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.

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

sorry what a asnns

explanation: