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Mismatch of Impedance 4
- Thread starter Dagapa
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davidbeach
Electrical
How much mismatch, and how precise does the answer need to be? Which is more useful, an underestimation of the impedance or an over estimation? What are you using the numbers for, full sequence models, positive sequence models, models that only accept one impedance for a 3-phase bank, or models that accept 3 impedances for a 3-phase bank?
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- #3
The impedance of the two transformers are 1.69 at Tap 18 and the impedance of one transformer is 8.47 at Tap 18. I would like to know the consequences of using these transformers with their respective impedance. Would the difference in impedance have an impact on the life of the transformer(s)?
regards,
regards,
davidbeach
Electrical
Impact in what way? The impact on fault currents is vastly different than the impact on load flow. Again, what level of precision? For fault current it may be sufficient to call them all 1.69 and for load flow to call them all 8.47. Or just average them. Or weight the results some other how. Is this for hand calcs or model input?
On the other hand, that seems like a rather radical mismatch. 1.69 and 2.1 might be seen in real life; not sure how 1.69 and 8.47 are even possible. Given those numbers I'm going to guess that you've also got radically different kVA ratings. What are the impedances when all placed on the same kVA base? That's much more meaningful than comparing two impedance numbers on two different bases.
On the other hand, that seems like a rather radical mismatch. 1.69 and 2.1 might be seen in real life; not sure how 1.69 and 8.47 are even possible. Given those numbers I'm going to guess that you've also got radically different kVA ratings. What are the impedances when all placed on the same kVA base? That's much more meaningful than comparing two impedance numbers on two different bases.
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Thank you so much to both davidbleach and MattherDB for their responses.
The primaries are connected in delta and the secondary loads are also in delta. These are connections for a Furnace transformers. One of the transformer was indicating higher temperature (at around 12degrees higher)reaching a max of 72degrees of winding temperature. Eventually one day one of the transformer broke down with just the Gas evolution and the Gas stream indications on the annunciation boards. The HV side was found to be solidly earthed. This happened on one of the lower impedance transformer.
During the operation (prior to the failure), even on attaining full load current of >630amps, MW was not shooting up. Could this impedance have played a part for it? The transformers are all idential of 4MVA each, 11kV/185volts, with 18taps of equal voltage regulations.
I have been assigned to find out if the difference in the impedance was a part of the consequences? Therefore, am seeking the advices of the experts here on this forum. Your kind comments or advices would be highly appreciated.
regards,
The primaries are connected in delta and the secondary loads are also in delta. These are connections for a Furnace transformers. One of the transformer was indicating higher temperature (at around 12degrees higher)reaching a max of 72degrees of winding temperature. Eventually one day one of the transformer broke down with just the Gas evolution and the Gas stream indications on the annunciation boards. The HV side was found to be solidly earthed. This happened on one of the lower impedance transformer.
During the operation (prior to the failure), even on attaining full load current of >630amps, MW was not shooting up. Could this impedance have played a part for it? The transformers are all idential of 4MVA each, 11kV/185volts, with 18taps of equal voltage regulations.
I have been assigned to find out if the difference in the impedance was a part of the consequences? Therefore, am seeking the advices of the experts here on this forum. Your kind comments or advices would be highly appreciated.
regards,
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- #7
davidbeach
Electrical
1.69% at 4MVA? For a furnace? That just doesn't sound right. 1.69% impedance sounds like a 50 or 75kVA pole mount transformer. I would not have pictured anything that large.
You should do a detailed per-phase calculation; a good model will help. If nothing else, you could put it together in Mathcad. Look at the individual phase loadings; look at the impact of load variations and the degree of load sharing. If you have access to some sort of EMTP program that will make the modeling easier than in Mathcad.
You should do a detailed per-phase calculation; a good model will help. If nothing else, you could put it together in Mathcad. Look at the individual phase loadings; look at the impact of load variations and the degree of load sharing. If you have access to some sort of EMTP program that will make the modeling easier than in Mathcad.
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- #8
Loading of a delta=delta bank. By way of explanation let me start with single phase loading.
If you put a single phase load on a balanced delta-delta bank, you will find that the current is equal in all three of the transformers.
The in phase transformer will supply 50% of the load and the other two transformer will supply 50% of the load with one at a PF of 50% leading and the other with a power factor of 50% lagging.
With a balanced three phase load, the leading and lagging currents will cancel.
With an unbalanced three phase load, the leading and lagging currents will cancel for the balanced part of the load, but all three transformers will contribute to the unbalanced part of the load.
