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March 30, 2017

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In a simple model of the wind speed associated with hurricane Emily, we assume there is calm eye 10.0 km in radius. The winds, which extend to a height of 5550 m, begin with a speed of 208.0 km/hr at the eye wall and decrease linearly with radial distance down to 0 km/hr at a distance of 150.0 km from the center. Assume the average density of the air from sea level to an altitude of 5550 m is 0.891 kg/m3. Calculate the total kinetic energy of the winds.

  • physics - ,

    find KE as a function of radius slice dr
    from r = 10^10^3 to r = 150*10^3

    208 km/hr /3.6 = 57.8 m/s

    Velocity = m r + b linear
    57.8 = m (10*10^3) + b
    0 = m( 150*10^3) + b
    so solve for slope and intercept
    m = -.413 * 10^-3
    b = 61.9
    so
    v = 61.9 - .413*10^-3 r

    the volume at radius r = 2 pi r dr(5550)

    so the mass at radius r = .891*5550* 2 pi r dr

    so the KE at radius r = (1/2) (.891*5550* 2 pi r dr) (61.9-.413*10^-3 r)^2
    when you square that and multiply by r dr you will get terms like
    a r + b r^2 + c r^3
    which will integrate to
    a r^2/2 + b r^2/3 + c r^4/4
    evaluate that integral at r = 10*10^3 and at 150*10^3 and you have it

  • physics - ,

    Where did 3.6 come from?

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