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 ................ ................... .................. ................... .................. ................. . 8-3 Squeeze Cementing Terminology .................. .................. .................. .................. .................. ................ 8-4 Hesitation ................ .................. .................. .................. .................. .................. .................. ................. ..8-5 .. 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 this process. The bottom of casing strings (shoe) and liner tops may also require squeezing to correct problems. These types of jobs may not require perforating prior to squeezing.
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.
Topic Areas
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. Pump pressure is allowed to rise in order to "squeeze" cement slurry into the desired area. The cement is allowed to harden into a plug which blocks fluid movement. Drill-out of set cement inside the casing is normally required.
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: •
Existing production perforations may be abandoned by squeeze cementing. Also, casing leaks due to pipe corrosion may be sealed with
•
The purposes of squeeze cementing How to calculate a basic squeeze cementing job
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 squeeze cementing
•
planning considerations for squeeze jobs.
Purposes of Squeeze Cementing Some of the more common reasons for performing a squeeze job are
•
to provide a seal for places purposely not cemented during primary cementing — for example, squeezing liner tops
•
to reduce the gas/oil ratio by shutting off some of the gas-producing perforation (isolating the gas and oil zones)
•
to improve the oil/water ratio by shutting off some 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 correct a defective primary cementing job caused by channeling or insufficient fill-up
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to repair casing holes caused by corrosion, perforation, etc. (Fig. 8.1)
Cementing 1
Squeeze Cementing
Figure 8.2 – Whole cement slurry does not enter the formation
In squeeze jobs, forcing the formation to fracture is 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 Cementing Terminology In order to understand the different methods used in squeeze cementing, it is necessary to first be familiar with the following terms: •
Cement dehydration
•
Pump-in pressure and injection rate
•
Low- and high-pressure squeeze
•
Block squeezing
The volume per minute at which the fluid will be pumped during the squeeze job is called the injection 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 slurry is composed of cement, additives and water. When slurry reaches a permeable formation, 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 squeeze job, 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 squeeze job.
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Block squeezing requires 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.
the necessity of using a spacer (in case of fluid incompatibility) •
•
the maximum pressure to be used, which cannot exceed the pressure limitations of the workstring, casing, BOPs, and other equipment to be used
•
the type of cement to be used, in that fluidloss additives are used to ensure that a small amount of cement filter cake will form against the formation, while the slurry in the casing remains fluid enough to reverse out
•
the amount of cement to be used, which depends 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 annulus to the pressure required to reverse out the maximum height of cementing the workstring.
Hesitation Squeeze
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.
the difference between the depths of the perforations and the packer (if used), which should from 100 to 150 ft to allow enough volume to continue the squeeze after the cement has cleared the workstring
•
Figure 8.3 – Block 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.
the bottomhole static temperature, which affects the setting time of the cement
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, which will affect the pressure to reverse out, and
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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 cementing job 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 being performed. 5. If the pressure drops off during a squeeze job, this is called ______________. To correct this condition, a _______________ squeeze may be performed. 6. The ____________ pressure to be used during a squeeze job cannot exceed the limitations of the equipment 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):
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):
1 Volume of cement (bbl). 2 Pressure to reverse one barrel of slurry from workstring.
A Drillpipe/tubing size
3 Minimum water requirements.
B Packer depth
4 Displacement volume to spot cement one barrel above packer.
C Top of perforations D Bottom of perforations
5 Pressure to reverse cement when spotted.
E Casing size
6 Pressure to reverse cement from workstring when cement reaches top perforation. 7 Pressure to reverse cement from workstring when cement reaches bottom perforation.
A
2 4
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d i u l F l l e t W n e m e C
5
B C D
E
Figure 8.3 – Well schematic showing squeeze calculations and parameters.
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d i u l F l l e W
d i u l F l l e W
t n e m e C
6
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7
8 9
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. You will need a copy of the Halliburton Cementing Tables (the Red Book ) to use during the samples.
