A feed solution of 10000 lbm at 130°F containing 47.0 lb FeSO4 per 100 lb water is cooled to 80°F, where FeSO4•7H2O crystals are removed. The solubility of the salt is 30.5 lb FeSO4 per 100 lb water. The average heat capacity of the feed solution is 0.70 BTU/lbm-°F. The heat of solution at 18°C is –4.4 kcal/gmol FeSO4•7H2O, Calculate the yield of crystals and make a heat balance. Assume that no water is vaporized.
A feed solution of 10000 lbm at 130°F containing 47.0 lb FeSO4 per 100 lb water is cooled to 80°F, where FeSO4•7H2O crystals are removed. The solubility of the salt is 30.5 lb FeSO4 per 100 lb water. The average heat capacity of the feed solution is 0.70 BTU/lbm-°F. The heat of solution at 18°C is –4.4 kcal/gmol FeSO4•7H2O, Calculate the yield of crystals and make a heat balance. Assume that no water is vaporized.
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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A feed solution of 10000 lbm at 130°F containing 47.0 lb FeSO4 per 100 lb water is cooled to 80°F, where FeSO4•7H2O crystals are removed. The solubility of the salt is 30.5 lb FeSO4 per 100 lb water. The average heat capacity of the feed solution is 0.70 BTU/lbm-°F. The heat of solution at 18°C is –4.4 kcal/gmol FeSO4•7H2O, Calculate the yield of crystals and make a heat balance. Assume that no water is vaporized.
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