Exercise 1 Cementing Calculations — Stinger Cementation
The surface (20”) casing of a well is normally cemented to surface (continue pumping cement until it is seen at surface). In order to determine the volume of slurry required one calculates the annular space between the conductor (30”) and the surface string (20”) and between the surface string and the openhole. The volume of rathole is added to the above and the slurry volume is translated via the yield of the cement recipe to the number of sacks of cement required for the entire job.
The volume of mixwater required is specified in the slurry recipe in terms of cu ft. per sack of cement and will be determined on the basis of a required cement strength, setting time and allowable free water content.
The time required for the cement job will include the mixing and pumping time (assuming that the slurry is not batch mixed), the time to displace the cement from the cement stinger (since this type of job would normally be carried out using a stinger cementation technique) and 1 hr. contingency time to allow for operational problems during the job. The operation duration will be used to design the slurry so that the cement is set as soon as possible after the job is complete.
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5" d. p
1500′
1530′
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26" Hole |
a. No. sxs cement
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= 2.0944 ft3/ft = 400 x 2.0944 = 838 ft3 |
Slurry volume between the 20" casing and 30" casing: 20" casing/30" casing capacity annular volume
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= 1.5053 ft3/ft = 1100 x 1.5053 = 1656 ft3 |
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1656 ft3 3312 ft3 |
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plus100% excess Total |
Slurry volume between the casing and hole: 20" csg/ 26" hole capacity annular volume
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Slurry volume in the rathole
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b. Mixwater Requirements
Mixwater requirements for class C cement with 6% Bentonite
= 1.36 ft3/sk
Mixwater required = 2326 x 1.36
= 3163 ft3
c. Displacement Time
Total Displacement time = Time to mix and pump cement + time to displace cement
Total Volume of Cement = 4372 ft3
= 779 bbl
Displacement vol. = vol to displace down drillipipe leaving 1 bbl under displaced
d. p. capacity = 0.01776 bbl/ft
Displacement to 1500 ft = 0.01776 x 1500
= 26.6 bbl
(underdisplace by 1 bbl ) = 25.6 bbl
Total Volume to mix and displace = 779 + 25.6 = 804.6 bbls
Total time @ 5 bbl/min = 804.6/5
= 160.9 = 2.7 hrs
%
Exercise 2 Cementation Calculations — Two Stage Cementation
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77 lb/ft 72 lb/ft |
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1000′ .1500′ |
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20" Shoe, |
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TOC |
6300′
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17 1/2" Hole |
6940′
7000′
7030′
a. No. sxs cement Stage 1:
Slurry volume between the casing and hole:
13 3/8" csg/ 17 1/2" hole capacity = 0.6946 ft3/ft
annular volume = 700 x 0.6946
= 486 ft3
plus20% excess = 97 ft3
Total = 583 ft3
Slurry volume below the float collar:
Cap. of 13 3/8, 72 lb/ft csg = 0.0.8314 ft3/ft
shoetrack vol. = 60 x 0.8314
Total = 50 ft3
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= 1.6703 ft3/ft = 30 x.6703 = 50.11 ft3 = 10.02 ft3 = 60 ft3 = 693 ft3 |
Slurry volume in the rathole Cap. of 17 1/2" hole rathole vol.
plus 20%
Total
Yield of class G cement for density of 15.9 ppg = 1.18 ft3/sk TOTAL No. SXS CEMENT STAGE 1: 693/1.18 = 587 sxs
Stage 2:
20" csg/ 13 3/8" csg = 1.0194 ft3/ft
annular volume = 500 x 1.0194
= 508 ft3
TOTAL SLURRY VOL. STAGE 2 : 508 ft3
Yield of class G cement for density of 13.2 ppg = 1.89 ft3/sk
TOTAL No. SXS CEMENT STAGE 2: 508/1.89 = 269 sxs
b. Mixwater Requirements Stage 1:
mixwater requirements for class G cement for density of 15.9 ppg
= 0.67 ft3/sk
Mixwater required = 587 x 0.67
Stage 2:
mixwater requirements for class G cement for density of 13.2 ppg
= 1.37 ft3/sk
Mixwater required = 270 x 1.37
c. Hydrostatic Head Stage 1:
Mud Hydrostatic (0 — 6300 ft) + Cement Hydrostatic (6300 — 7030 ft)
= 6300 x 10 x 0.052 + 730 x 15.9 x 0.052 = 3880 psi
Stage 2:
Mud Hydrostatic (0 — 1000 ft) + Cement Hydrostatic (1000 — 1500 ft)
= 1000 x 10 x 0.052 + 500 x 13.2 x 0.052 = 863 psi
A knowledge of the hydrostatic pressure exerted by the cement slurry when it is place will ensure that the formation fracture pressure will not be exceeded during the cement job.
