236 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS
Calculate the time taken for the liquid level to fall to 5 cm below the top of the
tube, neglecting the effects of bulk flow in the vapour. 1 kmol occupies 22.4m^3 .Molec-
ular weights of acetone, dibutyl phthalateD58 and 279 kg/kmol respectively. Liquid
densities of acetone, dibutyl phthalateD764 and 1048 kg/m^3 respectively. Vapour pres-
sure of acetone at 315 KD 60 .5kN/m^2. Diffusivity of acetone vapour in air at 315 KD
0 .123 cm^2 /s.
Solution
Considering the situation when the liquid has fallen to a depthhcm below the top of the
tube, volume of acetone evaporatedD/ 4
0. 6
2 h 1. 15
D 0. 283 h 1. 15
cm^3.
At this time, the amount of dibutyl phthalate is:
2 ð 1. 048 / 278
D 0 .00754 mol
and the amount of acetoneD[2 0. 283 h 1. 15
]0. 764 / 58 D 0. 0306 0. 00372 h
∴ Mole fraction of acetoneD
0. 0306 0. 00372 h
0. 00754 C 0. 0306 0. 00372 h
D
8. 23 h
10. 24 h
Partial pressure of acetoneD 60. 5
(
8. 23 h
10. 24 h
)
kN/m^2
Molar concentration of acetone vapour at the liquid surface
D
(
60. 5
101. 3
)
ð
(
273
315
)
ð
(
1
22400
)(
8. 23 h
10. 24 h
)
D 2. 31 ð 10 ^5
(
8. 23 h
10. 24 h
)
mol/cm^3
Rate of evaporation of acetone:NAD
dh
dt
ð
0. 764
58
D 0. 0132 dh/dt
mol/cm^2 s
DD/h
ðmolar concentration at surfaceD 0. 123 /h
2. 31 ð 10 ^5
(
8. 23 h
10. 24 h
)
∴ 0. 0132
dh
dt
D
1
h
ð 2. 84 ð 10 ^6
(
8. 23 h
10. 24 h
)
and:
(
10. 24 h
8. 23 h
)
hdhD
dt
4650
The time for the liquid level to fall from 1.15 cm to 5 cm below the top of the tube is
obtained by integrating this equation:
∫ 5
1. 15
(
10. 24 h
8. 23 h
)
hdhD
1
4650
∫t
0
dtD
∫ 5
1. 15
(
h 2. 02
16. 6
h 8. 23
)
dhD
1
4650
∫t
0
dt
and: tD79500 s
79 .5ks³22 h