I need some help understanding this question, using the methods my professor uses. As far as I can tell, we should plot a graph of depth vs time, with percent solids removed for each (i.e. at 40 minutes and 0.5m depth, 55% of solids were removed). However, he then uses the table given below the problem with delta r and Zi to find the theoretical efficiency and I have no idea what those numbers are or how he got them (the numbers in the table are from an example problem and do not correspond to this homework question).

(Again, numbers here are not related to the above problem, I just need to understand what this is).

A settling column analysis is run on a type-2 suspension with the following results. The entries are suspended solid concentrations at stated times. Depth, m Time, 120 164 279 361 0.5 1.0 1.5 2.0 2.5 3.0 3.5 0 820 820 820 820 820 820 820 40 369 442 631 672 713 722 738 80 238 369 476 558 590 615 656 min. 160 107 213 287 353 402 460 492 200 66 164 230 287 344 394 418 240 41 115 180 238 262 320 360 280 33 90 148 187 230 262 303 426 492 533 574 Determine the theoretical efficiency (%) of a settling basin with a depth of 2.5 m, a volume of 2200 m, and an inflow of 13,200 m3/day. r 0.07 0.1 Zi 2.58 1.75 1.2 0.78 0.3 0.1 r * Z 0.1806 0.175 0.12 0.078 0.03 = 0.5836 0.1 0.1 Total Removal Fraction = 0.43 + (0.5836/3.0) = 0.625 = 62.5% A settling column analysis is run on a type-2 suspension with the following results. The entries are suspended solid concentrations at stated times. Depth, m Time, 120 164 279 361 0.5 1.0 1.5 2.0 2.5 3.0 3.5 0 820 820 820 820 820 820 820 40 369 442 631 672 713 722 738 80 238 369 476 558 590 615 656 min. 160 107 213 287 353 402 460 492 200 66 164 230 287 344 394 418 240 41 115 180 238 262 320 360 280 33 90 148 187 230 262 303 426 492 533 574 Determine the theoretical efficiency (%) of a settling basin with a depth of 2.5 m, a volume of 2200 m, and an inflow of 13,200 m3/day. r 0.07 0.1 Zi 2.58 1.75 1.2 0.78 0.3 0.1 r * Z 0.1806 0.175 0.12 0.078 0.03 = 0.5836 0.1 0.1 Total Removal Fraction = 0.43 + (0.5836/3.0) = 0.625 = 62.5%