Incident Energy Calculations- Lack of Utility Data Issues....

[Brad45]

Hello to all,

I am in the process of doing some Arc Flash Risk Assessment (AFRA) work and, as I'm sure anyone who is familiar with these might expect, I am having issues getting the utility data from the providers. Does anyone have any idea if the utility company has any legal obligation to provide these specs to the client? I have letters of authorization to pursue the data on our clients behalf, but still am not having much luck inspiring the utility companies to be very forthcoming with this info.

Anyone know of any possible actions to expedite this process?

Also, to allow the AFRA to move forward without this data, I am running my incident energy calcs using infinite bus all the way down to around 2 MVA and just about everything in between with wide range of X/R for each MVA level. I run all these and save a scenario report for each and then filter them to show the worst case and use this for my data/ labels until actual data is provided. My thinking is that whatever the actual data is, I will be covered as the calculations are utilizing the worst case scenario, would you agree?

Thanks in advance to anyone who can offer any guidance or input,

Cheers-

 

[wbdvt]

I have had issues with the utility not wanting to provide the data and have had to cite IEEE 1584 and OSHA requirements to them. I have also had to explain why infinite bus values cannot be used.

The most extreme case was where the utility was not going to provide the info after all my explanations and I had to file a complaint with the state PUC. I had to educate the PUC customer advocate and the PUC engineer on arc flash and why the utility values were so important. After the PUC Deputy Director had a conference with the utility, the utility is providing the values.

 

[Jraef]

The big utility around here will, if you don't know to ask differently, provide you with a report that is based on infinite bus, without specifically saying so. They just give you a value. Then if you challenge them on it (and whine like all get-out) they will finally give you the real data. Basically, someone has to go find their records and check the validity, which means assigning an Engineer to it, but they don't want to spend the time if they don't have to. The result is often that people are installing much beefier equipment than necessary. That's no skin off of the nose of the utility, so they don't really care. I once had a project where, based on their report, the AFC came to 44kA, and the user had already bought MCCs rated for 42kA, it was going to cost a bundle to reduce it; either change the transformer or add a reactor (or add cable by wrapping it around the building in a trench, which is what I proposed). I looked again at the report they provided and figured out that the primary 12.47kV available fault current was way too high, which turned out to be because they had just defaulted to infinite bus on the 35kV feed. Got them to give us real numbers, turned out the AFC at our terminals was only 28kA...

 

[ptonsparky]

I haven’t asked the POCO South of me for a while, but they would only give the values verbally. Nothing written.

 

[Sahib]

One method I found in an Indian book to find the fault current is as follows and I am not sure whether it is acceptable here. Measure voltage V at service equipment location. Connect a resistance at service equipment location. Let resistance load current be I and change in voltage be dV. Then fault current If=V*I/dV. The derivation is as follows. ( dV/dV )*(V*I/V*I)=(dV/I)*(V*I)/(dV*V)=Z*(V*I)/(dV*V), where Z is circuit impedance. Then Z*(V*I)/(dV*V)=V*I/dV*(V/Z)=V*I/dV*If=1. Or If=V*I/dV.

 

[wbdvt]

One method I found in an Indian book to find the fault current is as follows and I am not sure whether it is acceptable here. Measure voltage V at service equipment location. Connect a resistance at service equipment location. Let resistance load current be I and change in voltage be dV. Then fault current If=V*I/dV. The derivation is as follows. ( dV/dV )*(V*I/V*I)=(dV/I)*(V*I)/(dV*V)=Z*(V*I)/(dV*V), where Z is circuit impedance. Then Z*(V*I)/(dV*V)=V*I/dV*(V/Z)=V*I/dV*If=1. Or If=V*I/dV.

I fail to see how that will equate to a utilities available fault current on the primary side. Here is a actual situation:
Utility available fault current:
3ph: 1.055kA, 3.08 X/R
SLG: 0.741kA 3.12 X/R
12.47kV Primary wye connected
Transformer: 3x167kVA, 2%Z
Secondary . 480/277 V wye connected

Please show me how your method arrives at the fault current.