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Squeeze Problem One Well Parameters A
2 7/8 in., 6.5 lb/ft EUE Tubing
Drillpipe/tubing size
2 7/8 in., 6.5 lb/ft EUE
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 Displacement/well fluid
75 sk Fresh water (8.33 lb/gal)
1 Volume of Cement (bbl) B
Packer at 5000 ft
C
Perf top at 5094 ft
D
Perf bottom at 5094 ft
E
7 in., 20 lb/ft Casing
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 the Red 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 (extracted from the Class G section of the Red Book ) shows this to be 1.15 cu.ft/sk.
Figure 8.5 – Parameters for Sample Problem 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 the Red 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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the number of feet that one barrel will fill inside the tubing, which is 172.76 ft/bbl.
2 Pressure to Reverse One Barrel of Slurry from Workstring
3. Multiply the differential pressure found in Step 1 by the value found in Step 2 to obtain the pressure required to reverse out one barrel of slurry from the workstring:
2
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.
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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.)
Figure 8.8 – Hydrostatic pressure data from Red Book.
1. Referring to the “Calculations and Formulae” section of the Red Book (shown in Fig. 8.8), look up the psi/ft (hydrostatic pressure gradient) of the 15.8 lb/gal cement and the 8.33 lb/gal water. Then find the difference 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 (Table 211) for 2 7/8 in., 6.5 lb/ft EUE tubing, find
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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 the capacity 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 water unless the slurry is otherwise designed)
•
Displacement fluid (the capacities of the tubing and the casing)
•
5000 ft × 0.00579 bbl/ft = 28.95 bbl 3. Then figure the capacity of the 7 in., 20 lb/ft casing (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 primeup and wash-up volumes, which would also need to be accounted for.
4. You may have to reverse out the excess cement remaining in the tubing (after unsetting the packer). This reverse circulation 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 Data” section of the Red 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 the cement 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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4 Displacement Volume to Spot Cement One Barrel Above Packer
5 Pressure to Reverse when Cement Spotted After the cement is spotted one barrel above the packer, the packer will be set to avoid circulating any cement behind the packer. When the 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
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5
The pressure needed to reverse out this cement in the tubing can be calculated by multiplying the barrels of cement in the tubing, which in this case is the total cement volume, by the differential pressure per barrel of fluid (as calculated 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 to pump 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 your Red Book to 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 Reaches the Top Perforation
94 ft × 0.0404 bbl/ft = 3.80 bbl 2. 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 to multiply 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):
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11.56 bbl × 67.00 psi/bbl = 775 psi t n e m e C
6
Figure 8.11 – Pressure to reverse when cement 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 of cement 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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(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):
7 Pressure to Reverse Cement from Workstring when Cement Reaches Bottom Perforation
9.87 bbl × 67.00 psi/bbl = 661 psi Note: The effects of friction have been disregarded in all these calculations.
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7
Figure 8.12 – Pressure to reverse when cement 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 in the casing. This calculation has already been performed in Calculation 3 (5.49 bbl). 2. Now subtract this volume from the total volume of cement: 15.36 bbl – 5.49 bbl = 9.87 bbl 3. The pressure needed to reverse out this cement in the tubing can now be calculated by multiplying the barrels in the tubing
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8 Pressure to Reverse Out at the Completion of the Job
9 Amount of Cement Pumped Through the Perforations Using information obtained from previous calculations, you can calculate the amount of cement pumped through the perforation during this squeeze job. 1. The total volume of slurry (15.36 bbl) minus the volume of slurry in the casing (5.49 bbl) minus the volume of slurry now in the tubing is the volume of cement in the formation:
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15.36 bbl – 5.49 bbl – 4.95 bbl = 4.92 bbl 2. Divide this by the yield of the cement, 1.15 cu.ft/sk, to obtain the amount of cement in the formation: 27.62 ft3 ÷ 1.15 ft3/sk = 24 sk
8 9
Results of Calculations The following table presents the results of the nine calculations for Squeeze Problem One: Squeeze Cementing Calculation Results Description 1 Volume of cement (bbl).
