• Author / Uploaded
• Pravin Nair

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Example 3: Extraction of Acetic Acid in a Single Stage Extractor A single-stage extraction is performed in which 400 kg of a solution containing 35 wt% acetic acid in water is contacted with 400 kg of pure isopropyl ether. Calculate the amounts and compositions of the extract and raffinate layers. Solve for the amounts both algebraically and by the lever-arm rule. What percent of the acetic acid is removed? Use equilibrium data given below.

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Solution Extract (yA1, yB1, yC1) Feed (xAo, xBo, xCo)

(yA2, yB2, yC2) Solvent

Raffinate (xA1, xB1, xC1)

Given Lo=400 kg and V2= 400 kg xAo = 0.35, xBo =0.65, yA2 = 0.0, yC2 = 1.0 xCo = 1- xAo-xBo = 1-0.35-0.65=0.0 yB2 = 1- yA2-yC2 = 1-0-1= 0.0 Find: i) xA1, xB1, xC1, yA1, yB1, and yC1 ii) V1, L1 (algebraically and by the lever-arm rule), iii) Percentage removal of the acetic acid 3

Recall the steps:

Steps: 1.L0 (Feed) and V2 (Solvent) are known. 2. We calculate M, xAM, and xCM by using Eqs (12.5-12) – (12.5-14). 3. Draw line L0, V2, M in the Figure. 4. Using trial and error a tie line is drawn through the point M, which locates the compositions of L1 and V1. 5. The amounts of L1 and V1 can be determined by substitution in Eqs (12.5-12) – (12.5-14) or by using lever-arm rule. 4

Step 1: Locate point Lo and V2 Point Lo = (xAo, xCo ) = (0.35,0) Point V2 = (yA2, yC2 ) = (0,1) Step 2: Locate M-point Eq (12.5-12): Lo + V2 = L1 + V1 = M 400 + 400= M =800 kg Eq (12.5-13): LoxAo + V2yA2 = L1xA1 + V1yA1 = MxAM 400 (0.35) + 400(0) = 800 xAM xAM = 0.175 Eq (12.5-14): LoxCo + V2yC2 = L1xC1 + V1yC1 = MxCM 400(0) + 400(1) = 800 xCM xCM = 0.5 5

Step 3 and Step 4 : Draw line L0, V2, M and draw tie line to locate V1 and L1 (try and error)

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Step 5: From the figure: xA1 = 0.25, xC1 = 0.03, then xB1 = 0.72 From the figure: yA1 = 0.11, yC1= 0.85, then yB1 = 0.04 Eq (12.5-13): L1xA1 + V1yA1 = MxAM L1(0.25) + V1(0.11) = 800 (0.175) 0.25 L1 + 0.11 V1 = 140

(1)

Eq (12.5-12): L1 + V1 = 800

(2)

Solve Eqs (1) and (2), L1 = 371 kg and V1 = 429 kg

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By using level-arm rule: L1/M = L1/800 = MV1/V1L1 L1 = (4.1/9.4) x 800 L1 = 349 kg Then V1 = 800 – 349 = 451 kg Amount of Acid in the Feed = 400 x 0.35 = 140 kg Amount of Acid in Raffinate = 371 x 0.25 = 92.75 kg % Acid Removal = (140-92.75) /140 = 33.75%

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Next Topic •

Continuous multistage countercurrent extraction

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