 

[Ingenieur]

I fail to see how that will equate to a utilities available fault current on the primary side. Here is a actual situation:
Utility available fault current:
3ph: 1.055kA, 3.08 X/R
SLG: 0.741kA 3.12 X/R
12.47kV Primary wye connected
Transformer: 3x167kVA, 2%Z
Secondary . 480/277 V wye connected

Please show me how your method arrives at the fault current.

is the 3 ph fault
3 ph bolted to ground or L-L?

isn't the L-G > 2L-G ?
L-G : I = 3 I1 = 3 (Vf/(Z0+Z1 +Z2+3Zf))
L-L: I = -j sqrt3 I1 = -j sqrt3 (Vf/(Z0+Z1 +Z2+3Zf))

 

[Ingenieur]

One method I found in an Indian book to find the fault current is as follows and I am not sure whether it is acceptable here. Measure voltage V at service equipment location. Connect a resistance at service equipment location. Let resistance load current be I and change in voltage be dV. Then fault current If=V*I/dV. The derivation is as follows. ( dV/dV )*(V*I/V*I)=(dV/I)*(V*I)/(dV*V)=Z*(V*I)/(dV*V), where Z is circuit impedance. Then Z*(V*I)/(dV*V)=V*I/dV*(V/Z)=V*I/dV*If=1. Or If=V*I/dV.

basically the same method as here
http://forums.mikeholt.com/showthread.php?t=188923&page=2&p=1885011#post188501

If = V/Z (where Z = Vdrop/I)

it will give a good approximation of fault current based on the stiffness of the supply voltage

 

[Sahib]

I fail to see how that will equate to a utilities available fault current on the primary side. Here is a actual situation:
Utility available fault current:
3ph: 1.055kA, 3.08 X/R
SLG: 0.741kA 3.12 X/R
12.47kV Primary wye connected
Transformer: 3x167kVA, 2%Z
Secondary . 480/277 V wye connected

Please show me how your method arrives at the fault current.

The method involves measurement of voltage and current on the secondary side using a known resistance connected phase to neutral. The resulting voltage drop/current gives the circuit impedance. Then the no load voltage/circuit impedance gives SLG from which primary side SLG may be found out using transformer %Z. Similarly, 3ph fault current by using 3 equal resistances connected phase to phase.

 

[Sahib]

basically the same method as here
http://forums.mikeholt.com/showthread.php?t=188923&page=2&p=1885011#post188501

If = V/Z (where Z = Vdrop/I)

it will give a good approximation of fault current based on the stiffness of the supply voltage

The 'Vdrop' is actually vectorial but we use an arithmetical difference. I think by taking into account the power factor measured on the primary side of the transformer, the Vdrop may be made more accurate. Any supporting calculations? Thanks.

 

[wbdvt]

The method involves measurement of voltage and current on the secondary side using a known resistance connected phase to neutral. The resulting voltage drop/current gives the circuit impedance. Then the no load voltage/circuit impedance gives SLG from which primary side SLG may be found out using transformer %Z. Similarly, 3ph fault current by using 3 equal resistances connected phase to phase.

It sounds like you are determining basically the impedance of the transformer and then using that to get fault current which would be an infinite bus. Still not the same as getting the information from the utility.

As I stated before, please use my numbers and show the math.

 

[Sahib]

It sounds like you are determining basically the impedance of the transformer and then using that to get fault current which would be an infinite bus. Still not the same as getting the information from the utility.

You are correct. But when a resistance load is applied to transformer secondary, its actual no load voltage reduces though slightly. By using transformation ratio the reduction on primary side may be found out and same procedure may be used to find fault current on primary side.

As I stated before, please use my numbers and show the math.

To verify your numbers requires additional numbers ie voltage drop due to a known resistance load current.

 

[Ingenieur]

just some round numbers to illustrate the method
the utility data posted can't be used because it does not have the test data: pre load V, post load V, R load and load current

assume:
277/480 sec
test R = 10 Ohm
w/o load V measured 480
with load V measured 479
I measured ~48 A

V drop = 1 V
Z = 1/48 = 0.0208 Ohm
that is the total system drop, prim AND sec

so I fault = 480/0.0208 = 23 kA at the xfmr sec terminals

 

[wbdvt]

So it is not that far off from the infinite bus fault of ~30kA at the txf secondary terminals.

The other issue is who is going to allow this method on the secondary of their transformer as it seems the plant would have to be shutdown, utility in to open primary side so the test set up can be done de-energized. Then when energized what arc rated PPE are you wearing to do voltage measurements? 40 cal suit?

 

[Ingenieur]

So it is not that far off from the infinite bus fault of ~30kA at the txf secondary terminals.

The other issue is who is going to allow this method on the secondary of their transformer as it seems the plant would have to be shutdown, utility in to open primary side so the test set up can be done de-energized. Then when energized what arc rated PPE are you wearing to do voltage measurements? 40 cal suit?

no
the prim is closed in
yes, the only load on the sec is the test load

we have a test set-up
we do it at the main swgr
a 200 A cb
a 200 A fuse as back-up
contactor with remote switch
a 10 Ohm ngr
we use recording meters
clamp onto main lugs cb load side with jumper cables

we do
L-L
L-G