Drilling Calculations CD Complete Course

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(1)Drilling Calculations Course. © Randy Smith Training Solutions Ltd. July 2002.

(2) Drilling Calculations Course CONTENTS Section 1:. Units of Measurement. Section 2:. Background Mathematics. Section 3:. Fluid Circulation Calculations. Section 4:. Cementing Calculations. Section 5:. Pressure Control. Section 6:. Hoisting Calculations. Section 7:. Buoyancy Effects. Section 8:. Miscellaneous Calculations. Appendix:. Course Consolidation Exercises. © Randy Smith Training Solutions Ltd. July 2002.

(3) Drilling Calculations Course. Section 1: Units of Measurement. Calculations would not exist without measurement. Section 1 covers the most commonly used systems of measurement together with basic symbols and common Conversion Factors.. © Randy Smith Training Solutions Ltd. July 2002.

(4) Units of Measurement. Science today is totally dependent on measuring systems. A system was developed by a group of people to fit their needs, much like a language. Today only two systems survive – the Imperial and Metric. What do we measure ? Everything that exists on earth and in space has physical, chemical and biological properties known as MATTER – which is measurable.. The most common measurements taken are: Length Area Volume Mass (weight) Density Pressure Time. Some are Derived units: Density is derived from Mass Area and Volume are derived from units of Length. There are many more eg: Light frequency; radioactivity; heat; viscosity and reflection. © Randy Smith Training Solutions Ltd. July 2002.

(5) IMPERIAL SYSTEM. LENGTH:. inches, feet, yards and miles are the most common 12 inches 3 feet 1760 yards 5280 feet. Exercise:. = = = =. 1 foot 1 yard 1 mile 1 mile. Convert 2845 inches to yards, feet and inches. First, divide by 12 to obtain feet and inches: 2845 = 237 feet 1 inch 12 Second, divide the feet by 3 to get yards and feet: 237 3. = 79 yards 0 feet. Therefore, 2845 inches = 79 yards 0 feet 1 inch.. To simplify the system for Rig use, yards are not used and inches are often changed to tenths of a foot. The Rig Tape is calibrated in feet and tenths.. © Randy Smith Training Solutions Ltd. July 2002.

(6) The same units as length with the addition of the word “square” in AREA: front – square inches, square feet etc. As with length, problems arise when converting from one unit to another. 144 square inches 9 square feet 3,097,600 square yards 27,878,400 square feet. Exercise: inches.. 1.. 2.. = = = =. 1 square foot 1 square yard 1 square mile 1 square mile. Convert 92,846 square inches to square yards, sq.feet and sq. First, divide by 144. =. 92846 144. =. 644 sq ft, 110 sq inches. Second, divide 644 by 9 =. Therefore: 92,846 sq inches. © Randy Smith Training Solutions Ltd. 644 9. =. 71 sq yds, 5 sq ft. =. 71 sq yds, 5 sq ft, 110 sq ins. July 2002.

(7) VOLUME:. The same units as length, but prefixed by cubic – Cubic inches, cubic feet etc. 1728 cubic inches 27 cubic feet. = =. 1 cubic foot 1 cubic yard. The common term for Mass is WEIGHT. Weight is measured in MASS: ounces, pounds, hundred weights and tons.. 16 ounces 112 pounds 20 hundred weight 2240 pounds. = = = =. 1 pound 1 hundred weight 1 ton 1 ton. 1 ton is also called a LONG TON.. DENSITY: Density is the weight of a given volume of substance and is measured in pounds per cubic foot. Density distinguishes different substances, whereas weight does not take size into account.. (A block of wood will not weigh the same as a block of gold as their densities are different).. © Randy Smith Training Solutions Ltd. July 2002.

(8) The U.S. oilfield unit is measured in pounds per gallon. Gallon is a liquid volume measurement and is therefore used in measuring liquid density.. The gallon is different in the U.S. and U.K. The U.K. gallon of water weights 10 pounds, whereas the U.S. gallon of water weighs 8.34 pounds. The U.S. gallon is standard in the Oilfield. (A cubic foot of water weight 62.4 pounds). The density measurements are therefore calculated in ppg (pounds per us gallon) and pcf (pounds per cubic ft). Conversion means changing gallons to cubic feet or vice versa.. Exercise: 1.. Convert 8.34 ppg to pcf:. First, how many us gallons in a cubic foot? =. 2.. Second, multiply. 8.34 x 7.4809. =. 7.4809 U.S. gal / cubic ft 62.4 pounds/cubic ft. PRESSURE: Pressure is the force applied over a given area and is measured in pounds per square inch.( psi) PSI has always been the common unit, therefore conversion problems do not exist. With very high pressures, the pound may be changed to TONS. In the case of pressure being expressed in TONS/square ft we need to convert both measurements: Tons to pounds, and square feet to square inches.. © Randy Smith Training Solutions Ltd. July 2002.

(9) 20 tons per sq ft. =. (20 x 2240) pounds per 144 sq inches 44800 pds per 144 sq inches. 44800 144 pounds per sq in. =. 311 lbs per sq in. To make conversion easier, a table of Units and Conversion Factors is included at the end of Section 1.. © Randy Smith Training Solutions Ltd. July 2002.

(10) THE METRIC SYSTEM The Metric system covers all units of measurement, but makes use easier as it is based on units in multiples of ten. LENGTH. The fundamental unit is the METRE 1 metre. 1000 millimetres 100 centimetres 1000 metres. = = =. =. 39.37 inches. 1 metre (milli = one thousandth) 1 metre (centi = one hundredth) 1 kilometre (Kilo = a thousand times). To use the Metric system, and understanding of DECIMAL places is essential. 1 10 1/10 1/100 1/1000. Exercise:. in decimal in decimal in decimal in decimal in decimal. = = = = =. 1.0 10.0 0.1 0.01 0.001. How can 0.04 be expressed in words or as a fraction.. Counting from the decimal point, move to the right, until the decimal point is to the right of the last number. 1 jump = 1/10, 2 jumps = 1/100 Therefore, 0.04 can be expressed as 4/100 or four hundredths.. © Randy Smith Training Solutions Ltd. July 2002.

(11) Exercise:. Express 0.00328 in words or as a fraction.. 1st 2nd 3rd 4th 5th. = = = = =. tenth hundredth thousandth ten thousandth hundred thousandth. There are 5 jumps to the right. Therefore, 0.00328 is 328/100,000 or three hundred and twenty eight, one hundred thousandth. Most measurements go down to thousandths. 3_ 1000. =. 0.003. 25_ 1000. =. 0. 025. These are commonly used when measuring small parts of a unit.. 0.025 of a metre is 25 millimetres or 2.5 centimetres.. © Randy Smith Training Solutions Ltd. July 2002.

(12) DECIMAL POINT MOVEMENT: 1 place to the right 2 places to the right 3 places to the right 4 places to the right 5 places to the right 6 places to the right. = = = = = =. one tenth one hundredth one thousandth one ten-thousandth one hundred thousandth one millionth. = = = = = =. 0.1 0.01 0.001 0.0001 0.00001 0.000001. AREA: 1 1 1 1. sq metre sq metre sq kilometre hectare. = = = =. 100cm x 100 cm 1000mm x 1000mm 1000m x 1000m 100m x 100m cm mm m. MASS: (Weight). = = =. = = = =. 10,000 sq cms 1,000,000 sq mm 1,000,000 sq m 10,000 sq m. centimeter millimeter metre. The gram is the basic metric unit of weight 1000 grams 1000 milligrams 1000 kilograms. © Randy Smith Training Solutions Ltd. = = =. 1 kilogram 1 gram 1 metric ton. July 2002.

(13) VOLUME:. The metre is again the standard but it is called a CUBIC metre. The metric system commonly uses cubic centimeters or cubic metres to express volume and the LITRE when using liquids. 1 cubic metre 1000cc 1000 litres. = = =. 100 x 100 x 100 1 litre 1 cubic metre. =. 1,000,000 cubic cms. PRESSURE: The metric unit of pressure is kilograms/sq centimeters, and the smaller units of grams/sq centimetres. DENSITY:. Defines the weight of a given volume of a substance.. In the metric system, density is measured in kilograms/cubic metre or grams/cubic centimetre. On the rig, drilling fluid is often measured in pounds/cubic foot, Specific Gravity or pounds per gallon. Specific gravity is similar to Density is as much as the mud weighing 1gm/cc (water) has a Specific Gravity of 1. A S.G. of 2 means that the substance has a density twice that of water (of 2gms/cc). The Mud Balance gives 3 units of density measurement: Pounds/cubic ft Specific gravity (gms/cc) Pounds per gallon. © Randy Smith Training Solutions Ltd. July 2002.

(14) COMMON SYMBOLS AND ABBREVIATIONS. Inches Feet Cubic inches Cubic feet Square inches Square feet Pounds Ounces Pounds per cubic foot Pounds per gallon Pounds per square inch Millimetres Centimetres Metres Square metres Cubic centimetres Kilometre Grams Kilograms per sq centimetre Barrel. © Randy Smith Training Solutions Ltd. = = = = = = = = = = = = = = = = = = = =. ins or “ ft or ‘ cu ins or ins3 cu ft or ft3 sq. ins or ins2 sq. ft or ft2 lbs oz pcf or lbs/ft3 ppg or lbs/gall P.S.I. mm cm m m2 cc or cms3 km gm kg/cm2 bbl. July 2002.

