Voltage Rise and Protection Coordination for a 5 MW PV Plant on a Radial 11 kV Feeder
A 5 MWp solar PV plant connects to the end of an 8 km, 11 kV radial feeder. The feeder conductor is 185 mm² ACSR with r = 0.35 Ω/km and x = 0.40 Ω/km. The source is a 132/11 kV, 20 MVA transformer with 8% impedance and an off-load tap currently at +2.5%; the 11 kV bus at the source is held at 1.02 pu. The inverter is rated 5.0 MVA and can operate between 0.95 leading and 0.95 lagging. The plant exports 5 MW at unity power factor at noon. The local daytime load is 1 MW at 0.9 lagging and can be treated as concentrated at the PV bus. The utility limits the 11 kV bus to ±6% of nominal. The feeder protection is a 400 A recloser at the source with two reclose attempts at 1 s and 5 s, and the PV plant has a reverse-power relay and IEEE 1547 anti-islanding. A 2 MVAr switched reactor is available at the PV bus.
1. Calculate the voltage rise at the PV bus and determine the reactive power needed to comply with the ±6% limit.
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Use ΔV ≈ (P R + Q X)/V at the PV bus, with R = 0.35 × 8 = 2.8 Ω and X = 0.40 × 8 = 3.2 Ω. The local load is 1 MW at 0.9 lagging, so Q_load = 1 × tan(acos 0.9) = 0.484 MVAr; net export P = 5 − 1 = 4 MW. At unity inverter power factor, Q_net = −0.484 MVAr. ΔV = (4 × 2.8 + (−0.484) × 3.2)/11 = (11.2 − 1.55)/11 = 0.877 kV, an 8.0% rise. With the source at 1.02 pu (11.22 kV), the PV bus is about 12.10 kV, or 10.0% above nominal; either way it exceeds the ±6% limit. To hold 6% (0.66 kV), 0.66 = (11.2 + 3.2 Q_net)/11 gives Q_net = −1.23 MVAr, so the inverter must absorb 1.23 + 0.484 = 1.71 MVAr, slightly more than its 1.64 MVAr capability at 0.95 pf.
2. Recommend an operating arrangement using the available reactor, inverter reactive capability, curtailment, and tap position. Support the choice with numbers.
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The 2 MVAr reactor at the PV bus gives Q_net = −0.484 − 2 = −2.484 MVAr; ΔV = (11.2 − 7.95)/11 = 0.295 kV (2.7%), so the PV bus is about 11.52 kV (4.7% above nominal), compliant. Inverter absorption at 0.95 pf alone gives Q_net = −1.64 − 0.484 = −2.124 MVAr; ΔV = (11.2 − 6.80)/11 = 0.400 kV (3.6%), also compliant if the inverter can sustain 1.64 MVAr. Curtailing to 3.5 MW with the 2 MVAr reactor gives ΔV = (9.8 − 7.95)/11 = 0.168 kV, compliant but loses 1.5 MW. The +2.5% tap is already raising voltage; moving to −2.5% lowers the source by about 5%, but affects all customers and may violate low-voltage limits at night. Recommended: install the 2 MVAr reactor as primary control, use inverter Q(V) absorption up to 1.0 MVAr as fine control, and keep curtailment as a last resort. Verify with a load-flow study at minimum daytime load.
3. Assess the protection and reclosing risks introduced by the PV plant, including reverse power, anti-islanding, and fault contribution.
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The inverter contributes about 5 MVA/(√3 × 11 kV) = 262 A during a fault, small compared with feeder fault current, so it will not materially change overcurrent grading. But the 400 A recloser at the source sees reverse flow at noon; the reverse-power relay must discriminate normal export from a genuine reverse-flow condition and should have a time delay to ride through cloud transients. Anti-islanding must trip the PV before the first reclose at 1 s. IEEE 1547 permits tripping within 2 s, which is too slow for a 1 s reclose. Use transfer trip or a reclose-blocking signal from the recloser to the PV controller, or lengthen the first reclose to more than 2 s. Verify that the PV does not hold up voltage on the islanded section after the recloser opens. The 2 MVAr reactor must be interlocked with the PV so it is not left connected at night and does not cause overvoltage.