Drilling Calculations CD Complete Course

Drilling Calculations

Course

Randy Smith Training Solutions Ltd July 2002

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



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.


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


IMPERIAL SYSTEM LENGTH: inches, feet, yards and miles are the most common 12 inches = 1 foot 3 feet = 1 yard 1760 yards = 1 mile 5280 feet = 1 mile Exercise: 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 = 79 yards 0 feet 3

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.


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


VOLUME: The same units as length, but prefixed by cubic Cubic inches, cubic feet etc 1728 cubic inches = 1 cubic foot 27 cubic feet = 1 cubic yard MASS: The common term for Mass is WEIGHT. Weight is measured in ounces, pounds, hundred weights and tons. 16 ounces = 1 pound 112 pounds = 1 hundred weight 20 hundred weight = 1 ton 2240 pounds = 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).


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: Convert 8.34 ppg to pcf: 1. First, how many us gallons in a cubic foot? = 7.4809 U.S. gal / cubic ft 2. Second, multiply 8.34 x 7.4809 = 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.


20 tons per sq ft = (20 x 2240) pounds per 144 sq inches 44800 pds per 144 sq inches

44800 = 311 lbs per sq in 144 pounds per sq in

To make conversion easier, a table of Units and Conversion Factors is included at the end of Section 1.


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 = 39.37 inches 1000 millimetres = 1 metre (milli = one thousandth) 100 centimetres = 1 metre (centi = one hundredth) 1000 metres = 1 kilometre (Kilo = a thousand times) To use the Metric system, and understanding of DECIMAL places is essential.

1 in decimal = 1.0 10 in decimal = 10.0 1/10 in decimal = 0.1 1/100 in decimal = 0.01 1/1000 in decimal = 0.001

Exercise: 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.


Exercise: Express 0.00328 in words or as a fraction. 1st = tenth 2nd = hundredth 3rd = thousandth 4th = ten thousandth 5th = 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.


DECIMAL POINT MOVEMENT: 1 place to the right = one tenth = 0.1 2 places to the right = one hundredth = 0.01 3 places to the right = one thousandth = 0.001 4 places to the right = one ten-thousandth = 0.0001 5 places to the right = one hundred thousandth = 0.00001 6 places to the right = one millionth = 0.000001 AREA: 1 sq metre = 100cm x 100 cm = 10,000 sq cms 1 sq metre = 1000mm x 1000mm = 1,000,000 sq mm 1 sq kilometre = 1000m x 1000m = 1,000,000 sq m 1 hectare = 100m x 100m = 10,000 sq m

cm = centimeter mm = millimeter m = metre

MASS: (Weight) The gram is the basic metric unit of weight

1000 grams = 1 kilogram 1000 milligrams = 1 gram 1000 kilograms = 1 metric ton


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 = 100 x 100 x 100 = 1,000,000 cubic cms 1000cc = 1 litre 1000 litres = 1 cubic metre 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


COMMON SYMBOLS AND ABBREVIATIONS Inches = ins or Feet = ft or Cubic inches = cu ins or ins3

Cubic feet = cu ft or ft3 Square inches = sq. ins or ins2 Square feet = sq. ft or ft2 Pounds = lbs Ounces = oz Pounds per cubic foot = pcf or lbs/ft3 Pounds per gallon = ppg or lbs/gall Pounds per square inch = P.S.I. Millimetres = mm Centimetres = cm Metres = m Square metres = m2

Cubic centimetres = cc or cms3 Kilometre = km Grams = gm Kilograms per sq centimetre = kg/cm2 Barrel = bbl


COMMON SYMBOLS AND ABBREVIATIONS + = Plus 2 + 6 = 8 - = minus 7 2 = 5 x = multiplied by 3 x 4 = 12 = divided by 10/2 = 5 > = greater than 6 > 5 < = less than 5 < 6 = plus or minus 60% 1% : = the ratio 1 : 4 = therefore A + B = C A = C - B 42 = square of 4 4 x 4 = 16 = square root 4 = 2 11 = parallel to = perpendicular = triangle = square = pi % = percent 320 = degree 4 = inches 4 = feet a-2 = negative exponent 3 = cube root 364 = 4


UNITS AND CONVERSION FACTORS Multiply by to obtain To obtain by Divide DEPTH/LENGTH: cm 0.39370 in 0.3281 ft