(15) COMMON SYMBOLS AND ABBREVIATIONS +. =. Plus. 2+6 =. 8. -. =. minus. 7–2 =. 5. x. =. multiplied by. 3x4 =. 12. ÷. =. divided by. 10/2 =. 5. >. =. greater than. 6. >. 5. <. =. less than. 5. <. 6. ±. =. plus or minus. 60% ±. :. =. the ratio. 1:4. ∴ ∴A. = therefore C-B. A+B =. C. 42. =. square of 4. 4x4 =. 16. √. =. square root. √4. =. 2. 11. =. parallel to. ⊥. =. perpendicular. ∆. =. triangle. =. square. π. =. pi. %. =. percent. 320. =. degree. 4”. =. inches. 4’. =. feet. a-2. =. negative exponent. 3√. =. cube root. =. © Randy Smith Training Solutions Ltd. 1%. 3√64 = 4 July 2002.

(16) UNITS AND CONVERSION FACTORS. DEPTH/LENGTH:. Multiply To obtain. → ←. cm. 0.39370 in 0.3281 ft 0.01 m ____________________________ 25.40005 mm 2.54000 cm 0.08333 ft ____________________________ 30.48006 cm 12.0 in 0.30480 m ____________________________ 100.0 cm 39.370 in 3.2808 ft 1.936 yd ____________________________ 3.280.83 ft 1.000 m 0.62137 mi ____________________________ 5,280.0 ft 1,609.34 m 1,609.34 km ____________________________. in. ft. m. km. mi. © Randy Smith Training Solutions Ltd. by by. → ←. to obtain Divide. July 2002.

(17) UNITS AND CONVERSION FACTORS. Multiply To obtain. AREA:. VOLUME/CAPACITY:. © Randy Smith Training Solutions Ltd. → ←. by by. → ←. to obtain Divide. cm2 0.15499 in2 ________________________________ 6.4516 cm2 in2 ________________________________ ft2 929.0341 cm2 0.092903 m2 ________________________________ 1,549.9969 in2 m2 10.76387 ft 2 ________________________________ acres 43,560.0 ft2 4,480.0 yd2 4,46.873 m2 0.00405 km2 0.0015625 mi2 _________________________________ km2 247.104 acres _________________________________ mi2 640.0 acres 2.5899 km2 _________________________________ cm3 1,000.00 mm3 0.01 litre 0.6102 in3 0.0002642 gal 0.00003531 ft3. July 2002.

(18) UNITS AND CONVERSION FACTORS. Multiply To obtain. VOLUME/CAPCITY (cont). in3. litre. gal (U.S.). gal (imp) bbl (U.S.). © Randy Smith Training Solutions Ltd. → ←. by by. → ←. to obtain Divide. 16.38716 cm3 0.4329 gal 0.1638 litre 0.5787 ft3 _______________________ 1,000.0 cm3 1,000 ml 61.2705 in3 1.57 qt 0.26417 gal (U.S.) 0.3531 ft3 _______________________ 3.785.0 cm3 231.0 in3 4.0 qt (U.S.) 3.7853 litre 0.83268 gal (imp) 0.13368 ft3 0.2381 bbl (42) _______________________ 1.20095 gal (U.S.) _______________________ 158.984 litre 42.0 gal (U.S.) 5.61458 ft3 0.9997 bbl (imp). July 2002.

(19) UNITS AND CONVERSION FACTORS. Multiply To obtain. VOLUME/CAPCITY (cont). → ←. by by. → ←. to obtain Divide. 159.031 litre 42.112 gal (U.S.) ________________________________ 3 1,728.0 in3 ft 28.31684 litre 7.4809 gal (U.S.) 0.1781 bbl (42) 0.2831 m3 _________________________________ m3 264.17 gal (U.S.) 35.314 ft3 6.290 bbl (42) 1.3079 yd3 _________________________________ acre/ft 325.850.0 gal (U.S.) 43.560.0 ft3 7,758.4 bbl (42) 1,613.33 yd3 1,233.49 m3 _________________________________ DENSITY/CONCENTRATION Gm/cc (s.g.) 350.51 lb/bbl (42) 62.42976 lb/cu ft 8.34544 lb/gal (U.S.) 0.036127 lb/cu in ___________________________________. © Randy Smith Training Solutions Ltd. bbl (imp). July 2002.

(20) UNITS AND CONVERSION FACTORS. Multiply To obtain. → ←. by by. → ←. to obtain Divide. DENSITY/CONCENTRATION (cont) lb/gal (U.S.). WEIGHT/MASS. © Randy Smith Training Solutions Ltd. 42.0 lb/bbl (42) 7.4809 lb/cu ft 0.119826 gm/cc (S.G.) ____________________________________ lb/cu ft 5.6146 lb/bbl (42) 0.13368 lb/gal (U.S.) 0.016018 gm/cc (s.g.) _____________________________________ Grain 0.06479 gm 0.229 oz _____________________________________ gm 15.43236 grain 0.3528 oz 0.220 lb _____________________________________ oz 437.5 grain 28.34952 gm 0.0625 lb _____________________________________ kg 35.274 oz 2.2046 lb _____________________________________ lb 453.59237 gm 16.0 oz 0.4536 kg. July 2002.

(21) UNITS AND CONVERSION FACTORS. Multiply To obtain. WEIGHT/MASS (cont). MUD WEIGHT. MUD WEIGHT To PRESSURE GRADIENT. ANNULAR VELOCITY. FLOW RATE. © Randy Smith Training Solutions Ltd. → ←. by by. → ←. to obtain Divide. ton (short) 2.000 lb ton metric 0.90718 ton (metric __________________________________ ton (long) 2.240.0 lb 1.12 ton (short) 1.1605 ton (metric) __________________________________ PPG x 119.8 Kgm3 x 0.00835 lbs per gallon Kg/m3 __________________________________ PPG x 0.052 psi/ft Pressure Gradient SG x .433 psi/ft 3 b/ft ÷ 144 psi/ft Kg/m3 x 0.000434 Or ÷ 2303 psi/ft Kg/m3 x 0.00982 K/Pa/m __________________________________ Ft/min x 0.3048 m/min M/min x 3.2808 ft/min __________________________________ Gal/min x 0.003785 m3/m Barrels/min x 0.159 m3/m M3/min x 6.2905 bbl/min 3 x 264.2 gal/min M /min __________________________________. July 2002.

(22) UNITS AND CONVERSION FACTORS. Multiply To obtain. RESISTIVITY. PRESSURE. © Randy Smith Training Solutions Ltd. → ←. by by. → ←. to obtain Divide. ohms/cm2cm 0.01 ohms m2 m __________________________________ Ohms/m2m 100. Ohms m2m __________________________________ psi 70.3067 gm/cm2 0.0703070 kg/cm2 0.0689474 bar 0.0680458 atm __________________________________ atm 14.6960 psi 1.3323 kg/cm2 1.1325 bar __________________________________ 14.22333 psi kg/cm2 0.980665 bar 0.967842 atm __________________________________ bar 106 dynes/cm2 14.5038 psi 1.1972 kg/cm 0.98624 atm __________________________________. July 2002.

(23) UNITS AND CONVERSION FACTORS. Multiply To obtain. by by. → ←. to obtain Divide. → ←. F = 1.8 oC + 32. TEMPERATURE:. 0. 0. 0. 0. 0. 0. K =. C = 5/9 (oF – 32). 0. Fahrenheit. 0. Rankine. R =. C + 273. 0. F + 460. Centigrade. 0. Kelvin. Water boils. 212.. 672.. 100.. 373.. 680F. 68.. 528.. 20.. 293.. 600F. 60.. 520. 15.56.. 288.56.. Water freezes. 32.. 492.. 0.. 273.. O0F. 0.. 460.. -17.8.. 255.. Absolute zero. -460.. 0.. -273.. 0.. © Randy Smith Training Solutions Ltd. July 2002.

(24) Drilling Calculations Course. Section 2: Background Mathematics. This section covers the basic maths involved in Drilling Calculations. How to calculate Percentages; Areas; Volumes; Capacities and how to use Fractions.. © Randy Smith Training Solutions Ltd. July 2002.

(25) Fractions What is a Fraction? A fraction is a part of a whole. Two and a half inches is equal to two inches plus one half of an inch. This can be represented in two ways.. First:. 2½. OR. as a decimal. =. 2.5. Second:. 5/8. OR. as a decimal. =. 0.625. To find 0.625. 5 is divided by 8. Certain conversions leave five, six, seven and above numbers after the decimal point e.g.. 0.28463215. This is clumsy and should be reduced for most purposes to four figures e.g.. © Randy Smith Training Solutions Ltd. 0.2846. July 2002.

(26) As most calculations are performed on the calculator it is easy, and accurate, to use four figures. Using four figures in a hand calculation is clumsy and leads to error. Therefore, use the calculator often. Measuring in feet and inches presents problems when tallying pipe. To ease the situation, feet and tenths of a foot are used. You will have noticed the Pipe Measuring Tape is calibrated in feet, and tenths of a foot. Diameters are most commonly measured in feet and inches because they are usually taken on their own. In contrast, length is measured in feet and tenths of a foot for ease of addition. When diameters are involved in calculations, for instance in volumes, the inches or vulgar fraction has to be converted to decimal. e.g.. Cylindrical Tank 6ft 4 inches diameter To convert: There are 12 inches to 1 ft, Therefore, 4 inches = 4/12. Therefore 4 ÷ 12. =. 0.3333. Diameter in decimals. =. 6.3333 ft. The calculation is recurring therefore four decimal places are used.. © Randy Smith Training Solutions Ltd. July 2002.

(27) What is a Decimal Place? When asked to calculate to four Decimal Places your inputs should have four numbers to the right of the decimal point. e.g.. What is 8.32567418 to 3, 4 and 5 decimal 3 decimal places 4 decimal places 5 decimal places. = = =. 8.326 8.3257 8.32567. Notice that the first three decimal places are 8.325, but the answer above is 8.326. The technique of Rounding-Off is being used. If the next number is five or greater, then increase your last decimal place by one. e.g.. 8.32748. To “round-off” to 4 decimal places, look at the first decimal place. Being 8 it is greater than five, therefore increase 4 to 5 = 8.3275 Examples “Round off” to 4 decimal places 9.382416 9.221134 9.18796 9.25256. © Randy Smith Training Solutions Ltd. = = = =. 9.3824 9.2211 9.188 9.2526. July 2002.

