5. 4. Primary Cementing Calculations

Section 5

Primary Cementing Calculations Table of Contents Introduction................................................................................................................................................5-3 Topic Area..............................................................................................................................................5-3 Learning Objectives ...............................................................................................................................5-3 Unit A: Capacity Calculations ...................................................................................................................5-3 Fill-Up Calculations ...............................................................................................................................5-4 Displacement..........................................................................................................................................5-4 Bull-Plugged Pipe ..................................................................................................................................5-5 Open-Ended Pipe....................................................................................................................................5-5 Unit A Test:............................................................................................................................................5-7 Unit B: Annular Volume Calculations.......................................................................................................5-8 Unit B Test ...........................................................................................................................................5-11 Unit C: Slurry Weight and Volume Calculations ....................................................................................5-12 Density .................................................................................................................................................5-12 Specific Gravity....................................................................................................................................5-13 API Gravity ..........................................................................................................................................5-13 Absolute Density vs. Bulk Density ......................................................................................................5-13 Absolute Volume..................................................................................................................................5-13 Unit C Quiz: .........................................................................................................................................5-17 Unit D: Primary Cementing Calculations Example .................................................................................5-18 Given Information for Primary Cementing Calculations .....................................................................5-20 1 Calculations for Pressure Required to Lift Pipe...............................................................................5-20 2 Calculations for Amount of Cement ................................................................................................5-22 3 Sacks of Cement ..............................................................................................................................5-25 4 Calculations for Amount of Mixing Water ......................................................................................5-25 5 Calculations for Amount of Fluid to Displace Top Plug .................................................................5-26 6 Calculations for Pressure to Land the Plug ......................................................................................5-27 7 Calculations for Resulting Force......................................................................................................5-28 Intermediate Casing Job One ...............................................................................................................5-33 Unit D Quiz ..........................................................................................................................................5-36 Answers to Unit Quizzes .........................................................................................................................5-37

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Primary Cementing Calculations

Use for Section Notes…

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Introduction Primary cementing is the cementing operation performed immediately after the casing has been run downhole. The materials, tools, equipment, and techniques to be used vary depending on the hole conditions, depth of the well, and the people planning the job. Successful primary cementing presents a constant challenge and requires up-to-date knowledge and technology.

C. Slurry Weight and Volume Calculations D. Primary Cementing Calculations Example

Learning Objectives Upon completion of this section, you should be familiar with:

Topic Area The units in this section are: A. Capacity Calculations B. Annular Volume Calculations

•

How to calculate the amount of cement for a Primary job.

•

How to calculate the mixing water required for the cement.

•

How to calculate the pressure required to land the top plug

Unit A: Capacity Calculations find the volume units you want across the top. Read the capacity factor where the columns intersect.

Capacity is a term frequently used interchangeably with volume. As used in the oilfield, it is the volume that a certain length of pipe will hold. Knowing the shape of the pipe is round, the volume can be calculated by hand.

Sample Problem

In this unit, we will discuss fill-up and displacement calculations.

What is the capacity, in gallons, of 1000 ft of 27/8 in., 10.4 lb/ft internal upset drill pipe? Use the sample of Section 210 shown in Figure 5.1 to aid your calculations.

Capacity calculations are one of the many types of mathematical problems that can be greatly simplified by using the Halliburton Cementing Tables (Red Book). Section 210 lists capacity factors for various sizes of drill pipe, tubing and casing. Currently, these are listed in terms of gallons per foot, barrels per foot, and cubic feet per foot. See Figure 5.1 for a sample of a capacity table for drill pipe.

Solution Find the appropriate capacity factor (in gal/ft) in Figure 5.1. Then multiply by the length of the drill pipe. Capacity Factor = 0.1888 gal/ft Capacity = 0.1888 gal/ft × 1000 ft = 188.8 gal

To use the Capacity tables, locate the correct table for the type of pipe you’re dealing with: drill pipe, tubing, or casing. Next, locate the size and weight of pipe in the two left columns. (Note: For tubing, there are four columns.) Then

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Figure 5.1

Fill-Up Calculations

Displacement

Fill-up of pipe is defined as the length of pipe a specified volume will fill. Fill-up factors are also listed in Section 210 of the Red Book.

Displacement is defined as the volume of fluid displaced by an object when it is placed in that fluid. In the oilfield, the terms displacement and displace may be confused. The word displace often refers to pumping the fluid inside the pipe out of the pipe, as in displacing cement with fresh water. To do this, the volume of fluid pumped is usually equal to the capacity of the pipe. This is totally different from the definition of displacement as given in the first sentence of this paragraph.

Sample Problem How many feet of 3-1/2 in., 15.50 lb/ft internal upset drill pipe will 25 barrels of oil fill? Use Figure 5.1 to aid in your calculations. Solution: Fill-up Factor = 152.05 ft/bbl Fill-up = 152.05 ft/bbl × 25 bbl = 3801.25 ft

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Bull-Plugged Pipe

Open-Ended Pipe

Figure 5.2 illustrates the volume of fluid displaced when bull-plugged pipe is run in the hole. This volume is equal to the outside diameter’s flat surface area multiplied by the length of the pipe:

When the pipe is open-ended (that is, some opening permits the pipe to fill up on the inside as it is lowered into the well), it will displace less fluid than the bull-plugged string. As shown in Figure 5.3, open-ended pipe will displace a volume equal only to the volume of steel placed in fluid. This displacement can be calculated by multiplying the cross-sectional area by the length:

Displacement = OD Area × Length or Displacement = 0.7854 × OD × OD × Length

Displacement = Cross-sectional Area × Length 3 in. 3 in.

1.5 in.

1.5 in.

Dry 1,000 ft Dry

1,000 ft Bottom BullPlugged Outside View

Sectional View

Bottom Open

Figure 5.2

Outside View

Sectional View

Figure 5.3

Sample Problem Sample Problem

What is the displacement, in barrels, for the string of bull-plugged pipe shown in Figure 5.2?

What is the displacement, in barrels, for the open-ended pipe illustrated in Figure 5.3?

Solution: 3 in. = 0.25 ft

Solution:

Displacement = 0.7854 × 0.25 ft × 0.25 ft × 1000 ft = 49.09 ft3

OD = 3 in. = 0.25 ft ID = 1.5 in. = 0.125 ft OD Area = 0.7854 × 0.25 ft × 0.25 ft = 0.0491 ft2

Conversion factor = (located in section 240, page 85) 0.1781 bbl/ft3

ID Area = 0.7854 × 0.125 ft × 0.125 ft = 0.0123 ft2

Displacement = 49.09 ft3 × 0.1781 bbl/ft3 = 8.74 bbl

Cross-sectional Area = 0.0491 ft2 – 0.0123 ft2 = 0.0368 ft2

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NOTE: In the above examples, it has been assumed that the tubular goods were flush joint; that is, no allowance was made for internal upsets, external upsets, or couplings. Section 130 of the Red Book contains factors that allow for upsets and couplings, as shown in Figure 5.4.

