# Multicomponent Distillation Material Balance Solution Given data: Feed flow rate, \( F = 300 \) kmol/h Feed composition: \( x_{A,F} = .40 \), \( x_{B,F} = .35 \), \( x_{C,F} = .25 \) Distillate composition: \( x_{A,D} = .90 \), \( x_{B,D} = .08 \), \( x_{C,D} = .02 \) Bottoms composition: \( x_{A,B} = .05 \), \( x_{B,B} = .30 \), \( x_{C,B} = .65 \) Let: \( D \) = distillate flow rate (kmol/h) \( B \) = bottoms flow rate (kmol/h) ## (a) Molar Flow Rate of the Distillate Stream ### Step 1: Overall Mass Balance \[ F = D + B \] \[ 300 = D + B \] ### Step 2: Component A Balance \[ F x_{A,F} = D x_{A,D} + B x_{A,B} \] \[ 300 \times .40 = D \times .90 + B \times .05 \] \[ 120 = .90D + .05B \] ### Step 3: Solve the Two Equations Substitute \( B = 300 - D \) into the component A balance: \[ 120 = .90D + .05(300 - D) \] \[ 120 = .90D + 15 - .05D \] \[ 120 - 15 = (.90 - .05)D \] \[ 105 = .85D \] \[ D = \frac{105}{.85} \approx 123.53 \text{ kmol/h} \] ## (b) Molar Flow Rate of the Bottoms Stream \[ B = 300 - D = 300 - 123.53 = 176.47 \text{ kmol/h} \] ## (c) Molar Flow Rate of Component A in the Distillate \[ \text{A in distillate} = D \times x_{A,D} = 123.53 \times .90 = 111.18 \text{ kmol/h} \] ## (d) Molar Flow Rate of Component C in the Bottoms \[ \text{C in bottoms} = B \times x_{C,B} = 176.47 \times .65 = 114.71 \text{ kmol/h} \] ## (e) Recovery of Component A in the Distillate (in %) Recovery is defined as the fraction of feed A that is recovered in the distillate, times 100%: \[ \text{Recovery} = \frac{\text{A in distillate}}{\text{A in feed}} \times 100\% \] \[ = \frac{111.18}{120} \times 100\% = 92.65\% \] ## (f) New Distillate Flow Rate if Feed Increases by 15% (Same Product Compositions) New feed rate: \( F' = 1.15 \times 300 = 345 \text{ kmol/h} \) Repeat the same overall and component A balance: \[ F' = D' + B' \] \[ 345 = D' + B' \] Component A balance: \[ 345 \times .40 = D' \times .90 + B' \times .05 \] \[ 138 = .90D' + .05B' \] Substitute \( B' = 345 - D' \): \[ 138 = .90D' + .05(345 - D') \] \[ 138 = .90D' + 17.25 - .05D' \] \[ 138 - 17.25 = .85D' \] \[ 120.75 = .85D' \] \[ D' = \frac{120.75}{.85} = 142.06 \text{ kmol/h} \] # Final Answers (a) Distillate flow rate = **123.53 kmol/h** (b) Bottoms flow rate = **176.47 kmol/h** (c) Molar flow of A in distillate = **111.18 kmol/h** (d) Molar flow of C in bottoms = **114.71 kmol/h** (e) Recovery of A in distillate = **92.65%** (f) New distillate flow rate (with 15% higher feed) = **142.06 kmol/h**