Max % of allowable currrent unbalance in lighting transformer

[Phil Corso]

Cool, thanks. I've tried an example calculation, and the three delta-side line currents do not add up to zero. A delta transformer doesn't connect to the neutral. So if this delta primary were supplied from another WYE system, how would the WYE system supply this imbalance?

Example calculation:
Ias = 100A
Ibs = 90A
Ics = 80A

Iap = 41.2
Ibp = 36.8
Icp = 39.1

**This does not add up to zero, but adds up to 3.75A that would ordinarily be on the neutral.

Carultecl...

My calcs are based on a step-down, D-Y Xfmr, serving a 3-ph, 4w panel!

**What are the pf angles in your example?

Will do calc base on your sec'y load-currents assuming unity-pf!

Phil

 

[Ingenieur]

using the OP's numbers and assuming the pf is balanced across phases
N = 9.4 / -155 deg A
delta imbalance = 2.37 / -155

 

[GoldDigger]

When you have a delta primary the unbalance between L to N loads is converted to an inequality of L-L currents in the primary. Those will sum, if added as vectors, to line currents which do not require any L to N connections.
When you feed that delta from a wye source there will still be no L to N currents involved.
It is magical.
If instead you connected that delta primary to a delta secondary with a wye primary the neutral current will magically reappear.
The smoke normally confined within the transformer is, in this case, combined with (current) mirrors.

mobile

 

[Smart $]

...
The smoke normally confined within the transformer is, in this case, combined with (current) mirrors.

:lol:

 

[Phil Corso]

... If instead you connected that delta primary to a delta secondary with a wye primary... mirrors.

I must admit it's a real thigh-slapper, but I addressed OP data: Supply Xfmr => Delta-Wye, Prim'y/Sec'y Voltages => 480-208Y/120!

Phil

 

[Carultch]

Carultecl...

My calcs are based on a step-down, D-Y Xfmr, serving a 3-ph, 4w panel!

**What are the pf angles in your example?

Will do calc base on your sec'y load-currents assuming unity-pf!

Phil

Assume unity power factor.

Am I wrong in thinking that the currents feeding a DELTA primary transformer must add up as vectors to zero? Can the missing current be explained by current that circulates among the three windings?

Most other circuits require current in all lines, add up as vectors to zero, to complete the round trip path of electricity.

 

[Phil Corso]

using the OP's numbers and assuming the pf is balanced across phases N = 9.4 / -155 deg A
delta imbalance = 2.37 / -155

Now, try it with vectors! Perhaps this reminder will help... In = - [Sum of vectors, Ias, Ibs, & Ics ] !

Phil

 

[Ingenieur]

Now, try it with vectors! Perhaps this reminder will help... In = - [Sum of vectors, Ias, Ibs, & Ics ] !

Phil

That was done with phasors

 

[Phil Corso]

Assume unity power factor. Am I wrong in thinking that the currents feeding a DELTA primary transformer must add up as vectors to zero? Can the missing current be explained by current that circulates among the three windings? Most other circuits require current in all lines, add up as vectors to zero, to complete the round trip path of electricity.

1) You are not wrong in stating line-currents from utility source to delta-side winding is zero!

2) What do you mean "Missing" current!

3) Regarding "Most other lines add up as vector to zero.. " is only true of the 480V source line-currents.

4) It isn't true of the three (3) 208Y/120 phase lines to the panel. If it were true the neutral-current would be zero... even with unity-pF loads!

5) It isn't true of the currents within the closed-delta! The zero-sequence current IS the circulating-current!

Phil

 

[Phil Corso]

That was done with phasors

Would be so kind enough to list them as complex numbers?

Phil

 

[Phil Corso]

Will do calc base on your sec'y load-currents assuming unity-pf! Phil

Using your 100, 90, 80 panel-loads, ignoring D-Y phase-shift, and using A-B-C, phase-sequence, then:

1) Sec'y neutral-current is 17.3A.

2) Delta circulating-current 2.52A.

3) Primary line-currents, A, B, C, are 45.8A, 30.1A, and 39.1A, respectively.

Phil

 

[Ingenieur]

Would be so kind enough to list them as complex numbers?

