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Section 8
Squeeze Cementing
Table of Contents
Introduction................................................................................................................................................8-3
Topic Areas ............................................................................................................................................ 8-3
Learning Objectives ...............................................................................................................................8-3
Unit A: Squeeze Cementing Background ..................................................................................................8-3
Purposes of Squeeze Cementing ............................................................................................................8-3Squeeze Cementing Terminology .......................................................................................................... 8-4
Hesitation ...............................................................................................................................................8-5
Planning Squeeze Cementing Jobs.........................................................................................................8-5
Unit A Quiz ............................................................................................................................................ 8-6
Unit B: Squeeze Cementing Calculations..................................................................................................8-7
Squeeze Problem One ............................................................................................................................8-9
Squeeze Problem Two..........................................................................................................................8-17
Unit B Quiz .......................................................................................................................................... 8-22
Answers to Unit Quizzes ......................................................................................................................... 8-23
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Use for Section Notes
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Introduction
If all primary cementing jobs were completely
successful, there would rarely be a need for
squeeze cementing. However, if drilling mud
has been bypassed (channeling) during the
primary job, squeeze cementing may be required
to correct or remediate the problem.
Squeeze cementing is the process by which
cement is forced into the casing-hole annulus.Fluid returns are not normally expected at the
surface. Perforations in the pipe are often
required to obtain a flow path to the annulus.
The channel or area of poor bond, as well as the
perforations, are filled with cement. Pumppressure is allowed to rise in order to "squeeze"
cement slurry into the desired area. Thecement
is allowed to harden into a plug which blocks
fluid movement. Drill-out of set cement inside
the casing is normally required.
Existing production perforations may be
abandoned by squeeze cementing. Also, casing
leaks due to pipe corrosion may be sealed with
this process. The bottom of casing strings (shoe)
and liner tops may also require squeezing tocorrect problems. These types of jobs may not
require perforating prior to squeezing.
Topic Areas
In this section, these units will be included:
A. Squeeze Cementing Background
B. Squeeze Cementing Calculations
Learning Objectives
Upon completion of this section, you should be
familiar with:
The purposes of squeeze cementing
How to calculate a basic squeeze cementingjob
Unit A: Squeeze Cementing Background
To perform a successful squeeze job, it is first
necessary to become familiar with
the purposes of squeeze cementing
terminology associated with squeezecementing
planning considerations for squeeze jobs.
Purposes of SqueezeCementing
Some of the more common reasons for
performing a squeeze job are
to correct a defective primary cementing jobcaused by channeling or insufficient fill-up
to provide a seal for places purposely notcemented during primary cementing for
example, squeezing liner tops
to reduce the gas/oil ratio by shutting offsome of the gas-producing perforation
(isolating the gas and oil zones)
to improve the oil/water ratio by shutting offsome of the water-producing formation
(isolating the oil and water zones) to close-off an unproductive formation from
the wellbore
to prevent fluid migration
to repair casing holes caused by corrosion,perforation, etc. (Fig. 8.1)
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Figure 8.2 Whole cement slurry does notenter the formation
In squeeze jobs, forcing the formation to fractureis not the objective. If the formation fractures, it
will break down. Then whole cement slurry (not
just the filtrate) will be displaced into the
formation. Therefore, care must be taken so that
the pump pressure and the pressure exerted by
the weight of the fluid are not sufficient to force
a fracture of the formation. The pressure
required to force filtrate into the formation
without fracturing it is called the pump-in
pressure.
Figure 8.1 Squeezing to repair casing.
Squeeze CementingTerminology
In order to understand the different methods
used in squeeze cementing, it is necessary to
first be familiar with the following terms:
The volume per minute at which the fluid will be
pumped during the squeeze job is called theinjection rate. Both the pressure and rate should
be established by performing an injection test in
which well fluid is pumped into the formation to
determine at what rate and pressure the fluid will
be absorbed into the formation.
Cement dehydration
Pump-in pressure and injection rate
Low- and high-pressure squeeze
Block squeezing
Cement slurry is composed of cement, additives
and water. When slurry reaches a permeableformation, only the water (filtrate) will pass into
the cracks of the formation (Figure 8.2). Cement
dehydration is the process by which the cement
forms a cake and hardens on the face of the
formation.
