I want to know if my understanding of dynamic equation associated with the High Speed Shaft Brake is correct or not.
I think when the brake is applied, HSS governing equation is:
HSShftTq - T_gen - HSSBrakeTq = ( J_gen + J_brakedisk ) * alpha * GBratio
“J_gen” = generator inertia relative to the HSS
“J_brakedisk” = Brake disk inertia relative to the HSS
“alpha” = the LSS rotational acceleration
“T_Gen” = generator torque applied to the HSS
But still I don’t know how FAST calculates the braking energy required to stop the WT. In other words I’d like to know the equations that FAST uses to decide on the time that the WT comes to stop. In fact regarding FAST calculations, how long does it take for the HSSbrake to stop the WT?
Yes, your equation for the HSS dynamics in FAST is correct.
How long it takes for the generator to be stopped by the brake depends on the nature of the brake and aerodynamic torques. (The aerodynamic torque is transferred through HSShftTq in your equation.) The generator will decelerate until both the generator acceleration and generator speed are zero. Page 4 of the following paper explains in a bit more detail how the HSS brake model in FAST is implemented: nrel.gov/docs/fy04osti/35077.pdf.
First I have to thank you for your valuable response.
As you have mentioned in your comment, aerodynamic torque is transmitted to the HSS through HSShftTq. So if one wants to calculate the braking energy using FAST outputs, he can use the following equation. (In the following equation, by “Int” I mean Integral)
Where “HSShftPwr”= HSS power
“J_Gen”=generator moment of inertia with respect to HSS
“J_Brk”=brake disk moment of inertia with respect to HSS
“HSShftV0”=HSS angular speed when the brake is applied
“HSSBrTq”=HSS brake torque
“HSShftV”=Instantaneous HSS angular velocity
Is it right?
In fact above equation states that the braking power is equal to HSS power (which accounts for the aerodynamic effect) plus the kinetic energy of the HSS rotating parts (which is Generator and Brake disk and probably HSS part of Gearbox which has not been accounted for in the above equation).
Does FAST use such method as I mentioned above to calculate the braking torque necessary to bring the shaft speed to zero?(“FAST now calculates the braking torque necessary to bring the shaft speed to zero at each time increment. The value of this braking torque is used as long as it does not exceed the limits of the brake. This technique ensures that the brake locks the shaft during a shutdown maneuver, bringing about realistic dynamic loads and “ringing” behavior.” you stated in your paper)
What if one wants to calculate the required braking torque or energy, directly from aerodynamic torque exerted to the WT? I think in this case the following equation can be used:
Int[(AeroTqLSShftV-GetPwr)dt]+total kinetic energy with respect to the HSS=Braking energy
Where “total K.E. with respect to HSS”=0.5J_totalHSShftV0^2,
" J_total"=Moment of inertia of all rotating parts with respect to the HSS
Am I right?
I’m looking forward to your comments
Your equations for braking energy seem reasonable except that the GenPwr in FAST is an electical power (including mechanical-to-electrical conversion efficiency losses) and the equations you use assume mechanical power. However, I would guess that the brake is only every applied when the generator is offline, meaning that GenPwr likely equals zero in these equations.
That said, FAST does not use these equations or attempt to calculate the braking energy. Instead, the maximum brake torque in FAST is simply a user-specified value. The drivetrain will slow down and stop as long as external torques do not exceed this specified braking torque.
As you know, in variable speed pitch-regulated wind turbines, the pitch system (beside the mechanical brake) plays an important role in bringing the WT to stop. Is it possible to determine the amount of energy that is dissipated (during a braking operation involving both the mechanical brake and the pitch system resistance simultaneously) by means of the pitch system? In other words is it possible to determine the amount of resisting torque of pitch system in a braking process?
(In fact I want to separately specify the share of mechanical brake and the pitch system in bringing the WT to stop using FAST outputs)
Without modification, FAST does not directly output the shaft torque resulting from pitch-control actions. However, you should be able to derive the contribution of the pitch system by subtracting the energy extracted from the mechanical brake (as you proposed above) from the total change in kinetic energy of the rotor during the braking event.