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А |
d. Displacement Volumes Stage 1:
Displacement vol. = vol between cement head and float collar
= 0.1463 x 1000 (77 lb/ft casing) + 0.148 x 5940 (72 lb/ft casing)
= 1025 bbl
(add 2 bbl for surface line) = 1027 bbl
Stage 2:
Displacement vol. = vol between cement head and stage collar
= 0.1463 x 1000 (77 lb/ft casing) + 0.148 x 500 (72 lb/ft casing)
= 220 bbl
(add 2 bbl for surface line) = 222 bbl
1. INTRODUCTION
1.1 Functions of a Drilling Fluid
1.2 Types of Drilling Fluid
1.3 Historical Development of Drilling Fluids
1.4 Composition of Mud
2. FIELD TESTS ON DRILLING FLUIDS
2.1 Mud density
2.2 Viscosity
2.3 Gel Strength
2.4 Filtration
2.5 Sand Content
2.6 Liquid and Solid Content
2.7 pH Determination
2.8 Alkalinity
2.9 Chloride content
2.10 Activity(aw)
2.11 Cation Exchange Capacity
3. WATER BASED MUD
3.1 Clay Chemistry
3.2 Additives to WBM’s
3.3 Special Types of Water Based Muds
3.3.1 Inhibited Muds
3.3.2 Brine Drilling Fluid
4. OIL-BASED MUDS
4.1 Water in oil emulsions
4.2 Wettability control
4.3 Balanced activity
4.4 Viscosity control
4.5 Filtration control
5. SOLIDS CONTROL
5.1 Solids Control Equipment
5.2 Solids Control Systems
Having worked through this chapter the student will be able to:
General:
• List and describe the functions of drilling fluids and the properties which influence the capability of the fluid to achieve these functions.
• Describe the most important properties of drilling fluids.
• Describe the principle issues considered when programming a drilling fluid.
• List the various generic types of drilling fluid and the composition of these fluids.
Drilling Fluid Testing:
• Describe the equipment and procedures used to determine the: density; rheological properties; gel strength; filtration properties; sand content; liquids and solid content; pH; alkalinity; and CEC of a drilling fluid.
Water Based Muds:
• Describe the composition of water based muds.
• Define the terms: aggregation; dispersion; flocculation; and de-flocculation and describe the ways in which clays will end up in these conditions.
• Describe the additives used to increase/decrease the: viscosity; density and filtration of WBM.
• Describe the chemical formulation of inhibited WBM’s.
Oil Based Muds:
• Describe the chemical formulation of oilbased muds.
• Describe the ways in which the: wettability; activity; viscosity and filtration of OBM can be measured..
Solids Control:
• Describe the principal mechanisms used in solids removal
• Describe the operation of: a shale shaker; a desander and desilter and; a centrifuge.
• Describe the configuration of solids control equipment for weighted and unweighted muds.
Drilling fluid or drilling mud is a critical component in the rotary drilling process. Its primary functions are to remove the drilled cuttings from the borehole whilst drilling and to prevent fluids from flowing from the formations being drilled, into the borehole. It has however many other functions and these will be discussed below. Since it is such an integral part of the drilling process, many of the problems encountered during the drilling of a well can be directly, or indirectly, attributed to the drilling fluids and therefore these fluids must be carefully selected and/or designed to fulfil their role in the drilling process.
The cost of the mud can be as high as 10-15% of the total cost of the well. Although this may seem expensive, the consequences of not maintaining good mud properties may result in drilling problems which will take a great deal of time and therefore cost to resolve. In view of the high cost of not maintaining good mud properties an operating company will usually hire a service company to provide a drilling fluid specialist (mud engineer) on the rig to formulate, continuously monitor and, if necessary, treat the mud.