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Run-down of a wheel as a func. of time from full speed to standstill.

  1. Jul 18, 2013 #1
    1. The problem statement, all variables and given/known data
    Hello

    I need to find a an equation that describes the rundown of a wheel as function of the time.
    The wheel is running at an constant speed and is then disconnected and runs down to standstill.

    We know what the angular velocity of the wheel is at time 0, Wm. We know the inertia, J. with this we can calculate the kinetic energy in the wheel at time 0 as

    E = J * Wm^2 / 2

    Now we also now the Friction losses in kW at the constant speed, Pfr at time 0.

    What i want to know is how to combine these. I know that the frictional losses are subtracted somehow from the kinetic energy but i down now where to start.

    Any help appreciated.
     
  2. jcsd
  3. Jul 18, 2013 #2

    TSny

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    Hello. Can you determine the torque due to friction from the initial frictional power loss, Pfr, and the initial angular speed ωm?
     
  4. Jul 19, 2013 #3
    Yes these two things are know and given. The wheel is running at the grid network of 50Hz and at that speed Pfr is known and constant.

    I think that the friction losses are decreased with the angular frequency to the power of 2 or 3 but i am not really sure on this.

    Ie

    Pfr(w) = Pfr0 * w^3
     
  5. Jul 19, 2013 #4

    TSny

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    This equation is inconsistent with dimensions. Maybe it should be written P = bωn where b is a constant and the exponent n is 2 or 3. You should be able to write the constant b in terms of the initial frictional power loss Po and the initial rate of spin ωo.

    It looks like this will require some calculus in order to determine how ω changes with time. Do you see how to relate P and E using a derivative?
     
  6. Jul 23, 2013 #5
    Sorry its more [itex] P_{fr} = P_{fr_0} (\frac{w}{w_0})^3[/itex]

    If we look at newtons second equation we have

    [itex]
    \frac{}{}
    J * \frac{dw}{dt} + kw = 0
    [/itex]

    it gives that

    [itex]
    \frac{dw}{dt} + \frac{k}{J}w = 0
    [/itex]

    if we set the solution as [itex]w * e^{(k*t /J )}[/itex] we get

    [itex]
    \frac{d}{dt}w * e^{(k*t /J )} + \frac{k}{J}w * e^{(k*t /J )} = 0
    [/itex]

    solving this for [itex]w * e^{(k*t /J )}[/itex] we will have an solution as

    [itex] w(t) = w(0) * e^{(k*t /J )}[/itex] but i dont know how i should link ##k## and ##Pfr##.

    All i want is to graf w(t), that is that the speed is decreasing with time due to friction losses.
     
  7. Jul 23, 2013 #6

    TSny

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    OK, so we can write ##P_{fr} = P_{fr_0} (\frac{w}{w_0})^3 = aw^3## where ##a = \frac{P_{fr_0}}{w_0^3}##

    How did you get the torque ##\tau## to be ##kw##? The magnitude of the torque is related to the power as ##P_{fr} = \tau w##. Use ##P_{fr} = aw^3## to discover how the torque depends on ##w##.
     
  8. Jul 25, 2013 #7
    Ok if we do that we will get an equation a standard solution that looks something like

    ##\frac{d\omega}{dt} = -a\omega^3## -> ##\frac{d\omega}{\omega^3} = -adt ##

    ##\omega(t) = \frac{1}{\sqrt{2at + C_1}} ## where ##C_1 = \omega(0) = 1/w_0 ##

    And if i plot this as a fuction of time i will get a rolldown time of seconds while it should take minutes. So i looked at the dimensions of the newtons equation and they are not the same :/ should there be another term to make sure that they are the same dimensions?

    KR
     
  9. Jul 25, 2013 #8

    TSny

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    The frictional power is the rate at which the energy is decreasing due to friction: ##P = -dE/dt ##.

    What differential equation do you get if you substitute ##P = aw^3## and ##E = \frac{1}{2} J w^2##?
     
  10. Jul 25, 2013 #9
    Ok so ##dE/dt + P = 0## as ##dE/dt## has dimension Joule/s which is the same as Watt.

    So what we will have is

    ##\frac{d\omega^2J}{dt2} + a\omega^3 = 0##

    (not sure if im doing the next step correct or not) but if we take away one ##\omega## on each side we will get

    ##\frac{d\omega}{dt}*\frac{J}{2} + a\omega^2 = 0 ##

    and if we solve this we will get

    ## \omega(t) = \frac{\omega_0}{2*P_{fr0}/J\omega_0 + 1} ##

    am i doing this correctly?
     
