ch16 - Exercise 16.1 Subject Mass balance check on test...

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Exercise 16.1 Subject: Mass balance check on test data around a leaching unit. Given: Feed rate to extractor of 6.375 lb/h with 10.67 wt% moisture and 0.2675 g oil per g dry, oil-free flakes. Solvent rate to extractor of 10.844 lb/h. Extract of 7.313 lb/h with 15.35 wt% oil. Leached solids with 0.0151 g oil per g dry, oil-free flakes. Assumptions: Solvent is n-hexane. Extract contains no solids. Leached solids contain all of the moisture in the feed and are wet with solvent. Find: Check on mass balances around the leaching unit for oil and hexane. Mass ratio of oil to flakes in leached solids. Analysis: Make calculations with AE units, noting that g/g is the same as lb/lb. Calculate the flow rates of components in the feed: Moisture = 0.1067(6.375) = 0.680 lb/h moisture Dry flakes = 6.375 – 0.680 = 5.695 lb/h dry flakes Oil = [0.2675/(0.2675 + 1.00)] 5.695 = 1.202 lb/h oil Dry, oil-free flakes = 5.695 – 1.202 = 4.493 lb/h dry, oil-free flakes Calculate flow rates in the leached solids, assuming all dry, oil-free flakes are in the leached solids: Oil = 0.0151(4.493) = 0.068 lb/h oil Therefore, by mass balance around the leaching unit, Oil in the extract = 1.202 – 0.068 = 1.134 lb/h oil Compare this to data given for the extract: Oil in the extract = 0.1535(7.313) = 1.123 lb/h oil This number is 99% of the mass balance number, which is an excellent check. A similar check cannot be made for the hexane, but we can calculate the amount of hexane contained in the wet leached solids. Hexane = 10.844 – (7.313 – 1.123) = 4.654 lb/h hexane in the wet leached solids. Mass ratio of oil to wet leached solids = 0.068/(0.068 + 4.493 + 0.680 + 4.654) = 0.0069 Summary of mass balance in lb/h: Component Feed Solvent Wet, leached solids Extract Moisture 0.680 0.000 0.680 0.000 Oil 1.202 0.000 0.068 1.134 n-Hexane 0.000 10.844 4.654 6.190 Dry, oil and solvent-free flakes 4.493 0.000 4.493 0.000
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Exercise 16.2 Subject: Manufacture of insoluble barium carbonate by precipitation of barium sulfide in an aqueous solution with sodium carbonate, followed by washing with water to remove sodium sulfide. Given: Specifications given below for a 5-stage countercurrent flow system, where precipitation occurs in the first stage and washing is done in four stages of thickeners. Assumptions: No solids in the overflows. Liquid in the underflow and liquid in the overflow leaving a stage have the same composition. Stoichiometric amounts for the precipitation reaction. Find: (a) A schematic diagram of the process. (b) The flow rates of sodium carbonate and wash water required. The flow rates of barium carbonate and sodium sulfide produced. (c) The wt% of sodium sulfide leaving in the liquid in the underflow from each stage. The flow rate of sodium sulfide in the barium carbonate after drying. Analysis: Because the underflow liquid is given as a mass ratio of Na 2 S-free water to the solid BaCO 3 in the underflow, use Na 2 S compositions in the liquid as mass ratios, X , to water. (a) Schematic diagram of the process with given information. Final Overflow Washing Water 10 wt% Na 2 S 120,000 kg/d H 2 O Each underflow is two parts water/part BaCO 3 . Final Underflow 40,000 kg/d BaS solid Na 2 CO 3 (b) Na 2 CO 3 requirement and BaCO 3 and Na 2 S produced by stoichiometry: BaS (aq) + Na 2 CO 3 (s) = BaCO 3 (s) + Na 2 S (aq) Molecular Weight: 169.42 106.00 197.37 78.05 1 2 3 4
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