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
A cable is several wires that are factory assembled. A "circular inch" is the area of a circle 1" diameter. A "circular mil" is the area of a circle 1/1000" (1 mil) diameter. (so there are 1 million circular mils in a circular inch). This unit is used a lot in sizing larger wires, but I find it rather useful in conduit fill calcs.
The interior of 1/2" PVC Sched 80 conduit is 0.546" or 546 x 546 = 298,116 circular mils. (or 0.298 circular inch).
Hold on... is this leeegal?
Yes, a branch circuit can share a conduit with a Grounding Electrode Conductor. In that thread you see a great deal of arguing about how bonding works, but that's because they are using metal conduit. You're in PVC so you sidestep the discussion. Nobody challenges the idea of a branch circuit in a conduit with a GEC.
The grounding electrode and branch circuit need to go their separate ways. You can't do that in a conduit body (at least not this particular type). So this conduit body must be changed out to a junction box or more flexible conduit body. Remember the branch circuit needs a conduit "stub-up" down to 24" burial depth. That is probably incompatible with the bare GEC wire routing.
You are not allowed to cut and splice GEC runs, so you will need to pull it out of the conduit for this modification and then put it back in. You are also not allowed to assemble conduit around cable, by the way.
UF cable doesn't work...
With oval shaped cables, the wide dimension is used for conduit fill. Because it twists.
With 2 things in the pipe (cable and ground wire), we are allowed to use 30% of that or 89,435 circular mils.
The #4 bare ground wire is .232" diameter at its widest. That is 53,824 circular mils.
Leaving 35,611 circular mils to work with. Square root of that -- the largest cable that will fit is 0.188" across in the wide dimension. No cable exists anywhere near that small. Not gonna happen.
But let's crunch the numbers anyway.
A 12/3 UF cable is .626" diameter at its widest, so 391,876 cmil. So 445,700 cmil, divide by .3 and we need a conduit with 1.485 circular inches, or 1.218" ID on the conduit. So you'd need 1-1/4" conduit. Wow!
A 14/2 UF cable is .423" diameter at its widest, so similar math: 1" conduit.
...try it with THWN wire
Okay, cable in conduit is always a stupid idea, and the above is a picture postcard as to the reason why. Let's try it with THWN individual wires, at least through the pipe. Since we have 3 or more things, it'd be 40% or 119,246 cmil. The #4 bare ground wire is still 53,824 cmil, leaving us 65,422 cmil to play with.
A #12 THWN copper wire is 0.127" or 16,129 cmil.
16,129 x 4 wires (hot hot neutral ground) = 64,516 cmil. And we skate under by a nose! If it were closer we could use a solid ground wire and strip it bare.
OK, so we've got our 12/3 multi-wire branch circuit through. Now we have to bury it.
Routing to the shed
Since you said "direct burial", you must trench down to 24" of cover (so 25" depth). Really. As discussed above, you will need a junction box or more versatile conduit body there. Loose THWN wires can't be buried, so after you're through the hole, you will need to splice the THWN wires to a 12/3 UF cable that can be direct buried... and that means, it will need to be a junction box, not a conduit body.
You cannot use the 12" exception for GFCI protected circuits, because that only works on plain 120V circuits, and your planned/future multi-wire branch circuit does not qualify.
If you run conduit to the shed, the news gets brighter. You need conduit for the stub-ups at both ends anyway, so if you simply stay in conduit in the trench, then you can just continue the THWN wires in the conduit. No splicing and you only need 18" burial depth.
Rigid conduit only requires 6" burial depth (12" under vehicle pathways), the stuff is hella expensive at over $2/foot Of course, you're only going 10 feet. It will take a friendly hardware store who threads pipe and a couple of runs to see them once you have lengths worked out.