Maximum Safe Depth Calculation for a Submarine with a Thick Window

[k3N70n]

Homework Statement

A research submarine has a 10.0 -diameter window 8.90 thick. The manufacturer says the window can withstand forces up to 1.10×10^6 . What is the submarine's maximum safe depth?

The pressure inside the submarine is maintained at 1.0 atm.

Homework Equations

p=F/A
p=p(not) + ρgd
A=Pi*r^2

The Attempt at a Solution

p = Fmax/A = Fmax/A = 1.1*10^6/(Pi * 0.05^2) = 1.40*10^8 Pa

p=p(not) + ρgd
1.40*10^8 = 101300 + 1030*9.81*d --> d= 13851m

I must be making some sort of error in understanding the theorey of this question. Any help would be greatly appriciated. Thank you kindly.

Kenton

 

[Andrew Mason]

Homework Statement

A research submarine has a 10.0 -diameter window 8.90 thick. The manufacturer says the window can withstand forces up to 1.10×10^6 . What is the submarine's maximum safe depth?

The pressure inside the submarine is maintained at 1.0 atm.

Homework Equations

p=F/A
p=p(not) + ρgd
A=Pi*r^2

The Attempt at a Solution

p = Fmax/A = Fmax/A = 1.1*10^6/(Pi * 0.05^2) = 1.40*10^8 Pa

p=p(not) + ρgd
1.40*10^8 = 101300 + 1030*9.81*d --> d= 13851m

You do not have to include the atmospheric pressure in your analysis. This is because at the surface the pressure is the same on the inside as outside so the net force is 0. As the submarine descends, the force on the window is due to water depth:

$$F = \rho gdA$$

$$d = F_{max}/\rho gA$$

I assume the maximum force is 1.10e6 N. Your answer should be about 10 m. deeper.

AM

 

[k3N70n]

Well I worked it out that way and got 13861 as you said though that is still wrong.

I just emailed my prof and it he gave me this hint

The given maximum force would be the net force taking into account the atmospheric pressure inside and the water outside." It looks like you are only considering the inward force from the ocean water, not the outward force due to the air pressure inside.

I'm not sure exactly how to proceed from here. Though I'll be working on it.

 

[civil_dude]

I agree with Andrew, that the sub's internal pressure should cancel out with the atmospheric pressure. Is there something amiss?

 

[Andrew Mason]

Well I worked it out that way and got 13861 as you said though that is still wrong.

I just emailed my prof and it he gave me this hint

I'm not sure exactly how to proceed from here. Though I'll be working on it.

You can add the sea-level atmospheric pressure (1 atm) to the water pressure to find the total outside pressure at at depth and then subtract the inside atmospheric pressure (1 atm) to find the net pressure on the window (and multiply by the window area to find the inward force on the window). The result is that the net force is just the pressure from the water x area of the window ($\rho gdA$).

I get a slightly different answer than you do, using your figures:

$$d = F/\rho gA = 1.10e6/1030*9.81*3.14*.05^2 = 13868 m$$.
To an accuracy of three significant figures the correct answer would be 13900 m.

AM

 

[k3N70n]

Thanks. I really wasn't understanding last night. I good nights rest was the best answer. Thanks for all your help.

Kenton