How to size isolation transformer to reduce fault current ? 1

[james64]

Hi all,

I am experiencing an issue with a low-fault current panel X that is connected to an upstream feeder breaker with a capacity of 600V/400A. The client wishes to increase the fault current in the current panel X, but replacing the panel is a challenging task. To resolve the issue, we have decided to add an isolation transformer. My question is whether we should consider the upstream feeder capacity, or just the impedance of the isolation transformer to reduce the fault current in panel X. After checking the connected load for this panel, it is only around 100kW with 600V.

 

[waross]

With a large installation, the upstream supply is often considered for its contribution to reducing the fault current.
This poses two challenges;
1. Determining the upstream impedance.
2. The possibility that the upstream impedance may change as improvements are made to the supply system over the years.
For such a small system the upstream impedance may be considered as zero as it will have little effect on the available fault current of the small isolation transformer.
It is also safely conservative as the calculations will not be affected by possible system upgrades.
And use the defined "Available Short Circuit Current" rather than agonizing over the actual peak offset fault current.
"Available Short Circuit Current" is a defined quantity, the result of dividing the transformer percent impedance voltage into therated current.
Devices such as switches, breakers and fuses that have an ASCC rating have been tested with fault currents that include DC offsets and the test currents are greater than the actual ASCC.
Standardizing on the easy to calculate ASCC makes selection of field devices easy and much less subject to error than using actual peak, offset fault currents.
The peak asymmetrical offset fault current may approach 2.82 times or 282% of the ASCC. The actual peak depends on the X/R ratio of the transformer and the point on wave of the fault.
Using ASCC you may safely avoid all that possible confusion in all but the most extreme cases.

Fixed a bad typo.
--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[james64]

Thank you for your comment, Waross. I'm curious about the source of the 2.82% mentioned.

 

[che12345]

Mr. james64 (Electrical)(OP)24 Jul 23 21:14
".....#1. I am experiencing an issue with a low-fault current panel X that is connected to an upstream feeder breaker with a capacity of 600V/400A. ....#2. The client wishes to increase the fault current in the current panel X, but replacing the panel is a challenging task.....#3. To resolve the issue, we have decided to add an isolation transformer. .... #4. My question is whether we should consider the upstream feeder capacity, or just the impedance of the isolation transformer to reduce the fault current in panel X. After checking the connected load for this panel, it is only around 100kW with 600V".
I am confused. Your question seems to be confused between "load current, usually in A" and "fault current, usually in kA".
1. Your upstream breaker is 400 A and system is (600 V ?) Note: 600 V is not widely used.
2. Why your client wishes to increase the "fault current" ? Does he means increase the "load current capacity" ? Adding more loads to this panel.
3. What is the issue? Adding an isolating transformer for what purposes? It can only lowered the "fault current", but that is contradict to what your client wanted.
4. Why do you have to reduce the fault current if the existing panel X is adequately short-circuit / fault current rated, in kA ?.
4.1 100 kW at 3phase 600 V work out to be slightly above 100 A. The upstream breaker rated 400 A, which can handle much more loads. Note: adding "load current" does NOT alter the "short-circuit/fault current" of the panel, which is dependent on the mechanical structure to withstand the mechanical force during fault.
Che Kuan Yau (Singapore)









x

 

[james64]

Thank Che Kuan for your question.
See below

I agree with fault current kA, load current shall be A.
1. Your upstream breaker is 400 A and system is (600 V ?) Note: 600 V is not widely used.
A) This voltage is used in Canada.

2. Why your client wishes to increase the "fault current"? Does he means increase the "load current capacity" ? Adding more loads to this panel.
A) After checking the panel schedule, it requires at least 35kA. But interrupting rate of the panel has only 14kA breakers without main breaker. So coordination study was not done properly as per the initial design.

3. What is the issue? Adding an isolating transformer for what purposes? It can only lowered the "fault current", but that is contradict to what your client wanted.
A) To reduce fault current at the panel, I am proposing an isolation transformer. I do not understand the contradiction here. Can you elaborate on it? Question is here how to size or specify isolation transformer to reduce fault current?

4. Why do you have to reduce the fault current if the existing panel X is adequately short-circuit / fault current rated, in kA ?.
A) Refer to question 2 for this answer.

4.1 100 kW at 3phase 600 V work out to be slightly above 100 A. The upstream breaker rated 400 A, which can handle much more loads. Note: adding "load current" does NOT alter the "short-circuit/fault current" of the panel, which is dependent on the mechanical structure to withstand the mechanical force during fault.
A) I understand your point.

 

[waross]

I'm curious about the source of the 2.82% mentioned.

You should be. It was a bad typo.
It should be 2.82 times or 282%. (Or 283% depending on rounding.)
The peak value of a sine wave is 1.42 (Root two) of the RMS value.
A fully offset asymmetrical current will be 2 x 1.41 (Two times root two)

As I understand the original question, your client wanted to either increase the fault current rating of the panel or reduce the available Short Circuit Current to within the rating of the panel.

