Page 351 - Chemical Process Equipment - Selection and Design

314 SOLID-LIQUID SEPARATION

EXAMPLE 11.4 Filtering period is
Filtration and Washing of a Compressible Material
A kaolin slurry has the properties if = 0.25 + O.O035(vf - 0.0423) + 70.0(Vj - 0.0018).

c = 200 kg solid/m3 filtrate, Daily production rate,
y = 0.001 N sec/mz, 2.78(E - 7) N hr/m2,
ps = 200 kg/m3, R, = (no of batches/day)(filtrate/batch)
a= 87(E10)(1+ P/3.45)0.7 m/kg with P in bar,
E 1 - 0.460(1+ P/3.45)0.12.
- 245
-
The equations for a and E are taken from Table 11.8. 1.25 + 0.0035(vf - 0.0423) + 70(Vj - 0.0018).
Filtration will proceed at a constant rate for 15min, the
pressure will rise to 8 bar and filtration will continue at this pressure The tabulation shows that R, is a max when V, = 0.127.
until the end of the operation. Filter cloth resistance is
Rf = 1(1010) m-l. The down time per batch is 1 hr. +
v, Rd
a. Find the maximum daily production of filtrate. 0.12 1.3507
b. The filtrate will be blown and then washed with a volume of 0.126 1.3526
water equal to the pore space of the cake. Find the maximum 0.127 1.2533 1.3527 (rnax)
daily production of filtrate under these conditions. 0.1 28 1.3526
0.1 29 1.3525
0.130 1.3522
Part (a)
Basis 1 m2 of filtering surface. At P = 8 bar, or 8(105) Pa Part (b)
a = 87(1010)(1 + 8/3.45)0.7 = 2.015(1012) m/kg,
= 200(0'47) = 0.07095,
E= 1 - 0.46(1+ 8/3.45)'.l2 = 0.47, Amount of wash liquid = p,(l- E) 2500(0.53)
yca = (0.001/3600)(200)(2.015)(1012) = 1.12(10s) N hr/m4.

The filtration equation (11.2) is wash rate = filtering rate at the conclusion of the filtration
dV- AAP AP
-
dt y(Rf + aCV/A) = (0.001/3600)[1010 + 2.Ol5(lO1')(20O)V]
- AP t, = wash time = 0.709vf[2780 + 1.12(108)V,]
-
2780 + 1.12(1OS)V' 8(105)
= V,(0.000246 + 9.9265),
The rate when t = 0.25h and AP = 8(105) Pa, 245
R, = l+if+t,
8(lO5) 8(105)
24Vf
-
=2780+ 1.12(108)Qt=2780+0.28(10s)Q - [1+ O.O035(V - 0.0423) + 7010(V; - 0.0018)
= 0.1691 m3/m2 hr.
+ V,(0.000246 + 9.926vf)I.
The amount of filtrate at this time is
The optimum operation is found by trial:
V, = Qt = 0.1691(0.25) = 0.0423 m3.
Vf = 0.105,
The integral of the rate equation at constant P is
%= 1.0805,
2780(5 - 0.0423) + 0.56(1OS)(V; - (0.00423)2] t, = 0.1095,
= 8(lO5)(tf - 0.25). R, = 1.1507 (max), daily production rate.

on the parameters of the correlating equation COMPRESSIBILITY-PERMEABILITY (CP) CELL
MEASUREMENTS
a = a0(AP)". (1 1.24)
The probable success of correlation of cake resistivity in terms of all
The measurements were obtained with a small filter press. Clearly, the factors that have been mentioned has not been great enough to
the resistivity measured at a particular rate is hardly applicable to have induced any serious attempts of this nature, but the effect of
predicting performance at another rate or at constant pressure. pressure has been explored. Although the a's can be deduced from