Displacement = 0.0368 ft2 × 1000 ft = 36.8 ft3 Conversion Factor = 0.1781 bbl/ft3 Displacement = 36.8 ft3 × 0.1781 bbl/ft3 = 6.55 bbl

Figure 5.4 – Red Book data.

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Unit A Test: For items 1 and 2, fill in the blanks. For items 3 to 5, use your Red Book as a reference and find the solutions to check your progress in Unit A. 1. The volume that a certain length of pipe will hold is known as the pipe's ______________________. 2. ____________________________is defined as the volume of fluid displaced by an object when it is placed in that fluid. 3. What is the capacity, in barrels, of 10,000 ft of 7 in., 23.0 lb/ft casing?

4. How many feet of 4-1/2 in., 16.6 lb/ft internal upset drill pipe will 25 barrels of oil fill?

5. What is the displacement, in barrels, for this open-ended pipe? OD = 4.5 in. ID = 2.0 in. Length = 2000 ft

Now, check your answers against the Answer Key at the back of this section.

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Unit B: Annular Volume Calculations Annular volume is the volume contained between the outside of the pipe and the open hole (Figure 5.5) or between the outside of the drill pipe or tubing and the inside of the casing (Figure 5.6). Annular volumes can be determined by calculating the cross-sectional area between the open hole or casing ID and the drill pipe OD and multiplying this by the length.

Figure 5.6 Note: Diameters and lengths must be in the same unit of measurement. Rather than calculating cross-sectional areas to determine annular volumes, you can refer to the Red Book, which provides factors that you can multiply by the length of the annulus to more simply arrive at volumes. There are two sections in the Red Book that you should become familiar with regarding annular volumes:

Figure 5.5

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•

Section 122 is a set of tables of annular volume and annular fillup factors with tubular goods in various size holes. It also lists the factors for multiple strings of tubular goods in various hole sizes.

•

Section 221 lists factors for pipe strings inside other pipe. There are tables for tubing, drill pipe, and casing inside of larger casing. Like Section 122, there are also tables for multiple tubing strings inside casing.

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Figure 5.7 Annular Volume = 0.0479 bbl/ft × 1000 ft = 479 bbl

Sample Problem

You can also use these tables to determine the length that a given volume of fluid will fill.

Find the annular volume, in barrels, between 6000 ft of 2-3/8 in. (2.375 in.), 4.7 lb/ft tubing and a 6-1/8 in. hole. Use Section 122-A of the Red Book (see Figure 5.7).

Sample Problem

Solution:

How many feet of annular space between 4-1/2 in., 10.5 lb/ft casing and a 6 in. hole will 100 bbl of cement fill? Use Section 122 of the Red Book.

Annular Volume Factor = 0.0310 bbl/ft (from the Red Book) Annular Volume = 0.0310 bbl/ft × 6000 ft = 186 bbl

Solution:

Sample Problem

Annular Fill-up = 65.3597 ft/bbl × 100 bbl = 6535.97 ft

What is the annular volume, in barrels, for 1000 ft of 5-1/2 in., 17 lb/ft casing inside 9-5/8 in., 36 lb/ft casing? Use Section 221 of the Red Book.

Another way of calculating annular volume between a string of casing and the ID of the hole is by using Section 210 of the Red Book. This contains the factors for different size cylinders or holes with a length of 1 foot. Annular volume can be calculated by subtracting the annular

Annular Fill-up Factor = 65.3597 ft/bbl

Solution: Annular Volume Factor = 0.0479 bbl/ft

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volume factors for one foot of height and then multiplying by the desired annular height.

Solution:

Sample Problem

Volume Factor for 2-3/8 in. tubing = 0.0055 bbl/ft

Find the annular volume, in barrels, between 23/8 in., 4.7 lb/ft tubing and a 6-1/8 in. hole. Use Section 210 of the Red Book.

Annular Volume Factor = 0.0364 bbl/ft – 0.0055 bbl/ft = 0.0309 bbl/ft

Volume Factor for 6-1/8 in. hole = 0.0364 bbl/ft

Annular Volume = 6000 ft × 0.0309 bbl/ft = 185.4 bbl The differences between this way of determining annular volume and using Section 122 can be explained by the methods used when rounding off of decimals in establishing the tables.

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Unit B Test Calculate the answers for the following: 1. What is the annular volume, in gallons, between 5000 ft of 9-5/8 in., 53.50 lb/ft casing and 12-1/4 in. hole? Use the Red Book to find the appropriate factor.

2. What is the annular volume in cubic feet, for 7675 ft of 2-7/8 in., 6.5 lb/ft tubing inside 7 in., 26 lb/ft casing? Use the Red Book to find the appropriate factor.

Now, look up the suggested answers in the Answer Key at the back of this section.

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Unit C: Slurry Weight and Volume Calculations The density of a base fluid is affected by the addition of additives. This density should be collected for additive concentration when determining hydrostatic pressure.

Density in lb/ft3 Solids Gold Mercury Lead Iron Aluminum Wood Ice

To help you with density calculations, this unit will introduce you to several terms. After completing this unit, you will be familiar with: • • • • • •

Density in g/cc

the relationships among density, weight, and volume API gravity specific gravity absolute density vs. bulk density absolute volume slurry weight and volume calculations

1206.2 846.0 712.5 485.0 165.6 50.0 56.9

19.3 13.5 11.4 7.7 2.6 0.8 0.9

125.0 64.3 62.5 50.0 46.8

2.00 1.03 1.00 0.80 0.75

Liquids Sulfuric Acid Sea Water Fresh Water Kerosene Gasoline Gases

Density

Air 0.075 Oxygen Nitrogen Carbon Monoxide Hydrogen

Density can be defined as the weight of a substance per unit volume. In the English system, one cubic foot is a unit of volume, and one pound is a unit of weight. In the metric system, a unit of volume is cubic centimeter (cc) and weight can be measured in grams (g). If you measured the weight of a specific volume of iron, wood, lead, and water, you would find they have widely different weights. Therefore, a term is needed to refer to the weight of a unit-volume of substance -- that term is density. For example, a cubic foot of water weighs 62.4 lb, so the density of water is 62.4 lb/ft3.

0.0075 0.084 0.0737 0.0734 0.0053

0.0012 0.00134 0.00118 0.00117 0.000085

For practical purposes, the densities of gases are compared with air at atmospheric pressure instead of with water. Using air as a comparative reference, the vapor densities of the gases listed above are: Air Oxygen Nitrogen Carbon Monoxide Hydrogen

If you know the density and the volume of a material, you can calculate its weight: Weight = Density × Volume

1.00 1.120 0.983 0.979 2.004

Some densities of common materials are listed here for your reference.

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Specific Gravity Specific gravity (abbreviated Sp Gr) is the weight of a volume of material divided by the weight of the same volume of material taken as a standard. For solids and liquids, the standard is water; for gases, the standard is air. Another definition of specific gravity is the ratio of the density of a substance to the density of water or air. The density of water is 8.33 lb/gal. It is a simple matter to convert density to specific gravity (or vice versa).

