if 3phase bank is connected in ungrounded wye 13.8kv /centered tapped delta 120/240v, what will happen when 1) one primary fuse is blown but the corresponding transforer is intact? will the voltages rise at the secondary, by howmuch, and why? 2) fuse blown as well as the corresponding transformer?
Wye-Delta issues
- Thread starter Basra123
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rcwilson
Senior Member
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- Redmond, WA
Re: Wye-Delta issues
My thinking is:
One blown fuse on the ungrounded wye primary will place phase-phase volts, 13.8 kV, across the two remaining HV windings. Each winding will have 13.8/2 kV =6.9 kV instead of 13.8/1.732 = 7.96 kV.
The 240 V secondary on the unfaulted phases will drop to 208 V [208= 240 x (6.9/7.96)]. But these two voltages will be in phase and will add to 416 V across the third phase. With the open primary fuse, there is nothing to keep the voltage triangle from collapsing to single phase and the two "good" phases from adding together.
Phase-to-ground voltages depend on which transformer blew. They could be 120V, 120V, 360 V. Or, if the open fuse is on the 120/240V grounded transformer, they could be 208V, 208V, 0V.
The ?bad? transformer will have 416 V on its secondary and 13.8 kV on its primary winding that is floating.
With a bad transformer??? The same voltages could occur. But actual voltage will depend on many things: Did the HV winding fail to ground? How is the primary utility system grounded? How much fault current is fed back from the other good transformers LV windings to the faulted winding? etc, etc.
Bottom line - ungrounded wye's are not wise.
My thinking is:
One blown fuse on the ungrounded wye primary will place phase-phase volts, 13.8 kV, across the two remaining HV windings. Each winding will have 13.8/2 kV =6.9 kV instead of 13.8/1.732 = 7.96 kV.
The 240 V secondary on the unfaulted phases will drop to 208 V [208= 240 x (6.9/7.96)]. But these two voltages will be in phase and will add to 416 V across the third phase. With the open primary fuse, there is nothing to keep the voltage triangle from collapsing to single phase and the two "good" phases from adding together.
Phase-to-ground voltages depend on which transformer blew. They could be 120V, 120V, 360 V. Or, if the open fuse is on the 120/240V grounded transformer, they could be 208V, 208V, 0V.
The ?bad? transformer will have 416 V on its secondary and 13.8 kV on its primary winding that is floating.
With a bad transformer??? The same voltages could occur. But actual voltage will depend on many things: Did the HV winding fail to ground? How is the primary utility system grounded? How much fault current is fed back from the other good transformers LV windings to the faulted winding? etc, etc.
Bottom line - ungrounded wye's are not wise.
rcwilson
Senior Member
- Location
- Redmond, WA
Re: Wye-Delta issues
I think you are right, Rattus. IMO, The wye point "neutral" will float to a point determined by the voltage drops of the currents in the three legs of the wye.
If the facility load is predominately 3-phase motors, most of the load will be balanced and there's few problems until a fault occurs, or the load gets unbalanced.
But my head starts hurting when I try to visualize where that unbalanced current flows in a 3-phase, 3-wire, wye connection.
I think you are right, Rattus. IMO, The wye point "neutral" will float to a point determined by the voltage drops of the currents in the three legs of the wye.
If the facility load is predominately 3-phase motors, most of the load will be balanced and there's few problems until a fault occurs, or the load gets unbalanced.
But my head starts hurting when I try to visualize where that unbalanced current flows in a 3-phase, 3-wire, wye connection.
Re: Wye-Delta issues
RCWillson,
Thank you for the explination. However, it seems we would always have line to line voltages across any two windings in 3phase system regardless if the fuse blown or not? Also, why each winding will have half of the primary line to line as opposed to its normal phase voltage: line/1.732? when we lose one phase.
As to your questions about the blown transformer, this exists under the same conditions where the fuse also blown which means the winding are smoked as well as the fuse. In this case, what would be the single and 3phase voltages?
RCWillson,
Thank you for the explination. However, it seems we would always have line to line voltages across any two windings in 3phase system regardless if the fuse blown or not? Also, why each winding will have half of the primary line to line as opposed to its normal phase voltage: line/1.732? when we lose one phase.