The client wishes to increase the fault current in the current panel X

Possibly by changing to a different primary feeder or changing an upstream supply transformer to a transformer with greater ASCC.

I don't see any mention of adding load, but load may have been added elsewhere in the plant and as a result the main transformer was up-sized.
600 Volts?
Standard for over 40 years in British Columbia, Canada.
Available for much longer than that in Ontario.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[che12345]

james64 (Electrical)(OP)24 Jul 23 21:14
"...I am experiencing an issue with a low-fault current panel X that is connected to an upstream feeder breaker with a capacity of 600V/400A. The client wishes to increase the fault current in the current panel X,.... we have decided to add an isolation transformer. My question is whether we should consider the upstream feeder capacity, or just the impedance of the isolation transformer to reduce the fault current in panel X. After checking the connected load for this panel, it is only around 100kW with 600V".
I look at it "in simplified form" as following, for your consideration.
1. If panel X is now loaded at about 100 A, but wish to finally up to 400 A at 600 V . The transformer shall be rated 1.732 x 600V x 400A = 415.68 kVA.
2. The panel X is rated short-circuit current = 14kA.
3. The issue is what should be the %Z of the transformer to limit the short-circuit to < 14kA ?
%Z = (I x 100%)/Isck = (400A x 100%)/14000A = 2.85 %
4. Info : The load now is about 100A. If your client wishes to increase it finally say up to 200A, the transformer can be reduced to half the kVA rating.
Che Kuan Yau (Singapore)











4tended to

 

[waross]

4. Info : The load now is about 100A. If your client wishes to increase it finally say up to 200A, the transformer can be reduced to half the kVA rating.
Che Kuan Yau (Singapore)

And at half of the KVA rating a %Z of twice 2.85% or %Z of < 5.7% is acceptable to hold the ASCC to < 14kA.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[waross]

By the way, Mr. Che; Don't confuse short circuit current with Available Short Circuit Current. (ASCC)
A discussion of fault current may be referring to Asymmetrical current.
Available Short Circuit Current is the symmetrical fault current. (ASCC)
Your formula pertains to ASCC, not asymmetrical fault current.



--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[che12345]

@ Mr. waross (Electrical)24 Jul 23 22:17
".....The peak asymmetrical offset fault current may approach 2.82 times or 282% of the ASCC. The actual peak depends on the X/R ratio of the transformer and the point on wave of the fault....."

1. With due respect to your learned advice, please advise what the peak asymmetrical offset fault current may approach 2.82 times or ...... has to do with the question asked? This information is irrelevant.
2. I consider the issue is what should be the kVA and the percentage impedance ratings of the transformer to lower the short-circuit current such that the panel X which is rated 14kA would be able to take up more loads.
Che Kuan Yau (Singapore)

 

[james64]

Thank both waross and che12345. My concern is what is appropriate transformer impedance for reducing fault current preventing trip panel branch breakers. To be honest I am not quite sure. What kind of fault should be applied for this calculation ?

 

[che12345]

Mr. james64 (Electrical)(OP)28 Jul 23 10:01
".... My concern is what is appropriate transformer impedance for reducing fault current preventing trip panel branch breakers. To be honest I am not quite sure. What kind of fault should be applied for this calculation ?"
1. Your panel X is rated 14kA. Up stream breaker is rated 400A. Based on these data I assumed that panel X busbars are rated for 400A. Attention: could be lower.
a) The transformer shall be rated 1.732 x 600V x 400A = 415.68 kVA.
b) the %Z of the transformer to limit the short-circuit to < 14kA
%Z = (I x 100%)/Isck = (400A x 100%)/14000A = 2.85 %
2. Panel X can be loaded up to 400A (if the busbar is adequately sized), add MCCB
rated > 14kA for additional loads.
Che Kuan Yau (Singapore)

 

[waross]

Thank both waross and che12345. My concern is what is appropriate transformer impedance for reducing fault current preventing trip panel branch breakers. To be honest I am not quite sure. What kind of fault should be applied for this calculation?

Asymmetric fault currents are used in design to evaluate mechanical forces.
Symetrical fault current or ASCC (Available Short Circuit Current) is used to match equipment. (When the equipment is tested and rated, the asymmetrical fault current is considered so that you don't have to worry about it.
Use the simple, symmetrical fault current.
You do not have to consider ASCC when adding loads.
If there are changes to the supply network that incase the ASCC of the supply to the panel, you must consider the rating of the panel.
What happens when the supply ASCC exceeds the panel rated ASCC?
In the event of a fault, the breaker may explode trying to clear the fault.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[waross]

Confusion:

I am experiencing an issue with a low-fault current panel X that is connected to an upstream feeder breaker with a capacity of 600V/400A. The client wishes to increase the fault current in the current panel X, but replacing the panel is a challenging task.

