The discussion revolves around determining the maximum safe depth for a research submarine based on the specifications of its window, including its diameter, thickness, and the force it can withstand. The context involves fluid mechanics and pressure calculations in a saltwater environment.
Some participants have provided guidance on the calculations and the need to ensure consistent units. There is an ongoing examination of the assumptions regarding pressure inside and outside the submarine, with some participants expressing uncertainty about their calculations.
There are mentions of needing to convert atmospheric pressure from atm to Pascals for consistency in calculations. The discussion also highlights the importance of using the correct density value for seawater in the calculations.
jason_r said:A research submarine has a 20.0 cm-diameter window 8.90 cm thick. The manufacturer says the window can withstand forces up to 1.10×10^6 N. What is the submarine's maximum safe depth?
The pressure inside the submarine is maintained at 1.0 atm.
The sub is in salt water
alphysicist said:Hi jason_r,
What have you tried so far?
jason_r said:I''ll show you what I've got so far
please let me know if I am doing this right, because i don't have a key or anything.
thanks
P=P_o + pgh
1.10 x 10^6= 1 + (1030)(9.8)(h)
alphysicist said:There are two problems with this line. You have 1.10 x 10^6 on the left side; that is the force the window experiences, but in this equation you need the pressure that's on the window. So how can you find the pressure on the window if you know the force on the window?
Also, the first term on the right side: you seem to be saying that P_o is 1 atm, which is definitely true; however, to match the other term you need P_o to be in units of Pascals.
jason_r said:ok, so..
this is what i have now:
P=P_o + pgh
35014087.48=(P_o(air) - P_o(sub)) + (1030)(9.8)(h)
h=3.47 x 10^3
(does the pressure of air on top of the sea cancel the pressure of air in the sub?)
Also can you confirm that i have the correct anser?
thanks
alphysicist said:Yes, those pressures will cancel if you calculate it in one step, and that answer looks right to me. (As long as you have the right value for the density of seawater.)