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

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
.
.

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.
Given: Tank diameter = 10 ft, with a stagnant gas layer above liquid benzene of 0.2-in.
thickness. For benzene, liquid density = 0.877 g/cm
3
, MW = 78.11, vapor pressure = 100 torr,
and the diffusivity in air = 0.08 cm
2
/s.
Assumptions: All mass-transfer resistance is in the thin gas layer of constant thickness. Steady-
state with a benzene mole fraction in the air adjacent to the liquid given by Raoult's law, and the
benzene mole fraction in the air at the other side of the gas layer equal to zero, assuming the
evaporated benzene is continuously swept away as in Example 3.2. Ideal gas.