Page 196 - Petroleum Production Engineering, A Computer-Assisted Approach

Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 192 3.1.2007 9:07pm Compositor Name: SJoearun

13/192 ARTIFICIAL LIFT METHODS
where DHp m is mechanical power losses, which is usually Calculate gas apparent molecular weight:
taken as 20 horsepower for bearing and 30 horsepower for MW a ¼ (0:65)(29) ¼ 18:85
seals.
The proceeding equations have been coded in the com- Calculated gas constant:
puter program CnetriComp.xls (on the CD attached to this 1,544
3

book) for quick calculation. R ¼ ¼ 81:91 psia-ft =lb m - R
18:85
Calculate polytropic head:

Example Problem 13.4 Assuming two centrifugal com- 1:09 þ 0:99 3:41 0:277 1
pressors in series are used to compress gas for a gas lift H g ¼ (81:91)(530) 2 0:277
operation. Size the first compressor using the formation
given in Example Problem 13.3. ¼ 65,850 lb f -ft=lb m
Calculate gas horsepower:
Solution Calculate compression ratio based on the inlet
and discharge pressures: Hp b ¼ 3,100 þ 50 ¼ 3,150 hp
r ffiffiffiffiffiffiffiffiffiffiffi ThecomputerprogramCentrifugalCompressorPower.xlscan
1,165 be used for solving similar problems. The solution given by
r ¼ ¼ 3:41
100 the program to this example problem is shown in Table 13.3.
Calculate gas flow rate in scfm:
13.5 Selection of Gas Lift Valves
32,000,000
q ¼ ¼ 22,222 scfm Kickoff of a dead well requires a much higher gas pressure
(24)(60)
than the ultimate operating pressure. Because of the kickoff
Based on the required gas flow rate under standard condi- problem, gas lift valves have been developed and are run as
tion (q), estimate the gas capacity at inlet condition (q 1 )by part of the overall tubing string. These valves permit the
ideal gas law: introduction of gas (which is usually injected down the annu-
lus) into the fluid column in tubing at intermediate depths to
(14:7) (560) unload the well and initiate well flow. Proper design of these
q 1 ¼ (22,222) ¼ 3,329 cfm
(250) (520) valve depths to unsure unloading requires a thorough under-
standing of the unloading process and valve characteristics.
Find a value for the polytropic efficiency based on q 1 :
13.5.1 Unloading Sequence
E p ¼ 0:61 þ 0:03 log (3,329) ¼ 0:719
Figure 13.7 shows a well unloading process. Usually all
valves are open at the initial condition, as depicted in
Calculate polytropic ratio (n 1)=n:
Fig. 13.7a, due to high tubing pressures. The fluid in
tubing has a pressure gradient G s of static liquid column.
1:25 1 1
R p ¼ ¼ 0:278 When the gas enters the first (top) valve as shown in
1:25 0:719
Fig. 13.7b, it creates a slug of liquid–gas mixture of less-
Calculate discharge temperature by density in the tubing above the valve depth. Expansion
of the slug pushes the liquid column above it to flow to

T 2 ¼ (530)(3:41) 0:278 ¼ 745 R ¼ 285 F: the surface. It can also cause the liquid in the bottom
Estimate gas compressibility factor values at inlet and hole to flow back to reservoir if no check valve is installed
discharge conditions: at the end of the tubing string. However, as the length of
the light slug grows due to gas injection, the bottom-hole
z 1 ¼ 1:09 at 100 psia and 70 8F pressure will eventually decrease to below reservoir
z 2 ¼ 0:99 at 341 psia and 590 8F pressure, which causes inflow of reservoir fluid. When
Calculate gas capacity at the inlet condition (q 1 ) by real the tubing pressure at the depth of the first valve is
gas law: low enough, the first valve should begin to close and the
gas should be forced to the second valve as shown in
(1:09)(14:7) (530)
q 1 ¼ (22,222) ¼ 3,674 cfm Fig. 13.7c. Gas injection to the second valve will gasify
(0:99)(100) (520) the liquid in the tubing between the first and the second
Use the new value of q 1 to calculate E p : valve. This will further reduce bottom-hole pressure and
cause more inflow. By the time the slug reaches the depth of
E p ¼ 0:61 þ 0:03 log (3,674) ¼ 0:721
the first valve, the first valve should be closed, allowing
Calculate the new polytropic ratio (n 1)=n: more gas to be injected to the second valve. The same
1:25 1 1 process should occur until the gas enters the main valve
R p ¼ ¼ 0:277 (Fig. 13.7d). The main valve (sometimes called the master
1:25 0:721
valve or operating valve) is usually the lower most valve in
Calculate the new discharge temperature: the tubing string. It is an orifice type of valve that never

T 2 ¼ (530)(3:41) 0:277 ¼ 746 R ¼ 286 F closes. In continuous gas lift operations, once the well is
fully unloaded and a steady-state flow is established, the
Estimate the new gas compressibility factor value: main valve is the only valve open and in operation
z 2 ¼ 0:99 at 341 psia and 286 8F (Fig. 13.7e).
Because z 2 did not change, q 1 remains the same value of
3,674 cfm. 13.5.2 Valve Characteristics
Equations (13.12) and (13.16) describing choke flow
Calculate gas horsepower: are also applicable to the main valve of orifice type.

(3,674)(100) 1:09 þ 0:99 3:41 0:277 1 Flow characteristics of this type of valve are depicted
Hp g ¼
(229)(0:721) 2(1:09) 0:277 in Fig. 13.8. Under sonic flow conditions, the gas
passage is independent of tubing pressure but not casing
¼ 3,100 hp pressure.