Drilling Cementing PDF

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Drilling Cementing Drilling Engineering (1) Azad University – Marvdasht Campus M. Moarefian Reference: Applied Drilling Engineering Chapter 3 Cementing ‹ Cementing Processes • Casing • Liner • Squeeze • Plug ‹ Density of Mixtures ‹ Cementing Equipment 1 Types of Cementing Processes I. Primary Cement...

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Drilling Cementing Drilling Engineering (1) Azad University – Marvdasht Campus M. Moarefian

Reference: Applied Drilling Engineering Chapter 3

Cementing ‹ Cementing Processes • • • •

Casing Liner Squeeze Plug

‹ Density of Mixtures ‹ Cementing Equipment

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Types of Cementing Processes I.

Primary Cementing

II.

Secondary Cementing

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Primary Cementation Cementing of casing in a well bore. To provide axial support for the casing. To provide a seal across permeable and impermeable formations Seal annulus at the bottom of the casing To provide corrosion protection.

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Secondary Cementation Plugging back (P&A) prior to side tracks To cure lost circulation (While drilling) Sealing perforation prior to workover Repair unsuccessful primary cementing job

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Types of Cementing Processes I. Primary Cementing 1. 2. 3. 4.

Full String (Casing) Liners Large Pipe Stage

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Primary Cementing Steel Casing

Borehole

Cement

Steel Liner

Full String Cementing

Liner Cementing 6

Types of Cementing Processes

I.

Secondary Cementing 1. Plug-back 2. Squeeze

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Secondary Cementing 1. Squeeze Cementing

1. Casing (Perfs.) 2. Open-hole (lost-circulation)

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NEW SLURRY

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Secondary Cementing

2. Plugging 1. Open-hole plug back A. Sidetracking B. Abandonment 2. Casing

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Cementing Equipment

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Guide Shoe Float Shoe Built in check valve Adjustable check valve Float collar

one to three joints above the shoe

Centralizers Rigid

Casing - casing

Flexible

Casing - open hole

Scratches Rotating Reciprocating 14

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Cementing

After Cementing (check) 18 17

Mixing Cement

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Wiper Plugs Wiper plugs are equipped with rubber-cupped fins which wipe mud from the walls of the casing ahead of the cement and clean the walls of casing behind the slurry. Examples of wiper plugs are shown in the next slide. The top plug also serves as a means of determining when the cement is in place. 20

Cement plugs bottom plug top plug

Cementing head

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Cementing Head •Permit the cement plugs to be released at the propre time •Provide a connection between cementing pumping equipment and casing which can be closed to allow testing the cement lines. •Permit pressure to be held in casing after cementing to backup the floating equipment.

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Single Stage

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Preparation for cementing program Input data: Size and depth of the shoe A Directional survey of the hole Estimate the formation strengths and possible loss zones Hole contents and fluid properties Pressure regime in the well Top of cement Casing size and weight 27

Calculation •Total volume of slurry, spacer, and preflushed required (caliper) •Amount of dry cement required •Volume of mix-water (slurry volume and density) •Quantity of additives •Volume and pump speed of displacing fluid •Expected pressure differential (pump and casing test pressures) •Total volume drilling fluid returns (casing+slurry+spacer) •Calculate whether the casing will float or not after displacing the cement 28

•Hydrostatic head at various depths in the casing annulus •Total time required for the job (mixing, release plug & displace) •Compare that time with the thickening time •Compare the max. of following loads with the design loads Hook load Crown load Block line safety factor (if more lines needed) tone miles while running the casing

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Successful Cementation Good slurry design Adequate and timely testing slurry Proper blending of cement and additives The use of centerlizers and scratchers Casing running procedures Reciprocate or rotate of the casing string Proper drilling fluid properties adequate displacement rate 30

mass of dry cement + mass of mixing water Slurry Density = ----------------------------------------------------------volume of cement + volume of water

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Mixing Cement (basis is 1 sk. of cmt.) ‹ The density-volume formula: W w + Wc + Wb + ... + Wn = Wmix ρ w v w + ρ c v w + ρ n v b + ... + ρ n v n = ρ mix v mix

‹ Slurry Yield (volume of mixture/sk): ⎛ gal ⎞ vmix ⎛ cu.ft.⎞ vw + vc + vb + ... + vn = vmix ⎜ ⎟= ⎜ ⎟ ⎝ sk ⎠ 7.48 ⎝ sk ⎠ 32

