First,
let’s look at the pressure associated with Challenger Deep, the deepest point
of the Mariana Trench. Challenger Deep is at a depth of 35,756 feet (10843m).
This is more than a mile deeper than Mount Everest is tall! At this depth, the
density of sea water actually increases by 4.96% due to compression because of
the great weight of the sea water above. The normal density of sea water is
1.025 x 10

^{3}kg/m^{3}. The density of sea water at Challenger Deep is (1.025 x 10^{3 }kg/m^{3}) x (1.0496) = 1.076 x 10^{3}kg/m^{3}.
We
can determine the pressure at Challenger Deep using the equation:

P = ρgh + pressure of the atmosphere
where P= pressure, ρ= density of sea water, g=
9.80m/s

^{2}, h=10843m, and the pressure of the atmosphere= 1.013 x 10^{5}N/m^{2}
So
P = (1.076 x 10

^{3}kg/m^{3})(9.80m/s^{2})(10843m) + 1.013 x 10^{5 }N/m^{2}= 1.144 x 10^{8}N/m^{2}
To
put this in perspective, this pressure is the equivalent of three SUV’s sitting
on your big toe!

Clearly,
Cameron’s Deepsea Challenger submarine had to be well outfitted to withstand
these high pressure conditions. The Deepsea Challenger looked like this:

http://deepseachallenge.com/the-latest/page/4/
The
submarine was 7.3m long and had a mass of 11.8 tons (10705kg). Cameron sat in a
tiny compartment made out of thick steel. The remainder of the vertical column
was made out of syntactic foam called Isofloat, a solid composed of hollow
microspheres.

For
the sake of calculations, I approximated the submarine to be a cylinder with
diameter of 2.0m (I couldn’t find a measurement for the diameter of the
submarine, but by looking at scaled images of it, I concluded that its width
was a little over ¼ of its length). The radius is 2.0m/2 = 1.0m.

I
decided to determine the buoyant force on the submarine when it was submerged
in the sea water at Challenger Deep. The buoyant force is the net force on the
submarine exerted by fluid pressure and acts upward.

F

↓_{1}Cylinder

_{2}

F

_{B }= F_{2}-F_{2 }= ρ_{F}gA(h_{2}-h_{1}) = ρ_{F}gAΔh
The
density of the fluid (compressed sea water) = 1.076 x 10

^{3}kg/m^{3}, g= 9.80m/s^{2}, A = πr^{2}= π(1.0m)^{2}= π m^{2}, Δh = 7.3m
So:

F

_{B}= (1.076 x 10^{3}kg/m^{3})(9.80m/s^{2})(π m^{2})(7.3m) = 2.4 x 10^{5}N
Finally,
despite its capabilities to withstand intense pressure, the Deepsea Challenger
actually compressed and lost three inches (0.0762m) in length under the
pressure at Challenger Deep. Normally, when an object is submerged in a fluid
(which exerts a pressure on the object in all directions), the object is
subjected to inward forces from all sides and its volume will decrease. I read
in the textbook that when there is a volume change (as when an object is
submerged in a fluid), we use the Bulk Modulus B. However, the only data I
could find was that the length of the Deepsea Challenger changed, not its
radius. Therefore, although the volume of the submarine changed due to the
change in length, we can also just examine this problem through the Young’s Modulus
E since the submarine was compressed (force acted inwardly on the object) with
just a change in its length. This compression is due to the force of the
pressure at Challenger Deep. We can use the Young’s Modulus E to approximate
the magnitude of this force. The submarine’s volume was composed of 70%
Isofloat and this would therefore be the most appropriate Young’s Modulus E value
to use. However, I could not find a value for the Young’s Modulus E of Isofloat
so I will instead use the second most abundant material composing the
submarine, steel. For this calculation, we assume that the whole submarine is
made of steel.

Young’s
Modulus:

ΔL
= (1/E)(F/A)L_{0}or F/A = E(ΔL/L

_{0}) or E = (F/A) / (ΔL/L

_{0})

where L

_{0}is the original length of the object, A is the cross-sectional area, ΔL is the change in length due to the applied force F, and E is the Young’s Modulus.

So
for the Deepsea Challenger:

L_{0}= 7.3m

ΔL = 0.0762m

A= πr

^{2}= π(1.0m)

^{2}= π m

^{2}

Young’s Modulus E = 200 x 10

^{9}N/m

^{2}(for steel)

So
the applied force of the pressure that causes the change in length can be
calculated using:

F=E(ΔL/L_{0})A = (200 x 10

^{9}N/m

^{2})(0.0762m/7.3m)(π m

^{2}) = 6.6 x 10

^{9 }N (a very large force!).

Note:
The Young’s Modulus E of Isofloat is likely smaller than the Young’s Modulus E
of steel so the applied force of pressure is realistically smaller than this
calculated force.

Calculating
the approximate Bulk Modulus of the Submarine:

When
an object is submerged in a fluid (which exerts a pressure on the object in all
directions), the object is subjected to inward forces from all sides and its
volume will decrease. I read in the textbook that when there is a volume change
(as when an object is submerged in a fluid), we use the Bulk Modulus B
(although we didn’t discuss it in class).
Since
the length of the submarine was reduced by 0.0762m at Challenger deep, the
volume (V= πr

Volume
of Deepsea Challenger at Surface of Ocean= V^{2}h) of the submarine also changed:_{0 }= π(1.0m)

^{2}(7.3m) = 22.9m

^{3}

Volume of Deepsea Challenger at Challenger Deep = V

_{f }= π(1.0m)

^{2}(7.3m-0.0762m) = 22.7m

^{3}

ΔV= V

_{f}-V

_{0}= -0.239 m

^{3}

Pressure
at Surface of Ocean = P

= 1.076 x 10_{0}= ρgh + pressure of the atmosphere^{3}kg/m

^{3})(9.80m/s

^{2})(0.00m) + 1.013 x 10

^{5 }N/m

^{2}= 1.013 x 10

^{5}N/m

^{2}

Pressure at Challenger Deep = P

_{f}= 1.144 x 10

^{8}N/m

^{2}

ΔP = P

_{f}– P

_{0 }= 1.143 x 10

^{8}N/m

^{2}

Bulk
Modulus:

ΔV/V_{0}= -(1/B)ΔP or B = -ΔP/(ΔV/V

_{0})

where ΔV is the change in volume, V

_{o}is the original volume, ΔP is the change in pressure, and B is the bulk modulus.

The minus sign means that the volume decreases with an increase in pressure.

Using
our knowledge of the situation, we can calculate the Bulk Modulus B of the
submarine. Since the submarine was composed of mostly Isofloat (70% of the
volume of the submarine was Isofloat), we can say that the Bulk Modulus we
calculate is most closely related to the Bulk Modulus of Isofloat.

B
= -(1.143 x 10

^{8}N/m^{2}) / (-0.239 m^{3}/22.9m^{3}) = 1.1 x 10^{10}N/m^{2}
This
Bulk Modulus for Isofloat is large, but it is smaller than the Bulk Modulus of
steel (140 x 10

^{9}N/m^{2}). This supports our above prediction that the Young’s Modulus of steel is larger than the Young’s Modulus of Isofloat and consequently caused our applied force of the pressure on the submarine to be slightly larger than it should realistically be.**Referenced Article:**

**Video Clip:**

http://www.youtube.com/watch?v=Y2tm40uMhDI

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