Page 275 - APPLIED PROCESS DESIGN FOR CHEMICAL AND PETROCHEMICAL PLANTS, Volume 1, 3rd Edition

Mechanical Separations 245

feed must not 'jet" into the vessel, and should be baffled and:
to prevent impingement in the main liquid body, keeping
turbulence to an absolute minimum to none. Baffles D � l/2(Q/vct) 112 � 1/2(0.187 /0.005) 11 2
can/ should be placed in the front half of the unit to pro- D = 3.057 ft
vide sloto flow of the fluids either across the unit or
up/ down paths followed by the larger stilling chamber, Length, L = 50 = 5(3.0) = 15 ft
before fluid exits. (See Figure 4-13)
Interface Level: Assume: Hold interface one foot below
Example 4-4: Decanter, using the method of top of vessel to prevent interface from reaching the top
Reference (32] oil outlet.
A plant process needs a decanter to separate oil from Then, h = 0.5 ft
water. The conditions are: r = 3.0/2 = 1.5 ft
I= 2(r 2 - h 2)112 = 2[(1.5) 2 -- (0.5) 2)112 = 2.828 ft
Aoil = (1/2) (rr) (1.5) 2 - 0.5[(1.5) 2 - (0.5) 2]'/2 - (1.5)2
Oil flow = 8500 lb/hr arc sin (0.5/1.5)
p = 56 lb/cu ft
= 3.534 - 0.707 - 0.765
µ = 9.5 centipoise = 2.062 sq ft
Water flow = 42,000 lb/hr
p = 62.3 lb/cu ft Note: In radians: Arc sin (0.5/1.5) (19.47 I l80)1T =
µ = 0.71 centipoise 0.3398

Units conversion: Awarer= Jt(l.5) 2 - l\,il = 7t(2.25) - 2.06 = 5.01
sq ft
Q,,i1 = (8500) (56) (3600) = 0.0421 cu ft/sec P = 2(!.5)[arc cos (0.5/1.5)] = 3.69 sq ft
.Uoil = (9.5) (6.72 X 10- 4) = 63.8 X 10-·l lb/ft-sec Area interface, A1 = (2.828) (15) = 42.42 sq fr
Q,a,cr = 42,000/ (62.3) (3600) = 0.187 cu ft/sec
µw = (0.71)(6.7;! X 10- 1) = 4.7'7 X 10- 4 lb /ft-sec Secondary settling: Continuous phase water droplets
to resist the oil overflow rate if it gets on wrong side of
interface.
Checking dispersed phase, Equation 4-36:

Vwatcr � Q,il/ A1 = 0.0421 I 42.42 = 0.0009924 ft/ sec
EJ- 0.042� [ (56)(4.77Xl0- 4) ]0.3
0.187 (62.3) (63.8 x 10- 4) Then, from settling-velocity equation:
= 0.010009
d = [(18)(6.38 x 10- )(0.0009924)/(32.17)(62.3 - 56)] /2
1
3
d = 0.0007498 ft, (216 um)
Therefore, light phase is always dispersed since 0 is less
than 0.3. Checking coalescence time:
Settling rate for droplets of oil through water: Assume Hj, = height of dispersion band= 10% ofD =
Assume droplet size is d = 0.0005 ft ( 150 µm), as earli- 0.3 ft
er discussed. Time available to cross the dispersed band

voil = (32.17)(0.0005) 2 (56 - 62.3)/((18)(4.77 X 10- 4)] = 1/2(1-1 0 A 1 /Qo) should be> 2 to 5 min
= -0.005 ft/sec = l/2[(0.3)(42.42)/(0.0421)]
= 150 sec, which is 2.5 min
The ( - ) sign means the oil rises instead of settles.
Overflow rate: Should be acceptable, but is somewhat low.
Assume I (Figure 4-12) is 80% diameter, D, of vessel
0
and that L/D = 5. Then D i1 = 4(2.062) I (2.828 + 3.69) = l .265 ft
V ;1 = 0.0421/ (2.062) = 0.0204 ft/sec
0
Then, C&:/ Ar< Vct
A1 = IL = (0.80) (SD) = 4D 2, then, N = (0.0204) (l.265) (56) = 226 5
_
2
Qj4D < Vct Re oil 6.38 x 10- 3