Figure 8.13 – Pressure to reverse out at the completion of the job.
2 Pressure to reverse one barrel of slurry from workstring.
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 tubing now, given that 24 bbl of displacement fluid were pumped behind it: 28.95 bbl – 24 bbl = 4.95 bbl 2. 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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Result 15.36 bbl 67.00 psi/bbl
3 Minimum water requirements.
72.32 bbl
4 Displacement volume to spot cement one barrel above packer.
12.59 bbl
5 Pressure to reverse cement when spotted.
1029 psi
6 Pressure to reverse cement from workstring when cement reaches top perforation.
775 psi
7 Pressure to reverse cement from workstring when cement reaches 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
Cementing 1
Squeeze Cementing
Squeeze Problem Two In this job, you are to squeeze off a channel that runs from perforations below a packer to a point above the packer. Note that the first seven questions you will answer are based on the same calculations as in Squeeze Problem One, with two additional calculations (8 and 11):
1 2 3 4 5 6 7 8 9 10 11
Volume of cement Pressure to reverse one barrel of slurry from workstring Minimum water requirements Displacement volume to spot cement one barrel above packer Pressure to reverse cement when spotted Pressure to reverse cement from workstring when cement reaches top perforation Pressure to reverse cement from workstring when cement reaches bottom perforation Maximum pump pressure when cement slurry has reached the bottom perforation Pressure to reverse out at the completion of the job Amount of cement pumped through the perforations Maximum pump pressure if the channel is full of cement
Well Parameters Drillpipe Size Packer Depth
2 7/8 in., 6.516 lb/ft EUE Tubing 6200 ft
Top of Perfs
6240 ft
Bottom of Perfs
6250 ft
Casing Size Cement Type Cement Additives Cement Volume Displacement/ Well Fluid
A
2 7/8 in., 6.5 lb/ft EUE Tubing
B
Packer at 6200 ft
C
Perf top at 6240 ft
D
Perf bottom at 6250 ft
E
7 in., 20 lb/ft J-55 Casing
Figure 8.14 – Parameters for Squeeze Problem B.
7 in., 20 lb/ft, J-55 Class H 0.4% Halad-413 150 sk Fresh Water (8.33 lb/gal)
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1 Volume of Cement 2 Material Name
Material (lb)
Class H Cement 0.4% Halad413
Water Totals
Absolute Volume (gal)
Factor (gal/lb)
94 ×
0.0382 =
3.5908
0.376 ×
0.0811 =
0.0305
×
=
35.819 = 130.20
8.33 ×
Mixing Water Required (gal) 4.3 0 +
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+ 4.3 gal 7.92
+ 4.3 gal
Total mixing water must be entered under absolute gallons before totaling. Find the weight of the mixed cement by using this formula: Total Pounds/Total Absolute gallons = pounds/gallons Find the cement yield in cubic feet per sack by using this formula: Total Absolute gallons / 7.4805 gal/cu.ft. (constant) = cu.ft./ sack The mixing water per sack is the sum of t he gallons in the far right column Cement Density (lb/gal) Cement Yield 3 (ft /sk) Mixing Water required
130.20 /
7.92 =
16.4
lb/gal
7.92 /
4.4805 =
1.06
ft /sk
4.3
3
Gal/sk
Figure 8.15 – Determining pressure to reverse 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 Requirements
2 Pressure to Reverse One Barrel of Slurry From Workstring
1. Cement Mix Water:
Hydrostatic pressure gradients (from “Calculations and Formulae” section of Red Book ) 16.4 lb/gal cement 8.33 lb/gal water
⇒ ⇒
150 sk × 4.3 gal/sk = 645 gal 645 gal ÷ 42 gal/bbl = 15.36 bbl 2. Displacement Fluid:
0.8519 psi/ft 0.4330 psi/ft
Tubing: 6200 ft × 0.00579 bbl/ft = 35.90 bbl Casing: 6250 ft – 6200 ft = 50 ft 50 ft × 0.0404 bbl/ft = 2.20 bbl
Differential Hydrostatic Pressure Gradient =
3. Reversing Fluid: 35.90 bbl