(28) Exercise:. (Round off to 4 decimal places if necessary). 1). Convert. 6-2/8; 3-4/16; 5-7/13; 8-2/6 to decimals.. 2). Convert. 42ft 7 inches to decimals.. 3). Convert. 10ft 6-1/2 inches to decimals.. © Randy Smith Training Solutions Ltd. July 2002.

(29) Areas The use of area is found in many places around the rig. Force on a unit area Area of deck space Surface area of pits. Area is expressed as a square – a square inch, square centimetre, square foot, square mile, etc. A square inch is the area taken up by a square, of 1-inch long sides. There are 3 common shapes that can easily have their areas calculated. A shape with 4 sides, each side at 90° to the other – Rectangle. Area =. Length. x Breadth. 5. 4. 5 x 4 = 20 sq. ins. © Randy Smith Training Solutions Ltd. July 2002.

(30) 6. 6 x 3 = 18 sq. ins 3. © Randy Smith Training Solutions Ltd. July 2002.

(31) A shape with 3 sides, angles between each side are variable – Triangle. Area = Base x ½ vertical height. height. base. ½ height. Area = base x ½ height. © Randy Smith Training Solutions Ltd. July 2002.

(32) A shape with 4 sides, none of the angles are 90° - Trapezium Area = Sum of Parallel Sizes x ½ distance between them. a ht. (a+b) x ½ ht. b. Cut trapezium into 3 parts. © Randy Smith Training Solutions Ltd. July 2002.

(33) Another common shape, but not readily calculated, is the Circle. The area is a relationship between radius, or diameter and circumference. The Radius is the distance from the centre to the edge. The Diameter is the distance from the edge to edge via the centre.. Radius Diameter. =. 2 x Radius. Radius. =. ½ Diameter. The Circumference is the distance round the edge of the circle. This has a fixed relationship with the diameter. The diameter of any circle will go round the circumference 3.1416 times. This value is constant and is called PI (π).. To calculate Circumference using diameter, multiply Diameter by π. Or. Circumference. =. π x diameter. Circumference. =. 2 π x radius. © Randy Smith Training Solutions Ltd. July 2002.

(34) To find the formula for calculating area we can divide the circle into slices like a cake.. Circumference = 2 π x Radius. For instance the circle has been divided into 32 equal portions - each like a triangle. Unpeeling the circle we get the shape below.. Radius 2πr The base. =. circumference. Each triangle has an area of. © Randy Smith Training Solutions Ltd. =. π D or 2 π r. 1/2ht x base.. July 2002.

(35) Height. =. radius. ½ Height. =. Base. =. or. Diameter 2. Radius 2. 2πr 32. To calculate for 32 triangles – =. 32 x r x 2 π r 2 32. 32 cancels out.. =. r x 2 π r 2 32. 2 cancels out.. =. r x π r. =. π r2. © Randy Smith Training Solutions Ltd. July 2002.

(36) If using Diameter –. Area of Circle. =. 32 x D x π D 4 32. =. D x π D 4. =. π D2 4. =. π r2 or π D2 4. Exercise: Find Area of Circles with the following: a) b) c) d) e) f). Diameter Diameter Radius Diameter Radius Circumference. © Randy Smith Training Solutions Ltd. = = = = = =. 12” 7” 4” 7 ½” 3 ¼” 24”. July 2002.

(37) To aid calculation, remember. Therefore, Area. =. π 4. =. .7854. .7854 x D2. π D2 One major application of 4 is the calculation of Annular Area and Volume. The Annular Area is the area between two concentric circles. For instance hole to pipe or OD of pipe to ID of pipe.. ANNULAR AREA. The Annular area is calculated by subtracting the small circle from the larger circle. With D. =. Annular Area. diameter of large circle d = diameter of small circle. =. © Randy Smith Training Solutions Ltd. π D2 4. -. π d2 4. July 2002.

(38) π Because 4 is common to both the above formula can be rewritten: π 4 or. (D2 - d2). .7854 (D2 - d2). Example: Find Annular Area when D = 10” and d = 5”. π. Area =. 4. (102 – 52). =. .7854 (100 – 25). =. .7854 x 75. =. 58.9 sq. inches. © Randy Smith Training Solutions Ltd. July 2002.

(39) Formulas and Problems Up to this point the formulas used show division, multiplication and brackets. This can lead to problems unless two basic rules are practiced. First: the use of brackets. Whenever brackets appear in a formula the calculation inside must be done prior to using the values outside. e.g.. .7854 (D2 – d2). Calculate the bracket first = = =. .7854 (102 – 52) .7854 (100 – 25) .7854 (75). The value outside can now be multiplied with that inside. Second:. Solving the equation. This means rearranging a formula to get the unknown value on one side and the known value on the other side.. Find a:. a+b=c. Move b across and change + to – i.e.. a=c–b. find b:. a+b=c. Move a across and change + to – i.e.. b=c–a. © Randy Smith Training Solutions Ltd. July 2002.

(40) find a:. a–b=c. Move b across and change – to + a= c+b. In multiplication the technique is different. Values are moved diagonally.. Find a:. a b. =. c. Move b diagonally across = sign a=cxb. Find b:. a b. =. c. Move b up to c and c to b © Randy Smith Training Solutions Ltd. July 2002.

(41) a c =b. a=bxc. Find b:. =. a b. (c + d). a = b (c + d). (c If. a +. =. b. 10 (3 + 2). =. b. 10 5. =. b. 2. =. b. d). a = 10 c=3 d=2. What is b:. Pressure. =. Depth. Solve the Equation to find. © Randy Smith Training Solutions Ltd. x. Mud Weight. x. 0.052. a) Depth b) Mud Weight. July 2002.

(42) a). Pressure Mud Weight x 0.52. =. Depth. b). Pressure Depth x .052. =. Mud Weight. Solving an Equation with squares requires the use of Square Roots. Example:. Area =. π D2 4. Find D:. Area x 4 π. to eliminate the square you must square root the other side.. D. =. Area x 4 π. Square roots are commonly found on calculators today. The square root of The square root of. © Randy Smith Training Solutions Ltd. 4 is 2 64 is 8. (2 x 2 = 4) (8 x 8 = 64). July 2002.

(43) Volumes and Capacities. With an understanding of how to calculate areas it is a straight forward procedure to calculate the volume of a container. Volume is the amount of space in a container. Capacity is the amount of a substance that can be placed in that container expressed in units relating to both substance and container. When talking about the capacity of a tank or hole we use barrels, and think of common rig substance like oil, mud or cement. To calculate volume we multiply the surface area by the height.. Example: A tank of 12” long x 6” wide x 8” deep =. 12 x 6 x 8. =. 576 cubic inches. This means 576 cubes of 1” x 1” x 1” would fit into a tank 12” x 6” x 8”.. When calculating volume all units must be the same.. © Randy Smith Training Solutions Ltd. July 2002.

(44) Example: Find capacity in cubic inches of a tank 1’ 2” x 8” x 3’ 6” 1’ 2” = 14” 3’ 6 = 42”. Capacity. =. 14 x 8 x 42. =. 4704 cubic inches. We have assumed vertical walls. If the tank had sloping walls the following volume calculations would be used.. 50. 30 10. 20. Side View. Plan View. © Randy Smith Training Solutions Ltd. 10. July 2002.

(45) The area of side A can be found using the formula for a trapezium. Area =. Sum of Parallel sides x half distance between them. Area =. 10 (50 + 30) x 2. =. 400 sq. inches. Then calculate capacity as:. Area x. sum of parallel sides on wall B 2. 20 + 10 2. =. 400. x. =. 400. x. 30 2. =. 400. x. 15. =. 6000 sq. inches. © Randy Smith Training Solutions Ltd. July 2002.

(46) Calculating volumes of Cylinders or the Annulus the formula is: Area x height. Volume. π D2 4. =. x. height. Make sure all units are the same. π 4(D2 – d2) x height. Annular Volume π 4. =. .7854. Example: Find a volume of cylinder in cubic feet/foot of depth if diameter is 10”. Volume =. .7854 (102) 144. x 1ft. = .5454 x 1ft = .5454 cubic feet/foot of depth. The 144 is used to convert square inches into square feet (1 square foot = 144 square inches).. © Randy Smith Training Solutions Ltd. July 2002.

(47) Example: Calculate Annular Volume if. Volume =. D = 10”. d = 6”. depth = 1ft. .7854 (102 – 62) x 1 144. = .349 cubic feet/foot. The use of cubic feet is not as common as barrels. To calculate the volume in barrels, we need to convert feet to barrels. 1 barrel = 5.6146 cubic feet.. Applying this to the formula:. Volume in barrels/ft. =. .7854 (D2) 144 x 5.6146. x 1. Calculating out .7854, 144 and 5.6146 we can simplify the formula to. Volume in bbls/ft. =. D2 1029. Or (D2 – d2) 1029 © Randy Smith Training Solutions Ltd. July 2002.

(48) Percentage Calculations Calculating percentages involves simple multiplication, division and rearranging formula. Percent is the number of parts of 100. Example 1 What is 10% of 200 logs? 1%. ∴ 10%. =. 200 100. =. 2 logs. =. 2 x 10. =. 20 logs. Example 2 How many % is 35 logs of 200 logs? 1%. =. 200 100. 1%. =. 2 logs. 1 log = ∴ 35 logs. = =. © Randy Smith Training Solutions Ltd. ½% ½ x 35 17.5% July 2002.

