CHEMICAL ENGINEERING

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