Answer:
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