(49) Example 3 If 42 logs. = 75% of the total, how many logs are there?. 42 logs. =. 75%. =. 42 75. =. 42 75. =. 56 logs. 1%. ∴ 100%. x. 100. Each of the above examples tackles the problem differently. Example 1 - What was the value of 10% Example 2 - What was the % Example 3 - What was the value of 100%. The above examples, although different, use the same formula. P=RXB P= R= B=. Percentage:-the actual value equaling chosen % Rate in decimals:- the part of a 100 to be found ie in 4% of 50, 4% is rate. Base:- the number of which some percentage is to be found.. © Randy Smith Training Solutions Ltd. July 2002.

(50) Example 1 (Repeat) What is 10% of 200 logs? The question asks you to find a number that equals a %, being 10% here. P=. RxB. Rate is the parts of a 100 to be found. In this case 10 parts (10%). Remember Rate is expressed in decimals. 10% of 100% = .1 P=. .1 x B. Base is the number of which some percentage is to be found. In this case we want to find 10% of 200. P=. .1 x 200. =. 200 logs. =. 20 logs. 10% of 200 logs. Example 2 (Repeat) What % is 35 logs of 200 logs? The question asks for an actual percentage. This being the Rate P. © Randy Smith Training Solutions Ltd. =. RxB. (R is unknown). July 2002.

(51) Percentage means the actual number. In this case 35 logs. 35. =. Rate x Base. Base is the whole. In this case 200. 35. =. Rate x 200. Rate =. 35 200. Rate =. .175 whole (1.75 of 1.0). Convert decimal to % by multiplying by 100. .175 x 100 =. 17.5%. To convert % to decimal ÷ 100 To convert decimal to % x 100 Example 3 (Repeat) If 42 logs = 75% of the total, how many logs are there? P. =. Percentage is number of logs. © Randy Smith Training Solutions Ltd. RxB =. 42. July 2002.

(52) Rate is parts of 100 to be found in decimals. 42. =. 75% ÷ 100. =. .75 x B. = .75. Base is equal to total or 100% Base =. =. 40 .75 56 logs. Examples What is 42% of 381? P P. = = =. R x B .42 x 381 160.02. = = =. R x B R x 281 48 281. What % of 281 is 48 P 48 R. = .1708 x 100. © Randy Smith Training Solutions Ltd. =. 17.08%. July 2002.

(53) 225 is 15% of what? P. =. R x B. 225. =. .15 x B. B. =. 225 .15. =. 1500. To remember formula use following diagram. P P R. P. =. RxB. R. =. P B. B. =. P R. © Randy Smith Training Solutions Ltd. B R. July 2002.

(54) Drilling Calculations Course. Section 3: Fluid Circulation Calculations. This section covers the most commonly used calculations involved with Fluid Hydraulics used by Drill Crews.. © Randy Smith Training Solutions Ltd. July 2002.

(55) Fluid Circulation Calculations. Annular Volume Calculations Using the formula (D2 - d2) x depth, ft 4 x 144 The annular volume in cubic feet can be obtained. For answer in barrels use: (D2 - d2) x depth, ft 1029. D d. = =. large diameter smaller diameter. Volumes use:(non annular). (inside diameter of the hole) (outside diameter of the string). D2 1029. x. Depth, ft. With varying string diameters, casing and open hole it is good policy to draw a fully-labelled diagram before calculation. Example: Calculate Annular Volume in barrels of an 8000 ft hole, 12 ¼ inside diameter with 5” drill pipe.. © Randy Smith Training Solutions Ltd. July 2002.

(56) 0’ Annular Volume. =. [(12.25)2 - (5)2] x 8000’ 1029 =. .1215 x 8000’. =. 972.3 bbls. 8000’. Convert to gallons.. Annular Volume in gallon. © Randy Smith Training Solutions Ltd. =. 972.3 x 42. =. 40,836.6 gallons. July 2002.

(57) Example: 9⅝ 8½ 5” 600ft of 6”. Casing set at 9000 ft. ID = 8.84” Open hole to 11,000 ft Drill pipe 19.5 lbs/ft ID = 4.276” x 2 ½” Drill collars. Calculate. a) b). annular volume in bbls, cu ft and gallons volume of mud inside string in bbls. Volume of Casing annulus =. (8.842 - 52) x 9,000’ 1029. =. 464.83 bbls. Volume OH to Collar annulus =. (8.52 - 62) 1029. =. 21.14 bbls. Vol of OH to Pipe annulus =. Total. x. 600’. (8.52 - 52) x 1400 1029. =. 64.28 bbls. =. 64.28 + 21.14 + 464.83. =. 5520.25 bbls. © Randy Smith Training Solutions Ltd. July 2002.

(58) In Gallons. =. 550.2542. =. 23110.5 galls. In cubic ft = =. b). 550.25 x 5.6146 3089.4 cubic ft. Capacity of drill string =. =. (id) 2 x length 1029. =. (4.2762) 2 1029. =. 184.79. =. 188.4 bbls. x. cap. of pipe + cap. of collars (id) 2 x length 1029. 10,400. + +. (2.52) 2 1029. x. 600’. 3.64. Example: 10,000ft well. Drill pipe is 5”, 19.5 lbs/ft ID 4.276” 600 ft collars 9” x 3” One stand = 90 ft Calculate barrels of mud required to:a) b). Fill hole after 10 stands of drill pipe have been pulled Fill hole after each stand of collars is pulled. © Randy Smith Training Solutions Ltd. July 2002.

(59) c) d). Total mud required to keep hole full when pulling out Quantity of mud displaced running in the hole with extra 300’ of 8” x 2 ¾ “ collars. a). Volume of steel in 10 stands of drill pipe = (52 - 4.2762) 1029. x. 900’. = 0.065. x. 900’. = 5.87 barrels. b). Volume of steel in 1 stand for drill collars = (92 - 32) 1029. x. 90’. = .06997. x. 90’. = 6.3 barrels. c). Total mud to fill hole = Drill pipe disp/ft x 9400’ + drill collar disp/ft x 600 = .0065 = 61.1. x 9400’ + .06997 +. x 600. 41.98. = 103 barrels. © Randy Smith Training Solutions Ltd. July 2002.

(60) d). Running in the hole = drill pipe (disp/ft x 9100). drill collars + (disp/ft x 600) +. new drill collar (disp/ft x 300). = (.0065 x 9100). + (.06997 x 600) +. (82 - 2.752) x 300 1029. = 59.15. +. 41.98. = 101.13. +. 0.548 x 300. +. 16.46. = 117.58 barrels. 100 COLLARS B B L S. 50 PIPE. I 50. © Randy Smith Training Solutions Ltd. I STANDS. 100. July 2002.

(61) Pump Outputs. Pump output calculations are simply volumes Practical tests for Pump Output per stroke can be made, manufacturers calculation can be used or you can calculate based on stroke length, liner size and an Efficiency factor.. Example: Find pump output/stroke on Triplex with 12” stroke and 6” liners at 95% Efficiency. Triplex has 3 cylinders. Volume of Cylinder. =. Π D2 4. Volume of 3 cylinders. =. 3. =. 3 (.7854 (62) x 12). =. 3 (.7854 x 36 x 12). =. 3 x 339.29. =. 1017.88 cubic inches. (Π = .7854) 4. © Randy Smith Training Solutions Ltd. x. length. (Π D2 x length) 4. July 2002.

(62) Convert to barrels. =. 1017.88 1728 x 5.6146. 1728 cubic inches 5.6146 cubic ft. = =. 1 cubic ft 1 bbl. Out put. =. .1049 bbls/stroke. .1049bbl at 100% Efficiency 1% Efficiency. =. .1049 100. 95% Efficiency. =. .1049 x 95 100. =. .0996 bbls/stk. Annular Velocities and Circulation Times Knowing hole volumes and pump output the annular velocity for a section of hole and the time for circulation can be calculated.. Annular velocity (ft/min). =. Pump output (bbls/min) Annular Volume (bbls/ft). Barrels can be substituted for galls, cubic feet etc. © Randy Smith Training Solutions Ltd. July 2002.

(63) Bottoms up time (mins). =. Annular Volume (bbls) Pump Output (bbls/min). Again, barrels can be substituted for other units.. Drill string + Annular Volume + Pit Volume Total Circulation Time = Pump Output (bbls/min). Hydraulics Calculations Observing the size of pumps, pressure rating of unions, safety chains on hoses, safety clamps on pipe, torque required for tool joints and packing required for swivel, we can conclude that mud is circulated round the system 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 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 system that creates a back pressure against which the pump must work. Friction within the system causes pressure. The pressure at the pumps is the sum of all the 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:. Assuming pressure at pump is 3000 psi:. © Randy Smith Training Solutions Ltd. July 2002.

(64) Pressure inside Kelly Pressure inside near bit sub Pressure in drill collar annulus Pressure at flow line. = = = =. 2950 2200 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 approximately 60-65% of Pump Pressure. Pressure losses can be divided into sections thus: 1. 2. 3. 4.. Surface Lines Drill String Drill Bit Annulus. P S I. 1. © Randy Smith Training Solutions Ltd. 2. 3. 4. July 2002.

(65) The horsepower required to circulate a known quantity of mud at a certain pressure can be calculated using the formula.. Hydraulic Horsepower (HHP). P V. = =. P x V 1714. Pump Pressure (psi) Pump Output (gallons/minute). 1714 is a constant Due to mechanical inefficiency, the output horsepower is always less than input horsepower. Mechanical Efficiency =. HHP Output Mechanical Horsepower Input. Most pumps have a Mechanical Efficiency of approximately 85%. The same principle applies to volumetric output of a pump – called Volumetric Efficiency.. Exercise: Find Hydraulic Horsepower of pump pumping 350gpm at 2,800 psi. HHP. =. P x V 1714. =. 2800 x 350 1714. =. 572 HHP. © Randy Smith Training Solutions Ltd. July 2002.

