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Q.5: 1.75 kg/s of a solid containing 0.55 kg soluble material (Liquid), is to be washed with 2.8 kg/s pure solvent in a counter-current unit. The final residue is to contain not more than 0.1% solute when dried. Find the number of stages given that the amount of solution in under flow varies with solute concentration according to the relation: W-0.2x+ 0.3. How did the value of B = Solution: Variable under flow 1.2. become? S. - 0.55, B. 1.2& W, 0.55 kg/s

Q.5: 1.75 kg/s of a solid containing 0.55 kg soluble material (Liquid), is to be washed with 2.8 kg/s pure solvent in a counter-current unit. The final residue is to contain not more than 0.1% solute when dried. Find the number of stages given that the amount of solution in under flow varies with solute concentration according to the relation: W-0.2x+ 0.3. How did the value of B = Solution: Variable under flow 1.2. become? S. - 0.55, B. 1.2& W, 0.55 kg/s

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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0.5: 1.75 kg/s of a solid containing 0.55 kg soluble material (Liquid), is to be washed with 2.8 kg/s pure solvent in a counter-current unit. The final residue is to contain not more than 0.1% solute when dried. Find the number of stages given that the amount of solution in under flow varies with solute concentration according to the relation: W-0.2x+ 0.3. How did the value of B = Solution: Variable under flow 1.2. become? S, 0.55, B, 1.2 & W, 0.55 kg/s
Transcribed Image Text:0.5: 1.75 kg/s of a solid containing 0.55 kg soluble material (Liquid), is to be washed with 2.8 kg/s pure solvent in a counter-current unit. The final residue is to contain not more than 0.1% solute when dried. Find the number of stages given that the amount of solution in under flow varies with solute concentration according to the relation: W-0.2x+ 0.3. How did the value of B = Solution: Variable under flow 1.2. become? S, 0.55, B, 1.2 & W, 0.55 kg/s
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