During a low-pressure squeezejob, enough
pressure is applied to form a filter cake of
dehydrated cement on the formation. In other
words, the pump-in pressure or the pressure
necessary to place cement against the formation
will not cause the formation to fracture.
However, if the formation will not absorb filtrate
at the pump-in pressure, (because of blocked
perforations or low formation permeability),
more pressure may be applied. This will result in
a fractured formation - whole slurry will fill the
fractures. This is considered a high-pressure
squeezejob.
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Block squeezingrequires that perforations be
made at the interval to be squeezed. Then,
cement is forced into this interval (Figure 8.3).
Block squeezing is generally used to isolate the
producing zone before completing a well.
Figure 8.3 Block Squeeze
Hesitation Squeeze
At some point during a squeeze job, you will
have attained the pressure planned for the job.
That pressure is then held. If it drops off (that is,
bleed off is occurring), you know that your
cement is continuing to dehydrate.
In this case, a hesitation squeeze may be
conducted. Time is allowed for the cement to
begin to set. The pressure is applied again. If
bleed off continues, more time is allowed. This
is repeated as many times as is necessary; the
only limitation is the thickening time of the
cement. If too long a period is allowed, the
workstring may be cemented up.
Planning Squeeze Cementing
Jobs
Before any type of squeeze job is undertaken,
information must be obtained and choices must
be made, including
the types of well fluids to be used, whichwill affect the pressure to reverse out, and
the necessity of using a spacer (in case of
fluid incompatibility)
the bottomhole static temperature, whichaffects the setting time of the cement
the difference between the depths of the
perforations and the packer (if used), whichshould from 100 to 150 ft to allow enough
volume to continue the squeeze after the
cement has cleared the workstring
the maximum pressure to be used, whichcannot exceed the pressure limitations of the
workstring, casing, BOPs, and other
equipment to be used
the type of cement to be used, in that fluid-loss additives are used to ensure that a small
amount of cement filter cake will form
against the formation, while the slurry in thecasing remains fluid enough to reverse out
the amount of cement to be used, whichdepends on the volume of the workstring
(volume of cement should not exceed the
capacity of the tubular goods) and the length
of the interval to be squeezed (a rule of
thumb is to use 2 sk/ft).
testing all wellhead equipment and annulusto the pressure required to reverse out the
maximum height of cementing the
workstring.
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Unit A Quiz
Fill in the blanks with one or more words to check your progress in Unit A.
1. Common reasons for performing a squeeze job are to ___________ a defective primary cementingjob or to __________ holes in casing caused by corrosion. In addition, the ______________ ratio is
improved.
2. When the filtrate enters the formation, cement ____________ to form a cake on the formation.
3. The pressure required to force filtrate into the formation without ___________ it is called the___________ pressure. This pressure is established by pumping _________________ into the
formation to be squeezed.
4. If the formation fractures during a squeeze job, then a _________ pressure squeeze job is beingperformed.
5. If the pressure drops off during a squeeze job, this is called ______________. To correct thiscondition, a _______________ squeeze may be performed.
6. The ____________ pressure to be used during a squeeze job cannot exceed the limitations of theequipment being used.
7. Cement used for squeeze job usually contains ______________ additives.
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Unit B: Squeeze Cementing Calculations
Before beginning a squeeze cementing job,
several calculations must be performed. The
types of calculations to make depend on the
nature of the job. Following are the basic
squeeze problem calculations (Fig. 8.3,4):
1 Volume of cement (bbl).
2 Pressure to reverse one barrel of slurry from
workstring.
3 Minimum water requirements.
4 Displacement volume to spot cement one
barrel above packer.5 Pressure to reverse cement when spotted.
6 Pressure to reverse cement from workstring
when cement reaches top perforation.
7 Pressure to reverse cement from workstring
when cement reaches bottom perforation.
8 Pressure to reverse out at the completion of
the job.
9 Amount of cement pumped though the
perforations.
