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Mathematical modeling of sewage treatment problem
Question 1. It is assumed that the factory adopts similar precipitation method, that is, sewage treatment is precipitated according to binomial distribution. The obtained treatment rate is P = 10%/ hour, and the residual rate is H = 1-P = 90%.

If there are 24 hours in a day, it will leave (90%) 24 dirt = 7.98%.

Let the remaining half length of t be H T = 50%, and T = 6.5788 hours is deduced.

Question 2. Let its capacity be V, and in order to keep the balance of the tank contents, the inflow should always be equal to the outflow, that is, the treated water and extractive solution that flow out every hour *** 100KG.

Suppose there is water in the container containing dirt V2, water V 1, and treated dirt V3(V3+V2+V 1=V). The amount to be removed this time is treated dirt V3, water containing dirt (100-V3)/(V2/(v/kloc-0).

100KG of sewage flows in every hour, of which DV 1 = 100× 0.4 and DV2 = 100× 0.6. Since the precipitation probability is binomial, it is assumed that there is water containing Q2 and water Q 1 in the original container, and the sewage Q3 is treated for one hour.

After n hours, V2 = Q2 * 0.9N+DV2 * (0.9 (n-1)+0.9 (n-2) ...+0.91).

After simplification, V2 = Q2 * 0.9N+DV2 * (0.9 (1-0.9 (n-1))/(1-0.9))

Data analysis: take lim(N tends to infinity) v2 = dv2 * 9 =100 * 0.6 * 9 = 540kg.