Sample Problem

Formulas for conversion are: Degrees API Gravity = Specific Gravity =

141.5 - 131.5 Sp Gr

141.5 API + 131.5

Absolute Density vs. Bulk Density

What is the specific gravity of a 10 lb/gal brine? Solution

Absolute density is the mass per unit volume. Absolute density considers only the actual volume occupied by a material. Bulk density is mass per unit bulk volume -- which includes the actual volume of the material plus the volume of trapped “air.”

density of substance Sp Gr = density of standard

Sp Gr =

API readings are standardized at a temperature of 60°F. If taken at any other temperature, readings must be converted to 60°F to be accurate. Tables for this conversion and for conversion of API gravity to specific gravity can be found in the API Standard 2500 bulletin and in various engineering handbooks.

10 lb/gal 8.33 lb/gal

Sp Gr = 1.2

Absolute Volume

If you were given the specific gravity of a liquid as 1.2, the density can be calculated:

Absolute volume is the volume per unit mass. Here is an example of absolute volume. Let's assume we have a container (Figure 5.8) that measures one cubic foot (1 ft high, 1 ft deep, 1 ft wide). This container is filled with golf balls. There are void spaces between the golf balls filled with trapped air. We want to know the volume in gallons that is occupied by the golf balls only.

1.2 × 8.33 lb/gal = 10 lb/gal

API Gravity Baume’s gravity is a scale that uses salt water as a reference rather than fresh water. It is used in refineries to determine the gravity of acids and alkalis only. API gravity is used in most other instances in the oil field. Water is used as the standard. Water's API gravity is 10 degrees. As the specific gravity increases, the API gravity decreases. Crude oil with a 42° API gravity has an actual specific gravity of 0.82. American crude is commonly run from 0.768 to 0.966 specific gravity, which would range from 52.6° to 10.5° on the API gravity scale.

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The volume occupied by the golf balls can best be determined by first calculating the volume occupied by the void spaces. For this, we could add water to the container and measure it. After obtaining the number of gallons of water needed to fill the void spaces, we can subtract from 7.4805 gal/ft3 to determine the volume occupied by the golf balls alone.

One cubic foot 7.4805 gal. 1 ft

1 ft 1 ft

The volume of golf balls determined in this way is called the absolute volume. While this example is exaggerated, it does help explain the concept. We work with sand, cement, etc. rather than golf balls, but the balls can be seen as a magnification of sand or cement particles. Void spaces exist in sand; the volume we add to fracturing fluids is actually the absolute volume of the sand.

This container below is filled with golf balls.

1 ft

1 ft

Calculating absolute volume for sand or cement can be simplified by using the tables for "Physical Properties of Cementing Materials and Admixtures" provided in the Technical Data section of the Halliburton Cementing Tables (Figure 5.9).

1 ft

Figure 5.8 – The volume of golf balls is less than the total volume because of the spaces between them.

Figure 5.9 – Red Book data.

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Total lb ÷ Total abs gal = lb/gal

The left-hand column lists the material, the second column gives the bulk weight in lb per ft3. The third column gives the specific gravity, and the fourth column provides absolute volume in gal/lb.

12.33 lb ÷ 1.1824 gal = 10.427943 lb/gal

Sample Problem What is the absolute volume (gallons) and weight (lb per gallon) of this slurry?

Note: When using fluids in absolute volume calculations, you will not find a gal/lb factor in the Red Book. However, you can calculate this factor if you know the fluid density in lb/gal. Simply divide 1 by the fluid density. For example, water is 8.33 lb/gal, so:

• • •

Base Fluid is water @ 8.33 lb/gal Additive is 3% KC1 4 lb of sand is added

Solution

gal/lb = 1 ÷ 8.33 lb/gal = 0.12 gal/lb

ABSOLUTE

Sample Problem

MATERIAL

FACTOR

MATERIALS

(lb)

(gal/lb)

What is the absolute volume (gallons) and weight (lb per gallon) of this slurry?

Water

8.33

×

0.1200

=

0.2499

×

0.0443

= 0.0110705

• •

Sand

4.0

×

0.0456

=

3% KC1

Base Fluid is water @ 8.33 lb/gal 4 lb of sand is added.

TOTALS

(gallons) 1.0

0.1824 1.1934705 gal

Total lb ÷ Total abs gal = lb/gal

Solution We will use the chart below to help in our calculation of absolute volume. First, list the materials and their weights in the first two columns. Then, using the Red Book table, list the absolute volume factors (gal/lb) in the third column. FACTOR

VOLUME

MATERIALS

(lb)

(gal/lb)

(gallons)

Water

8.33

0.1200

Sand

4

0.0456

• •

Multiply the materials (lb) by the factor (gal/lb) to obtain the absolute volume and add these values to the table. For the totals, add the materials (lb) together and add the absolute volumes together: VOLUME

MATERIALS

(lb)

(gal/lb)

(gallons)

Water

8.33

×

0.1200

=

1.0

Sand

4.0

×

0.0456

=

0.1824

12.33 lb

Class H Cement Water @ 8.33 lb/gal

Solution For this cementing problem, we will add a column to the chart for mixing water requirements. Proceed as in the last two sample problems until you need to fill in the mixing water requirements for the Class H cement. This quantity you can find in the Technical Data section of the Red Book in the table for “Water Requirements.” For Class H cement, the requirements are 4.3 gal/sk. Enter this under the appropriate column:

ABSOLUTE FACTOR

Note: Generally, when working with sand and water (or base fluid) slurries, the calculations are based on 1 gallon of base fluid and the weight of that 1 gallon of fluid.

What is the absolute volume (gallons) and weight (lb per gallon) of this slurry?

TOTALS

MATERIAL

12.5799 lb ÷ 1.1934705 gal = 10.5 lb/gal

Sample Problem

ABSOLUTE MATERIAL

TOTALS

12.579 lb

VOLUME

1.1824 gal

Find the weight of the mixed slurry using this formula:

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ABSOLUTE

MIXING

MATERIAL FACTOR VOLUME MATERIALS

(lb)

Cement

94

Water

(gal/lb) ×

0.0382

=

ABSOLUTE

WATER

(gallons)

(gallons)

3.6

4.3

MATERIAL FACTOR VOLUME MATERIALS Cement

8.33

Water

TOTALS

TOTALS

Enter 5.2 gal for the absolute volume for water. Divide the water factor by the absolute volume to determine the materials (lb) for water. Then calculate the totals:

MIXING

(lb)

(gal/lb)

(gallons) 4.3

94

×

0.0382

=

3.6

35.82

=

8.33

×

4.3

129.82 lb

WATER

(gallons)

7.9 gal

Total lb ÷ Total abs gal = lb/gal 129.82 lb ÷ 7.9 gal = 16.4 lb/gal Find the yield of cement (ft3 per sack) by using this formula: Total abs gal ÷ 7.4805 gal/ft3 = ft3/sk 7.9 gal/sk* ÷ 7.4805 gal/ft3 = 1.06 ft3/sk *NOTE: When working with cement slurries, the calculations are usually based on one sack of cement and the weight of that sack.

