Re: marina panel breakdown
Let me start by saying that, if you are talking about 69 separate slips, then a pair of 200 amp panels will not get the job done. Let me explain.
First, what version of the code applies? As of the 2002 version, the shore power receptacles have to be a minimum of 30 amps. The 1999 version allowed 20 amp shore power receptacles.
Secondly, do you mean to say that the 48 (30 amp) receptacles serve different slips than the 21 (20 amp) receptacles? What I am asking is whether a single boat moored at this marina has a choice of whether to plug into the 30 or the 20? If you have both receptacles at a single slip, then you need only count the larger of the two. Specific example: Suppose your marina had exactly 48 slips. Suppose that you want to install a 30 amp at every slip, and install a 20 amp at 21 of the 48 slips. In this specific example, for the purposes of load calculations, you can ignore the existence of the 20 amp receptacles. The total load would be 48 times 30 times 50%, or 86,400 VA (360 amps).
Finally, let?s suppose that you have a total of 69 slips, with 48 served at 30 amps and 21 served at 20 amps, and that the 1999 code applies. Suppose further that you want to use two panels of the smallest size that satisfies code. Then you have to use the demand factor (1999 Table 555-6, 2002 Table 555.12) for the number of receptacles connected to each panel. Specific example: Suppose you split the load in half. You put 24 (30 amp) and 11 (20 amp) on one panel, and you put 24 (30 amp) and 10 (20 amp) on the other panel. The load on the first panel is calculated as follows:
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- <font size="2" face="Verdana, Helvetica, sans-serif">24 times 30 times 120 = 172,800
11 times 20 times 120 = 26,400
Subtotal = 112,800
Demand factor for 35 receptacles = 60%
112,800 times 60% = 67,680
67,680 divided by 240 = 282 amps.
Similarly, the load on the other panel is 276 amps.</font>
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