# Simple Question on Pressure

1. Nov 6, 2004

### Spectre5

I have been working on this problem, but I am not getting the books answers....

A submarine has 20-cm diameter windows that are 8.0-cm thick. They can withstand a force up to 1.0 x 10^2 N. What is the maximum safe depth (depth where the windows will not break). The pressure in the submarine is kept at 1.0 atm.

Ans = 1003 m

2. Nov 6, 2004

### Tide

What exactly have you tried so far?

3. Nov 8, 2004

### Spectre5

anyone?? This seems so simple but I don't get the correct answer...here is what I have done:

Given:
diameter of window = 20 cm = .2 m
thickness of window = 8 cm = .08 m
pressure in the sub = 1 atm = 101.3 KPa
Max force the window can withstand = 1.0 x 10^6 N

So...Here is what I did:

$$P=\frac{F}{A}$$

$$P=\frac{1.0\times10^6 N}{\pi(\frac{.2 m}{2})^2}$$

$$P=31.831\times10^6 Pa$$

And since the pressure in the the sub resists it....

$$P = 31.831\times10^6 Pa - 101.3\times10^3 Pa$$

$$P = 31.7297\times10^6 Pa$$

Now using the equation for hydrostatic pressure...

$$P = P_o+\rho gh$$
where:
P = the pressure (calculated above)
P_o = pressure above the surface (101.3 KPa here)
rho = mass density (given in book as 1030 kg/m^3 for seawater)
g = gravitational constant (9.81 m/s^2)
h = depth (what we are looking for!)

Thus:

$$P = P_o+\rho gh$$

$$31.7297\times10^6 Pa = 101.3\times10^3 Pa+(1030 kg/m^3)(9.81 m/s^2)h$$

$$h = 3130 m$$

but the answer is 1003 m

I never used the thickness of the glass....I don't see where it would be needed...but then again I messed up somewhere?! Please help!

4. Nov 9, 2004

### Spectre5

anyone?? I know this is not very hard

5. Nov 9, 2004

### Spectre5

does anyone know where I went wrong?? Or how I should/could use the thickness (or is it not needed).....

6. Nov 9, 2004

### KingNothing

I think it may have something to do with the varying net pressure on the window in relation to the edges (the center can support less than the edges)...but I'm not sure of that, and I don't know how to do those calculations.

7. Nov 9, 2004

### Spectre5

hm...that does make sense....but my teacher has not gone over anything like that and I don't see it in my book (at least not in the section that this problem is in)

That would make sense also, becuase then the thicknesw would (I think) somehow come into play...hm...anyonw know how to go about computing this?

8. Nov 10, 2004

### PerennialII

The other way around it to include the edges of boundary supports would be to apply plate bending theory contra to the uniaxial tension theory you've applied til now. Or I suppose we could simplify it to a beam bending problem, to some extent at least, do such topics come about in the topics you're working on ?

9. Nov 10, 2004

### Staff: Mentor

I gave your work a once over: Looks OK to me. (However, the atmospheric pressure within the sub balances the atmospheric pressure above the ocean---so you don't have to include it. Doesn't change the answer much though.) Perhaps the book (or one of your starting numbers) is just wrong?

(Don't worry about pressure variations over a 20 cm window at that depth. Or about beam bending in the glass.)

10. Nov 10, 2004

### Spectre5

hm....maybe the answer is just wrong then...I will just leave it at what I have....About the bending in the glass and all...it is not discussed at all in the section...I also kind of thought that at this depth the variation would not have that much of an effet...and the effect that it does have, I don't think would account for the 2000+ meter difference in my answer :/

I will just assume the book is wrong :) lol...until I get the solution from my teacher and see if I missed something

11. Nov 10, 2004

### Spectre5

btw, I also checked the number AGAIN with the book....I used the same numbers they gave in the problem..