8. Squeeze Cementing

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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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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


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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Squeeze Cementing

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


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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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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Squeeze Cementing

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


when spotted.2050 psi

6 Pressure to reverse cementfrom workstring when cementreaches top perforation.

1932 psi


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

Documents

8. Squeeze Cementing