Page 273 - Offshore Electrical Engineering Manual

260 CHAPTER 7 Protection and Discrimination

Fuse B1 operates at t = 0.01 s at 2400 A, from the fuse characteristic, with fault
current obtained from computer results.
From the GEC Guide, the relay Z operating time is:
t + 0.45 t + 0.15 = 0.164 s
The PSM is 2400/450 = 5.33. Therefore the relay operating time at a PSM of
5.33 and a TMS of 1 is 1.8 s, from the relay characteristic. The required TMS is
0.164/1.8 = 0.09 ≈ 0.1.
In this case, a check was made against the relay characteristic. It was found that
the curve for TMS = 0.1 gives an operating time of only 0.14 s at a PSM of 5.33. As
this is too fast, the time setting was adjusted to 0.2. Relays U, T and Y may be set in
a similar manner.

RELAY F: RANGE 10%–200%, CT RATIO 1500/1; 10 VA 5P10
This relay must coordinate with earth fault relays Z and U and with the largest outgo-
ing fuse in the accommodation switchboard, fuse U, 400 A.
(This assumes that coordination with fuse U assures coordination with relay.)
Set relay F at three times fuse U rating, i.e., 3 × 400 A = 1200 A. Then the nearest
current setting is:
0.8 × 1500 = 1200 A
From the SC studies, the maximum earth fault current is 55.458 kA. At this cur-
rent, fuses Z and U operate in 0.01 s, from the fuse characteristics. From the GEC
Guide, the relay F operating time is:
t + 0.4 + 0.15 = 0.164s
From the SC studies, the average earth fault current at 0.164 s is 51.44 kA.
As a PSM, this is 51,440/1200 = 42.9.
From the relay characteristics, the relay operating time at a PSM of 42.9 and a
TMS of 1 is 0.25 s. Therefore the required TMS is 0.164/0.25 = 0.66 ≪ 0.7.
It is necessary to carry out a relay saturation check. The relay impedance at the
2
0.8 setting is given by (relay VA)/(0.8) = 0.39 Ω. Allowing 1 Ω for lead resistance
gives 1.39 Ω.
Thus the secondary burden is 1.39 × (1 A) = 1.39 VA. Therefore:

I maxunsat = 10 VA × (10/1.39) × (1500/1)
= 107.9 kA (which is above the maximum fault current).

Note that if a saturation current significantly lower than the fault current is
obtained, it is advisable to select a different CT. The burdens associated with elec-
tronic relays are much lower than those used above for induction disc types, and
because of this are less likely to give rise to saturation problems.