Now consider again the single phase load with equal transformers;
The two out of phase transformers will form an open delta transformer with the open side being a virtual transformer with regulation voltage drop and impedance voltage drop equal to a single transformer.
So we have a single transformer in parallel with a virtual transformer formed by the open delta.
These transformers will share the load in the inverse ratio of their impedances. (The combined or total impedance of the open delta transformer.)
Transformers with a lower impedance will hog the load and transformers with a higher impedance will shun the load.
Now if we look at this as a combination of three single phase loads on unequal transformers we will see:
The part of the load connected to the high impedance transformer will be hogged by the lower impedance virtual transformer and shunned by the higher impedance in phase transformer.
The part of the load connected to either of the lower impedance transformers will be hogged by the in phase transformer and shunned by the virtual transformer because of the higher impedance transformer forming part of the virtual transformer.
Generally when we are calculating the effect of impedance mismatches we may be looking at ratios of less than 2:1, often less than 1.5:1
With an impedance mismatch of 5:1 the contribution of the high impedance transformer is so little that it is hardly worth the time and effort to place and connect it.
Yes, both of the lower impedance transformers can be expected to overheat on full load.
Bill
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"Why not the best?"
Jimmy Carter
If you put a single phase load on a balanced delta-delta bank, you will find that the current is equal in all three of the transformers.
The in phase transformer will supply 50% of the load and the other two transformer will supply 50% of the load with one at a PF of 50% leading and the other with a power factor of 50% lagging.
With a balanced three phase load, the leading and lagging currents will cancel.
With an unbalanced three phase load, the leading and lagging currents will cancel for the balanced part of the load, but all three transformers will contribute to the unbalanced part of the load.
Now consider again the single phase load with equal transformers;
The two out of phase transformers will form an open delta transformer with the open side being a virtual transformer with regulation voltage drop and impedance voltage drop equal to a single transformer.
So we have a single transformer in parallel with a virtual transformer formed by the open delta.
These transformers will share the load in the inverse ratio of their impedances. (The combined or total impedance of the open delta transformer.)
Transformers with a lower impedance will hog the load and transformers with a higher impedance will shun the load.
Now if we look at this as a combination of three single phase loads on unequal transformers we will see:
The part of the load connected to the high impedance transformer will be hogged by the lower impedance virtual transformer and shunned by the higher impedance in phase transformer.
The part of the load connected to either of the lower impedance transformers will be hogged by the in phase transformer and shunned by the virtual transformer because of the higher impedance transformer forming part of the virtual transformer.
Generally when we are calculating the effect of impedance mismatches we may be looking at ratios of less than 2:1, often less than 1.5:1
With an impedance mismatch of 5:1 the contribution of the high impedance transformer is so little that it is hardly worth the time and effort to place and connect it.
Yes, both of the lower impedance transformers can be expected to overheat on full load.
Bill
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"Why not the best?"
Jimmy Carter
You have mentioned that load is connected in delta. What we are interested is in the connection of the secondary winding of transformer. I believe 1.69 & 8.47 may be values after failure. Please check the original test reports and ascertain the impedance values. Usually this impedance mismatch will be limited to 2-5 % in a bank. If it is more, there will be excessive circulating current inside delta winding.
Some years back, I came across a case. An auto transformer bank with stabilizing tertiary. One transformer failed and was replaced with another unit having the same H-L impedance, but the impedances to tertiary were different. On loading it was found, tertiary was carrying high circulating current.
Some years back, I came across a case. An auto transformer bank with stabilizing tertiary. One transformer failed and was replaced with another unit having the same H-L impedance, but the impedances to tertiary were different. On loading it was found, tertiary was carrying high circulating current.
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I believe that you have a delta transformer connection on the secondary.
I had an overheating problem with a wye:delta transformer feeding electrodes in a glass furnace.
Fortunately I had about 4 electrodes per phase. I broke the delta and connected two electrodes on each side of the break. That is instead of four electrodes connected to 2a2 and 3a1, I connected two electrodes to 2a2 and two electrodes to 3a1. Instead of the unbalance creating circulating currents and extra heat in the transformer, any unbalance was passed through the molten glass between 2a2 and 3a1, doing useful work.
With only three electrodes in total, I expect that that impedance mismatch will cause heating and circulating currents.
A delta transformer secondary abhors any inequality in either voltage or phase angle.
Bill
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"Why not the best?"
Jimmy Carter
I had an overheating problem with a wye:delta transformer feeding electrodes in a glass furnace.
