Alternator Failure in Icing with a Finite Battery
Night IFR in a single-engine aeroplane, 28 V system with a 70 A alternator and a 24 Ah battery. The alternator fails; the ammeter shows discharge and the low-voltage light is on. Current loads are: pitot heat 10 A, propeller heat 14 A, airframe de-ice boots 20 A, nav/strobe lights 10 A, avionics 12 A, transponder 3 A, panel lighting 2 A, cabin light 2 A. Total 73 A. The battery is 80% charged and the battery can be used to 20% capacity. The aircraft is in icing conditions at 9,000 ft, 35 nm from a suitable airport. The pilot has a handheld radio and a flashlight. The destination airport has an ILS and a 2,000 m runway. The aircraft is not certified for flight into known icing beyond the boot system.
1. Calculate the battery endurance at the full load and at a reduced essential load.
Read the worked answer
Usable battery capacity = 24 Ah × (80% − 20%) = 14.4 Ah. At the full 73 A load, endurance = 14.4/73 = 0.197 h, about 12 minutes. That is not enough to hold in icing and fly 35 nm. If the load is reduced to pitot heat 10 A + avionics 12 A + transponder 3 A = 25 A, endurance = 14.4/25 = 0.58 h, about 34 minutes. Keeping the boots (20 A) as well gives 45 A and about 19 minutes. These are nominal figures; battery voltage falls and avionics may drop out before the calculated time, so the real margin is smaller.
2. What load-shedding order and diversion decision do you recommend?
Read the worked answer
Shed in this order: cabin light 2 A, panel lighting if not needed, prop heat 14 A, then nav/strobe if visibility allows. Do not shed pitot heat, because a blocked pitot tube gives false airspeed and can lead to stall or overspeed; keep the transponder and one com. The boots are the difficult load: in icing they are operationally important, but 20 A halves the remaining endurance. The better decision is to declare an emergency, turn toward the nearest suitable airport and descend to warmer air, keeping pitot heat and avionics, rather than trying to hold in icing on battery power. If the boots must be kept, accept a much shorter diversion and land before the battery is exhausted.
3. Explain why pitot heat is treated as non-sheddable and what instrument errors follow if it fails.
Read the worked answer
Pitot heat is non-sheddable because the airspeed indicator is the primary source of stall margin and performance information. Without heat, ice can block the pitot tube; the airspeed then behaves like an altimeter-climbing shows increasing airspeed, descending shows decreasing-while the static port may also ice. The pilot could misread a stall or overspeed. In this scenario, losing pitot heat while already in icing is a reason to leave the conditions immediately. The altimeter and VSI would also be unreliable if the static source ices, and the attitude indicator may be vacuum- or electrically driven; the electrical failure may remove the standby instruments too.