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Mass Transfer
Mass Transfer Operations
GATE 2013
Hard

A crosscurrent cascade of N ideal stages is used to treat a feed stream of molar flow rate E. The feed stream contains a solute which is to be recovered by a pure solvent having a molar flow rate S. The solvent is divided equally between these N stages. The linear equilibrium curve relating the mole fractions x and y* of the solute in the raffinate and the extract respectively, is given by y*=mx. Assume dilute conditions. The ratio of the solute mole fraction in the original feed to that in the exit raffinate stream i.e. (x0/xN) is given by

  • [1+(mS/NE)]^N
  • B[1+(NS/mE)]^N
  • C[1+(NE/mS)]^N
  • D[1+(mE/NS)]^N

Explanation

For each stage with solvent S/N: mass balance E·x(n−1)=E·xn+(S/N)·yn, and yn=m·xn (equilibrium). So E·x(n−1)=xn[E+mS/N], giving x(n−1)/xn=1+mS/(NE), the same ratio for every stage. Applying this over all N stages: x0/xN=[1+mS/(NE)]^N.

Source: GATE CH 2013 (2013)

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