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Unit C Quiz: For items 1 3, fill in the blanks. For items 4 and 5, calculate the answer. 1. In dealing with specific gravity, the weight of a volume of a material is ________________________ by the weight of the same volume of a material taken as a_______. 2. Absolute density considers only the___________ ____________ occupied by the material. 3. Volume per unit mass defines ______________ _______________. 4. Calculate the weight (lb/gal) of the following slurry: • base fluid is 10 lb/gal brine • sand at 10 lb/gal (0.0456 gal/lb)

5. Using Class H cement (94 lb/sk) and 4.3 gallons mixing water, what is the cement slurry density (lb/ gal) and yield (ft3/sk)?


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Unit D: Primary Cementing Calculations Example Before a primary cementing job can begin, several calculations must be done. In this unit, you will learn these calculations:

Primary Cementing Calculations 7 critical calculations that need to be made with every surface casing job

Also included is a new slurry weight and yield worksheet.

1

7

1

Critical Circulating Pressure – Pressure required to lift (pump) the casing out of the hole

2

Cement volume – The volume of cement required to fill the required footage of the annulus plus the shoe track capacity.

3

Sacks of Cement – Converting the required volume of cement into sacks.

4

Mixing water required for given slurries.

5

Displacement fluid required to the top plug from surface to the top of the shoe track.

6

Pressure to land the plug – Differential pressure required to pump the plug to the top of the shoe track.

7

Resulting Force – The calculated hook load at the top of the casing once the plug has landed.

F A

Well Parameters A B C D E F G

B 5

C

2 3 4

Pipe Size Well Fluid Hole Size Pipe Depth Shoe Track Length Required Cement fill-up Excess volume required (percent)

Calculation Guidelines • Do not apply any decimal places for pressure or sacks

6

• Apply one decimal place for lb/gal • Apply 2 decimal places for psi/bbl cubic feet pounds feet area gallons barrels

E

• For psi/ft use same number of decimal places as in Red Book “Hydrostatic Pressure And Fluid Weight Conversion Tables”

D

Figure 5.10 – Primary Cementing Setup

• Show one decimal place for bbl/min • Rounding numbers – if the last number is to be dropped, round up if ≥ 5, down if < 5.

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Worksheet for Slurry Weight and Volume Calculations MATERIAL NAME

Material (lb)

.

Water

Absolute Volume (gal)

Factor (gal/lb)

Mixing Water Requirements (gal)

X

=

X

=

+

X

=

+

X

=

+

X

=

+

X

=

+

=

lb/gal

gal

TOTALS ==>

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 = lb/gal Find the cement yield in cubic feet per sack by using this formula: Total Absolute gallons / 7.4805 gal/ft3 (constant) = ft3/sack The mixing water per sack is the sum of the gallons in the far right column

Cement Density (lb/gal) ==>

lb/gal

Cement Yield (ft3/sk) ===>

ft3/sk

Mixing Water required ===>

gal/sk

Figure 5.11 – Absolute Volume Worksheet

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Primary Cementing Calculations 7 critical calculations that need to be made with every surface casing job

1

Critical Circulating Pressure – Pressure required to lift (pump) the casing out of the hole

2

Cement volume – The volume of cement required to fill the required footage of the annulus plus the shoe track capacity.

3

Sacks of Cement – Converting the required volume of cement into sacks.

4

Mixing water required for given slurries.

5

Displacement fluid required to the top plug from surface to the top of the shoe track.

6

Pressure to land the plug – Differential pressure required to pump the plug to the top of the shoe track.

7

Resulting Force – The calculated hook load at the top of the casing once the plug has landed.

In the Essential Red Book course, you were introduced to the basic calculations and concepts needed in preparing for a primary cementing job. In this unit you will be shown the steps for calculating a surface casing cement job. Later, you’ll be asked to actually make the calculations for a similar job on your own.

Given Information for Primary Cementing Calculations Refer to the well parameter information as you work through the calculations for the surface casing cementing job. Figure 5.12 – Surface Casing Cementing Job

1 Calculations for Pressure Required to Lift Pipe

Surface Casing Job One When pipe is run into a hole, the result in a ram effect. This ram effect increases as the running speed and the diameter of the pipe increases. In some cases, the ram effect will break down lowpressure zones. Sand may slough off and bridge the annulus. If the casing is stuck in the hole, you cannot pull it out without parting it.

This casing job consists of a single slurry, which is to be circulated from total depth to surface.

Well Parameters Pipe Size

9 5/8 in 36 lb/ft

Well Fluid

8.7 lb/gal

Hole Size

12.25 in

Pipe Depth

300 ft

Shoe Track Length

40 ft

Required Cement Fill-Up

To surface

Excess Volume Required (percent)

100%

These calculations are performed as a precaution. They are done before mixing any cement. Just in case the annulus has bridged, you need to know how much pressure would be required to lift the pipe. This pressure could

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d) The buoyancy factor (Step 2a) multiplied by the weight per foot of casing (Step 2b) times the length of the casing (Step 2c) equals the weight of the pipe hanging in fluid:

possibly lift the pipe out of the hole, so you need to chain the pipe down during the operations. 1. First, the area of the casing must be found. a) In the “Calculations and Formulae” section of the Red Book (see also Figure 5.13), look up the formula for the area of a circle. Area = 0.7854 × D

0.8671 BF × 36 lb/ft × 300 ft = 9364.68 lb Ð (The arrow indicates the direction of this force.)

2

3. Now you have enough information to calculate the amount of pressure to apply to the casing at the surface, pumping downward through the casing below the casing shoe, to start the lift (or to balance the pipe). The larger the diameter of the pipe the less pressure is required to lift the string. This is why you need to chain down large diameter casing during the pumping operation (chain it to the substructure or a leg of the derrick, but not to the rotary table).

Figure 5.13

The downward force of the pipe (step 2d) divided by the area of the pipe (step 1b) is the pressure needed to start the lift:

b) Plug the diameter (9 5/8 in.) into the formula for area of a circle (area of the casing):

9364.68 lb Ð ÷ 72.76 in. = 129 psi

9.625 in. × 9.625 in. × 0.7854 = 72.76 in.2

When applying pressure to start circulation, caution should be taken to prevent the pipe from blowing out of the hole and causing damage. The casing should be chained down and all personnel except the operator, should be cleared off the rig floor.

2. The next step is to calculate the weight of the pipe when it is hanging in fluid (the downward force of the pipe in the wellbore.) a) First, look up the buoyancy factor for the wellbore fluid you are working with. Keep in mind that this buoyancy factor relates to the fact that open-ended pipe weighs less in a fluid than it does in air. The weight of this fluid is 8.7 lb/gal. Referring to the “Displacement” section of the Red Book (see also Figure 5.14), you will see that the buoyancy factor is 0.8671 for a fluid of that weight. b) You also need to know how much your casing weights in air. From the casing stamp, you can find that this casing weigh 36 lb/ft. c) Since it is not known at what depth the annulus might become bridged, use the overall length of your casing (300 ft) for these calculations.