(66) Calculation of Mud Weight and S.P.M. Effect on Pump Pressure By changing Mud Weight or Pump S.P.M., we fundamentally alter the system hydaulics. Pressure loss changes cause a change in pump pressure. The effect can be calculated using simple formula:. New Pressure. =. Old Pressure x. (New SPM)2 (old SPM)2. Example: What is pump pressure if an SPM of 60 giving 2500 psi is changed to 70 SPM?. New Pressure. =. 2500. x. (70)2 (60)2. =. 2500. x. 1.361. =. 3403 psi. =. 3200 psi. Example: Pressure at 80 SPM. New Pressure. © Randy Smith Training Solutions Ltd. What is pressure at 70 SPM. =. 3200. x. (70)2 (80)2. =. 3200. x. .8752. =. 2450 psi. July 2002.

(67) Changing Mud Weight will affect pump pressure in the following way:. New pump pressure. =. Old pump pressure x. New Weight Old Weight. Example: Pump pressure = 2800 psi with 10.5 ppg mud. What will pressure be if weight is increased to 11.0ppg ? New Pump Pressure. =. 2800. x. 11.0 10.5. =. 2800. x. 1.047. =. 2931 psi. Both formulae can be written: P2. =. P1. x. (SPM2) 2 (SPM1) 2. P2. =. P1. x. W2 W1. P2. =. New Pressure. SPM2. =. New SPM. P1. =. Old Pressure. SPM1. =. Old SPM. W1 W2. = =. Old Mud Weight New Mud Weight. © Randy Smith Training Solutions Ltd. July 2002.

(68) Nozzle Size Calculation Nozzle sizes refer to either the diameter of the hole in 32nds of an inch, or the cross sectional area in square inches. You may find the need to calculate square inch area from 32nds or vice versa.. Example: A bit is to have 2 x 15’s and 1 x 14 nozzles. What is the Total Cross Sectional Area of the nozzles in square inches? First: Calculate the area of 2 x 15’s. 15. Area. =. 15/32 nds of an inch, diameter. π D2 4. =. 2 x. =. .7854 x (15/32) 2 x 2. Convert fraction to decimal =. .7854 x ( .4687)2 x 2. =. .1725 sq inches. =. .345 sq inches. © Randy Smith Training Solutions Ltd. x 2. July 2002.

(69) Second:. Calculate for 14 nozzles. =. .7854 x (14/32) 2. =. .7854 x .1914. =. .1503 sq inches. Total Cross Sectional Area = =. .1503 + .345. .4953 square inches. Example Convert Total Cross Sectional Area of three nozzles in 32nds of an inch, with each nozzle to be as close in size as possible. Cross Sectional Area Three nozzles One nozzle Area. = = = =. .3137 square inches .3137 .10456 sq inches approximately .7854 x D2. Solve the equation to get D i.e.. © Randy Smith Training Solutions Ltd. 2. D. =. Area .7854. July 2002.

(70) .10456 .7854. D =. =. .133138. =. .3649 of an inch. Convert .3649 of an inch into 32nds, this is done by writing: ? 32 = .3649 OR ? = .3649 x 32 = 11.67 32 nds This was not a complete number - it has .67 of a 32nd. But from this we can see that the nozzles are approximately 11’s or 12’s. The .67 is almost 2/3, meaning the average nozzle size is 2/3rds of the way toward 12. This means that the nozzles are 12, 12, 11 To check back if 12, 12, 11 is right, we follow method shown in previous example:. Area. = .7854 x 122 x 2 + .7854 x 112 32 32 =. .22089 + .0928. =. .3137 sq inches. Therefore 12, 12, and 11 nozzles is correct. © Randy Smith Training Solutions Ltd. July 2002.

(71) Drilling Calculations Course. Section 4: Cementing Calculations. The Drill Crew should have an understanding of what is involved in the calculations in order to check the cementing programme. The most commonly used calculations are used in Single Stage, Multiple Stage and Plug jobs.. © Randy Smith Training Solutions Ltd. July 2002.

(72) Cementing Calculations. Single stage jobs, Multiple stage jobs and plugs are drilling practices which require cement to be placed downhole. Not just placed anywhere, but accurately positioned in order to perform a specific task. This requires accurate calculation that will be checked by 3 or 4 persons – one being the Driller, but most likely the Toolpusher. The calculations are slightly different in each case i.e. Single, Multiple and Plug jobs, but they all require skill at calculating Annular Volumes.. To recap:Capacity of Cylinder. =. OR. Capacity of Annular Space. OR. ∏D2 4. (Answer in cubic inches, ft etc). D2 1029. (Answer in barrels per ft). =. ∏ (D2 – d2) 4 (D2 – d2) 1029. Number of strokes required to pump =. © Randy Smith Training Solutions Ltd. Volume______ Pump Output/Stroke. July 2002.

(73) Single Stage Job Given 12-¼” diameter hole from surface to 5000 ft Casing O.D. 9-⅝ run from surface to 5000 ft. Float Collar set 40ft up inside casing (9.00” I.D.). Exercise: Calculate number of barrels of cement required to cement to surface. Volume of Slurry. =. Annular Volume. +. Volume of Casing. x 40ft. = (12 ¼2 - 9-⅝2) x 5000’ 1029. +. (92). x 40’. = (12.252 – 9.6252) x 5000 1029 1029. +. 81. x 40. = 279 + 3.15 = 282.15 bbls. Exercise: If Class D cement at 16.4 ppg is used and each sack of cement yields 1.06 cubic feet, how many sacks will be required? Convert Volume to cu ft and divide by yield. =. 282.15 x 5.6146 1.06. © Randy Smith Training Solutions Ltd. =. 1494 sacks. July 2002.

(74) Having mixed all the cement how many pumps strokes will be required to displace cement into position if Pump output = .109 bbls/stroke ? This is the amount of strokes required to pump top plug into place using mud. Volume inside casing to float shoe. Strokes to bump top plug. =. (id)2 x (5000’ – 40’) 1029. =. (9)2 x 4960 1029. =. 390 bbls at .109 bbls/stk. =. 390 .109. =. 3582 strokes. The calculations required in the above examples were: Slurry Volume Number of sacks Pump Strokes to bump top plug The following example puts these all together with the addition of an earlier casing string set in the hole, and multiple 7” string. Example: Hole depth = 11250 ft. From Electric logs, the 81/2 hole was found to have 9.2” average diameter. 95/8” casing was set at 7.100ft using N80, 47.00 lbs/ft with 8.681” I.D.. © Randy Smith Training Solutions Ltd. July 2002.

(75) 7”, N80, 35.00 lbs/ft casing will be run from surface to 3000’ I.D. = 6.004” From 3000’ to 9000’ 7”, N80, 32 lbs/ft with 6.094” I.D. From 9000 to TD7”, N80, 29lbs/ft with 6.184 I.D. Float Collar is 60ft above Shoe Cement Yield per sack = 1.21 cubic feet Pump Output = .201 bbls/stroke Calculate:a) b) c). Slurry Volume required to bring cement to surface Number of sacks required Pump Strokes to bump top plug. Hint:- Always draw a diagram a) Slurry Volume = Casing Capacity x 80ft + Annular Cap. X 11250ft Annular capacity = open hole/casing annulua + casing/casing annulus = (8.6812 - 72) x 7100’ + (9.22 - 72) x 11250 - 7100 1029 1029 = .0256 =. x 7100’. 181.8. + .0346. x 4150. + 143.7. = 325.5 bbls. © Randy Smith Training Solutions Ltd. July 2002.

(76) =. 6.1842. x 60. Slurry Volume. =. 325.5. +. b). =. Cubic ft of slurry Yield/sack in cubic ft. =. 327.73. =. 1520 Sacks. Casing Capacity. =. .0372. =. 2.23 bbls. x 60. Sacks of Cement. x 1.21. 2.23 =. 327.73 bbls. 5.6146. Capacity of Casing =. (Cap of 7” N80) (35 lbs to 3000). +. (Cap of 7” N80) (32 lbs x 6000’). +. (Cap of 7” N80) (29 lbs x 2190). =. (6.0042) x 3000’ 1029. +. (6.0942) x 6000’ 1029. +. (6.1842) x 2190’ 1029. =. 105. +. 216.5. +. 81.4. =. 402.9 bbls. Strokes required = =. © Randy Smith Training Solutions Ltd. 402.9 .201 2004 Strokes. July 2002.

(77) Two Stage (Multiple) Cementing: In these jobs we need to calculate: a) b) c) d) e). Slurry for 1st stage Slurry for 2nd stage Strokes to bump top plug of 1st stage Strokes to bump closing plug of 2nd stage Pump strokes for mud between top plug 1st stage and opening plug of 2nd stage (if opening plug is displaced type).. Example: Two stage job using Displacement type Opening Plug for Stage Collar. TD = 10,000ft of 12 ¼” diameter hole. 95/8” Casing with 8.7555” ID from surface to T.D. Stage collar set at 5,000ft Float collar 50’ inside casing.. Calculate: a) b) c) d). Slurry volume for 1st stage Volume of mud to be pumped between top plug and opening plug Slurry column for 2nd stage Volume of mud to be pumped behind closing plug of 2nd stage. © Randy Smith Training Solutions Ltd. July 2002.

(78) a). Slurry volume = Annular Capacity x 5000’ + Casing Capacity x 50’ (12.252) - 9.625” x 5000 1029 =. 279. +. 1st stage Slurry Volume. b). = =. d). x 50. 3.7 =. 282.7 bbls. Volume equals capacity between Stage Collar and Float Collar.. Volume. c). (8.7552) 1029. 4950 x (8.7552) 1029 368.bbls. Slurry Volume for 2nd stage = Annular Capacity x 5000’. =. (12.25 2- 9.6252) 1029. =. 279 bbls. x. 5000’. Volume of mud behind closing plug. Casing Volume. © Randy Smith Training Solutions Ltd. =. 8.7552 x 5000’ 1029. =. 372 bbls. July 2002.