Phil

re = cos magnitude
img = sin magnitude
complex = re + j img
ang = arctan img/re
mag = (re^2 + img^2)^1/2

 

[Phil Corso]

re = cos magnitude img = sin magnitude complex = re + j img ang = arctan img/re mag = (re^2 + img^2)^1/2

Foolish me... I thought you were a "REAL" engineer! If you had asked the same of me I would have gotten a copy of my work to you! Phil

 

[Ingenieur]

Foolish me... I thought you were a "REAL" engineer! If you had asked the same of me I would have gotten a copy of my work to you! Phil

I did it in my head, not on paper

 

[Phil Corso]

I did it in my head, not on paper

No wonder then!:happysad:

 

[Carultch]

Foolish me... I thought you were a "REAL" engineer! If you had asked the same of me I would have gotten a copy of my work to you! Phil

Karl Gauss believed it was counterproductive to call them "imaginary numbers", as it makes them seem fake and deters people from believing that they have useful applications. He recommended the terms direct, inverse, and lateral numbers. Direct = positive, inverse = negative, and lateral = imaginary.

 

[Carultch]

1) You are not wrong in stating line-currents from utility source to delta-side winding is zero!

2) What do you mean "Missing" current!

3) Regarding "Most other lines add up as vector to zero.. " is only true of the 480V source line-currents.

4) It isn't true of the three (3) 208Y/120 phase lines to the panel. If it were true the neutral-current would be zero... even with unity-pF loads!

5) It isn't true of the currents within the closed-delta! The zero-sequence current IS the circulating-current!

Phil

Ok, that explains it. Zero sequence current.

By "missing" current, I mean that if you put an ammeter clamp around all three current-carrying conductors, you will not measure zero amps, or even a time average of zero amps, as you do with other systems in my familiarity (examples below). The A+B+C current does not add up to zero, so my instinctive feeling is to wonder what happened to the amps that are not accounted by the three known CCC's.

Examples:
2-wire DC. Current goes out on the positive, and all of it returns on the negative. Positive current + negative current = 0
1-ph/2-wire AC. Current goes out on the line conductor, and returns on the neutral. Neutral current = negative of line current. It adds up to zero.
1-ph/3-wire AC. Current goes out on the black line and returns on the red line. If any imbalance occurs, it returns on the neutral. Black + Red + neutral currents = 0.
3-ph/4-wire AC. Current goes out on the A-phase and the return current on the B & C phases add up to the opposite. If any imbalance occurs, it returns on the neutral. A + B + C + N = 0.

 

[GoldDigger]

Ok, that explains it. Zero sequence current.

By "missing" current, I mean that if you put an ammeter clamp around all three current-carrying conductors, you will not measure zero amps, or even a time average of zero amps, as you do with other systems in my familiarity (examples below). The A+B+C current does not add up to zero, so my instinctive feeling is to wonder what happened to the amps that are not accounted by the three known CCC's.

Examples:
2-wire DC. Current goes out on the positive, and all of it returns on the negative. Positive current + negative current = 0
1-ph/2-wire AC. Current goes out on the line conductor, and returns on the neutral. Neutral current = negative of line current. It adds up to zero.
1-ph/3-wire AC. Current goes out on the black line and returns on the red line. If any imbalance occurs, it returns on the neutral. Black + Red + neutral currents = 0.
3-ph/4-wire AC. Current goes out on the A-phase and the return current on the B & C phases add up to the opposite. If any imbalance occurs, it returns on the neutral. A + B + C + N = 0.

I can guarantee that if you put a clamp around all three wires going to a delta primary you will measure 0 amps unless there is a ground fault in the wiring.
There is nowhere else for the current to go.

If you clamp each wire individually, you will not be able to perform the vector sum to confirm that.

 

[Carultch]

I can guarantee that if you put a clamp around all three wires going to a delta primary you will measure 0 amps unless there is a ground fault in the wiring.
There is nowhere else for the current to go.

That's what I initially thought too.

However, according to the results in my example from post #20 that Phil confirmed in post #31, those three currents do not add up to zero. There is somewhere else for current to go, when the current feeds a delta primary as its load. It can circulate among all three delta coils.

 

[GoldDigger]

That's what I initially thought too.

However, according to the results in my example from post #20 that Phil confirmed in post #31, those three currents do not add up to zero. There is somewhere else for current to go, when the current feeds a delta primary as its load. It can circulate among all three delta coils.

The current circulating among the delta coils does NOT appear on any of the lines going into the transformer. That is what circulating means.
I do not think that Phil meant what you though he meant. He assumed unity PF.

If they do not add up to zero, it is a measurement error.

Don't forget that the phase angle for each winding current is independent since you have different line to neutral loads, so you cannot just use 0, 120, 240 for the vector angles.

Measure the currents with a scope and add the vectors. You should get zero.