Sample review question for test

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single phase service feeds 7 Rv park pedestals, each with a 30 amp 120volt receptacle and a 20 amp 120volt receptacle. What is minimum size grounded conductor? I get #4 copper?....
 
I did VA X 55% derate divide by 240volts..... hmmmmm

Try this:
3600 va ea. for 20 & 30A
7 x 3600 =?
7 pedestals/sites =55% demand factor per table 551.73
7 x 3600 = ? ? x 55% = ??
??/240v= __A

__A per 310.16 = # AWG

You fill in the answers for ?-??-__ & # AWG just so you do some of the work!:thumbsup:
 
I agree with 3600 volt amps x 7x .55, which comes to 13,860 va. But dividing by 240 assumes the load is split evenly across both legs, 3.5 campers per leg. That can't be right. Your best case is four on one leg and three on the other. So I'd multiply 13,860 by 4/7 and 3/7 and figure 7920 on one leg and 5940 on the other. Then 7920/120 = 66 amps. Seems to bump wire size from #6 to #4 at 75 deg C.

Ben
 
I agree with 3600 volt amps x 7x .55, which comes to 13,860 va. But dividing by 240 assumes the load is split evenly across both legs, 3.5 campers per leg. That can't be right. Your best case is four on one leg and three on the other. So I'd multiply 13,860 by 4/7 and 3/7 and figure 7920 on one leg and 5940 on the other. Then 7920/120 = 66 amps. Seems to bump wire size from #6 to #4 at 75 deg C.

Ben

The question is about the grounded conductor, which if there were four amps on one hot and three on the other, there would only be one amp on the grounded conductor. So I am not sure where you are going with this.
 
The question is about the grounded conductor, which if there were four amps on one hot and three on the other, there would only be one amp on the grounded conductor. So I am not sure where you are going with this.
Current on the neutral doesn't matter. 551.73 requires neutral conductors to have an ampacity not less than the ungrounded conductors. Calculated loads cannot be balanced, so Ben is correct.
 
Current on the neutral doesn't matter. 551.73 requires neutral conductors to have an ampacity not less than the ungrounded conductors. Calculated loads cannot be balanced, so Ben is correct.

I disagree, well the part about dividing by 120V. Since 551.71 requires an RV park to provide a minimum of 20% of the sites with a 50A 240/120V receptacle, and since this park has 7 sites you would have to have at least one, maybe two sites with a 50A 240/120V receptacle. You would therefore have to divide by 240V.

Given that requirement, it would also affect the total calculated load since you have to use 9600va for the sites with 50A receptacles. We don't know how many sites have the 50A receps.

But we (I) don't know what the test question is looking for, or if it was all the information that was given. It may well be just the part that the OP posted with no consideration for the 50A. I hate test questions that can't be fully answered per the NEC requirements.
 
I disagree, well the part about dividing by 120V. Since 551.71 requires an RV park to provide a minimum of 20% of the sites with a 50A 240/120V receptacle, and since this park has 7 sites you would have to have at least one, maybe two sites with a 50A 240/120V receptacle. You would therefore have to divide by 240V.

Given that requirement, it would also affect the total calculated load since you have to use 9600va for the sites with 50A receptacles. We don't know how many sites have the 50A receps.

But we (I) don't know what the test question is looking for, or if it was all the information that was given. It may well be just the part that the OP posted with no consideration for the 50A. I hate test questions that can't be fully answered per the NEC requirements.
I agree.

I considered the 50A receptacle requirement before posting, but decided not to include it based on the way the question was worded. Granted, OP may have paraphrased the question, but I took it as 7 sites with 20/30A recepts... didn't say whether 7 was total sites or just 20/30A sites. I took it as the latter on one feeder, 50A receptacle sites on another feeder.
 
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