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Merz-Price: protected winding (13.8 kV)

Short answer

Protected winding = 1 − (I_op×Rn/Vph). I_op = 0.10×5229.6 = 523 A, Vph = 7967 V, Rn=2 Ω. x unprotected = 523×2/7967 = 0.1313, so protected = 86.9%.

A 13.8 kV, 125 MVA, star-connected alternator has a synchronous reactance of 1.4 per unit per phase and negligible resistance. It is protected by a Merz-Price balanced current system which operates when out-of-balance current exceeds 10% of the full load current. If the neutral point is earthed through a resistance of 2 Ω, determine what proportion of winding is protected against earth fault.

An earth fault at fraction x of the winding from the neutral drives a current x·V_ph through the neutral resistance (and the faulted part of the winding). The relay sees this current and operates only if it exceeds the setting.

Data: 13.8 kV, 125 MVA, star; X_s = 1.4 pu; R negligible; R_n = 2 Ω; setting = 10% of full-load current.

Step 1: Basic quantities

V_ph = 13 800 / √3 = 7967.43 V
I_FL = 125 × 10⁶ / (√3 × 13 800) = 5229.62 A
I_op = 0.10 × 5229.62 = 522.96 A
Z_base = 13.8² / 125 = 1.5235 Ω
X_s = 1.4 × 1.5235 = 2.1329 Ω

Step 2: Unprotected fraction (neglecting winding reactance)

x × V_ph / R_n = I_op
x = 522.96 × 2 / 7967.43 = 0.1313
Protected = 1 − 0.1313 = 0.8687 → 86.87%

Step 3: Including the reactance of the faulted part

The faulted part has reactance x·X_s (as an approximation, since synchronous reactance does not divide exactly in proportion to turns):

x × 7967.43 = 522.96 × |2 + j x × 2.1329|
Solving numerically: x = 0.1326
Protected = 86.74%

The reactance changes the answer only slightly because x is small.

Answer: about 86.9% of the winding is protected against earth faults (86.7% if winding reactance is included); the 13.1% nearest the neutral is unprotected.

✓ Answer checked by Claude Opus 5.5 (high effort)

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