(Worked) Examples in interface mass transfer,.doc

(Worked) Examples in interface mass transfer,.doc - Mass...

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Mass Transfer / Set 2 (Worked) Examples in interface mass transfer, Mass transfer coefficients, overall coefficients and transfer units Example 1: An exhaust stream from a semiconductor fabrication unit contains 3 mole% acetone and 97 mole % air. In order to eliminate any possible environmental pollution, this acetone-air stream is to be fed to a mass-transfer column in which the acetone will be stripped by a countercurrent, falling water stream at 293 K. The tower is to be operated at a total pressure of 1.013 x 10 5 Pa. If the combined Raoult-Dalton equilibrium relation may be used to determine the distribution of acetone between the air and the aqueous phase, determine (i) the mole fraction of acetone within the aqueous phase which would be in equilibrium with the 3 mole% acetone gas mixture, and (ii) the mole fraction of acetone in the gas phase which would be in equilibrium with 20 ppm acetone in the aqueous phase. The vapour pressure of acetone at 293 K is 5.64 x 10 4 Pa. Solution: Dalton’s law gives the partial pressure of a component A in the gas-phase as follows: P y p A phase gas A where y A is the mole fraction of A in the gas-phase and P is the total pressure in the gas-phase. Raoult’s law gives the partial pressure of component A in the gas-phase as follows: A A phase liquid A P x p where x A is the mole fraction of A in the solution and P A is the vapour pressure of pure A. In a perfectly ideal system , where ideal liquid and ideal vapor are assumed, Raoult's law is combined with Dalton’s law to give the following relationship: A A A P x P y which will be used to solve Example 1. (i) Mole fraction of acetone within the aqueous phase which would be in equilibrium with 3 mole% acetone gas mixture can be expressed as follows: 0539 . 0 Pa 10 64 . 5 Pa) 10 013 . 1 )( 03 . 0 ( 4 5 A A A P P y x (ii) Mole fraction of acetone in the gas phase which would be in equilibrium with 20 ppm acetone in the aqueous phase is calculated in two stages. First 20 ppm acetone in the aqueous phase is converted to mole fraction of acetone in the aqueous phase as follows: 6 10 207 . 6 mol ) 18 / 980 , 999 ( mol ) 58 / 20 ( mol ) 58 / 20 ( water g 980 , 999 acetone g 20 acetone g 20 phase aqueous in acetone ppm 20 A x
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  • Spring '16
  • albhilil
  • Mole

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