2007-10-28 06:30:54 · 5 answers · asked by MeDIGURU 1 in Science & Mathematics ➔ Physics
Answer 1 is good, but maybe this will help too. The pressure at the surface is about 1 atmosphere. The pressure increases by 1 atm for every 10.3m you descend, so at 120m, you have a pressure of 120/10.3 + 1 = 12.6 atmospheres. That's about 186psi.
2007-10-28 06:42:21 · answer #1 · answered by Experimentor 2 · 0⤊ 0⤋
P = F/A; where P is pressure, F is force, and A is the area the force is acting on. The area in this case is the surface area of the sub. But the good news, because of Pascal's Law of pressure, we can simply specify an area (like one square foot) knowing the pressure there will be pretty much the same everywhere else on the sub's surface. So, let's call A = 1 ft^2.
Now, what's the force? It's the force F = force of a column of water h feet high above the sub with area A = 1 ft^2 plus the force of the air on that column of water (about 14 psi (1 atmosphere) X 144 in^2 (the 1 ft^2 area) ~ 2000 pounds).
For the force of the sea water F = rho V = rho h A; so the force varies with the depth and surface area. But as P = F/A = rho h, we see the pressure varies only with depth for a given weight density. Sea water density rho ~ 64 lb/ft^3; so plug in to find the pressure due to water only. Note mass density Rho = rho/g in case you have the mass density number and not the weight density. Also 1 m ~ 3.3 feet for unit consistency with pounds and square feet.
Thus, the total pressure P = F/A = rho h + 2000 pounds/ft^2 = 64 lb/ft^3 (120 m * 3.3 ft/m) + 2000 lb/ft^2 ~ 25000 + 2000 = 27000 pounds/ft^2. Roughly 10 tons per square foot of pressure; for more precision, you can do the math.
This enormous pressure on the hull of a submarine at depth is why they are so very vulnerable to overpressure crushing when a depth charge or torpedo explodes near them That is, they are already near crushing; that little extra pressure from the explosion puts them over the limit...smoosh.
2007-10-28 07:06:27 · answer #2 · answered by oldprof 7 · 0⤊ 0⤋
Pressure under a fluid is depth multiplied by the density of the fluid. Assume water has a density of 1g/ml the the pressure exerted at this depth is 12000g/sqcm or 12kg/sqcm in addition to the atmospheric pressure above.
2007-10-28 06:36:34 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Basically, P= P0 + (rho)g (delta h )
rho = mass/volume . and delta h =120 m, P0 = atmospheric pressure =1.01 x 10^5 Pascal. V can be equal to A(dS) , also P can be equal to F/A
2007-10-28 09:06:08 · answer #4 · answered by y 2 · 0⤊ 0⤋
About 6 pounds per square centimeter.
2007-10-28 06:38:37 · answer #5 · answered by Professor Sheed 6 · 0⤊ 1⤋