Page 108 - Petroleum and Gas Field Processing

Rearrange Eq. (20) to solve for Q o :
2 !
D L 2
Q ¼ 0:7 D L
o
t
Substitute L o ¼ 10.5 and D ¼ 36 in the above equation and solve
for Q o :
Q ¼ 3:176 bbl=day
o
Comparing the above actual oil and gas capacities to the design production
rates of Example 3 shows that the separator can handle almost five times the
design gas rate of Example 3. This is because the cross-sectional area for gas
flow is much more than needed for settling. In such situations, it is evident
that the assumption of allowing each phase to occupy one-half of the
separator effective volume is not a good assumption. The size of the
separator could certainly be reduced if the design was based on allowing
the gas to flow through a smaller area than the 50% of the total area. It is
recommended that the reader solve this problem assuming the gas to occupy
a smaller volume than 50% of the separator volume. Care must be taken in
modifying the above design equations used.
Geometrical relationships relating the cross-sectional area occupied
by the gas, the height of the gas column, and the vessel diameter could
be derived. For the cases where the cross-sectional area of the gas flow,
A g , is not equal to one-half of the vessel cross-sectional area, A; that is, the
height of the gas, H g , is not equal to the radius, D/2, Lockhart [6]
presented the following simplified, but reasonably accurate, relationships.

(a) To determine H g /D as a function of A g /A:
0 A g /A 0.2:
2 3

H g A g A g A g
¼ 2:481 12:29 þ31:133
D A A A
0.2 A g /A 0.8:
H g A g
¼ 0:8123 þ 0:0924
D A
0.8 A g /A 1.0:
2

H g A g A g
1 ¼ 2:481 1 12:29 1
D A A
3

A g
þ 31:133 1
A

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