(79) Field Calculations would be further complicated by previous casing string and multiple casing string. Cement Plugs The sketch shows the situation that should exist after plug is pumped into position. Note the height of the cement in annulus equals height in pipe. Also same heights for water. This gives equal hydrostatic heads thus reducing contamination when pipe is pulled.. MUD. WATER. CEMENT. This is called the Balance Method. Example: Set a plug 500ft long in a 12 ¼” hole with open ended 5” Drill Pipe (4.276” ID) from 10,000’ – 9500’ Pump 8 bbls of water ahead of the plug. Pump output = .105 bbls/stroke. © Randy Smith Training Solutions Ltd. July 2002.

(80) We need to calculate Volume of Slurry Volume of water behind to balance the plug Number of strokes or volume of mud to displace water into position First: Calculate the number of barrels of slurry to fill hole for 500ft without pipe.. a). Volume. Second:. =. (12.252) 1029. =. 72.9 bbls. x 500. Calculate height of 8 bbls ahead of water in annulus.. Annular Volume. =. 12.252 – 52 1029. =. .1215 bbls/ft. Ht in Annulus. =. ____8____ .1215. =. 65 ft. Third:. Calculate barrels of water to fill 65ft in drill pipe. Drill Pipe Capacity =. 4.2762 1029. b). =. Barrels of water. © Randy Smith Training Solutions Ltd. =. 0.1776 bbls/ft. 65 x .01776 = 1.15bbls. July 2002.

(81) Fourth:. Calculate height filled by 72.9 bbls in Annulus and drill pipe.. .1215 bbls/ft .01776 bbls/ft. = =. Annulus Volume Drill Pipe Volume. =. 72.9_____ (.1215 + .01776). =. 72.9_ .13926. =. 523.5 ft. =. 65 ft. Height filled by 72.9 bbls. Ht. filled by 8bbls of water. Therefore depth of top water = =. 10,0000 – (523.5 + 65’) 9411.5 ft. Volume of mud to displace water after cement. Strokes. =. 167.15 .105. © Randy Smith Training Solutions Ltd. =. 9411.5 ft x .01776. =. 167.15 bbls. =. 1591 Strokes. July 2002.

(82) To recap: 1.. Calculate volume of slurry without pipe. 2.. Calculate height of water in Annulus. 3.. Calculate water to give same height in Drill Pipe. 4.. Calculate height of slurry with pipe. 5.. Add height of slurry with pipe. 6.. To height of water in Annulus. 7.. Subtract this value from base of cement plug. 8.. Multiply this value by pipe capacity to get mud volume to be pumped behind plug. © Randy Smith Training Solutions Ltd. July 2002.

(83) Drilling Calculations Course. Section 5: Pressure Control. This section covers a number of the more basic calculations as a preliminary to attending Pressure Control Schools.. © Randy Smith Training Solutions Ltd. July 2002.

(84) Pressure Control. Pressure Calculations What is Pressure? Pressure is the force acting on an area. By force, we mean weight and by area we mean square inches, square centimetres etc. Therefore, pressure is the force in pounds acting on one square inch, or the force in kilograms acting on one square centimetre. Pressure is most commonly measured in psi (pounds/square inch). If 10 pounds was resting on a plate 2 inches by 2 inches, what pressure would be acting on one square inch of plate? A 2” x 2” plate has an areas of 4 square inches, over 1 square inch 10 4. = 2.5 pounds/square inch. We mostly talk of pressure in relation to liquids, i.e. pump pressure, hydrostatic pressure. Hydrostatic pressure depends on depth. Any substance will exert more pressure if it is taller or deeper. It may not exert more overall weight, because this depends on the base area. A column of liquid 10ft high will exert more pressure than the same column 5ft high in fact, twice as much.. © Randy Smith Training Solutions Ltd. July 2002.

(85) Pressure resulting from a column of liquid. Pressure at any point is Directly Proportional to. Depth below the Surface.. By Depth it is meant Vertical Depth.. The pressure is the same at the bottom of the two columns although they have different measured depths.. © Randy Smith Training Solutions Ltd. July 2002.

(86) Why is a dam thicker at its base ?. Pressure is calculated by multiplying the density of a fluid by its depth.. Example: Water weighs 62.4 pounds/cubic foot. What pressure is exerted at a depth of 20 ft?. Pressure. =. Weight. x. =. 62.4 x. 20. =. 1248 pounds/sq ft. Depth. There are 144 square inches in a square foot, therefore: 1248 144. =. 8.67 psi. Using oilfield units of pounds per gallon, we must have a conversion factor to get psi values.. © Randy Smith Training Solutions Ltd. July 2002.

(87) Explanation: 1 cubic foot of water weight is 62.4 lbs.. A cubic foot of drilling mud of 10 pounds per gallon would weigh : 10. x. 7.4808. =. 74.808 pounds. (There are 7.4808 gallons to 1 cubic foot). If 1 cubic foot of 10ppg mud weighs 74.808 pounds. Then 1 cubic foot of 1ppg mud weighs 7.4808 pounds. This means that a 1ft cube of 1ppg mud exerts 7.4808 pounds on a square foot.. 12”. 12” 12”. © Randy Smith Training Solutions Ltd. July 2002.

(88) On 1 square inch it would exert. 7.4808 144 =. 0.52 psi/ft of depth. If the hole was 10,000 ft deep the pressure at the bottom would be 10,000 x .52. =. 5,200 psi. The formula can be written: Pressure (psi). =. Mud weight (ppg) x Depth (ft) x 0.052. Exercise: Calculate pressure of fluid: a) b) c). 10,000 ft of 8.5ppg mud 7,200 ft of 11.4ppg mud 14,280 of 10.7ppg mud. What if we are using Specific Gravity or pounds/cubic foot units ? following conversion factors are used:. Pressure (psi) Pressure (psi). = =. The. Mud weight (S.G.) x Depth (ft) x .433 Mud weight (pcf) x Depth (ft) x .007. Using the same mud weight, it can be seen that pressure will increase with depth.. © Randy Smith Training Solutions Ltd. July 2002.

(89) On the rig, one of the functions of a drilling fluid is to hold back formation fluids. These formation fluids will exert pressure according to their depth and density. This pressure, both for formation fluids and drilling fluids, is called Hydrostatic Pressure. When formation fluids exert a pressure that is a function of Depth and Density, they are said to be NORMAL. NORMAL formation fluid pressure is approximately .465 psi/ft that is the pressure exerted by a column of salt water of 100,000 parts per million salinity.. 0’. ft. 5000’. 1000. 2325 psi. 4560. ABNORMAL formation fluid pressure is when fluid exerts a pressure greater than .465 psi/ft. This occurs when fluid cannot escape the formation due to a seal forming, and as further overburden pressure is exerted at the surface, the fluids take up this weight equivalent.. © Randy Smith Training Solutions Ltd. July 2002.

(90) As long as the mud pressure is enough to balance these formation pressures, we can drill ahead safely. When we go under-balance, the conclusion is a kick,(flow of formation fluids into the well bore) or if uncontrolled, a blow-out. When the kick is taken, it becomes necessary to increase mud weight to balance the formation. How do we know the mud weight required to kill the kick ? When a kick takes place, formation fluids enter the wellbore or annulus because this is the line of least resistance. Our drill string is therefore full of 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) gauge will be equal to the imbalance between Mud hydrostatic in the pipe and formation fluid pressure. Convert this pressure to mud weight (ppg) and add to known mud weight in pipe. 700. 6279psi. 6979. © Randy Smith Training Solutions Ltd. July 2002.

(91) If Drill Pipe gauge reads 700 psi, mud weight is 10.5 ppg and depth is 11,500 ft. We can calculate mud weight required to kill the well. First, rearrange formula to get mud weight:. Pressure. =. mud weight x Depth x .052. Rearrange to mud weight. =. Pressure (SIDPP) Depth x .052. Remember to add on existing mud weight.. Kill mud weight. Kill Mud Weight. =. mud weight +. (SIDPP) (Depth x .052). =. 10.5. +. (700) 11,500 x .052. =. 10.5. +. (700) (598). =. 10.5. +. 1.17. =. 11.67 ppg. In the annulus, the kick fluid has contaminated the hydrostatic head of mud.. © Randy Smith Training Solutions Ltd. July 2002.

(92) The diagram below shows an influx of gas, exerting a hydrostatic pressure of .1 psi/ft gradient and extending 300 ft up inside the annulus. If 11.67 ppg mud will kill the well, then the formation pressure is: =. 11.67 x 11,500 x .052. =. 6978 psi. What casing pressure reading will be observed at surface?. Mud pressure. = = =. 10.5 x (11500 – 300) x .052 10.5 x 11200 x .052 6115 psi 700. Gas Pressure. = .1psi/ft x 300 = 30psi. Total Pressure of mud + gas in annulus = 6115 + 30 = 6145 psi. Difference. 833. 10.5 ppg. = 6978 - 6145 = 833 psi -11200’. Therefore, Casing pressure gauge will show 833 psi 11,500’. © Randy Smith Training Solutions Ltd. July 2002.

(93) Exercise:. 10,000 well – Gas influx in annulus is 300 ft high at .07 psi/ft Old Mud Weight 10.5 ppg. Find Kill Mud Weight and shut in Casing Pressure (SICP) for the following if: a) b) c). SIDPP SIDPP SIDPP. © Randy Smith Training Solutions Ltd. = = =. 650 psi 820 psi 300 psi. July 2002.

