Available fault Current - Engineer Question

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Davebones

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We have a 1500KVA ,14,4KV to 480 V main transformer . The utility says we have 33000 amps of fault current there . The transformer feeds a 4000 amp switchgear about 150' away . My question is roughly how much fault current is available at the main switchgear . I understand to actually determine the fault current you have to calculate wire sizes and distance etc... Roughly what kind of drop in AFC you expect to see in a situation like this...
 

zog

Senior Member
Location
Charlotte, NC
We have a 1500KVA ,14,4KV to 480 V main transformer . The utility says we have 33000 amps of fault current there . The transformer feeds a 4000 amp switchgear about 150' away . My question is roughly how much fault current is available at the main switchgear . I understand to actually determine the fault current you have to calculate wire sizes and distance etc... Roughly what kind of drop in AFC you expect to see in a situation like this...

You already have the first 3 steps here, so use your 33kA and start at step 4

Basic Point-to-Point Calculation Procedure

The application of the point-to-point method permits the determination of available short-circuit currents with a reasonable degree of accuracy at various points for either 3? or 1? electrical distribution systems. This method assumes unlimited primary short-circuit current.

Step 1. Determine the transformer full load amperes from either the nameplate or the following formulas:

3? Transformer IFL = KVA x 1000
EL-L x 1.732

1? Transformer IFL = KVA x 1000
EL-L

Step 2. Find the transformer multiplier.

Multiplier = 100
%ZTransformer

Transformer impedance (Z) helps to determine what the short circuit current will be at the transformer secondary. Transformer impedance is determined as follows: The transformer secondary is short-circuited; voltage is applied to the primary, which causes full load current to flow in the secondary. This applied voltage that cause full rated load current to flow through the secondary windings, divided by the rated primary voltage, is the impedance of the transformer.


Step 3. Determine the transformer let-thru short-circuit current.

ISC = IFL x Multiplier

Motor short-circuit contribution, if significant, may be added to the transformer secondary short-circuit current value as determined in Step 3. Proceed with this adjusted figure through Steps 4, 5 and 6. A practical estimate of motor short-circuit contribution is to multiply the total motor current in amperes by 4.


Step 4. Calculate the "f" factor.

3? Faults f = (1.732 x L x I)/(C x EL-L)


Where:
L = Length (feet) of circuit to the fault.
C = Constant from Table (attached). For parallel runs, multiply C values by the number of conductors per phase.
I = Available short-circuit current in amperes at beginning of circuit.

Step 5. Calculate "M" (multiplier).

M = 1/(1 + f)

Step 6. Calculate the available short-circuit symmetrical RMS current at the point of fault.

ISC sym RMS = ISC x M
 
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