Does the client want to add loads, or has the client realized that the supply to the panel is greater than the ASCC rating of the panel?
These two issues are not related.
Mr.Che's advice covers one possible solution.

For your 100 kW load, you may round the current off to 100 Amps.
In the absence of accurate PF information, loading is often calculated at 0.8 PF. that rounds off to 125 Amps. (125 KVA)
You may choose to add capacity for future added loads.
You may choose to supply the full 400 Amps capacity to the panel.

FIRST CHOOSE A TRANSFORMER CAPACITY FROM ONE OF THE ABOVE CHOICES.
When you have chosen the transformer capacity apply MR. Che's formula to determine the minimum allowable %Z.
For a given ASCC rating, as the KVA rating of the transformer increases, the minimum allowable ASCC increases.
Note that the transformer regulation (The voltage drop under load) is less than the %Z but is related to the %Z.
When selecting your transformer, it may be well to look at the rated regulation and consider the added voltage drop under load to be expected.
(The rated regulation is a guide. The actual voltage drop may be higher or lower depending on the PF of the load.)
And finally, the calculated %Z is a minimum value. A higher value of %Z is acceptable but be aware that with a higher %Z, a greater voltage drop under load may be expected.

Why is %regulation different than %Imp (%Z)?
It has to do with power factor and X/R ratios.
%Imp is related to fault currents which typically have a PF of close to 0.0%.
%Regulation is calculated at a typical power factor (0.8 PF as I recall).



--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[waross]

This may save a little money:
Consider adding a series reactor rather than a transformer to limit the ASCC to below 14kA.
A reactor will be much less material than a transformer for the same %Z, but you have to check the price.
Reactor Specification
Part Number:
CRX0412AC
Centurion R Line and Load Reactors
3ph 412A 60Hz 600V 3.0%Z CU
Phase:
3
Rated Current*:
412A
Load Voltage:
600V
Termination (Std):
Mounting Pads
Frequency:
60Hz
Frame Style (NPL):
4S (Refer to Installation Sheet)
Termination Style (NPL):
5 (Refer to Installation Sheet)
Watts Loss (NPL):
435W
Approval:
CSA, UL, CE
% Impedance:
3.0%
Inductance:
0.07mH
Enclosure:
Enclosure Type:
Core & Coil
Outline Ref (NPL):
CR-4
Reactor Dimensions:
12.5in/318mm(H), 13.75in/350mm(W), 10.38in/264mm(D)
Reactor Mounting (W/D/Size):
9in/229mm(W), 7.3in/186mm(D), 0.44in/11.18mm X 1in/25.4mm(Slot)
Enclosure Kit:
No
Net Weight (NPL):
116 lb / 53 kg
Engineering Drawing (PDF):
EDB:CRX0412AC
Engineering Drawing (DXF):
DXF:CRX0412AC
Engineering Drawing (STEP):
STEP:CRX0412AC
E-Quotes Lite Configuration Link:
mailto:?subject=HPS CRX0412AC Product Information&body=

https://engprd01.hammondpowersolutions.com/EQv2-Lite/CRX0412AC

Brochure:
PDF
Frequently Asked Questions:
FAQ
Troubleshooting Guide:
Troubleshooting
Lead Time*:
Typically Built To Order
*Please contact HPS or Partner for current product lead-time
* Indicates User Specified Value
FinishCancelSave

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[cranky108]

It really is just a matter of impedance. Calculate the impedance you need, and specify a device with that impedance. Also note that most larger device have a manufactured tolerance, so add a little more impedance for that factor.
Transformer, reactor, or other wise it is just impedance you need.

 

[waross]

This may save even more money:
You haven't told us the ASCC of the source.
The supply conductors add impedance to the circuit and reduce the ASCC at the panel.
I have been able to use equipment with an ASCC rating a little below the ASCC of the source by showing that the impedance of the supply conductors would reduce the ASCC at the panel to below the rating of the panel.
I have also seen this taken one step further:
A minimum length of supply cable was specified to connect equipment.
The equipment in question was unit substations at 13,000 Volts.
The saving in cost of not having to go to higher ASCC rated unit subs was substantial.
All it took was a minimum of 100 feet of supply cable.
The cable was run past the subs and doubled back.
Your existing cables may be already limiting the ASCC at the panel to below 14 kA.
You may be able to solve your issues by adding some cable in series with the existing feed cables.

The red-neck solution:
More for interest than a suggested solution;
Wireless reactors;
The ASCC could be limited by passing the feeder conductors one or more times through window CTs.
Unfortunately, I cannot supply the calculations for this scheme.
At one time, wireless reactors were used to match impedances of paralleled transformers.
But that is from a text book that is over 100 years old.

My memory is writing checks that my body can't cash!

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[mparenteau]

Not ideal, but is there a possibility of a series rating application at the panelboard?

Mike