As to your questions about the blown transformer, this exists under the same conditions where the fuse also blown which means the winding are smoked as well as the fuse. In this case, what would be the single and 3phase voltages?
rcwilson
Senior Member
- Location
- Redmond, WA
Re: Wye-Delta issues
Basra123 - Imagine a voltage triangle with three equal sides of 13.8 kV each. Since it is a 3-wire system, there is no ground or neutral reference point in the middle of the triangle. Any neutral point is imaginary.
Now imagine connecting three rubber bands together at one end and stretching them to connect the three points of the triangle. If they are equal strength, they will form a "wye" with their common point in the middle of the triangle. It will look just like our normal phase-neutral and phase-phase voltage triangle with 7.69 kV p-n and 13.8 kV p-p.
During normal operation, the floating wye point is pulled to the middle of the voltage triangle by the balanced currents and voltage drops in the three transformers. Just like the rubber bands. To a casual observer it looks like there are normal phase-neutral voltages on each transformer.
When the fuse blows, one rubber band is broken and the other two snap to one side of the triangle. The common or ?neutral? point is now at the midpoint of the one side.
Another way of looking at it - With only two connections to the system, the load has to be single phase. What else could it be? Remember, there is no neutral like ther would be on a 120/208 V single-phase panel.
Regarding the failed fuse, smoked winding scenario ? the voltages will be zero because fault will be back fed by the other two phases and their fuses will blow. Or, if both sides of the fault point get blown clear so there is no short circuit, the voltages will be what I explained above. If it is a ground fault at the halfway point of the HV winding, that winding will pull the neutral point to one side of the triangle, increasing the voltage on the other two windings to 10 kV so the secondary voltage may go to 304V or something else.
Sorry, there are too many variables to accurately calculate what the voltage will be if the two other primary fuses don?t blow.
Basra123 - Imagine a voltage triangle with three equal sides of 13.8 kV each. Since it is a 3-wire system, there is no ground or neutral reference point in the middle of the triangle. Any neutral point is imaginary.
Now imagine connecting three rubber bands together at one end and stretching them to connect the three points of the triangle. If they are equal strength, they will form a "wye" with their common point in the middle of the triangle. It will look just like our normal phase-neutral and phase-phase voltage triangle with 7.69 kV p-n and 13.8 kV p-p.
During normal operation, the floating wye point is pulled to the middle of the voltage triangle by the balanced currents and voltage drops in the three transformers. Just like the rubber bands. To a casual observer it looks like there are normal phase-neutral voltages on each transformer.
When the fuse blows, one rubber band is broken and the other two snap to one side of the triangle. The common or ?neutral? point is now at the midpoint of the one side.
Another way of looking at it - With only two connections to the system, the load has to be single phase. What else could it be? Remember, there is no neutral like ther would be on a 120/208 V single-phase panel.
Regarding the failed fuse, smoked winding scenario ? the voltages will be zero because fault will be back fed by the other two phases and their fuses will blow. Or, if both sides of the fault point get blown clear so there is no short circuit, the voltages will be what I explained above. If it is a ground fault at the halfway point of the HV winding, that winding will pull the neutral point to one side of the triangle, increasing the voltage on the other two windings to 10 kV so the secondary voltage may go to 304V or something else.
Sorry, there are too many variables to accurately calculate what the voltage will be if the two other primary fuses don?t blow.
Re: Wye-Delta issues
RC,
let's call h1 the top of first winding, h2 for top of the second and h3 for top of the third, connected in wye with star point floating. so when one fuse open say, h1, shouldn't the star point float to the top of that winding, ie. h1? in this case line voltage would be impressed between h1 and say h2 and h1 & h3 which is 1.732 the phase voltage. Thus the secondary will rise accordingly on the centere-tapped and the other transformer. What I am missing here?? help!
As regard to one tranformer blown, other than the center tapped, along with primary fuse, shouldn't that automatically be an open wye open delta and no change in the 120/240 output?
RC,
let's call h1 the top of first winding, h2 for top of the second and h3 for top of the third, connected in wye with star point floating. so when one fuse open say, h1, shouldn't the star point float to the top of that winding, ie. h1? in this case line voltage would be impressed between h1 and say h2 and h1 & h3 which is 1.732 the phase voltage. Thus the secondary will rise accordingly on the centere-tapped and the other transformer. What I am missing here?? help!
