3 Phase System Line to Ground fault Fault current calculation

[sweetguy_in16]

Six, 6.6kV ,3phase, alternators are connected to a common set of bus bars.
Each has positive negative and zero sequence reactances of 0.90 ohm, 0.72 ohm, 0.30 ohms respectively . An Earth fault occurs on one bus bar . Determine the value of fault current if all alternator neutrals are solidly grounded.

HINT: Fault Current,I =3V/(Z0+Z1+Z2) for single generator when single Line to ground fault occurs.

where, V is the Phase voltage (i.e. ( 6.6kV / 1.732 = 3.8106 kV ))


Can anyone help me on how to solve this problem. Please.

 

[subtech]

It's been a while, but if memory serves...
Total current = I1 + I2 + I0
You have the impedances. Do one at a time.
I'll do a little book work and see if I can get back up to speed.
Post again if you are still stumped.
I'll try to help.

 

[piercas]

Based on the given information, the fault current calculation for a single generator with a line to ground fault can be calculated using the formula I = 3V / (Z0 + Z1 + Z2), where V is the phase voltage and Z0, Z1, and Z2 are the positive, negative, and zero sequence reactances, respectively.

In this case, we have six generators connected to a common set of bus bars, each with their own sequence reactances. Since the fault is on one bus bar, we can assume that only one generator is affected by the fault. Therefore, we can use the above formula to calculate the fault current for one generator.

First, we need to calculate the phase voltage, which is equal to the line voltage divided by the square root of 3. In this case, the line voltage is 6.6kV, so the phase voltage would be 6.6kV / √3 = 3.8106 kV.

Next, we need to calculate the total sequence reactance for one generator. This can be done by adding the individual sequence reactances together. In this case, the total sequence reactance would be 0.90 + 0.72 + 0.30 = 1.92 ohms.

Finally, we can plug these values into the formula to calculate the fault current:

I = 3V / (Z0 + Z1 + Z2)
= 3(3.8106 kV) / (1.92 ohms)
= 5.911 kA

Therefore, the value of fault current for a single generator with a line to ground fault would be 5.911 kA. However, since there are six generators connected to the bus bars, the total fault current would be six times this value, which is equal to 35.466 kA.

It is important to note that this calculation assumes that all alternator neutrals are solidly grounded. If this is not the case, the fault current value may be different. It is always important to consider the specific conditions and parameters of a system when calculating fault currents.