use differential equations to analyze a simplifiedmodel of dissipation of heavy crude oil spilled at a rate of S ft^{3}/sec into a flowing bodyof water. The flow region is a canal, namely, a straight canal of rectangularcross section, w feet widebyd feet deep, having a constant flow rate of v ft/sec; the oil is presumed to float in a thin layer of thicknesss (feet) on top of the water, withoutmixing.ProblemStatement
Theoil that passes through the cross section window in a short time ∆t occupies a box of dimensions s by wby v∆t. To make the analysis easier, ill presume that the canal isconceptually partitioned into cells of length L ft. each, and that within each particular cell the oilinstantaneously disperses and forms a uniform layer of thickness s_{i}(t) in cell i (cell 1 starts at the point of thespill). So, at time t, the ith cellcontains s_{i}(t)wL ft^{3 }ofoil. Oil flows out of cell i at arate equal to s_{i}(t)wv ft^{3}/sec,and it flows into a cell i at therate s_{i-1}(t)wv; it flowsinto the first cell at S ft^{3}/sec.
(a) Formulatea system of differential equations and initial conditions for the oil thicknessin the first three cells. Take S = 50gallons/min, which was roughly the spillage rate for the Mississippi Riverincident, and take w = 200 ft, d = 25 ft, and v = 1 mi/hr. Take L = 1000ft.
(b) Solvefor s_{1}(t).
(c) Ifthe spillage lasts for T seconds, what is the max. oil layer thickness in cell1?
(d) Solvefor s_{2}(t). What is themax. oil layer thickness in cell 2?
(e) Probablythe least tenable simplification in this analysis lies in regarding the layerthickness as uniform over distances of length L. Reevaluate your answer to part (c) with L reduced to 500 ft. By what fraction does the answer change?
I know that the oil in a cell will equal the oil going in - oil going out. Therefore s_{1}(t)wL = 0.1114ft^{3}/sec (S) -s_{1}(t)wv ft^{3}/sec....
Im just not sure that this answers part (a) and can I just solve for s_{1}(t) from this???