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17 December, 05:54

Ear "popping" is an unpleasant phenomenon sometimes experienced when a change in pressure occurs, for example in a

fast-moving elevator or in an airplane. If you are in a two-seater airplane at 3000 m and a descent of 100 m causes your

ears to "pop," what is the pressure change that your ears "pop" at, in millimeters of mercury? If the airplane now rises to

8000 m and again begins descending, how far will the airplane descend before your ears "pop" again? Assume a U. S.

Standard Atmosphere.

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Answers (1)
  1. 17 December, 06:09
    0
    a) dh_HG = 6.72 mm

    b) dz' = 173 m @8000 m

    Explanation:

    Given:

    Original height h_o = 3000 m

    Descent = 100 m

    Find:

    what is the pressure change that your ears "pop" at, in millimeters of mercury?

    If the airplane now rises to 8000 m and again begins descending, how far will the airplane descend before your ears "pop" again?

    Solution:

    - Assuming density of air remains constant from 3000 m to 2900 m. From Table A3:

    p_air = 0.7423*p_SL = 0.7423*1.225 kg/m^3

    p_air = 0.909 kg/m^3

    - Manometer equation for air and mercury are as follows:

    dP = - p_air*g*descent dP = - p_HG*g*dh_HG

    Combine the pressures dP:

    dh_HG = (p_air / p_HG) * descent

    dh_HG = 0.909*100 / 13.55*999

    dh_HG = 6.72 mm

    - Assuming density of air remains constant from 8000 m to 7900 m. From Table A3:

    p_air = 0.4292*p_SL = 0.4292*1.225 kg/m^3

    p_air = 0.526 kg/m^3

    - Manometer equation for air are as follows:

    @8000m @3000m

    dP = - p'_air*g*dz' dP = - p_air*g*dz

    dz' = p_air / p'_air * dz

    dz' = 0.909 / 0.526 * 100

    dz' = 173 m
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