Pressure on a submarine window problem

In summary: In this case, since pressure is in Pa and force is in N, the units must add up on both sides of the equals sign. So the units of pressure would be Pa and N, and the units of force would be N and Pa.
  • #1
jamdr
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.
 
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  • #2
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.
 
  • #3
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.
 
  • #4
If you are unfamiliar with integration in polar coordinates, you should definitely use Doc Al's procedure!
 
  • #5
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.
 

1. What causes pressure on a submarine window?

The pressure on a submarine window is caused by the weight of the water above it. As a submarine descends deeper into the ocean, the water above it becomes denser, creating a higher pressure against the window.

2. How does pressure affect the strength of a submarine window?

The higher the pressure on a submarine window, the stronger it needs to be in order to withstand the force. As pressure increases, the window must be made of thicker, stronger materials to prevent it from cracking or breaking.

3. What happens if a submarine window breaks due to pressure?

If a submarine window breaks due to pressure, it can lead to catastrophic consequences for the submarine and its crew. The rapid influx of water can cause the submarine to sink or become unbalanced, and the crew may be at risk of drowning or being exposed to extreme pressure and cold temperatures.

4. How do scientists design submarine windows to withstand pressure?

Scientists use materials with high strength and durability, such as titanium or reinforced glass, to design submarine windows that can withstand the high pressure of the ocean. They also use mathematical models and simulations to test and optimize the design for different depths and pressures.

5. Can the pressure on a submarine window be equalized?

Yes, the pressure on a submarine window can be equalized by filling the window with a fluid that has the same density as the surrounding water. This helps to distribute the pressure evenly and prevent the window from breaking. However, this method is only effective up to a certain depth, and at deeper depths, the window must still be designed to withstand high pressure.

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