Braking is essential for the safe operation of the HOLI 300 small wind turbine. In this section we explain how to calculate the braking torque. We also find a brake type able to supply this torque. Important Note

Preliminary content from design report

The content of this article is taken from the December 2013 preliminary design report. It represents intention of design at that stage but does not necessarily show the final version of the HOLI 300 turbine design.

The contest regulations require the wind turbine system to provide a manual blocking of the rotor and a manual emergency stop. In order to fulfill the general safety requirements as described here the turbine has to be protected against overspeed as well. Overspeed can occur by different failures. A precondition for overspeed are high wind speeds. If these occur and the furling system fails to limit the rotor speed as described in the respective article, the disk brake gets activated. Except of a failure in the furling system, overspeed can also be caused by:

• failure of main shaft

The disk brake is triggered by the microcontroller, which determines the overspeed state by means of sensors. In case of overspeed, the brake is configured to stop the turbine in a short amount of time which will be specified in the following paragraph.

### Braking basics

The requirements for the braking system of the wind turbine are principally determined by the necessary braking torque. This torque is calculated from three parts. The necessary torque to break down the inertia of the system, the torque to break down the energy input out of the wind and the frictional resistance. The frictional resistance can be neglected. The inertia of the system can be calculated as follows:

It is assumed that a fault in the safety system or a furling delay leads to a 50 % overspeed condition compared to the cut-out wind speed with a rotor speed of $n=476.94\,\frac{1}{\textrm{min}}\cdot1.5\thickapprox715\,\frac{1}{\textrm{min}}$ and the brake should completely stop the rotor in a time of $t=3\,\textrm{s}$. This assumption is made in addition to IEC 61400-2 Load Case G to fulfill the customer’s high safety requirements. Also, it ensures a safe shutdown if the yaw axis is blocked and as a result, the furling system cannot move the rotor out of the wind. With an assumed blade inertia $J_{\textrm{b}}=0.305\,\textrm{kgm}^{2}$, generator inertia $J_{\textrm{g}}=0.006\,\textrm{kgm}^{2}$ and an estimated inertia taking into consideration other hub components as spinner and shaft $J_{\textrm{o}}=0.006\,\textrm{kgm}^{2}$, the total hub inertia is $J_{\textrm{h,t}}=4J_{\textrm{b}}+J_{\textrm{g}}+J_{\textrm{o}}=1.232\,\textrm{kgm}^{2}$. The brake torque $M_{\textrm{B,J}}$ is calculated taking the angular velocity $\omega$angular velocity into account. $M_{\textrm{B,J}}=J_{\textrm{h,t}}\omega=J_{\textrm{h,t}}\frac{2\pi\triangle n}{\triangle t}=30.75\,\textrm{Nm}$

Theoretically, the braking torque to break down the energy input at a constant level is rising but by taking into account a with the rotational speed decreasing efficiency, this phase has been neglected. The related equation is as follows: $M_{\textrm{B,nec}}=M_{\textrm{B,J}}+M_{\textrm{B,E}}=\frac{(\sum J)\cdot n_{\textrm{B}}}{k\cdot t_{\textrm{B}}}+\frac{P_{\textrm{input}}}{n_{\textrm{nom}}}\cdot k$

where $P_{\textrm{input}} =$ energy input $J =$ inertia of the system $n_{\textrm{B}} =$ rotational speed at moment of brake activation $n_{\textrm{nom}} =$ nominal rotational speed at moment of operation $t_{\textrm{B}} =$ braking time $k=\frac{60}{2\pi} =$ conversion factor

For this project an emergency braking time of about 3 seconds has been set. Furthermore, a power input at a maximum of 500 W has been assumed. The brake is activated once the overspeed limit has been triggered. The required torque to brake the rotor in 3 seconds is 42.67 Nm. This torque can be lowered by assuming higher braking times. The torque gives the range the brake should be able to cover.

### Disc Brake

To fulfill the the safety requirements to be able to react to failures in each component, a fail safe brake system was chosen. The disc brake operates with a brake caliper activated by a spring to ensure failsafe mode. The spring will be tensioned, so that the brake caliper is kept open electromagnetically. These electro magnets need an opening energy supply of 110 W and a permanent energy supply of 25 W to keep the brake caliper open. The energy supply should be provided by the generator or the battery. The voltage can be adapted.

### Summary

The advantage of this system is that the system is independent of human control and covers every safety requirements set for this contest. A disadvantage of the fail safe brake system is the complexity it adds to the project. However, this disadvantage is compensated by the advantages in protection of the turbine operator and the turbine components itself. Another disadvantage is the high amount of energy that is constantly needed to keep the brake in an open, operating state. For this reason, the authors are still evaluating other fail-safe alternatives for the emergency brake system, which might have a lower energy consumption.