# Hazardous Materials Expert

## The Scenario

You are a sought-after world expert in managing hazardous materials, especially at huge scales like found after oil spills and nuclear meltdowns. We’re going to follow a year in your life as you fly out to locations all around the world to solve major environmental disasters using your scientific and mathematical skills.

## Math Scenario 1: Radiation Containing Concrete Dome

You are called out to Enewetak Atoll, also known as Bikini Atoll, the site of 67 of the United State’s nuclear tests [1].

After the tests, the contaminated soil from the islands was gathered and dumped in a big 100 metre diameter unlined crater, which was then covered with a thick concrete dome plug.

You’ve been called in to create a new, completely lined dome to contain the waste, before it catastrophically leaks out into the environment

First job is to work out the size of the underground dome you need. The underground dome is a hemisphere, and needs to store about 350000 cubic metres of radioactive material.

You’ll need the volume of a sphere formula, but halved since we’re only filling half of a sphere:

*V = 0.5 * 4/3 * PI * r^3*

*2V * 3 / 4 / PI = r^3*

*r = cube root of (1.5V / PI)*

*r = cube root of (1.5 * 350000 / PI)*

*r = 55.08 m*

So the underground dome will need to have a radius of 55.08 metres, which means the existing 50 metre radius crater will need to be enlarged.

## Math Scenario 2: Best Shape for Containment

After fixing the Bikini Atoll situation, your services for nuclear waste containment are in especially high demand. One of your new clients is running on a budget and needs to work out what shape of concrete containment tank is going to be best for their purposes.

They want the minimum amount of container wall material for the maximum storage volume.

You evaluate three common tank designs: a cube, a sphere and a cylinder. For the cylinder, you already know that the cylinder length needs to be two times the radius for the maximum volume to surface area ratio.

First you can find the general volume to surface area ratios for all these shapes. Then we can work out the specific ratios for a certain target volume – let’s use 350000 cubic metres from the previous scenario.

### Cube:

Then we can write a statement that relates the surface area to the volume capacity of the tank. First we calculate the surface area:

*Cube surface area = 6 * L * L*

*Cube surface area = 6L^2*

Then the volume:

*Cube volume = L * L * L*

*Cube volume = L^3*

So the volume to surface area ratio is:

*V/SA ratio = volume / surface area*

*V/SA ratio = L^3 / 6L^2*

*V/SA ratio = L / 6*

### Cylinder:

The volume of the cylinder is:

*V = PI * r^2 * l*

The surface area of the cylinder is:

*SA = 2 * circle end areas + body area*

*SA = 2 * PI * r^2 + 2 * PI * r * l*

The ratio of volume to surface area is:

*V/SA ratio = volume / surface area*

*V/SA ratio = PI * r^2 * l / [2 * PI * r^2 + 2 * PI * r * l]*

*V/SA ratio = PI * r^2 * l / [2 * PI * r (r + l) ]*

*V/SA ratio = rl / [ 2(r + l) ]*

But we know for maximum volume to surface area we need *l = 2r*:

*V/SA ratio = r * 2r / [ 2(r + 2r) ]*

*V/SA ratio = 2r^2 / 6r*

*V/SA ratio = r / 3*

### Sphere:

Volume first:

*V = 4/3 * PI * r^3*

Surface area:

*SA = 4 * PI * r^2*

The ratio of volume to surface area is:

*V/SA ratio = volume / surface area*

*V/SA ratio = 4/3 * PI * r^3 / 4 * PI * r^2*

*V/SA ratio = r/3*

We can work out the specific values of *r* and *L* for these shapes for a volume of 350000 cubic metres:

*Cube volume = 350000 = L^3 , so L = 70.47 m*

*Cylinder volume = 350000 = PI * r^2 * 2r, so r = 38.19 m*

*Sphere volume = 350000 = 4/3 * PI * r^3, so r = 43.72 m*

Now we can substitute these into the general volume to surface area ratios we’ve calculated:

**Cube:** V/SA ratio = L / 6 = 70.47 / 6 = 11.75

**Cylinder:** V/SA ratio = r / 3 = 38.19 / 3 = 12.73

**Sphere:** V/SA ratio = r/3 = 43.72 / 3 = 14.57

We want the biggest volume for the smallest surface area, so we want the biggest volume to surface area ratio – which is the spherical tank.

## Math Scenario 3: Concrete Tank Wall

After a catastrophic fire, a tanker ship carrying toxic chemicals is to be sunk and contained within a concrete tank for all eternity. The tank will have hemisphere ends and a cylindrical main section.

Once the ship has been contained in the tank, salt water will be pumped in to equalize the pressure – your jobs is to calculate how many litres of sea water must be pumped in.

The ship has a volume of 177000 cubic metres. The tank cylinder section has a radius of 30 metres (as do the hemispherical ends) and the cylindrical section is 230 metres long.

*Volume sea water = volume tank – volume ship*

*Volume sea water = sphere volume + cylinder volume – 177000*

*Volume sea water = 4/3 * PI * r^3 + PI * r^2 * L – 177000*

*Volume sea water = 4/3 * PI * 30^3 + PI * 30^2 * 230 – 177000*

*Volume sea water = 113097 + 650310 – 177000*

*Volume sea water = 586407 m^3*

## Real-Life Example: The Runit Dome at Bikini Atoll

Unfortunately, there is a real-life example that the first scenario was loosely based on – The Runit Dome at Bikini Atoll. You can read more here about the mess that has been created by inadequately storing large amounts of radioactive material.

Unfortunately, only the top of the dome is concrete, and the concrete is starting to crack and fall away. The crater in which the radioactive material sits is unlined – so the surrounding soil is now contaminated.

All up there’s about 111,000 cubic yards of radioactive material under this dome. That’s enough material to fill a cube measuring 48 yards on each side, or about 44 metres:

[1] https://www.theguardian.com/world/2015/jul/03/runit-dome-pacific-radioactive-waste