full output power--And you need to size the wiring, transformers, AC inverter, etc. to this required current flow.

For a typical induction motor, roughly the Cos (phase angle) = ~0.70 (~70% of the current is used to turn the motor, the other 30% is "wasted" doing nothing other than pulling more current in the wiring).

If you motor is a 2,000 Watt input, then its "VA" rating would be (roughly):

- 2,000 Watts / 0.70 = 2,857 VA

And the current draw would be:- 2,857 VA / 240 VAC = 11.9 Amps

So--All the wiring/breakers would need to be sized to supply 11.9 amps continuously (more complicated than that, there is starting surge of 7,500 Watts -- Really probably 7,500 VA).Now, lets say you want a 3% maximum voltage drop (240 VAC * 3% = 7.2 volt drop) when running (you may need to run 5% drop--We will see what the math looks like):

- 7.2 volt drop, 11.9 amps, 650 feet
- http://www.calculator.net/voltage-drop-calculator.html

6 AWG:Voltage drop:

6.11Voltage drop percentage:

2.55%Voltage at the end:

233.89So, a 650 foot 6 AWG copper cable wire run would seem to be a good starting point.

If you cannot fit 6 AWG down to the pump, perhaps you use a smaller AWG in the well, and a larger AWG from the house to the well.

In your cause 300 feet of 8 AWG in well:

Voltage drop:

4.49Voltage drop percentage:

1.87%Voltage at the end:

235.51and that leaves 7.2 volts - 4.5 volt 8 awg drop = 2.7 volt drop for the 350 foot run:

4 AWG (350 foot from inverter to well):

Voltage drop:

2.07Voltage drop percentage:

0.86%Voltage at the end:

237.93I have to go right now--But that is where I would start.

-Bill

107✭✭Thank you bill I did talk to the well company that put in the pump back in 1992 and they told me 6AWG would do it. I would just rather a second opinion if you know what I mean. expecially since that sounds too small for me I did the same Math as you and think it should be 4 AWG or larger expecially since I would rather not change out the wire that is already in the well.

again thank you for the info and taking the time to give the reply.

System #2 is a off grid water system @ 1590w (6 Kyocera 265's) on a 6 module DPW top of pole mount. Feeding a Granfas deep well pump and pump controller at 580 feet. 2 x 2800 gallon above ground poly storage tanks, and 1 x 1200 gallon underground cistern and a Granfas 24v booster pump feeding a 90 gallon carbon fiber pressure tank.

Vag woodstove for heat.

Follow our journey at

https://www.facebook.com/ShawnpHarvey

175✭✭✭8,813✭✭✭✭✭It may be worthwhile to have an electrictian use a peak reading meter to get your accurate starting amps. because your

transformerhas a hard limit of what it can pass till the core shuts down on you. You can have all the power in the world, but an undersize transformer will not let it start.|| Midnight Classic 200 | 10, Evergreen 200w in a 160VOC array ||

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3,123✭✭✭✭primaryside unless it is overengineered.All that happens when you draw excess current from a transformer is that the resistance of the primary and secondary winding cause the voltage to drop and the temperature of the transformer core to increase. In the short term heating is not an issue, and in the event of a short circuit the short circuit current is determined by the available POCO current at the primary and the impedance of the transformer windings. The secondary current will be modified from the POCO available current by the turns ratio of the transformer.