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Thread: Computing short circuit current of Open Delta 3 phase transformers

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    Computing short circuit current of Open Delta 3 phase transformers

    For a 3 phase open delta transformers:




    When you are computing the short circuit current between the 208v high leg (point B) and neutral. Do you only use one 75kVA or do you add the other transformer to produce for example 75kVAx1.5 (or other value) = 112.5kVA? What exact values you use?

    I'm thinking if one needs to add because between point B and neutral, there are more than 1 winding (or 1 transformer) involved. What do you think?

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    I'm thinking the total transformer impedance in this calc is 1.5 x the 75 kVA impedance. It's one winding in series with a half winding.
    Alternate current
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    Quote Originally Posted by JoeStillman View Post
    I'm thinking the total transformer impedance in this calc is 1.5 x the 75 kVA impedance. It's one winding in series with a half winding.
    First, kudos to tersh for starting a new thread, this is discussion that started in another thread.

    I have always wondered exactly how one should figure impedance on such systems. Your answer is what I assumed would be correct but never really got any verification on this in the past when asking about it.

    Now to throw in a few variances that still have me wondering at times..

    Say drawing in OP were full delta with three of the same transformers.

    in that case you have parallel paths for current if you have a fault from high leg to neutral. An additional transformer surely increases total kVA, does it change impedance? My guess is yes. Probably still 1.5 times one coil but then figured similar if not same as figuring resistance of two resistors in parallel.

    If my suspicion on that is correct it probably answers some of my other variances -

    different sized transformer for the high leg, full delta with large lighting pot and two smaller pots for the rest of the delta, even kind of wondered if a wye bank of transformers should consider the fact you have three separate paths with three separate impedances in parallel if you have a bolted three phase fault. I think a lot of guys just look at the impedance on the nameplate of one unit in that situation and plug that into their fault current calculator though it probably isn't correct for a line to line fault - would apply to a line to neutral fault though.
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    Quote Originally Posted by kwired View Post
    First, kudos to tersh for starting a new thread, this is discussion that started in another thread.

    I have always wondered exactly how one should figure impedance on such systems. Your answer is what I assumed would be correct but never really got any verification on this in the past when asking about it.

    Now to throw in a few variances that still have me wondering at times..

    Say drawing in OP were full delta with three of the same transformers.

    in that case you have parallel paths for current if you have a fault from high leg to neutral. An additional transformer surely increases total kVA, does it change impedance? My guess is yes. Probably still 1.5 times one coil but then figured similar if not same as figuring resistance of two resistors in parallel.

    If my suspicion on that is correct it probably answers some of my other variances -

    different sized transformer for the high leg, full delta with large lighting pot and two smaller pots for the rest of the delta, even kind of wondered if a wye bank of transformers should consider the fact you have three separate paths with three separate impedances in parallel if you have a bolted three phase fault. I think a lot of guys just look at the impedance on the nameplate of one unit in that situation and plug that into their fault current calculator though it probably isn't correct for a line to line fault - would apply to a line to neutral fault though.
    Here is the actual nameplate I took picture in the actual transformers used (before they were put in the pole that day).







    What do you make of it? Assuming infinite bus, can you help compute for the short circuit current of the transformers load side (ignoring all conductors)?

    I'm using rough 75,000/208/0.0259 = 13,921A short circuit infinite bus assumption.. but if the impedance has to be multiply by 1.5X then it becomes:

    75,000/208/0.03885= 9281A only?

    But in IEEE-1584-2002. They used test setup with open space (below), but when barrier like breaker top was put inside, 208v can arc flash even with 2000A bolted short circuit current at 12mm gap.


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    If the transformer shown in #4 is the actual in the open post this is not a 3 phase transformer in open delta but a single phase transformer. In the nameplate it is not mention any grounding in the low voltage windings.
    Click image for larger version. 

Name:	Single phase noopen delta.jpg 
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    Quote Originally Posted by Julius Right View Post
    If the transformer shown in #4 is the actual in the open post this is not a 3 phase transformer in open delta but a single phase transformer. In the nameplate it is not mention any grounding in the low voltage windings.
    Click image for larger version. 

Name:	Single phase noopen delta.jpg 
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    It's the actual transformer. There were two like it. This is the 1st one (I traced the 3 phases late last year):




    The identical transformer at the back has the grounding at the centertap:




    The following was the bottom view of it where the centertap ground wire was connected to the multiple neutral network which has rods besides the poles:




    The reason it was really an open delta 3 phase was because in 2015, the electrician measured the 3 phases to neutral. One was 208v, the others were 120v.





    X1 (right most wire) to neutral was measured as 208 Volts.
    X2 to neutral was measured as 120 volts.
    X3 (left most wire) to neutral was measured as 120 volts.

    But between X1-X2, X1-X3, X2-X3.. they are all 240 volts.

