Phase to ground fault current

To get the real secondary fault current, you really need to know the available fault current on the primary.
You can do a worst case "infinite bus" calculation for the secondary fault current.
First find the secondary full load current 1500000/480/√3 = 1804 amps
Divide that by the percentage impedance in decimal form
1804/.0575 = 31,374 amps available fault current at the secondary terminals of the transformer, but in the real world, the actual fault current will be less, because the primary current is limited, not infinite.
 
don_resqcapt19 said:
To get the real secondary fault current, you really need to know the available fault current on the primary.
You can do a worst case "infinite bus" calculation for the secondary fault current.
First find the secondary full load current 1500000/480/√3 = 1804 amps
Divide that by the percentage impedance in decimal form
1804/.0575 = 31,374 amps available fault current at the secondary terminals of the transformer, but in the real world, the actual fault current will be less, because the primary current is limited, not infinite.
Click to expand...

What about phase to ground fault currents?


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I don't remember and am not going to look it up.
There are online calculators and Bussmann has a free app called FC².
 
hhsting said:
Not in some cases. Therefore the interest


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Click to expand...
After secondary conductor lengths of 50' or so the L-L will exceed the L-G.
Usually the only time L-G is considered is for close coupled equipment, like unit substations.

My experience is L-G currents are ignored as often as X/R ratios are.
 
hhsting said:
Not in some cases. Therefore the interest
Click to expand...
Excerpt from IEEE Violet Book, Chapter 2. (Diagrams omitted)

Bolted line-to-line faults, are more common than three-phase faults and
have fault currents that are approximately 87% of the three-phase bolted fault current.
This type of fault is not balanced within the three phases and its fault current is seldom
calculated for equipment ratings because it does not provide the maximum fault current
magnitude. The line-to-line current can be calculated by multiplying the three-phase value
by 0.866, when the impedance Z1 = Z2. Special symmetrical component calculating techniques
are not required for this condition.

Line-to-line-to-ground faults, are typically line-to-ground faults that have
escalated to include a second phase conductor. This is an unbalanced fault. The magnitudes
of double line-to-ground fault currents are usually greater than those of line-to-line
faults, but are less than those of three-phase faults. Calculation of double line-to-ground
fault currents requires the use of symmetrical components analysis. The impedance of the
ground return path will affect the result, and should be obtained if possible.

Line-to-ground faults, are the most common type of faults and are usually
the least disturbing to the system. The current in the faulted phase can range from near
zero to a value slightly greater than the bolted three-phase fault current. The line-to ground
fault current magnitude is determined by the method in which the system is
grounded and the impedance of the ground return path of the fault current. Calculation of
the exact line-to-ground fault current magnitudes requires the special calculating
techniques of symmetrical components. However, close approximations can be made
knowing the method of system grounding used. On ungrounded distribution systems, the
line-to-ground fault currents are near zero. Line-to-ground fault current magnitudes in
distribution systems with resistance grounded system neutrals can be estimated by
dividing the system line-to-neutral system voltage by the total value of the system ground to-
neutral resistance. Line-to-ground fault current magnitudes in distribution systems with
a solidly grounded system will be approximately equal to the three-phase fault current
magnitudes.
_________________________________________________________________________________________________________
Are you requesting a SCC calc that provides ALL of these fault types SCC values in a plan review?
 
hhsting said:
What is phase to phase and phase to ground fault current for 1500kva transformer 5.75% impedance 480/277V three phase system?
Click to expand...
The fault current calculator on Mike Holt's main site under free stuff will calculate this - it will be assuming infinite line side capacity, just enter the transformer size, impedance and secondary voltage.
 
Assuming this is a delta-wye transformer, a line to ground fault at the transformer will have fault current that is about 115% of the calculated three-phase bolted fault current at the transformer. However, as noted, the fault current will decrease pretty quickly as the distance to the fault increases.
For a wye-wye transformer (commonly used by utilities), it will be less than this - you need to know the utility zero sequence impedance.
 
D. Castor said:
Assuming this is a delta-wye transformer, a line to ground fault at the transformer will have fault current that is about 115% of the calculated three-phase bolted fault current at the transformer.
Click to expand...
Is that 115% as in 2 / sqrt(3)? I.e. the imepedance of the L-G fault at the transformer will be 1/2 as much (half as many transformer coils are involved) but the L-G voltage is 1/sqrt(3) times the L-L voltage?

Cheers, Wayne
 
D. Castor said:
Assuming this is a delta-wye transformer, a line to ground fault at the transformer will have fault current that is about 115% of the calculated three-phase bolted fault current at the transformer. However, as noted, the fault current will decrease pretty quickly as the distance to the fault increases.
For a wye-wye transformer (commonly used by utilities), it will be less than this - you need to know the utility zero sequence impedance.
Click to expand...
Is the IEEE document not stating it can be ignored?
 
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