Transformer Available fault Current

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augie47

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Tennessee
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State Electrical Inspector (Retired)
Charts have me doubting my math skills.
Would one our more learned members step me thru the formula to determine the available fault current for a 150 kva 480Y/277 transformer, 3.1% impedance and assuming infinite primary.
My math gave me 5900 but a chart I have shows 11,000. One or both is/are wrong. :D
 

iceworm

Curmudgeon still using printed IEEE Color Books
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North of the 65 parallel
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EE (Field - as little design as possible)
Charts have me doubting my math skills.
Would one our more learned members step me thru the formula to determine the available fault current for a 150 kva 480Y/277 transformer, 3.1% impedance and assuming infinite primary.
My math gave me 5900 but a chart I have shows 11,000. One or both is/are wrong. :D

here you go augie:

3ph bolted fault (single phase fault is different)

fla = (150000/480)/(sqrt(3))

Isc = (FLA/.031)

ice
 

jim dungar

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Wisconsin
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PE (Retired) - Power Systems
Charts have me doubting my math skills.
Would one our more learned members step me thru the formula to determine the available fault current for a 150 kva 480Y/277 transformer, 3.1% impedance and assuming infinite primary.
My math gave me 5900 but a chart I have shows 11,000. One or both is/are wrong. :D

I'll give it a try.

FLA/%Z = (150/.480/1.732)/.031 = 180/.031 = 5820A
 
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GearMan

Member
Location
WI
Not sure if your next step is to establish the short circuit rating of secondary equipment, if so, don't forget about motor contribution. It can add up to 4x the FLA of the xfmr with heavy motor loads.
 

hurk27

Senior Member
I'm coming up with 6008 amps using a 100/3.1% multiplier but without knowing the primary voltage its a guess, I think the chart is showing a worst case of all three phases faulted together which would give you about 10406 amps rounded up to 11ka

Here is a great PDF from Bussman on calculating fault current:

Short Circuit Current Calculations

Just substitute your info to the ones in the formulas
 

augie47

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Tennessee
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State Electrical Inspector (Retired)
Not sure if your next step is to establish the short circuit rating of secondary equipment, if so, don't forget about motor contribution. It can add up to 4x the FLA of the xfmr with heavy motor loads.

Thanks, but loads are 100% lighting.

I'm coming up with 6008 amps using a 100/3.1% multiplier but without knowing the primary voltage its a guess, I think the chart is showing a worst case of all three phases faulted together which would give you about 10406 amps rounded up to 11ka

Here is a great PDF from Bussman on calculating fault current:

Short Circuit Current Calculations

Just substitute your info to the ones in the formulas

Thanks. I think that might be thr origin of my chart but the impedance was not shown.

My concern is some lighting contactors with 5k withstand.... but the conductors should take care of the problem with that
low a number at the transformer.
 

Julius Right

Senior Member
Occupation
Electrical Engineer Power Station Physical Design Retired
I think due to the fact that transformation ratio 480/277 it is
the same as 480/sqrt(3)[phase-to-neutral primary voltage] the misunderstanding is .Usually the transformer symbol will include the secondary connection too[let's say: 480Y/277D].
The short-circuit impedance on secondary winding could be 3.1/100*277^2/150000=0.015857 ohm and so Isc=277/0.015857/sqrt(3)= 10086 A
Admitting 9% plus for voltage, frequency and other factors one can get 11000 A rated [for a certain time interval].
 

topgone

Senior Member
I think due to the fact that transformation ratio 480/277 it is
the same as 480/sqrt(3)[phase-to-neutral primary voltage] the misunderstanding is .Usually the transformer symbol will include the secondary connection too[let's say: 480Y/277D].
The short-circuit impedance on secondary winding could be 3.1/100*277^2/150000=0.015857 ohm and so Isc=277/0.015857/sqrt(3)= 10086 A
Admitting 9% plus for voltage, frequency and other factors one can get 11000 A rated [for a certain time interval].

There you go. Actually, it's 10% possible overvoltage, hence 10,086 X 1.10 = 11,094 approx = 11,000A.
 

augie47

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Location
Tennessee
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State Electrical Inspector (Retired)
Does your chart show both line to line as well as line to neutral fault current?

It's now my trashed chart but as I recall it only showed L-L

Post 9 & 10 sure confused my issue. :D

I was happy with Jim's 5820.. now I,m back to 11,000 ????

Not sure if I need to clarify... it's a 12kv pad with a 480Y/277 secondary.

