MWBC 2 Pole GFCI Breaker / vector analysis

[MD84]

Could someone please help me understand some of the fundamentals behind the operation of a 2 Pole GFCI operating on a 208/120V system. I will be installing (4) 120V receptacles for diesel heaters. I am considering either (2) circuits with each on a single pole GFCI Breaker running a neutral for each or (1) MWBC and sharing the neutral. I am leaning towards the MWBC.

I decided to do some vector analysis to satisfy my curiosity on the 2 Pole option. I was curious what my neutral current would be if 3/4 receptacles were in operation. I found that if I used a value of 20A for one phase and 10A for the other phase I would get about 17A on the neutral. As I understand the GFCI would see this as balanced. My question is what would be the phase angle of the neutral in relation to the other phases? Would the current vectors equal a sum of 0? Please provide an example vector analysis.

 

[iwire]

I can't answer your question directly but I would not use a two pole GFCI for this. If you get one faulty diesel heater both circuits will shut down


I would lean towards two standard single pole breakers and individual GFCI devices at each outlet.

 

[JFletcher]

What is the actual current draw per heater?

 

[iwire]

MD84, I added 'vector analysis' to your thread title to attract the forum members that do that sort of thing.

 

[MD84]

Thanks iwire. I agree with you on those points. This is going to be a bid job so the customer may want to go the most economical route. I would agree with you about increasing the reliability by using separate circuits. The best for reliability would be four dedicated circuits. Of course that would bring on some additional cost.

These will be nema 5-15r cord ends on a 20' 12/3 So cord coming out of a bell box. The customer will plug in their trucks to these. Typically the block/water jacket heaters installed on the trucks are 1kW. Some I have seen go up to 1500W but that is not as common. Typical current for one truck will be 8-10a.

 

[gar]

151213-1252 EST

ND84:

Intuitively think about how the GFCI breaker works. All three current carrying conductors pass thru one current transformer. The one CT instantaneously sums the three currents.

There are two circuits passing thru the CT. Turn one circuit off and the GFCI performs just like a single pole GFCI. The hot and neutral currents are identical, but flow in opposite directions and thus their net instantaneous is zero.

Turn the first circuit off and turn the second on. Same as above

Turn both circuits on and now consider the two separate components of the neutral. These components perform as they did when the individual circuits were on alone. Thus, the whole combination is balanced and there is net zero current thru the CT.

Vector diagram --- Suppose each circuit has a 1 A resistive load, then the neutral current is 1 A for a precise 120 degree difference between the two vectors. These results derive from the fact that mathematically the sum of two different sine waves of the same frequency (period), but of different amplitudes and phase angles instantaneously sum to a sine wave. Once source of the proof is at http://2000clicks.com/mathhelp/GeometryTrigEquivPhaseShift.aspx .

From a practical perspective I would not use a two pole breaker. Consider troubleshooting and problem solving as a reason to argue for single pole breakers in circuits that are 120 V fed from a multiphase source.

.

 

[Little Bill]

I also wouldn't put the heaters on a 2-P GFCI.
Plus, if I'm not mistaken, at least one mfg (SqD) says their 2-P GFCIs are not listed/rated for 208V.

 

[JFletcher]

I would go 4 15A circuits with GFCI receptacles/breakers. 8A+ engine heaters x2/circuit is going to be more than what you can continuously load a 20A breaker for, and going to #10 and 25/30A breakers may wind up more money. Sure, you could use a 2p GFCI with a MWBC, but if they are 1500W block heaters, you will have more than 20A on the neutral with 120/208V service.

 

[kwired]

I also wouldn't put the heaters on a 2-P GFCI.
Plus, if I'm not mistaken, at least one mfg (SqD) says their 2-P GFCIs are not listed/rated for 208V.

AFCI's won't work on a 208 Y system, GFCI's do. All the GFCI is looking for is a net zero in the CT and you will get it with 120/208 MWBC under "normal" conditions.

 

[don_resqcapt19]

I would go 4 15A circuits with GFCI receptacles/breakers. 8A+ engine heaters x2/circuit is going to be more than what you can continuously load a 20A breaker for, and going to #10 and 25/30A breakers may wind up more money. Sure, you could use a 2p GFCI with a MWBC, but if they are 1500W block heaters, you will have more than 20A on the neutral with 120/208V service.

How would we have more than 20 amps on the neutral?

 

[GoldDigger]

How would we have more than 20 amps on the neutral?

With line to neutral loads on three wires of the MWBC, even ideal balancing will have the full current of one heater on the neutral, but still not 20A. But if the customer plugs three heaters on one phase line or two heaters alone onto one phase, you will overload the neutral.
With hard wiring you are safe unless one heater's thermostat shuts off before the others.

