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 (18.5 KiB) Viewed 31 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 FeSO4.7H₂O 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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