Galley Inverter Tripping: kVA, Power Factor and an Extra Load
A 115 V, 400 Hz, single-phase inverter on a twin turboprop is rated at 1.5 kVA and trips at 13.0 A rms output current. Its load is a galley heater and fan drawing 1.2 kW at 0.75 power factor lagging. The inverter has been tripping in high ambient temperatures; output voltage holds at 114 V rms under load. The operator wants to add a further 200 W resistive load to the same circuit. The maintenance team proposes fitting a power factor correction capacitor across the load terminals and has 10 µF, 20 µF and 40 µF, 250 V AC, 400 Hz-rated capacitors in stores. The inverter manual requires the load to remain below rated kVA and below rated current at all times. No spare inverter is available and the next flight is a 3-hour sector with the galley in use.
1. Calculate the apparent power and output current, and explain why the inverter trips although the real power of 1.2 kW is below the 1.5 kVA rating.
Read the worked answer
Apparent power S = P / pf = 1200 / 0.75 = 1600 VA. Current I = S / V = 1600 / 114 = 14.0 A (13.9 A at 115 V). Rated current = 1500 / 115 = 13.0 A - identical to the trip threshold.
So the load is about 7% over the current limit although the real power is only 1200 W. The inverter's output devices, filter components and internal wiring are rated by current, and the reactive component adds to the in-phase component in quadrature: I = √(I_active² + I_reactive²) = √(10.5² + 9.3²) = 14.0 A. The heater element uses real power; the fan motor's magnetising current does not, but it still heats the inverter.
Note also that the current is computed at the measured 114 V, not the nominal 115 V, so for the same VA the current is slightly higher - the margin is thinner than a nominal-voltage calculation suggests. The 1200 W figure is a red herring; kW is not the limiting rating here.
2. Calculate the capacitance needed to correct the power factor to about 0.95, select a capacitor from stores, and determine whether the extra 200 W can then be added.
Read the worked answer
Required reactive power: Q_c = P (tan φ₁ − tan φ₂). φ₁ = cos⁻¹ 0.75 = 41.4°, tan φ₁ = 0.882. φ₂ = cos⁻¹ 0.95 = 18.2°, tan φ₂ = 0.329. Q_c = 1200 × (0.882 − 0.329) = 1200 × 0.553 = 664 var. C = Q_c / (2π f V²) = 664 / (2π × 400 × 114²) = 664 / (2513 × 12 996) = 664 / 3.27 × 10⁷ ≈ 2.0 × 10⁻⁵ F = 20 µF.
Select the 20 µF, 250 V AC, 400 Hz-rated part (250 V AC gives ample margin over 114 V; the 10 µF would under-correct and the 40 µF would over-correct toward a leading factor).
Resulting load: S = 1200 / 0.95 = 1263 VA, I = 1263 / 114 = 11.1 A - a comfortable margin.
With the extra 200 W and the same 20 µF capacitor: capacitor output Q = 2πfCV² = 2513 × 20 × 10⁻⁶ × 12 996 ≈ 653 var. New load reactive = 1400 × 0.882 = 1235 var; net Q = 582 var; S = √(1400² + 582²) = 1516 VA; I = 13.3 A. That exceeds the 13.0 A trip. So the correction fixes the existing overload but does not create headroom for the additional 200 W - the operator must be told no, or a load-shedding scheme introduced.
3. The proposal is to bolt the capacitor permanently across the inverter output terminals. Assess the risk and state how it should be installed.
Read the worked answer
With no real load the capacitor still draws leading current: I_c = V · 2πfC = 114 × 2513 × 20 × 10⁻⁶ = 5.7 A. That is within 13 A, but it is a purely capacitive load the inverter may not regulate stably, and it gives a badly leading power factor at low load - some inverters trip on overvoltage or loss of regulation under leading loads.
Other risks: a permanently connected capacitor stays charged after the heater circuit is isolated, presenting a shock hazard during maintenance; if the capacitor fails short it becomes a direct short across the inverter output; and a capacitor switched separately from the inductive fan motor can resonate with it near the 400 Hz supply frequency, producing high circulating currents.
Correct installation: mount the capacitor on the load side of the galley contactor so it is switched with the load it corrects; use a part rated for continuous 400 Hz AC operation (not a DC or 50/60 Hz electrolytic); protect it per the manual and provide a discharge path; secure it against vibration; and after modification verify with a power quality analyser at worst-case voltage and full load that current stays below 13.0 A and the power factor lies within the inverter's published limits. Confirm in the technical record that the extra 200 W was not added.