As regard to one tranformer blown, other than the center tapped, along with primary fuse, shouldn't that automatically be an open wye open delta and no change in the 120/240 output?
rcwilson
Senior Member
- Location
- Redmond, WA
Re: Wye-Delta issues
When the H1 fuse blows it is an open circuit. There is no connection = no voltage = no amps. There is nothing to make the common star point float to the H1 voltage. You are correct that with an open circuited winding, the H1 and the star point voltage will be equal, because there is no voltage across the winding. But it doesn?t float up to H1.
With balanced load on the other two windings, their voltage drops will be equal and will add up to the H2 ? H3 voltage. With only two voltage leads and no neutral, it has become a single-phase system, so all voltages have the same phase angle and add arithmetically, not vectorially. That?s why the voltage will be about 50% of the phase-to-phase voltage.
With a blown fuse, the transformer system no longer has any connection to the H1 voltage. So there is no way for the star point to drift, float or jump to an external voltage.
A ?blown? transformer is still connected into the secondary circuit, unless it blows up with enough force to blast all wiring clear. So the system will go to zero volts, eventually.
If you are thinking of removing a transformer to make it an open delta system, that will work. But you will need to use phase-phase 13.8 kV HV windings, and connect to all three phases, or get a neutral from the utility and use the 7.69 kV windings.
There is no way to get three phases with one fuse blown and no neutral connection.
When the H1 fuse blows it is an open circuit. There is no connection = no voltage = no amps. There is nothing to make the common star point float to the H1 voltage. You are correct that with an open circuited winding, the H1 and the star point voltage will be equal, because there is no voltage across the winding. But it doesn?t float up to H1.
With balanced load on the other two windings, their voltage drops will be equal and will add up to the H2 ? H3 voltage. With only two voltage leads and no neutral, it has become a single-phase system, so all voltages have the same phase angle and add arithmetically, not vectorially. That?s why the voltage will be about 50% of the phase-to-phase voltage.
With a blown fuse, the transformer system no longer has any connection to the H1 voltage. So there is no way for the star point to drift, float or jump to an external voltage.
A ?blown? transformer is still connected into the secondary circuit, unless it blows up with enough force to blast all wiring clear. So the system will go to zero volts, eventually.
If you are thinking of removing a transformer to make it an open delta system, that will work. But you will need to use phase-phase 13.8 kV HV windings, and connect to all three phases, or get a neutral from the utility and use the 7.69 kV windings.
There is no way to get three phases with one fuse blown and no neutral connection.
Re: Wye-Delta issues
rc, true, if we lose one line, we are left with a single phase system so we cannot use the 1.732 factor. Now for something completely different:
If we look carefully at the diagram of the 3-wire wye-delta system, we see that the phasing of the transformers is such that the secondary voltages of the two driven transformers cancel so that the secondary voltage of the third transformer is 0V, not 416V.
This is almost sure to start a 100 post discussion, but before anyone jumps in, they should draw out the schematic with polarity marks and dope it out for themselves.
[ September 19, 2005, 09:03 PM: Message edited by: rattus ]
rc, true, if we lose one line, we are left with a single phase system so we cannot use the 1.732 factor. Now for something completely different:
If we look carefully at the diagram of the 3-wire wye-delta system, we see that the phasing of the transformers is such that the secondary voltages of the two driven transformers cancel so that the secondary voltage of the third transformer is 0V, not 416V.
This is almost sure to start a 100 post discussion, but before anyone jumps in, they should draw out the schematic with polarity marks and dope it out for themselves.
[ September 19, 2005, 09:03 PM: Message edited by: rattus ]
sandsnow
Senior Member
- Location
- Southern California
Re: Wye-Delta issues
Thanks for finally sharing the answer and making it so simple
Thank you very much. In a thread long ago, it was said you could operate a wye ungrounded as a three wire system. Fair enough, but I wanted to know why no one did it-everyone started explaining what a delta or what a 3phase 3 wire was used for.
Thanks for finally sharing the answer and making it so simple
Re: Wye-Delta issues
RC,
you mentioned this "The 240 V secondary on the unfaulted phases will drop to 208 V [208= 240 x (6.9/7.96)]. But these two voltages will be in phase and will add to 416 V across the third phase. Phase-to-ground voltages depend on which transformer blew. They could be 120V, 120V, 360 V. Or, if the open fuse is on the 120/240V grounded transformer, they could be 208V, 208V, 0V.
