High impedance bus differential - CT Saturation Calculation 1

[bspace123]

Hello

I am looking at the manual for a MFAC34 high impedance bus differential relay. In the manual it states the following:

"The minimum setting voltage to ensure stability is
VS = IF (RCT + 2RL) N
The relay plug setting voltage VR must be set to the nearest tap above VS.
The minimum CT knee point voltage must be VK = 2VR

Where RCT = CT resistance, RL = lead resistance (one way between relay and CT) and N I assume is 1 for 3ph faults and 2 for ground faults (not defined in the manual).

I cannot make sense of this. My understanding is that the relay must not pickup for maximum external faults (i.e. "through-faults") and must not saturate for internal faults (the statements sort of contradict each other, if we assume a switchboard supplied by a single energy source with only outgoing resistive or inductive loads).

If all CT's are identical and wired in parallel with the diff relay, we assume 30kA (primary) fault current and the CT on that branch completely saturates, then the maximum voltage the diff relay will see is (where RCT = 4ohms, RL=0.5ohms, CT ratio = 1200/1):
Vs = (30000/1200) * (4 + 2*0.5) = 125V. (for an earth fault you'd multiply this by 2).

I am used to setting the diff relay pickup to 1.5x this voltage (safety factor), so the setting will be 187V.

For an internal fault, we need to ensure the CT will not saturate (implying for the above mentioned condition this would probably never happen...). For this to occur, the equation Vk = (30000/1200) * (4 + 0.5) = 112.5V (for an earth fault you'd multiply this by 2).

Again I'd adopt a 1.5x safety factory implying min. CT knee point will be 225V.

Why is the MFAC calling for 2xVs, when Vs considers 2xRL (resistance to/from the CT)? Under phase fault conditions, the CT burden will be one-way lead resistance to the diff relay?

 

[j_p_c]

Hi bspace123,

According to the manual, N is the turns ratio of the CT, the manual is formatted so that the N is hidden at the end of the line above its definition. It looks like the relay is missing a division sign before the N which would make the equation work. That or they intend the Vs quantity to meet Volt-turns.

Actually I looked at two other application guides for this relay and they're riddled with errors, they're some of the worst material I've ever seen written for P&C work. Any how, you already accounted for this by dividing by your CT ratio above, but thought I'd mention it.

I'm totally confused on the application of the MFAC34. Their literature is contrary to what I'm used to and what other relay vendors for high impedance applications note for a one way lead length for 3p faults. One of the MFAC manual notes applications examples for ground fault and three phase and both examples show 2RL! Normally I'd say go for safety and use the higher value (And follow their manual), but you're right it's weird.

At least in the low impedance world saturation is primarily a concern during through faults, relays experiencing internal faults should be able to ride out very bad saturation and still respond securely. My hunch is it should be similar for this application.

 

[stevenal]

I fail to understand your "safety factor" applied to the internal fault calculation. If you want dependable operation for in-zone faults, your pickup should be lower not higher.

See IEEE C37.234 Annex A. In short: Calculate voltage produced by maximum through fault with one CT saturated and multiply by 1.6 security factor. Then calculate Operation current at that voltage for an internal fault to ensure relay picks up dependably.

 

[bspace123]

Thanks for the responses.

You're right, the manual is inconsistent and in my opinion wrong.

The safety factor for the relay pickup is what you define as security factor. I used 1.5 instead of 1.6.

The safety factor used for CT knee point was just arbitrary, to ensure the CT will not saturate during any internal or external faults, which really defeats the purpose of applying the security factor mentioned above.

 

[stevenal]

If you apply a "safety factor" of something higher than 1.0 to a calculated voltage pickup for an internal fault, you are ensuring the relay will not ever pickup.

I would suggest that saturation for an internal fault is normal and expected when the only return path is through the high impedance relay input and the MOV.

You might want to have a look the SEL-587Z relay. You should be able to find it at selinc.com after registering. Different relay, but the math should be similar.

 

[DiscoP]

Agree with Stevenal.

The CTs will saturate for an internal fault, so you want to make sure the relay operates before this happens.

That is why they recommend the CT's Vk be at least twice the setting voltage when using the MFAC

Maybe check the NPAG - availble on the GE Grid website (the manufacturer of the MFAC relay).

It doesn't mention any safety factors.