Page 142 - Petroleum and Gas Field Processing

Retention Time Constraint
The separator size must provide sufficient space for the oil and water such
that each phase is retained within the separator for the desired retention time.
The assumption that the liquid will occupy half of the separator volume,
which was used for two-phase separators, is also used here. However, in the
present case, both oil and water occupy that volume. Therefore, the volume
occupied by the liquid phase (both oil and water), V l , in a separator having a
diameter, D (in.), and effective length, L (ft), is given by

D 2
V l ¼ 0:5 L ft 3
4 12
3
Because 1 barrel (bbl) ¼ 5.61 ft ,
4
2
V l ¼ 4:859 10 D L bbl ð14Þ
The volume of separator occupied by oil, V o , is the product of the oil flow
rate, Q o , and the oil retention time, t o .If Q o is in barrels per day (BPD)
and t o is in minutes, then

t o Q o
V o ¼ bbl ð15Þ
24 60
Similarly, the volume of separator occupied by water, V w , is the product of
the water flow rate, Q w , and water retention time, t w :

t w Q w
V ¼ bbl ð16Þ
w
24 60
Because V l ¼ V o þ V w ,
4 2 Q o t o þ Q w t w
4:859 10 D L ¼
24 60
Therefore,
2 2
D L ¼ 1:429ðQ o t o þ Q w t w Þ in: ft ð17Þ
Again, using diameters smaller than the maximum diameter determined
from the above water droplet settling constraint, Eq. (12) is used to
determine possible diameter and length combinations that satisfy the
retention time constraint.
The procedure for determining the diameter and length of a three-phase
horizontal separator can, therefore, be summarized in the following steps:
1. Determine the value of A w /A from Eq. (11).
2. Use Figure 6 to determine the value of H o /D for the calculated
value of A w /A.

Copyright 2003 by Marcel Dekker, Inc. All Rights Reserved.