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You will find Class G slurry properties on page 3 of the Class G Section in the Technical Data Section of the Red Book. We will work through the absolute volume calculations as an example. Refer to the gray pages in the “Technical Data: section of your Red Book (see also Figure 5.9). You will see that the table labeled “Physical Properties of Cementing Materials and Admixtures” contains much of the information needed to calculate the slurry’s absolute volume (as well as its weight that will be calculated later). To aid in these calculations, the Worksheet for Slurry Weight and Volume Calculations has been developed (Figure 5.11). This worksheet will be completed as the following calculations are worked out. 1. First, find the bulk weight of API cements listed in Figure 5.9 – 94 lb/ft3. The factor for absolute volume of API cements is 0.0382 gal/lb. 2. Then calculate the correct figures for each of the additives and the water being used. a) Calculate how much the calcium chloride will weigh by multiplying the weight of cement by 2%: 0.02 × 94 lb = 1.88 lb b) Look up the factor for absolute volume of Calcium Chloride, which is 0.0612 lb/gal. Figure 5.14

c) Look up the absolute volume factor for Flocele, which is 0.0845 gal/lb. 3. To find the absolute volumes of the cement, Calcium Chloride and Flocele, multiply the numbers in the materials column by the numbers in the factor column:

2 Calculations for Amount of Cement

94 lb × 0.0382 gal/lb = 3.6 gal 1.88 lb × 0.0612 gal/lb = 0.12 gal. 0.25 lb × 0.0845 gal/lb = 0.02 gal

To find the amount (sacks) of cement needed for this surface casing cementing job, you need to know the type of cement, its weight and its yield. Class G cement with 2% calcium chloride and ¼ lb/sk Flocele has been chosen for this surface casing job. Using this information, you can perform the calculations necessary to eventually find out how much cement you will need.

5 • 22

Cementing 1


Worksheet for Slurry Volumes Worksheet for Slurry Weight and Volume Calculations Casing Job One API Cement Weight = 94 lb/sack 1 sack = 1 cubic foot Material Name

Material (lb)

Class G Cement 2% Calcium Chloride .25lb Flocele

Figure 5.15 – Class G data from Red Book 4. To find the mixing water requirements for a sack of cement with its additives, refer to the gray pages in the “Technical Data” section of your handbook (see also Figure 5.15). These requirements are 5.0 gal., 0 gal and 0 gal for the cement, Calcium Chloride and Flocele respectively. Add these figures together for the water’s absolute volume: 5 gal + 0 + 0 gal = 5 gal

Water Totals

Factor (gal/lb)

94 ×

Absolute Mixing Volume Water (gal) Required (gal) 0.0382 = 3.5908 5.00

1.88 ×

0.0612 =

0.11506

0.25 × ×

0.845 = =

0.02113

41.65 ×

8.33 =

5.00

137.78

8.7270

5.00

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 = lb/gal Find the cement yield in cubic feet per sack by using this formula: 3

3

Total Absolute gal ÷ 7.4805 gal/ft (constant) = ft /sk The mixing water per sack is the sum of the gallons in the far right column

5. To obtain the weight of the water required, multiply the weight per gallon of water by the volume of water (step 4): 8.33 gal/lb × 5 gal = 41.65 lb

Cement Density

137.78 ÷

8.7270 =

15.8

lb/gal

Cement Yield

8.7270 ÷

7.4805 =

1.17

ft /sk

5.00

gal/sk

Mixing Water Required

6. Total the materials and absolute volume columns:

5 • 23

3

Cementing 1


8. Then find the yield by dividing the total absolute volume in gallons (Step 6) by the constant which can be found in the table “Conversion Constants” (Figure 5.16) – 7.4805 gal/ft3 8.7270 gal ÷ 7.4805 gal/ft3 = 1.17 ft3/sk Now assume that you have circulated through the surface casing down to a depth of 300 ft and that you have a well-conditioned hole. It is necessary to calculate the volume of slurry needed for the job. This is a combination of the slurry needed to fill annular space and to fill the shoe joint. 9. To determine how much slurry is needed to fill the annulus, the ft3/ft for the annulus is multiplied by the length of the annulus. a) First, refer to the “Volume and Height Between Tubing, Casing, Drill Pipe, and Hole” section in the handbook. (To use this table, you need to find the table with the correct number of strings. For example, see the four different tables for tubing with an OD of 3 ½ in). Look on the table for 9 5/8 in (see also Figure 5.17). The ft3/ft for this annulus is 0.3132. b) You know that the length of the annulus is 300 ft. Multiplying this by the ft3/ft value found in Step 9a results in the volume needed to fill the annular space. 300 ft × 0.3132 ft3/ft = 93.96 ft3. c) Excess cement – knowing we need 100% excess (double the calculation volume) we calculate the following:

Figure 5.16

93.96 ft3 × 2 = 187.92 ft3

7. To find the weight of the cement slurry in pounds per gallon, divide the total pounds by the total absolute volume in gallons: 137.78 ÷ 8.7270 gal = 15.8 lb/gal

5 • 24

Cementing 1


Figure 5.17 10. Now, the amount of slurry in the shoe joint (or track) needs to be calculated in the last step, to determine how much slurry is in the shoe joint, the capacity factor of the joint is multiplied by its length.

3 Sacks of Cement 1. Now that you know the cubic feet needed (Step 11), you use the yield (Step 8) to calculate the number of sacks needed:

a) To find this capacity factor, turn to the “Capacity” section of the Cementing Table (See also Figure 5.18) You need to know the OD (9 5/8 in.) and the weight (36 lb/ft) of your casing. The capacity factor is 0.4340 ft3/ft.

205.28 ft3 ÷ 1.17 ft3/sk = 175 sk

4 Calculations for Amount of Mixing Water

b) The length of the shoe track was given as 40 ft.

After you know the number of sacks of cement needed, you need to calculate the volume of water required on location to mix the slurry. This water should always be fresh water (unless slurry is designed for salt or seawater).

c) Multiply the capacity factor for the shoe joint (10a) by the length of the shoe joint (10b) for the capacity for the shoe joint: 0.4340 ft3/ft × 40 ft = 17.36 ft3

Your worksheet shows that you will need 5 gal of water per sack of cement. This is multiplied by the number of sacks in order to obtain the total number of gallons of mixing water needed. This unit of measurement needs to be converted

11. Add the volume of slurry to fill the annulus (step 9c) to the volume of slurry to fill the shoe joint (step10c) to determine the total numbers of barrels of cement that you need: 187.92 ft3 + 17.36 ft3 = 205.28 ft3

5 • 25

Cementing 1


Figure 5.18 bottom plug is not run, the top plug will wipe down the walls and mud will collect in the shoe joint between the top plug and the cement slurry.

to barrels since tanks on trucks are marked off in barrels. 1. First, find out how much mixing water you need in gallons: 2. Since there are 42 gal in a barrel, convert to barrels:

When all of the cement has been mixed, the top plug will be pumped down to a depth of 260 ft. The number of barrels needed to do this is the same as the capacity of the casing to the float collar.

875 gal ÷ 42 gal/bbl = 20.83 bbl

0.0773 bbl/ft × 260 ft = 20.1 bbl

5 gal/sk × 175 sk = 875 gal

Therefore, it will take this amount of water to mix your cement. You will need more water on location, considering the water used for cleanup, spacer, etc.

