Question 3: (15 marks) 10000 lb at 130°F containing 47.0 lb FeSO 100lb total water is a feed. Then, the feed is cooled t

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Question 3: (15 marks) 10000 lb at 130°F containing 47.0 lb FeSO 100lb total water is a feed. Then, the feed is cooled t

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Question 3 15 Marks 10000 Lb At 130 F Containing 47 0 Lb Feso 100lb Total Water Is A Feed Then The Feed Is Cooled T 1
Question 3 15 Marks 10000 Lb At 130 F Containing 47 0 Lb Feso 100lb Total Water Is A Feed Then The Feed Is Cooled T 1 (18.52 KiB) Viewed 29 times
Question 3: (15 marks) 10000 lb at 130°F containing 47.0 lb FeSO 100lb total water is a feed. Then, the feed is cooled to 80°F where FeSO4.7H;O crystals are removed. The solubility of the salt is 30.5 lb FeSO4 100 lb total water. The average heat capacity of feed solution is 0.70 btu lbm F. The heat of solution at 18°C is -4.4. kcal/gmol (-18.4 kJ/gmol) FeSO4.7H₂O. Calculate the yield of crystals and make a heat balance. Assume no water vaporized. (MW: FeSO4.7H₂O = 278.02, FeSO4= 152.0, 7H₂O = 126.0)
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