Voltage drop- shared neutral

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clint2000

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Hi
I would say I have a good understanding of electricity and design. But for some reason I have not really understood the phenomenon of the balanced or canceled neutral current on a shared single phase system with a neutral and the 2 phase conductors. I recognize that the neutral doesn?t balance on a three phase feeder or circuit. I understand the theory of the 120/240 connected loads but I am not sure if this applies to a recreational vehicle campground with 120 volt 30 amp power receptacles. In particularly when sizing the feeder from the service equipment to the distribution panel. There would be (6) 120v receptacles, NEC value 3600va each (21.6kw) connected to each of the single phase legs in the single phase 120/240 volt distribution panel (at about 15 locations). Then the branch circuit wiring would be installed to each 120 volt 30 amp receptacle location. When I did the voltage drop calculations, I came up with the following:

4/0 AL 120v 21.6kw (6 receptacles) 3%vd equaled max distance of 100?
4/0 AL 240v 43.2kw (12 receptacles) 3%vd equaled max distance of 200?

I know 120 volt isn?t really applicable for the 240v panel but what I recognized was twice the power could be distributed twice the distance at 240volt. Does this mean that theoretically the neutral conductor of the 120/240 feeder not be counted in the distance of the voltage drop calculation because theoretically it would be zero if the loads were balanced? Would it be correct to consider the (12) 120 volt 30 amp receptacle loads (6) on each leg as a 240 volt load (43.2 kw)for voltage drop calculation purposes at the distribution panels? The loads from the RV?s would be a mix of liner and non linear loads such as water heaters, coffee makers, toasters, electric plug in heaters, microwaves, air conditioning and general incandescent lighting.

This leads me to another question that if the feeder was unbalanced in which case it almost always is in this application, would there be any calculation for the reduced neutral current?

Thank you in advance for your comments.
 
Re: Voltage drop- shared neutral

Step 1 is to add up the load. That is 12 times 120 times 30, or 43.2 KVA.

Step 2 is to calculate the current. That is 43,200 divided by 240, or 180 amps.

Step 3 is to calculate the voltage drop, in volts. You use the two-way distance. You don't need to distinguish between "current going out one phase and returning on the neutral" and "current going out one phase and returning on the other phase." You just take the distance from panel to load, and double it. I came up with about 7 volts, using a spreadsheet and cheating just a bit.

Step 4 is to calculate the voltage drop, in percentage of rated voltage. That is 7 divided by 240, and multiplied by 100 (to get %). My result is 2.9%, so I agree with your numbers.

Step 5 is to disregard any possible imbalance. If the two phases are exactly balanced, then there is no neutral current, and the 3% voltage drop will take place along one phase conductor and back along the other. If one phase has full load and the other phase has no load (the maximum possible imbalance), then the 3% voltage drop will take place along one phase conductor and back along the neutral. Finally, if there is a partial imbalance, then there will be voltage drop along all three conductors, but it will not exceed the calculated 3%.
 
Re: Voltage drop- shared neutral

Charlie

My intuition tells me that you calculated a total imbalance scenario, and not a completely balanced one. I can't identify the flaw in my logic, so please help me out.

Consider the source point A, the line side of the 120 v receptacle B, the neutral side of the receptacle C, and the neutral connection at the source D. The voltage drop from A to D would be AB+BC+CD. Your calculation makes drops AB=CD, which would require current in the neutral. By previous agreement, there is no neutral current.

If CD=0 then you have actually reduced overall voltage drop line to neutral. Where did I go wrong?

Jim T
 
Re: Voltage drop- shared neutral

You need to remember how simple circuits are: What goes out must come back, and where the wires meet everything adds to zero.
 
Re: Voltage drop- shared neutral

You are describing a single phase 120 volt circuit, not a single phase, multi-wire, 120/240 volt circuit. Let me ask you to draw the following.
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">Draw three parallel lines, about 1 inch apart, and about 3 inches long.</font>
<font size="2" face="Verdana, Helvetica, sans-serif"></font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">From top to bottom, label the left-hand ends of the three lines as "A," then "D," then "F."</font>
<font size="2" face="Verdana, Helvetica, sans-serif"></font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">From top to bottom, label the right-hand ends of the three lines as "B," then "C," then "E."</font>
<font size="2" face="Verdana, Helvetica, sans-serif"></font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">Connect point "B" to point "C," placing a resistor between the two points to represent a receptacle outlet. Do the same between points "C" and "E."</font>
<font size="2" face="Verdana, Helvetica, sans-serif">Now go back and read your description of the voltage drops.

If current is indeed flowing in the path "A-B-C-D," as you described, then you do have neutral current. In that event, the voltage drop from "A-B" would be the same as the voltage drop from "C-D," as you correctly pointed out.

But for a balanced load, I was describing current flow as going along the path "A-B-C-E-F." In that event, there is no neutral current, and the voltage drop from "A-B" would be the same as the voltage drop from "E-F."

Does that clarify things for you?
 
Re: Voltage drop- shared neutral

Charlie

Once I drew out your example, it became clearer. You're the best. What I wasn't considering is that in the multiwre circuit, slightly more current will flow than in the individual line to neutral circuit.

Life is good now. Thanks a lot.

Jim T
 
Re: Voltage drop- shared neutral

I think all the above is correct, but I question one thing. Shouldn't the voltage drop percentage be based on 120V, since all the connected loads are 120V?

The absolute voltage drop will be up to 7 volts regardless of which wires are doing the dropping. And it could be half of this (on a 120V circuit) if things go well. The problem is 7V is kind of a lot of drop for a 120V circuit, but I've been in campgrounds where its been worse than this.
 