5 Calculations for Amount of Fluid to Displace Top Plug The first step in the cementing process is to run a bottom plug to wipe the casing clean of mud buildup. The cement following the bottom plug will rupture the plug’s diaphragm. Then it will move through the center of the plug, out the bottom of the shoe, and up the annulus. If a

5 • 26

Cementing 1


6 Calculations for Pressure to Land the Plug Calculations for pressure to land the plug should be made on each job. You need to know the pressure required to put the cement in place. Any pressure buildup in excess of this pressure might indicate channeling or bridging in the annulus. A loss of pressure could mean a loss of cement into the formation. Figure 5.12 shows a balanced section of hole. The shoe is at 300 feet and the plug landing point is 260 feet. The shoe joint and the annular space opposite it are full of the same weight cement. Therefore that part of the well is balanced and no calculations are required. You do need to make a few calculations concerning the unbalanced portion of the hole indicated in Figure 5.12. The annular space from 260 ft back to surface is filled with 15.8 lb/gal cement (this density was determined during the amount of cement calculations, Step 7 in Calculations for Amount of Cement). Also, you have displaced the plug with 8.7 lb/gal fluid in the casing. Because the fluids have different densities, a differential pressure results. You must convert the lb/gal of the cement and displacement fluid to psi/ft in order to make the calculation (Figure 5.19).

Figure 5.19 1. Refer to the Hydrostatic Pressure and Fluid Weight Conversion tables in the “Calculations and Formulae” section of your Red Book (see also Figure 5.19). To find the psi/ft for the weight of the cement (15.8 lb/gal), To minimize errors, here is the most accurate way to make this calculation: Working form the bottom of the casing up to the surface, calculate the hydrostatic pressure outside, then calculate the hydrostatic pressure inside. Remember, work from the bottom up, for the entire casing string.

5 • 27

Cementing 1


calculations – Step 1) for a fluid weighing 15.8 lbs/gal.

Hydrostatic Outside Cement 300 ft × 0.8208 psi/ft

=

Total: 300 ft

246 psi

The hydrostatic pressure in a casing job is the pressure in the annular space. The plug will close off the pressure in the casing when it lands on a float collar. If you do not land the plug or run a float collar or back pressure valve, but only stop the plug in the casing, then you will close in the cementing head and the pressure to land the plug will remain on the casing until the cement has set.

246 psi

Hydrostatic Inside Cement 40 ft × 0.8208 psi/ft Well Fluid 260 ft × 0.4519 psi/ft Total 300 ft

=

33 psi

=

117 psi 150 psi

Differential Pressure (At top of shoe) = 246 psi – 150 psi = 96 psi

To calculate the hydrostatic pressure at a given depth, multiply that depth by the psi/ft of the fluid in the annulus – in this case, cement slurry:

Always re-total the depth. If you don’t end up with the same depth then something is wrong. NOTE: You must slow the pump rate down to a half-barrel per minute to read this on a chart or gauge because high rates create friction pressure (which can give you a false pressure reading).

300 ft × 0.8203 psi/ft = 246 psi

7 Calculations for Resulting Force

When the plug has been pumped to its landing position and 96 psi is readable on the gauge, this job has been done successfully. Remember that you will not be able to correctly read the pressure to land a plug on all jobs. Losing cement into the formation or developing channels in the cement will alter the pressure reading. Release the pressure when the plug lands; this will prevent a microannulus. Keep in mind that most companies will require more pressure than what is required to land the plug in order to test the plug’s seal. Depending on the situation, some customers will pressure test casing after the plug lands.

Precautions must be taken before pumping the plug down when you are working on the surface casing. You need to find out if the casing will remain still, or if the plug landing pressure will pump the casing out of the hole – if so, you will need to chain the pipe down (in addition to chaining the head to the elevators). As you did in Calculation 1, you will calculate the difference between the upward force on the pipe and the downward force on the pipe. 1. You have calculated the area of the surface casing (Step 1b - Pressure to Lift the Pipe Calculations) to be 72.76 in.2. In addition, you know the pressure to land the plug is 96 psi (Step 6 – Pressure to Land the Plug Calculations). Multiplying the two values will give you the amount of upward force in pounds.

Calculations for Hydrostatic Pressure at a Given Depth You need to be able to calculate the hydrostatic pressure at any depth. For the purpose of this problem, you will determine the hydrostatic psi at 260 ft. Will this formation support the amount of pressure exerted by the column of cement down the annulus to the given point?

72.76 in2 × 96 psi = 6984.96 lb

In this example problem, you will have only one hydrostatic pressure to work with since you have cement from the top to the bottom of the well. That pressure is 0.8208 psi/ft, which you determined earlier (Pressure to Land the Plug

5 • 28

2. In a previous problem, you looked up the buoyancy factor for the 8.7 lb/gal displacement fluid (Step 2a – Pressure to Lift the Pipe Calculations). Referring to the “Displacement” section again, look up the buoyancy factor for the cement used (see also Figure 5.14) The weight of cement is 15.8 lb/gal as determined in Step 7 of the

Cementing 1


amount of cement calculations. To find the buoyancy factor for this weight, you must look up 15.8.

3. In the same way, calculate the downward force contributed by the cement: 40 ft × 0.7586 BF × 36 lb/ft = 1092.38 lbÐ

Now the length and weight of the piece of pipe in question are needed. Cement is inside the pipe in the shoe joint (40 ft) and displacement fluid is inside the pipe above the float collar (260 ft). The weight of the pipe is 36 lb/gal.

4. Adding these two downward forces together (Step 2 and 3) will give you the total downward force:

Using the formula for downward force, determine the downward force contributed by the displacement fluid:

5. Subtract the upward force (Step 1) from the downward force (Step 4) to obtain the resulting force:

Length of pipe × buoyancy factor for the displacement fluid inside the pipe × pipe weight per foot = pounds of downward force

8116.06 lbÐ + 1092.38 lbÐ = 9208.44 lbÐ

9208.44 lbÐ - 6984.96 lbÏ = 2223.96 lbÐ Although this is considered a downward force, it will be necessary to chain down the casing as a precautionary measure.

260 ft × 0.8671 BF × 36 lb/ft = 8116.06 lbÐ

5 • 29

Cementing 1


Surface Casing Job Two This casing job consists of LEAD and TAIL slurries in which cement is to be circulated from total depth to surface.

Well Parameters Pipe Size

13 3/8-in. Casing

17 1/2-in. Hole

13 3/8 in., 68 lb/ft

Well Fluid

8.5 lb/gal

Hole Size

17.5 in.

