Fault Calculation
Fault Calculation
(OP)
Gents,
Just want to get your opinion regarding the fault calculation on 33kV feeder line
if the load are neglected and do not contribute into the fault and are fed by 4 x 33kV feeder cables 5km long and terminated to 33kV generator terminal bus at one end , should the impedance of cable (5km) be neglected as well in fault calculation assuming the fault occurred at one of the feeders near the 33kV bus?
many thanks,
Danilo






RE: Fault Calculation
RE: Fault Calculation
RE: Fault Calculation
RE: Fault Calculation
The impedance of such a cable is 0.08+j0.1 ohm/km [positive/negative/phase].
Four parallel cables of 5 km will be Zcables=0.1+j0.125 ohm
The total current [ 4 cable in underground] has to be about 1900-2000 A then the generator S=100 MVA. X"d=0.08-0.3%. X"d=0.3/100*33^2/100 =0.03267 ohm
In my opinion, you may neglect the generator subtransient reactance and take only the cable.
RE: Fault Calculation
is 3.267 and of course you cannot neglect it.
RE: Fault Calculation
thanks for your replies.
I am thinking of neglecting the impedance of the cables during the fault condition at one of the cables when the generator CB is OFF and consider the cable impedance only when the generator is ON. I just dont know if my assumption is correct. what do you think?
regards,
Danilo
RE: Fault Calculation
You have to introduce all the source at maximum output.
The resistance of cables and overhead lines will be at 20oC.
2.5 Minimum short-circuit currents
You have to introduce all the sources at minimum output.
The resistance of cables and overhead lines will be at maximum rated[90 or 70oC].
If you will follow IEEE Std 141/1993 see ch.4 short-circuit current calculations.
IEEE 551/2006 IEEE Recommended Practice for Calculating Short-Circuit Currents in Industrial and Commercial Power Systems ch.3.13 Significant cable lengths.
In no case these standard recommend to neglect the cable impedance.
RE: Fault Calculation
Just to clarify that fault occured at one of the 33kV cable feeders terminated at 33kV bus (Utility) and the other end to 33kV generator bus where the CB is OFF.
Immediately after the fault, can I neglect the impedance of the faulted cable?
regards,
Danilo
RE: Fault Calculation
RE: Fault Calculation
RE: Fault Calculation
Gents,
please see attached SLD..all loads are neglected. If CB is off and on, what is the equivalent circuit diagram? Cable is 5km long each.
thanks,
Danilo
RE: Fault Calculation
If the CB is closed and the generator is running there will be fault current in A, B, and C but you'll have to calculate which direction it flows.
RE: Fault Calculation
when CB is on and assuming the ff:
cable impedances = 1 p.u/cable
utility impedance = 0.5 p.u
generator impdance = 1.5 p.u
the fault current is 0.393 p.u. could you suggest how do you derive the generator contribution on the fault current? does part of the fault current
from generator still restrict going through D cable?
RE: Fault Calculation
Why do you want to calculate the fault current in the damaged cable? It will not serve any purpose. You should always calculate the bus fault current for any meaningful evaluations/analysis.
RE: Fault Calculation
RE: Fault Calculation
With the generator off;
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Fault Calculation
A fault adjacent to the utility bus will be limited by the utility immpedance.
A fault on the generator bus will limited by the utility impedance and by the impedance of 4 cables in parallel.
But when the generator is on, referring on the above diagram, I would assume that fault occurred at the load side of imaginary CB at Cable D. Without changing the original fault location which is at cable D, I would assume that , Fault current at D = Fault contribution from utility + Fault contribution from Generator.
If there is an imaginary circuit breaker for each cable near the utility bus and each cable having the same impedance, therefore, each CB would have a fault current equivalent to 1/3 of Fault contribution from generator. Could there be an error with this analysis?
RE: Fault Calculation
However, you may use the superposition method [see IEC 60909-1] and calculate the current in any position of a short-circuit-[you have to take the vectorial sum of the two currents into consideration].
As I see, the minimum short-circuit current through the cables will be at the utility end fault -if the utility transformer is the same size with the generator step-up transformer- and in case of the fault at the generator side the short-circuit current through cables will about double.
If you neglect the cable impedances the calculated current could be 8-10% more.
By the way, I wonder how the p.u. values are calculated. What is the base apparent power ?
The cable p.u. is very high-in my opinion. The 1.5 p.u. of generator is including step-up transformer too?
However, the utility impedance does not include the utility transformer, I think.
RE: Fault Calculation
The assumed MVA Base is 100 MVA.
Assuming the location of the fault at Cable D is at utility end and if the cable impedances of A,B & C are significant, does it make sense to include the cable impedances contribution into fault current?
Davidbeach mentioned that "If the CB is open there will be fault current down the page on lines A, B, and C and back up into the fault on D. All impedances matter". Does it mean that even at fault condition, the cable impedance of the faulted cable matter when it looped back to the utility through cables A,B & C?
How can this be related to the analysis of Waross that "With the generator off;
A fault adjacent to the utility bus will be limited by the utility immpedance."?
RE: Fault Calculation
RE: Fault Calculation
RE: Fault Calculation
if for instance Zdu is zero when fault occurred at the terminal of end of the cable, what sort of analysis should follow?
RE: Fault Calculation
of short-circuit you intend to analyze: phase-to-phase [3 phases],phase-to-ground ,phase-to-phase-to
ground. For the last two you need to introduce zero impedance [considering the Z2 [negative impedance] it will be still Z2=Z1].The system neutral could be solid grounded, resistance or reactance grounded or ungrounded.
RE: Fault Calculation
So are we saying that with the generator off and a fault on "D" close to the utility bus there will be fault current from the utility bus directly to fault on D but there will also be additional fault current going down lines A, B, & C and back up to fault on D? So total fault contribution at D will be fault current directly from utility bus (majority of fault current) and fault current down A,B&C (minority of fault current)
I always assumed that for a fault cacls you looked at the upstream impedance (utility impedance in this case) and used the shortest impedance path for calculating max fault current. But I can see now from this conversation that the "total fault current" will be a function of current divider and although path directly from utility bus to fault on D represents the shortest impedance you still technically need to include the small amount of current flowing down A, B, & C in order to get the "total" amount of fault current at D.
Is this how most commercial software analysis programs calculation fault current for this particular situation?
RE: Fault Calculation
This current will be limited by the impedances of cables "A", "B" and "C", and the part of cable "D" back up to the fault.
The sum of both these currents will not exceed to full fault current at the bus.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Fault Calculation