Following are the well parameters needed for the
calculations (Fig. 8.3):
ADrillpipe/tubing size
BPacker depth
CTop of perforations
DBottom of perforations
ECasing size
A
B
C
D
E
2
4
5
WellFluid
WellFluid
Cement
Figure 8.3 Well schematic showing squeeze calculations and parameters.
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WellFluid
WellFluid
6
WellFluid
7 8
9Cement
Figure 8.4 - Well schematic showing squeeze calculations and parameters.
The remainder of this section presents two
sample squeeze problems and shows, step by
step, how to calculate all the needed data. Youwill need a copy of theHalliburton Cementing
Tables (theRed Book) to use during the
samples.
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Squeeze Problem One
Well Parameters
Drillpipe/tubing size 2 7/8 in., 6.5 lb/ftEUE
Packer Depth 5000 ft
Top of perforations 5094 ft
Bottom of perforations 5136 ft
Casing size 7 in., 20 lb/ft
Cement type Class G
Cement volume 75 sk
Displacement/well fluid Fresh water (8.33
lb/gal)
A
B
C
D
E
2 7/8 in., 6.5 lb/ft
EUE Tubing
Packer at 5000 ft
Perf top at 5094 ft
Perf bottom at
5094 ft
7 in., 20 lb/ftCasing
1 Volume of Cement (bbl)
To convert the given volume of cement from
sacks to barrels, you must determine the yield of
the slurry.
1. Using the Technical Data section of theRed Book, determine the yield of a neat
Class G slurry, given its weight of 15.8
lb/gal. The table shown in Fig. 8.6 (extractedfrom the Class G section of theRed Book)
shows this to be 1.15 cu.ft/sk.Figure 8.5 Parameters for SampleProblem A. 2. Multiply the volume of cement in sacks by
the slurry yield to determine the volume in
cubic feet.
75 sk 1.15 cu.ft/sk = 86.25 cu.ft
Now, convert cubic feet to barrels using the
conversion constant found in the Technical
Data section of theRed Book:
86.25 cu.ft 0.1781 bbl/cu.ft = 15.36 bbl
Figure 8.6 Class G data from Red Book.
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2 Pressure to Reverse One Barrel of
Slurry from Workstring
2
WellFluid
Figure 8.7 Determining pressure to
reverse on barrel of slurry from drillpipe.
The following is a quick way to calculate the
pressure required to reverse out the slurry from
the workstring. These calculations will be based
on a column equal in height to one barrel of
fluid in the workstring. (Your answer will be in
psi/bbl.)
1. Referring to the Calculations andFormulae section of theRed Book(shown
in Fig. 8.8), look up the psi/ft (hydrostatic
pressure gradient) of the 15.8 lb/gal cementand the 8.33 lb/gal water. Then find thedifference between these two values: 0.8208
psi/ft and 0.4330 psi/ft:
0.8208 psi/ft 0.4330 psi/ft = 0.3878 psi/ft
2. Then, using the Capacity section (Table211) for 2 7/8 in., 6.5 lb/ft EUE tubing, find
the number of feet that one barrel will fill
inside the tubing, which is 172.76 ft/bbl.
3. Multiply the differential pressure found inStep 1 by the value found in Step 2 to obtain
the pressure required to reverse out one
barrel of slurry from the workstring:0.3878 psi/ft 172.76 ft/bbl = 67.00 psi/bbl
Later on, you will be calculating different values
of cement left in the tubing, according to where
you are in your job. You will then use the value
determined in the step above to calculate the
total pressure required to reverse out the cement.
Figure 8.8 Hydrostatic pressure data fromRed Book.
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First, determine the capacity of the tubing. It
extends from the surface to the packer, a
total of 5000 ft. Look up the capacity factor
for the 2 7/8 in., 6.5 lb/ft EUE tubing in the
Capacity section (Fig 8.9). This factor is
0.00579 bbl/ft. Multiply the length of the
tubing by its capacity factor to obtain thecapacity of the tubing:
3 Minimum Water Requirements
The minimum fluid (water) requirements for a
squeeze job include the volumes needed for the
following:
Cement mixing water (always fresh waterunless the slurry is otherwise designed)
Displacement fluid (the capacities of thetubing and the casing)
5000 ft 0.00579 bbl/ft = 28.95 bbl
3. Then figure the capacity of the 7 in., 20 lb/ftcasing (Table 214) from the packer to the
lowest perforation (5136 ft 5000 ft = 136
ft).
