UNIT 4
Density and Pressure
Example 1
Example 2
Example 3
Example 4
Example 5
Example 6
Example 7

 

Density and Pressure in Fluids
Example 1

example1

In a movie a stuntman scuba diver at 10.3m underwater gets his air hose cut and looses his air supply so he ascends to the surface. He lungs are full with 5.00 X 10-6 m3 of air. It is a natural reaction to hold one's breath in such a situation, but the diver is trained to exhale to expel the expanding air on the way up. Fatal lung injury (pulmonary barotrauma) occurs when the lungs are expanded by more than 10%.
(a) How much can the diver ascend safely and not experience pulmonary barotrauma if he holds his breath?
(b) How many times would the air in his lungs expand when it reaches the surface?
Atmospheric pressure = 1.01 X 105 Pa
Density of water = 1.00 X 10-3 kgm3
Acceleration due to gravity = 9.81 ms-2

example1

(a) Let final unsafe depth = hf
Initial depth of diver = 10.3m
Initial Volume of lungs = 5.00 X 10-6 m3

Final volume of lungs = (5.00 X 10-6 m3) / 1.1
Final volume of lungs = (4.55 X 10-6 m3)

Initial pressure on Lungs = Atmospheric pressure + Pressure due to the water
Initial pressure on Lungs = (1.01 X 105 Pa) + ρgh
Initial pressure on Lungs = (1.01 X 105 Pa) + (1.01 X 105 Pa)
Initial pressure on Lungs = (2.02 X 105 Pa)

Unsafe pressure on Lungs = Atmospheric pressure + Pressure due to the water
Unsafe pressure on Lungs =1.01 X 105 Pa + ρgh
Unsafe pressure on Lungs =1.01 X 105 Pa+ ρghf ---Eq1
PV = Constant if T is constant:

Let Pi= Pressure at 10.3 m deep
Vi = Volume at 10.3 m deep
Pf = Pressure at unsafe depth
Vf = Volume at unsafe depth

PiVi = PfVf
Pf= [(PiVi) / (Vf)]
Pf= {[(2.02 X 105 Pa) X (4.55 X 10-6 m3)] / (5.00 X 10-6 m3)}
Pf= 1.82 X 105 Pa

Eq 1 becomes 1.01 X 105 Pa + ρghf = 1.84 X 105 Pa
ρghf = [(1.82 X 105 Pa) - (1.01 X 105 Pa)]
hf = [(1.82 X 105 Pa) - (1.01 X 105 Pa)] / (ρg}
hf = 8.26m
Assent = 10.3 m - 8.25m
The diver can safely ascend 2.05 m without exhaling

(b) Let Pi= Pressure at 10.3 m deep
Vi = Volume at 10.3 m deep
Pf = Pressure at surface
Vf = Volume at surface
Vf= [(PiVi) / (Pf)]
Vf= {[(2.02 X 105 Pa) X (5.00 X 10-6 m3)] / (1.01 X 105 Pa)}
Vf= 1.0 X 10-5 m3
(Vf) / (V1) = 2 times.


   
<
>
Concept by Kishore Lal. Programmed by Kishore Lal... Copyright © 2015 Kishore Lal. All rights reserved.