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I am installing a well pump. The pump is single phase 240 V, 12.3 amps and has a soft start. Supply voltage from the Mfg: 1 x 200-240 V –10%/+6%, 50/60 Hz.

It's 200 feet (61 m) from the panel to the well, and the pump will be 300 feet (91 m) down. So 500 feet (152 m) total. Without knowing any better I installed 10 AWG ground burial wire for the first 200 feet.

Turns out the manufacturer (and science) says I should use an 8 AWG wire for 500 feet. Since a 10 AWG wire is fine for a 200 foot distance with the pump, am I OK if I switch to 8 AWG the rest of the way or do I need to dig it up and run 8 AWG the whole way? Ground temp is ~54 degrees Fahrenheit.

If I am understanding correctly my concern should be more about voltage drop than current carrying capacity as it's only 12.3amps. So will the voltage be adequate? Thanks!

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  • \$\begingroup\$ Use one of the online calculators such as southwire.com/calculator-vdrop. Work out the voltage drop on each section and add them together. \$\endgroup\$
    – Transistor
    May 25 at 6:27
  • \$\begingroup\$ Is there any information as to the maximum distance for AWG 8? \$\endgroup\$ May 25 at 6:29
  • \$\begingroup\$ Manufacturer say 520' is max distance for AWG 8, 330' for AWG 10. Doing some reverse calculations on the wire gauges and distances they are indicating a voltage loss of 4.2% is acceptable. \$\endgroup\$ May 25 at 6:46
  • \$\begingroup\$ You'd probably get a lot more insight if you asked this on the DIY stack instead of here, especially with regard to code compliance. \$\endgroup\$ May 26 at 16:54

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Seems like a 4% drop at rated operating current is acceptable, at least according to UK/IEC standards. Using the Wiki AWG table:

I get 200ft * 1mΩ/ft * 2 = 0.4Ω plus 300ft * 0.6282mΩ/ft * 2 = 0.377Ω, so total 0.777Ω round trip.

At 12.4A 240V nominal that's 9.6V drop or just about 4%.

With 520ft of AWG 8 it would be more like 3.2% drop.

But local codes may restrict what can be used, if you ask a similar question on DIY you'll likely get a more accurate answer as to codes and practices in your particular jurisdiction, but I suspect you'll be okay. That and $2 will get you a coffee, of course. You're exceeding the maximum that the manufacturer recommends by about 25% if you go that way.

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  • \$\begingroup\$ Thank you. Well installs are not inspected in my area, so code isn't an issue. More concerned about functionality and longevity of the device. \$\endgroup\$ May 25 at 7:04
  • \$\begingroup\$ The functional concern would be possible stalling and the pump perhaps burning out. Is that likely? Hard to say. \$\endgroup\$ May 25 at 7:07
  • \$\begingroup\$ Weigh the cost of a new pump to the cost of good cable - then when you have to buy a 3rd pump… \$\endgroup\$
    – Solar Mike
    May 25 at 7:14
  • \$\begingroup\$ Is "That and $2 will get you a coffee, of course." a north American idiom for something? What does it mean here? \$\endgroup\$
    – jonathanjo
    May 26 at 8:19
  • \$\begingroup\$ @jonathanjo It means that what I suspect cannot be relied upon over the recommendations of the manufacturer. Since a cup of coffee is available for $2 the opinion cannot be counted upon to add anything. A pump and the associated labo[u]r is considerably more than $2, and OP should be fully cognizant that there is some unknown amount of risk. \$\endgroup\$ May 26 at 15:18
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One way to look at it is how much resistance is the wire adding to the circuit?

  • 8 AWG = 628 µΩ per foot (2.060 mΩ per m)
  • 10 AWG = 999 µΩ per foot (3.276 mΩ per m)

400 feet of 10 AWG (200 feet for supply, and another 200 for return) has a resistance of 399.6 mΩ. 600 feet of 8 AWG (300 each way) has a resistance of 376.8 mΩ. Thus the total resistance for the mixed-wire scenario is 776.4 mΩ.

1000 feet of 8 AWG (500 each way) would have a resistance of 628 mΩ.

In theory if the pump draws 12.3 A, the mixed-gauge wire will drop 9.55 V. If it were all 8 AWG, the voltage drop would be 7.72 V. The pump won't have any problems with 230-232 V.

Power dissipation (loss) in the wires is I²R. The 10 AWG wire would lose about 60.5 watts (or 151 mW per foot). The 8 AWG would lose about 57 watts (95 mW per foot). Total power lost is 117.5 W. Contrasted with entirely 8 AWG is 95 W (95 mW per foot). Switching to all 8 AWG would save you about 22.5 W of lost power.

Given the cool temperature of the ground and the long distance it is spread across, I don't think you would need to worry about exceeding any temperature ratings of insulation.

The reality however is that the pump is an inductive load, so it's not going to be able to pull its maximum current with the long run of wire. Ignoring startup or stall current, you could think of it as being a resistive load of (240/12.3) 19.5 Ω. Combined with a resistance of, let's call it 1 Ω, means the current will be a little less at (240/20.5) 11.7 A.

I am not an electrician, so check with one for any local codes you have to meet. I would say based on my basic calculations that you'll probably want to weigh the cost of ~22 watts of wasted power against the cost of larger gauge wire. How often and how long the pump runs (its duty cycle) at maximum capacity will be key.

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  • \$\begingroup\$ The 90+ m down will need to withstand gravity. \$\endgroup\$
    – greybeard
    May 26 at 9:31
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One solution to this is use a lower-resistance cable for the extra portion, so that the resistance of the existing 200 ft plus the new 300 ft is less than the permitted resistance of 520 ft of 8 AWG.

So, don't do the extra portion in 8 AWG, use 6 AWG.

You say:

  • Manufacturer says 520 ft of 8 AWG is permitted, which is a resistance of 520 mΩ.
  • You have already installed 200 ft of 10 AWG, which is 320 mΩ.

Thus you have 200 mΩ left to be inside the spec, and 300 feet to go. At that length, 6 AWG is 192 mΩ.

The totals would therefore be:

  • 200 ft of 10 AWG + 300 ft of 6 AWG = 512 mΩ.

Which is a little under what the manufacturer says is permitted.

Resistances per Wikipedia American Wire Gauge:

  • 6 AWG is 0.64 mΩ/ft
  • 8 AWG is 1 mΩ/ft
  • 10 AWG is 1.6 mΩ/ft.
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