Fortunately I had about 4 electrodes per phase. I broke the delta and connected two electrodes on each side of the break. That is instead of four electrodes connected to 2a2 and 3a1, I connected two electrodes to 2a2 and two electrodes to 3a1. Instead of the unbalance creating circulating currents and extra heat in the transformer, any unbalance was passed through the molten glass between 2a2 and 3a1, doing useful work.
With only three electrodes in total, I expect that that impedance mismatch will cause heating and circulating currents.
A delta transformer secondary abhors any inequality in either voltage or phase angle.
Bill
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"Why not the best?"
Jimmy Carter
- Thread starter
- #14
If three single phase transformers of 8.4, 1.3 and 1.3 impedance is connected in delta on primary and also connected in delta on the secondary, what would be the percentage of load sharing on each transformer? Other parameters are all identical for all three transformers.
kVA = 4MVA x 3Nos.
Nos of tap positions (OLTC) = 18
Voltage ratios: 11kV / 185volts.
kVA = 4MVA x 3Nos.
Nos of tap positions (OLTC) = 18
Voltage ratios: 11kV / 185volts.
- Moderator
- #15
To keep it simple, consider a single phase case.
Two transformers are to be connected in parallel.
Transformer #1- 100 KVA, 8.4% Z
Transformer #2- 100 KVA, 1.3% Z
First the transformers must be coverted to the same %Z base.
100 KVA @ 8.4% Z = 100 x (1.3/8.4) = 15.5 KVA at a 1.3% Z base.
The full load capacity of these transformers when connected in parallel will be 115.5 KVA
At that loading the 1.3% Z transformer will be 100% loaded. Any further increase in loading above 115.5 KVA will result in the 1.3% Z transformer being overloaded.
Note that a delta secondary connection abhors any unbalanced conditions. In the event that the uneven primary loading causes supply voltage drops and possibly phase angle errors, circulating currents will flow in the secondary delta.
KVA ratings are based on the maximum safe current. Even though the voltage causing the circulating current may be only a few percent, the current is limited not by rated voltage but by three times the impedance of the transformer. In the case of unequal impedances the circulating current will be limited by the sum of the impedances, 1.3 + 1.3 + 8.4 = 11% Z.
This implies that an unbalance of 11% will cause a circulating current flow equal to full load current.
The point is that even though the KVA of the circulating current is quite low, the current must be added to the load current to determine loading and overloading.
If you are on a "Soft" distribution system, the effect of circulating currents may reduce the contribution of the high impedance transformer even further.
Solution.
Consider a transformer with a much lower % impedance.
Mitigation.
If the high impedance transformer is in stock, compare the cost of a lower impedance transformer with the cost of adding electrodes.
With an even number of electrodes per phase, you may "Break" the delta. The circulating current will be greatly reduced and the small remaining circulating current will do useful work, generating heat in the furnace.
With a "Broken Delta", you may adjust the tap setting of the high impedance transformer to increase the loading. With a closed delta adjusting the tap setting may increase the circulating currents.
Kludge or Red-Neck fix.
The statement above notwithstanding,
With a known stable load, you may be able to adjust the tap setting of the high impedance transformer to reduce the circulating currents and allow higher overall KVA loading.
Caution. Do not attempt this with variable loads. It will have a sweet spot at one load level only.
Look for a balance between primary voltages and currents. Currents take precedence.
Watch the temperatures of the transformers. Try for more or less equal transformer temperatures.
It may not be possible to completely eliminate circulating currents, but it may be possible to increase the loading of the high impedance transformer while keeping the circulating current at an acceptable low level.
Bill
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"Why not the best?"
Jimmy Carter
Two transformers are to be connected in parallel.
Transformer #1- 100 KVA, 8.4% Z
Transformer #2- 100 KVA, 1.3% Z
First the transformers must be coverted to the same %Z base.
100 KVA @ 8.4% Z = 100 x (1.3/8.4) = 15.5 KVA at a 1.3% Z base.
The full load capacity of these transformers when connected in parallel will be 115.5 KVA
At that loading the 1.3% Z transformer will be 100% loaded. Any further increase in loading above 115.5 KVA will result in the 1.3% Z transformer being overloaded.
Note that a delta secondary connection abhors any unbalanced conditions. In the event that the uneven primary loading causes supply voltage drops and possibly phase angle errors, circulating currents will flow in the secondary delta.
KVA ratings are based on the maximum safe current. Even though the voltage causing the circulating current may be only a few percent, the current is limited not by rated voltage but by three times the impedance of the transformer. In the case of unequal impedances the circulating current will be limited by the sum of the impedances, 1.3 + 1.3 + 8.4 = 11% Z.
