250MCM for 21.7 amps? Voltage drop calc must be wrong?

[GoldDigger]

Although on a balanced 3 ph system the 3 line i's mathematically cancel at the neutral and no i flows to the X0 (zero seq reactance) that does not mean a loads i evaporates
the sum is 0 but each current has value, a composite of other ph loads
if balanced ia + ib = -ic and so on
it returns on a different phase so a V drop results
total roundtrip length must be consider
with ph considered sqrt3 vs 2
I still use 2 lol

You are trying to think in the right direction, but I think that you missed.
If the voltage is zero, whether it is the result of no current or three currents that add up to zero the effect on load current and power loss will be exactly the same.
If you are looking at a single load which will be operated independently and may be the only load on that three phase branch (whether single phase or part of an MWBC) I agree that you are justified in doing the two wire calculation.

 

[mivey]

You are trying to think in the right direction, but I think that you missed.
If the voltage is zero, whether it is the result of no current or three currents that add up to zero the effect on load current and power loss will be exactly the same.
If you are looking at a single load which will be operated independently and may be the only load on that three phase branch (whether single phase or part of an MWBC) I agree that you are justified in doing the two wire calculation.

What he was pointing out was that your lecture #2 was wrong. It is not the same as the 1X one-way drop but with balanced loads is simply the three-phase 1.73X case.

 

[Ingenieur]

What he was pointing out was that your lecture #2 was wrong. It is not the same as the 1X one-way drop but with balanced loads is simply the three-phase 1.73X case.

This^^^^

if balanced the load current for line a returns thru b/c
b thru a/c
c thru a/b

and thereby induces V drop and according to KVL loop it must have effect, ie drop
it is not 2 because it is divded into 2 phases effectively 2 parallel return paths
that lowers Z and therefore Vdrop
when all the the phase angles etc are considered that pesky sqrt 3 shakes out
Basically 2 x sin120 which is sqrt 3
I still use 2 lol
the reason I use 2 is as the unbalance increases it converges on 2
only makes <1/2% for a 3% drop

 

[GoldDigger]

What he was pointing out was that your lecture #2 was wrong. It is not the same as the 1X one-way drop but with balanced loads is simply the three-phase 1.73X case.

I gave careful thought to what I wrote in item 2 and stand by it.
If dealing with balanced line to line loads I agree that the limiting case is the standard three phase load calculation which has a sqrt(3) in it (although that depends on the details of how you describe the voltages.) The improvement over the two wire calculation comes from the ratio between the phase current and the line current among other things.
But I was careful to specify balanced line to neutral loads. And in that case the load I is exactly the wire I. And the voltage drop corresponds to one wire length, not sqrt(3) times the single wire length.
The % VD takes the line to neutral VD and divides by the line to neutral voltage.
Any other description is either misleading or flat out wrong.
You can also look at it as neither the source nor the loads can distinguish between the delta and wye cases in terms of physical results. Any apparent differences come from not using either both line current and line to neutral voltage OR both phase current and phase voltage.

 

[Ingenieur]

Wye xfmr-wye R load with neut
480/277
each line and neut is 2 ohm
each load is 10 ohm

loop thru 2 windings 2 loads
480/24 = 20 A
drop = 2 x 20 x 2 / 480 = 16.7%

loop thru 1 winding and 1 load return neutral
assume no neutral current, where it I don't know lol
277/12 = 23.1 A
drop = 23.1 x 2 / 277 = 16.7%

loop thru 1 winding and 1 load return neutral
assume 480/2 or 240 V (just for the heck of it)
assume no neutral current, where it is I STILL don't know lol
240/12 = 20 A
drop = 20 x 2 / 240 = 16.7%

makes no difference

 

[GoldDigger]

But if you assume balanced line to line loads the you have less than 16.7%

By changing the line current but keeping the same wire size for your examples and mixing balanced with unbalanced loads you are totally confusing the issue.
If you assume the same power consumption in the load for each case you might get more informative results.
But that would fold in the fact that for the same load and wire higher voltages will have a lower %VD.

 

[mivey]

I gave careful thought to what I wrote in item 2 and stand by it.

OK then. I did not think about it much more than just a quick thought about a balanced load. I'll give it more thought when I have time.

