use differential equations to analyze a simplifiedmodel of dissipation of heavy crude oil spilled at a rate ofProblemStatementS ftinto a flowing bodyof water. The flow region is a canal, namely, a straight canal of rectangularcross section,^{3}/secwfeet widebydfeet deep, having a constant flow rate ofvft/sec; the oil is presumed to float in a thin layer of thicknesss(feet) on top of the water, withoutmixing.

Theoil that passes through the cross section window in a short time ∆toccupies a box of dimensionssbywbyv∆t.To make the analysis easier, ill presume that the canal isconceptually partitioned into cells of lengthLft. each, and that within each particular cell the oilinstantaneously disperses and forms a uniform layer of thicknesssin cell_{i}(t)i(cell 1 starts at the point of thespill). So, at time t, theithcellcontainssft_{i}(t)wL^{3 }ofoil. Oil flows out of celliat arate equal tosft_{i}(t)wv^{3}/sec,and it flows into a celliat theratesit flowsinto the first cell at_{i-1}(t)wv;Sft^{3}/sec.

(a) Formulatea system of differential equations and initial conditions for the oil thicknessin the first three cells. TakeS= 50gallons/min, which was roughly the spillage rate for the Mississippi Riverincident, and takew= 200 ft,d= 25 ft, andv =1 mi/hr. TakeL =1000ft.

(b) Solvefors_{1}(t).

(c) Ifthe spillage lasts for T seconds, what is the max. oil layer thickness in cell1?

(d) SolveforsWhat is themax. oil layer thickness in cell 2?_{2}(t).

(e) Probablythe least tenable simplification in this analysis lies in regarding the layerthickness as uniform over distances of lengthL.Reevaluate your answer to part (c) withLreduced 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???