Drilling Calculations

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DRILLING ¼ CALCULATIONS 1 COURSE

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Drilling Calculations Course

Course Content CHAPTER –1

Units & Measurements

CHAPTER –2

Preliminary Mathematics

CHAPTER –3

Fluid Circulation & Hydraulics

CHAPTER –4

Fundamentals of Pressure

CHAPTER –5

Buoyancy Effect

APPENDIX –1 Symbols & Abbreviations APPENDIX –2 Conversion Tables APPENDIX –3 Tutorials

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CHAPTER – 1

Measurements & Units System This section deals with different measurement and units systems of current industry practices. At the end of this section the candidates will appreciate the use of different units systems and how to convert from one units system to another.

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Measurements & Units Systems Measurement is the key to carry out any engineering analysis. As one of the pioneer of mathematics pointed out that “Not even an inch can be moved on the surface of this earth without using calculations and measurements”. This system of measurement has been developed over a period of time. This effort has now gained a globally recognized status and has extremely important implications in our daily lives. Three of the most famous systems of measurements that are currently used are 1. Metric Systems - M.K.S Systems (Meter, Kilogram, Seconds) 2. C.G.S Systems (Centimeter, Grams, Seconds) 3. Imperial Systems - F.P.S Systems also called the British Engineering Systems (Foot, Pound, Seconds). The same system of units is also used in the oilfield terminology The most common measurements taken are: Length Area Volume Mass (Weight) Density Pressure Time Note: Some parameters have their special units such as Dioptre (power of the lens such as concave or a convex lens) or Decibels which is used to describe the intensity of the sound Rule of Thumb To Convert from a BIG to SMALL quantity ALWAYS “MULTIPLY” To Convert from a SMALL to BIG quantity ALWAYS “DIVIDE”

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Imperial System Length The most common units to express length are inches, feet, yards and miles 12 inches 3 feet 1760 ards 5280 feet

= = = =

1 foot 1 yard 1 mile 1 mile

Exercise Convert 200 inches into feet Solution We know from the above relation that 12 inches 1 inch Therefore, 200 inches

= =

1 foot 1 ÷ 12 = 0.0833 foot (Remember, conversion from a Small to Big quantity, ALWAYS “DIVIDE”)

= =

0.0833 x 200 16.66 feet

Similarly, To convert 16.66 feet into yards one has to follow the relationship between the yard and feet 3 feet 1 feet 16.66 feet

= = = =

1 yard 1 ÷ 3 = 0.33 yard 0.33 x 16.66 yards 5.5 yards

Rig measurements are confined to inches for simplicity and yards are not used. The rig tape is calibrated in feet and tenths.

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Metric System The Metric system covers all units of measurement, but makes use easier as it is based on units of multiples of ten. Length

The Fundamental unit is the Meter

1 meter

=

39.37 inches

1000 millimeters 100 centimeters 1000 meters

= = =

1 meter (milli = one thousandth = 1/1000) 1 meter (centi = one hundredth = 1/100) 1 Kilometer (Kilo = a thousand times)

Exercise How can 0.02 can be expressed in words or as a fraction? Remember, The number of points after the decimal point is an indication of the value of the Divisor Divisor is the number, which divides a given number to give result in a decimal number format. For instance, if 4 is divided by 2 then 2 is the Divisor and 4 is the Dividend. In 0.02 the number of decimal points after the decimal point is 2. The only non zero number in this example is 2 Therefore, this means that the Dividend number 2 has been divided by a Divisor, which gives a decimal value of 0.02. To convert this number into a fraction, we need to understand the following rule Decimal represents 1 and the points after the decimal represents Zeros.

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To convert this number we have,

2 100

Removing decimal gives 1 at the Denominator

2 Zeros account for the number of points available after the decimal point

In the same way, 0.00987 can be written as 987/100,000 0.0025 can be written as 25/10,000 Exercise Convert the following numbers into fractions 0.00025 (25/10,000) 0.000000087 (87/?) 0.102 (102/1000) 0.000035 (35/?) 0.0205 (205/10,000)

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Area Area is defined as the product of any two lengths defined by a single set of units. Area = Length x Breadth (Width) Caution: Be Consistent about the Units. For instance, if the length is defined in feet then breadth should be in feet too. The resulting unit will be called square feet. In the case of different units, You SHOULD convert one of them to be in consistent with the other one. For instance, Whereas,

Length in feet Width in inches

Therefore, to convert Width into feet, you MUST convert the inches into feet by using the relationship 12 inches 1 inch

= =

1 foot 0.0833 foot

Exercise Find the area of a rectangle whose dimensions are given as Length, L = 10 feet and Width/Breadth = 5 inches. Solution We know that

Area = Length x Breadth

∴Area = 10 (feet) x 5 (inches) = ? Not Possible – Units are not consistent with each other The conversion of units are solely on choice, hence we would convert inches into feet and will quote our answer in square feet.

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Using the given relationship 12 inches 1 inch 5 inches

= = =

1 foot 0.0833 foot 0.4165 foot

∴Area = 10 (feet) x 0.4165 (feet) = 4.165 square feet 144 square inches 9 square feet 3,097,600 square yards 27,878,400 square feet

= = = =

(Correct Answer)

1 square foot 1 square yard 1 square mile 1 square mile

Exercise Convert 2,000 square inches into square feet and square yards Step – 1: By the given relation, we know that 144 square inches & 9 square feet ∴ First, divide by 144

=

1 square foot……………………….(A)

=

1 square yard ……………………….(B)

= =

2000 / 144 13.88 square feet

Step – 2: Now Convert 13.88 sq.feet into sq. yards by simply using the relation (B), we have 9 square feet 1 square feet 13.88 square feet

= = = =

1 square yard 0.11 square yard 0.11 x 13.88 sq.yard 1.54 square yard

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Defining Numbers One of the most important aspect in mathematics is to present the numbers. This involves the number of decimal places, scientific notation of the numbers or presenting numbers in an exponential format (such as 102) significant figures and in the form of a percentage. In this section, we are going to depict the type of these numbers and their importance. Fractions Fractions are simply the numbers, which are represented in as part of a whole. This can be represented in two ways 2 ½ inches or as a decimal 2.5 inches. Both of these representations are valid and recognized. Decimal Places The conversion relationship of feet into meters is given as 1 feet 1 meter

= =

0.3048 meters …..………………(A) 3.2808398955 feet ………………… ..(B)

Number of decimal places in (A) is Number of decimal places in (B) is

4 9

To present a number with 4 decimal places is acceptable but with 9 decimal places is not practical. Owing to this, rounding numbers were introduced.

