Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Braking torque question

  1. Dec 19, 2012 #1
    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
     
    Last edited: Dec 19, 2012
  2. jcsd
  3. Dec 19, 2012 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    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.
     
  4. Dec 19, 2012 #3
    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
     
  5. Dec 20, 2012 #4
    Here is the calculation file in xls format
     

    Attached Files:

  6. Dec 20, 2012 #5

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    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).

    The first one looks right.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Braking torque question
  1. Brake Power Question (Replies: 3)

  2. Prony brake question (Replies: 1)

Loading...