Reversing fluid
Therefore, each of these volumes needs to be
calculated and then added together.136 ft 0.0404 bbl/ft = 5.49 bbl
Note: This volume does not include both prime-
up and wash-up volumes, which would also
need to be accounted for.
4. You may have to reverse out the excesscement remaining in the tubing (after
unsetting the packer). This reversecirculation occurs around the setting depth
of the packer (5000 ft). So the volume in the
tubing to reverse out is the same as the
displacement volume you have already
calculated for the tubing (Step 2), which is
28.95 bbl.
1. With the help of the Technical Datasection of theRed Book, you can calculate
how much mixing water you will need (see
Fig. 8.6). Because you are using Class G
cement with a weight of 15.8 lb/gal, the
water requirement is 5.0 gal/sk. Multiplying
the amount of cement needed, in sacks, by
the water requirement gives you your total
mixing water:
Keep in mind that you should use the type
and weight of fluid in the annulus for
reversing fluid.5.0 gal/sk 75 sk = 375 gal
5. Now, the minimum amount of fluid required
for this job can be calculated by adding thecement mixing water (Step 1), the two
displacement fluid volumes (tubing and
casing- Steps 2 and 3), and the reversing
volume (Step 4):
To convert to barrels:
375 gal 42 gal/bbl = 8.93 bbl
2. As for the fluid needed for displacement,you will need to calculate the capacities of
the appropriate parts of both the tubing and
casing. Since there is no way of knowing
when a well will squeeze (pressure will not
bleed off), you must have enough fluid to
displace all of the slurry to the lowest
perforation.
8.93 bbl + 28.95 bbl + 5.49 bbl + 28.95 bbl
= 72.32 bbl
NOTE: As a good practice, we recommend that
you have double the volumes needed to reverse.
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Figure 8.9 Tubing capacity data from Red Book.
Figure 8.10 Casing capacity data from Red Book.
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5 Pressure to Reverse when Cement
Spotted
4 Displacement Volume to Spot
Cement One Barrel Above Packer
After the cement is spotted one barrel above the
packer, the packer will be set to avoid
circulating any cement behind the packer. Whenthe packer is set and you attempt to begin
pumping fluid into the perforations, you may
experience problems establishing an injection
rate, in which case you would unset the packer
and reverse all the cement out of the tubing.
Therefore, you need to calculate how much
pressure is required to do this.
4
5
WellFluid
Cement
The pressure needed to reverse out this cement
in the tubing can be calculated by multiplyingthe barrels of cement in the tubing, which in this
case is the total cement volume, by the
differential pressure per barrel of fluid (ascalculated in Calculation 2 , 67.00 psi/bbl):
15.36 bbl 67.00 psi/bbl = 1029 psi
Figure 8.10 Displacement Volume.
To avoid damaging the formation that can be
caused by pumping all the well fluid ahead of
the cement into the perforations, the packer is
left unset while the cement is spotted one barrel
above it. This allows well fluid to circulate out
of the well.
To determine the volume of displacement topump behind the slurry to spot it one barrel from
the packer, you subtract the total cement volume
plus one barrel from the tubing capacity
(determined during Calculation 3 ):
Tubing Capacity 28.95 bbl
Cement Volume - 15.36 bbl
Volume Above Packer - 1.00 bbl
Displacement Volume 12.59 bbl
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Use yourRed Bookto find the capacity
factor (bbl/ft) of the 7 in., 20 lb/ft casing,
then multiply this value by the length of
casing between the packer and the top
perforation (5094 ft 5000 ft = 94 ft).
6 Pressure to Reverse Cement from
Workstring when Cement Reachesthe Top Perforation
6
WellFluid
Cement
94 ft 0.0404 bbl/ft = 3.80 bbl2. Now, subtract this volume from the total
volume of cement to determine how much
slurry remains in the tubing:
15.36 bbl 3.80 bbl = 11.56 bbl
3. The final step in this calculation is tomultiply the volume of cement remaining in
the tubing by the differential pressure per
barrel of fluid (as calculated in Calculation
2 , 67.00 psi/bbl):
11.56 bbl 67.00 psi/bbl = 775 psi
Figure 8.11 Pressure to reverse whencement reaches the top perforation.