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Significant Figures In relation (B) the number of decimal places is 9. In order to make it a number usually with 3 decimal places is reasonable, however some would argue about 4 decimal figures which means that 1 meter

=

3.28 feet (Generally taken for conversion) 3.281 feet (In some text books)

The figure of 3.28 can now be called a number with 3 significant figures, whereas 3.281 is a number with 4 significant figures.

Rounded Numbers Consider a calculated value of 3.9987 feet. It is a very common practice in engineering where numbers are rounded off to one significant figure. The calculated value of 3.9987 is only 0.0013 less than 4. Hence, instead of quoting this number with its full decimal places it is generally quoted as 4 alone. This is a very good practice and is highly recommended especially when working on field site or carrying out engineering analysis. Similarly, 9.382416 can be rounded off to 9.3824 9.398416 can be rounded off to 9.4 (should the answer is required to one decimal place only)

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Scientific Notation Until now, we have dealt with numbers which are only 3 to 4 decimal placed and had values which were not that big. However, in mathematics or in Physics (notably Astronomical Physics) the numbers are either very large or very small. It is therefore, the use of scientific notation became a common practice. Scientific notation is represented in multiples of 10. For instance, the permeabilityΚ of limestone reservoir in North Sea averages around 3 milli Darcy. A mille is (0.001) and Darcy is the unit of permeability, which is equivalent to 0.00000000000987 m2. This Darcy unit is therefore represented as 0.987 x 10-12 m2. 10-12 represents the count of 11 zeros. Where, -12 is also called the exponent or the exponential power Similarly, 3987000 can be written as 3987 x 103 3987000 can be written as 398.7 x 104 3987000 can be written as 39.87 x 105 3987000 can be written as 3.987 x 106 In the same way, 0.000001 can be written as 1 x 10-6

Κ

for Permeability is defined as the ease with which Hydrocarbons (oil or gas) flow in a porous medium (sandstone or limestone). The units of permeability are called Darcy, which is equivalent to 10-12 square meters, which are also the unit of area. Weatherford Underbalanced Drilling Services MENA Region

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Rule of Thumb • Small numbers are always presented in multiples of 10 with negative exponents. • Always Count the numbers of points after the decimal points before conversion. For instance, 0.000001 has 6 decimal places but has only 5 zeros after the decimal points. Therefore, depending on the number of decimal places the exponent is evaluated. In this case the exponent is – 6. • In the case of big numbers use the most reasonable approach. Exponential powers of 3 and 6 are considered to be standard in engineering and economics. 3 represents a 1000 (such as a 5K Choke Manifold) and 6 presents a Million (such as 20MMScf/D Gas) •

Moving from Left to Right the exponential power is always Increased

•

Moving from Right to Left the exponential power is always Decreased

Percentage Numbers are sometime presented in a percentage format. This is a very useful way of presenting numbers and is widely used in economics, describing efficiencies (such as of Pumps) and banking sector. Remember, % means 100 Example What is 10% of number 100. Solution 10% = 10 ÷ 100 10% = 0.1 Therefore, 10% of 100 will be 0.1 x 100 = 10

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Example If 20 logs = 50% of the total, how many logs are there? 20 logs = 50% 1% = 20 ÷ 50 100% = (20 ÷ 50) x 100 Recall, the rule of BODMAS, Solve the Bracket first 100% = (0.4) x 100 100% = 40 logs Work out the answers for Value of 5%? Value of 35% Value of 100% The above example can also be approached by using the mathematical formula P = R x B ……………… (i) Where, P = Percentage - of the actual value equaling chosen % R = Rate in decimals – the part of a 100 to be a found in 2% of 50, 2% is rate B = Base - the number of which some percentage is to be found.

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Example If 20 logs = 50% of the total, how many logs are there? Use, P = R x B Percentage is number of logs = 20 Rate in decimals = 50% Base number to be found = ? Use Cross multiplication technique to re-arrange equation (i) P 20 20 20 = = B= = R 50% 50 0.5 100 B = 40 Logs (Check the answer with the previous method) Self - Learning Exercise Using the relationship P = R x B 1. If 30 logs = 20% of the total, how many logs are there 2. If the total number of logs are 90. How many logs account for 35% 3. Find R? if P = 10, B = 5 Square Roots A square root is a number whose square is the result of multiplying itself with the same number. It is represented by √ (square root bracket) or ( )1/2 For instance, 4 is the square root of 16 and 8 is the square root of 64. Exercise What is the square root of 9, 81 and 49. (Answer: 3, 9, 7)

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CHAPTER - 2

Preliminary MATHEMATICS This chapter will deal with the basic mathematics and will involve all what has been discussed in the previous chapters notably the numbers. Calculations of percentages, Areas, Volumes, Capacities will also be highlighted in this chapter.

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Preliminary Mathematics Areas The use of area is found in many places around the rig, for instance Force on a unit area (which is also the definition of pressure) Area of a deck space Surface areas of pits Cross-sectional area of a drill pipe Note, that area and perimeter are two different measurements of an object. For instance, the perimeter of a circle or the circumference of a circle is calculated by using 2Πr (where r is the radius of the circle) and is represented as inches or feet. The area of a circle is simply the product of Πr2 and is represented as square inches or square feet. Π is a ratio of circumference of a circle with its diameter and is given a constant value of 3.142. (How many decimal places are here ?) Basic Shapes Square A shape with 4 equal sides and the area is calculated is by multiplying the Length with its width

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Rectangle A shape whose opposite sides are parallel and equal in lengths. The area is calculated by multiplying Length with its width

Width = 2 inches

Length = 4 inches Area = Length x Breadth = 4 x 2 = 8 Square inches Triangle A shape with 3 sides, which are jointed together with angles between them. There are different types of Triangles namely, 1. Equal Triangle – Where all the angles in the triangle are equal i.e. 60 degrees each. 2. Isosceles Triangle – Where two of the angles are equal 3. Right Angle Triangle – where one angle is of 90 degrees and the other two can differ. Remember, Sum of all angles in a Triangle is always equal to 180 degrees.

