If I go 100 feet below the ocean surface, what is the weight of the water above my head?

Posted on March 17, 2008. Filed under: Uncategorized |

I have realized that it is finding answers to such questions that has led to the coining of the term ‘Curiosity kills’. But it’s a CK promise to answer reasonably sane questions. So here goes.

Since you have specified yourself as the test specimen in this experiment, we’ll start by throwing you in the ocean. Then we’ll figure out a way to measure the water above your head. Scrap idea. Note to self: Curiosity kills, but not literally.

The weight of water per cubic feet is 28.316847kg

- The density of fresh water is  62.4 pounds per cubic foot (28.3 kg/ 0.03 m3).

- Seawater, however, is denser: 64 pounds per cubic foot (29 kg/0.03 m3).

Since you have chosen the ocean to immerse yourself, I am going to consider seawater weight for our calculations.

Usually, a person’s weight is slightly less than the weight of the displaced amount of water. For example, a person who weighs 80kg displaces 79dm2 of water, which weighs 79kg, that is, he has about 1kg of negative buoyancy. So considering you displace 79kg of water when you are just immersed, at 100 feet you will displace around the same amount of water. But this again changes with your level of buoyancy which depends on the amount of air in your lungs

Here’s the part where you start reading slowly because the sentences are strung together with little complex words.

Now, if we define pressure as weight per unit area, the force exerted on a body immersed in a liquid is due to the weight of the liquid above the point where the pressure is being determined. Therefore the pressure of a liquid is directly proportional to its depth and the liquid exerts the same amount of pressure in all directions.

We take some good looking equations here to prove the point. Non-physics, non-maths and general ‘non’ people can skip to the answer which is 44.44ibf/in-sq.

For others:

Pressure = weight / area

= [density of fluid x volume x acceleration due to earth’s gravity] / [Area in ft-sq x lbm-ft / lbf-sec-sq (i.e. g1)]

Volume is the cross sectional area x height. Substituting this above,

P = phg / g1

Therefore, if the density of water is 64 lbm/ft-cube, and at a depth of 100ft,

=(64 lbm/ft-cube)x (100ft)x(32.17ft/sec-sq / 32.17lbm-ft/lbf-sec-sq)

=(6400 lbf/ft-sq) x (1 ft-sq / 144 in-sq)

= 44.44 lbf/in-sq.

If your are going ‘huh?’ at the end of the equation, I’ll make it simple. It’s a lot of pressure / weight at 100 feet and you don’t want to be there unless you can bet on discovering a hidden treasure.

All pressure and no brain makes you a dull boy. So how about you tell me how the china and other fragile items from shipwrecks remain intact for several years despite the pressure on the bottom of the ocean?

At this point, realization would ideally dawn that CK not only has good answers but also good questions. Read on.

Solids do not have air pockets in them to be effected by external pressure. Therefore, things like fragile china only get slightly smaller when equal pressure is applied on all sides. But for substances like wood, which is porous, you can forget you’ll ever recognize a sunken chest.

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