(94) Calculations for Circulating Heavy Mud. 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 the invading fluid. With the heavy mud ready to pump, we need to calculate: a) b) c). a). Initial pump pressure Pump pressure with heavy mud at bit When to adjust choke to get smooth transition between a) and b). Initial Pump Pressure: This is pressure 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. Initial Pump Pressure. b). =. 800 + 700. =. 1500 psi. Final Pump Pressure: This is the pressure required to circulate once heavy mud has reached bit. This calculation uses formula for Pressure v Mud Weight change.. © Randy Smith Training Solutions Ltd. July 2002.

(95) New Mud Wt New Pressure. =. Old Pressure x Old Mud Wt. With the heavy mud inside the drill string, the pump pressure required will be greater. As the heavy mud is pumped down, the hydrostatic pressure in the drill string increases until the heavy mud reaches the bit, at which point mud hydrostatic equals formation pressure. If the pump was stopped the SIDPP should equal zero. Therefore, the pump no longer has to overcome any pressure imbalance. The pressure required to circulate will be the pressure at a slow pump rate plus some extra due to the heavier mud. This can be expressed in the fomula:. Final Circulating Pressure = Slow Pump Pressure x. New mud wt Old mud wt. Example: Slow pump rate test gave 800 psi at 45 SPM with 10ppg mud. Kill mud weight. =. Final Circulating Pressure. 11.2 ppg. =. =. © Randy Smith Training Solutions Ltd. 11.2 800. x. 10. 896 psi. July 2002.

(96) c). 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 Circulation Pressure.. Example: T.D. is 10,000 ft. Initial Circulating Pressure Final Circulating Pressure = 700 psi 5” Drill Pipe, 4.276” I.D. 600 ft 8” x 3” Collars Pump Output, .2 bbls/strokes. =. 1200 psi. Calculate Pump Pressure every 100 strokes.. First:. Calculate capacity of Drill String in barrels =. Drill Pipe Capacity. =. 4.2762 1029. =. 167. =. 172.2 barrels. © Randy Smith Training Solutions Ltd. x. +. 9400’. Drill Collar Capacity. +. 32 1029 x 600’. +. 5.2. July 2002.

(97) Second:. Calculate number if strokes from the Surface to Bit:. Surface t bit strokes. Third:. =. = 861 strokes. 172.2 .2. Calculate Pressure change every 100 strokes Pressure drop. =. Initial C. Pressure - Final C. Pressure. =. 1200 - 700. =. 500 psi. Pressure must drop 500 psi in 861 strokes. Every stroke pressure drops. 500 861. Every 100 strokes pressure drops =. 500 861. x. 100. =. .58. x. 100. =. 58psi/100 strokes. With the table on the following page, the Choke Operator can make the necessary adjustments. A graph can be used in place of the table. © Randy Smith Training Solutions Ltd. July 2002.

(98) 0. Strokes. =. 1200 psi. 100. Strokes. =. 1142 psi. 200. Strokes. =. 1084 psi. 300. Strokes. =. 1026 psi. 400. Strokes. =. 968 psi. 500. Strokes. =. 970 psi. 600. Strokes. =. 852 psi. 700. Strokes. =. 794 psi. 800. Strokes. =. 736 psi. 861. Strokes. © Randy Smith Training Solutions Ltd. =. 700 psi. July 2002.

(99) Calculating the Effects of Gas Expansion. Any type of kick is dangerous, but some are more dangerous than others: Formation fluids can either be Gas, Oil or Water. Oil and Water are liquids, therefore volume is unaffected by pressure: with gas the greater the pressure, the greater the compression. One barrel of gas at the bottom of the well 10,000 ft deep with a mud weight of 9ppg will expand to 320 bbls at atmospheric pressure. Gas behaviour under pressure is defined mathematically in “Boyles Law”.. “If the temperature of a gas is kept constant, then the volume will be inversely proportional to the pressure”.. Boyles Law states:. This means, if the pressure is reduced by one half, then the volume will double. Boyles Law is expressed:. V1 V2. V1 V2 P1 P2. =. P2 P1. = = = =. © Randy Smith Training Solutions Ltd. or. V1P1. =. P2 V2. Original Volume New Volume Original Pressure New Pressure. July 2002.

(100) Example A gas invasion of 15 barrels is taken at 8500 ft. The bottom hole pressure is 4,500 psi. What will be the gas volume at the Casing Shoe set at 5,000 ft if mud weight is 10 ppg.. V1 V2. =. P2 P1. Solve the equation to find V2.. V2. =. V2. =. V1 x P1 P2. 15 x 4500 (10 x 5000 x .052). =. 67500 2600. =. 26 barrels. © Randy Smith Training Solutions Ltd. July 2002.

(101) M.A.A.S.P Calculations M.A.A.S.P is the Maximum Allowable Annular Surface Pressure, which should be read as the maximum pressure gauge, before something breaks down. 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: a) b) c). Casing B.O.P’s Formation below the casing. Most often the formation below the Shoe is the weakest point. An excess of pressure would cause the formation to fracture with a resultant loss of mud. To find the fracture point a Leak-off Test is run after drilling out the shoe. With the rams closed, a small amount of mud is pumped into the well, after a short wait, the process is repeated. By plotting volume pumped against Pump Pressure, a straight line will not rise, but level off. This is when the formation is taking mud: The pressure at this point is the Leak Off Pressure. 5 4 3. 2 1. 500. 1000 psi. © Randy Smith Training Solutions Ltd. July 2002.

(102) The Leak Off Pressure can then be used to calculate Pressure.. Formation Fracture. Formation Fracture Pressure = Leak off Pressure + Mud Hydrostatic Pressure. Example: Shoe Depth 5000’ Leak off Pressure =. 1500 psi. Mud Weight. 9.5 ppg. =. Fracture Pressure = = =. 1500 + (9.5 x 5000’ x .052) 1500 + 2470 3970 psi. Therefore, with a Mud Weight of 9.5 ppg, the maximum surface pressure allowed (MAASP) is 1500 psi. When this value is reached, the pressure at the shoe is equal to the Formation Fracture Pressure. If Mud Weight is changed when drilling ahead, the MAASP will change. The following formula can be used:. MAASP. =. Shoe Depth x (Frac. Gradient – Mud Gradient). Example: Shoe Depth 400 ft, mud weight 10.5 ppg Leak off pressure was 1400 psi with 10ppg mud in the hole. © Randy Smith Training Solutions Ltd. July 2002.

(103) Calculate Formation Fracture pressure and convert to gradient.. First:. Frac. Pressure. =. 1400 + (10 x 6400’ x .052). =. 4728 psi Frac. Pressure Shoe Depth. Convert to gradient:. Gradient. Second:. 4728 6400. =. .74 psi/ft. Calculate Mud Gradient of mud in the hole. Mud Gradient (psi/ft). Third:. =. =. 10.5 x .052 x 1ft. =. .546 psi/ft. Apply gradients and Shoe Depth to formula for MAASP MAASP. © Randy Smith Training Solutions Ltd. =. 6400 x (.74 - .546). =. 6400 x .193. =. 1235 psi July 2002.

(104) By increasing the Mud Weight from 10ppg (when test was taken) to 10.5 deeper down, the MAASP had dropped from 1400 psi to 1235 psi. In most cases, a safety factor is used to allow for errors when operating the choke. The safety factor is applied to the formation fracture gradient.. Example: Shoe Depth Mud Weight Leak off Pressure. 7200 ft 11.5 ppg 1200 psi. Calculate MAASP if 90% of formation fracture gradient is used.. Formation Fracture Pressure. =. 1200 + 5505 psi. Formation Fracture Gradient. =. 5505 7200. =. .765 psi/ft. Mud Gradient. =. 11.5 x .052. =. .598 psi/ft. Fraction Gradient @ 90%. =. .765 x .9. =. .6885 psi/ft. © Randy Smith Training Solutions Ltd. (11.5 x 7200 x .052). July 2002.

(105) MAASP. =. Shoe Depth x (Fracture Gradient - Mud Gradient). =. 7200. x ( .6885 - .598). =. 7200. x. =. 651 psi. .905. If mud weight was increased then MAASP would decrease.. © Randy Smith Training Solutions Ltd. July 2002.

(106) Drilling Calculations Course. Section 6: Hoisting Calculations. This section covers the basic theory and calculations behind Lifting Machines; Wire Rope Design factors and Ton Mile accumulations.. © Randy Smith Training Solutions Ltd. July 2002.

(107) Hoisting Calculations. Hoisting Systems There comes a point where an object cannot be manhandled, usually due to weight, size or distance to be moved. Here we need a human energy saving device, commonly called a machine. A machine is normally any device that can be used to gain some kind of advantage. The amount of advantage is called Mechanical Advantage.. Mechanical advantage. =. Weight of Load moved Effort used to move load. A force of 50 lbs is used to lever a stone slab weighing 200 lbs.. The advantage would be:. 200 50. =. 4. Lifting systems can be categorized into 4 main types: 1. 2. 3. 4.. © Randy Smith Training Solutions Ltd. Levers Wheels and axles Inclined Planes Pulleys. July 2002.

(108) Pulleys are used to lift heavy loads vertically. A load of 500 lbs can be lifted using a 4-line pulley with:. 500 4. =. 125 lbs Pull. To calculate Pull required, divide Load number by the lines strung in Derrick.. Pull. =. load x co-efficient of Friction Number of lines. The fast line having an accumulation of friction losses has the greatest tension of all lines strung. The more lines strung, the greater the co-efficient of friction. Below is a table of constants that can be applied to the formula. Fast Line Tension =. Weight of Load. x. Constant. Fast Line Constants No. of lines strung 4 6 8 10 12 14 © Randy Smith Training Solutions Ltd. Constant .271 .1882 .1469 .1224 .1062 .0948 July 2002.