0.01 m ____________________________

in 25.40005 mm 2.54000 cm

0.08333 ft ____________________________

ft 30.48006 cm 12.0 in 0.30480 m ____________________________

m 100.0 cm 39.370 in 3.2808 ft 1.936 yd

____________________________ km 3.280.83 ft

1.000 m 0.62137 mi

____________________________ mi 5,280.0 ft 1,609.34 m 1,609.34 km ____________________________


UNITS AND CONVERSION FACTORS Multiply by to obtain To obtain by Divide AREA: cm2 0.15499 in2 ________________________________ in2 6.4516 cm2 ________________________________ ft2 929.0341 cm2 0.092903 m2 ________________________________ m2 1,549.9969 in2 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 _________________________________

VOLUME/CAPACITY: cm3 1,000.00 mm3 0.01 litre 0.6102 in3 0.0002642 gal 0.00003531 ft3


UNITS AND CONVERSION FACTORS Multiply by to obtain To obtain by Divide VOLUME/CAPCITY (cont) in3 16.38716 cm3

0.4329 gal 0.1638 litre 0.5787 ft3 _______________________

litre 1,000.0 cm3 1,000 ml

61.2705 in3 1.57 qt 0.26417 gal (U.S.) 0.3531 ft3 _______________________

gal (U.S.) 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) _______________________

gal (imp) 1.20095 gal (U.S.) _______________________

bbl (U.S.) 158.984 litre 42.0 gal (U.S.) 5.61458 ft3 0.9997 bbl (imp)


UNITS AND CONVERSION FACTORS Multiply by to obtain To obtain by Divide VOLUME/CAPCITY (cont) bbl (imp) 159.031 litre

42.112 gal (U.S.) ________________________________

ft3 1,728.0 in3 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 ___________________________________


UNITS AND CONVERSION FACTORS

Multiply by to obtain To obtain by Divide

DENSITY/CONCENTRATION (cont) lb/gal (U.S.) 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.) _____________________________________ WEIGHT/MASS 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




WEIGHT/MASS ton (short) 2.000 lb (cont) ton metric 0.90718 ton (metric __________________________________ ton (long) 2.240.0 lb

1.12 ton (short) 1.1605 ton (metric)

__________________________________ MUD WEIGHT PPG x 119.8 Kgm3 Kg/m3 x 0.00835 lbs per gallon __________________________________ MUD WEIGHT PPG x 0.052 psi/ft To PRESSURE Pressure Gradient GRADIENT SG x .433 psi/ft b/ft3 144 psi/ft Kg/m3 x 0.000434 Or 2303 psi/ft Kg/m3 x 0.00982 K/Pa/m __________________________________ ANNULAR VELOCITY Ft/min x 0.3048 m/min M/min x 3.2808 ft/min __________________________________ FLOW RATE Gal/min x 0.003785 m3/m Barrels/min x 0.159 m3/m M3/min x 6.2905 bbl/min M3/min x 264.2 gal/min __________________________________




RESISTIVITY ohms/cm2cm 0.01 ohms m2 m __________________________________

Ohms/m2m 100. Ohms m2m __________________________________ PRESSURE 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

__________________________________ kg/cm2 14.22333 psi

0.980665 bar 0.967842 atm

__________________________________ bar 106 dynes/cm2

14.5038 psi 1.1972 kg/cm 0.98624 atm

__________________________________




TEMPERATURE: 0F = 1.8 oC + 32 0K = 0C + 273 0C = 5/9 (oF 32) 0R = 0F + 460 0Fahrenheit 0Rankine 0Centigrade 0Kelvin 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.



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.


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: 2 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. 0.2846


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.


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 = 8.326 4 decimal places = 8.3257 5 decimal places = 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 = 9.3824 9.221134 = 9.2211 9.18796 = 9.188 9.25256 = 9.2526


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.


Areas The use of area is found in any places around the rig. Force o

Area of

Surfac Area is expressed as a squsquare mile, etc. A square inch is the area t There are 3 common shape A shape with 4 sides, each

Area = L

4 5 x 4 =

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deck space

e area of pits

are a square inch, square centimetre, square foot,

aken up by a square, of 1-inch long sides.

s that can easily have their areas calculated.

side at 90 to the other Rectangle

ength x Breadth

5

20 sq. ins

July 2002

6

6 x 3 = 18 sq. ins

3


A shape with 3 sides, angles between each side are variable Triangle Area = Base x vertical height

height

base


height

Area = base x height

A shape with 4 sides, none of the angles are 90 - Trapezium Area = Sum of Parallel Sizes x distance between them.

a

ht (a+b) x ht

b

C 3 parts

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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 = 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 (). To calculate Circumference using diameter, multiply Diameter by

Circumference = x diameter Or Circumference = 2 x radius


To find the formula for calculating area we can divide the circle into slices like a cake.

