Voltage drop for boat slip

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newguy

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Help me out with this guys. When running a 240v feeder to a 4 circuit marine panel with (2) 30 amp 120v receptacles in it, do you calculate the voltage at 120 or 240 to figure the v/drop? If only one receptacle (120v)is used and is drawing near full current the drop could be high. If both are being used at the same time then the drop would be applied over the entire 240v and would be less. This confusion results in a big difference in wire size. Thnaks
 
I would Concentrate on line current even if the circuit operates at 120 it is going to cause X amount of current to flow on Li or L2 and Neutral.


Vd= Iline. Rline

R= k.length/CMA k is approx 12.7 at 75 degree C

Or you can refer to Table 8 of chapter 9 and use a proportion for the length of conductor you will install. Don't forget every circut length includes the feed and return. Or 2 times the distance.
 
If you have a single 120 volt load the voltage drop will be twice that of two identical loads oposite hots of a multiwire branch circuit. You would have to design for the worst case of a single 120 volt load. If that requires conductors that are too large, you could install run a 240 volt feed and install a 240 to 240/120 volt transformer at the load end.
Don
 
Well, thats what I thought. I just saw a calc by a engr. who figured the feeder drop at 240v. which I thought could be wrong but wasnt sure. I would be concerned of the drop with only one boat using the powerpost.
 
Let?s be clear about this. The amount of voltage drop, in units of volts, is the same, whether you are supplying a single receptacle (at 120 volts) or two receptacles (at 120 volts each). The question here is whether that number of volts should be computed as a percentage of 120 volts or as a percentage of 240 volts. The answer is 120. The reason that the answer is 120 is that you do not have a 240 volt receptacle available at this particular slip.

For the sake of discussion, let us assume that one volt is dropped along a one-way length from the source to the receptacle.

If only one receptacle is being used, you will have one volt dropped along the hot wire to the receptacle, and one volt dropped along the neutral wire back to the source. That is a total of two volts dropped, and that is a 1.7% VD.

If both receptacles are being used, you will have one volt dropped along the hot wire to the first receptacle, and one volt dropped along the other hot wire (from the other receptacle) back to the source. The neutral wire will not be carrying current in this case. You still have a total of two volts dropped, and that is still a 1.7% VD.

Finally, suppose that neither of the 120 volt receptacles is in use, but that a single 240 volt receptacle is in use, and carrying the same 30 amps of current. Here again, you will have one volt dropped along the hot wire to the receptacle, and one volt dropped along the other hot wire (from the same receptacle) back to the source. Here again, you have a total of two volts dropped. But now that is 2 out of 240, or only a 0.8% VD.

When you have both 120 volt and 240 volt receptacles located at the same slip, you treat the situation as if only one was present, and that one being the one that draws the higher amount of power (in units of VA).
 
Charlie,
If both receptacles are being used, you will have one volt dropped along the hot wire to the first receptacle, and one volt dropped along the other hot wire (from the other receptacle) back to the source. The neutral wire will not be carrying current in this case. You still have a total of two volts dropped, and that is still a 1.7% VD.
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But each 120 volt load will only see a 1 volt drop, not the 2 volt drop that the single 120 volt load would see.
Don
 
Okay, thanks for everybody's answer but maybe I'm just too slow to figure this out. Here is the example, one 240v, 50a. feeder 165' long to a power post with 2, 120v 30amp recepts, nothing else. To properly size the wire, should the drop be figured on the 50amp 240v feeder, or the fact each recept in the powerpost is only 30a/120v. The egr. shows #6 wire (50a/240v), I think it should be #4(30a/120). Boss says shut up and put in whats on drawing. boss is always right! Help!
 
charlie [SIZE=3 said:
[/SIZE]
If both receptacles are being used, you will have one volt dropped along the hot wire to the first receptacle, and one volt dropped along the other hot wire (from the other receptacle) back to the source. The neutral wire will not be carrying current in this case. You still have a total of two volts dropped, and that is still a 1.7% VD.

