Wiresize of wire for a 700 foot run?

I will assume your load is AC, not DC.

If you are in the USA, then the installation likely will need to meet the requirements of the National Electrical Code (NEC).

The NEC prohibits using anything smaller than 14 AWG for branch circuits and feeder circuits. So don't even consider 16 or 18 AWG. There are exceptions that don't apply to your situation (such as inside a light fixture).

The recommended maximum voltage drop for reasonable efficiency of the wiring system is 5%. Typically 2% is allowed for feeders and 3% for branch circuits, although you can apportion them any way you like.

You didn't mention whether your load was single-phase or three phase. Either configuration affects the voltage drop.

You also didn't mention what the power factor (PF) of the circuit is. That affects voltage drop as well. A resistance heater has a PF of 1.00. A fully-loaded motor can be assumed to have a PF of 0.80.

Wire ampacity by AWG, insulation type, and temperature is listed in the NEC (starting with Table 310-16). That ampacity may have to be derated based on ambient temperature and number of conductors carried in the same conduit or raceway with your circuit conductors.

Anyway, to get you into the ballpark, let's assume a PF of 0.85. The NEC publishes Table 9 which applies to a PF of 0.85. Also, you mentioned 110V and 220V. Those are voltages that you might see labeled on your loads, but they are meant for connection to circuits that have a nominal rating of 120V or 240V. It is recognized that there will be some voltage loss between the source voltage and the load, so it is customary to calculate the losses using the nominal voltage values. I'll refer to the circuits as 120V or 240V from here forward.

If your load is 120V, 10A single phase, a 3% voltage drop corresponds to a voltage drop of:

Vdrop = 3% x 120V = 3.6V

That corresponds to a wiring impedance of:

Ztotal = Vdrop/I = 3.6V / 10A = 0.36 ohm.

Since both the neutral and the hot conductor carry current and show a voltage loss, that means that each wire can have no more impedance than:

Zconductor = 0.36 ohm / 2 = 0.18 ohm (per 700 feet)

The above value is based on a circuit length of 700 feet. Table 9 of the NEC is based on 1000 feet. So to use Table 9, we will adjust the above value as though the circuit were 1000 feet long:

Zconductor = 0.18 ohm x 1000ft/700ft = 0.257 ohm (per 1000 ft)

Table 9 indicates the following conductor sizes:

copper: 3 AWG

aluminum: 1 AWG

A 240V, 10A load carries twice the power of a 120V, 10A load. You can run the same calculations as above. The allowable voltage drop will be twice the above value, and thus the allowable impedance will also be double, or:

Zconductor = 0.514 ohm

Table 9 indicates the following minimum conductor sizes:

copper: 6 AWG

aluminum: 4 AWG

For a 240V, 10A, 3-phase load, the line-to-neutral voltage is 138.6V (=240V / sqrt(3)). The allowable voltage drop is:

Vdrop = 3% x 138.6V = 4.158V

The corresponding line-to-neutral impedance is:

Ztotal = Vdrop / I = 4.158V / 10A = 0.4158 ohm

Since the neutral carries no current (and might not even be present), it does not contribute to the voltage drop, so you don't divide the above value in half.

Zconductor = Ztotal = 0.4158ohm

Adjust to 1000 feet:

Zconductor = 0.4158 x 1000ft/700ft = 0.594 ohm

From Table 9, the minimum conductor sizes are:

copper: 6 AWG

aluminum: 4 AWG

========

None of the above sizes account for any ampacity derating that might be required.

The above process is somewhat tedious. There are online calculators that can be used, such as the following one. You can try the above numbers and find good agreement. For "select parallel runs", select "single set of conductors". You can choose different parameters if you want to allow the voltage drop to approach 5%, which will decrease your conductor size:

http://www.nooutage.com/vdrop.htm

And in case it isn't obvious, the higher the system voltage, the smaller the wire size.

In most places in the US you need a minimum of #16AWG to carry 10A over even the shortest distance. So that is the smallest we can consider.

Copper Wire Table

AWG.......ohm/1000ft

..6............0..3951

..8............0.6282

..9............0.7921

10............0.9989

12............1.588

14............2.525

16............4.016

I believe the National Electric Code (NEC) recommends a maximum voltage drop at 5% at full load or 5.5V at 110V

If you are going to a load 700 feet away you will go through 700 feet in each of 2 wires or 1400 ft

with #16AWG that will be 5.6 ohms. IR drop for 5.6 ohms at 10 amps is 56 volts.

We would need to lower our resistance by a factor of 10 to bring the voltage drop to 5.6 V

AWG 6 wire will give us 0.3951 x1.4 = 5.53V drop Which is essentially 5% of 110V

AWG 9 would give a 5% drop for 220V But that is less readily available. AWG 8 should be used.

normally 3 is subtracted from the copper gauge to get the Aluminum Gauge

so in Al, #3 AWG is needed and #2 AWG would probably be used for 110V

and #6 AWG would be needed for 220V

Romex use a 6/3 wg good for 50 amps and that is the breaker rated for a range plug. Installing wire larger than needed is cheaper than a fire and is more efficient than a higher resistance (smaller size) wire.