Fault Calculation

[Danilo917]

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

 

[davidbeach]

Why would you not include the cable impedance?

 

[jghrist]

No.

 

[pwrengrds]

Only if your willing to live with what could be a >10-20% error depending on generator and wire size.

 

[7anoter4]

If there are 4 parallel cables of 33 kV I presume they are 3*300 mm^2 copper conductors.
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.

 

[7anoter4]

Sorry! My mistake. It is not 0.08-0.3% but 8-30% Then actually the maximum generator reactance
is 3.267 and of course you cannot neglect it.

 

[Danilo917]

Gents

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

 

[7anoter4]

If you will follow IEC 60909-0 ch. 2.3 Method of calculation 2.4 Maximum short-circuit currents
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.

 

[Danilo917]

Thanks again for the replies.

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

 

[jghrist]

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.

This is the first mention of a utility bus. I thought the cables were served by the generator. Do you have two contributions to the fault, a utility at one end and a generator at the other? Where on the cable did the fault occur?

 

[mgtrp]

It sounds like it might help if you were to sketch and upload a single line diagram of your arrangement.

 

[Danilo917]

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

 

[davidbeach]

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.

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.

 

[Danilo917]

Thanks

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?

 

[krisys]

Danilo917 (Electrical)

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.

 

[Danilo917]

Thanks. i am trying to figure out if current divider principle can be applied to short circuit analysis. by just looking at the diagram ,the fault contribution from the generator appears to be divided among the 4 cables which includes the faulted cable D.Otherwise, the fault contribution from the generator appears to be the same as the fault current at 33kV utility bus secondary side. am i missing something here?

 

[waross]

You can't anticipate where on the cable a fault may occur.
With the generator off;

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.​

With the generator on-line;

Calculate the combined generator contribution and utility contribution to a fault adjacent to each bus. The fault will be fed by all four cables.​

The current due to a mid-line fault will be between these limits.

As a sanity check,​

you may wish to calculate the fault currents for a fault at the quarter points of the faulted line in the case when both the generator and utility contribute. There may be an instance where a mid line fault has the highest fault current. This could be the case for some combinations of cable impedance and relative capacities of the generator and the grid. I doubt it but as I said, "sanity check".


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[Danilo917]

I agree With the generator off;
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?

 

[7anoter4]

In my opinion, since you don't know the actual fault point position in the cable you may take the two extreme positions-a fault at the generator end and a fault at utility end[as it is already recommended by waross and krysis ].
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.

 

[Danilo917]

Gents,thanks for the replies.

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."?