Pressure on a window of a submarine

  • Thread starter indietro
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  • #1
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Homework Statement


a research submarine has a 20 cm diameter window 8 cm think. the manufacturer says the window van withstand forces up to 1x106N. What is the submarine's maximum safe depth?


Homework Equations


p = po + [tex]\rho[/tex]gd


The Attempt at a Solution


p = F/A = 1X106 / 0.03 = 3.2x107 Pa

so now do i use p = po + [tex]\rho[/tex]gd ??

but if i do what would be my [tex]\rho[/tex]?
[tex]\rho[/tex] = m/v = m/[tex]\pi[/tex]r2h
how would i find mass?

thanks in advance for help!
 

Answers and Replies

  • #2
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ok so i solved it with the help of a different post but i had a question:
she does p' = p1 + [tex]\rho[/tex]gd
p' - p1 = [tex]\rho[/tex]gd

and then she says that p1 = 1x105 Pa > atmospheric pressure. I was just wondering why?

can someone explain how
[tex]\Delta[/tex]p = p1 - po relates or is used in conjunction with p' = p1 + [tex]\rho[/tex]gd? is [tex]\Delta[/tex]p = p'
 
  • #3
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anyone??
 
  • #4
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i really just cant seem to wrap my head around pressure concepts, if someone knows how to explain it clearly or knows a website where i can see examples and stuff that would be great!
 
  • #5
PhanthomJay
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ok so i solved it with the help of a different post but i had a question:
she does p' = p1 + [tex]\rho[/tex]gd
p' - p1 = [tex]\rho[/tex]gd

and then she says that p1 = 1x105 Pa > atmospheric pressure. I was just wondering why?
it gets confusing with so many different letter variables like [tex] p, p_1, p_o, p', \rho [/tex] (and Pa for Paschals to boot), so let's define the variables:

let

[tex]p_o [/tex] = atmospheric pressure = 1.01(10^5) Paschals
[tex] p [/tex] = total pressure at depth d
[tex] \rho [/tex] = density of water = 1000kg/m^3)

Now when the top of the water is not sealed off (that is, it is exposed to the atmosphere),
[tex]p = p_o + \rho gd[/tex], that is , the total pressure at depth d is the sum total of the atmospheric pressure and the water pressure at that depth.
 

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