Tutorials_2_DDT oil and gas industry exp
- 1. Drawdown Test Tutorial-2 By Eng. MAbbas
- 3. Exercise (1) • A constant-rate drawdown test was run in an oil well with the following data: Ø= 0.039 Bo=1.136 rbbl/stb Ct=17 10-6 psi-1 Qo=250 STBOPD µO=0.8 cp h=69 ft rw=0.198 ft Determine: 1-Formation permeability 2-Skin Factor 3-Pressure drop due to skin 4-Flow efficiency and damage ratio 5- Reservoir pore volume 6-Determine the geometry and drainage area of the test well. Time(hrs) Pwf(psi) 0 4412 0.12 4272 0.2 4201 0.3 4122 0.4 4057 0.5 3995 0.6 3953 0.7 3912 0.8 3884 0.9 3857 1 3822 1.94 3699 2.79 3653 4.01 3636 4.82 3616 5.78 3607 6.94 3600 8.32 3593 14.4 3573 Time(hrs) Pwf(psi) 17.3 3567 20.7 3561 24.9 3555 29.8 3549 35.8 3544 43 3537 51.5 3532 61.8 3526 74.2 3521 89.1 3515 107 3509 128 3503 154 3497 185 3490 222 3481 266 3472 319 3460 383 3446 460 3429 By Eng. MAbbas
- 4. 2800 3000 3200 3400 3600 3800 4000 0 50 100 150 200 250 300 350 400 450 500 Pressure ,Pwf,psi Flowing Time,hrs DRAW DOWN TEST,Cartesian data plot 3530 psi By Eng. MAbbas
- 5. 3400 3450 3500 3550 3600 3650 3700 1 10 100 1000 Pressure ,Pwf,psi Flowing Time,hrs DDT,Semi log data plot) 3652 psi By Eng. MAbbas
- 6. 100 1000 0.1 1 10 100 1000 (Pi-Pwf),psi flowing time ,hrs DDT,log-log data plot WBS By Eng. MAbbas
- 7. From log-log plot data: Time at the beginning of transient period=12 hrs Time at the end of transient period=150 hrs From semi-log plot data: P1hr=3652psi The slop of the middle time m =-70 psi/cycle 1-Formation permeability = 2- Skin Factor By Eng. MAbbas
- 8. 3-Pressure drop due to skin factor: 4- Damage ratio is estimated as 2.3 = 1- 386.9 891 =0.43=43% By Eng.MAbbas
- 10. = 5.456( −70 −0.222 ) + 𝐸𝐸 2.303(3652 − 3530) −70 EXP = 31.08 From Table 2.1 CA=31.08 corresponds to a well in the center of circle ,square or hexagon The Drainage area can be calculated as the following = (3,315,367 ∗ (5.615 43460(0.39)(69) ) =15.9 acres A 6) By Eng. MAbbas
- 11. Wellbore Storage- Unloading Bottomhole Rate Surface Rate Rate Time 0 By Eng. MAbbas
- 12. Wellbore Storage Definition wb wbc V C Fluid Expansion wb wb c wb wb A . g g . A C 65 25 615 5 144 Changing liquid level By Eng. MAbbas
- 13. Example-2(Wellbore Storage) The following data is given for an oil well that is scheduled for a drawdown test: • volume of fluid in the wellbore = 180 bbl • tubing outside diameter = 2 inches • Wellbore diameter = 7.675 inches • average oil density in the wellbore = 45 lb/ft3 • h = 50 ft, φ = 15 %, rw = 0. 25 ft, μo = 2 cp k = 30 md, s = 0 ,ct = 20 × 10−6 psi−1, co = 10 × 10−6 psi−1 If this well is placed under a constant production rate, Calculate : • the wellbore storage coefficient C. • How long will it take for wellbore storage effects to end? By Eng. MAbbas
- 14. Solution: 1- Calculate the wellbore storage factor caused by fluid expansion: 2- Calculate the wellbore storage factor caused by falling fluid level: By Eng. MAbbas
- 15. 3- Calculate the total wellbore storage coefficient: 4- Time required for wellbore storage influence to end: Solution: By Eng. MAbbas
- 16. Any Questions?????????? By Eng. MAbbas