Rotary Drilling Cementing (basis is 1 sk of cmt.) ‹ Density of Each Component in Slurry: e.g. Density of Barite = Specific Gravity of Barite * Density of Water

ρ b = 4 .23 * 8 .33 lb/gal ρ b = 35.2 lb/gal 33

Rotary Drilling Cementing (basis is 1 sk. of cmt.) ‹ Density of Cement Slurry:

ρmix

∑ρ ν = ∑ν i

i

i

i

ρmix ν mix = ν mix

⎛ mass ⎞ ⎟ ⎜ ⎝ volume ⎠

i

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Problem Calculate the density and yield of a cement slurry consisting of: 65% Class “A” cement 35% Pozmix cement, 6% bentonite BWOC and 10.9 gal/sk of water. 35

Problem (i) Initial tabulations and calculations: Weight Specific Density Component lbs/sk Gravity lbs/gal Class “A” 94 3.14 8.33*3.14 = 26.16 Pozmix 74 2.46 8.33*2.46 = 20.49 Bentonite 2.65 8.33*2.65 = 22.07 Water 1.00 8.33*1.00 = 8.33

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Problem (ii) Determine the properties of one sack of dry cement mixture; 65% Class “A” and 35% Pozmix: Cement Class A, 0.65 sk

Pozmix, 0.35 sk

Weig ht (lbs)

Vol. (gal)

0.65 * 94 = 61.1

61.1 = 2.336 26.16

0.35 * 74 = 25.9

87.0

25.9 = 1.264 20.49

3.600 37

Problem (iii) Determine density and yield of final slurry: Component

Weight

Cmt. mix, 1 sk

Bentonite,

6%

Water, 10.9 gal

(lbs)

87.0 0.06 * 87 = 5.22 10.9 * 8.33 = 90.80 183.02

Vol. (gal) 3.600 5.22 22.07

= 0 . 2365

10.9 14.7365 38

Problem

ρ mix

183 .02 = = 12 .42 lb/gal 14 .74

←

14.74 gal/sk Yield = 3 7.48 gal/ft

= 1.97 ft / sk 3

← 39

Thickening time is the time required to keep the cement pumpable Thickening time =Mixing time + Surface time + Displacement time + Plug release time + Safety time (0.5-1 Hr) Mixing time is the time required to mix cement with water and additives Mixing time =

Volume of dry cement (sack) -----------------------------------mixing rate (sack/min)

Displacement Time is the time required to displace cement (by mud) From inside the casing to the annulus Amount of fluid required to displace top plug Disp. Time = ------------------------------------------------------Displacement rate The volume of cement to be displaced exclude the volume inside the shoe track 40

Cementing Calculations ‹Given: ‹Casing = 7”, 23 lb/ft casing to be set at 8000’ ‹Hole OD = 9” ‹Cement the bottom 2000’ of casing ‹Class “H” neat, 15.6 ppg

‹80’ casing between float collar and float shoe 41

Cementing Calculations ‹Determine: ‹Volume of cement required ‹Volume of mixing water ‹Volume of mud displacement

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Cementing Calculations ‹Solution: ‹Yield of cement = 1.18 cu.ft/sk (Halliburton) ‹Cement volume ‹Annular Capacity = .1745 cu ft/ft (Halliburton) ‹Annular Vol = .1754 cuft/ft X 2000’ = 349 cu ft ‹Casing Capacity = .2210 cu ft/ft ‹Casing Volume = .2210 X 80 = 17.68 cu. ft. ‹Cement volume = 349 + 17.68 cu. Ft = 366.68 cu. ft. 43

Cementing Calculations ‹Sacks of cement = 366.68 cu.ft/1.18 cu.ft/sk ‹Sacks of cement = 310.75 sacks ‹Volume of mixing water = ‹5.2 gal/sack X 310.75 sacks = 1615.88 gal ‹1615.88 gal/42gal/bbl = 38.47 bbl 44

Cementing Calculations ‹Volume of Mixing Water ‹Volume of mixing water = ‹5.2 gal/sack X 310.75 sacks = 1615.88 gal ‹1615.88 gal/42gal/bbl = 38.47 bbl

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Cementing Calculations ‹Mud displacement ‹Casing capacity = .0396 bbl/ft ‹Depth to FC = 8000 – 80 = 7920’ ‹Volume of mud displacement = ‹.0396 X 7920 = 313.63 bbl

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