This implies that an unbalance of 11% will cause a circulating current flow equal to full load current.
The point is that even though the KVA of the circulating current is quite low, the current must be added to the load current to determine loading and overloading.
If you are on a "Soft" distribution system, the effect of circulating currents may reduce the contribution of the high impedance transformer even further.
Solution.
Consider a transformer with a much lower % impedance.
Mitigation.
If the high impedance transformer is in stock, compare the cost of a lower impedance transformer with the cost of adding electrodes.
With an even number of electrodes per phase, you may "Break" the delta. The circulating current will be greatly reduced and the small remaining circulating current will do useful work, generating heat in the furnace.
With a "Broken Delta", you may adjust the tap setting of the high impedance transformer to increase the loading. With a closed delta adjusting the tap setting may increase the circulating currents.
Kludge or Red-Neck fix.
The statement above notwithstanding,
With a known stable load, you may be able to adjust the tap setting of the high impedance transformer to reduce the circulating currents and allow higher overall KVA loading.
Caution. Do not attempt this with variable loads. It will have a sweet spot at one load level only.
Look for a balance between primary voltages and currents. Currents take precedence.
Watch the temperatures of the transformers. Try for more or less equal transformer temperatures.
It may not be possible to completely eliminate circulating currents, but it may be possible to increase the loading of the high impedance transformer while keeping the circulating current at an acceptable low level.
Bill
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"Why not the best?"
Jimmy Carter
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- #17
Unequal transformer temperatures in similar transformers are an indication of circulating currents. Given similar transformers, the current will be greatest in the transformer with the highest effective applied voltage. (The term "effective applied voltage" is a simplified label for the actual effect. The voltage causing circulating currents is effected by phase angle errors as well as voltage differences.) The circulating current will tend to add to the load current in the transformer with the highest applied voltage. The circulating current will tend to reduce the currents in the transformers with lower applied voltages.OP said:
In your operation before failure, the two low impedance transformers would be expected to have approximately equal temperatures.
One of two similar transformers running hotter is an indication of either impending failure or of a supply voltage issue. As excess heat tends to shorten the life of a transformer, you may have been seeing a combination of unbalanced voltages and impending failure.
With a resistive load, the KVA generally equals the kW. All of the current through the load should generate heat.
A poor power factor or KVA noticeably higher than kW is an indication that not all the current is flowing through the load. That is, part of the current is circulating current.
The effect of adding a high impedance transformer will be that the available KVA will be less than the sum of the three individual transformers.
Note: The effect of increasing the voltage of the lower impedance transformer will be best at one combination of primary voltage unbalance and secondary load. Any change of primary voltage unbalance or load may result in increased circulating current.
Adding even one electrode and breaking the delta will break the circulating current.
Consider,
1> Adding one higher impedance transformer will not cause circulating currents.
2> A delta secondary is prone to circulating currents.
3> Adding a third transformer of equal impedance may not, and probably will not, eliminate the circulating currents. It may increase the circulating currents.
4> The circulating currents (kW less than KVA) are reducing the ability of the low impedance transformers to provide full kW output to the load.
5> You have two indications of circulating currents. 1: Unequal temperatures. 2: KVA much greater than kW.
Conclusion. The combination of small increase in capacity when the third transformer is added and the noticeable circulating currents may be reducing the useful capacity of the transformer bank below the capacity of two transformers running in open delta.
Suggestion. Try running in open delta with only the two low impedance transformers in the circuit. Compare the kW loading possible on open delta with the present maximum kW loading with three transformers.
Comment: The circulating currents may be a much greater issue than the mismatched impedances. In the probable event that the circulating currents are caused by voltage unbalances on the primary supply, the effect may be greater at some times than at other times. It is possible that at times, the hottest transformer was even hotter than it was when it was observed.
Bill
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"Why not the best?"
Jimmy Carter
- Moderator
- #18
I strongly suspect that even were your transformers to be of equal impedance, %Z, you would still have problems.
One transformer hotter than a similar transformer and KVA much greater than kW with a resistive load are indications of serious incoming power issues.
Bill
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"Why not the best?"
Jimmy Carter
One transformer hotter than a similar transformer and KVA much greater than kW with a resistive load are indications of serious incoming power issues.
Bill
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"Why not the best?"
Jimmy Carter
"The impedance of each single phase unit in a three phase bank shall have to be identical as far as possible and variation of impedance of any single phase unit shall be within 2 % of the impedance of other units." This is from the specification of a major utility. So using 1.3 % and 8.4 % impedance transformers to form a three phase bank is inviting disaster from circulating currents.
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