Julius or Topgone, think one of you guys can talk down to my level ???
 
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iceworm

Curmudgeon still using printed IEEE Color Books
Location
North of the 65 parallel
Occupation
EE (Field - as little design as possible)
... I was happy with Jim's 5820. ...
For an infinite primary, bolted 3ph fault, I'm happy with Jim's answer as well.

QUOTE=augie47;1461560]...Post 9 & 10 sure confused my issue.

... Julius or Topgone, think one of you guys can talk down to my level ???[/QUOTE]

I think they are looking at single phase to neutral/ground fault current. But I'm hoping they get down to my level as well.:dunce:

I still don't know what a 480Y/277D winding looks like.:ashamed1:

ice
 

jahounou

Member
Location
Washington DC
Senior Electrical Engineer, PE

Senior Electrical Engineer, PE

Your Math is correct, the fault current is in fact approximately 5900A.

Jean-Jacques Ahounou, PE
 

mivey

Senior Member
It's now my trashed chart but as I recall it only showed L-L

Post 9 & 10 sure confused my issue. :D

I was happy with Jim's 5820.. now I,m back to 11,000 ????

Not sure if I need to clarify... it's a 12kv pad with a 480Y/277 secondary.

Julius or Topgone, think one of you guys can talk down to my level ???
Fault formulas:
I_3ph = V_lg / (Z1+Zf)
I_LG = V_lg / ( (Z0+Z1+Z2)/3 + Zf )
I_LL = sqrt(3) * V_lg / ( (Z1+Z2) + Zf )

if we assume the fault impedance (Zf) is zero, the zero sequence is blocked, and that the positive and negative sequence impedance is the same (Z1 = Z2), then we have

I_3ph = V_lg / Z1
I_LG = V_lg / ( 2*Z1/3 ) = 1.5 * V_lg / Z1 = 1.5 * I_3ph
I_LL = sqrt(3) * V_lg / ( 2*Z1 ) = 0.866 * I_3ph

Jim's calculation was the standard 3-phase fault calculation. When we have a delta-wye transformer, the source zero sequence (Z0) gets blocked and the line-to-ground fault can actually be higher than the three-phase fault near the secondary terminals. I_LG drops back down below I_3ph as we move down the feeder and pick up Z0 along the way (usually larger than Z1).

The end result is that it looks like your table may have been LG fault values or it could just be as simple as the impedance was lower for the given table. With ranges from 1.8% to 5% possible, we use 1.8% for 150 kVA pads for worst-case fault calculations. Using 1.8% gives 10 kA. A 10% over-voltage with 1.8% gives 11 kA .
 

mivey

Senior Member
My concern is some lighting contactors with 5k withstand.... but the conductors should take care of the problem with that low a number at the transformer.
I would not bet on that. How comfortable are you that the 3.1% impedance will not change? If it is a utility transformer, it is possible that the impedance could be lower if they change the transformer. That is why we give the worst case fault to the customer.

That said, the 5820A is a good number for a 3-phase fault (6402A for a 10% over-voltage). But you really should consider the line-ground fault if you have a delta-wye transformer. I suspect you have a wye-wye, and the zero-sequence source impedance should make the 3-phase fault the worst case but you really should check since you theoretically could have up to 150% of the 5820A (or 6402A) value.
 

Julius Right

Senior Member
Occupation
Electrical Engineer Power Station Physical Design Retired
If the transformer is 150 KVA 12/0.48 KV 3.1% vsc then indeed Isc [at 0.48 kV side] will be 5.82 KA[three-phase metallic].But ?in extreme situation- as per IEC 60909 considering Xo/X=0.1[for Y/Yz [zigzag grounded secondary ]the phase-to-ground short-circuit :
Ik1"=1.1*sqrt(3)*VL_L/(Z1+Z2+Zo) Z1=Z2=3.1/100*.48^2/0.15=0.047616 ohm
Ik1"=1.1*sqrt(3)*0.48/(0.047616*2.1)=9.146 KA
 

augie47

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Location
Tennessee
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State Electrical Inspector (Retired)
Thanks.
The job has a design engineer so I have the E/C obtaining documentation. I often try to come up with a ballpark figure to look for potential problems before having someone verify. That was the case here.
My concern was the panel being adjacent to the transformer and the contactors being a different manufacturer than the OCP.
Since the inquiry, they have changed to a "same manufacturer" contactor and have documentation of 65k with the upstream breaker they are using.
 
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