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1. Based on the figure below: ng(mol) 0.983 DA 0.017 W(V) 100 mol 10.983 DA 10.017 W(v) n;(mol) nz(mol) AIR nalmol) 0.960 DA 0.040 W(v) 0.977 DA COND. 0.983 DA 0.017 Wtv) 0.023 W(v) DA = dry air W = water nylmol W1)) Solve for the Following: a. ni or Fresh Feed (F) b. n3 or Water Condensed (W) c. ns or Recycle (R)

1. Based on the figure below: ng(mol) 0.983 DA 0.017 W(V) 100 mol 10.983 DA 10.017 W(v) n;(mol) nz(mol) AIR nalmol) 0.960 DA 0.040 W(v) 0.977 DA COND. 0.983 DA 0.017 Wtv) 0.023 W(v) DA = dry air W = water nylmol W1)) Solve for the Following: a. ni or Fresh Feed (F) b. n3 or Water Condensed (W) c. ns or Recycle (R)

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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1. Based on the figure below: ng(mol) 0.983 DA 0.017 W(v) ny(mol) 0.960 DA .- 0.040 W(v): 100 mol 10.983 DA 0.017 Wv) ng{mol) AIR nalmol) 0.977 DA 0.023 W(v) COND. 0.983 DA 0.017 Wtv) DA = dry air W = water nylmol W1)] Solve for the Following: a. ni or Fresh Feed (F) b. n3 or Water Condensed (W) c. ns or Recycle (R)
Transcribed Image Text:1. Based on the figure below: ng(mol) 0.983 DA 0.017 W(v) ny(mol) 0.960 DA .- 0.040 W(v): 100 mol 10.983 DA 0.017 Wv) ng{mol) AIR nalmol) 0.977 DA 0.023 W(v) COND. 0.983 DA 0.017 Wtv) DA = dry air W = water nylmol W1)] Solve for the Following: a. ni or Fresh Feed (F) b. n3 or Water Condensed (W) c. ns or Recycle (R)
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