Fault Duty and Relay Setting with a New Generator
A 2.5 MVA, 13.8 kV/480 V, 5.5% impedance transformer is protected on the 13.8 kV side by a 51 relay with a 150:5 CT. The relay has a 5 A nominal current and available taps of 1, 2, 3, 4, 5, 6, 8, and 10 A. The transformer supplies 480 V switchgear rated 65 kA symmetrical. The upstream 13.8 kV breaker has a 250 MVA interrupting rating. A new 2 MVA synchronous generator with X'' = 0.15 pu is to be connected to the 13.8 kV bus through a 6% reactor. The plant wants to keep the existing 13.8 kV breaker and set the 51 relay to allow full transformer loading while coordinating with the generator. The generator is not expected to export more than 1.5 MW.
1. Calculate the 480 V three-phase fault current with and without the new generator. Is the 65 kA switchgear rating adequate?
Read the worked answer
On the 480 V base, Ibase = 2.5×10⁶/[√3(480)] = 3,007 A. Without the generator, the transformer alone gives
If = 3,007/0.055 = 54.7 kA.
This is below the 65 kA switchgear rating. For the generator, convert to the 2.5 MVA base: Xg'' = 0.15(2.5/2) = 0.1875 pu and the reactor is Xr = 0.06(2.5/2) = 0.075 pu. The generator branch is 0.2625 pu. In parallel with the transformer,
Xeq = (0.055)(0.2625)/(0.055 + 0.2625) = 0.0455 pu, If = 3,007/0.0455 = 66.1 kA.
The new fault level exceeds the 65 kA switchgear rating by about 1.7%. The existing 13.8 kV breaker is not the limiting device; the 480 V switchgear is. The design should not be approved without mitigation, even though the margin is small, because fault-current calculations have tolerances and motor contribution was neglected.
2. Select a 51 relay tap and explain the coordination considerations.
Read the worked answer
Transformer full-load current on the 13.8 kV side is
IFL = 2.5×10⁶/[√3(13,800)] = 104.6 A.
With a 150:5 CT (ratio 30), the secondary full-load current is 104.6/30 = 3.49 A. A 5 A tap gives a primary pickup of 5(30) = 150 A, or 150/104.6 = 1.43 times full load. That allows normal overload and cooling but stays above the continuous rating. A 4 A tap would be 120 A, only 1.15 times full load, too close for reliable load swing and CT error. The 5 A tap is preferred. The relay must have enough time delay to ride through transformer inrush, typically 8-12 times full load for about 0.1 s, and must coordinate with the upstream 13.8 kV relay. For a 480 V fault, the relay sees about 66,100(480/13,800) = 2,300 A primary, or 2,300/30 = 76.7 A secondary, which is 15.3 times the 5 A tap. The time dial should be set from the coordination study, not from pickup alone.
3. What is the effect of the 6% reactor on fault current and voltage regulation? Would removing it help?
Read the worked answer
The reactor adds impedance in the generator branch, reducing its fault contribution. Without the reactor, Xg'' = 0.1875 pu and the parallel equivalent is 0.055(0.1875)/(0.2425) = 0.0425 pu, giving If = 3,007/0.0425 = 70.8 kA - worse than the 66.1 kA with the 6% reactor. To hold the 480 V fault at 65 kA, the required equivalent is Xeq = 3,007/65,000 = 0.0463 pu. Solving the parallel combination gives a generator branch of about 0.291 pu, so the reactor must be about 0.291 - 0.1875 = 0.104 pu on 2.5 MVA, or 0.104(2/2.5) = 0.083 pu on the generator base - roughly an 8.3% reactor. The reactor also causes a reactive voltage drop when the generator exports, so the AVR and generator capability must be checked. Removing the reactor would increase fault duty and is not a fix.