CHEMICAL ENGINEERING

232 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

Thus a plot oft/hh 0

 againsthh 0

 will be a straight line of slopeswhere:

sD+LCBm/ 2 MDCACT or DD+LCBm/ 2 MCACTs

The following table may be produced:

t (ks) 1.6 11.1 27.4 80.2 117.5 168.6 199.7 289.3 383.1
hh 0
(mm) 2.5 12.9 23.2 43.9 54.7 67.0 73.8 90.3 104.8
t/hh 0

s/mð 10 ^6 0.64 0.86 1.18 1.83 2.15 3.52 2.71 3.20 3.66

0 20 40 60 80 100 120

(h−h 0 ) mm

0.5

1.0

1.5

2.0

2.5

3.0

3.5

Slope = 3.04× 10 −^7 s/m^2

t / (

h−

h^0

) s/m

×^10

−^6

Figure 10b.

These data are plotted as Fig. 10b and the slope is:

sD 3. 54  0. 5

10 ^6 / 100 ð 10 ^3 D 3. 04 ð 10 ^7 s/m^2

CTD 1 / 22. 4 

 273 / 321

 D 0 .0380 kmol/m^3

MD154 kg/kmol

CAD 37. 6 / 101. 3 

 1 / 22. 4 

 273 / 321

 D 0 .0141 kmol/m^3

+LD1540 kg/m^3

CB 1 D 0 .0380 kmol/m^3