As with the previous question, problems with
injection rates may occur when cement reaches
the top perforation. In this case, we can only
reverse out the cement that is left inside the
tubing, and not the cement that has exited below
the packer and is now inside the casing.
The first calculation we need to make, therefore,
is the volume of cement remaining in the tubing.
1. To do this, you calculate the volume ofcement in the casing and subtract that
volume from the total volume of cement.
Determining the volume of cement in the
casing requires you to calculate the capacity
of the casing from the packer depth (5000 ft)
to the depth of the top perforation (5049 ft).
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7 Pressure to Reverse Cement from
Workstring when Cement ReachesBottom Perforation
WellFluid
7
Figure 8.12 Pressure to reverse whencement reaches bottom perforation.
Related to the calculation just completed, this
calculation will be based on the fact that cement
has made it to the bottom perforation, but no
cement has entered the perforations. Therefore,
we need to once again calculate the volume of
cement that remains in the tubing.
1. First, find the volume of cement that is inthe casing. This calculation has already been
performed in Calculation 3 (5.49 bbl).
2. Now subtract this volume from the totalvolume of cement:
15.36 bbl 5.49 bbl = 9.87 bbl
3. The pressure needed to reverse out thiscement in the tubing can now be calculated
by multiplying the barrels in the tubing
(Step 2) by the differential pressure per
barrel of fluid (as calculated previously in
Step 3 of the calculation for pressure to
reverse out one barrel, 67.00 psi/bbl):
9.87 bbl 67.00 psi/bbl = 661 psi
Note: The effects of friction have beendisregarded in all these calculations.
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9 Amount of Cement Pumped
Through the Perforations
8 Pressure to Reverse Out at the
Completion of the Job
Using information obtained from previous
calculations, you can calculate the amount of
cement pumped through the perforation duringthis squeeze job.
WellFluid
8
9
1. The total volume of slurry (15.36 bbl) minusthe volume of slurry in the casing (5.49 bbl)
minus the volume of slurry now in thetubing is the volume of cement in the
formation:
15.36 bbl 5.49 bbl 4.95 bbl = 4.92 bbl
2. Divide this by the yield of the cement, 1.15cu.ft/sk, to obtain the amount of cement in
the formation:
27.62 ft3 1.15 ft3/sk = 24 sk
Results of Calculations
The following table presents the results of the
nine calculations for Squeeze Problem One:
Squeeze Cementing Calculation Results
Description Result
1 Volume of cement (bbl). 15.36 bbl
2 Pressure to reverse one
barrel of slurry from workstring.
67.00 psi/bbl
3 Minimum water
requirements.
72.32 bbl
4 Displacement volume to
spot cement one barrel abovepacker.
12.59 bbl
5 Pressure to reverse cement
when spotted.
1029 psi
6 Pressure to reverse cement
from workstring when cementreaches top perforation.
775 psi
7 Pressure to reverse cementfrom workstring when cementreaches bottom perforation.
661 psi
8 Pressure to reverse out at
the completion of the job.
332 psi
9 Amount of cement pumped
though the perforations.24 sk
Figure 8.13 Pressure to reverse out at the
completion of the job.
For the following calculations, you have to
assume that some arbitrary volume of
displacement fluid was pumped behind the
cement at the time squeeze pressure is reached.
The volume chosen here is 24 bbl.
