Feeder Voltage Drop and Capacitor Bank Justification
A 12.47 kV, three-phase, four-wire overhead feeder supplies a 3.6 MW, 1.74 Mvar lagging load at a distance of 2.5 miles from the substation. The line impedance is 0.35 + j0.75 Ω per mile. The substation bus is held at 12.47 kV. The utility limits the voltage at the customer service to no less than 95% of nominal. The planner proposes a 1.2 Mvar shunt capacitor bank at the customer bus. The existing conductor is 336.4 kcmil ACSR with a continuous rating of 400 A. The load is balanced and can be modeled as constant power.
1. Compute the per-phase voltage drop and percent regulation without the capacitor. Does the customer bus meet the 95% limit?
Read the worked answer
Use the per-phase line drop approximation. First find the load current:
S = √(3.6² + 1.74²) = 4.00 MVA, I = 4.00×10⁶/[√3(12,470)] = 185.2 A.
For 2.5 miles, R = 0.35(2.5) = 0.875 Ω and X = 0.75(2.5) = 1.875 Ω. With cos φ = 3.6/4.0 = 0.90 and sin φ = 0.4359,
ΔVφ = I(R cos φ + X sin φ) = 185.2[(0.875)(0.90) + (1.875)(0.4359)] = 297 V.
The phase-to-neutral base is 12,470/√3 = 7,199 V, so the drop is 297/7,199 = 4.13%. The customer bus is about 95.9% of nominal, just above the 95% limit. Because the load is constant power, the current will rise slightly as voltage falls, so the exact result is marginally worse than this first-order estimate.
2. Compute the voltage at the customer bus with the 1.2 Mvar bank. Does the bank meet the 95% limit?
Read the worked answer
With the 1.2 Mvar bank, the net reactive power is 1.74 - 1.20 = 0.54 Mvar. Then
S = √(3.6² + 0.54²) = 3.64 MVA, I = 3.64×10⁶/[√3(12,470)] = 168.6 A.
cos φ = 3.6/3.64 = 0.989 and sin φ = 0.149. The new drop is
ΔVφ = 168.6[(0.875)(0.989) + (1.875)(0.149)] = 193 V,
or 193/7,199 = 2.68%. The customer bus rises to about 97.3% of nominal, so the bank meets the 95% limit with margin. The feeder current is also well below the 400 A conductor rating. The bank should be placed at the customer bus, not at the substation, so the reactive current is removed from the line segment causing the drop.
3. The capacitor bank is switched daily. What protection or operational concerns should be checked?
Read the worked answer
Check light-load overvoltage: the same bank that helps at peak load can push the bus above limits when the 3.6 MW load is off. Verify capacitor switching inrush and breaker or vacuum-switch restrike capability. Check resonance between the bank and the feeder/source inductance; a 1.2 Mvar bank on a 12.47 kV feeder can interact with harmonic sources. Confirm the bank rating for the actual operating voltage, unbalance protection, and overcurrent protection. Also check whether the utility requires a fixed or switched bank and whether the capacitor should be split into steps. The 95% limit is a peak-load criterion; the design must satisfy both peak and light-load voltage.