- #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.
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.