pete 411 well drilling

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PETE 411

Well Drilling

Lesson 31

Plugback Cementing

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Plugback Cementing

Case I: No Spacer

Case II: Equal Height Spacers

Case III: Spacer Ahead of Cmt. (only)

Case IV: Two Unequal Spacers

Mixtures and Solutions

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Read:Applied Drilling Engineering

Ch. 3. Cementing

HW #16Due November 22, 2002

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Balanced Cement Plug

Fig. 3.11- Placement technique used for setting cement plug.

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Cementing (Open-Hole Plugging)

1. Plug-back for abandonment

2. Plug-back for fishing or hole deviation

Open-hole plugging is usually performed with “slick” drillpipe or tubing.

In some cases, reciprocating scratchers may be run to enchance cement bonding.

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Types of Balanced Plugs

Case I: No water or other fluid of different density from that in the hole is run ahead or behind the cement slurry.

Case II: Water or other fluid of different density from that hole is run ahead and behind cement slurry. The volume of fluid ahead and behind slurry is calculated so that height in casing is same as height inside the string.

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Displacement

Case III: Water or other fluid of different density from that in the hole is run ahead of cement slurry and hole fluid only is used as displacing fluid.

Case IV: Water or other fluid of different density from that in the hole is run ahead and behind cement slurry. In this case, the heights of fluid in annulus and drill string are not equal.

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Case I

ft

ft capacity, pipe drillT

ft

ft capacity,annular C

3

3

C

Height of plugafter pulling pipe

Height of plug with

pipe in place

T

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Case I

T)H(C

T*HC*HV

placein pipeh wit

plugcement of heightH

ft slurry, of volumeV

ft

ft capacity, pipe drillT

ft

ft capacity,annular C

3

3

3

TC

VH

CT

H Final

Height

10

Example Balanced Plug - Case I

Set a balanced cmt. plug from 8,500-9,000 ft, with no fluid spacers.

1. Open hole diameter = 10 3/4”

2. Assume no washout

3. Use 5”, 19.50 #/ft DP, open ended

4. Use class H cement, 15.6 #/gal

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Example - Case I

(a) Calculate volume of cement slurry required:

required slurry of ft 315.15

ft) 500(ft12

75.10

4LD

4V

3

22

2H

DH

L

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Example - Case I

(b) Calculate actual height of plug when DP is in place at 9,000 ft.

If

then H)TC(V

ft/ftin capacity drillpipeT

ft/ftin capacity annularC3

3

TC

VH

CT

= Height of Plug,with Pipe in place

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Example - Case I

(b) cont’d

In this case,

ft/ft 1*ft144

575.10

4C 2

22

( Halliburton Book )

ft/ft 0.49394 3

ft/ft 0997.0T 3

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Example - Case I

(b) cont’d

ft/ft)0997.049394.0(

ft 15.315

TC

VH

3

3

ft 530.9 H

place in pipe withplug, ofheight

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Example - Case I

(c) Determine the quantity of mud displacement inside the DP that will ensure a balanced plug.

Balance requires that the pressures be equal inside the DP and in the annulus, at 9,000’.

MAMAMDMD

MACAMDCD

AD

)h0.052(ρ)h0.052(ρ

PPPΔP

PP

PD PAhMD = hMA

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Example - Case I

ft 8,469.1

530.9-9,000

drillpipe inside mud ofheight

annulus in mud ofheight

drillpipe inside mud ofheight

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Example - Case I

33

ft 61.5

bbl*ft 4.844

VDispl = 150.4 bbl (of mud)

Volume of mud displacement(behind the cement slurry)

= 8,469 ft * 0.0997 ft3/ft

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Example - Case I

Also required:

Class H cement req’d

Mix water req’d

sk/ft 18.1

ft 15.3153

3

sks 1.267

gal/bbl 42

gal/sk 5.2*sks 1.267

bbl 1.33quiredReWater

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mud

water

cement

water

mud

Case II

hW


hWD = hWA

C

V

T

V WAWD

CVV WAWD

T

Height of plug with

pipe in place

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Example, Balanced Plug - Case II

Set a balanced plug, 500 ft high, with its bottom at 9,000 ft. Use water spacers of equal height inside DP and in annulus.

Volume of annular water spacer = 10 bbl

Open hole diameter = 10 3/4”. No washouts

5” DP, 19.50 #/ft, open ended.

