# ch03 - Exercise 3.1 Subject: Evaporation of a mixture of...

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Exercise 3.1 Subject: Evaporation of a mixture of ethanol (AL) and ethyl acetate (AC) from a beaker into still air within the beaker. Given: Initial equimolar mixture of AL and AC, evaporating into still air at 0 o C and 1 atm. Vapor pressures and diffusivities in air of AL and AC at 0 o C. Assumptions: Well-mixed liquid and Raoult's law. Negligible bulk flow effect. Air sweeps across the top of the beaker at a rate such the mole fractions of AL and AC in the air at the top of the beaker are zero. Find: Composition of the remaining liquid when 50% of the initial AL has evaporated. Analysis: All of the mass-transfer resistance is in the still air layer in the beaker, which increases in height, z , as evaporation takes place. Apply Fick's law to both AL and AC with negligible bulk flow effect. Thus, from Eq. (3-16), the molar flux for ethanol through the gas layer in the beaker is as follows, where D i is the diffusivity of component i in air. N D dc dz D c dy dz N dz D c dy N dz D c dy y y AL AL AL AL AL AL AL AL 0 AC AC AC 0 Rearranging, (1) Similarly, (2) AL AC = - = - = - = - Dividing Eq. (1) by (2), N N D y D y y y AL AC AL AL AC AC AL AC = = × × - - 6 45 10 9 29 10 6 6 . . (3) where y AL and y AC are mole fractions in the vapor at the vapor-liquid interface. By material balance, the molar flux of component i is equal to the rate of decrease in moles, n i , , of component i in the well-mixed liquid in the beaker per unit of mass-transfer area. Thus, N dn Adt N dn Adt AL AL AC AC (4) (5) = - = - By Raoult's law, at the gas-liquid interface, using Eqs. (2-19) and (3) in Table 2.3, y P P x P P n n n n n n y P P x P P n n n n n n s s s s AL AL AL AL AL AL AC AL AL AC AC AC AC AC AC AL AC AC AL AC (6) (7) = = + ± ² ³ ´ µ = + ± ² ³ ´ µ = = + ± ² ³ ´ µ = + ± ² ³ ´ µ 162 101 323 101 . .

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Exercise 3.1 (continued) Analysis: (continued) Substituting Eqs. (4) to (7) into (3), and rearranging, dn dn n n n n n n n i AL AC AL AC AL AL AC AC (8) Integrating Eq. (8) from the start of the evaporation, letting be the initial values, (9) 0 0 = ± ² ³ ´ µ ± ² ³ ´ µ ± ² ³ ´ µ ± ² ³ ´ µ = ± ² ³ ´ µ 162 323 9 29 6 45 0 722 0 . . . . ln . ln As a basis, assume 100 moles of original mixture. Thus, n n n n AL AC AL AC 0 0 0 50 mol and mol When mol, i.e. half of the original, Eq. (9) gives mol = = = = 50 25 19 2 . Therefore, the mole fractions in the well-mixed liquid when 50% of the AL has evaporated are, AL AC 0.566 0. 25 25 19.2 19.2 25 19.2 434 x x = + = = + =
Exercise 3.2 Subject: Evaporation of benzene (B) at 25 o C and 1 atm from an open tank through a stagnant air layer of constant thickness.

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## This note was uploaded on 04/21/2008 for the course CHE 31 taught by Professor Gilchrest during the Fall '08 term at Lehigh University .

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ch03 - Exercise 3.1 Subject: Evaporation of a mixture of...

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