Braking system design for 20kW direct drive wind turbine

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Rajamani
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Everybody,

I am designing a braking system for a direct drive (no gear box) wind turbine. I am not sure how to calculate the braking time and number of brakes required. I made this in two methods, which gave different values. What is the exact method for calculating the number of brakes and braking time?
Here is my problem

Generator capacity = 20kW
Rotational speed = 100 rpm

Method-1 -finding number of brakes (Equate torque by rotor and braking system)
Total torque at rotor Tt = Aerodynamic torque Ta + Rotor torque Tr
Aerodynamic torque (P=2.pi.N.Ta/60) Ta = 1909.8 N-m

Rotor torque = Tr = I.α
Mass MI of rotor I = 650 kg-m^2
Time to stop the rotor t = 0.5 sec
Alpha = ω2-ω1/t
ω1 = 2.pi.100/60 = 10.47 rad/sec
ω2 = 0 rad/sec (stopped condition)
Alpha = 10.47-0 / 0.5 = 20.9 rad/s^2
Rotor torque = Tr = 6500*20.9 =13585 N-m

Tt=1909.8+13585 = 15494.9 N-m

Braking force = 10600N
Brake pad distance from centre = 0.3 m
Braking torque = 3180 N-m
No of brakes required = 15494.9 / 3180 = 4.87 (5 brakes)

I need five brakes to stop the rotor in 0.5 sec

Method-2 - finding time required to stop by fixing number of brakes
Kinetic energy of rotating body = ½.I.omega^2
=0.5*650*10.47^2 = 35626 N-m
Adding aerodynamic torque = 35626+1909.8=37536 N-m (Is adding torque right?)

For bringing the rotating body to rest work done by the braking system should be equal
Wd = F.D
D = 37536/(10600*5)=0.708 (assume 5 brakes)
0.708 = pi.(0.3*2).N -> pi.dia.Number of revolutions
N=0.376 revolutions
Revolutions in 1 sec = 100/60=1.66 rev
braking time = 0.375/1.66 = 0.23 sec

Thanks
Raj
 
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Discussion

The second method is problematic, and does not work in that way. You would have to consider the total rotation angle during the stopping process, and multiply the torque with that value to add it to the kinetic energy.
 
mfb thanks for your attention.
You are right. The torque has to be converted to work done. I found the force from the torque and multiplied it with the distance traveled (I considered 1 sec so pi.D.N/60). Even now the results of both methods are having huge difference.
First method says, 5 brakes are required for stopping the rotor in 5 sec.
Second method says, 5 brakes are required for stopping the rotor in 0.33 sec.

I want to know which method is correct?

I do not know how to attach the excel calculation file. Hence I just copied and pasted the revised calculation.

Method-1 Equating torque
Power - P = 20000 W
Rated rpm - n = 100 rpm
Torsional moment from rated Power (Aerodynamics) P=2.pi.n.Ta/60 -> Ta = 1909.85 N-m
CoG of blade Rcog - Rcog = 1.55 m
Mass MI of rotor - Ib = 650 kg-m^2
Initial Angular Velocity - w2=2.pi.n/60 -> 10.47 rad/sec
Final Angular Velocity - w1 = 0 rad/sec
time required to stop rotor - t = 0.5 sec
a=w2-w1/t -> a = 20.94 rad/sec^2
Torque by Rotor Inertia Tr=Ib.a -> Tr = 13613.56 N-m
Total Braking torque required - Tt=Tr+Ta -> 15523.42 N-m
Force by braking system - Fb = 10600 N
Braking radius Rb = 0.3 m
Braking Torque by one brake - Tb=F.Rb -> 3180 N-m
No of brakes (Fraction) - Nb = 4.88 Nos
No of brakes (Rounded) - Nb = 5 Nos

Method-2 -> Equating energy
Kinetic energy by the rotating body E=0.5.I. w^2 -> Er = 35640.23 N-m
Force by Blades to generate Aerodynamic Torque Ta=Fb.RCog -> Fb = 1232.16 N
Distance moved by Blades force in one sec Db = pi.D.(n/60) -> Db = 16.23 m
Work Done by Blades on generator - Wb=Fb.Db -> 20000 N-m
Total Work done on rotor - Wt=Er+Wb -> 55640.23 N-m
Force by braking system - Fb = 10600 N
No of brakes - Nb = 5 Nos
For bringing the rotating body to rest, workdone by the braking system should be equal to workdone on rotor Wt = Fb.D.Nb -> D = 1.04 m
pi.d.Nr -> Nr= number of revolutions - Nr = 0.55 rev
Revolutions in 1 sec - = 1.66 rev/sec
braking time - t = 0.33 sec
 
Rajamani said:
mfb thanks for your attention.
You are right. The torque has to be converted to work done. I found the force from the torque and multiplied it with the distance traveled (I considered 1 sec so pi.D.N/60).
You cannot just assume some arbitrary value for the angle. You have to solve an equation system in some way for this approach (unless you fix deceleration time and calculate the required number of brakes here as well).

I want to know which method is correct?
The first one looks right.