Page 205 - Petroleum Production Engineering, A Computer-Assisted Approach

Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 201 3.1.2007 9:07pm Compositor Name: SJoearun

GAS LIFT 13/201

Equation (13.57) indicates that the P tro also depends on p vo1 ¼ (900) 1 þ 2,347 ¼ 953 psia
the optional string load S t for double-element valves. The 40,000
S t value can be determined on the basis of manufacturer’s The dome pressure at the valve depth is calculated on the
literature. basis of Eq. (13.42):
The procedure for setting and testing valves in a shop is
as follows: P d ¼ 0:7438(953) 0 þ (0:2562)(424 ) ¼ 817 psia
The valve closing pressure at the valve depth is calculated
. Install valve in test rack. with Eq. (13.43):
. Adjust spring setting until the valve opens with S t psig P vc ¼ 817 þ (0) 0:7438ð Þ ¼ 817 psia
applied pressure. This sets S t value in the valve.
. Pressureupthedomewithnitrogengas.Coolvalveto60 8F. The dome pressure at 60 8F can be calculated with a trial-
and-error method. The first estimate is given by idea gas
. Bleed pressure off of dome until valve opens with P tro
psig applied pressure. law:

Example Problem 13.7 Design gas lift valves using the P d at 60 F ¼ 520P d ¼ (520)(817) ¼ 725 psia
T d (126 þ 460)
following data:
Spreadsheet programs give z 60F ¼ 0:80 at 725 psia and
Pay zone depth: 6,500 ft 60 8F. The same spreadsheet gives z d ¼ 0:85 at 817 psia
Casing size and weight: 7 in., 23 lb. and 126 8F. Then Eq. (13.58) gives
3
Tubing 2 ⁄ 8 in., 4.7 lb. (1.995 in. ID)
Liquid level surface: P d at 60 F ¼ (520)(0:80)P d (817) ¼ 683 psia:

Kill fluid gradient: 0.4 psi/ft (126 þ 460)(0:85)
Gas gravity: 0.75
Bottom-hole temperature: 170 8F Test rack opening pressure is given by Eq. (13.57) as
Temperature surface flowing: 100 8F P tro ¼ 683 þ 0 ¼ 918 psia:
Injection depth: 6,300 ft 0:7438
Minimum tubing pressure at injection
Following the same procedure, parameters for other valves
point: 600 psi
are calculated. The results are summarized in Table 13.5.
Pressure kickoff: 1,000 psi
The spreadsheet program GasLiftValveDesign.xls can be
Pressure surface operating: 900 psi used to seek solutions of similar problems.
Pressure of wellhead: 120 psi
Tubing pressure margin at surface: 200 psi
Casing pressure margin: 0 psi
13.6 Special Issues in Intermittent-Flow Gas Lift
Valve specifications given by Example Problem 13.6 The intermittent-flow mechanism is very different from that
of the continuous-flow gas lift. It is normally applicable in
Solution Design tubing pressure at surface ( p hf,d ): either high-BHP–low PI or low-BHP–low PI reservoirs. In
these two reservoir cases, an excessive high drawdown is
120 þ 200 ¼ 320 psia
needed, which results in a prohibitively high GLR to pro-
Design tubing pressure gradient (G fd ): duce the desired quantity of oil (liquid) by continuous gas
lift. In many instances, the reservoir simply is not capable of
(600 320)=6,300 ¼ 0:044 psi=ft
giving up the desired liquid regardless of drawdown.
Temperature gradient (G t ): The flow from a well using intermittent gas lift techniques
is called ‘‘ballistic’’ or ‘‘slug’’ flow. Two major factors that
(170 100)=6,300 ¼ 0:011 F=ft define the intermittent-gas lift process must be understood:
1 R1:0 0:2562 ¼ 0:7438 1. Complex flowing gradient of the gas lifted liquids from
T:E:F: ¼ R=(1 R) 0:2562=0:7438 ¼ 0:3444 the well.
2. Contribution of the PI of the well to the actual deliver-
Depth of the top valve is calculated with Eq. (13.49):
ability of liquid to the surface.
1,000 0 120 Figure 13.19 shows the BHP of a well being produced by
D 1 ¼ 1,000 0 ¼ 2,347 ft
0:40 intermittent-flow gas lift.
40,000 The BHP at the instant the valve opens is indicated by
Point A. The pressure impulse results in an instantaneous
Temperature at the top valve: 100 þ (0:011) (2,347) pressure buildup at Point B, which reaches a maximum at

¼ 126 F
Design tubing pressure at the top valve: 320 þ (0:044) C after the initial acceleration of the oil column.
(2,347) ¼ 424 psia Figure 13.20 shows the intermittent-flowing gradient,
which is a summation of the gradient of gas above the
For constant surface opening pressure of 900 psia, the slug, the gradient of the slug, and the gradient of the lift
valve opening pressure is calculated with Eq. (13.9): gas and entrained liquids below the slug.
Table 13.5 Summary of Results for Example Problem 13.7
Design Surface Valve Dome Valve Dome Test
tubing opening opening pressure closing pressure rack
Valve Valve Temperature pressure pressure pressure at depth pressure at 60 8F opening
no. depth (ft) (8F) (psia) (psia) (psia) (psia) (psia) (psia) (psia)
1 2,347 126 424 900 953 817 817 683 918
2 3,747 142 487 900 984 857 857 707 950
3 5,065 156 545 900 1,014 894 894 702 944
4 6,300 170 600 900 1,042 929 929 708 952