Page 224 - Packed bed columns for absorption, desorption, rectification and direct heat transfer

214

The experimental data [214, 312] for the interfacial area of Nutter rings
obtained in absorption of COa in NaOH solution are presented in Fig. 27 versus
the liquid superficial velocity L,

ZSZ —-
NR#1 a=168 rrWrn 3
2
NR#1 75 a=105m /m»
2 3
NR#2 5 a=83 m /m

80 SO 100 120 140 160 180 200 221
s
3
Specific Liquid Load, m /(m .h)
Fig. 27. Interfecial area of Nutter rings measured by Duss et al. [214,312].
It is mentioned that the analysis of the data points shows that there is a
strong influence of the pressure drop in the area over the loading point (AP > 4
mbar/m). Under this point the value of a e for a given packing is a function only
of the liquid superficial velocity. Unfortunately, in the figure no information
about the gas velocity or the pressure drop is given. It is to be expected that the
great difference between the points which reaches up to 38%, obtained for one
and the same packing and one and the same liquid superficial velocity, is owing
to different gas velocities.
The dependence of a e on the pressure drop is presented in Fig. 28.
As noted by the authors [214], the results indicate that the increase of
the effective surface area with the pressure drop above 4 mbar/m can be fitted
best by assuming that the gain in area per pressure unit is constant for all three
2 3
ring sizes, i.e. 3.9 m /(m mbar), and it is not proportional to the packing specific
area as one might expect.
The measured interfacial area is higher than the geometrical area for all
3
data points at liquid loads above 20 m /(m h). We think that the last contradicts
with the fact that the smaller the geometrical area, the higher the ratio of the
effective to geometrical area of the packing. The results of Kolev et al. [95] and
of Nakov et al. [216] for RSR, Fig. 18, and for IMTP, Fig. 20, also contradict to
the contention that the effective and the geometrical surface areas become equal
at one and the same liquid superficial velocity. From Fig. 18 it can be seen, for
example, that for metal RSR No 0.5, a e is equal to a at liquid superficial
2
3
velocity X ft=200 m /(m h). For RSR No 3 the respective value is I*=10