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    Quote Originally Posted by Julius Right View Post
    If the transformer shown in #4 is the actual in the open post this is not a 3 phase transformer in open delta but a single phase transformer. In the nameplate it is not mention any grounding in the low voltage windings.
    Click image for larger version. 

Name:	Single phase noopen delta.jpg 
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    The question is what is the impedance involved of a high leg to neutral fault on the secondary when two of those single phase transformers are used to build an open delta bank? Yes the nameplate tells you impedance of that particular unit by itself. If you are connecting them so the secondaries are an open delta, and you have a high leg then the mid point of one unit is grounded and the high leg is the opposite corner of the delta from that grounded midpoint. You can only ground one point of the system or you will have fault current between any multiple grounded points.
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    Vbn=Vab+Vca/2 Vab=240 Vac=-120+207.85i Vbn=180+103.92i=207.85<30
    Zbn=Zab+Zan=1.5*Zab
    Po=noload losses=43 W ; Pload=full load losses=758 W
    Pcu=copper losses=Irat^2*Rcu ; Pload=Irat^2*Rcu+Po ; Rcu=(Pload-Po)/Irat^2
    Irat=312.5 A ; Rcu=0.007322 ohm
    Zab=Vab^2/S*imp%/100=0.240^2/0.075*2.59/100=0.019891ohm
    Xab=√(Zab^2-Rcu^2)=0.018495 ohm
    Zab=0.007322+0.018495i ohm
    Iscbn=Vbn/1.5Zab=5459.09-4327.2i [6966.1<-38.4o]

    Click image for larger version. 

Name:	Open delta short circuit current.jpg 
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ID:	22538

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    Quote Originally Posted by Julius Right View Post
    Vbn=Vab+Vca/2 Vab=240 Vac=-120+207.85i Vbn=180+103.92i=207.85<30
    Zbn=Zab+Zan=1.5*Zab
    Po=noload losses=43 W ; Pload=full load losses=758 W
    Pcu=copper losses=Irat^2*Rcu ; Pload=Irat^2*Rcu+Po ; Rcu=(Pload-Po)/Irat^2
    Irat=312.5 A ; Rcu=0.007322 ohm
    Zab=Vab^2/S*imp%/100=0.240^2/0.075*2.59/100=0.019891ohm
    Xab=√(Zab^2-Rcu^2)=0.018495 ohm
    Zab=0.007322+0.018495i ohm
    Iscbn=Vbn/1.5Zab=5459.09-4327.2i [6966.1<-38.4o]

    Click image for larger version. 

Name:	Open delta short circuit current.jpg 
Views:	32 
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ID:	22538
    So what is the short circuit current of the open delta above? I guess below 10kA?

    I understand higher impedance means the short circuit current in open delta is smaller. But in case there would be a transient while the short circuit occurs, would it increase the short circuit current by maybe 4 to 5 times?

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    Let's say the short circuit current of the above open delta transformers is 6970 Amps.

    Given there is 23 feet of AWG 1 conductors to the region or parts where bolted short occurs. How do you include the conductor impedance to the source impedance to get an overall decreased scc? And how much is the decreased bolted short circuit current? How do you compute for it?

    I just read the full paper called "Investigation of Factors Affecting the Sustainability of Arcs Below 250 V" by Michael J. Lang, Member, IEEE, and Kenneth Jones, Member, IEEE at Library Genesis.

    The contents are horrifying. Here they used setups representative of real world equipments.



    quoting their finding a bit:

    "Tests performed with gaps of 12.7 and 50.8 mm were used to determine the effects of gap and X/R ratio on arc sustainability and incident energy with the barrier in place. Testing at progressively lower currents revealed the barrier configuration’s ability to reliably sustain arcs for more than 1 s with a 12.7-mm gap at 4 kA and 208 V. The 32-mm gap performed intermittently at the lower values."

    Here is the abstract:

    "Abstract—Recent testing with various electrode configurations and insulating barriers suggests that 250-V equipment omitted from arc flash hazard analyses has the potential for burn injury. Research into the sustainability of arcs at these voltages shows that assumptions about the magnitude of these hazards need to be revised. This research enhanced the work of previous efforts by focusing on the sustainability of arcs with fault currents lower than 10 kA. Gap lengths between electrodes, electrode shape, electrode material, and voltage variations are studied for their effects on arc sustainability. A modified barrier design representative of the space around panelboard bus bars is also studied."

    The following is their conclusion:

    "CONCLUSION

    The testing discussed in this paper shows that sustained arcs are possible at 208 V even at relatively low fault currents but are dependent on several factors including voltage variations, conductor material, the configuration of conductors, and the presence of insulating barriers. The challenge to industry is to advance the research identified in the references, do additional testing on a variety of low-voltage equipment, and incorporate those findings into improved standards. These test strategies must consider all practical locations within the equipment where arcs may occur; within all equipment is the possibility for different electrode orientations.
    Enhanced models for various equipment"

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