avail. short circuit calc

[Toros]

Hi
the San Pedro inspector told to my electrician to provide AIC rating on existing recently modified panel which is fed through a transformer (150 kva, 480v to 120/208v, 3ph. 4w)
we do not know what is avail. S.C at the Service unfortunately
what do you think?????

 

[augie47]

Am I correct in that he is wanting you to show the SCA on the secondary of a "in-house" 150 kva transformer ?
As a first step, I would just assume infinite primary and calculate based on the 150 kva impedance. That would be the worst case scenario.
According tom the charts I have, in the worse case you would be looking at around 43k.

 

[Jraef]

Hi
the San Pedro inspector told to my electrician to provide AIC rating on existing recently modified panel which is fed through a transformer (150 kva, 480v to 120/208v, 3ph. 4w)
we do not know what is avail. S.C at the Service unfortunately
what do you think?????

Do you not know because the facility generates it's own power, or because your facility buys bulk power and owns its own substations, and you cannot contact your on-site power engineer? Or is it because you are unaware of the fact that SCE can TELL you the AFC at the primary of the service drop by simply calling them and asking?

 

[kingpb]

Am I correct in that he is wanting you to show the SCA on the secondary of a "in-house" 150 kva transformer ?
As a first step, I would just assume infinite primary and calculate based on the 150 kva impedance. That would be the worst case scenario.
According tom the charts I have, in the worse case you would be looking at around 43k.

Not sure where you are getting the number from but that would be an unattainable impedance for the transformer.

43kA @ 208V is 15.5MVA_sc

150KVA / Z = 15.5MVA, solving for Z would be 0.0097 or 0.97%.

 

[lielec11]

Not sure where you are getting the number from but that would be an unattainable impedance for the transformer.

43kA @ 208V is 15.5MVA_sc

150KVA / Z = 15.5MVA, solving for Z would be 0.0097 or 0.97%.

I agree, for a copper wound transformer that size you're looking at around 4% worst case. At that level, the SCA is only 10.4kA.

 

[augie47]

I didn't do the math prior to posting. The number, however, comes from Bussmann's 2008 SPD PG 244

Voltage Full % Short
and Load Impedance?? Circuit
Phase kVA Amps (Nameplate) Amps?
25 104 1.5 12175
37.5 156 1.5 18018
120/240 50 208 1.5 23706
1 ph.* 75 313 1.5 34639
100 417 1.6 42472
167 696 1.6 66644
45 125 1.0 13879
75 208 1.0 23132
112.5 312 1.11 31259
150 416 1.07 43237
120/208 225 625 1.12 61960
3 ph.** 300 833 1.11 83357
500 1388 1.24 124364
750 2082 3.50 66091
1000 2776 3.50 88121
1500 4164 3.50 132181
2000 5552 4.00 154211
2500 6940 4.00 192764
75 90 1.00 10035
112.5 135 1.00 15053
150 181 1.20 16726
225 271 1.20 25088
300 361 1.20 33451
277/480 500 602 1.30 51463
3 ph.** 750 903 3.50 28672
1000 1204 3.50 38230
1500 1806 3.50 57345
2000 2408 4.00 66902
2500 3011 4.00 83628
Single-phase values are L-N values at transformer terminals. These figures

 

[jim dungar]

1.07% impedance might be a 'design value' for utility supplied oil-filled 'pole-top' transformers. Although, most of the utility data I've seen shows 1.5-2% as more likely.

 

[augie47]

1.07% impedance might be a 'design value' for utility supplied oil-filled 'pole-top' transformers. Although, most of the utility data I've seen shows 1.5-2% as more likely.

That makes sense and that is probably what they are referencing. I just had the SPD handy so I took a look. I knew the number was high compared to any I had seen.
A case where "no good deed goes unpunished" .......................

then again, the OP did not state what type transformer or if it were utility owned.......

 

[kwired]

I didn't do the math prior to posting. The number, however, comes from Bussmann's 2008 SPD PG 244

Voltage Full % Short
and Load Impedance?? Circuit
Phase kVA Amps (Nameplate) Amps?
25 104 1.5 12175
37.5 156 1.5 18018
120/240 50 208 1.5 23706
1 ph.* 75 313 1.5 34639
100 417 1.6 42472
167 696 1.6 66644
45 125 1.0 13879
75 208 1.0 23132
112.5 312 1.11 31259
150 416 1.07 43237
120/208 225 625 1.12 61960
3 ph.** 300 833 1.11 83357
500 1388 1.24 124364
750 2082 3.50 66091
1000 2776 3.50 88121
1500 4164 3.50 132181
2000 5552 4.00 154211
2500 6940 4.00 192764
75 90 1.00 10035
112.5 135 1.00 15053
150 181 1.20 16726
225 271 1.20 25088
300 361 1.20 33451
277/480 500 602 1.30 51463
3 ph.** 750 903 3.50 28672
1000 1204 3.50 38230
1500 1806 3.50 57345
2000 2408 4.00 66902
2500 3011 4.00 83628
Single-phase values are L-N values at transformer terminals. These figures

Sorry, but that just looks to me like some math geek drank too much Red Bull or Monster:lol:

 

[topgone]

What @augie47 said. Assume infinite source at the point of connection. Read out the transformer percent impedance of your 150 kVA transformer and divide the full-load amps of the transformer by the transformer impedance--> you now have the available short-circuit current available at your trafo secondary. IMO, it's around 23kA @ 208V, assuming a %Z = 1.8.
Hope I understand what your problem was.

 

[jim dungar]

Sorry, but that just looks to me like some math geek drank too much Red Bull or Monster:lol:

What does that say about me, if I understood it?:huh:

 

[kwired]

What does that say about me, if I understood it?:huh:

Are you waiting for a reply before you tell us if you did understand it?

 

[Fulthrotl]

Hi
the San Pedro inspector told to my electrician to provide AIC rating on existing recently modified panel which is fed through a transformer (150 kva, 480v to 120/208v, 3ph. 4w)
we do not know what is avail. S.C at the Service unfortunately
what do you think?????

is this a subpanel fed from your transformer?

san pedro is DWP. port of long beach is SCE. terminal island is mostly DWP,
with a bit of SCE.

DWP can give you a fault current letter of the AIC available at the main,
and long beach, terminal island and the port won't inspect a job without
said letter on site.

the last letter i got from SCE shows 1 set of 700 mcm aluminum feeders
going 90 feet from a 300 kva poco xfmr fed off a 12 KV primary, and the
short circuit current was calculated at 32,804 amps (symmetrical).

in the absence of the letter, i was told i had to series rate everything for
200K aic.