HW 4 - p 5.6 - Countercurrent Washing of Al2(SO4)3 from Processed Bauxite Ore - 6 pts
Aluminum sulfate, commonly called alum, is produced as a concentrated aqueous solution from bauxite ore by reaction with aqueous sulfuric acid, followed by a three-stage, countercurrent washing operation to separate soluble aluminum sulfate from the insoluble content of the bauxite ore, followed by evaporation. In a typical process, 40,000 kg/day of solid bauxite ore containing 50 wt% Al2O3 and 50% inert is crushed and fed together with the stoichiometric amount of 50 wt% aqueous sulfuric acid to a reactor, where the Al2O3 is reacted completely to alum by the reaction
Al2O3 + 3H2SO4 Al2(SO4)3 + 3H2O
The slurry effluent from the reactor, consisting of solid inert material from the ore and an aqueous solution of aluminum sulfate is then fed to a three-stage, countercurrent washing unit to separate the aqueous aluminum sulfate from the inert material. If the solvent is 240,000 kg/day of water and the underflow from each washing stage is 50 wt% water on a solute-free basis, compute the flow rates in kilograms per day of aluminum sulfate, water, and inert solid in each of the two product streams leaving the cascade. What is the percent recovery of the aluminum sulfate? Would the addition of one more stage be worthwhile?
Al2O3 + 3H2SO4 Al2(SO4)3 + 3H2O
The slurry effluent from the reactor, consisting of solid inert material from the ore and an aqueous solution of aluminum sulfate is then fed to a three-stage, countercurrent washing unit to separate the aqueous aluminum sulfate from the inert material. If the solvent is 240,000 kg/day of water and the underflow from each washing stage is 50 wt% water on a solute-free basis, compute the flow rates in kilograms per day of aluminum sulfate, water, and inert solid in each of the two product streams leaving the cascade. What is the percent recovery of the aluminum sulfate? Would the addition of one more stage be worthwhile?
posted by Dr. B at 2:37 PM
6 Comments:
basil 5:20 PM:
I think you are talking about HW #4. HW #4 is due Tue, 10/24, at 5 PM to me (in the UO Lab). Did I mess that up somewhere ?
on the HW4 page it says "Monday, Oct 24 at 5 PM (bring it to the UO Lab)". The 24th is Tuesday
kibosh 7:40 PM:
OK, I fixed it. Tuesday, 10/24 at 5PM in UO the Lab
The problem says that 50% of each underflow stream is pure water. Since the amount of Inert solids doesn't change and the concentration of solute does, then would that mean that R is not constant. I thought that R is assumed to be constant in section 5.1. How do you proceed to solve with a non constant R?
karen 9:15 PM:
Yes, you are correct. R is not constant. You need to write and solve the material balance equations. This makes it a typical 310 problem. I solved the water balances first because they are easier and can be solved one at a time, probably by inpection. The alum balances are a bit more work because you must solve all three, one for each stage, simultaneously.
dr. phil:
OK, I compared this problem to the example and I thought about this some more and here is what I think. The only real difference between Ex 5.1 and this HW problem is that in this problem some water comes in with the feed. I think you could actually IGNORE this water and solve the problem just as in Ex 5.1. The only error in doing so woul be ignoring all the extra water in the feed. But all the extra water in the feed goes out in the overflow from the feed stage (stage 3) anyway.
What I said before is true, R is not constant. R in stage 3 is different. So, if you just grab eqn 5.9 and solve it using the matrix functions in Excel, you will get the correct answer to this problem, but your water flow rate in the overflow from stage 3 will be too small...by the amount of water that comes in with the feed from the reactor.
I hope this clears everything up.
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