  11. Jul 25, 2013 #10

    TSny

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    Careful with the factors of 2. Note that ##\frac{d}{dt}\omega^2 = 2\omega\frac{d\omega}{dt}##. So, ##\frac{dE}{dt} = \frac{1}{2}J\frac{d}{dt}\omega^2 = J\omega\frac{d\omega}{dt}##.

    After you set up ##P = -\frac{dE}{dt}##, separate variables and integrate to find ##\omega(t)##.

    [Note that your result ## \omega(t) = \frac{\omega_0}{2*P_{fr0}/J\omega_0 + 1} ## doesn't make sense because your right hand side doesn't depend on time. You know that ##\omega## should decrease with time.]
     
  12. Jul 25, 2013 #11
    Sorry forgot the ##t## there in the nominator. There is supposed to be a ##t## there like this.

    ## \omega(t) = \frac{\omega_0}{2*P_{fr0}*t/J\omega_0 + 1}##
     
  13. Jul 25, 2013 #12

    TSny

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    OK. After correcting for the factor of 2, I think there's still an error in part of your expression where you have the ratio ##P_{fr0}*t/J\omega_0##. Check to make sure you have the correct exponent on the ##\omega_0## here.
     
  14. Jul 26, 2013 #13
    Ok this is how i have reasoned when looking at this.

    ##\frac{dE}{dt} + P = 0 ##

    we know how the equation is set up

    ##E = \frac{J\omega^2}{2}; P = \frac{P_{fr0}}{\omega_0^3}*\omega^3 → a * \omega^3##

    we replace E and P with their dependency of ##\omega## and cross out one ##\omega## on each side, it is this step that im am hesitant on whether you can really do this

    ##\frac{J}{2}\frac{d\omega^2}{dt} + a*\omega^3 = 0 → \frac{J}{2}\frac{d\omega^{2→1}}{dt} + a*\omega^{3→2} = 0 → \frac{J}{2}\frac{d\omega}{dt} + a*\omega^2 = 0##

    simplify a bit

    ## \frac{d\omega}{dt} + \frac{2*a}{J}*\omega^2 = 0## call ##\frac{2*a}{J}## for ##\alpha##

    ## \frac{d\omega}{dt} + \alpha*\omega^2 = 0##

    move over to the other side

    ## \frac{d\omega}{dt} = -\alpha*\omega^2##

    ## \frac{d\omega}{\omega^2} = -\alpha*dt##

    integrate

    ## ∫\omega^{-2}d\omega = -\alpha*∫dt##

    ## -\omega^{-1} = -\alpha*t + C##

    change singes

    ## +\omega^{-1} = +\alpha*t - C##

    ## \omega(t) = \frac{1}{\alpha*t - C}##

    where we find the constant

    ##\omega(0) = \omega_0##

    ## \omega(0) = \frac{1}{0 - C} → C = -1/\omega_0##

    giving us

    ## \omega(t) = \frac{1}{\alpha*t 1/\omega_0}##

    if we multiply with \omega_0 / \omega_0 on both sides we will have

    ## \omega(t) = \frac{1}{\alpha*t + 1/\omega_0} * \frac{\omega_0}{\omega_0}##

    ## \omega(t) = \frac{\omega_0}{\omega_0\alpha*t + 1} ##

    if we substitute the ##\alpha## with what is has been then we will get

    $$ \omega(t) = \frac{\omega_0}{\frac{2P_{fr0}*t}{J*\omega_0} + 1} $$

    the problem is that with

    ## P_{fr0} = 170 000 watt ##
    ## w_0 = \pi * 50 ## ( 4 pole machine)
    ## J = 4000 kgm^2 ##
    ## t = seconds ##

    we will get a decrease that is too fast. The slowdown should take around one hour when it takes minutes. If however we put the power in kW instead of W then rate is much closer to the reality though in this case it more the other way around as it takes 2x longer.
     
  15. Jul 26, 2013 #14

    TSny

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    As noted earlier, ##\frac{d}{dt} \omega^2 = 2\omega \frac{d\omega}{dt}##. So ##\frac{d}{dt}(\frac{1}{2}J\omega^2) = J\omega \frac{d\omega}{dt}##. (The 2's have cancelled.)

    Then, $$\alpha = \frac{a}{J} = \frac{P_{0fr}}{J \omega_0^3}$$ Note that ##\omega_0## is raised to the third power here.

    When you substitute this into ##\omega(t) = \frac{\omega_0}{\omega_0 \alpha t +1}##, you end up with $$\omega(t) = \frac{\omega_0}{\frac{P_{0fr}}{J \omega_0^2}t +1}$$

    So, you have the square of ##\omega_0## in the denominator.
     
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