Page 120 - Petroleum Production Engineering, A Computer-Assisted Approach

Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap09 Final Proof page 114 21.12.2006 2:16pm

9/114 EQUIPMENT DESIGN AND SELECTION
AEDL F BK ¼ A p (p i p o ) ¼ (8:30)(12,800 4,800) ¼ 66,317 lb f
DF ¼ : (9:33)
L p
4
I ¼ (2:875) (2:441) 4 ¼ 1:61 in: 4
64
Example Problem 9.3 The following data are given for a
cement squeezing job: W air ¼ 6:5lb f =ft
7
Tubing: 2 ⁄ 8 in., 6.5 lb/ft (2.441-in. ID) W fi ¼ (15)(7:48)(4:68=144) ¼ 3:65 lb f =ft
Casing: 7 in., 32 lb/ft (6.094-in. ID)
Packer: Bore size D p ¼ 3.25 in., set at 10,000 ft W fo ¼ (0:88)(62:4)(6:49=144) ¼ 2:48 lb f =ft
Initial condition: Tubing and casing are full of
30 API oil (S.G. ¼ 0.88) W ¼ 6:5 þ 3:65 2:48 ¼ 7:67 lb f =ft
Operation: Tubing is displaced with 15 ppg
2
2
cement with an injection pressure DL BK ¼ r F BK
5,000 psi and casing pressure 8EIW
1,000 psi. The average temperature 2 2
drop is 20 8 F. ¼ (1:6095) (66,317) ¼ 3:884 ft
(8)(30,000,000)(1:61)(7:67)
1. Calculate tubing movement if the tubing is not
restrained by the packer, and discuss solutions to the 1. Tubing is not restrained by the packer. The tubing
possible operation problems. shortening is
2. Calculate the tubing force acting on a restraining packer.
DL ¼ DL T þ DL P þ DL B þ DL BK
Solution ¼ 1:38 þ 5:65 þ 2:898 þ 3:844 ¼ 13:77 ft:
Temperature Effect: Buckling point from bottom:
F BK
6
Dl T ¼ bLDT avg ¼ (6:9 10 )(10,000)(20) ¼ 1:38 ft L BK ¼
W
Piston Effect: 66,317
¼
7:67
DP i ¼ (0:052)(10,000)[15 (0:88)(8:33)] þ 5,000 ¼ 8,646 ft
¼ 8,988 psi
To keep the tubing in the packer, one of the following
Dp o ¼ 0 psi measures needs to be taken:
2
A p ¼ 3:14(3:25) =4 ¼ 8:30 in: 2 a. Use a sleeve longer than 13.77 ft
b. Use a restraining packer
2
A i ¼ 3:14(2:441) =4 ¼ 4:68 in: 2 c. Put some weight on the packer (slack-hook) before
treatment. Buckling due to slacking off needs to be
2
A o ¼ 3:14(2:875) =4 ¼ 6:49 in: 2 checked.
2. Tubing is restrained by the packer. The force acting on
DF ¼ [Dp i (A p A i ) Dp o (A p A o )] the packer is
¼ [(8,988)(8:30 4:68) (1,000)(8:30 6:49)] AE DL (6:49 4:68)(30,000,000)(13:77)
DF ¼ ¼
¼ 30:727 lb f L (10,000)
¼ 74,783 lb f :
LDF (10,000)(30,727)
DL P ¼ ¼ ¼ 5:65 ft
AE (6:49 4:68)(30,000,000) Summary
Ballooning Effect: This chapter presents strength of API tubing that can be
used for designing tubing strings for oil and gas wells.
DP i,avg ¼ (10,000=2)(0:052)[15 (0:88)(8:33)] þ 5,000 Tubing design should consider operating conditions in
¼ 6,994 psi individual wells. Special care should be taken for tubing
strings before a well undergoes a treatment or stimulation.
DP o,avg ¼ 1,000 psi
References
2
R ¼ 6:49=4:68 ¼ 1:387
guo, b., song, s., chacko, j., and ghalambor, a. Offshore

2
2L Dp i avg R Dp o avg Pipelines. Burlington: Gulf Professional Publishing,
DL B ¼
10 8 R 1 2005.
2
hasan, r. and kabir, c.s. Fluid Flow and Heat Transfer in
2(10,000) 6,994 1:387(1,000) Wellbores, pp. 79–89. Richardson, TX: SPE, 2002.
¼ ¼ 2:898 ft
10 8 1:387 1 ramey, h.j., jr. Wellbore heat transmission. Trans. AIME
April 1962;14:427.
Since the tubing internal pressure is higher than the exter-
nal pressure during the cement squeezing, tubing string
buckling should occur.
Problems
p i ¼ 5,000 þ (0:052)(15)(10,000) ¼ 12,800 psi
9.1 Calculate the collapse resistance for a section of 3-in.
p o ¼ 1,000 þ (0:88)(0:433)(10,000) ¼ 4,810 psi API 9.20 lb/ft, Grade J-55, non-upset tubing near the
surface of a 12,000-ft string suspended from the sur-
r ¼ (6:094 2:875)=2 ¼ 1:6095 in: face in a well that is producing gas.