 

[kwired]

With line to neutral loads on three wires of the MWBC, even ideal balancing will have the full current of one heater on the neutral, but still not 20A. But if the customer plugs three heaters on one phase line or two heaters alone onto one phase, you will overload the neutral.
With hard wiring you are safe unless one heater's thermostat shuts off before the others.

I think the question was intended to be how do you put more current on the neutral then you have on any individual phase conductor of a properly wired multiwire circuit?

 

[GoldDigger]

I think the question was intended to be how do you put more current on the neutral then you have on any individual phase conductor of a properly wired multiwire circuit?

And the answer is that if all loads have the same phase angle (same power factor and all inductive for example) then you cannot have more current on the neutral than on a phase wire for single or three phase. For two phase five wire you could actually get up to sqrt(2) times the line current appearing on the neutral. (Only two phase lines at 90 degrees loaded).

 

[kwired]

And the answer is that if all loads have the same phase angle (same power factor and all inductive for example) then you cannot have more current on the neutral than on a phase wire for single or three phase. For two phase five wire you could actually get up to sqrt(2) times the line current appearing on the neutral. (Only two phase lines at 90 degrees loaded).

That is my understanding as well.

 

[JFletcher]

How would we have more than 20 amps on the neutral?

Apologies, I messed up the math. Not sure where I went wrong with it but I initially came up with 21A and change figuring 1500W loads.

 

[gar]

151213-2145 EST

MD84:

From your first post.

I decided to do some vector analysis to satisfy my curiosity on the 2 Pole option. I was curious what my neutral current would be if 3/4 receptacles were in operation. I found that if I used a value of 20A for one phase and 10A for the other phase I would get about 17A on the neutral. As I understand the GFCI would see this as balanced. My question is what would be the phase angle of the neutral in relation to the other phases? Would the current vectors equal a sum of 0? Please provide an example vector analysis.

I get 17.32 . An interesting result from
sq-root of ( (10*0.5*sq_root 3)^2 + (20-10*0.5)^2 ) )
sq-root of ( ( 25*3) + 225 ) = sq-root (100) * sq-root 3 = 10*1.732 = 17.32

sine 60 deg = 0.5*sq_root 3 = 1.723/2 = 0.866
cosine 60 deg = 0.5
sine 30 deg = 0.5

I am giving you verification on your current. So from your vector diagram you should be able to figure out the angles of interest. Note: you have a right triangle with sides 10*0.5*sq_root 3 = 8.866 and 20-10*0.5 = 15 .

.

 

[mivey]

20@0deg + 10@120deg = 17.32@30deg

or

20@120deg + 10@0deg = 17.32@90deg

etc.

 

[MD84]

Thank you for all the feedback.

So I understand intuitively that the three conductors passing through the GFCI CT are instantaneously summed. This should equal 0 for normal operations. What I am having trouble with is the angle of the neutral.

When adding the three vectors as described above the sum would not be 0.

I think the answer is that the vector for the neutral would be considered a negative vector or the angle would be opposite. This makes intuitive since to me since the current would be returning to the source on the neutral.

So for the examples above would the angle actually be 210 instead of 30 or 270 instead of 90? Or could one instead consider it -30 or -90 respectively?

 

[david luchini]

Thank you for all the feedback.

So I understand intuitively that the three conductors passing through the GFCI CT are instantaneously summed. This should equal 0 for normal operations. What I am having trouble with is the angle of the neutral.

When adding the three vectors as described above the sum would not be 0.

I think the answer is that the vector for the neutral would be considered a negative vector or the angle would be opposite. This makes intuitive since to me since the current would be returning to the source on the neutral.

So for the examples above would the angle actually be 210 instead of 30 or 270 instead of 90? Or could one instead consider it -30 or -90 respectively?

Rather than the sum of all three conductors equalling 0, I would think of it as the difference between the line conductors and the neutral conductor...Ia + Ib - In = 0.

 

[mivey]

Thank you for all the feedback.

So I understand intuitively that the three conductors passing through the GFCI CT are instantaneously summed. This should equal 0 for normal operations. What I am having trouble with is the angle of the neutral.

When adding the three vectors as described above the sum would not be 0.

Correct. We follow Kirchhoff's Current Law. One way to say it is:
The sum of all currents entering a node equals zero (or equivalently, the sum of all currents leaving a node equals zero).

For normal conditions, the super-node only has A, B, C, & N lines crossing the boundary (i.e. 4 possible current paths). For a ground fault, we also have a "G" line (an additional current path).

So for normal conditions, the neutral current into the node at any given instant is the sum of all other currents leaving the node or what I gave in my previous example. To get the neutral current leaving the node, we would indeed add 180 deg to the prior result, as you surmised (210 deg and 270 deg).

For a ground fault, we sum A, B, C, & N to get the current for "G". If G <> zero (or some other threshold), then we trip for a ground fault.