Page 273 - APPLIED PROCESS DESIGN FOR CHEMICAL AND PETROCHEMICAL PLANTS, Volume 1, 3rd Edition

Mechanical Separations 245

feed must not ‘)et” into the vessel, ,and should be baffled and:
to prevent impingement in the main liquid body, keeping
turbulence to am absolute minimum to none. Baffles D 2 1/2(Q/vd)l/* 2 1/2(0.187/0.005)1/2
can/should be pllaced in the front half of the unit to pro- D = 3.057 ft
vide slow flow od the fluids either across the unit or
up/down paths €olbwed by the larger stilling chamber, Length, L = 5D = 5(3.0) = 15 ft
before fluid exits. (See Figure 413)
Interface Level: Assume: Hold interface one foot below
top of vessel to prevent interface from reaching the top
oil outlet.

A plant process nleecls a decanter to separate oil from Then, h = 0.5 ft
water. The conditions are: r = 3.0/2 = 1.5 ft
I = 2(r2 - h2)lln = 2[(1.5)’ - (Q.5)2]1’z = 2.828 ft
A,,il = (1/2)(~)(1,3)* - 0,5[(1.5)‘ - (0.5)2]1’2 - (1.5)’
Oil flow = 8500 Ib/hr arc sin (0.5/1.5)
p = 56 Ibi’cu ft = 3.534 - 0.707 - 0.765
p = 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) = (19.47/180)~ =
p = 0.71 centipoise 0.3398

Units conversion: Awater = ~(1.5)‘ - = ~(2.25) - 2.06 = 5.01
sq ft
Q,l = (8500) (56) (3600) = 0.042 1 cu ft/sec P = 2(1.3) [arc cos (0.5/1.5)] = 3.69 sq ft
= (9.5) (6.72 X 1V4) = 63.8 X lb/ft-sec Area interface, AI = (2.828) (15) = 42.42 sq fi
aster = 42,000/(65!.3) (3600) = 0.187 cu ft/sec
:&. = (0.71) (6.72 X lW4) = 4.77 X lb/ft-sec Secondary settling: Continuous phase water droplets
to resist the oil overflow rate if it gets on wrong side of
Checking dispersed phase, Equattion 436: interface.
v,,,,, 5 Qil/AI = 0.0421/42.42 = 0.0009924 ft/sec
(56) (4.77 x w4)
Then, from settling-velocity equation:
(62.3) (63.8 X
= 0.010009
d = [(18) (6.38 X lW3) (0.0009924)/(32.17) (62.3 - 56)]’/‘
d = 0.0007498 ft, (216 pm)
Therefore, light phase is always dispersed since 0 is less
than 0.3. Checking coalescence time:
Settling rate for dxoplets of oil through water: Assume HD = height of dispersion band = 10% of D =
Assume droplet size is d = 0.0005 ft (150 pm), as earli- 0.3 ft
er discussed. Time available to cross the dispersed band

Voll = (32.17) (~0.00~35)2(56 62.3)/[(18) (4.77 X = 1/2(HD A,/%) should be > 2 to 5 rnin
-
= -0.005 ftisec = 1/2[(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 412) is 80% diameter, D, of vessel
Then Doil = 4(2.062)/(2.828 + 3.69) = 1.265 ft
V,, = 0.0421/(2.062) = 0.0204 ft/sec

Then, Q/AI < vd
) (5D) = 4D2, then,
Q/4D‘ < Vd