are flash study

Status
Not open for further replies.

steve66

Senior Member
Location
Illinois
Occupation
Engineer
Doing an arc flash study. The utility services the site with 34.5KV, and they step it down to 4160V with an on-site substation.

I got the following info. from the utility:

The fault currents at the 34.5KV Tap are as follows:
3LG = 15475A
2LG = 7908A
LG = 10474A
LL = 13400A
34.5KV BUS
THEVENIN IMPEDANCE (OHM)
Z(+) = 0.17454+j1.30862 Z(-) = 0.17506+j1.30891 Z(0) = 0.75464+j3.12885

Fault currents at the 4.36KV Tap are as follows:
3LG = 6761A
2LG = 7018A
LG = 6887A
LL = 5855A
4.36KV BUS
THEVENIN IMPEDANCE (OHM)
Z(+ ) = 0.00787+j0.38293 Z(-) = 0.00788+j0.38293 Z(0) = 0.00507+j0.36191

I'm trying to figure out which numbers to plug into SKM, and trying to make sure all the numbers make sense. On the 34.5KV side, I converted the thevenin impedance numbers to per unit numbers using:
Zpu = Z/Z base
and;
Zbase = Vbase^2/MVA base

Then I entered the 34.5 KV short circuit currents, and the Per unit Z's were close to my calculated numbers (close, but not exact). Like this:

Utility Data.jpg Utility zoom.png


But then I tried the same thing on the 4160 V side by making a 4160V utility source. At the lower voltage, the thevenin impedances don't seem to match the fault currents. I tried going back and forth, and even using a different voltage as the base voltage, but nothing seems to match.

Did the utility make a mistake on the numbers they provided, or am I doing something wrong?
 
Last edited:

jim dungar

Moderator
Staff member
Location
Wisconsin
Occupation
PE (Retired) - Power Systems
Utilities are known for providing too little or too much information.

Basically, if they give transformer secondary information I use that in my software. Their secondary information may not be dependent on the transformer that is actually installed (which is why you cannot get the numbers to work). These values should hold even if the transformer gets changed.

I do look at the primary side values for a sanity check. I have been known to run multiple scenarios just to decide which source configurations to use in my study to find the worst case AF incident energy.

I am lazy, I use the Thevinen values and let the software make the conversions.
 

steve66

Senior Member
Location
Illinois
Occupation
Engineer
Utilities are known for providing too little or too much information.

Basically, if they give transformer secondary information I use that in my software. Their secondary information may not be dependent on the transformer that is actually installed (which is why you cannot get the numbers to work). These values should hold even if the transformer gets changed.

I do look at the primary side values for a sanity check. I have been known to run multiple scenarios just to decide which source configurations to use in my study to find the worst case AF incident energy.

I am lazy, I use the Thevinen values and let the software make the conversions.

That makes sense. So the 34.5KV transformer might be pretty close (or the same) as the installed transformer, while the 4160V transformer may not be the same as the specs.

For being lazy, I like entering the 3LG and LG values if they are available.
 

jeremy.zinkofsky

Senior Member
Location
nj
Doing an arc flash study. The utility services the site with 34.5KV, and they step it down to 4160V with an on-site substation.

I got the following info. from the utility:



I'm trying to figure out which numbers to plug into SKM, and trying to make sure all the numbers make sense. On the 34.5KV side, I converted the thevenin impedance numbers to per unit numbers using:
Zpu = Z/Z base
and;
Zbase = Vbase^2/MVA base

Then I entered the 34.5 KV short circuit currents, and the Per unit Z's were close to my calculated numbers (close, but not exact). Like this:

View attachment 15639 View attachment 15640


But then I tried the same thing on the 4160 V side by making a 4160V utility source. At the lower voltage, the thevenin impedances don't seem to match the fault currents. I tried going back and forth, and even using a different voltage as the base voltage, but nothing seems to match.

Did the utility make a mistake on the numbers they provided, or am I doing something wrong?


How are you modeling the service XFMR? If you have placed the service XFMR in your model with the correct Voltage, VA, and %Z then you only have to plug in the Utility data for the primary side. If you are using the secondary side data, then I would omit modeling the XFMR and only show a Utility connection point. Also, Do you need to be exact? Quite often engineers are only interested in the "worst case" scenario. The most conservative results come from making the Utility an infinite bus and inputting ridiculously high numbers for the Utility contribution. This will create a Full-Load condition on the service XFMR and yield the maximum fault current that is possible.
 

steve66

Senior Member
Location
Illinois
Occupation
Engineer
How are you modeling the service XFMR? If you have placed the service XFMR in your model with the correct Voltage, VA, and %Z then you only have to plug in the Utility data for the primary side. If you are using the secondary side data, then I would omit modeling the XFMR and only show a Utility connection point.

Yes, but I was trying to do a sanity check, and see if the data the utility gave me on the line side of the transformer was reasonably close to matching what I should get on the load side. I was also trying to see if the thevenin impedances matched the fault currents.

Also, Do you need to be exact? Quite often engineers are only interested in the "worst case" scenario. The most conservative results come from making the Utility an infinite bus and inputting ridiculously high numbers for the Utility contribution. This will create a Full-Load condition on the service XFMR and yield the maximum fault current that is possible.

That works for fault or short circuit studies, but you don't want to use infinite utility capacity for an arc flash study. Sometimes the lower fault currents trip breaker much slower, so the energy released is actually higher. So for arc flash, you want to start with a reasonable estimate of what the utility or source can provide.
 
Status
Not open for further replies.
Top