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

Pressure on a submarine window problem

  • Thread starter jamdr
  • Start date
13
0
I have this fluids problem I've been working on for a while, but I can't seem to get the correct answer. The problem is:

A circular window with radius 25 cm in a submarine can withstand a maximum force of 1.23E6 N. If the interior of the submarine is maintained at a pressure of 1 atm, approximately how deep the submarine can dive without rupturing the window? (You can assume sea water density of 1000 kg/m3.)

The pressure underwater can be found using P = P[0] + pgh, where P[0] is 1 atm (101325 Pa). So what I did was this:

1.23E6 N = 101325 Pa + 1000*9.8*h

and solved for h. The answer I got was 115 m. This is wrong, and I think it's because I didn't take into account the radius of the window. Since the pressure would be different at different points on the window, I can't just treat the window as a point. I think I need to integrate the pressure over the height of the window, but how would I do this? Would it look something like this:

[tex]\int_{0}^{0.25} \left(\rho*g*h\right) dh[/tex]

Thanks for any help you can give me.
 

arildno

Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,948
130
No it won't be correct to do so either.

Force=pressure*area,

hence, you need to integrate the pressure over the circular area in order to find the net force.
 

Doc Al

Mentor
44,803
1,061
approximate, don't integrate

Your main mistake, as arildno explained, is confusing force with pressure ( P = F/area).

First figure out the maximum pressure that the window can withstand. I would ignore pressure variation over the height of the window--just assume a uniform pressure. (Don't bother trying to actually integrate the pressure over the window surface including its variation with height: too hard!)

Once you have that maximum pressure, then figure out how deep the sub can go.
 

arildno

Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,948
130
If you are unfamiliar with integration in polar coordinates, you should definitely use Doc Al's procedure!
 

enigma

Staff Emeritus
Science Advisor
Gold Member
1,738
9
Don't forget that the pressures add up. You need to account for the pressure on the inside pushing outwards as well.

As was mentioned above, anytime you have an equality sort of like:

X Pa = Y N

You know you've done something wrong. The units need to add up on both sides of the equality sign, or else they're not equal.
 

Related Threads for: Pressure on a submarine window problem

  • Posted
Replies
2
Views
2K
Replies
4
Views
3K
Replies
9
Views
4K
Replies
7
Views
293
Replies
7
Views
14K
  • Posted
Replies
1
Views
3K
Replies
3
Views
3K
  • Posted
Replies
10
Views
7K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top