Saturday, December 9, 2017

Venice and the Importance of Buoyancy

In class on Friday we talked about thermal expansion and how the earth's rising temperature leads to the rising sea level. The increase in temperature is of course due to human emissions of greenhouse gases, such as carbon dioxide. Greenhouse gases trap heat near the earth's surface and are a vital part of our atmosphere, but as more of these gases are released into the atmosphere, more heat is trapped. This poses serious threats to the environment and humans alike.

In Venice, which is built on a series of islands off the coast of northeastern Italy, flooding has historically been a problem. Venice is only about a meter above sea level, so when there is a high enough tide, parts of the city are submerged in water. The rising sea level makes this problem even more serious because it increases the frequency and severity of the floods. 

In 1987, the Venice Water Authority began to take steps to protect Venice and the other populated areas around the lagoon in which Venice is located. One of the steps was to start planning the implementation of gates at each of the three inlets from the Adriatic Sea into the lagoon. In 2003, after years of planning, the project began. 

The project is know as "MOSE", which stands for "MOdulo Sperimentale Elettromeccanico" (Experimental Electromechanical Model". When the project is finished, there will be a total of 78 gates (41 at one inlet, 18 at another, and 19 at the other). The gates are made of steel and mostly hollow. The smallest gate is 18.5 by 20 by 3.6 meters and the largest one is 29.5 by 20 by 4.5 meters. Each gate is housed in a structure on the seafloor and attached to two hinges. When "resting", the gate is filled with seawater. When there is a high tide, some of the seawater is expelled by pumping air into the gate, causing it to rise through buoyancy.  

What I found so fascinating about these gates is that, despite all of the complicated engineering that goes into their construction and implementation, as well as the importance that they serve, they are raised simply through buoyancy. This is not only simple, but probably more efficient than using something like hydraulics to lift something that weighs about 300 tons. Buoyancy is so simple, in fact, that I was able to calculate how much seawater, and how much air, is present within one of the gates when it is raised and approximately in static equilibrium. I was able to do this only using what we learned in class about torque, force, and buoyancy (I also used a bunch of triangles and basic trigonometry). 

The gate I decided to do my calculations for was one of the gates at the Malamocco inlet. This inlet is both the deepest and has the largest gates. The dimensions of the gate in question are 29.5 x 20 x 4.5 and the inlet is about 14 m deep. The gates are designed to stop a tide of up to 3 meters, so if there was no high tide, the gate would be 17 m above the seafloor. This was the situation I considered because I did not have to take into account the different heights of water on each side of the gate and how this would effect the static equilibrium. 

First I calculated the volume of air and volume of steel in the gate when it was empty. To do this, I used the density of air (1.225 kg/m^3) and steel (8050 kg/m^3) and the approximation of the mass of the gate of 300 tons and the total volume calculated from the dimensions.. The conversion from tons to kg is 1 ton=907.185 kg. The calculations for this are shown below:

300 tons * (907.185 kg/1 ton) = 2.72 * 10^5 kg

V-total = 29.5 m * 20 m *4.5 m = 2655 m^3

1.225 kg/m^3 * V-air + 8050 kg/m^3 * (2655 m^3 - V-air) = 2.72 * 10^5 kg

V-air = 2622 m^3
V-steel = 33.4 m^3

Then I figured out what forces and torques act on the gate in static equilibrium. A free-body diagram showing the torques, forces, angles of each torque with respect to the object, axis of rotation, and the distance the torques are from the axis of rotation are shown below:

The torque and force from gravity and buoyancy are shown acting on the center of mass of the gate, which I treated as a rectangular prism, even though the actual shape is slightly different. The axis of rotation is the hinge, for which there is no torque because the force acts at a radius of 0 with respect to the axis of rotation (note that the force from the hinge accounts for both hinges so the force for each hinge is half of what I calculated). 𝜃-1 = 31.7° and 𝜃-2 = 148.3°. The equations that show the sum of all the forces and all the torques are shown below:

ΣF = 0 N = F-H +⍴Vg - mg where V is the seawater displace by the gate and ⍴ is the density of the seawater
Σ𝝉 = 0 N*m = 𝝉-B + 𝝉-g = ⍴Vg*sin(31.7°)*10 m - mg*sin(148.3°)*10 m

For the buoyancy force, I had to figure out how much seawater was displaced because the gate is not completely submerged. I did so by using similar triangles and trigonometry. The diagram below shows the dimensions of the gate and the triangles I used to find the area of the gate that is out of the water:

I split the part of the gate that is out of the water into a rectangle and a triangle. The rectangle has sides of 3.53 m and 4.5 m while the triangle has a base of 2.78 m and a height of 4.5 m. I then calculated the area and multiplied by the width of the gate to find the volume that is out of the water:

(3.53 m * 4.5 m + 0.5*2.78 m*4.5 m) * 29.5 m  = 654.9 m^3

Subtracting this value from the total volume of the gate gave me the volume of the gate that is underwater and therefore how much seawater is displaced:

2655 m^3 - 654.9 m^3 = 2000.1 m^3 = V of seawater displaced

Plugging this back in the equation for buoyancy force and using the density of seawater (1029 kg/m^3), I got:

F-B = ⍴Vg = 1029 kg/m^3 * 2000.1 m^3 *9.8 m/s^2 = 2.02 * 10^7 N

With the buoyancy force, I then calculated the buoyancy torque:

𝝉-B = (2.02*10^7 N) *10 m * sin(31.7°) = 1.06 *10^8 N*m

Then I used the fact that the sum of the torques is 0 N*m to solve for the volume of air and seawater in the gate. To do so, I rewrote the mass of the gate as follows:

m = ⍴-steel*V-steel + ⍴-air*V-air + ⍴-seawater*V-seawater

and because the original V-air (2622 m^3)= V-air +V-seawater:

m = ⍴-steel*V-steel + ⍴-air (2622 m^3 -V-seawater) + ⍴-seawater*V-seawater

Now the sum of the torques can be written as follows:

Σ𝝉 = 0 N*m = 𝝉-B + 𝝉-g = 𝝉-B - (⍴-steel*V-steel + ⍴-air (2622 m^3 -V-seawater) + ⍴-seawater*V-seawater)g*sin(148.3°)*10 m

-1.06 * 10^8 = -(8050 kg/m^3*33.4 m^3 + 1.225 kg/m^3(2622 m^3 - V-seawater) + 1029 kg/m^3*V-seawater)g*sin(148.3°)*10 m

V-seawater = 1736 m^3
V-air = 885.6 m^3

And just for fun, I calculated the average density of the gate:

⍴-avg = (m-steel + m-air + m-seawater)/(V-air + V-seawater + V-steel) = 780 kg/m^3

Even though 885.6 m^3 is a lot of air, relative to the size of the gate, it is surprisingly small and only takes up about a third of the space in the gate. This shows how significant a difference of a factor of almost 1000 in densities can be and how a very simple concept can be used as a solution to a very large scale problem.


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