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Mass Transfer - Engineering

Q1:

In a single stage extraction process, 10 kg of pure solvent S (containing no solute A) is mixed with 30 kg of feed F containing A at a mass fraction xf = 0.2. The mixture splits into an extract phase E and a raf-finate phase R containing A at xB = 0.5 and xR = 0.05 respectively. The total mass of the extract phase is (in Kg)

A 6.89

B 8.89

C 10

D 8.25

ANS:B - 8.89

To solve this problem, we can use the material balance equation for a single-stage extraction process: 𝐹=𝐸+𝑅F=E+R Where:

  • 𝐹F is the total mass of the feed,
  • 𝐸E is the total mass of the extract phase,
  • 𝑅R is the total mass of the raffinate phase.
We are given the following information:
  • Total mass of feed (𝐹F) = 30 kg
  • Mass fraction of solute A in the feed (𝑥𝐹xF​) = 0.2
  • Mass fraction of solute A in the raffinate phase (𝑥𝑅xR​) = 0.05
First, we need to calculate the mass of solute A in the feed: 𝑚𝐴𝐹=𝑥𝐹×𝐹mAF​=xF​×F 𝑚𝐴𝐹=0.2×30=6 kgmAF​=0.2×30=6kg Next, we need to calculate the mass of solute A in the raffinate phase: 𝑚𝐴𝑅=𝑥𝑅×𝑅mAR​=xR​×R 𝑚𝐴𝑅=0.05×𝑅mAR​=0.05×R Since the solvent S does not contain solute A, all of the solute A in the feed is transferred to the raffinate phase: 𝑚𝐴𝑅=𝑚𝐴𝐹=6 kgmAR​=mAF​=6kg Now, we can use the fact that the total mass of the raffinate phase (𝑅R) is equal to the total mass of the feed minus the mass of solute A in the raffinate phase: 𝑅=𝐹−𝑚𝐴𝑅R=F−mAR​ 𝑅=30−6=24 kgR=30−6=24kg Finally, since the extract phase contains all the solute A removed from the feed, the mass of solute A in the extract phase (𝑚𝐴𝐸mAE​) is equal to the mass of solute A in the feed minus the mass of solute A in the raffinate phase: 𝑚𝐴𝐸=𝑚𝐴𝐹−𝑚𝐴𝑅mAE​=mAF​−mAR​ 𝑚𝐴𝐸=6−6=0 kgmAE​=6−6=0kg Therefore, the total mass of the extract phase (𝐸E) is equal to the mass of the solvent S, which is 10 kg10kg. So, the correct answer is 10 kg10kg.