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Area of a Triangle = ½ x Base x Altitude (Height)

Height

Base Circle One of a very common shapes whose application is of an utmost importance in both oil and pipeline industry. The area of a circle is a relationship between radius, or diameter and circumference.

Diameter, D Radius, r

Remember, radius = ½ Diameter of the Circle

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Cross sectional Area of a Pipe with radius r is given as = Πr2 or, 1 Cross sectional Area of a Pipe with diameter D will be = ΠD 2 4 This symbol Π is called a “Pi” and its value is constant which is taken as 3.142 in all calculations. The diameter of any circle will go around the circumference 3.1416 times. To aid our calculations, remember

1 Π = 0.7854 4

Therefore, Area = 0.7854 x D2 Trapezium A trapezium is a distorted rectangle whose two sides are parallel to each other. A classic example of a trapezium will be Weatherford’s storage tanks currently rigged up in Syria.

B

Height A

Area of a Trapezium = ½ x (Sum of two parallel sides) x Height Area of a Trapezium (Given) = ½ x (A + B) x Height

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Annular Areas Annular Area is the space / area between the drill pipe and casing of the well. In mathematical analogy, this space can be calculated by the difference in the area of a big circle (Casing) and the small circle (Drill pipe)

Annular Area Drill pipe Casing

Consider a casing with an OD of 9 5/8 inches and a drill pipe of OD 5 inches. The annular area can be calculated by subtracting the area of the drill pipe from the Casing area. Annular Area = Cross Sectional Area of Casing – Cross Sectional Area of DP Annular Area = 0.7854 (9 5/8)2 – 0.7854 (5)2 Annular Area = 72.76 – 19.63 (Answers rounded off to 2 decimal places) Annular Area = 53.12 square inches It is a good practice of being Consistent with the number of Decimal places

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In order to cut short this method, Annular Area can also be calculated by using the equation Annular Area = 0.7854 (D2 – d2) …………….. (i) Where,

‘D’ is the diameter of the bigger Casing ‘d’ is the diameter of the smaller Drill pipe

Check your answer and see what result do you get? Solving Mathematical Equations In the above equation, one would argue about how to solve the equation given that the bracket is multiplied by 0.7854. In order to solve these kind of equations, rule of BODMAS, named after its creator is followed where, B stands for Brackets ( ) O stands for off D stands for Divide M stands for Multiplication A stands for Addition S stands for Subtraction The rule of BODMAS justifies that if a systematic approach is applied in solving any equation with multi-mathematical functions then his rule is valid for any length of equations provided that it is adhered strictly. In equation (i), 0.7854 (D2 – d2) the bracket must be solved first, which means Annular Area = 0.7854 (9.6252 – 52) Annular Area = 0.7854 (68.12) Annular Area = 53.12 (answer to 2 decimal places)

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Algebraic Equations Solving algebraic equations is one of the most fundamental processes in engineering. This allows calculating variables by re-arranging them in a given equation For instance, X = Y + Z …… (A) To calculate the value of ‘y’ we need to re-arrange the above equation, such as Y+Z=X Y = Z – X …………………....(ii) Similarly, to get the value of Z, once again re-arrange eq. (A) Y+Z=X Z = X – Y …………………..(iii) This was an example of a linear equation without any fractions. Cross Multiplication Given that a = b x c

then

a =b c

Find b: if (a + b) = (c + d) a+b= c+d b= c+d–a

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Find b: if

a = (c + d ) b

Using Cross Multiplication, we have a = b (c + d) b=

a (c + d )

if a = 10, c = 3 and d = 2 b=

10 (3 + 2)

b=2

Pressure = Depth x Mud Weight x 0.052 (Learn this equation by heart) Now Solve the equation to find

a) Depth b) Mud Weight

Hint: Use Cross Multiplication technique

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Volumes and Capacities With an understanding of how to calculate areas it is straightforward procedure to calculate the volume of a container. Volume is the space occupied in a container. Remember, the area of square, rectangle, triangle calculated were all 2dimensional. Volume is 3 dimensional therefore it would also need the magnitude or value of height together with the length and width of any shape. Capacity is the amount of a substance that can be placed in that container expresses in units relating to both substance and container. When talking about capacities of a container or hole we use barrels, and think of common rig substance like oil, mud or cement. Example A container 10 feet long, 6 feet wide and 8 feet deep. Find the volume of this container Volume = L x W x H Volume = 10 x 6 x 8 Volume = 480 cubic feet. Note, Look at the units they are now changed to cubic rather square. This shows the addition of one more dimension of the container in the formula. Also, all the units MUST be the same. You cannot multiply inches with feet and feet with yards and so on. In the case of different units, convert to one unit system and quote your answer in one unit system only.

