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Kinetic Energy of a Hurricane

  1. Nov 7, 2013 #1
    1. The problem statement, all variables and given/known data
    In a simple model of the wind speed associated with hurricane Emily, we assume there is calm eye 18.8 km in radius. The winds, which extend to a height of 4000 m, begin with a speed of 303.0 km/hr at the eye wall and decrease linearly with radial distance down to 0 km/hr at a distance of 128.8 km from the center. Assume the average density of the air from sea level to an altitude of 4000 m is 0.969 kg/m3. Calculate the total kinetic energy of the winds.


    2. Relevant equations
    K = 1/2 mv2
    dK = 1/2 dmv2
    dA = 2πr dr

    3. The attempt at a solution
    I tried finding the integral of dK in order to find the total KE.
    This gives:
    ∫dK = ∫1/2 v2ρ*h*2π*r dr

    Since v is a function of r, I converted the radii and velocities to meters and found a linear fit that represented v(r), which was:
    v = -2.76r + 355137
    And since ρ, h, and 2π are constants, they can be taken out of the integral, which gives:
    KE = ρhπ ∫[v(r)]2 r dr
    Since v(r) is already restricted from the outside of the eye to the outside of the entire hurricane, as far as I know it doesn't need to be further restricted in the integral.

    I keep getting wrong answers from this, though, so I'm not quite sure what I'm doing wrong.
     
  2. jcsd
  3. Nov 7, 2013 #2

    TSny

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    This doesn't seem to be correct. For example, at r = 18,800 m it doesn't give the correct speed in m/s. Did you convert hours to seconds?
     
  4. Nov 7, 2013 #3
    Ah no, thanks for pointing out my mistake. This gives a new v(r):
    -0.000765r + 98.5554
    But this is still wrong when I plug it into the integral (I get 1.636e+17 J).
     
  5. Nov 7, 2013 #4

    TSny

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    What did you use for the limits of integration?
     
  6. Nov 7, 2013 #5
    None--I was told that since v(r) already takes into account the restriction from 18.8 km to 128.8 km, the limit on the integration didn't need to be set. But I'm not sure how correct this is.
    Also, I've been using 128.8 km for the r when plugging in; should I be using 128.8 km - 18.8 km = 110.0 km instead?

    Edit: Just tried that, still wrong.
     
  7. Nov 7, 2013 #6
    I just tried setting the limits on the integration, which gave me the correct answer, 1.465E+17J. I was just given some bad information for the integral I guess (it's been a while since I've done integrals myself so I just trusted what I heard). Thanks for the help!
     
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