(109) Example: Hook load is 280,000 lbs. Blocks are strung with 10 lines. Calculate Fast Line Load tension. Fast Line Load. = = =. Weight of Load 280,000 34,272 lbs. x x. Constant .1224. This value is used in calculating the Design Factor of the system. Design Factor is the ratio of Nominal Wire Rope Breaking Strength to the Fast Line Load.. Design Factor. =. Nominal Rope Breaking Strength Fast Line Load. 1-3/8 Improved Plow Steel Drilling line has a rated strength of 167,000 pounds.. The recommended minimum design factor is 3. Therefore, with 1-3/8 line, we must not have a fast line load of more than 167,000 3. =. 55,666 lbs. if Fast Load is 55,666 lbs. With 10 lines strung up, what is Hook Load? Fast Line Load. © Randy Smith Training Solutions Ltd. =. Weight of Load. x. Constant. July 2002.

(110) Solve equation to:. Weight of Load. =. Fast Line Load Constant. Constant from Table. =. .1224. Weight of Load. =. 55,666 .1224. =. 454,790 lbs. Therefore, the Hook Load must not go above 454,799 pounds.. These calculations show us how loads to string ups can be evaluated.. A light load with a 10 or 12 line string up gives high Design Factors. For instance, a Hook Load of 160,000 lbs using 12 lines gives a Design Factor of 9.9 This means that it will take a long time to run up the Ton-miles to cut-off. Field experience confirms that the slow accumulation of ton miles will wear out the wire due to the higher number of bending cycles.. © Randy Smith Training Solutions Ltd. July 2002.

(111) Example: When making a Connection the string gets stuck. The blocks are strung with 8 lines of 1-3/8 Improved Plow Steel Wire Rope (breaking strength of 167,000 pounds). A Design Factor of 3.5 is used. After working pipe, the String is calculated to be stuck at 10,280 ft. (5”, 19.5 lbs/ft pipe is being used). Calculate the maximum over-pull that can be used.. Design Factor. =. Nominal Breaking Strength Fast Line Load. Solve the equation to get Fast Line load.. Fast Line load. © Randy Smith Training Solutions Ltd. =. Nominal Breaking Strength Design Factor. =. 167,000 3.5. =. 47,714 pounds. July 2002.

(112) Hook Load with the Fast Line Load of 47,714 pounds using 8 lines.. Hook Load. =. =. Fast Line Load Constant 47,714 .1469. =. Therefore, Maximum Hook Load. Weight of String. Maximum Overpull. 324,800 pounds. =. 324,800 pounds. =. Length. x. Weight/ft. =. 10,280. x. 19.5. =. 200,460 lbs. =. 324,800. -. 200,460. =. 124,340 pounds. If 5”, 19.5lb/ft drill is Grade E, Premium can this pull be made safely?. © Randy Smith Training Solutions Ltd. July 2002.

(113) The minimum Tensile Strength of Grade E 5” pipe is 311,400 pounds. Therefore, this pull cannot be made.. Max Overpull. =. 311,400. -. =. 110,940 pounds. 200,460. Wear on the line has to be monitored and measured , in addition to visual checks a record of use is kept . The unit of measurement is the ton mile. Ton-Mile calculations. -. What is a Ton-mile ?. A Ton-Mile of work is said to be done when we pull 1 ton for 1 mile. When we pull 1 ton of pipe out of a hole 1 mile deep, that 1 ton is getting less, the more pipe pulled. On average, we only pull half a ton. Therefore, we have done ½ ton-mile of work.. Example: Drill Collars : Drill Pipe: Block and Hook weigh:. 900 ft long weigh 100,000 pounds in mud 14,100 ft Long Weigh 250,000 pounds in mud 45,000 lbs. Calculate ton-miles to pull out of hole.. (1 short ton =. © Randy Smith Training Solutions Ltd. 2,000 lbs. 1 mile =. 5280 ft). July 2002.

(114) First:. Calculate ton-miles for drill pipe. Wt of pipe. 250,000 2,000. =. 125 tons. Distance moved is:. 14,100’ 5,280. =. 2.67 miles. Pulling out we have an average of:. 62.5 tons pulled 2.67 miles. Second:. 125 2. =. 62.5 tons. =. 62.5 x 2.67. =. 166.8 Ton-Miles. Calculate Ton Miles for Drill Collars. The collars are pulled 14,100’ before they reach the surface. This is 2.67 miles.. The weight is:. 100,000 2,000. =. 50 tons. Therefore, 50 tons are pulled 2.67 miles. © Randy Smith Training Solutions Ltd. =. 50 x 2.67. =. 133.5 Ton-miles July 2002.

(115) Then pulling 900ft of collars out, we pull the average of 50 2. =. 25 tons. 900 ft. =. .17 miles. Therefore, 25 tons are pulled .17 miles. Ton mile for Drilling String. Third:. Block weighs. =. 25 x .17. =. 4.25 Ton-miles. =. 166.8 + 133.5 + 4.25. =. 304.55 Ton-miles. Calculate Ton-Miles for Blocks. 45,000 2,000. =. 27.5 Tons. Distance traveled is 15,000’ up and 15,000’ down 30,000’ 5,280. © Randy Smith Training Solutions Ltd. =. 5.68 miles. July 2002.

(116) Therefore, 27.5 tons for 5.68 miles miles. =. 27.5 x 5.68. =. 156.2 Ton-. Total Ton Miles to pull out of the hole =. 304.55. +. =. 460.7 Ton-Miles. 156.2. Exercise: Hole depth 11,000 ft. Drill pipe 5”, 19.5 lbs/ft. Mud Weight 11ppg. 800 ft of Drill Collars at 147 lbs/ft. Travelling Block eight 40,000 lbs. 1 mile 1 short ton. = =. 5,280 ft 2,000 pounds. Calculate Ton-Miles for a complete Round Trip?. Ton Miles for Drill Pipe pulling out:. Weight of pipe. =. Buoyancy factor. 10,200 x 19.5. =. 198,900 x Buoyancy Factor. =. .8328 165,652lbs. © Randy Smith Training Solutions Ltd. July 2002.

(117) Wt in Mud. =. Average weight. 198,900 x .8328 = 82.38 2 =. 41.41 Tons moving 1,93 Miles. 2,000 pounds. =. 82.83 Tons. 41.41 Tons. =. 41.41 x 1.93. =. 79.9 Ton-Miles. Ton miles for drill collars to reach surface:. Wt of Drill Collars in mud. =. 800 x 147 x .8328 2,000. =. 48.9 Tons moving 1.93 miles. =. 48.9 x 1.93. 94.5 Ton-Miles. =. 48.9 Tons. Ton Miles for Collars to be removed:. Average weight. =. Distance pulled. =. 24.5 tons moving .15 miles © Randy Smith Training Solutions Ltd. 48.9 Tons 2. 800 5,280. =. 24.5 Tons. =. .15 miles. =. 3.7 Ton Miles July 2002.

(118) Ton Miles for Blocks:. Travel twice 11,000. =. 22,000’ 5,280’. =. 4.16 miles. Weight. =. 40,000 2,000. =. 20 Tons. 20 Tons moving 4.16 miles. =. 8325 Ton Miles. Total Ton Miles to pull out. =. 83.2 + 3.7 + 94.5 + 79.9. =. 261.3 Ton Miles. =. 522.6 Ton Miles. For running in hole, is the same again. Total Round Trip. © Randy Smith Training Solutions Ltd. =. 2 x 261.3. July 2002.

(119) Drilling Calculations Course. Section 7: Buoyancy Effects. This section covers the calculations used to measure string weight when immersed in mud and the number of Collars required to give selected Weight on Bit.. © Randy Smith Training Solutions Ltd. July 2002.

(120) BUOYANCY. Archimedes first made scientific observations of Buoyancy. He stated that a body 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 and 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 10ppg 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 steel has 7.9 times the weight of an equal volume of water. To convert mud weight in ppg to specific gravity devide by 8.33. Fresh water has a specific gravity of 1 and a weight in ppg of 8.33 , Therefore 10 ppg mud has a specific gravity of 10/ 8.33 = 1.2. © Randy Smith Training Solutions Ltd. July 2002.

(121) Apply the values to formula to get the Buoyancy Factor.. Buoyancy Factor. = 1-. Mud Weight ppg ÷ 8.33 Specific Gravity of Steel. Example: If mud weight is 10 ppg calculate Buoyancy Factor. Buoyancy Factor. =. 1-. 10 ÷ 8.33 7.9. =. 1-. 1.2 7.9. =. 1-. .1519. =. .848. To find Hook Load in mud, first calculate dry weight, then multiply dry weight by Buoyancy Factor.. © Randy Smith Training Solutions Ltd. July 2002.

(122) Example Calculated The Immersed Weight of 10,000 ft of 5”, 19.5 pounds/ft drill pipe Buoyancy Factor. =. .848. Immersed Weight. =. (10,000 x 19.5) x .848. =. 195,000 x .848. =. 165,360 pounds. Buoyancy factor tables are found in most rig handbooks, but keep a copy of the formula in your notebooks just in case. The 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. First: Calculate the Buoyancy Factor =. 1 -. =. 1 -. =. © Randy Smith Training Solutions Ltd. 11.5 ÷ 8.33 7.9 .1747 .825. July 2002.

(123) Second: Calculate the immersed weight per ft of drill collar =. 112 x .825. =. 92.4 lbs/ft. Third: Divide 50,000lbs by 92.4 lbs/ft to get length of collar string. = 50,000 92.4. =. 541 ft. Fourth: Divide by collar string by 30’ lengths to get the number required =. © Randy Smith Training Solutions Ltd. 541 30. =. 18 collars. July 2002.

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