 

[Ingenieur]

But if you assume balanced line to line loads the you have less than 16.7%

By changing the line current but keeping the same wire size for your examples and mixing balanced with unbalanced loads you are totally confusing the issue.
If you assume the same power consumption in the load for each case you might get more informative results.
But that would fold in the fact that for the same load and wire higher voltages will have a lower %VD.

no you don't, those are balanced loads
just because the neutral I is 0 doesn't mean an associated loads current doesn't return to the source
it doesn't dissipate lol

'confusing' can mean 2 things
1 I was not clear
2 you do not understand (when most would)
just saying

it is better to keep things constant when making a comparison
saying there is no current in the neutral so therefore has no V drop is a contradiction of KVL
the current thru load a returns thru b and c, so has a V drop associated
the loop under study must have current flow around the complete loop
you can't make a valid analysis saying that current flows into the load but not out of it (violates KCL), or not account for its return path in the other phases

the proper way is a 6x1 V, 6x6 Z and 6x1 I matrices
V = vab, vac, vbc, van, vbn, vcn
I = ia, ib, ic, ian, ibn, icn
etc.
invert and get the Y admittance matrix
since V is known solve for I
with that the drop can be determined

or a single line equivalent
but a simultaneous solution is better

but we don't need do to that
we know the factor is 1.72 (or 2) depending on phases, regardless of whether it is ph-ph, ph-n, balanced or unbalanced

 

[mivey]

I gave careful thought to what I wrote in item 2 and stand by it.

OK then. I did not think about it much more than just a quick thought about a balanced load. I'll give it more thought when I have time.

Ok. I see what you mean. For three L-N balanced loads and the same currents, the L-N drop is half the drop you get with a single L-N load (or close to the one-way drop).

 

[GoldDigger]

no you don't, those are balanced loads
just because the neutral I is 0 doesn't mean an associated loads current doesn't return to the source
it doesn't dissipate lol

'confusing' can mean 2 things
1 I was not clear
2 you do not understand (when most would)
just saying

it is better to keep things constant when making a comparison
saying there is no current in the neutral so therefore has no V drop is a contradiction of KVL
the current thru load a returns thru b and c, so has a V drop associated
the loop under study must have current flow around the complete loop
you can't make a valid analysis saying that current flows into the load but not out of it (violates KCL), or not account for its return path in the other phases

the proper way is a 6x1 V, 6x6 Z and 6x1 I matrices
V = vab, vac, vbc, van, vbn, vcn
I = ia, ib, ic, ian, ibn, icn
etc.
invert and get the Y admittance matrix
since V is known solve for I
with that the drop can be determined

or a single line equivalent
but a simultaneous solution is better

but we don't need do to that
we know the factor is 1.72 (or 2) depending on phases, regardless of whether it is ph-ph, ph-n, balanced or unbalanced

Fascinating!
I am not saying that there is no return current flow, just that when you have a balanced wye the presence of zero net current in the neutral results in zero voltage drop across the neutral.
You could equally well say that the return current flows through the other two loads in the wye but that the load in question is not affected by their voltage drop there since they are outside the two terminals between which we measure voltage and voltage drop.

P.S. There is no sqrt(3) in the wye calculations, other than that the line to line voltage is sqrt(3) times the line to neutral voltage.
The appearance of sqrt(3) is most important when working with power (VA) calculations rather than voltage drop and when relating the phase current to the line current. For a single unbalanced line to line load the line current is identical to the phase current but is out of phase with the line voltages.

 

[Ingenieur]

Fascinating!
I am not saying that there is no return current flow, just that when you have a balanced wye the presence of zero net current in the neutral results in zero voltage drop across the neutral.
You could equally well say that the return current flows through the other two loads in the wye but that the load in question is not affected by their voltage drop there since they are outside the two terminals between which we measure voltage and voltage drop.

P.S. There is no sqrt(3) in the wye calculations, other than that the line to line voltage is sqrt(3) times the line to neutral voltage.
The appearance of sqrt(3) is most important when working with power (VA) calculations rather than voltage drop and when relating the phase current to the line current. For a single unbalanced line to line load the line current is identical to the phase current but is out of phase with the line voltages.

Amazing!
and what I am saying is the N is 0 because it is balanced
only the imbalance flows in it
the load current returns on the phases and incurs loss
if you size on one way length the actual drop will be twice that

yes there is no drop across N in a balanced ckt
because there is no current
but there is associated load current in the other 2 phases and a V drop that reduces the V across the ph-N load
that is the KVL loop
not along the N since the associated loads current is on the other phases
draw it out
delete the N since no current
Draw a loop thru 2 windings and their 2 loads


PS you are stating the obvious

 

[mivey]

when you have a balanced wye the presence of zero net current in the neutral results in zero voltage drop across the neutral.