1. Find out how much cement is in the tubingnow, given that 24 bbl of displacement fluid
were pumped behind it:
28.95 bbl 24 bbl = 4.95 bbl2. Now multiply the amount of cement in the
tubing by the pressure required to reverse
one barrel out of the tubing (67.00 psi/bbl)
4.95 bbl 67.00 psi/bbl = 332 psi
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2
WellFluid
1 Volume of Cement
MaterialName
Material(lb)
Factor(gal/lb)
AbsoluteVolume
(gal)
MixingWater
Required(gal)
Class HCement 94 0.0382 = 3.5908 4.3
0.4% Halad-413
0.376 0.0811 = 0.0305 0 +
= +
Water 35.819 = 8.33 4.3 gal +
Totals 130.20 7.92 4.3 gal
Total mixing water must be entered under absolute gallonsbefore totaling.Find the weight of the mixed cement by using this formula:Total Pounds/Total Absolute gallons = pounds/gallonsFind the cement yield in cubic feet per sack by using thisformula:Total Absolute gallons / 7.4805 gal/cu.ft. (constant) = cu.ft./ sackThe mixing water per sack is the sum of the gallons in the farright column
Cement Density(lb/gal)
130.20 / 7.92 = 16.4 lb/gal
Cement Yield(ft
3/sk)
7.92 / 4.4805 = 1.06 ft3/sk
Mixing Waterrequired
4.3 Gal/sk
Figure 8.15 Determining pressure toreverse one barrel of slurry from workstring.
Based on the worksheet,
150 sk 1.06 cu ft/sk = 159 cu ft
159 cu. ft 0.1781 bbl/cu. ft = 28.32 bbl
3 Minimum Water Requirements2 Pressure to Reverse One Barrel ofSlurry From Workstring
1. Cement Mix Water:
Hydrostatic pressure gradients (from
Calculations and Formulae section ofRed
Book)
150 sk 4.3 gal/sk = 645 gal
645 gal 42 gal/bbl = 15.36 bbl
2. Displacement Fluid:16.4 lb/gal cement 0.8519 psi/ft
8.33 lb/gal water 0.4330 psi/ftTubing: 6200 ft 0.00579 bbl/ft = 35.90 bbl
Casing: 6250 ft 6200 ft = 50 ft
50 ft 0.0404 bbl/ft = 2.20 bblDifferential Hydrostatic Pressure Gradient =
3. Reversing Fluid: 35.90 bbl0.8519 psi/ft 0.4330 psi/ft = 0.4189 psi/ft
Feet/Barrel Factor For Workstring = 172.76 lb/ft0.4189 psi/ft 172.76 ft/bbl = 72.37 psi/bbl Minimum water required:
15.36 bbl + 35.90 bbl + 2.02 bbl + 35.90 bbl
= 89.18 bbl
NOTE: Always plan to reverse with 2 times the
tubing capacity. Therefore in this problem plan
for an additional 35.9 bbls.
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4 Displacement Volume to Spot
Cement One Barrel Above Packer
6 Pressure to Reverse Cement from
Workstring when Cement ReachesTop Perforation
Tubing Capacity 35.90 bbl
Cement Volume - 28.32 bbl
Volume Above Packer - 1.00 bbl
Displacement Volume 6.58 bbl
Capacity of casing from packer (6200 ft) top
perforation (6240 ft):6240 ft 6200 ft = 40 ft
40 ft 0.0404 bbl/ft = 1.62 bbl5 Pressure to Reverse Cement
When Spotted 28.32 bbl 1.62 bbl = 26.70 bbl remaining intubing
26.70 bbl 72.37 psi/bbl = 1932 psi28.32 bbl 72.37 psi/bbl = 2050 psi
4
5
WellFluid
Cement
6
WellFluid
Cement
Figure 8.17 Pressure to reverse cement.Figure 8.16 Calculations 4 and 5.
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7 Pressure to Reverse Cement from
Workstring when Cement ReachesBottom Perforation
Capacity of casing from packer to bottom
perforation = 2.02 bbl (from Calculation 3 )28.32 bbl 2.02 bbl = 26.30 bbl
26.30 bbl 72.32 psi/bbl = 1903 psi
WellFluid
7
Figure 8.18 Pressure to reverse cement.
8 Maximum Pump Pressure when
Cement Slurry Has Reached theBottom Perforation
It is necessary to determine the maximum
pressure that can be applied at the surface for
this squeeze job. The pressure applied at the
surface may be transmitted through the channel
to the outside of the casing above the packer. If
the pressure applied is high enough, it may
collapse the casing.
