- #1
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
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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