Seems so simple until you mix in line voltages and get all discombobulated.

Ingenieur: you are over-thinking it.

P.S. There is no sqrt(3) in the wye calculations, other than that the line to line voltage is sqrt(3) times the line to neutral voltage.

Which throwing in the 1.73X was my quick-trigger response until I dropped back to think about it.

 

[mivey]

Draw a loop thru 2 windings and their 2 loads

The two windings being loaded rather than three windings being loaded is indeed a more fun case where we do have neutral current. However, GoldDigger was talking about all three windings being loaded equally.

 

[GoldDigger]

Amazing!
and what I am saying is the N is 0 because it is balanced
only the imbalance flows in it
the load current returns on the phases and incurs loss
if you size on one way length the actual drop will be twice that

yes there is no drop across N in a balanced ckt
because there is no current
but there is associated load current in the other 2 phases and a V drop that reduces the V across the ph-N load
that is the KVL loop
not along the N since the associated loads current is on the other phases
draw it out
delete the N since no current
Draw a loop thru 2 windings and their 2 loads


PS you are stating the obvious

As to the bolded statement above, all that I can suggest is that you try the experiment and make the measurement. The voltage drop actually seen by a line to neutral load in a balanced configuration will be that corresponding to a single length of wire.
There are voltage drops all over the 4 wire circuit, but the only drop seen by a single line to neutral load is the drop on its own hot wire.
You may find it easier to understand this if you look at a 120/240 single phase three wire configuration instead. At least that way you will not be tempted to throw in a factor of 1.73.

Finally, if I am in fact stating the obvious, why do you persist in disagreeing with it? :happysad:

 

[GoldDigger]

Which throwing in the 1.73X was my quick-trigger response until I dropped back to think about it.

Happens to me all the time, but as a defense mechanism I have tried to make my first analysis in a way where I do not need to think about whether the 1.73 is there or not.

Not always possible, of course, since sometimes that sqrt(3) just cannot be avoided.

 

[mivey]

but there is associated load current in the other 2 phases and a V drop that reduces the V across the ph-N load

But you can't parcel out the drops from the other phases and assign it to only one phase. Each phase is going to have a drop associated with the RMS values, not some instant in time. You will wind up with 3X the one-way RMS volt drop for all three phases if you somehow try to sum them to one phase but we can't do that. We wind up with a 1X drop per phase.

 

[Ingenieur]

As to the bolded statement above, all that I can suggest is that you try the experiment and make the measurement. The voltage drop actually seen by a line to neutral load in a balanced configuration will be that corresponding to a single length of wire.
There are voltage drops all over the 4 wire circuit, but the only drop seen by a single line to neutral load is the drop on its own hot wire.
You may find it easier to understand this if you look at a 120/240 single phase three wire configuration instead. At least that way you will not be tempted to throw in a factor of 1.73.

Finally, if I am in fact stating the obvious, why do you persist in disagreeing with it? :happysad:

You are trying but can't really grasp it
keep at it!
you may find it easier if you draw it out

because the 'obvious' is not the subject
the 'obvious' is the use on the sqrt 3

 

[Ingenieur]

But you can't parcel out the drops from the other phases and assign it to only one phase. Each phase is going to have a drop associated with the RMS values, not some instant in time. You will wind up with 3X the one-way RMS volt drop for all three phases if you somehow try to sum them to one phase but we can't do that. We wind up with a 1X drop per phase.

1.73 X per phase-neutral load

 

[mivey]

1.73 X per phase-neutral load

If you want to convert to phase-phase.

 

[GoldDigger]

1.73 X per phase-neutral load

100% absolutely not.
If you take the load current as I then the current in the corresponding phase line is also I, and the voltage drop is IR where R is the one way resistance of that conductor.
There is no place that you can logically shoehorn in the 1.73 no matter how hard you try.

The reason that the 1.73 shows up for delta (line to line load) calculations is that the load current is NOT equal to the line current when the loads are balanced. Or at least that is one way of looking at it.

PS: If you have an unbalanced (single) line to neutral load, the VD will be based on distance times 2; still no place for 1.73.