• Support PF! Buy your school textbooks, materials and every day products Here!

Calculating pressure in high altitude?

  • Thread starter dmk90
  • Start date
  • #1
8
0

Homework Statement


What is the change in pressure on a special-op soldier who must scuba dive at a depth of 20 m in seawater one day and parachute at an altitude of 7.6 km the next day? Assume that the average air density within the altitude
range is 0.87 kg/m3.

Homework Equations


p=p0 + ρgh

The Attempt at a Solution


I understand how to calculate pressure using the above formula for something at the surface, but how do I apply it to a high altitude? In the solution manual, they just ignore atmospheric pressure, so I am confused. At that high altitude, doesn't it mean the person is not affected by the usual atmospheric pressure of 1 atm?
 

Answers and Replies

  • #2
lightgrav
Homework Helper
1,248
30
Pressure is less at the top of the sea than at the bottom of the sea ...
we live on the bottom of a "sea of air" that's only about 8km deep.
Pressure is zero at the "top" of the atmosphere, the edge of space.
 
  • #3
19,927
4,098
In the solution manual, they just ignore atmospheric pressure, so I am confused. At that high altitude, doesn't it mean the person is not affected by the usual atmospheric pressure of 1 atm?

I doubt that the solution manual ignored the atmospheric pressure at the surface of the earth. Please show us how the manual solution got the pressure at 7.6 km.

Chet
 
  • #4
8
0
This is from the solution manual:

The gauge pressure at a depth of 20 m in seawater is
p1swgd = 2.00 x 105 Pa
On the other hand, the gauge pressure at an altitude of 7.6 km is
p2airgh=6.48 x 104 Pa
Therefore, the change in pressure is
Δp = p1 - p2 = 1.4 x 105 Pa
 
  • #5
19,927
4,098
This is from the solution manual:

The gauge pressure at a depth of 20 m in seawater is
p1swgd = 2.00 x 105 Pa
On the other hand, the gauge pressure at an altitude of 7.6 km is
p2airgh=6.48 x 104 Pa
Therefore, the change in pressure is
Δp = p1 - p2 = 1.4 x 105 Pa
The result for 7.6 km from the solution manual is incorrect. They have the sign wrong. It should be p2= - ρairgh= - 6.48 x 104 Pa. The absolute pressure at 7.6 should be 1. x 105 - 6.48 x 104 Pa. This is where the atmospheric pressure at the surface comes in.

Of course, this error makes the final answer incorrect also.

Chet
 
  • #6
8
0
Isn't 105 only applies to things at roughly sea level? Wouldn't something at 7.6 km up not be subjected to this pressure?
 
  • #7
19,927
4,098
Isn't 105 only applies to things at roughly sea level? Wouldn't something at 7.6 km up not be subjected to this pressure?
Sure. It''s less. That's what I calculated.

Chet
 

Related Threads on Calculating pressure in high altitude?

Replies
7
Views
1K
Replies
3
Views
655
  • Last Post
Replies
3
Views
604
  • Last Post
Replies
5
Views
6K
  • Last Post
Replies
3
Views
5K
Replies
1
Views
1K
Replies
5
Views
824
Replies
2
Views
836
Replies
2
Views
2K
  • Last Post
Replies
0
Views
1K
Top