voltage drop calc for shore power outlets at small marina

[Scott K]

I am trying to figure wire size for shore power outlets at a small marina. I understand how to size the service but having trouble with the branch circuits. Each pedestal has two 30 amp 120 volt receps. What value should I use for amps in the voltage drop calc? Is it 30? Also, I would like to run a 240v 3 wire circuit big enough to feed more than one pedestal at different intervals. Is there a calc for this? Any help is much appreciated. Thanks!

 

[augie47]

On the individual outlets, I don't know of any published data that would reference anything but full load. I've worked on several marinas and found that the actual load can often come close to the circuit size.
I would think you would be safe using the values in Table 555.12 when you look at feeders,.
There are a lot of special requirements for marinas and each Code cycle seems to bring significant changes. Be sure to check the Code cycle being enforced.

 

[Scott K]

voltage drop calc for shore power outlets at small marina

Thanks Augey, Does this sound right to you? If I have a pedestal with 2- 30 amp 120 volt receps. approx. 350' from the panel, if I run a 240v 3 wire circuit can I use 7.2 volts for allowable voltage drop in the calc or Should I use 3.6 volts since there would only be 1 recep per phase? Also would this be considered a continuous load? Thanks again for your help on this!

 

[charlie b]

7.2 volts is 3% of 240. 3.6 volts is 3% of 120. It is the same calculation either way. I would use the 30 amps per receptacle times 2 receptacles, for a 60 amp total load, without including a 1.25 factor for continuous loading.

 

[charlie b]

Also, I would like to run a 240v 3 wire circuit big enough to feed more than one pedestal at different intervals. Is there a calc for this?

I would do it one section at a time. Suppose you have 5 pedestals, each with a pair of 30 amp receptacles. Suppose they are spaced 100 feet apart, with the first one at 300 feet from the panel. I would do a voltage drop as follows:

1. Start with (5 pedestals) x (2 receptacles) x (30 amps) x (80% demand factor per table 555.12). Calculate the VD for this situation at a distance of 300 feet, and select the cable size that works.
2. Next calculate (4 pedestals) x (2 receptacles) x (30 amps) x (90% demand factor per table 555.12). Calculate the VD for this situation at a distance of 100 feet, and select the cable size that works.
3. Next calculate (3 pedestals) x (2 receptacles) x (30 amps) x (90% demand factor per table 555.12). Calculate the VD for this situation at a distance of 100 feet, and select the cable size that works.
4. Next calculate (2 pedestals) x (2 receptacles) x (30 amps) x (100% demand factor per table 555.12). Calculate the VD for this situation at a distance of 100 feet, and select the cable size that works.
5. Next calculate (1 pedestal) x (2 receptacles) x (30 amps) x (100% demand factor per table 555.12). Calculate the VD for this situation at a distance of 100 feet, and select the cable size that works.

 

[GoldDigger]

7.2 volts is 3% of 240. 3.6 volts is 3% of 120. It is the same calculation either way. I would use the 30 amps per receptacle times 2 receptacles, for a 60 amp total load, without including a 1.25 factor for continuous loading.

But if a 30A 120V load gives you a 3.6V drop (counting the drop in line and neutral), then a balanced pair of 120V loads or a single 240V load will also give you only a 3.6V drop. So running the calculator for a 3? drop at 240 will undersize the wire for the VD of 120V loads.


Tapatalk!

 

[Scott K]

I would do it one section at a time. Suppose you have 5 pedestals, each with a pair of 30 amp receptacles. Suppose they are spaced 100 feet apart, with the first one at 300 feet from the panel. I would do a voltage drop as follows:

1. Start with (5 pedestals) x (2 receptacles) x (30 amps) x (80% demand factor per table 555.12). Calculate the VD for this situation at a distance of 300 feet, and select the cable size that works.
2. Next calculate (4 pedestals) x (2 receptacles) x (30 amps) x (90% demand factor per table 555.12). Calculate the VD for this situation at a distance of 100 feet, and select the cable size that works.
3. Next calculate (3 pedestals) x (2 receptacles) x (30 amps) x (90% demand factor per table 555.12). Calculate the VD for this situation at a distance of 100 feet, and select the cable size that works.
4. Next calculate (2 pedestals) x (2 receptacles) x (30 amps) x (100% demand factor per table 555.12). Calculate the VD for this situation at a distance of 100 feet, and select the cable size that works.
5. Next calculate (1 pedestal) x (2 receptacles) x (30 amps) x (100% demand factor per table 555.12). Calculate the VD for this situation at a distance of 100 feet, and select the cable size that works.

Thank you for responding to my question but wouldn't there be a problem with overcurrent protection this way? In your example above we would have to start with a 250 amp breaker (and please correct me if I'm wrong) by the end of step 5. we end up with a #6 AWG conductor that would be fed from the 250 amp breaker if I'm looking at this right.

 

[augie47]

I agree with Charlieb in post #4. Same answer since on 240 you double your load and also there is no need to consider the load as continuous.
As far as his suggestion on voltage drop, he may teach you and I something, but I agree with you in that your conductor will need to have a minimum size based on your overcurrent device (except for the last run where you can apply tap rules)

 

[BAHTAH]

Pedestal VD Calc

Pedestal VD Calc

Don't know how this compares with Charlie's method but I use the Load-Center-Length method. Using the (2) 30 Amp receptacles at each pedestal and first pedestal at 300ft and balance of units 100ft apart. I figured the 30 Amps at 80% (24*2 per pedestal).
First I figure the Amp/ft of each pedestal from the source: Pedestal (1) =300ft * 48 amps= 14400. I do this for each pedestal and get: Ped(1)=14400, Ped(2)=19200. Ped(3)=24000, Ped(4)=28800, Ped(5)=33600. Then I divide the Total Amp/ft by the Total Amp Load to get the Load-Center-Length. LCL=120,000 Amp/ft / 240 Amps. LCL = 500 FT. Then I use the LCL in a single Voltage Drop Calculation. 240 Amps @ 500Ft using 400kcm Copper I get a 3.23% Drop.

 

[Scott K]

Don't know how this compares with Charlie's method but I use the Load-Center-Length method. Using the (2) 30 Amp receptacles at each pedestal and first pedestal at 300ft and balance of units 100ft apart. I figured the 30 Amps at 80% (24*2 per pedestal).
First I figure the Amp/ft of each pedestal from the source: Pedestal (1) =300ft * 48 amps= 14400. I do this for each pedestal and get: Ped(1)=14400, Ped(2)=19200. Ped(3)=24000, Ped(4)=28800, Ped(5)=33600. Then I divide the Total Amp/ft by the Total Amp Load to get the Load-Center-Length. LCL=120,000 Amp/ft / 240 Amps. LCL = 500 FT. Then I use the LCL in a single Voltage Drop Calculation. 240 Amps @ 500Ft using 400kcm Copper I get a 3.23% Drop.

Thank you for the input. I was wondering if it was permissible to use the 24 amps instead of 30 for the calac's. as the recep. should only be loaded to 80%. This will make a big difference in the wire size. My concern is that I am trying to create a bid for this project and obviously need to get it right. This is actually a rewire and the way it is done now the conductors are seriously undersized ( #2 awg feeding 4 pedestals which is 8- 30 amp 120v receps. with the longest run being approx. 380'.) Of course when I express this to the board (condominium) I get the age old response...."We've never had a problem with it this way". Pulling my hair out!