Customer has a building currently fed by a single 2500kva transformer. They planned ahead for load increase and there is a pad for a second 2500kva transformer. These are connected to the dual feed utility on the primary side with S&C switches with the PMH-9 connections, one source transfer and one manual. I know nothing about the internals of a S&C switch. If the transformer are matched 2500kva 5% in parallel doesn't that give me basically a 5000kva unit at 2.5% impedance and my available fault current goes from 59,454 amps to 229,941 amps?
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question on paralleled transformer short circuit current 1
- Thread starter v6racer
- Start date
With two identical transformers operated in parallel, the secondary side fault current will double. The effective impedance of your equivalent 5000 kVA transformer is still 5%.
Well, it won't quite double, due to impedance losses in cabling, etc, but it will be close to double. Are you sure they plan to operate the transformers in parallel?
Well, it won't quite double, due to impedance losses in cabling, etc, but it will be close to double. Are you sure they plan to operate the transformers in parallel?
- Thread starter
- #3
That is what they tell me and the spare pad with conduit goes back the the main bus. Looks like we will need to upgrade the main switch gear. Their previous short circuit study they had done showed 53,000 amps available fault current and the GE AV5 panels boards and Power Break circuit breakers rated at 100ka. I requested that they have someone more qualified do a modernization and life extension survey.
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1
- #4
GE AV5 panels boards and Power Break circuit breakers rated at 100ka.
New Available Short Circuit Current = 106kA.
That close, consider the impedance of the cabling.
You may also consider adding shorted CTs to the feeders to increase the impedance.
You may investigate the possibility of installing air core reactors to reduce the ASCC at the switchgear.
My very old text books showed devices called wireless reactors.
These were used to parallel transformers with unequal impedances.
A wireless reactor was basically a stack of laminations mounted on a cross arm near a distribution transformer with one the secondary lead passed through the lamination stack.
There are a couple of other red-neck tricks to increase the impedance of the feeder cables.
Bottom line;
I would go with the 100kA rated switchgear and investigate methods to reduce the ASCC at the switchgear by 6%.
--------------------
Ohm's law
Not just a good idea;
It's the LAW!
New Available Short Circuit Current = 106kA.
That close, consider the impedance of the cabling.
You may also consider adding shorted CTs to the feeders to increase the impedance.
You may investigate the possibility of installing air core reactors to reduce the ASCC at the switchgear.
My very old text books showed devices called wireless reactors.
These were used to parallel transformers with unequal impedances.
A wireless reactor was basically a stack of laminations mounted on a cross arm near a distribution transformer with one the secondary lead passed through the lamination stack.
There are a couple of other red-neck tricks to increase the impedance of the feeder cables.
Bottom line;
I would go with the 100kA rated switchgear and investigate methods to reduce the ASCC at the switchgear by 6%.
--------------------
Ohm's law
Not just a good idea;
It's the LAW!
I have the following proposal for your consideration. It would be more expensive, but with higher security.
1. For security reasons, It would be better that the second trafo to be connected to another new board. A N.O. four-pole coupler interconnecting the existing and the proposed new board. Essential loads are to be distributed from the existing board and the proposed new board. Should any one trafo fails, part of the essential loads are maintained by the other healthy trafo. By shedding non-essential loads and closing the coupler, all essential loads can be put back to service.
2. The breakers are mechanically and electrically interlocked, only two breakers out of three can be closed.
3. If the system is 3ph 4W with Neutrals separately grounded, use 4-ploe coupler.
Che Kuan Yau (Singapore)
1. For security reasons, It would be better that the second trafo to be connected to another new board. A N.O. four-pole coupler interconnecting the existing and the proposed new board. Essential loads are to be distributed from the existing board and the proposed new board. Should any one trafo fails, part of the essential loads are maintained by the other healthy trafo. By shedding non-essential loads and closing the coupler, all essential loads can be put back to service.
2. The breakers are mechanically and electrically interlocked, only two breakers out of three can be closed.
3. If the system is 3ph 4W with Neutrals separately grounded, use 4-ploe coupler.
Che Kuan Yau (Singapore)
Then the low voltage it is 480 V for a 2500 kVA 5% transformer and the three-phase [solid] short-circuit it is 59.45kA.
This it is-in my opinion-if the System short-circuit [apparent] power is infinite so the System impedance is 0.
If the short-circuit current was only 53 kA that means the System short-circuit power was only 222 MVA.
The System impedance is then [according to IEC 60909-0] ZQ=1.1*0.48^2/222=0.001142 ohm.
The transformer impedance it is 0.48^2/2.5*5%=0.00461 ohm.
In first case-one transformer and system- total impedance is 0.001142+0.00461=0.00575
I"k3=1.1*0.48/sqrt(3)/0.00575=53.016 kA
In the second case-two transformer and system- total impedance is 0.001142+0.00461/2=0.00345
I"k3=1.1*0.48/sqrt(3)/0.00345=88.36 kA
This it is-in my opinion-if the System short-circuit [apparent] power is infinite so the System impedance is 0.
If the short-circuit current was only 53 kA that means the System short-circuit power was only 222 MVA.
The System impedance is then [according to IEC 60909-0] ZQ=1.1*0.48^2/222=0.001142 ohm.
The transformer impedance it is 0.48^2/2.5*5%=0.00461 ohm.
