wire sizing issues

Shawn-H

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.

BB.

Shawn, I will give it a shot--But this is something you may need to talk with a well person/licensed electrician (familiar with well installations) about.

First, 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: 6.11
Voltage drop percentage: 2.55%
Voltage at the end: 233.89

So, 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: 4.49
Voltage drop percentage: 1.87%
Voltage at the end: 235.51

and 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: 2.07
Voltage drop percentage: 0.86%
Voltage at the end: 237.93

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

-Bill

Shawn-H

BB. wrote: »

Shawn, I will give it a shot--But this is something you may need to talk with a well person/licensed electrician (familiar with well installations) about.

First, 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: 6.11
Voltage drop percentage: 2.55%
Voltage at the end: 233.89

So, 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: 4.49
Voltage drop percentage: 1.87%
Voltage at the end: 235.51

and 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: 2.07
Voltage drop percentage: 0.86%
Voltage at the end: 237.93

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

-Bill

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.

scrubjaysnest

Not a bad stab BB, the only way to get it closer would be pull the pump and hope the pf is stamped on the name plate data.

mike95490

I suspect your 2Kw consumption estimate. But maybe with a larger motor, you have less overall losses. My 240VAC 1/2 hp pump logs 1Kw consumption, which includes the poor power factor of the average pump motor. I have about 300' of 6ga from the inverter to the controller, and then 100' #10 submarine cable from there to the pump.
It may be worthwhile to have an electrictian use a peak reading meter to get your accurate starting amps. because your transformer has 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.

inetdog

mike95490 wrote: »

It may be worthwhile to have an electrictian use a peak reading meter to get your accurate starting amps. because your transformer has 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.

Mike, I think that you are under a misapprehension on this point. The core will saturate and cause excessive current if you try to increase the primary voltage beyond the design level or if you reduce the frequency below the design frequency (both of those changes increase the integrated peak flux level during a half cycle of the input waveform.) The design is such that you can run the voltage up by as much as 10% above the nominal level (because of utility voltage standards among other things) but more than that will get you in trouble. This is also the reason that voltage adjustment taps on a transformer should always be on the primary side 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.