I'll go with don on this one. Two equal loads @ 120V on opposite lines of a 120/240V system = the same as one 240V load at that same amperage. What doubles is the voltage for 2 120 volt loads that for practicle purposes are looked at as if they were in series.
Click to expand...
 
Wes G said:
figure either 30A at 120V or 60A at 240V
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No, because it's 2 x 120v at 30A.

Otherwise if you figured both separately you get 120v * 30A = 3600VA for the first outlet, then 120v * 30A = 3600VA for the second, for a total of 7200VA. 240v * 60A = 14,400VA total, and that's not the same.
 
don_resqcapt19 said:
Charlie, But each 120 volt load will only see a 1 volt drop, not the 2 volt drop that the single 120 volt load would see.
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My bad. You are right. For a single receptacle, the 2 volts dropped will leave 118 volts across the receptacle. For the pair of receptacles, the 2 volts dropped will leave 238 volts across the pair, or 119 volts across each receptacle.

I did a VD calculation with an old spreadsheet. For a 165 foot run of #6 wire with a 30 amp load, the VD for a single receptacle was 4%, and the VD for two receptacles was 2%. The same run with #4 wire gives 3$ VD for the single receptacle. Since the "single receptacle" case is the more limiting, I would go with the #4.
 
Charle, I went back and reread your post. It seems like we are on the same page, it should be figured at 120v, because there are no 240 loads even though the feeder is 240v. There is only one 120v load per phase. In this case the spec'd wire size appears too small.
 
charlie b said:
For a single receptacle, the 2 volts dropped will leave 118 volts across the receptacle. For the pair of receptacles, the 2 volts dropped will leave 238 volts across the pair, or 119 volts across each receptacle.
Click to expand...

That's assuming the loads are the same. If you had the "A" phase receptacle being 1/3 of the "B" phase receptacle as far as power consumption (shared neutral), then is the voltage drop prortional? At first glance I would say yes, but when you consider part of the load is now going back on the neutral wire, that increases the overall resistance. What would the correct means of calculating that load? (single phase power).

Now, what if you were sending two phases of a wye service and a neutral? How would a 35-amp load on A-phase sharing a neutral with a 50-amp load on B-phase work out on voltage drop? Let's assume a 200' run with #4 copper.
 
hardworkingstiff said:
That's assuming the loads are the same. If you had the "A" phase receptacle being 1/3 of the "B" phase receptacle as far as power consumption (shared neutral), then is the voltage drop prortional? At first glance I would say yes, but when you consider part of the load is now going back on the neutral wire, that increases the overall resistance. What would the correct means of calculating that load? (single phase power).

Now, what if you were sending two phases of a wye service and a neutral? How would a 35-amp load on A-phase sharing a neutral with a 50-amp load on B-phase work out on voltage drop? Let's assume a 200' run with #4 copper.
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You figure each conductor individually. The standard VD formula does this. For the neutral current:
I_N = sqrt(I_A^2 + I_B^2 - I_A x I_B)​
Then, I'm thinking, VD_AN = VD_A + VD_N and VD_BN = VD_B + VD_N... but not certain.

[Edited to add] Scratch the grayed text above. It can't be right... justification is that say both A and B line conductors were carrying same amount of current. The neutral would then also carry an equal amount of current. If simply adding the VD of each conductor were correct, each circuit would have a total VD of 2X an individual conductor... and that simply is not so, as there would at least be some reduction in circuit VD. I'll have to rethink at least this part, if not my whole premise...
 
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hardworkingstiff said:
Now, what if you were sending two phases of a wye service and a neutral? How would a 35-amp load on A-phase sharing a neutral with a 50-amp load on B-phase work out on voltage drop? Let's assume a 200' run with #4 copper.
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35A v 50A on single (opposite) phase xfrms would net 15A on the shared neutral, since phases cancel 1-to-1.

Likewise, if 2 of the 3-phase circuits cancel 2/3-to-1, shouldn't this net (35 * 2/3 - 50) = 27A on the shared neutral. Therefore (35+27) = 62A accross circuit A, and (50+27) = 72A across circuit B.

