Question

Lubricating oil is to be treated continuously in a tower using a selective solvent to remove naphthenic material. The parameter R is defined as R = bbl solvent used/bbl oil treated and the treatment cost, without recovery of the solvent, in pesos per year is given by 18,600 R2 – 111,000R + 217,000. It costs Php 0.40 to separate each barrel of solvent circulated through the tower and 300 bbl/day of oil will be processed for 320 days per year. The most economical value of R to be used in bbl solvent per bbl oil is .

a.              1.92                     b.               2.89                     c.              3.44                    d.              5.77

Answer 1

RE bbl of solvent Used obl of Oil treated = R (bbl of oil treated) bbl of solvent used now, 300 bbl of oil is treated day for 320 days year { bbl of solvent used 300 bbl. 320 days day yeax RE bbl of solvent used. (96000 bb! year I bbl of solvent → 0.4 php To separate is used. is used. » To separate separate R[96000] bb) of solvent 0.4 (96000). Php cost of separation of solvent (in php is year) 0.4 (96000) R = 38400R

This cost of separation is the Recovery cost of solvent. NOW cost without recovery fin phe): year 18600 R² - 111000R +217000 Total cost including Recovery cost (in phe year becomes 18600R? illooor + 38400R t 217000 > 18000R? 72600R + 217000 Total cost (TC) (in php year, now, Total cost should always always be minimum, first derivative of TC should be Hence to zero. kept equal V d (70) = dR 18000 (2R) - 72600 = 0

11 18600 (2R) 72600 = ) R= 1,951 h 1.92 Hence; at minimum. lor this value So, this is of Ri TC will be the optimum value value) economical of Ro correct. is so a option
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