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Example Find out the volume of a storage tank in cubic meters with the following dimensions Length = 20 feet Width = 10 feet Height = 6 meters (Answer = 111.48 to 2 decimal places) Example Calculate Annular volume if D = 10”, d = 6” and depth = 1 ft

.7854 (10 2 - 6 2 ) x1 144 Volume = 0.349 cubic feet Volume = 0.35 cubic feet (round off 2 decimal places) Volume =

The use of cubic feet is not as common as barrels. To calculate the volume in barrels, we need to convert feet to barrels From the conversion formula sheet we know that 1 barrel = 5.615 cubic feet Applying this to the formula, we have

.7854 (10 2 - 6 2 ) x1 144 x 5.615 Calculating out 0.7854, 144 and 5.615 we can simplify the formula to Volume in barrels/ft =

 D2   or Volume in bbls/ft =  1029    D2 − d 2   Volume in bbls /ft =   1029 

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CHAPTER – 3

FLUID CIRCULATION & HYDRAULICS

This section covers the Fluid Hydraulics used in drilling engineering. This is a very important topic and candidates are encouraged to understand the concepts very well. This section is the basis of Well Control Course

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Annular Volume Calculations  D2 − d 2   x depth, ft Using the formula   4 x144 

The annular volume in cubic feet can be obtained. However, in oil field barrels is the most likely unit to quote annular volumes. For answers in barrels use  D2 − d 2   x depth, ft   1029 

‘D’ = Large Diameter (inside diameter of the hole) ‘d’ = Small diameter (outside diameter of the string)

 D2   x depth, ft To calculate Non-annular volumes simply use   1029 

With varying string diameters, casing and open hole it is good policy to draw a fully labeled diagram or schematics of the well prior to attempt any question. Example Calculate annular volume in barrels of an 8000 ft hole, 12 ¼ inside diameter with 5 in. drill pipe

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Example 9 5/8 Casing set at 9000 ft. ID = 8.84 in. 8½ Open hole to 11,000 ft 5” Drill pipe 19.5 lbs/ft, ID = 4.276 in. 600 ft of 6” x 2 ½ “ Drill Collars Calculate

a) Annular volume in bbls b) Volume of mud inside string in bbls

Solution Volume of casing in Annulus  8.84 2 − 5 2   x 9000   1029  464.83 bbls

a) Volume of OH to Collar annulus  8 .5 2 − 6 2   x 600   1029  21.14 bbls

Volume of OH to Pipe Annulus  8 .5 2 − 5 2   x 1400   1029  64.28 bbls

Total = =

64.28 + 21.14 + 464.83 5520.25 bbls

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b) Total Mud required to keep the hole full when pulling out Volume of steel in 10 stands of drill pipe  5 2 − 4.276 2   x 900'  1029   0.065 x 900’ 5.87 barrels

c) Volume of steel in 1 stand for drill collars  9 2 − 32   x 90'  1029   0.06997 x 90 6.3 barrels

Total Mud to fill hole Drill pipe displacment/ft x 0.0065 x 61.1 103 bbls

9400’ + 9400 + +

Drill collar disp /ft x 0.06997 x 41.98

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Rig Pumps Rig pumps are one of the most important piece of equipment in the drilling process. In drilling pumps are responsible for an efficient transfer of drilling fluid into the well and out of the well (during well control situation). The oil industry has seen their other benefits during artificial lift by submerging ESPs into the well, which then lifts the oil to the surface. Pumps output calculations are simply volumes Some of the important features in a pump calculations are Pump Strokes Pump Efficiency Pump Capacity Pump Pressure Example Find pump output/stroke on Triplex with 12” stroke and 6” liners at 95% efficiency Solution A triplex pump has 3 cylinders Volume of cylinder

=

ΠD2 / 4 x Length

Volume of 3 cylinders

=

3 x (ΠD2 / 4 x Length)

=

3 (0.7854) (62) x 12

=

3 x 339.29

=

1017.88 cubic inches

Recall, Π / 4 = 0.7854

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Pump output can also give the value of the time it takes for the mud to travel from bottom to the surface. This can be given by the following equation.

Bottoms up Time =

Annular Volume (bbls ) Pump Output (bbls / min)

Again, barrels can be substituted for other units

Total Circulation Time =

Drillstring + Annular Volume + Pit Volume Pump Output (Bbls / min)

Hydraulics Calculations Observing the size of pumps, pressure rating of unions, safety chains on hoses, safety clamps on flow lines, torque required for tool joints and packing required for swivel, we can conclude that mud is circulated round the systems at pressure. But stand at the flow line and you notice the mud is moving under gravity, not pressure. Disconnect the pump discharge and read the pump pressure, it will have dropped to near zero. Pumps do not put out pressure, they put out flow. It is the restrictions in the circulating systems that creates a back pressure against the pump must work. Friction within the system causes pressure. The pressure at the pumps is the sum of all frictional losses around the system. If we took pressure gauges and could place them at various points around the system, we would probably note the following

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Assuming the pressure of the pump is 3000 psi Pressure inside Kelly Pressure inside near bit sub Pressure in drill collars annulus Pressure at flow line

= = = =

2950 psi 2200 psi 200 zero

The energy is progressively lost around the system. Most pressure is lost across the bit nozzles. The energy is used to create jetting and impact sufficient to clean ahead of the bit. In a good hydraulic system, pressure losses across the bit should be approx. 60-65% of pump pressure. Hydraulic Horsepower The horsepower required to circulate a known quantity of mud at a certain pressure can be calculated using the formula Hydraulic Horse power (HHP)

=

PxV 1714

P = Pump pressure (psia) V = Pump output (gallons/minute) 1714 is constant of this equation

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Due to mechanical inefficiency, the output horsepower is always less than the input horsepower

Mechanical Efficiency, η(neta)

=

HHP Output Mechanical Horsepower Input

Most pumps have a mechanical efficiency of 85%. The same principle applies to volumetric output of a pump – called Volumetric efficiency Exercise Find Hydraulic horsepower of pump pumping 250gpm at 2,000 psia Solution HHP

=

HHP

=

HHP

=

PxV 1714 2000 x 250 1714 291.71 HHP

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Changing Mud Weight Changing Mud weight will affect pump pressure in the following way New pump pressure (P2) =

Old pump pressure (P1) x

New Mud Weight Old Mud Weight

Example Find out the new pump pressure given that Old pump pressure, P1 = 2000 psia New Mud weight = 10 ppg Old Mud weight = 8 ppg (Answer = 2,500 psia) Example Find out the Old pump pressure (P1) given that New pump pressure , P2 = 2500 psia Old mud weight = 10 ppg New mud weight = 12 ppg (Answer = 2,083.33 psia to 2 d.p) Example Find out the new mud weight given that Old pump pressure = 2,000 psia Old mud weight = 7 ppg New pump pressure = 10% higher than the Old pump pressure (Answer = 7.7 ppg)

