The bends during flight. Anyone who scuba dives is advised not to fly within the

Question

The bends during flight. Anyone who scuba dives is advised not to fly within the next 24 h because the air mixture for diving can introduce nitrogen to the bloodstream. Without allowing the nitrogen to come out of solution slowly, any sudden air-pressure reduction (such as during airplane ascent) can result in the nitrogen forming bubbles in the blood, creating the bends, which can be painful and even fatal. Military special operation forces are especially at risk. What is the change in pressure on such a special-op soldier who must scuba dive at a depth of 19 m in seawater one day and parachute at an altitude of 7.7 km the next day? Assume that the average air density within the altitude range is 0.81 kg/m3.

Explanation / Answer

Pressure at depth   h    is P1 = Po +seawater*g*h Now pressure at altitude h'  is P2 = Po  - air*g*h' Now the change in pressure: P1 -P2  =  ( Po +seawater*g*h )  - (Po  -air*g*h') P1-P2    =   seawater*g*h + air*g*h'   ------(1) seawater = 1030kg/m3 air = 0.87 kg/m3 h = 16m h'= 8.1km = 8100 m g = 9.8m/s2 Plug the values in the equation (1) and solve for the pressure difference . P1-P2 = 1030 *9.8 * 16 + 0.87*9.8* 8100 = 230564.6  = 2.31x105 Pressure at depth   h    is P1 = Po +seawater*g*h Now pressure at altitude h'  is P2 = Po  - air*g*h' Now the change in pressure: P1 -P2  =  ( Po +seawater*g*h )  - (Po  -air*g*h') P1-P2    =   seawater*g*h + air*g*h'   ------(1) seawater = 1030kg/m3 air = 0.87 kg/m3 h = 16m h'= 8.1km = 8100 m g = 9.8m/s2 Plug the values in the equation (1) and solve for the pressure difference . P1-P2 = 1030 *9.8 * 16 + 0.87*9.8* 8100 = 230564.6  = 2.31x105 Plug the values in the equation (1) and solve for the pressure difference . P1-P2 = 1030 *9.8 * 16 + 0.87*9.8* 8100 = 230564.6  = 2.31x105

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