Q 2, A feed, F, of 100 mole/h containing 50% A, 30% B, and 20% C is distilled in

Question

Q 2, A feed, F, of 100 mole/h containing 50% A, 30% B, and 20% C is distilled in a distillation column. The distillate and bottoms compositions are shown in figure (1). a) What is the percent recovery of A in the distillate? b) What is split fraction (SF) of component C? c) What is split ratio (SR) of component A? d) What is the percent purity of component B in the bottoms product? e) Compute the separation power (SP) of component (A) relative to (B) Distillate Comp mole/h 48 10 10 Colame Total 68 Ci Comp mole/h 50 30 Total 100 Bottoms Comp mole/h 20 Total 32

Explanation / Answer

Part a

Percent recovery of A =( recovered in distillate*100)/( A in feed)

= 48*100/50 = 96%

Part b

Split fraction of component C = C in distillate/C in feed

= 10/20 = 0.5

Part c

Split ratio of A = A in distillate/A in bottom

= 48/2 = 24

Part d

% purity of B in bottom = B in bottom/Total moles in bottom

= 20/32 = 0.625

Part e

Separation power of A to B

= Split ratio of A/split ratio of B

= 24/(10/20) = 48

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