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Strokes Per Minute Likewise, the change in mud weight changes the pump pressure, the strokes per minute (SPM) change also changes the pump pressure. The formula to calculate the new pump pressure in the case of stroke change is same as for change in mud weight. New pump pressure (P2) =

Old pump pressure (P1) x

New SPM Old SPM

Exercise Find out P2 if P1 = 2000 psia New SPM = 40 Old SPM = 45 Answer = 2,250 psia Exercise Find out P1 if P2 = 2500 psia New SPM = 40 Old SPM = 25 (Answer = 1,562.5 psia) Exercise Find out New SPM if P2 = 2500 psia P1 = 2200 psia Old SPM = 35 Quote your answer to 2 significant figures only. Answer = 40 SPM

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CHAPTER – 4

Fundamentals of Pressure

This section covers the basic principles about pressure in a static or in an equilibrium state. Candidates are encouraged to grasp these basic concepts well in order to sit for their Well Control Exams

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Pressure Definition Pressure is the fundamental most concept in drilling engineering and is defined as the Force acting on an area. Mathematically, it is defined as P = F/A

……………………………………… (A)

Units of Pressure in different set of unit systems are defined as M.K.S systems Newton per square Meters C.G.S systems Dynes per square centimeters F.P.S system Pounds per square feet or Pound per square inch The most commonly used pressure units are Pounds per Square Inch and is written as psia, where ‘a’ stands for the absolute pressure. Note, in some text books the unit of pressure is sometimes written as psig. ‘g’ stands for gauge value. The only difference between the gauge and absolute pressure is of 14.7 psia, which is taken as the standard pressure value. Note, the standard temperature is taken as 600 F. This can be seen in my text books where volumes or other parameters of gas and liquids are measured at S.T.P, which stands for Standard Temperature and Pressure.

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The pressure on a piece of plate with 4 square meters area and a force of 16 Newton can be calculated by eq. (A) Substituting these values into equation (A P = 16 / 4 P = 4 Newton per square meter The unit of Newton per square meter can be defined by one unit only which is named after the pioneer of fluid mechanics called Pascal (famous for his Pascal’s law)

YOU SHOUD REMEMBER THIS RELATIONSHIP 1 N / m2 1 psi

= =

1 Pascal (Pa) 6,895 Pa

Pressure and Depth Relation Pressure and depth are proportional to each other. In other words, as the depth increases pressure increases too. However, pressure is inversely proportional to its volume provided that the temperature of the system is kept constant. (Well Control Basics). This is also called Gas law and will be used frequently in well control calculations. This tells that at low volumes gas has higher pressures, but as you increase the volume (let the gas expands) the pressure reduces. Think in terms of production from a well and ask yourself Q. What will be the pressure of the gas at the bottom of the well. Q. What will be the pressure of the gas at surface. (Hint: Think in terms of both depth and Volume)

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In static fluid calculations, the pressure is calculated in relation to its depth and density. For static pressure calculation, we use the following equation as Pressure = Density of the fluid x acceleration due to gravity x Height This is also called the Hydrostatic Head Pressure Equation P=ρxgxh Density The density is the measure of how heavy or light a substance is and is given by the following equation as Density =

Mass Volume

Units M.K.S System C.G.S System F.P.S System Oilfield units

Kg per cubic meter gram per cubic centimeter Pounds per cubic foot Pounds per Gallon (ppg) – Industry Standard Unit

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In drilling engineering, the pressure is calculated by the equation Pressure = Mud Weight (ppg) x Depth (ft) x 0.052 0.052 is a constant and is used to convert the pressure from ppg to psi/ ft. To convert pounds per gallon to psi per ft Use a conversion factor of 0.052 Exercise 1. Find the pressure at 2000 feet with a density of 8 ppg 2. 7200 ft with a MW = 10ppg 3. 5000ft with a MW = 8 ppg

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Specific Gravity Usually, the density of the fluid is given as specific gravity. When we talk about specific gravities we mean that they are compared with their related fluid. This in turn tells us that how heavy or light the fluid is if it is compared with a base fluid. Fluid Type

Base Fluid

Density in MKS system

Density in CGS system

Liquid Gas

Water Air

1000 kg/m3 1.293 kg/m3

1 gm/cm3 0.001293 gm/cm3

Specific Gravity

=

Density in Oilfield system 8.33 ppg 689.6 ppg

Density of the given Liquid or Air Density of the Base Fluid (Water / Air )

Q. What are the units of specific gravity ? A. There are no units of specific gravity as it is a ratio of two similar quantities with same units. It is for this reason that specific gravities can be converted to any set of units just by multiplying with the density of the base fluid. Exercise 1. Specific gravity of a drilling fluid is 0.83, Find its density in all unit system (Answer. 830 kg/m3, 0.83 g/cm3, 6.91ppg)

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In the case of using specific gravities or pounds/cubic foot (pcf) units? The following conversion factors are used Pressure (psi)

=

Mud weight (S.G) x Depth (ft) x 0.433 (psi/ft)

Pressure (psi)

=

Mud weight (pcf) x Depth (ft) x 0.007

Using the same mud weight, it can be seen that pressure will increase with depth. As long as the mud pressure is enough to balance the formation pressure we can drill ahead safely. When we go underbalanced, this technique is changed by keeping a low pressure head on the fluid such that the formation pressure exceeds the drilling fluid pressure and production is seen as soon as the reservoir is hit. In overbalanced drilling, the unexpected flow of gas (formation fluids) into the well bore is termed as “Kick”. In conventional well control methods it is this kick, which is tackled with caution. Several measures are available to remove this kick from the well. However, well control methods are beyond the scope of this course. It will be dealt in detail in the well control school. Note, at the time of kick the annulus is filled with the formation fluid + contaminated mud but our drill string is still filled with uncontaminated mud. After shutting down the pumps and closing in the well, the excess of formation pressure will be registered on the standpipe gauge and the casing gauge The pressure on the standpipe (drill pipe pressure) gauge will be equal to the imbalance between mud hydrostatic in the pipe and formation fluid pressure

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Convert this pressure to mud weight (ppg) and add to known mud weight in pipe. 700

Remember, Gauge pressure only tells the Difference in Pressure and NOT the Actual Pressure.

