Saturday, July 31, 2010
If one wanted to "get rid" of a lot of oil quickly this might prove an effective way of doing so if the object is to protect beaches and ecosystems from being overwhelmed by a coating of oil The trade-off is that a lot of toxins may end up in the water column. Spreading out the toxins over a large volume of water reduces the harm to any particluar location. There may be organisms which can tolerate a small amount of toxins present but as with other pollutants biomagnification can increase the harm done if toxins enter the food chain.
The results of experiments such as these are used to determine the properties of molecules and are useful in identifying the presence of chemical compounds in a mixture. Two analytical tools used in chemistry and biology similar to diffusion are chromatography and electrophoresis.
Friday, July 30, 2010
Certainly errors were made in the Gulf of Mexico and it is likely that there are still lessons to be learned. Both the oil industry and the government need to review what happened and see what can be done to avoid making the same mistakes in the future.
Thursday, July 29, 2010
edit: For one dimension half of the particles will be within 0.674 σ of the center of the normal distribution. The fraction within 1 σ is 0.683, within 2 σ it is 0.954 and within 3 σ one will find 0.997 of all the particles. The particles slowly spread over time with σ (the diffusion length) acting as a scale factor. In two dimensions the radial distribution giving the density of particles as a function of the radius, r, only is also a normal distribution but with the one dimensional σ replaced by √2 σ.
Wednesday, July 28, 2010
Another problem is that diffusion is not a very efficient process. The fundamental law of diffusion is known as Fick's law which defines the diffusion coefficient. From Brownian dynamics and Einstein's dispersion relation (1908) we know that the root mean square (rms) dispersion of a number of particles is proportional to the square root of the time, t, and inversely proportional to the square root of the particle mass, M,
So diffusion would be more efficient for smaller particles of oil and molecules with lower molecular mass if γ, a drag coefficient, is relatively constant. A similar dependence of the rate on mass is known as Graham's Law*. One should also note that there are other physical processes such as agitation of the water and shear stresses which tend to disperse the particles.
*edit: Graham studied both effusion and diffusion. He showed that the square of the times of diffusion was proportional to the densities.
The dispersant initially used by BP was Corexit 9527 which is slightly toxic and the EPA has asked BP to switch to the less toxic Corexit 9500. Another dispersant used in oil spills is Dispersit. Before BP can do anything in the Gulf it has to file a plan with the government which needs to be approved. BP has posted its plans for the use of dispersants as part of its response to the oil spill on its website. These plans included the subsea injections of dispersants during the initial capping operation and topkill. The intent was to let nature take its course. They also note the need to monitor the water column.
Oil can also be broken up by mechanical means as was discovered by Joseph Plateau about 140 years ago. He altered the physical properties of oil in water and rotated the oil breaking it up into smaller particles. Physical forces act on the particles of oil and allow it to disperse through diffusion which is driven by the random motions of the particles like that in Brownian motion. One would expect the oil in suspension in the water column to spread both vertically and horizontally through diffusion and also by convection and currents.
There are still concerns about the persistence of the oil in the water column and its subsequent effect on the environment.
Tuesday, July 27, 2010
Monday, July 26, 2010
Monday, July 12, 2010
One can find one half the square root of 5 since it is the hypotenuse of a right triangle with sides 1 and 1/2. This length is added to one half the radius of the circle to find cos 2θ. Drawing a perpendicular at half this distance from the center gives two of the points of the pentagon at angles 2θ and 3θ. Bisecting this the angles between these points and the horizontal axis gives the points at angles θ and 4θ. The final point is at angle θ = 0°.
The image above shows the actual construction with additional information to make it easier to follow the construction.
Sunday, July 11, 2010
Saturday, July 10, 2010
The two lower side sheets which connect with the lower base sheet have a curved intersection which drops away from the upper level of the pentagon. If one looks closely one can see where these sheets intersect with the styrofoam at the bottom. The tree produced is very close to what was computed. The surface tension may have caused the soap films to adapt to balance the forces on them.
One has to be quick to capture these films with a weak soap solution. I had to hold the camera in one hand and dip the wire frame with the other. It took several attempts and a little coordination to successfully capture the trees before they disappeared.
Monday, July 5, 2010
Sunday, July 4, 2010
From the paths between the points, a-b, a-c-d, etc. we get a set of equations involving the unknown lengths and the directions (an e subscripted by one of the angles). From the way in which the paths were chosen the first three equations each introduce a new pair of unknown lengths. One can solve these equations for one pair of unknowns at a time since we can use the previously found functions for the lengths.
To solve for a pair of unknown lengths we can take the dot product of the equation with two independent vectors, in this case the unit vectors with the angles α and β, and it is useful to introduce the notation for a pair of dot products. (U V) is essentially a matrix and its product with a vector is a vector.
The resulting sets of equations for the unknown lengths are,
And by multiplying both sides by the inverse of the matrix on the left we get the linear functions for the unknown lengths.
Once the functions have been found we can then solve the last equation for the unknown angle, α.
If the points are the vertices of a pentagon, we find that α = 90° and we get a plot that looks more like a honey comb structure.
Friday, July 2, 2010
Bubble Science Kit - Amazon
Bubble Science Kit - Target
Bubble Builder - Scientifics