In first case-one transformer and system- total impedance is 0.001142+0.00461=0.00575
I"k3=1.1*0.48/sqrt(3)/0.00575=53.016 kA
In the second case-two transformer and system- total impedance is 0.001142+0.00461/2=0.00345
I"k3=1.1*0.48/sqrt(3)/0.00345=88.36 kA
Nayeem1993
Electrical
You now have a system of 5000kVA with percentage impedance of 5%. Paralleling with new transformer will double the capacity and fault current as well. If I was the designer, I would keep the sources same and divide the loads into two separate buses with an automated bus coupler which will close if one of the transformers trips out for any reason. In that way you can downsize the downstream protective devices for lower fault current capacity.
OP doesn't state the nature of the loads, viz., how critical are they? If running two separate supplies with N.O. bus tie breaker, would manual [and therefore simpler and less expensive] load switching suffice?
Also, interlocking should not be required in this situation, as parallels between sources during planned switching would only be present for brief periods [barring the presence of stupid, unreliable, untrustworthy or undertrained staff, of course].
CR
"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]
Also, interlocking should not be required in this situation, as parallels between sources during planned switching would only be present for brief periods [barring the presence of stupid, unreliable, untrustworthy or undertrained staff, of course].
CR
"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]
There must be some mistakes there. Using an infinite bus source, the fault level at the 480V of the 2,500 kVA should be 60.14 kA.
The OP did not say what the primary voltage was. If we will use your assumption of 53 kA from the system, the system short-circuit MVA shall be 1.732 X 53,000 X 13,800/ 1,000,000 = 1,267 MVA @ 13,8 kV.
Granting your system Short-circuit MVA is 222 MVA, the total short circuit MVA of the 2 X 2,500 kVA at 5%Z will be = 2 X 2.5/0.05) = 100 MVA. Solving the the fault current at the tied-up bus of the two 480V secondaries will be = (1/[1/222+1/100])/(1.732 X 480) = 82.926 kA.
I have the following opinion.
1. For higher security reasons, it would be better to have the new trafo feeding to another new board with a N.O. Coupler to the existing board. In this way there is no change on the existing board kA rating. The proposed new board can be of same kA, rated for 2.5MVA trafo.
2. Refrain from connect two 2.5MVA trafo in parallel. Two trafo in parallel would need a new study whether the incoming breakers, feeder breakers and busbar are adequately rated; for the higher short-circuit kA.
3. With each 2.5MVA trafo feeding to the respective board, coupled through a N.O. 4-pole Coupler, retains the same kA rating of the existing board. With higher security.
4. Attention: Ensure that the two boards are properly "Phased" , say both are A,B,C,. Otherwise all motors will run reverse when switch to the other trafo !
Che Kuan Yau (Singapore)
1. For higher security reasons, it would be better to have the new trafo feeding to another new board with a N.O. Coupler to the existing board. In this way there is no change on the existing board kA rating. The proposed new board can be of same kA, rated for 2.5MVA trafo.
2. Refrain from connect two 2.5MVA trafo in parallel. Two trafo in parallel would need a new study whether the incoming breakers, feeder breakers and busbar are adequately rated; for the higher short-circuit kA.
3. With each 2.5MVA trafo feeding to the respective board, coupled through a N.O. 4-pole Coupler, retains the same kA rating of the existing board. With higher security.
4. Attention: Ensure that the two boards are properly "Phased" , say both are A,B,C,. Otherwise all motors will run reverse when switch to the other trafo !
Che Kuan Yau (Singapore)
Thank you, Parchie for your comment. You are right. In order to get 59454 A, the short-circuit voltage has to be 5.06% [this error is permitted]. According to IEC 600076-1 Table 1 – Tolerances permitted- tolerance for short-circuit impedance is ±10% of declared value.
The short-circuit current noted [59.454 kA ] it cannot be for high-voltage terminals of the transformer but for low
voltage side only.
According to IEC 60909-0, for maximum value of the short-current a factor of c=1.1 is employed. For minimum short-circuit this value is 1. So, a more accurate approach gives I"k3 max=89 kA and minimum 80.9 kA.
I corrected the System short-circuit power to 232 [instead of 222 MVA].
The short-circuit current noted [59.454 kA ] it cannot be for high-voltage terminals of the transformer but for low
voltage side only.
According to IEC 60909-0, for maximum value of the short-current a factor of c=1.1 is employed. For minimum short-circuit this value is 1. So, a more accurate approach gives I"k3 max=89 kA and minimum 80.9 kA.
I corrected the System short-circuit power to 232 [instead of 222 MVA].
Parchie, 7anoter4; are the following comments correct.
1. The ASCC of the two transformers in parallel will be below 100 kA and will be acceptable.
2. Due diligence may call for the calculations to be verified, documented and stamped by an engineering study.
3. Possible future upgrades to the supply system may increase the ASCC above the 100kA safe limit.
--------------------
Ohm's law
Not just a good idea;
It's the LAW!
1. The ASCC of the two transformers in parallel will be below 100 kA and will be acceptable.
2. Due diligence may call for the calculations to be verified, documented and stamped by an engineering study.
3. Possible future upgrades to the supply system may increase the ASCC above the 100kA safe limit.
--------------------
Ohm's law
Not just a good idea;
It's the LAW!
- Thread starter
- #14
Thank you all for your comments and knowledge. Further investigation showed they have separate 6000 amps busses feeding each side of the switch gear so one transformer per side as you suspected. The utility input is 25,000 volts but that is not confirmed by the utility yet. The 6000 amp busses step down 4000 amps on one side and 2000 amps on the other which is unfortunate since the new equipment load will be on the 2000 amp side and it will need to be replaced with a 4000 amp bus for them to reach their equipment utilization goal.
mparenteau
Electrical
We shouldn't forget motor contribution...
Mike
Mike
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