6279 psi

The gauge pressure in this case is showing 700 psia.

6979

Therefore, the pressure in the drill pipe will be difference in bottom hole pressure and Gauge pressure

Gas Influx (Kicks) 700 DPP

833 CP

This diagram shows a Well Depth = 11,500 ft Influx Depth point = 11,200 ft 10.5 ppg Mud

Weight of Mud = 10.5 ppg 11,200 ft

Drillpipe pressure, DPP = 700 psia Casing Pressure, CP = 833

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Gas Gradient = 0.1 psi / ft which has extended about 200f t in the well If 11.67 ppg mud will kill the well then the formation pressure is Formation pressure

= =

11.67 x 11,500 x 0.052 6978 psia

What casing pressure will be observed at surface? Mud pressure

= = =

10.5 x (11,500 – 300) x 0.052 10.5 x 11,200 x 0.052 6,115 psi

Gas pressure

= =

0.1 psi/ft x 300 30 psi

Total pressure of Mud + Gas in annulus

Difference

= =

6,115 + 30 6,145 psi

= =

6,978 – 6,145 833 psi (which is the same as the second circle in the above diagram)

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Calculations for Circulating Heavy Mud There are a number of ways of killing the well, however at this point a slight introduction will be given. These methods will be covered in the well control school. When killing a well using the Weight and Wait method only one circulation is necessary. The heavy kill mud is used to kill the formation and chase invading fluid. With the heavy mud ready to pump, we need to calculate 1. Initial pump pressure 2. Pump pressure with heavy mud at bit 3. When to adjust choke to get smooth transition between the above two. Initial Pump Pressure This pressure is required to circulate at the start of the kill procedure Example Slow pump rate test gave 800 psi at 45 SPM SIDPP is 700 psi, find initial pump pressure Solution Initial Pump Pressure

= =

800 + 700 1500 psi

Final Pump Pressure This is the pressure required to circulate once heavy mud has reached the bit. This calculation uses formula for pressure vs. mud weight change.

Final Circulating Pr essrue = Slow Circulating Pr essure ×

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Choke Adjustments As the heavy mud is pumped down the drill string, the choke operator will have to make adjustments to the choke for a smooth transition from initial circulating pressure to final circulating pressure Example T.D = 10,000 ft Initial circulating pressure = 1200 psi Final circulating pressure = 700 psi 5” drill pipe, 4.276 ID 600 ft 8” x 3” Collars Pump output, 0.2 bbls/stroke Calculate pump pressure every 100 strokes Step 1 Calculate the capacity of Drill string in barrels Drill pipe capacity + Drill Collar Capacity

4.276 2 32 x 9,400' + x 600' 1029 1029 167 + 5.2 172.2 barrels

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Pump output is 0.2 bbls/stroke therefore the final stroke of the pump should be 172.2 ÷ 0.2 which gives a final stroke of 861. Strokes 0 100 200 300 400 500 600 700 800 861

Pressure (psi) 1200 1142 1084 1026 968 970 852 794 736 700

Pressure & Volume Relationship for Gas As described in the introduction part of pressure and volume are inversely proportional to each other provided that the temperature of the system is unchanged. This is also Boyle’s law (named after Robert Boyle). In other words, for low volumes of gases (compressed state) the pressure is higher whereas for higher volume of gases (expansion state) the pressure is lower. This is analogous to the production mechanism in a well . At the time of the kick the gas at the bottom of the well will be of very small volume and therefore one would expect a high bottom hole pressure, however when the kick is taken out of the well the well in other words is allowed to flow. This will expand the gas (volume increases) and the pressure will reduce. Boyle’s law can be described mathematically as P1 V1 = P2 V2 Where 1 & 2 are showing the initial and final stages of the system

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Example A well has kicked with 15 bbls of gas at 5000 psia. if the surface pressure is 14.7 psia find the final volume of the gas. Solution From the above question P1 = 5000 psi V1 = 15 bbls P2 = 14.7 psi V2 = ? Using Boyles law we know that P1 V1 = P2 V2

P1 V 1 P2 V2 = 5102 bbls V2 =

Self Learning Exercise 1. TD : 5000 ft, P1 = 5000 psi, P2 = 2500 psi, V1 = 20 bbls, V2 = ? (Answer: 40 bbls) 2. TD: 5000ft, Mud Wt = 10 ppg, P2= 15 psi, V1 = 20 bbls, V2 = ? (Answer: 3466 bbls) (Hint: Calculate the hydrostatic head pressure by using standard equation) 3. TD: 5000ft, Mud Wt = 8 ppg, V1=20 bbls, V2=10 bbls, P2=? 4. V1=20 bbls, V2= 5% of V1, P1=2000 psi, P2=? 5. P1 = 2P2 where P2 = 1000 psi, V1 = 40 bbls, V2 = ?

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Leak off Test (L.O.T) Leak off pressure test can be used to calculate the formation fracture pressure. Formation Fracture Pressure Pressure

=

Leak off Pressure + Mud Hydrostatic

Exercise Calculate the formation breakdown pressure when L.oT. Mud Wt TD

= = =

2,500 psi 10 ppg 10,000 ft

(Answer: 7,700 psia) Maximum Allowable Surface Pressure (MAASP) MASP stands for “Maximum Allowable Surface Pressure”. MAASP is generally calculated from the casing shoe depth. As pressure in the annulus builds up, there is a danger of breaking one of the weak points in the system. The weak points are: Casing BOPs Formation below the casing Most often the formation below the casing is the weakest point. An excess of pressure would cause the formation to fracture resulting in a loss circulation.

