Time-current grading for relays X, Y, Z

Time-current grading means setting each IDMT relay so that the relay nearest the fault trips first, and every relay towards the source waits one grading margin longer.

Assumptions: the given curve is for TMS = 1; the TMS of relay Z is 0.1; grading margin between adjacent relays = 0.5 s (CB time + relay overshoot + errors); CT secondary = 5 A; linear reading of the curve between points.

Formulas

Pick-up current (primary) = Current setting × CT primary
PSM = Fault current / Pick-up current
t = TMS × t(TMS=1, read from curve)

Relay Z (farthest from source)

Pick-up = 1.00 × 200 = 200 A
PSM     = 2000 / 200 = 10   → t(TMS=1) = 2.8 s
t_Z     = 0.1 × 2.8 = 0.28 s

Relay Y (back-up for Z)

For a fault at Z (2000 A), Y must wait 0.5 s longer than Z.

Pick-up = 1.25 × 200 = 250 A
PSM (fault at Z) = 2000 / 250 = 8 → 3.0 s
Required t_Y = 0.28 + 0.5 = 0.78 s
TMS_Y = 0.78 / 3.0 = 0.26
Fault at Y (4000 A): PSM = 4000/250 = 16 → 2.2 s
t_Y = 0.26 × 2.2 = 0.572 s

Relay X (back-up for Y)

Pick-up = 1.25 × 400 = 500 A
PSM (fault at Y) = 4000 / 500 = 8 → 3.0 s
Required t_X = 0.572 + 0.5 = 1.072 s
TMS_X = 1.072 / 3.0 = 0.357
Fault at X (5000 A): PSM = 5000/500 = 10 → 2.8 s
t_X = 0.357 × 2.8 = 1.0 s

Grading summary

Source ──X──────Y──────Z────── load
time:   1.0 s   0.57 s  0.28 s   (own-fault times)
For the same fault, each relay is 0.5 s slower
than the next relay downstream.

Answer: TMS_Z = 0.1, TMS_Y = 0.26, TMS_X ≈ 0.357 (set 0.36); the operating times for faults at their own points are 0.28 s, 0.572 s and ≈ 1.0 s. A fault at Z is cleared by Z in 0.28 s; Y would back it up at 0.78 s and X later still, so discrimination is achieved.