Example 4: Glycerol is a by-product of the biodiesel process and manifests in so

Question

Example 4: Glycerol is a by-product of the biodiesel process and manifests in solution with water. You have a feed of 80 mol% glycerol in water, and you are to separate water (W) from glycerol (G) by flashing at 200°C and 1 atm pressure. Calculate the recoveries of the glycerol and water into the relevant streams Data (glycerol-water system at 1 atm pressure): 10000 10000 128 04742 05025 482 03077 05295 I207.0 0.0945 0.4555 Chen, D. H. T & Thompson, A R. (1970) Isobaric vapor-liquid equilibria for the systems glycerol-water and glycerol water saturated with sodium chloride. Journal of Chemical and Engineering Data, 15 471-474.

Explanation / Answer

let F= Feed stream containing 80 mole % glycerol and 20 mole % water

when flashed it separates into two layers, one water rich layer and the other one glycerol rich layer. the plot of temperature (deg.c) vs x,y is drawn and shown below

at 200 deg.c, x=0.1 and y=0.48 for water ( from the plot), xw= 0.1 and yw=0.48

hence for glycerol 1-x=0.9 and y= 1-0.48=0.52 , xG= 0.9 and yG= 0.52

let F= feed = 1 mole = D+W, D =distillate and W= residue

1= D+W, D= 1-W

writing glycerol balance 0.8= D* 0.9+W*0.52

hence 0.8= (1-W)*0.9 +W*0.52

0.8= 0.9- W*0.9+W*0.52

W*0.38= 0.1, W= 0.1/0.38 =0.2631 and D= 1-0.2631= 0.7369

Percentage recovery of Glycerol =100*( 0.7369*0.9/0.8) = 82.9

percentage recovery of eater = 100* 0.2631*0.48/0.2 = 63.144%

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