Pipe Depth

2400 ft

Shoe Track Length

80 ft

Required Cement Fill-Up (TAIL)

400 ft′

Required Cement Fill-Up (LEAD)

2000 ft′

Excess Volume Required (percent)

120%

1

Critical Circulating Pressure

Area Of Pipe 2 13.375 in × 13.375 in × 0.7854 = 140.50 in. (Section 240 Page 85)

2,000 ft

Hook Load 2,400 ft × 68 lb/ft × 0.8701 = 142,000.32/lbs (Section 130 Table 132 Buoyancy Factor) Pressure to lift (pump) pipe out of the hole 2 142,000.32 lb ÷ 140.50 in. = 1010 psi

2,320 ft

2 Lead Tail Displacement

2,400 ft

Figure 5.20

Cement Volume

Lead Slurry Composition Halliburton Light Cement 2% Calcium Chloride 1/4 lb/sk Flocele

Given Sl. Wt 13.6 lb/gal 3 Sl. Vol. 1.55 ft /sk Water 7.62 gal/sk

Tail Slurry Composition Class H Premium 1% Calcium 1/4 lb/sk Flocele

Given Sl Wt 16.4 lb/gal 3 Sl. Vol 1.07 ft /sk Water 4.3 gal/sk

Tail 3 400 ft (openhole) × 0.6946 ft /ft

3

=

277.84 ft

277.84 ft × 2.2 (excess) 3 80 ft (shoes) × 0.8406 ft /ft

= =

611.25 ft 3 67.25 ft 3 678.50 ft (tail)

Lead 3 2000 ft (openhole) × 0.6946 ft /ft

=

1389.20 ft

=

3056.24 ft (lead)

3

3

1389.20 ft × 2.2 (excess)

3

3

3

NOTE: We are required to pump 120% excess cement on this job; 120% excess is equal to multiplying the perfect hole volume by 2.2.

5 • 30

Cementing 1


Surface Casing Two Lead Cement

Surface Casing Two Tail Cement

Worksheet for Slurry Weight and Volume Calculations


Material Name

Material (lb)

65% Class H Cement 35% Poz 6% Bentonite 2% Calcium Chloride .25lb Flocele Water Totals

Factor (gal/lb)

61.1 ×


25.9 × 5.22 × 1.74 ×

0.0488 = 0.0453 = 0.0612 =

1.26392 0.23647 0.10649

1.27 3.60 0.00

0.25

0.0845

0.02113

0.00

63.47 =

8.33 ×

157.68

7.62

Material Name

Material (lb)

Class H Cement 1% Calcium Chloride .25lb Flocele

Water

11.5820

7.62

Factor (gal/lb)

94 ×


0.94 ×

0.0612 =

0.05753

0.25 × ×

0.0845 = =

0.02113

35.819 =

8.33 ×

4.30

Totals 131.009

7.9695

4.30

Total mixing water must be entered under absolute gallons before totaling.

Total mixing water must be entered under absolute gallons before totaling.

Find the weight of the mixed cement by using this formula:

Find the weight of the mixed cement by using this formula:

Total Pounds ÷ Total Absolute gallons = lb/gal

Total Pounds ÷ Total Absolute gallons = lb/gal

Find the cement yield in cubic feet per sack by using this formula:

Find the cement yield in cubic feet per sack by using this formula:

3

3

3

3

Total Absolute gal ÷ 7.4805 gal/ft (constant) = ft /sk

Total Absolute gal ÷ 7.4805 gal/ft (constant) = ft /sk

The mixing water per sack is the sum of the gallons in the far right column

The mixing water per sack is the sum of the gallons in the far right column

Cement Density

Cement Density

Cement Yield Mixing Water Required

157.68 ÷

11.5820 =

13.6

lb/gal

11.5820 ÷

7.4805 =

1.55

ft /sk

Cement Yield

7.62

gal/sk


3

5 • 31

131.009 ÷

7.9695 =

16.4

lb/gal

7.9695 ÷

7.4805 =

1.07

ft /sk

4.3

gal/sk

3

Cementing 1


3

7

Sacks of Cement

Resulting Force

Force Down

Tail 3 3 678.50 ft ÷1.07 ft /sk

=

634 sk

Lead 3 3 3056.24 ft ÷ 1.55 ft /sk

=

1972 sk

2320 ft × 68 lb/ft × 0.8701 BF 80 ft × 68 lb/ft × 0.7494 BF 2400 ft Total Down

2


Tail 634 sk × 4.3 gal/sk

Lead 1972 sk × 7.62 gal/sk

= =

2726.2 gal ÷ 42 gal/bbl 64.91 bbl

= =

15026.64 gal ÷ 42 gal/bbl 357.78 bbl

137266.98 lb Ð 4,076.74 lb Ð 141343.72 lb Ð

Force Up 140.50 in × 661 psi

4

= =

Hook Load

=

92870.50 lb Ï 141343.72 lb Ð 92870.50 lb Ï 48473.22 lb Ð Downward Force When Plug Lands

Total 422.69 bbl

5

Displacement Fluid Volume

2400 ft (Total Pipe) - 80 ft (Shoe Track) = 2320 ft (Top of Shoe) 2320 ft × 0.1497 bbl/ft = 347.30 bbl

6

Pressure To Land The Plug ** Working From The Bottom Up**

Hydrostatic Outside Tail 400 ft × 0.8519 psi/ft Lead 2000 ft × 0.7065 psi/ft

=

341 psi

=

1413 psi

Total: 2400 ft

1754 psi

Hydrostatic Inside Tail 80 ft × 0.8519 psi/ft Well Fluid 2320 ft × 0.4416 psi/ft Total 2400 ft

=

68 psi

=

1025 psi 1093 psi

Differential Pressure (At top of shoe) 1754 psi – 1093 psi = 661 psi

Hint: Always re-total the lengths to make sure both outside and inside pipe segments are calculated. If the lengths are not the same then the calculations will be incorrect.

5 • 32

Cementing 1


Intermediate Casing Job One This casing job consists of a single tail slurry with 500 ft fill. Well Parameters

Well fluid Spacer Cement

Pipe Size

7 in. 32 lb/ft

Well Fluid

10.4 lb WBM

Hole Size

8 ¾ in.

Pipe Depth

9700 ft

Shoe track length

80 ft

Required cement fill-up (TAIL)

500 ft

Excess volume required (percent)

25%

GIVEN: Cement through casing as follows 30 bbl spacer @ 11.4 lb/gal

TOS (top of spacer)

Followed by: Premium Class G Cement containing 0.5% CFR-3 + 0.1% HR-5. Displace plug with 10.4 lb/gal WBM.

TOC (top of cement) 9,200 ft

Due to drillout and deepening displace with mud instead of water.