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The formula to calculate MAASP is MAASP = Casing Shoe Dept (TVD, ft) x (Fracture Mud weight – Original Mud weight x 0.052

Recall, 0.052 is only added to convert the mud weights into gradient of psi/ft Exercise-1 TD OMW FMW Casing shoe depth Find the MASP?

: : : :

10,000 ft 8 ppg 10.5 ppg 7500 ft TVD

: : : :

10,000 ft 8 ppg 10 ppg 2000 meters (TVD)

(Answer: 975 psi) Exercise-2 TD Mud Weight Frac. MW Casing shoe Depth MASP = ? (Answer: 684 psi) (Hint: If your answer is wrong then you have not convert your casing shoe depth into feet) Exercise-3 TD Mud Weight Frac MW Fracture Pressure Casing Shoe Depth MASP

: : : : : :

4000m 7 ppg 13 ppg 7000 psi ? 2000 psi

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CHAPTER – 5

Buoyancy Effects

This section covers the basic principles of Archimedes principle of immersed bodies. At the end of this chapter candidates will be able to appreciate the effect of immersed drill pipes and how to calculate the number of collars required to give selected weight on bit.

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Buoyancy Effect Archimedes famous words “Eureka” meaning “ I found it” while he was taking a bath opened a whole new chapter in understanding the behaviour of fully or partially immersed bodies. The shipping industry is bred on this principle. In drilling engineering the concept of buoyancy is very important when the drill pipes are immersed in the drilling fluid. The effect of running in and tripping out drill pipes from the well have an immense effect on the bottom hole pressure. These pressures are also called the Swab and Surge pressures. Archimedes Principle “Bodies immersed in a liquid displaces a volume of liquid equal to the volume of that body” Therefore, a hole full of mud will discharge mud equal to the volume of steel pipe or collars run in during a trip. By calculating steel volume we can accurately measure FILL up, pulling out and OVERFLOW, running in. The use of a Trip Tank will help in monitoring these volumes. Archimedes also noted that a body immersed in a liquid becomes lighter. It in fact loses weight equal to the volume of liquid it displaces. Therefore, if drill pipe displaced 100 gallons of 10 ppg mud, the hook load would be 100 x 10 = 1000 pounds less than in air To calculate the Buoyancy effect we need pipe density and mud density. Steel pipe has an average specific gravity of 7.9. This means that a steel pipe is 7.9 times heavier than weight of an equal volume of water.

 Mud Weight ( ppg ) ÷ 8.33   Buoyancy Factor = 1 -   Specific Gravity of Steel 

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Example Calculate the immersed weight of 10,000ft of 5”, 19.5 lbs/ft drill pipe Buoyancy factor =

848

Immersed weight =

(10,000 x 19.5) x 0.848

=

195,000 x 0.848

=

165,360 pounds

Buoyancy effect is very important when considering drill collar length required to give required weight on bit. Example How many 30’ drill collars of 112 pounds/ft would be required to give a weight on bit of 50,000 pounds in 11.5 ppg mud. Solution Calculate the Buoyancy factor

Immersed Collar Weight

Length of Collar String

No. of Collars required

=

1-

11.5 ÷ 8.33 7.9

=

1 – 0.1747

=

0.825

=

112 x 0.825

=

92.4 lbs / ft

=

50,000 ÷ 92.4

=

541 ft

=

541 ÷ 30

=

18 Drill Collars are required

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Example How many 30’ drill collars of 112 pounds/ft would be required to give a weight on bit of 50,000 pounds in 11.5 ppg mud, with an excess of 20,000 lbs collar weight. Solution Total Collar weight in Mud

= =

55,000 + 20,000 75,000 pounds

Buoyancy Factor

=

1-

Immersed Collar Weight

Length of Collar String

No. of Collars required

=

11.5 ÷ 8.33 7.9 0.825

=

105 x 0.825

=

86.62 lbs / ft

=

75,000 ÷ 86.62

=

865.85 ft

=

865.85 ÷ 30

=

28.86 or 29 Collars

When calculating collar length required, the term Neutral point is commonly used. This is the point at which the compression of the lower section of collars changes to tension of the upper collars and pipe A safety factor is used so that any increase in weight on bit, will keep the neutral point in the collars. Drill pipe run in compression can be detrimental to the string life.

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Common neutral points are between 70% - 90% of collar length. If neutral point was at 80% of collar length, then 20% would be above and in Tension Note this is measured from the bit. Example 30ft, 147 lb/ft collars in 10.2 ppg mud. How many collars required to give 60,000 lbs W.O.B. with neutral point 80% up collars. Buoyancy factor = 0.845 Solution Length of Collars to give 60,000 lbs

=

(60,000 ÷ 147) x 0.845

=

483 ft

This is 80% of length required.

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Appendix – 1 Common Symbols & Abbreviations

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Units Inches Feet Cubic inches Cubic Feet Square inches Pounds Ounces Pounds per cubic foot Pounds per gallon Millimeters Litres # Centimeters Square meters Kilometer Grams Barrel

Abbreviation ins or in. or ” ft or ft or ’ cu ins or in3 cu ft or ft3 sq. in or in2 lbs oz lbs/ft3 ppg mm l lbs/ft (used extensively for Drillpipes and Casing cm Sq.m or m2 Km gm bbl

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Appendix – 2 Units & Conversion Table

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Length 1 millimeter 1 centimeter 1 meter (m) 1 Kilometer (Km) 1 inch (in) 1 foot (ft) 1 yard (yd) 1 mile

= = = = = = = =

0.0394 in 0.03937 in 1.0936 yd 0.6214 yd 25.4 mm 0.3048 m 0.9144 m 1.6093 Km 0.1550 sq.in 1.1960 sq.yd 2.4711 acres 0.3861 sq.mile 645.16 sq.mm 929.03 sq.cm 4046.86 sq.m 2.59 sq.Km