1. This calculation is done by first subtractingthe pressure to reverse out cement to lowest
perforation and no cement in the formation
(1903 psi) from the collapse resistance of
the casing being used. This value can be
found in the Dimensions and Strengths
section of theRed Book. Refer to the correct
table for the 7 in, 20 lb/ft, J-55 casing being
used (Fig 8.20). The collapse resistance is
2270 psi.
2270 psi 1903 psi = 367 psi
2. Since there is a differential pressure fromthe packer to the perforation (no cement is in
the channel), you must also subtract this. To
calculate the differential, multiply the
distance from the packer to the lowest
perforation (6250 ft 6200 ft = 50 ft) by the
difference in the psi/ft for the cement and
well fluid (0.8519 psi/ft - 0.4330 psi/ft =
0.4189 psi/ft):
50 ft 0.4189 psi/ft = 21 psi
3. Now, subtract the differential pressure fromStep 1 to obtain the maximum pump
pressure:
368 psi 21 psi = 347 psi
Remember, this is the maximum pressure that
can be applied at the surface if the cement has
reached the perforations.
WellFluid
8
Figure 8.19 Maximum pump pressure.
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Figure 8.20 Casing data from Red Book.
Subtract the pressure to reverse the remaining
cement slurry from the tubing from the collapse
resistance of the casing to find the maximum
pump pressure at surface:
9 Calculation of Pressure to
Reverse Out at the Completion of theJob
2270 psi 789 psi = 14812 psiFinal Displacement Volume = 25 bbl
Cement Volume Remaining in Tubing:
WellFluid
9
10
11
35.90 bbl 25 bbl = 10.9 bbl
Reverse Pressure:
10.9 bbl 72.37 psi/bbl = 789 psi
10 Amount of Cement PumpedThrough the Perforations
Volume of Cement in Perforations:
28.32 bbl 10.9 bbl 2.02 bbl = 15.4 bbl
15.4 bbl 5.6146 cu ft/bbl= 86.46 cu ft
86.46 cu ft 1.06 cu ft/sk = 82 sk
11 Maximum Pump Pressure if the
Channel is Full of Cement
Assuming that 25 bbl of displacement fluid have
been pumped behind the volume of slurry, and
that the channel is full of cement, you know
there is no differential pressure from the packerto the perforations in this case. What is the
maximum pump pressure before the casing
collapses?
Figure 8.21 Calculations for 9, 10, and 11.
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Results of Calculations
The following table presents the results of the
nine calculations for Squeeze Problem Two:
Squeeze Cementing Calculation Results
Description Result
1 Volume of cement (bbl). 28.32 bbl
2 Pressure to reverse one
barrel of slurry from workstring.72.37 psi/bbl
3 Minimum water
requirements.89.18 bbl
4 Displacement volume to
spot cement one barrel abovepacker.
6.58 bbl
5 Pressure to reverse cement
when spotted.2050 psi
6 Pressure to reverse cementfrom workstring when cementreaches top perforation.
1932 psi
7 Pressure to reverse cement
from workstring when cementreaches bottom perforation.
1903 psi
8 Maximum pump pressure
when cement slurry hasreached the bottom perforation
347 psi
9 Pressure to reverse out atthe completion of the job
789 psi
10 Amount of cement
pumped through theperforations
82 sk
11 Maximum pump pressure
if the channel is full of cement1481 psi
Unit B Quiz
Fill in the blanks with one or more words to check your progress in Unit B.
1. The well parameters needed for squeeze calculations include:
____________________________________
____________________________________
____________________________________
____________________________________
____________________________________
2. To convert a given volume of cement from sacks to barrels, you must determine the
____________________________________________.
Now check your answers in the Answer Key at the back of this section.
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Answers to Unit Quizzes
Items from Unit A Quiz Refer toPage
1. correct, fill, oil/water 8-3
2. dehydrates 8-4
3. fracturing, pump-in, fluid 8-4
4. high 8-5
5. bleed-off, hesitation 8-5
6. maximum 8-5
7. fluid-loss 8-5
Items from Unit B Quiz Refer toPage
1. Drillpipe size
Packer depth
Top of perforations
Bottom of perforations
Casing size
8-7
2. yield of the slurry 8-9