What is the Maximum Safe Depth for a Submarine Window?

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SUMMARY

The maximum safe depth for a submarine window with a diameter of 20cm and thickness of 8cm, capable of withstanding forces up to 10^6N, is approximately 3153 meters. The pressure exerted on the window is calculated using the formula P = P(atmosphere) + ρ*g*h, where P(atmosphere) is 101,300 Pa, ρ is the density of seawater (1030 kg/m³), and g is the acceleration due to gravity (9.8 m/s²). The calculations confirm that the internal pressure of 1 atm must be considered when determining the external pressure at depth.

PREREQUISITES
  • Understanding of hydrostatic pressure principles
  • Familiarity with basic physics equations, specifically P = F/A
  • Knowledge of seawater density (1030 kg/m³)
  • Ability to perform calculations involving atmospheric pressure (101,300 Pa)
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  • Study hydrostatic pressure calculations in fluid mechanics
  • Learn about the structural integrity of materials under pressure
  • Research the effects of depth on submarine design and safety
  • Explore advanced topics in pressure vessel engineering
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Marine engineers, submarine designers, physics students, and anyone involved in underwater exploration or pressure vessel safety will benefit from this discussion.

baddin
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1. A research submarine has a 20cm diameter window, 8cm thick. The manufacturer says the window can withstand forces up to 10^6N. What is the submarine's maximum safe depth? The pressure inside the submarine is maintained at 1atm.



2. Homework Equations P = P(atmosphere) + ρ*g*h


3. I tried finding the pressure on the window, using P = F/A and got the maximum pressure the submarine window can handle as P = 10^6/(pi*(0.1)^2) = 3.18*10^7 Pa.
I then used P = P(atmosphere) + ρ*g*h
and substitued, P as 3.18*10^7, P(atmosphere) as 101,300, ρ(used density of seawater) as 1030, g as 9.8 and found h to be approximately 3140m.
The solutions found h to be 3153m.
I don't know where I have gone wrong...
 
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Did you take into account that the internal pressure is 1 atm?
 
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Thanks I didn't consider it before but thinking about it in force perspective so it made more sense. I did Net force = Force ourtside window - Force inside window. Therefore 10^6 = Fout - Fin. Then Fout = P(outside)*A and Fin = P(inside)*A. 10^6/A = P(outside) - P(inside). 10^6/A = ρ*g*h + P(atmosphere) - P(inside submarine). Now P(atmosphere) = 1atm=P(inside submarine), ρ.=1030 Therefore 10^6/A = ρ*g*h.
Then I h to be (10^6)/(pi*(0.1^2)*9.8*1030) = 3153m.
THANK YOU VERY MUCH!
 

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