Two feeders in parallel between the same two substations look like a simple reliability upgrade — if one feeder trips, the other carries the load. But a fault on one feeder can trick the relays into tripping both feeders, causing exactly the blackout the second feeder was meant to prevent. The fix is a classic piece of protection coordination: non-directional overcurrent relays (51) at the sending end and directional overcurrent relays (67) at the receiving end. This post explains how the scheme works, step by step.
The setup: two parallel feeders, four breakers
Two parallel feeders connect a Sending Bus (S) to a Receiving Bus (R). Each feeder has a circuit breaker at each end:
- Sending end: CB A on Feeder 1 and CB B on Feeder 2, each with a non-directional inverse-time overcurrent relay (ANSI 51), graded with a higher time delay, tA = tB ≈ 0.5 s.
- Receiving end: CB C on Feeder 1 and CB D on Feeder 2, each with a directional overcurrent relay (ANSI 67), with a low time setting, tC = tD ≈ 0.1 s. A 67 relay is polarised to trip only when fault current flows in its tripping direction — into the feeder, which is the reverse of the normal load flow.
Figure 1.1: Selective isolation of faulted Feeder 1 using sending-end non-directional relays (51) and receiving-end directional relays (67). (Diagram redrawn for this post.)
What happens when a fault strikes Feeder 1?
Suppose a fault F occurs on Feeder 1, close to the receiving end. The fault is fed from the sending bus through two paths:
- Direct path — IF1: from Bus S, through CB A, straight along Feeder 1 to the fault.
- Roundabout path — IF2: from Bus S, through CB B, along the healthy Feeder 2, through CB D, into Bus R — and then backwards through CB C along Feeder 1 to the fault.
That second path is the trap. The fault current IF2 flows through the healthy feeder's breakers (CB B and CB D) as well as through CB C. If every relay were non-directional, CB D would see this large fault current and trip — isolating the healthy Feeder 2 for no reason, and leaving the load with no supply at all.
Why the directional relays save the day
| Breaker | Relay | What it sees | What it does |
|---|---|---|---|
| CB C | 67 (t ≈ 0.1 s) | Fault current IF2 flowing from Bus R into Feeder 1 — its tripping direction | Trips first, at ≈ 0.1 s |
| CB D | 67 (t ≈ 0.1 s) | Fault current IF2 flowing from Bus S toward Bus R — the opposite of its tripping direction | Blocks (does not trip) |
| CB A | 51 (t ≈ 0.5 s) | Fault current IF1 flowing directly to the fault; after CB C trips, this is the only remaining fault feed | Trips at ≈ 0.5 s, clearing the fault from the sending end |
| CB B | 51 (t ≈ 0.5 s) | Fault current IF2 on its way to Bus R — but CB C clears the fault before its time delay expires | Does not trip |
The logic is simple once the directions are clear. A directional relay (67) only trips when current flows in its set direction. CB C's tripping direction is into Feeder 1 from Bus R — exactly the direction of the fault current returning to the fault — so it trips fast, in about 0.1 s. CB D's tripping direction is into Feeder 2 from Bus R, but the fault current passes through it in the opposite direction (toward Bus R), so it is in its blocking direction and stays closed. The sending-end relays (51) are deliberately slower (≈ 0.5 s) so that the fast receiving-end relay gets the first chance; CB A then acts as backup and clears the fault from the sending end.
The result: selectivity
Only the faulted Feeder 1 is isolated — from both ends — while the healthy Feeder 2 keeps carrying the load. There is no complete blackout. That is the whole point of the scheme: selectivity (discrimination), disconnecting only the faulted section and leaving the rest of the network energised.
Without the directional relays, both feeders would trip on the same fault, and the parallel-feeder arrangement would give no reliability benefit at all.
Key points to remember
- Sending end: non-directional inverse-time O/C relays (51), graded slow, t ≈ 0.5 s.
- Receiving end: directional O/C relays (67), set fast, t ≈ 0.1 s, tripping only into the feeder (reverse to normal load flow).
- A fault on Feeder 1 is fed two ways — directly (IF1) and around through Feeder 2 and Bus R (IF2).
- CB C trips first (0.1 s); CB D blocks because the current is in its reverse direction; CB A clears at 0.5 s.
- Result: only Feeder 1 is isolated; Feeder 2 stays in service.
Based on Power System Protection & Switchgear, Paper II, Q1(b), Exam 11 June 2026.