0.8519 psi/ft – 0.4330 psi/ft = 0.4189 psi/ft Feet/Barrel Factor For Workstring = 172.76 lb/ft 0.4189 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 Tubing Capacity Cement Volume Volume Above Packer Displacement Volume
6 Pressure to Reverse Cement from Workstring when Cement Reaches Top Perforation
35.90 bbl - 28.32 bbl - 1.00 bbl 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 bbl
5 Pressure to Reverse Cement When Spotted
28.32 bbl – 1.62 bbl = 26.70 bbl remaining in tubing 26.70 bbl × 72.37 psi/bbl = 1932 psi
28.32 bbl × 72.37 psi/bbl = 2050 psi
4
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d i u l F l l e W
5
t n e m e C
6
Figure 8.17 – Pressure to reverse cement.
Figure 8.16 – Calculations 4 and 5.
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(1903 psi) from the collapse resistance of the casing being used. This value can be found in the “Dimensions and Strengths” section of the Red 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.
7 Pressure to Reverse Cement from Workstring when Cement Reaches Bottom Perforation Capacity of casing from packer to bottom perforation = 2.02 bbl (from Calculation 3 ) 28.32 bbl – 2.02 bbl = 26.30 bbl
2270 psi – 1903 psi = 367 psi
26.30 bbl × 72.32 psi/bbl = 1903 psi
2. Since there is a differential pressure from the 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):
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50 ft × 0.4189 psi/ft = 21 psi 3. Now, subtract the differential pressure from Step 1 to obtain the maximum pump pressure: 7
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.
8
Figure 8.18 – Pressure to reverse cement.
8 Maximum Pump Pressure when Cement Slurry Has Reached the Bottom Perforation
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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 subtracting the pressure to reverse out cement to lowest perforation and no cement in the formation
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 the Job
2270 psi – 789 psi = 14812 psi
Final Displacement Volume = 25 bbl Cement Volume Remaining in Tubing: 35.90 bbl – 25 bbl = 10.9 bbl
11
Reverse Pressure: 10.9 bbl × 72.37 psi/bbl = 789 psi
10 Amount of Cement Pumped Through the Perforations d i u l F l l e W
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
9
11 Maximum Pump Pressure if the Channel is Full of Cement
10
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 packer to 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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Cementing 1
Squeeze Cementing
Results of Calculations The following table presents the results of the nine calculations for Squeeze Problem Two: Squeeze Cementing Calculation Results Description 1 Volume of cement (bbl). 2 Pressure to reverse one barrel of slurry from workstring. 3 Minimum water requirements.
Result 28.32 bbl
5 Pressure to reverse cement when spotted.
2050 psi
6 Pressure to reverse cement from workstring when cement reaches top perforation.
1932 psi
8 Maximum pump pressure when cement slurry has reached the bottom perforation
347 psi
9 Pressure to reverse out at the completion of the job
789 psi
11 Maximum pump pressure if the channel is full of cement
89.18 bbl 6.58 bbl
1903 psi
10 Amount of cement pumped through the perforations
72.37 psi/bbl
4 Displacement volume to spot cement one barrel above packer.
7 Pressure to reverse cement from workstring when cement reaches bottom perforation.
82 sk
1481 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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Cementing 1
Squeeze Cementing
Answers to Unit Quizzes Items from Unit A Quiz
Refer to Page
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 1. Drillpipe size
Refer to Page 8-7
Packer depth Top of perforations Bottom of perforations Casing size 2. yield of the slurry
8-9
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Cementing 1
