Derating -- diversity?

[iwire]

The actual installation has so many wires the adjustment factor is 50%, and it hurts! The tube is plenty big -- 20% fill.

So it seems it was an inexperienced designer to start with.

Groups of smaller raceways are often more cost efficient then one large raceway.

Once again, the real problem is other ckts. There are some #8s that need to carry 30 amps,

Once we get larger then 10 AWGs it you really want to run separate raceways due to the large derating penalties of a single raceway.

 

[peter d]

Once we get larger then 10 AWGs it you really want to run separate raceways due to the large derating penalties of a single raceway.

Utter nonsense. Just run a few 2" conduits out of the electric room and pick up an entire store's worth of circuits with them.

 

[rabtrfld]

Thank you, winnie. Here is a picture:

Of course X-Y is a vector difference so this applies to 3-phase also. The net current in the conduit is zero in both cases.

The upper diagram is a regular multiwire BC with shared neutral:

T310-16 Note 10(a): A neutral conductor which carries only the unbalanced current from other conductors, as in the case of normally balanced circuits of three or more conductors, shall not be counted when applying the provisions of Note 8.

The lower diagram has exactly the same current in the wires but is not entitled to an exception due to the word "neutral" in Note 10.

Sorry for whining about it, should just accept it and move on...

 

[charlie b]

Your diagrams are not telling us what you want them to tell us. Indeed, you just lost the argument that you were trying to make. It comes from the fact that in the top diagram, X and Y are equal, so X-Y gives you zero. But in the bottom diagram, X turns out to be two times Y, so X-Y is not zero. Please let me explain.

Let us assign values to the resistances in both diagrams, such that each load has a current of 10 amps. So in the upper diagram, you have 10 amps flowing in the top wire (L1), so X =10. You also have 10 amps flowing in the bottom wire (L2), so Y = 10. In the center wire (N), the current is therefore zero (10 ? 10). That is indeed what you get, when loads are balanced in a 120/240V single phase system, such as the one you have drawn.

? Top diagram summary: Three wires, one with 10 amps, another with 10 amps, the third with 0 amps.

Now look at the bottom diagram. You have 10 amps flowing through the left hand switch, so the thing you labeled ?X-Y? has a current of 10. You also have 10 amps flowing through the right hand switch, so the thing you labeled ?Y? has a current of 10. That means that the current in the top wire (L1) must have a current of 20.

? Bottom diagram summary: Three wires, one with 10 amps, another with 10 amps, the third with 20 amps.

I would easily accept the top diagram as including only two current-carrying conductors. But the bottom diagram clearly has three current-carrying conductors.

 

[winnie]

Charlie,

Your analysis of the current flow in the diagrams drawn, assuming equal resistive loads, is quite correct. In the first diagram the 'N' conductor doesn't carry current, whereas in the second diagram all conductors carry current.

I still think that rabtrfld raises a valid point. In the MWBC example, given the same conductor ampacities, we have two supply breakers, and could in theory deliver twice the power to loads. So if you assume equal resistive loads, the MWBC will show its natural advantages and hide the point that rabtrfld is making.

I believe that the term 'current carrying conductor' is a distraction from the discussion.

The real question that we need to ask is: What is the total heat produced by the set of wires being considered? What is the maximum heat that can be produced given the possible distributions of current flow in the conductors?

The equations given in the diagram for the current flowing in the central conductors are correct, and are the same in both cases.

In both cases the maximum heating will be when conductor X and conductor Y carry equal current.

In both cases, the minimum heating for a given current X will be when Y=X/2.

We have different expectations for the two circuits. Further, the heating given the _same_ loads will be different, with less heating in the conventional MWBC case. But the heating is the same given the same values of X and Y.

-Jon