First Point:
You can easily get lost if you do not convert everything to VA, do the additions, and convert back to amps at the end. Here you got lost in your second calculation. Did you note that the second answer, 792, was exactly twice the first answer, 396? A coincidence? Not at all. In your first calculation you went from VA to amps by dividing by 240. In your second calculation you left out the steps of multiplying by 120 then dividing by 240. That is how your second calculation came out twice as high as your first.
Second Point:
There is no ?derating? gong on here. The process of sizing cables takes into consideration two things: (1) How much ampacity do you need? (2) How much ampacity does a given conductor have? The concept of ?derating? is related to the second question, not the first. You are trying to figure out how much ampacity the feeder has to have, in order to supply the loads.
Table 555.12 gives you the ability to size a feeder smaller than it would have to be, if it had to carry 100% of the connected load. It recognizes that not everything will be running at 100% of its capacity at the same time. It gives you ?demand factors? to reduce the ?total connected load? to a ?calculated load,? from which you size your feeders.
Third Point:
When you add the 10 new receptacles, the total becomes 76. That puts you one spot further down Table 555.12. So I would calculate the load as follows:
120 volts times 30 amps times 66 receptacles equals 237,600 VA.
240 volts times 50 amps times 10 receptacles equals 129,000 VA.
237,600 plus 120,000 equals 357,600 VA.
357,600 times a demand factor of 30% equals 107,280 VA.
107,280 divided by 240 equals 447 amps.
Your 600 amp panel is adequate for this load.
You didn?t tell us how many parallel 350?s were used. But presuming you meant two sets of 350 MCM copper, that has an ampacity of 620 amps, and is adequate for this calculated load.
Final Point:
Voltage drop is not an easy calculation for a marina. If you take the worst possible circumstance, that of assuming all load is at the very end of the run, and if you get an acceptable value of VD, then you can be confident that there are no VD problems in the actual installation. That is not often the results, and it is not your results. I got a result of 8% VD, using that assumption.
So to get a realistic value of VD, you need to calculate VD from the MDP to the panel, assuming (correctly) that all the load goes through that feeder. Then you need to calculate the VD in steps. As you get further down the dock, the load gets less and less, so the current used to calculate VD gets less and less. However, the number of slips being powered from further down the dock is no longer 76, so you can?t use a demand factor of 30% any more. As you continue the VD calculation further down the dock, the available demand factor (for the slips that remain to be calculated) gets smaller, and the demand factor gets bigger.
The last time I did such a calculation, I used a spreadsheet. It helped with the math, but it was no easy task to set up. Good luck on this one.