Please note that the power legs are tied together to make 4 wire Delta. if we call the center-tapped leads L1, N, L2, and the other two transformer leads, power legs (P), are tied to make the fourth lead. then if I understand you right the secondary line voltages will be 208 between L1 & L2, and 208 between say L1 & P, and 416 between L2 & P. is that correct????
Also, the single phase voltages on the secondary must be 208/2=104 between L1 and N, and 104 between L2 and N, and 208+104=312 between P and N instead of 120,120, and 360.
Please assume the blown transformer is the non-grounded transformer and if no connection remain to the other two transformers...then the system should operate as open wye open delta, correct?
RC,
you mentioned this "The 240 V secondary on the unfaulted phases will drop to 208 V [208= 240 x (6.9/7.96)]. But these two voltages will be in phase and will add to 416 V across the third phase. Phase-to-ground voltages depend on which transformer blew. They could be 120V, 120V, 360 V. Or, if the open fuse is on the 120/240V grounded transformer, they could be 208V, 208V, 0V.
Please note that the power legs are tied together to make 4 wire Delta. if we call the center-tapped leads L1, N, L2, and the other two transformer leads, power legs (P), are tied to make the fourth lead. then if I understand you right the secondary line voltages will be 208 between L1 & L2, and 208 between say L1 & P, and 416 between L2 & P. is that correct????
Also, the single phase voltages on the secondary must be 208/2=104 between L1 and N, and 104 between L2 and N, and 208+104=312 between P and N instead of 120,120, and 360.
Please assume the blown transformer is the non-grounded transformer and if no connection remain to the other two transformers...then the system should operate as open wye open delta, correct?
Re: Wye-Delta issues
Rattus,
it seems correct that the sum is 0, not 416. so if L1, N, and L2 are the leads of the centered tapped transformer and P is the tied lead of the other two transformers, where the center-tapped transformer is in the middle and the other 2 transformers are on sides of it, then we would have 208 between L1 & 2 and 208 between L1 & P, and 0 between L2 & P.
As to the single phase voltages we would have 208/2=104 L1 & N, and 104 on L2 & N, and 208-104=104 between P & N. is that correct?
Rattus,
it seems correct that the sum is 0, not 416. so if L1, N, and L2 are the leads of the centered tapped transformer and P is the tied lead of the other two transformers, where the center-tapped transformer is in the middle and the other 2 transformers are on sides of it, then we would have 208 between L1 & 2 and 208 between L1 & P, and 0 between L2 & P.
As to the single phase voltages we would have 208/2=104 L1 & N, and 104 on L2 & N, and 208-104=104 between P & N. is that correct?
rcwilson
Senior Member
- Location
- Redmond, WA
Re: Wye-Delta issues
Rattus - thanks for pointing out my error. My humble apologies for leading Basra (& others) astray. Darn, I was having so much fun with the vector math that I jumped to the wrong conclusion.
You are correct that the two ?good? secondary windings have opposite polarities so their voltages will add to zero volts.
So, if one fuse blows, the 240 V phase voltages drop to 208, 208 and 0. The 120/240 leg will either be 0 volts or 104/208V.
Secondary loads will not be subjected to overvoltages.
Rattus - thanks for pointing out my error. My humble apologies for leading Basra (& others) astray. Darn, I was having so much fun with the vector math that I jumped to the wrong conclusion.
You are correct that the two ?good? secondary windings have opposite polarities so their voltages will add to zero volts.
So, if one fuse blows, the 240 V phase voltages drop to 208, 208 and 0. The 120/240 leg will either be 0 volts or 104/208V.
Secondary loads will not be subjected to overvoltages.
Re: Wye-Delta issues
Basra, yes, your magnitudes are correct. It would be beneficial though to dope out the phase angles of the secondary voltages (providing you have not done so already). You can assume any angle for the 13.8KV primary. Zero degrees is a convenient assumption.
Basra, yes, your magnitudes are correct. It would be beneficial though to dope out the phase angles of the secondary voltages (providing you have not done so already). You can assume any angle for the 13.8KV primary. Zero degrees is a convenient assumption.
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