Reactor example from Graffeo

[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

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.

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?

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

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.

thank you