Re: Voltage drop- shared neutral

Let's talk in terms of the three parallel lines I described earlier. The voltage between points A and F is 240 volts. That means that a total of 240 volts is dropped as current flows along the path A-B-C-E-F. We have calculated that the voltage drop along the conductors is 7 volts. That means that (1) 3.5 volts is dropped from A-B, (2) 233 volts is dropped from A-C-E, and (3) the last 3.5 volts is dropped from E-F. It also means that half of the 233 (i.e., 116.5 volts) is dropped from B-C, and the other half from C-E.

You can look at the voltage drop two ways:
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">If the wires were perfect conductors, the voltage from B-E would be 240. It is only 233. That is 7 volts short of perfect, out of an ideal 240 volts. That is 2.9%.</font>
<font size="2" face="Verdana, Helvetica, sans-serif"></font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">If the wires were perfect conductors, the voltage from B-C would be 120. It is only 116.5 (half of 233). That is 3.5 volts short of perfect, out of an ideal 120 volts. That is still 2.9%.</font>
<font size="2" face="Verdana, Helvetica, sans-serif">
 
Re: Voltage drop- shared neutral

Lets say 10A on L1 & L2 ( 0 neutral current )
Lets say that 10A creates 6V drop over length of conductor.

For L-N (each leg), 6V/120V = 5% drop
For L-L 12V/240V = 5%
Everybody?s happy.

Lets say we lose one phase
Now we have 10A on other phase, 10A on neutral 12V/120V = 10%vd
 
Re: Voltage drop- shared neutral

I was considering the unbalanced case. For calculations, we assume balance. But the first RV to pull up and connect will get 7V of drop (3.5 on AB, 3.5 on CD). When a second guy pulls up and hopefully balances things out, there is still 7V drop in the system (3.5 on AB and 3.5 on EF), but each RV only sees 3.5V of drop because the CD drop has gone to 0.

So you will always get 7V of drop in the system. But a given 120V user will see between 3.5V and 7V depending on how balanced the loads are. The good thing here is there are 6 receptacles and loading of those six was assumed to generate this level of voltage drop. So odds are good that the feeder load will be mostly balanced, and if its not the amps being drawn will be well below design so the voltage drop will still be less than 7V.
 
Re: Voltage drop- shared neutral

FWIW 7 volts drop (total) on a 120 circuit is not anything to worry about.

120 equipment is designed to run at least 10% low or 108 volts.
 
Re: Voltage drop- shared neutral

This is where I really miss Ed. he would have a diagram handy to show this.

My understanding is with 3 phase wye you can't balance the load with even loads between two lines because of the 120 electrical degrees between the phase lines.In other words the voltage of one phase is not in sync with the other phase.Example A and B if used are two different sources of power and 120 electrical degrees apart. Maybe some one here can help with the math but when A phase is equal to 120 volts B phase will only equal 69.3 volts for the neutral to be balanced both have to have equal voltages at the same time.

If it where delta 120/240 with a high leg you can. This is equal too single phase if you don't use the high leg. A phase to c phase of one transformer winding one power source with a center tap the peak to peak voltage happens at the same time. :)

[ September 24, 2005, 01:38 AM: Message edited by: ronaldrc ]
 
Re: Voltage drop- shared neutral

George -- Using Charlie's drawing description:

The 3 horizontal lines should be labeled on the left hand end: A, D & F

The 3 horizontal lines should be labeled on the right hand end: B, C & E

The top line is A-B, the center line is D-C, and the bottom line is F-E (all labeled left to right)

On the right hand ends, you have 2 loads, 1 drawn vertically between terminals B-C and 1 drawn vertically between C-E

On the left hand end, A=L1, D=Neutral & F=L2 (i.e. a 120/240V single phase supply)

On the right hand end it will look like a regular resistive load between "terminals" B-C (L1-N) and again between "terminals" C-E (N-L2).

I have to admit that I don't understand this either.

For 120/240V single phase: If you have 21.6 KW load (180A @ 120V) from L1-N and 21.6 KW load (180A @ 120V) from L2-N, then you have 43.2 KW total load (180A @ 240V) between L1 and L2.

Using the Vd calculator at: http://www.electrician.com/vd_calculator.html

#4/0 Al at 200' one way:

180A load at 240V = 7.2Vd = 3% per the calculator (or 7.2/240= 3%)

But assume that you blow a fuse serving L1. Half of the total load drops off, and you no longer have 240V available. You still have a 21.6 KW, 180A load between L2 and N.

180A load at 120V = 7.2Vd = 6% per the calculator (or 7.2/120= 6%) If you were to apply a voltmeter at the last (farthest) load, wouldn't the voltage L2-N read 120-7.2= 112.8V??

With a blown fuse on L1, I don't see how you would still have only a 3% Vd between L2-N. It looks to me like it would be 6%. This would be the maximum unbalanced condition. It would be the same if just 6 RV's were plugged in, all only on L2-N - all of the current would flow through L2 and N, and you should read 112.8V at the end of the line L2-N, right?

I agree with Bob that everything should still work OK at 112.8V.

I'm just trying to understand how Charlie's design would work, and which part of my understanding is wrong. It's step #5 that I get lost on. :confused:

I don't doubt that Charlie is right, it's just that I can't spot my logic error. Will someone please straighten me out and enlighten me? This is something I'd really like to understand.

Thanks!

Edited to fix my spelling and clarify.

[ September 24, 2005, 03:04 AM: Message edited by: tx2step ]
 
Re: Voltage drop- shared neutral

Imbalance does increase voltage drop. Worst case would be with 6 campers at full load on 1 phase with none on the other (7%). To exceed 5% with the example you'd need the equivalent of more than 4 full load campers with no load on the other phase. This seems unlikely, even with 6 campers all plugged in to one phase. I find it helpful when thinking about unbalance voltage effects (particularly 3-phase) to think of the neutral voltage as shifting from ground.
 
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