I am making biodiesel. The product stream leaving the reactor is 50% glycerol in
ID: 805911 • Letter: I
Question
I am making biodiesel. The product stream leaving the reactor is 50% glycerol in mixed with biodiesel. I feed this product stream to a series of separators. The first is a rough separator, which removes 76.4 moles a minute from the product stream. This waste stream is 60 mole % glycerol. The second separator is finer, and removes 4 moles a minute from the product stream, with a composition of 80 mole % glycerol. This product stream leaving these two separators has just 5 mole % glycerol. Determine the rate of production, and the rate of feed required to achieve the production rate. Also, determine the efficiency of the removal of glycerol from the reactor product stream (i.e., the compliment of the recovery of glycerol in the reactor product stream).
Explanation / Answer
Separator-1 removes 76.4 moles/min of glycerol and coverts the original product of 50 mole % glycerol into 40 mole % glycerol. This is fed into separator-2 which removes glycerol at 4 moles/min and coverts the product having just 5 mole% glycerol. Therefore, separator-2 is the rate limiting step.
If the 40 mole% glycerol product is fed into separator-2 at a rate that matches the removal rate, then the production rate can be determined.
entry concentration 40/100; exit conc. 5/100; feed rate say x.
after removal the exit conc. is 5% or .05 per 100 mole.
Therefore, the amount of product from separator-1 to be fed is, (x-4)/(x/0.4) = 0.05. Solving for x, x = 4.57 moles.
Feeding rate si 4.57 moles glycerol per min or 4.57/0.4 = 11.43 moles/min of the product from separator-1
This would be the production rate of biodiesel containing 5mol% glycerol.
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