Page 98 - Chemical Process Equipment - Selection and Design

70 TRANSFER OF SOLIDS

EXAMPLE 5.1
Conditions of a Coal Slurry Pipeline
Data of a pulverized coal slurry are

C, = 0.4,
D = 0.333 ft,
f= 0.0045 (Blasius’ eq. at N, = lo5),
s = 1.5.

Mesh size 24 48 100 Mixture
d(rnrn) 0.707 0.297 0.125 0.321
Weight fraction 0.1 0.8 0.1 1
u, (ft/sec) 0.164 0.050 0.010 0.0574
The terminal velocities are read off Figure 5.1, and the values of the
mixture are weight averages.
The following results are found with the indicated equations:
Item Eq. 24 48 100 Mixture
“c 5.1 7.94 5.45 3.02
k 5.8 1.36 2.89 9.38 3.39
5.6 20.6
6.27 1.25
AEIAPL 5.1 1 1.539
AEI4 5.13 1.296
Eq. (5.1): up = 34.6(0.4)(0.333)vwF7304.8
mrn

8.41~~
Eq. (5.6): u =--
Vmm - 125u,,
4 32.2(1.5 - 1) d,, - 0.0704drnrn
Eq. (5.8): CD =- --
3 u: 304.8 u; ’
069 [ - JTZ@~~BE-J~~
i
Ae
Eq. (5.11): --l+-
-
APL 0.4°.3 0.0574 304.8(3.39)’ Sphere diameter. ern
= 1.5391, Figure 5.1. Settling velocities of spheres as a function of the ratio 01
(0.0574)’(3.39) 1
Eq. (5.13): ~ A& - 0.0045(0.333)32.2(0.5) densities of the two phases. Stokes law applies at diameters below
- 1 + 0.272(0.4)[
approximately 0.01 cm (based on a chart of Lupple et ul., Chemical
APL Engineering Handbook, McGruw-Hill, New York, 1984, p. 5.67).
= 1.296.
With coal of sp gr = 1.5, a slurry of 40~01% has a sp gr = 1.2.
Accordingly the rule, AelAP, = sp gr, is not confirmed accurately
by these results. For particles of one size, Eqs. (5.7) and (5.8) combine to
APs/APL = 1 + ~OOC,[(U,D/U’)~~~~.~,
consistent units. (5.10)
The drag coefficient is The pressure drop relation at the critical velocity given by Eq. (5.1)
is found by substitution into Eq. (5.7) with the result

Ae/APL = 1 + 0 L[(l/ut)qgd(s - 1)/CDJ’3. (5.11)
69
c:3
For mixtures, a number of rules has been proposed for evaluating
the drag coefficient, of which a weighted average seems to be With Eq. (5.10) the result is
favored.
Ae/APL = 1 + 1/C:3. (5.12)
With the velocity from Eq. (5.6), Eq. (5.7) becomes
where the wi are the weight fractions of particles with diameters di. Ae/APL = 1 + 0.272Cu[fgD(s - ~)/u:G]’.~ (5.13)