12.47 kV Feeder Voltage Drop: Capacitor Versus Reconductoring
A 12.47 kV, three-phase overhead feeder serves a 3.0 MVA industrial customer at the end of a 5-mile line. The line's positive-sequence impedance is 0.30 + j0.60 Ω per mile and the substation bus is held at 12.47 kV. The customer's load is 0.85 power factor lagging and essentially constant kVA. The utility's planning limit is 5% voltage drop from the substation bus to the customer's service point. The existing conductor is also approaching its summer thermal limit, so any remedy that does not reduce current is viewed with suspicion. A 900 kvar fixed capacitor bank sits in stores, and a reconductoring option would cut line resistance by 40% while leaving reactance essentially unchanged and leaving current unchanged.
1. Compute the existing voltage drop as a percentage of nominal and the three-phase line losses at peak.
Read the worked answer
Line impedance (5 miles): Z = 5 × (0.30 + j0.60) = 1.5 + j3.0 Ω → |Z| = 3.354 Ω, X/R = 2.0
Load current: I = 3000 kVA / (√3 × 12.47 kV) = 3000 / 21.60 = 138.9 A at pf 0.85 → cos θ = 0.85, sin θ = 0.527
Exact per-phase drop (using the standard approximation): ΔV_phase ≈ I (R cos θ + X sin θ) = 138.9 × (1.5 × 0.85 + 3.0 × 0.527) = 138.9 × (1.275 + 1.581) = 138.9 × 2.856 = 397 V per phase
Percent drop: nominal phase voltage = 12,470/√3 = 7200 V ΔV% = 397 / 7200 = 5.51% → exceeds the 5% planning limit. Line-to-line at the customer: 12,470 − √3 × 397 = 12,470 − 688 = 11,782 V (94.5% of nominal), below the 95% ANSI Range A limit at the service point once the customer's own transformer drop is added.
Losses: P_loss = 3 I² R = 3 × 138.9² × 1.5 = 86.8 kW (about 3.4% of the 2550 kW delivered - high for a distribution feeder).
2. Evaluate the 900 kvar bank placed at the customer's end: recompute drop and losses, and explain why placement at the end matters.
Read the worked answer
Load reactive power: Q₀ = 3000 × 0.527 = 1580 kvar; P = 3000 × 0.85 = 2550 kW
With the bank: Q = 1580 − 900 = 680 kvar S = √(2550² + 680²) = 2639 kVA → I = 2639/21.60 = 122.2 A cos θ = 2550/2639 = 0.966, sin θ = 0.258
New drop: ΔV = 122.2 × (1.5 × 0.966 + 3.0 × 0.258) = 122.2 × (1.449 + 0.774) = 122.2 × 2.223 = 272 V per phase ΔV% = 272/7200 = 3.78% ✓ (within the 5% limit)
Losses: 3 × 122.2² × 1.5 = 67.2 kW → a reduction of 19.6 kW (23%).
Placement is the whole point. A capacitor compensates reactive current only on the line section between the bank and the source. At the customer's end, the entire 5 miles carries reduced reactive current and gets the full benefit. If the same 900 kvar were installed at the substation, the feeder would still carry the customer's full 1580 kvar and the drop would remain at 5.5% - the bank would reduce only transformer and upstream losses. Mid-feeder placement captures roughly half the benefit.
Cautions:
- 900 kvar is fixed; at light load it will overcorrect and raise voltage. Check the light-load case before locking in a fixed bank; a switched or two-step bank is usually correct for an industrial customer with a varying load.
- Confirm the customer is not already correcting power factor on their side of the meter - the utility must not pay for kvar the customer has already installed.
3. Compare the capacitor against the reconductoring option on voltage, losses, and thermal margin, and state which you would recommend and why.
Read the worked answer
Reconductoring (R reduced 40%, X unchanged): R = 0.6 × 1.5 = 0.90 Ω, X = 3.0 Ω, current unchanged at 138.9 A
- Drop = 138.9 × (0.90 × 0.85 + 3.0 × 0.527) = 138.9 × (0.765 + 1.581) = 138.9 × 2.346 = 326 V → 4.53% (passes, but with only 0.5% margin)
- Losses = 3 × 138.9² × 0.90 = 52.1 kW (a 34.7 kW reduction - better loss performance than the capacitor)
- Current unchanged → the conductor's thermal headroom improves only through the larger size itself
Capacitor (900 kvar at the customer end):
- Drop 3.78%, losses 67.2 kW, current reduced 12% (138.9 → 122.2 A)
Recommendation: install the capacitor bank first. Reasons:
- It is a fraction of the cost and can be in service this season, whereas reconductoring needs outages, permits and long lead time.
- It is the only option that reduces current, which directly buys thermal margin on the existing conductor during the summer peak - exactly the stated concern.
- It frees transformer and substation capacity as well, not just line capacity.
- It leaves 1.2% voltage margin rather than 0.5%, which matters because the drop calculation uses nominal load; actual peak load will exceed it.
Caveat: if the feeder's thermal limit is already being approached at 138.9 A and load is growing, the capacitor's 12% current reduction may not be enough on its own. In that case the staged answer is capacitor now, reconductor when growth justifies it - and note that the two together (122.2 A on a 40%-lower-R conductor) give about 2.6% drop and 37 kW loss, the best combined result.