Infinite Bus Calculation

[Brian LeBlanc 913]

Need help with an infinite bus calculation for PEPCO networked utility transformers. We have (2)1000KVA transformers (208V, 3phase secondary) connected to a common bus. I need the fault current at the bus.

 

[Elect117]

Is that all you have? You will need the available fault current from the utility at their point of connection. You also need the impedance of the transformers and a configuration. As in, are the transformer parallel fed or series? Is one a redundancy to the other?

 

[ron]

For the infinite bus calc, take the full load amps of each transformer and dive it by the transformer impedance.

For example, if the transformer impedance from the nameplate was 5.75%, it would be 5552/.0575 = 96555A

 

[jim dungar]

Is that all you have? You will need the available fault current from the utility at their point of connection. You also need the impedance of the transformers and a configuration. As in, are the transformer parallel fed or series? Is one a redundancy to the other?

by definition, you do not need the utility fault current if you are performing an Infinite bus short circuit calculation.
You do need to know the transformer impedance, but guessing low just make the infinite bus value higher. So if you want a SWAG worst case value, use 5.0% for large transformers, like those over 500kVA, and 1.5% for smaller units especially pole top units.

 

[Brian LeBlanc 913]

Is that all you have? You will need the available fault current from the utility at their point of connection. You also need the impedance of the transformers and a configuration. As in, are the transformer parallel fed or series? Is one a redundancy to the other?

Yeah that's all we have at this time, the transformers are not redundant. The contractor want to order gear but the utility fault current letter is not ready.

 

[ggunn]

by definition, you do not need the utility fault current if you are performing an Infinite bus short circuit calculation.

Right. The infinite bus model assumes infinite available fault current (AFC) on the primary side of the transformer. I have made the calculation both ways when I had the numbers from the utility and it made little if any difference to the AFC on the secondary. Maybe if the primary AFC were significantly lower if the xfmr is out on the end of a long MV supply line it would make a difference, but the infinite bus model will always be conservative.

 

[don_resqcapt19]

Right. The infinite bus model assumes infinite available fault current (AFC) on the primary side of the transformer. I have made the calculation both ways when I had the numbers from the utility and it made little if any difference to the AFC on the secondary. Maybe if the primary AFC were significantly lower if the xfmr is out on the end of a long MV supply line it would make a difference, but the infinite bus model will always be conservative.

For equipment selection, yes, but it can give you too low of a incident energy for arc flash calculations.

 

[ggunn]

For equipment selection, yes, but it can give you too low of a incident energy for arc flash calculations.

I have never done that analysis; are you saying that a lower AFC on the primary side will result in a higher incident energy for arc flash calculations? That seems counterintuitive to me.

 

[jim dungar]

I have never done that analysis; are you saying that a lower AFC on the primary side will result in a higher incident energy for arc flash calculations? That seems counterintuitive to me.

Incident energy is dependent on the amount of arcing current and the amount of time it exists.
Because most OCPD have an inverse time-current relationship, lower values of fault current can mean a protective device is slower to clear the fault. The longer arcing time means more incident energy is being released.

 

[topgone]

Incident energy is dependent on the amount of arcing current and the amount of time it exists.
Because most OCPD have an inverse time-current relationship, lower values of fault current can mean a protective device is slower to clear the fault. The longer arcing time means more incident energy is being released.

But isn't the arc duration the first figure one has to know prior to doing the arc flash analysis?

 

[jim dungar]

But isn't the arc duration the first figure one has to know prior to doing the arc flash analysis?

Yes.
The magnitude of the 3-phase bolted fault as well as the clearing time of the protective device are both some of the inputs to the complex equation whose result is incident energy. In general lower fault current means longer clearing time and thus higher incident energy. This is one reason branch circuits can have higher cal/cm^2 than their feeders and why we calculate down to every piece of equipment.

 

[kingpb]

Typically 208V does not have enough energy to maintain an arc. You are not going to tell the utility what transformer impedance to use, so assume worst case and use 1.5% as previously mentioned.

If, as stated, both transformers are connected in parallel to the same bus, then you would have 2 x the fault current as a single transformer. 1000KVA/0.015 = 66.7MVAsc (each), So, 2 x 66.7MVA = 133.3MVAsc; @ 208V 3ph that would be 370KAIC.

At that impedance, you have a problem. I can say I am not aware of 6000A gear @ 208V to accommodate both these feeders operating in parallel. Are you thinking of a sectionalizer? Also, I don't think the utility is going to allow them to operate in parallel.

Working backward, I think 200KAIC is your limit, so assuming that; you need transformers with an impedance of at least 2.8% or greater, if operating in parallel.

 

[jim dungar]

.....@ 208V 3ph that would be 370KAIC.

technically it would just be kA available.
AIC is a term that should only be used with protective devices as it relates to tested, not calculated, values.

The tested AIC of a device must exceed the available short circuit A.

 

[ggunn]

At that impedance, you have a problem. I can say I am not aware of 6000A gear @ 208V to accommodate both these feeders operating in parallel.

Wouldn't they be service conductors instead of feeders?

 

[kingpb]

Wouldn't they be service conductors instead of feeders?

From a code perspective that might be the case.

 

[Julius Right]

In my opinion, the standard impedance for 1000 kVA 208 V it is 5.75% [+/- 7.5%]

 

[kingpb]

In my opinion, the standard impedance for 1000 kVA 208 V it is 5.75% [+/- 7.5%]

That would be the IEEE std impedance if you did not specify something else when purchasing. You can always pay more and get a lower impedance, or higher impedance. Utilities typically do not buy std impedance transformers.

 

[eric stromberg]

There are about three different things going on here. Let's take them one at a time. The secondary current for a 1000 kVA transformer at 208/120 is 2778 amps. A %Z of 5.75% gives us 48,309 Amps. Two in parallel would give us 96.6kA.
incident energy? We don't often have the fuse size that the utility uses, and even if we do, the size is such that the incident energy is determined by the 2 second rule. If the utility has a tightly sized fuse, the clearing time might be a little less than the 2 second value. Higher fault currents can cause the fuse to clear quicker and thereby result in a lower incident energy. Utilities, often, are reticent to give a value at their transformer because there is typically switching in their system to allow the transformer to be fed from a different direction. One of their subs might have one value and another sub will have another value. Sometimes, a utility tie might be closed. You can experiment by running the model with different values for the utility. 300 MVA would be a fairly stiff system. You can run the model at 300, 200, 100, and see what you get.

 

[kingpb]

There are about three different things going on here. Let's take them one at a time. The secondary current for a 1000 kVA transformer at 208/120 is 2778 amps. A %Z of 5.75% gives us 48,309 Amps. Two in parallel would give us 96.6kA.
incident energy? We don't often have the fuse size that the utility uses, and even if we do, the size is such that the incident energy is determined by the 2 second rule. If the utility has a tightly sized fuse, the clearing time might be a little less than the 2 second value. Higher fault currents can cause the fuse to clear quicker and thereby result in a lower incident energy. Utilities, often, are reticent to give a value at their transformer because there is typically switching in their system to allow the transformer to be fed from a different direction. One of their subs might have one value and another sub will have another value. Sometimes, a utility tie might be closed. You can experiment by running the model with different values for the utility. 300 MVA would be a fairly stiff system. You can run the model at 300, 200, 100, and see what you get.

If you don't have the nameplate you must assume the -7.5% which would make the impedance 5.35%, and the available fault current would be slightly higher at 51.9kA.