Page 347 - Chemical Process Equipment - Selection and Design

310 SOLID-LIQUID SEPARATION

EXAMPLE 11.1 a = [l8,000(2)/C]AP/y = 36,000(0.5)(108)/135
Constants of the Filtration Equation from Test Data = 1.333(101') m/kg.
Filtration tests were performed on a CaCO, slurry with these
proper ties: At 0.8 bar,

C = 135 kg solid/m3 liquid, AP/p = 0.8(108),
p = 0.001 N sec/m2. Rf = 375(0.8)(108) = 3(101') m-*,

The area of the filter leaf was 500cm'. Data were taken of the a = 12,750(2)(0.8)(lO8)/135 = 1.511(101') m/kg.
volume of the filtrate (L) against time (sec) at pressures of 0.5 and Fit the data with Almy-Lewis equation, Eq. (11.24),
0.8 bar. The results will be analyzed for the filtration parameters:
a = kp",
0.5 bar 0.8 bar
(L) VIA t t/(v/A) t t/(V/A)
=
0.5 0.01 6.8 680 4.8 480 k = 1.511(1010)/0.80~26~ 1.604(101'),
1 0.02 19.0 950 12.6 630 :. a= 1.604(1010)P0~2664, m/kg, P in bar.
1.5 0.03 36.4 1213 22.8 760
2 0.04 53.4 1335 35.6 890
2.5 0.05 76.0 1520 50.5 1010
3 0.06 102.0 1700 69.0 1150 I I I I I I
3.5 0.07 131.2 1874 88.2 1260
4 0.08 163.0 2038 112.0 1400 2000 -
-
4.5 0.09 -
5 0.10 165.0 1650
The units of V/A are m3/m2. Equation (11.2) is 1500 -

W/A) -
-- AP
dt p(Rf + aCV/A) '
whose integral may be written

R,+ aC v t
AP/p mA=V/A'
%
Intercepts and slopes are read off the linear plots. At 0.5 bar,
AP/p = 0.5(105)/0.001 = 0.5(108), OO 0.02 0.04 0.06 0.08 0.10
Rf = 600AP/y = 3.0(10") m-', VIA-

Basic filtration Eq. (11.2) is solved for the amount of filtrate, 3045% of the retained filtrate has been found removed by
one-displacement wash. Figure 11.3(b) is the result of one such test.
--pR
V=- A AAP (11.10) A detailed review of the washing problem has been made by
pea( Q Wakeman (1981, pp. 408-451).
The equations of this section are applied in Example 11.3 to
Equations (11.8) and (11.10) are solved simultaneously for AP and the sizing of a continuous rotary vacuum filter that employs a
Q at specified values of V and the results tabulated so: washing operation.
v AP a i/a t COMPRESSIBLE CAKES
0 - - - 0
- - - - - Resistivity of filter cakes depends on the conditions of formation of
vfinal - - - Gina1 which the pressure is the major one that has been investigated at
length. The background of this topic is discussed in Section 11.3,
Integration is accomplished numerically with the Simpson or but here the pressure dependence will be incorporated in the
trapezoidal rules. This method is applied in Example 11.2. filtration equations. Either of two forms of pressure usually is
When the filtrate contains dissolved substances that should not taken,
remain in the filter cake, the occluded filtrate is blown out; then the
cake is washed by pumping water through it. Theoretically, an a = aOP" (11.11)
amount of wash equal to the volume of the pores should be or
sufficient, even without blowing with air. In practice, however, only a = ao(l + kP)". (11.12)