Transcript Petroleum Engineering 405 Drilling Engineering

PETE 411
Well Drilling
Lesson 31
Plugback Cementing
1
Plugback Cementing

Case I:

Case II: Equal Height Spacers

Case III: Spacer Ahead of Cmt. (only)

Case IV: Two Unequal Spacers

Mixtures and Solutions
No Spacer
2
Read:
Applied Drilling Engineering
Ch. 3. Cementing
HW #16
Due 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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T C
Case I
ft3
C  annularcapacity,
ft
ft3
T  drill pipe capacity,
ft
Height of
plug with
pipe in place
Height of plug
after pulling pipe
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Case I
T C
ft3
C  annular capacit y,
ft
ft3
T  drill pipe capacit y,
ft
V  volumeof slurry,ft3
H  height of cementplug
with pipe in place
H
Final
V  H *C  H *T
 H(C  T )
Height

V
H
CT
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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:
2
 2
  10.75  2
V  DH L  
 ft (500 ft)
4
4  12 
 315.15 ft
3
DH
L
of slurry required
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Example - Case I
(b) Calculate actual height of plug
when DP is in place at 9,000 ft.
If
C  annular capacity in ft / ft
3
T  drillpipe capacity in ft / ft
3
then V  (C  T)H
V
 H
CT
T C
= Height of Plug,
with Pipe in place
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Example - Case I
(b) cont’d
In this case,
  10.75 2  5 2  2
 ft * 1 ft/ft
C  
4
144

 0.49394 ft / ft
3
T  0.0997 ft 3 / ft
( Halliburton Book )
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Example - Case I
(b) cont’d
3
V
315.15 ft
H 

3
C  T (0.49394  0.0997) ft / ft
H  530.9 ft
 height of plug, with pipe in place
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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’.
PD  PA
ΔP CD  PMD  PCA  PMA
0.052(ρ MD )h MD  0.052(ρ MA )h MA
hMD = hMA
PD PA
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Example - Case I
 height of mud inside drillpipe
 height of mud in annulus
 height of mud inside drillpipe
 9,000 - 530.9
 8,469.1 ft
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Example - Case I
Volume of mud displacement
(behind the cement slurry)
3
= 8,469 ft * 0.0997 ft /ft
bbl
 844.4 ft *
3
5.61 ft
3
VDispl = 150.4 bbl (of mud)
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Example - Case I
Also required:
3
Class H cement
req’d
315.15 ft

3
1.18 ft / sk
 267.1 sks
Mix water req’d
267.1 sks * 5.2 gal/sk

42 gal/bbl
Water Re quired  33.1 bbl
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hWD = hWA
Case II
VWD VWA

T
C
T 
VWD  VWA  
C
mud
water
cement
water
mud
hW
Height of
plug with
pipe in place
Height of plug
after pulling pipe
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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:
V  315.15 ft 3 , vol. of cement slurry
C  0.49394 ft 3 / ft , annular capacity
T  0.0997 ft 3 / ft , drillpipe capacity
H
V
 530.9 ft,  height of plug
CT
with drillpipe in place
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Example - Case II
(c) Calculate height (length) of water spacer
in DP:
In annulus,
3
h WA
ft
10 bbl * 5.61
VWA
bbl


3
C
0.49394 ft / ft
 113.6 ft
 h WD  113.6 ft
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Example - Case II
(d) Volume of water spacer inside DP
V W,DP
T
 Vol. of spacer in annulus *  
 C
0.0997 ft 3 / ft
 10 bbls *
3
0.49394 ft / ft
V W,DP = 2.02 bbls
… for spacers of equal height
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Example - Case II
(e) A balanced plug
requires that
PD  PA
PCD  PWD  PMD  PCA  PWA  PMA
PD PA
 mud in drillpipe must extend to the surface.
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Example - Case II
(e) cont’d
 Height of mud in drillpipe
 9,000 - hCD  hWD
 9,000  530.9  113.6
 8,355.5 ft
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Example - Case II
Volume of mud required to displace
cement and spacers
ft 3
 8,355.5 ft * 0.0997
ft
3
= 833.0 ft
VDispl = 148.5 bbls
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Check
148.5 bbls  2.02 bbls
 150.5 bbls
 answer to previous problem - Case I
 OK.
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Pumping Sequence:
1. Water spacer for annulus:
10 bbls
2. Cement Slurry for Plug:
315.15 ft  56.2 bbls
3
3. Water spacer behind cement:
2.0 bbls
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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
Height of plug
after pulling pipe
PA  PD
0
PCA  PW A  PMA  PCD  PW D  PMD
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Case IV - General Case
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.
PA  PD
PCA  PW A  PMA  PCD  PW D  PMD
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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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Procedure in setting balanced plug
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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Procedure in setting balanced plug
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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Procedure in setting balanced plug
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.
 
π 2
V1 
d L ft 3
4
or, use Halliburto n tables.
2. Multiply this volume by excess factor
(50% excess
factor = 1.50)
V2  V1 * factor , ft
3
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Calculations for balanced plug - HINT
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.
V2  V1 * 1.5
If 50% excess is required
π 2 π 2
d2  d1 * 1.5
4
4
d2  d1 1.5  1.225 d1
Use d2 for calculations
This 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:
V
H
TC
h
V2
Volinside DP  Volannulus (based on d2 )


 3 
 ft 
 ft 3 


 ft 
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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
PDP  Pann
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Calculations for balanced plug - cont’d
7. Pann  Δ Pcmt  Δ Pspacer  Δ Pannulusmud
PDP  Δ Pcmt  Δ Pspacer  Δ PDP mud
Solve for Δ PDP mud
8. Convert this PDP mud to feet inside DP.
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Calculations for balanced plug - cont’d
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,
V2
N
Ysk
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Calculations for balanced plug - cont’d
11. If yield not shown, calculate from 1 v1
formula for mixtures. Solve for VW
in this formula. Add the V’s for yield.
12. Total mix water will be VW / sk times
number of sacks.
VW total = (VW / sk) * N
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