High Impedance Bus Differential Scheme

Hello

I am looking at the manual for a GE 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?