Pressure on a submarine window problem

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SUMMARY

The discussion centers on calculating the maximum depth a submarine can dive without rupturing a circular window with a radius of 25 cm, which can withstand a maximum force of 1.23E6 N. The correct approach involves understanding the relationship between pressure and force, specifically using the formula P = F/A. The participants emphasize that the pressure variation across the window can be ignored for this calculation, and instead, a uniform pressure assumption should be applied. The correct maximum depth calculation should consider the maximum pressure the window can withstand, which is derived from the force divided by the area of the window.

PREREQUISITES
  • Understanding of fluid mechanics principles, specifically pressure calculations.
  • Familiarity with the formula P = F/A for pressure and force relationships.
  • Basic knowledge of integration, particularly in polar coordinates.
  • Knowledge of the density of seawater (1000 kg/m³) and standard atmospheric pressure (1 atm = 101325 Pa).
NEXT STEPS
  • Review the principles of pressure calculations in fluid mechanics.
  • Study the derivation and application of the formula P = F/A in various contexts.
  • Learn about integration techniques in polar coordinates for more complex fluid problems.
  • Explore the effects of pressure variations on submerged structures and how to account for them in calculations.
USEFUL FOR

This discussion is beneficial for engineers, physicists, and students studying fluid mechanics, particularly those interested in submarine design and pressure-related calculations in aquatic environments.

jamdr
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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:

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

Thanks for any help you can give me.
 
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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.
 
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.
 
If you are unfamiliar with integration in polar coordinates, you should definitely use Doc Al's procedure!
 
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.
 

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