Protect a wind turbine from overspeed with three independent layers: a dump load that always gives the alternator something to push against, a furling tail that swings the rotor out of a strong wind, and a mechanical brake worked from the ground. A dump load must be wound for the voltage it sees: a 2 kW load on a 57.6 V charging bank needs 1.66 Ω, and a 230 V water-heater element, which is 26.5 Ω, would absorb only 125 W there.
Part of Advanced Wind Turbine
Overspeed protection stops a rotor from running away when the battery is full, the load is lost or the wind is too strong; the dump load is the electrical part of it.
Why An Unloaded Rotor Is The Danger
An alternator produces voltage in proportion to speed. If the load is disconnected, nothing resists the rotor, and it accelerates. The Wind Power page says an unprotected turbine in high winds will accelerate to destructive speed and may shake itself apart or throw a blade with lethal consequences. Centrifugal force grows with the square of speed: the root of a 1.5 kg blade at 381 RPM loads its bolts with 1,430 N (Carving Wooden Blades), and at three times the speed with 9 × 1,430 = 12,870 N.
Layer One: The Dump Load
The shunt charge controller diverts power to a dump load when the battery reaches its absorption voltage (14.4 to 14.8 V on a 12 V flooded bank, per Charge Controller) so the turbine is never unloaded. Size the dump load to absorb the turbine’s whole maximum output, as the hydro Load Control page does for water turbines (100 to 120% of the output).
A resistor of resistance R dissipates P = V² ÷ R. For a 2 kW dump at the full-charge voltage of each bank:
| Bank | Charging voltage | Resistance for 2 kW | Current |
|---|---|---|---|
| 12 V | 14.4 V | 0.104 Ω | 139 A |
| 24 V | 28.8 V | 0.415 Ω | 69 A |
| 48 V | 57.6 V | 1.66 Ω | 34.7 A |
The water-heater trap
The Load Control page suggests a water-heater element as a dump load. That works when the element is rated for the voltage across it. A 2 kW element made for 230 V has R = 230² ÷ 2,000 = 26.5 Ω. On a 48 V bank at 57.6 V it dissipates 57.6² ÷ 26.5 = 125 W, 6.3% of its rating, and at 48 V only 87 W. No switching speed can fix it, because the element’s resistance is too high to draw the current. Options:
- Elements wound for the bank voltage.
- Many elements in parallel: 26.5 ÷ 16 = 1.65 Ω, so sixteen 230 V, 2 kW elements in parallel on 57.6 V absorb about 2 kW in total (each 125 W). Wasteful of elements, but it shows the arithmetic.
- A heater tank wired in series-parallel groups so its resistance matches.
A dump load can fail. A burned-out element leaves the turbine unloaded at the moment the battery is full, which is why it is only the first layer. Fit an ammeter in the dump circuit so you can see that it works. An element must stay submerged in water; one run dry burns out (recalled).
Layer Two: Furling
The carrier’s furling tail is offset from the rotor axis; in high wind the aerodynamic force on the tail makes it fold, turning the rotor partly out of the wind, and it resets when the wind drops. The Wind Power page describes the same as side furling, a hinge so the turbine turns sideways at a set speed. It is passive and the standard method for small turbines.
The geometry does the work. A rotor turned by an angle θ from the wind sees only the component v × cos θ along its axis. Thrust goes with the square of that, cos²θ, and wind power with the cube, cos³θ. At 30° the thrust falls to 75% and the power to 65%; at 60°, to 25% and 12.5%; at 75°, to 6.7% and 1.7%. Real furling is messier, since the rotor’s flow changes when it is yawed (recalled), so use the figures as the direction of the effect. Set the furling speed above the speed where the alternator reaches its full output, so the tail does not fold while there is still power to harvest. For the 3 m rotor in these pages, full output of 2 kW comes at 10.97 m/s (Average Wind Speed & Power Curve).
Layer Three: Brake And Short Circuit
The carrier lists two further methods: a mechanical brake (a disc on the main shaft engaged by a pull cable) and electrical braking by short-circuiting the generator outputs, whose back-EMF creates a braking torque. Notes:
- A shorted alternator gives no braking torque at standstill, because there is no EMF to drive a current; it slows a spinning rotor and cannot hold a stopped one. Use the mechanical brake or a rotor lock for maintenance (a recalled property).
- The braking energy goes into stator copper as heat. Use the short only briefly and not as a parking brake.
- Mount the short-circuit switch and the brake handle at the base of the tower, where a person can reach them without climbing.
Worked example: how long a brake takes
Assume three 1.5 kg blades, each a slender rod 1.2 m long about the axis: J = 3 × 1.5 × 1.2² ÷ 3 = 2.16 kg·m². At 381 RPM, ω = 39.9 rad/s, so the stored energy is ½ × 2.16 × 39.9² = 1,717 J. A brake torque of 20 N·m (an assumed value) stops it in t = J × ω ÷ T = 2.16 × 39.9 ÷ 20 = 4.3 s. The wind keeps pushing, so the real time is longer. The hands that work the brake need a way to see the rotor stop.
Common Mistakes
- A dump load that cannot absorb the output — wrong voltage rating.
- One layer only — a single failure unloads the turbine.
- A brake that can only be reached by climbing.
Safety
Never approach a rotor that has not stopped
A runaway rotor throws blades and can bring down the tower. Stop it from the tower base with the brake, tilt or furl it, and wait until it has stopped before going near. The dump load heats water and wire: fit a thermostat and a pressure-relief valve to a heater tank, and keep the dump circuit fused. Name the person who works the brake and how long the rotor takes to stop.
What’s Next
- Learn next: Tilt-Up Design for Maintenance for lowering a stopped turbine
- Combine with: Load Control to compare the hydro approach