Plugging these values one-at-a-time into an unbalanced voltage-drop equation for 200ft of #4cu, 120vac @ Pf=1 in PVC gets me 112.3vac across circuit A, and 111.1vac across circuit B.
 
There is a reason I asked the question I did. I'm bidding a marina that will only have a 120/208 transformer available to supply three sets of docks. The other 2 sets will have 120/240 available.

I emphatically discussed with the developer/owner the correct way to wire boats is with a 120/240-volt service. I even pointed out 555.19(A)(3). What I learned though was that boats that plug into a 50-amp 120/240-volt receptacle almost always use that as two 120-volt supplies. The boats generally don't have 240-volt loads (until you get into the 70' and above range, and then they have load shedding technology, transformers and generators to deal with poor power).

Until I was reading this thread though, I never gave the voltage drop calculations for 2/3 of a 120/208 service much of a thought. The more I read on this site, the more I learn.

Thanks to all!
 
Do I have this right?

Do I have this right?

Assuming a 120/240-volt single phase service.

If you have a 200' run of #4 copper (type W cable, no conduit) and feed a receptacle to a boat that has a 50-amp load on each phase, the VD would be in the 2.57% range leaving 116.9 volts per phase to neutral. Now, if we were to drop the load on A-phase to 0-amps and leave the B-phase load at 50-amps, the voltage drop would go up to 5.13% leaving 113.84 volts for the load. If you brought A-phase load to 25-amps and leave B-phase at 50-amps then they would split the VD for the 25-amp load and B-phase would pick up all of the 25-amp VD it is using in excess of the shared VD, right? So A-phase would deliver 118.5 volts and B-phase would deliver 116.9 volts?
 
ramsy said:
35A v 50A on single (opposite) phase xfrms would net 15A on the shared neutral, since phases cancel 1-to-1.

Likewise, if 2 of the 3-phase circuits cancel 2/3-to-1, shouldn't this net (35 * 2/3 - 50) = 27A on the shared neutral. Therefore (35+27) = 62A accross circuit A, and (50+27) = 72A across circuit B.

Plugging these values one-at-a-time into an unbalanced voltage-drop equation for 200ft of #4cu, 120vac @ Pf=1 in PVC gets me 112.3vac across circuit A, and 111.1vac across circuit B.
Click to expand...

You may be right, but .....

Looking at the overall circuit, A-phase @ 35-amps, B-phase @ 50-amps would produce a neutral current of 44.4-amps.

What I can't grasp is how the proper VD calculations would be applied here. This is a little above me right now, and any education from the members would be greatly appreciated.
 
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hardworkingstiff said:
You may be right, but .....

Looking at the overall circuit, A-phase @ 35-amps, B-phase @ 50-amps would produce a neutral current of 44.4-amps.

What I can't grasp is how the proper VD calculations would be applied here. This is a little above me right now, and any education from the members would be greatly appreciated.
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I have no reference mat'l that covers this. My current theory...

View attachment 178

...calculates the total conductor VD for the A circuit using 35A & 7.88A, and the B circuit using 50A & 36.56A.

Whether this theory is correct or not, I would calculate for the highest possible VD... that is when the other circuit is not conducting, and the conducting circuit would be 2 x VD.
 
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hardworking,
I'm bidding a marina that will only have a 120/208 transformer available to supply three sets of docks.
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Be careful here. For the most part, unless you are supplying 3 phase receptacles, you will be limited to 120 volt receptacles on the 120/208 volt system.
555.19(A)(3) Branch Circuits Each single receptacle that supplies shore power to boats shall be supplied from a marine power outlet or panelboard by an individual branch circuit of the voltage class and rating corresponding to the rating of the receptacle.
FPN:Supplying receptacles at voltages other than the voltages marked on the receptacle may cause overheating or malfunctioning of connected equipment, for example, supplying single-phase, 120/240-volt, 3-wire loads from a 208Y/120-volt, 3-wire source.
Click to expand...
Don
 
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