9,620 ft

9,700 ft

Figure 5.21 – Intermediate Casing Job

5 • 33

Cementing 1


Worksheet for Slurry Weight and Volume Calculations MATERIAL NAME Class G Cement

Material (lb) 94 X

Factor (gal/lb) 0.0382

=

Absolute Volume (gal) 3.5908

Mixing Water Requirements (gal) 5.00

0.05% CFR-3 (dispersant)

0.47

X

0.0938

=

0.04409

+

0.1% HR-5 (retarder)

0.094

X

0.0750

=

0.00705

+

Water

X

=

+

X

=

+

X

=

+

41.65 = TOTALS ==>

8.33

136.214

X

5.00

gal

8.6419

5.00 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/ ft3 (constant) = ft3./sack The mixing water per sack is the sum of the gallons in the far right column

Cement Density (lb/gal) ==> Cement Yield (ft3/sk) ===> Mixing Water required ===>

136.214 /

8.6419 =

15.8

lb/gal

8.6419 /

7.4805 =

1.16

ft3/sk

5.00

gal/sk

5 • 34

Cementing 1


1

Cement Volume 3

Open Hole: 500 ft × 0.1503 ft /ft 3 25% Excess: 75.15 ft × 25% 3/ Shoe Track: 80 ft × 0.2025 ft /ft Total:

2

3

3

95 sks

= =

475 gal 11.31 bbl

Displacement Fluid Volume

9660 ft × 0.0360 bbl/ft

5

=


95 sk × 5 gal/sk 475 gal ÷ 42 gal/bbl

4

75.15 ft 3 18.79 ft 3 16.2 ft 3 110.14 ft

Sacks of Cement

110.14 ft ÷ 1.16 ft /sk

3

3

= = =

=

347.76 bbl

Pressure to Land The Plug

Unknown – Height of Spacer (for “worst case”) Height of Cement (for “worst case”) Discussion: In calculating the pressure to land the plug, we want to anticipate “worst case” which would be an increased pressure over plan. We planned our cement volume based upon hole washout (in this case 25%). We need to calculate the height of cement & spacer (Which are heavier than the mud) in a perfect hole situation which will give us a taller column & higher differential pressure. Height Of Cement 3

Open Hole Volume 25% Excess Volume

= =

Perfect Hole Fill-up Factor (7 in OD × 8 3/4 in hole) 3 3 93.94 ft x 6.652 ft/ft

=

75.15 ft (From 1) 3 18.79 ft 3 93.94 ft 3 6.652 ft/ft

=

624.89 ft cement

30 bbl spacer (given) 30 bbl × 37.3484 ft/bbl = ** Working from the bottom up**

1120.45 ft spacer

Height Of Spacer

Height Of Mud 9700 ft - 624.89 ft - 1120.45 ft

=

7954.66 ft

Hydrostatic Outside Cement: 624.89 ft × 0.8208 psi/ft Spacer: 1120.45 ft × 0.5922 psi/ft WBM: 7954.66 ft × 0.5403 psi/ft

= = =

Total: 9700 ft

513 psi 664 psi 4298 psi 5475 psi

Hydrostatic Inside Cement (in shoe): 40 ft × 0.8208 psi/ft WBM: 9660 ft × 0.5403 psi/ft

= =

Total: 9700 ft Differential Pressure (at top of shoe) 5475 psi - 5252 psi

33 psi 5219 psi 5252 psi

=

223 psi

5 • 35

Cementing 1


Unit D Quiz Fill in the blank with one or more words to check your progress in Unit D. 1. When using the Worksheet for Slurry Weight and Volume Calculations, the first thing you need to know is what ___________ of cement is being used. Then you can look up the ingredient’s ______________ and _____________ volume factors. 2. An absolute volume for each material is calculated by multiplying its _____ by its ___________. 3. To obtain the density of the cement mixture, divide the ________________ of the materials by the ________________________________________ of the materials. 4. To find the yield, the total absolute volume in gallons is divided by the constant, ___________ gal/ft³. 5. To calculate the total volume of slurry needed for this job, add the slurry needed to fill the _________________ and the slurry needed to fill the ____________________. Your unit of measure will be ___________ which needs to be converted to __________. 6. To calculate the number of sacks needed for this job, divide the amount of slurry in _____________ by the _______________________________ in cubic feet per sack. 7. To obtain the total amount of mixing water you will need, multiply the number of ______/sk water by the number of __________.Then convert to the unit of measurement used in tanks, ___________, by dividing gallons by _______gal/bbl. 8. You need to calculate the pressure needed to land the plug, since a pressure build up might indicate_________ and a loss of pressure may mean ______________________________________________. 9. For the pressure to land the plug, you will need to use the ____________ ___________________________________ conversion tables in the “Calculations and Formulas” section of you handbook. You will look up the _____/ft of the fluids (extrapolating, if necessary), then multiply these by the depth of the float collar (or the length of the _____________ part of the hole).


5 • 36

Cementing 1


Answers to Unit Quizzes Items from Unit A Quiz

Refer to Page

1. capacity 2. Displacement 3. Turn to Section 210 Pages 21 0.0393 bbl/ft × 10,000 ft = 393 bbl 4. Turn to Section 210, Table 212B, Page 9 25 bbl × 70.32 ft/bbl = 1758 ft 5. OD = 4.5 in. = 0.375 ft ID = 2.0 in. = 0.167 ft OD Area = 0.7854 × 0.375 ft × 0.375 ft = 0.1104 ft2 ID Area = 0.7854 × 0.167 ft × 0.167 ft = 0.0219 ft2 Cross-sectional Area = 0.1104 ft2 - 0.0219 ft2 = 0.0885 ft2 Displacement = 0.0885 ft2 × 2000 ft = 117 ft3 Conversion Factor = 0.1781 bbl/ft3 Displacement = 117 ft3 × 0.1781 bbl/ft3 = 31.52 bbl Items from Unit B Quiz

Refer to Page

1. Turn to Section 122-B 0.3132 ft3/ft × 5000 ft = 1566 ft3 1566 ft3 × 7.4805 gal/ft3 = 11714 gal 2. Turn to Section 221-B, Page 73 0.1697 ft3/ft × 7675 ft = 1302.45 ft3 Refer to Page

Items from Unit C Quiz 1. Divided, standard 2. Actual volume 3. Absolute volume

5 • 37

Cementing 1


4. ABSOLUTE MATERIAL

FACTOR

VOLUME

MATERIALS

(lb)

(gal/lb)

(gal)

Brine Water

10

Sand

10

TOTALS

= ×

0.0456

=

20 lbs

1.0 0.456 1.456 gal

Slurry Weight = 20 lbs ÷ 1.456 gal = 13.74 lbs/gal

5.


API Cement weight = 94 lb/sack 1 sack = 1 cubic foot Absolute MATERIAL

Material

Factor

Volume

(lb)

(gal/lb)

(gal)

NAME Class H Cement Water

94 X 35.819 =

TOTALS ==>

0.0382 =

3.5908

8.33 lb/gal

129.819

Mixing Water Requirements (gal) 4.30

4.30 gal 7.8908

4.30 gal

Find the weight of the mixed cement by using this formula: Total Pounds/ Total Absolute gallons = pounds/gallon. Find the cement yield in cubic feet per sack by using this formula: Total Absolute gallons / 7.4805 gal / ft3 (constant) = ft3/sack The mixing water per sack is the sum of the gallons in the far right column. Cement Density (lb/gal) ==> 3

Cement Yield (ft /sk) ===> Mixing Water req (gal/sk)===>

129.819 /

7.8908 =

16.5

7.8908 /

7.4805 =

1.05

4.30

Items from Unit D Quiz

Refer to Page

1. type, water requirement, absolute 2. gal/lb, weight 3. weight, gallons 4. 7.4805 5. annulus, shoe joint, ft3, sk 6. ft3, yield 7. gal, sk, bbl, 42 8. bridging, lost circulation 9. hydrostatic pressure & fluid weight, psi, unbalanced

5 • 38

Cementing 1

5. 4. Primary Cementing Calculations

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