Page 374 - Chemical process engineering design and economics

Separator Design 353

m F 2.0 (0.0 1488) -0
— _ _______ = 7 1 0S
————————— —————————————————————— ^^ ^ yj j
0.01488 - 7.44xlO~ 4
m SM
where m SM is the minimum solvent flow rate.
From Equation 6.32.4, the operating feed to solvent ratio,

nip rrip
— = 0.5 ——— = 0.5 (2.105) = 1 .053
ms m SM

m s = m F / 1.053 = 2218 / 1.053 =2106 Ib/h (955 kg/h)

is the operating solvent flow rate.
where m s
4
Substitute x 2K = 7.440X10" and m F/ m s = 1.053 into Equation 6.32.2 for the
methanol balance. The methanol mass fraction in the exit water stream,
yi K = 0 + 1.053 (0.01488 - IMQxW) = 0.01489

From Equation 6.32.13,

A E= 1.053 72 = 2.0.5265
Now, calculate the number of equilibrium stages from Equation 6.32.5.
0.1489-0
Ne
(17 0.52655) = ——————— (1 - 0.5265) + 0.5265
7.44x10" - 0
N e = 7.103

Rounding off N e> we obtain 4 equilibrium stages.
To calculate the extraction height from Equation 6.32.6, first calculate
HETS. HETS is correlated with interfacial tension in Figure 6.23. The interfacial
tension does not appear to be available for this system. Twifik [70] correlated
HETS with dimensionless groups, but his correlation also requires the interfacial
tension. We will use the data given in Table 6.31 for MIBK. Table 6.31 gives data
for several extractor diameters. Select the 12 in (0.305 m) diameter extractor,
which is expected to be close to the calculated diameter. For the 12 in (0.3048 m)
extractor there are several values of HETS at varying agitator speeds and through-
puts. Select the extractor that gives the maximum volumetric efficiency. The
minimum HETS is 5.6 in (0.142 m), and the total volumetric throughput is 1193

Copyright © 2003 by Taylor & Francis Group LLC