# Reactor example from Graffeo



## kduff70 (Aug 4, 2015)

[SIZE=10.5pt]I was going through Graffeo ex .26 on page 57-58 I don’t quit understand the equation he uses to solve for the fault current why is it 1 over the impedance? And is the 35,555A the current over the eight circuit break. And will it be 4444 A across each of the circuit breaker. I have attached the problem[/SIZE]

SKMBT_42015080407560.pdf


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## FeederFault (Aug 4, 2015)

Another way of doing it (which is the way I was taught) is to find the secondary bolted fault current using the equation I_Fault = MVA / (Sqrt(3)*VLL*Z%)

In this example the secondary bolted fault current would = (69*10^6) / (Sqrt(3)*12,470*.09) = 35,496 Amps. This is the fault current at the secondary terminals of the transformer.

The nominal current rating of the transformer is always a function of it's MVA rating and nominal voltage, so in this case it would = (69*10^6) / (Sqrt(3) * 12,470) = 3,195 Amps

From there they are finding the available secondary fault current using a combination of the above equation, which would include using the reciprocal of the impedance.


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## kduff70 (Aug 6, 2015)

FeederFault,

Thank you for clearing the equation part up for me. So the 35,555Amps is what all 8 circuit break would see therfore exceeding the 5000Amps Interrupting capacity. The reactors would not provide enough protection?


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## ZcoreX29 (Aug 6, 2015)

Only the circuit breaker that was upstream of that fault would see that 35kA. The others would not.


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## JB66money (Aug 7, 2015)

KDuff,

The reason that they used 1/.09, is because that is the per-unit value of the current. The term 1 is the per-unit voltage because it is assumed that the bus voltage and the base voltage are the same and since V(per-unit) = V(actual) / V(base) =1 p.u.. That is how 1 came about. and since the transformer impedance is given as 9%, it was converted to per-unit which is .09 per-unit. Therefore the per-unit fault current at that bus is iper-unit = 1/.09 p.u. and it was multiplied by 3200A, because the the 3200A was chosen to be the bus current.


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## kduff70 (Aug 10, 2015)

thank you


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