Adding a Standby Generator: Fault Duty and Islanded Coordination
A 13.8 kV utility service feeds a 5 MVA, 13.8/4.16 kV, 6.0% impedance transformer at an industrial plant. The utility's fault level at the point of delivery is 200 MVA. The existing 4.16 kV switchgear is rated 15 kA symmetrical interrupting at 4.16 kV. A 2.5 MVA, 4.16 kV standby generator with X″d = 0.15 per unit will be added and will run in parallel with the utility during peak shaving. The existing feeder overcurrent relays have a 400 A pickup, a time-delay curve that gives about 2.8 s at 2.3 kA primary, and an instantaneous element set at 6000 A primary. The generator manufacturer's damage curve allows 2300 A for 1.5 s.
1. Using a 5 MVA, 4.16 kV base, compute the three-phase fault current at the 4.16 kV bus with the utility only and with the generator paralleled, and check the switchgear rating.
Read the worked answer
Base current: I_base = 5000/(√3 × 4.16) = 694 A
Utility only:
- Utility source on 5 MVA base: 5/200 = 0.025 pu
- Transformer: 0.060 pu
- Total: 0.085 pu → I_sc = 694/0.085 = 8165 A (8.2 kA)
Generator contribution:
- X″d on 5 MVA base: 0.15 × (5/2.5) = 0.30 pu
- I_gen = 694/0.30 = 2313 A (2.3 kA)
Paralleled total: 8165 + 2313 = 10,478 A (10.5 kA)
Switchgear check: 10.5 kA < 15 kA ✓ - the symmetrical interrupting rating is adequate with about 30% margin.
Do not stop there. The momentary/close-and-latch duty must be checked separately using the peak current: with a combined X/R on the order of 8–10, the peak is roughly 2.4–2.5 × 10.5 kA ≈ 25–26 kA, which must be compared against the gear's peak withstand (typically 2.6 × rated, i.e., ~39 kA here - acceptable, but the calculation must be shown). Also confirm the worst-case fault is taken with the generator at rated voltage and the utility at maximum fault level simultaneously, and that the generator's decaying contribution is evaluated at contact parting time for the interrupting duty.
2. Evaluate the protection scheme for the islanded case, where the utility breaker is open and only the generator is on the bus. Is the existing feeder protection adequate?
Read the worked answer
Fault current in island mode: only the generator contributes: 2313 A at the bus, and less for any fault remote from the bus (feeder impedance reduces it further).
Check each element:
- Instantaneous (6000 A primary): never sees a fault in island mode - 2313 A is 2.6× below the setting. It is blind by design margin. ✗
- Time-delay (400 A pickup, ~2.8 s at 2.3 kA): the generator's damage curve permits 2300 A for only 1.5 s. The relay takes 2.8 s → the generator is damaged before the feeder relay trips. ✗
- The 400 A pickup itself is fine relative to generator rated current (2313 A ÷ 400 A ≈ 5.8× pickup), so pickup is not the problem - speed is.
Verdict: the existing scheme is not adequate for island operation. Options, in order of preference:
- Generator-mounted high-speed protection - 87G differential plus 51V voltage-restrained/controlled backup. The generator's own breaker clears generator-zone and nearby bus faults in well under 1.5 s, independent of the feeder TCC.
- Transfer trip / breaker-failure scheme so a feeder fault in island mode trips the generator breaker directly rather than waiting on the feeder relay.
- A separate relay setting group for island mode: lower instantaneous pickup to something above maximum load but below 2313 A - e.g., 1800 A primary - provided it is above cold-load inrush and motor starting current. This is only viable if the feeder's maximum load current leaves room between load and the minimum fault current; verify with the load data, and note the margin is thin (a fault at the far end of a long feeder may drop below 1800 A).
- Note that with the utility present, lowering the instantaneous element risks tripping on the transformer's magnetizing inrush, so a second setting group switched by the utility breaker status (or a 50/51 with torque control) is the practical implementation.
3. The plant plans to expand. How many additional identical 2.5 MVA generators can be paralleled before the existing 15 kA switchgear becomes inadequate, and what other check governs the answer?
Read the worked answer
Headroom: 15,000 − 8165 (utility) = 6835 A available for generator contributions.
Per generator: 2313 A → 6835 / 2313 = 2.95
So two additional units (three generators total = 6939 A of generation contribution) give: I_total = 8165 + 3 × 2313 = 8165 + 6939 = 15,104 A → 15.1 kA, which already exceeds the 15 kA rating by a hair.
Recalculate properly: 2 generators (one additional) → 8165 + 2 × 2313 = 12,791 A (12.8 kA) ✓; 3 generators → 15.1 kA ✗. So the answer is one additional unit (two generators total), keeping the bus at 12.8 kA and roughly 15% margin.
The margin is too thin to bank on - other governing checks:
- Duty is cumulative in the wrong direction: the generator contributions are vector-summed with the utility through the transformer, not simply added; the exact result depends on the combined X/R and whether the utility fault level is at its maximum. A rigorous study may give slightly more or less than the arithmetic sum.
- Momentary/peak duty scales the same way and must be rechecked against the gear's peak withstand.
- Coordination in island mode: more generation means higher islanded fault current, which actually helps the instantaneous elements and generator damage curve problem - a genuine trade-off, not a pure negative.
- Ground fault duty on the 4.16 kV system changes with each generator's neutral grounding method; if the units are high-resistance grounded, the ground-fault coordination must be redone for each configuration.
- Arc flash incident energy at the 4.16 kV bus rises with fault current and must be re-labelled after any expansion.