Page 144 - Petroleum and Gas Field Processing

Oil retention time: 15 min
Water retention time: 10 min
Solution
Using Eq. (9), determine H o,max :
3 2
1:28 10 t o ð
o
w Þd m
H o, max ¼ in:
o
3 2
1:28 10 ð15Þð1:04 0:89Þð500Þ
H o, max ¼ ¼ 36 in:
20
Use Eq. (11) to determine the ratio A w /A:
A w 0:5 Q w t w
¼
A Q o t o þ Q w t w

A w 0:5ð3000 10Þ
¼ ¼ 0:1
A ð3000 10Þþð8000 15Þ
From Figure 6, determine the ratio H o /D for A w /A ¼ 0.1:
H o
¼ 0:338
D
Therefore,
36
H o, max
D max ¼ ¼ ¼ 106:5 in:
H o D 0:338
This is the maximum allowable vessel diameter.
The gas capacity constraint, Eq. (13), yields
0:5

TZQ g g C d
DL ¼ 420
P d m ð o g Þ
0:5
!
8 g C d

DL ¼ 420 555 0:89
250 100ð l g Þ 1
Determine the gas and oil densities and substitute C d ¼ 0.65 in the above
equation:
2:7
g p 2:7 0:65 250 3
g ¼ ¼ ¼ 0:888 lb=ft
TZ 555 0:89
l ¼ w
1 ¼ 62:4 0:89 ¼ 55:54 lb=ft 3
Therefore, the gas capacity constraint is expressed by

DL ¼ 68:22 ðE1Þ

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