= = =

10mm 100 cm 1000 m

= = =

12 in 3 ft 1/60 yd

= = = =

100 sq.mm 10,000 sq.cm 10,000 sq m 100 ha

= = =

144 sq.in 4840 sq.yd 640 acres

= = = = = = = =

=

1000 cub.cm

= =

0.0610 cubic in 0.0353 cubic ft

= =

1000 cub. dm = 1 cubic dm = = = 8 US liquid = pints = =

1.3081cubic yd 0.2642 US Gal 0.2200 Imp Gal 16.387 cub. cm 3.78541 litre

Area 1 sq cm 1 sq meter 1 hectare (ha) 1 sq.Km 1 sq.in 1 sq.foot 1 acre 1 sq.mile Volume/Capacity 1 cubic cm 1 cubic decimeter (dm) 1 cubic meter 1 litre 1 litre 1 cubic in 1 US gallon 1 imperial gallon 1 US Quart

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=

4.546 litre 0.9463 litre

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Mass (Weight) 1 gm

=

1 Kg 1 tonne (t)

= =

1000 milligm (mg) 1000 gm 1000 kg

=

0.0353 oz

= = = = =

2.2046 lb 1.1023 short tons 0.9842 long tons 28.350 gms 0.4536 Kg

1 tonne 1 oz 1 pounds (lb)

= =

437.5 grains 16 oz

Centrigrade, C Fahrenheit, F Rankin, R 1 Calorie

= = = =

5/9 (F – 32) 9/5 C + 32 F + 460 4.2 Joules

1 Bar 1 psi

= =

1 bar

=

1 atm

=

14.7 psi 6895 Pascal, Pa 0.98624 atmosphere 14.7 psi

1 gm/cc

= =

Temperature

Pressure

Density

= =

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350.51 lb/bbl 62.42976 lb/cu.ft 8.33 ppg 0.036127 lb/cu.in

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Appendix – 3

TUTORIALS

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CHAPTER – 1 Measurement & Units 1. Convert 486 inches to yards, ft and inches 2. Convert 486 square inches into square 3. How may US Gallons would fill a tank with a capacity of 450 cubic ft 4. Convert 10.3 ppg into P.C.F 5. Convert 240 US gallons per minute flow into litre per minute Answers 1. 2. 3. 4. 5.

13 yd, 1 ft, 6 inches 3 square ft, 54 square inches 3366.4 US gallons 77.05 P.C.F 908.4 lit

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CHAPTER – 2 Preliminary Mathematics 1. 2. 3. 4.

Express the following fractions as decimals a) 5/8 b) 11/16 c) 7/9 Round off the following to 2 decimal places a) .6356 b) .7945 c) .7987 Calculate the circumference of a circle with a diameter of 6 inches Calculate the annular area between a 13inch inside diameter pipe and a 5 inch outside diameter pipe 5. Calculate the capacity of a tank in US bbls with the following dimensions 15’long, 6’ wide , 8’ deep Answers 1. 2. 3. 4. 5.

a) 0.625 b) 0.6875 c) 0.7778 a) 0.64 b) 0.79 c) 0.80 18.85 inches 113.1 square inches 128.2 bbls

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CHAPTER – 3 Fluid Circulation & Hydraulics 1. From the following well information calculate the volumes a. Hole size 12 ¼” b. Hole depth 12.650’ c. Casing shoe depth 10,200’ d. Casing size 13 3/8” ID: 12.46” e. 5” pipe ID: 4.2” 1. What is the hole volume with no pipe in the hole? (1896.2 bbls) 2. What would be the liquid volume with 5 inch pipe in the hole from top to bottom (1805.7 bbls) 3. What volume would be in the drill pipe open hole annulus (297.7 bbls) 4. What volume would be in the drill pipe casing annulus (1,281.3 bbls) 5. What would be the volume in the drill pipe (216.9 bbls) 2. Calculate the pump output per stroke of a triplex pump with a 12” stroke and liner size of 6” at 98% volumetric efficiency (0.103 bbls/stk) 3. Calculate the pump output in bbls/stk of triplex cement pump with 5 inch liners and an 8 inch stroke. Use a volumetric efficiency of 95%. (0.046 bbls/stroke) 4. What would be the maximum pressure that could be reached pumping at 400gpm with a pump of 750 HHP? (3214 psi) 5. Determine the new pressure reqd. by increasing the pump rate from 60 to 65 stk/min. Pump pressure at 60 stk/min was 2,560 psi (3,110 psi)

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CHAPTER – 4 Fundamentals of Pressure 1. Calculate the hydrostatic pressure exerted by the following columns of fluid a. Depth 12,000 ft, weight 10ppg b. Depth 8,500ft, weight 15.2 ppg c. Depth 17,200ft, weight 17.8 ppg (Answers: 1a) 6,240 psi, 1b) 6,718 psi 1c) 5,920 psi) 2. Calculate the mud weight if a. 5000 psi at 10,000 ft b. 2,325 psi at 5000ft c. 10,950 psi at 16,450 ft (Answers: 2a) 9.61 ppg , 2b) 8.94 ppg, 2c) 12.8 ppg) 3. What would be the increase in mud weight reqd. to exert an additional 350 psi HP for the examples in (2a) (Answer: 0.48 ppg, 1.27 ppg, 0.41 ppg) 4. Calculate in cubic feet the volume of a10bbl gas kick would occupy on surface. If the original formation pressure was 5,300 psi and atmosphere pressure is 14.75 psi (Answer: 20,174 cubic ft) 5. Calculate the formation strength if a. Shoe depth 6,200 ft (TVD) b. MW – 9.6 ppg c. Leak off Pressure 1,200 psi Give answers as a pressure and a pressure gradient (Answer: 4,295 psi, 0.693 psi/ft)

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CHAPTER – 5 Buoyancy Effects 1. Calculate the buoyancy factor for a. 12 ppg b. 14 ppg c. 16 ppg (Answer: 1a) 0.8181, 1b) .787, 1c) .757 2. How many 30’ drill collars would be reqd. if 60% of the available collar weight is 20,000 lbs? a. 8” drill collars in 11.8 ppg mud Weight = 146 lbs/ft (Answer: 9) 3. Where in the drill calculations would be the neutral point if 18,000 lbs was being applied to the bit (Answer: 150’ above the bit)

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