Use class H cement, 15.6 #/gal

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Example - Case II

(a) & (b) From previous example:

ft, 9.530TC

VH

capacity drillpipe ,ft/ft 0.0997T

capacity annular ,ft/ft 0.49394C

slurrycement of vol. ,ft 15.315V

3

3

3

place in drillpipe with

plug ofheight

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Example - Case II

(c) Calculate height (length) of water spacer in DP:

In annulus,

ft 6.113h

ft 113.6

ft/ft 49394.0bblft

5.61*bbl 10

C

Vh

WD

3

3

WAWA

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Example - Case II

(d) Volume of water spacer inside DP

C

T*annulus in spacer of .VolV W,DP

… for spacers of equal height

ft/ft 49394.0

ft/ft 0.0997*bbls 10

3

3

V W,DP = 2.02 bbls

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Example - Case II

(e) A balanced plug requires that

surface. the to extendmust drillpipe in mud

PD PA

AD PP

MAWACAMDWDCD PPPPPP

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Example - Case II

(e) cont’d

ft 5.355,8

drillpipe in mud ofHeight

WDCD hh-9,000

6.1139.530000,9

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Example - Case II

ft

ft0.0997*ft 8,355.5

3

Volume of mud required to displace cement and spacers

= 833.0 ft3

VDispl = 148.5 bbls

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Check

OK.

I Case - problem previous to answer

bbls 150.5

bbls 2.02bbls 5.148

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Pumping Sequence:

1. Water spacer for annulus:

10 bbls

2. Cement Slurry for Plug:

3. Water spacer behind cement:

2.0 bbls

bbls 2.56ft 15.315 3

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Pumping Sequence

4. Mud displacement behind second water spacer:

148.5 bbls

Total fluid pumped = 10 + 56.2 + 2 + 148.5

= 216.7 bbls

(at 10 bbl/min this would require ~22 min)

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Case III

Hole fluid density > density of water

Hydrostatic heads in DS and annulus must balance at top of cement slurry with DS in hole.

hW


Height of plug with

pipe in place

MDWDCDMAWACA

DA

PPPPPP

PP

0

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Case IV - General Case

MDWDCDMAWACA

DA

PPPPPP

PP

Hole fluid density is greater than water density.

Hydrostatic heads in DS and annulus must balance at top of cement slurry with DS in hole.

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Procedure in setting balanced plug

1. Run drillpipe in to depth where plug is to be set; in this case 9,000 ft. (open ended).

2. Circulate and condition mud one complete circulation to make sure system is

balanced.

3. Pump spacers and cement per calculations and displace w/proper amount of fluid

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4. Stop pumps; break connection at surface.

A. If standing full, plug is balanced.

B. If flowing back, a mistake in calculations has been made. Stab inside BOP, or have a slug of heavy mud ready to pump.

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5. Once the end of the drillpipe clears the plug, there is a good chance the pipe will pull wet. This is because pressures have gone back into a completely balanced mud system.

6. If pulling wet, slug pipe and pull out of hole.

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7. Even if plug is severely out-of-balance, never try to reverse cement out of hole.

8. Tag plug with DP at end of 8 hours. If too high, plug may have to be drilled out and another plug spotted. If too low, spot another plug to required height with DP just above top of first plug.

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Calculations to Design a Balanced Open Hole Cement Plug

1. Calculate cu. ft. of slurry required for plug in open hole.

2. Multiply this volume by excess factor (50% excess factor = 1.50)

tables. nHalliburto use or, ft Ld4

V 321

π

3 12 ft ,factor *VV

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When dealing with a washed-out hole, where an excess factor is required, it is usually easier to calculate a new, effective hole size, and use that instead of the excess factor.

Calculations for balanced plug - HINT

5112 .*VV If 50% excess is required

51.*d4

πd

4

π 21

22

112 d225.15.1dd Use d2 for calculationsThis is the effective dia.

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Calculations for balanced plug

3. Find height (h, ft) cement will occupy when

drillpipe is at bottom of plug during pumping:

ftft

ft

)donbased(VolVol

Vh 3

3

2annulusDP inside

2

CT

VH

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Calculations for balanced plug - cont’d

4. Find height (ft) water spacer ahead of cement will occupy in annulus. Use

d2 to calculate this (to account for the excess factor).

5. Find height (ft) water spacer behind cement will occupy in DP. Do

not use excess factor.

6. Pressures must balance at bottom of plug

annDP PP

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7.

8. Convert this mud to feet inside DP.

mud PΔfor Solve

PΔPΔPΔP

PΔPΔPΔP

DP

mud DPspacercmtDP

mud annulusspacercmtann

DPP


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9. Convert this footage to bbls inside DP for proper displacement.

10. To find sks cmt required, divide volume, V2, by yield/sk. This yield, Ysk, may be

in the Halliburton tables (or may not…).

Number of sx req’d, sk

2

Y

VN


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11. If yield not shown, calculate from formula for mixtures. Solve for in this formula. Add the V’s for yield.

12. Total mix water will be times number of sacks.

VW total = (VW / sk) * N

11vWV

sk/VW


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