current, CT ratio, service factor). This relay opens the motor contactor;
therefore, it should not be used for short-circuit protection. A fuse or circuit
breaker needs to be used to clear short circuits.
3) Plot the largest fuse (i.e., Device 2) or circuit breaker at the load end. This
type 9R fuse should protect the motor feeder cable. Therefore, it needs to be
below and to the left of the 2/0 AWG cable damage curve. This fuse protects the
motor from currents above the overload level.
j)
10 000 kVA transformer secondary main coordination.
1) Plot the 10 000 kVA through-fault-current protection curve and transformer
inrush point.
2) Plot the transformer secondary circuit breaker relay setting (i.e., Device 1) (see
Figure 15-19). This overcurrent relay is set to pick up at 2000 A (2000/5 CT
×
5AT). This setting allows the transformer forced-air rating of 12 500 kVA to be
utilized (or 1735 FLA). The pickup of this relay should be below the 250% of
the transformer full-load current required by NEC Article 450-3 for secondary
protection (2000 A pu/1388 FLA = 145%). The setting of this relay also protects
the 4160 V
bus
rated at 2000 A.
k)
10 000 kVA transformer primary relay coordination.
The next device to be plotted is
the medium-voltage feeder relay (i.e., Device 2) (see Figure 15-20). By examination
of the through-fault protection curve and the inrush point, the limits of the curve for
primary protection of the 10 000 kVA transformer can be determined. Keeping in
mind that a low pickup for this device is desirable for good cable protection, it is
good practice to keep the characteristic curve as far to the left as practical in order to
operate faster on faults. The pickup setting should be less than four times the
10 000 kVA transformer full-load amperes (or 5552 A at 4160 V), the maximum setting
permitted by NEC Article 450-3. The time-dial setting was chosen for a 0.3 s
coordination interval between the downstream relay (at maximum fault current) and
feeder circuit breaker relay (at instantaneous setting). The instantaneous setting of
7600 A (at 13.8 kV) or 25 212 A (at 4160 V) provides better protection of the
750 kcmil cable and is set above the calculated asymmetrical short-circuit current.
The 16% current margin is also satisfied over most of the relay curves.
In this example, an 800/5 A CT has been selected. The 800/5 A rating CT has a 160:1
ratio, and the tap setting times 160 produces the relay minimum pickup in terms of
the primary current (160
×
6AT = 960 A). The 4160 V system has now been
coordinated up to the 10 000 kVA transformer primary circuit breaker, and the 480 V
system has been coordinated up to the 1000 kVA transformer primary fuse. The next
area to coordinate is the 13.8 kV system up through the utility system.
l)
13.8 kV feeder relay coordination.
The next device to be plotted is the mediumvoltage feeder relay (i.e., Device 3) (see Figure 15-21). This feeder relay does not
provide primary protection for the 1000 kVA transformer; therefore, it can be set to
pick up for the feeder full-load current (800/5 CT
×
4AT = 640 A). This setup also
protects the feeder cable. The instantaneous element on this feeder relay is not used
because it does not allow selectivity with the 13.8 kV primary transformer fuses
downstream. The largest feeder circuit breaker relay should now be coordinated with
the upstream 13.8 kV main bus relays. The 4160 V substation feeder relay is also
shown on this curve as a comparison with the 480 V substations feeder relay. As can