Base Conductor Size
Start out by using Table 310.15(B)(16), and applying any required corrections, to determine what size conductors you'll need. For your situation, we'll assume we can use the 75°C column, that you want to use copper conductors, and there's no other corrections required. So in your case, if you want to install a 50 ampere panel, you'll need at least 8 AWG copper conductors. If you want a 60 ampere panel, you'll need 6 AWG copper conductors.
Voltage Drop
Once you have the base conductor size selected, you'll want to calculate the voltage drop across that size conductors for the length of the feeders. The first step here will be to use Table 8 from chapter 9 of the NEC, to determine the resistance of the conductors you've selected.
In your case, 8 AWG stranded copper wire has a resistance of 0.778 ohms per 1000 ft. 6 AWG stranded copper wire has a resistance of 0.491 ohms per 1000 ft.
Next you'll use the following formula, to calculate the voltage drop across the feeders.
V = L * 2 * R * A
Where:
- V = Voltage Drop
- L = Distance along the wire from one breaker to the next.
- R = Resistance per foot of wire.
- A = Current running through the conductor.
For a 50 ampere circuit, 130 ft. long, using 8 AWG stranded copper conductors, the calculation looks like this...
V = 130' * 2 * 0.000778 * 50 A
V = 260 * 0.000778 * 50 A
V = 0.20228 * 50 A
V = 10.114 V
10.114 V is 4.2% of 240 V. The NEC recommends having a voltage drop less than 3%. To achieve this, you're going to have to use larger conductors.
6 AWG stranded copper conductors have a resistance of 0.000491 ohms per foot, which means the voltage drop would only be 6.383 volts or 2.7%.
For a 60 ampere circuit 130' long, 6 AWG stranded copper conductors would have a voltage drop of 7.6596 volts or 3.2%. While 4 AWG stranded copper would be 4.8048 volts, or 2%.
Conductor Type
Once you know what size conductors you need, you'll have to determine what type of insulation the conductors should have. Since you're burying the conduit, you'll need a wire rated for wet locations. The popular choice in this situation, would be to use THWN wires.
Wire Size
Now that you know what size conductors, and what type of wires you'll use. Then next step is to determine the physical size of the wires, and how much space they'll take up in conduit. For this, you can use Table 5 from chapter 9 of the NEC. There you'll find that 6 AWG THWN wires have an area of 0.0507 square inches, while 4 AWG THWN wires have and area of 0.0824 square inches.
Conduit Fill
Using the size of one wire, you can figure out the area required for all four wires.
0.0507 * 4 = 0.2028 in.sq.
0.0824 * 4 = 0.3296 in.sq.
Use Table 1 from chapter 9 of the NEC, to determine the allowable conduit fill percent. Since you'll have more than 2 conductors, you can fill the conduit to 40%.
Conduit Type
If you know what type of conduit you're using, you can use Table 4 from chapter 9 of the NEC to look up the area fill values for various sizes of conduit.
Conduit Size
Since you've decided to use Schedule 80 PVC, you'll simply find that table in Table 4. Then look down the 40% fill column, until you find an area large enough for all your wires.
In your case four 6 AWG THWN conductors, will require 1" Schedule 80 PVC. While four 4 AWG THWN conductors, will require 1 1/4" Schedule 80 PVC.
Conduit Size Alt.
If you don't feel like calculating wire/conduit area, and all the wires are the same size, you could use Table C.9 from Annex C of the NEC to look up the conduit size required. There you'll find that you can fit five 6 AWG THWN wires throug 1" Schedule 80 PVC, and that you can fit six 4 AWG THWN wires though 1 1/4" Schedule 80 PVC.
tl;dr
- For 130' long 50 ampere feeder, use four 6 AWG stranded copper THWN conductors though 1" Schedule 80 PVC.
- For 130' long 60 ampere feeder, use four 4 AWG stranded copper THWN conductors through 1 1/4" Schedule 80 PVC.
NOTES:
- This answer contains some of the tables used in this answer.
- If you don't feel like doing any maths, you can surely find a calculator online to do all the work for you.
Best Answer
#14 is not adequate for a 175 foot, 50% circuit capacity load. Voltage drop is a function of resistance of the wire in series with the load. Extending a #14 circuit with lower resistance #12 is completely acceptable and does accomplish reducing the accumulated resistance that causes voltage loss on the extension.
5% is the total drop generally considered the maximum acceptable. Figure a distance to the gate of 100', add your 75' extension, you have 175', with an 7.5 amp load (a worm drive saw) you have a 3.5% loss just getting to the gate, then an additional 3% getting to your new receptacle. But if you reduce the resistance on the extension by using #12 you will get only about 1.5% loss on your extension, bringing you back to 5%. (Numbers get dramatically worse if you exceed the 50% circuit capacity that I used above.)
The 18" cover requirement is adequate for most conditions, be aware it is 18" of "cover", meaning the conduit is to be below 18" of material. Do an internet search for table 300.5 that could identify conditions that could change depth requirements.