Page 153 - Chemical Process Equipment - Selection and Design

6.11. FLUIDIZATION OF BEDS OF PARTICLES WITH GASES 125

EXAMPLE 6.16 (f) Operating gas velocity. The ratios of entraining and
Diensiow of a Flaidized Bed Vessel minimum fluidizing velocities for the two smallest particle sizes
A fluidized bed is to hold 10,OOO kg of a mixture of particles whose present are
true density is 1700 kg/m3. The fluidizing gas is at 0.3 m3/sec, has a
viscosity of O.iD17cP or 1.7(E-5) Nsec/m2 and a density of 0.049/0.0061= 8.03, for 30 pm,
1.2kg/m3. The size distribution of the particles is 0.0054/0.0061= 0.89, for 10 pm.

d(prn) 252 178 126 89 70 50 30 10 Entrainment of the smallest particles cannot be avoided, but an
~(wtfrac- 0.088 0.178 0.293 0.194 0.113 0.078 0.042 0.014
tion) appreciable multiple of the minimum fluidizing velocity can be used
u,(m/sec) 3.45 1.72 0.86 0.43 0.27 0.14 0.049 0.0054 for operation; say the ratio is 5, so that
The terminal velocities are found with Stokes' equation uf = 5umf = 5(0.0061) = 0.0305 m/sec.

(g) Bed expansion ratio. From Figure 6.10(c) with dp =
84.5 pm or 0.0033 in. and Gf/Gmf = 5,

(a) The average particle size is R=[ 1.16, by interpolation between the full lines,
1.22, off the dashed line.
I
dp = 1 (x,/d,) = 84.5 pm. Take R = 1.22 as more conservative. From Eq. (6.140) the ratio of
voidages is
(a) With dp = 84.5 and density difference of 1699 kg/m3, the
-
material appears to be in Group A of Figure 6.12. &A JEmf = 50.22 - 1.42.
(c) Minimum fluidization velocity with E&. (6.133)
From part (e), E& = 0.469 so that Emf = 0.469/1.42 = 0.330.
0.0093[84.5(E - 6)]1.82(1700 - l.Z)0.94 Accordingly, the ratio of bed levels is
4nf = [1.7(E - 5)]0.88(1.2)0.06
= 0.0061 mlsec, Lmb/Lmf = (1 - &,,?)/(I - E,&) =0.67/0.531 = 1.262.
and with Eqs. (6.134) and (6.135), Although the value of E~ appears somewhat low, the value of R
checks roughly the one from Figure 6.10(c).
1.2(1700 - 1.2)(9.81)[84.5(E - 6)p= 41.75, (h) Fluctuations in level. From Figure 6.10(d), with dp =
Ar=p [1.7(E - 5)]' 0.0033 in., the value of rn' =O.QZ, so that
Re, = 4(27.2)' + 0.0408(41.75) - 27.2 = 0.0313,
r = exp[0.02(5 - l)] = 1.083.
(i) TDH from Figure 6.10(i). At uf = umf - 4(0.0061) =
0.0244 m/sec, the abscissa is off the plot, but a rough extrapolation
Use the larger value, umf = 0.0061, as the conservative one. and interpolation indicates about 1.5 m for TDH.
(d) Minimum bubbling velocity, with Eq. (6.136), (j) Dimensions of the bed and vessel. With a volumetric flow
rate of 0.3 m3/sec, the required diameter is
umb = 33(84.5)(E - 6)[1.2/1.7(E - 5)]'.' = 0.0085 m/sec,
... u,b/U,,,f = 0.0085/0.0061 = 1.39. D = v0.3/(0.305)(~/4) = 3.54m.

From Eq. (6.139), With a charge of 10,000kg of solids and a voidage at minimum
bubbling of 0.469, the height of the minimum. bubbling bed is
-_
umb - 82[1.7(E - 5)]0-"(1.2>0."
umf 9.8:1[84.5(E - 6)]1.3(1700 - 1.9= 1'35' 10000
L= 1700(1- 0.469)(~/4)0' -
- I. 13 m.
which is in rough agreement.
(e) Voidage at minimum bubbling from Eq. (6.138): This value includes the expansion factor which was calculated
separately in item (g) but not the fluctuation parameter; with this
0.5
[1.7(E - 5)]' ] =0.1948, correction the bed height is
9.81[84.5(E - 6)]3(1700)2
:. E,~ = 0.469. Lb = 1.13(1.083) = 1.22m.
It is not certain how nearly consistent this value is with those at The vessel height is made up of this number plus the TDH of 1.5 m
minimum fluidization read off Figure 6.10(e). Only a limited or
number of characteristics of the solids are accounted for in Eq.
(6.138). vessel height= 1.22+1.5=2.72m.