Circumference = 2 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.

The base = circumference = D or 2 r

Radius

2 r

Each triangle has an area of 1/2ht x base.


Height = radius or Diameter

2 Radius

Height = 2

Base = 2 r 32

To calculate for 32 triangles = 32 x r x 2 r

2 32 32 cancels out. = r x 2 r

2 32 2 cancels out. = r x r = r2


If using Diameter = 32 x D x D 4 32

= D x D 4

= D2

4 Area of Circle = r2 or D2 4 Exercise:

Find Area of Circles with the following:

a) Diameter = 12 b) Diameter = 7 c) Radius = 4 d) Diameter = 7 e) Radius = 3 f) Circumference = 24


To aid calculation, remember = .7854 4 Therefore, Area = .7854 x D2

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 = diameter of large circle d = diameter of small circle Annular Area = D2 d2 4 - 4


Because 4 is common to both the above formula can be rewritten: (D2 - d2) 4 or .7854 (D2 - d2) Example: Find Annular Area when D = 10 and d = 5

Area = 4 (102 52) = .7854 (100 25) = .7854 x 75 = 58.9 sq. inches


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


find a: a b = c Move b across and change to + a = c + b In multiplication the technique is different. Values are moved diagonally.

a Find a: b = c

Move b diagonally across = sign

a = c x b a

Find b: b = c Move b up to c and c to b


a

a = b x c c = b

Find b: a = (c + d)

b a = b (c + d) a = b

(c + d) If a = 10 c = 3 d = 2 What is b: 10 = b (3 + 2)

10 = b 5

2 = b

Pressure = Depth x Mud Weight x 0.052 Solve the Equation to find a) Depth b) Mud Weight


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 = D2 4 Find D: Area x 4 to eliminate the square you must square root the other side.

D = Area x 4 Square roots are commonly found on calculators today. The square root of 4 is 2 (2 x 2 = 4) The square root of 64 is 8 (8 x 8 = 64)


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.


Example: Find capacity in c ic inches of a tank 1 2 x

1 2 = 14 3 6 = 42

Capacity = 14 x 8 x 42 = 4704 cubic inches We have assumed vertical walls. If the tanvolume calculations would be used.

30

Plan View

20

10

50

Randy Smith Training Solutions Ltd 8 x 3 6 ubk had sloping walls the following

View

10

Side

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The area of side A can be found using the formula for a trapezium.

Area = Sum of Parallel sides x half distance between them 10

Area = (50 + 30) x 2 = 400 sq. inches Then calculate capacity as:

Area x sum of parallel sides on wall B 2 20 + 10

= 400 x 2

= 400 x 30 2

= 400 x 15

= 6000 sq. inches


Calculating volumes of Cylinders or the Annulus the formula is:

Area x height D2

Volume = 4 x height

Make sure all units are the same Annular Volume 4(D2 d2) x height

4 = .7854

Example: Find a volume of cylinder in cubic feet/foot of depth if diameter is 10 .7854 (102)

Volume = 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).


Example: Calculate Annular Volume if D = 10 d = 6 depth = 1ft .7854 (102 62) x 1

Volume = 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: .7854 (D2)

Volume in barrels/ft = 144 x 5.6146 x 1 Calculating out .7854, 144 and 5.6146 we can simplify the formula to D2

Volume in bbls/ft = 1029

Or

(D2 d2) 1029


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% = 200 100 = 2 logs

10% = 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 = x 35 = 17.5%


Example 3

If 42 logs = 75% of the total, how many logs are there?

42 logs = 75%

1% = 42 75

100% = 42 75 x 100 = 56 logs 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 = R X B

P = Percentage:-the actual value equaling chosen % R = Rate in decimals:- the part of a 100 to be found ie in 4% of

50, 4% is rate. B = Base:- the number of which some percentage is to be found.


Example 1 (Repeat) What is 10% of 200 logs? The question asks you to find a number that equals a %, being 10% here.

P = R x B 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

10% of 200 logs = 20 logs Example 2 (Repeat) What % is 35 logs of 200 logs? The question asks for an actual percentage. This being the Rate

P = R x B (R is unknown)


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 = R x B Percentage is number of logs = 42


Rate is parts of 100 to be found in decimals

= 75% 100 = .75 42 = .75 x B

Base is equal to total or 100%

Base = 40 .75

= 56 logs Examples What is 42% of 381?

P = R x B P = .42 x 381

= 160.02 What % of 281 is 48

P = R x B 48 = R x 281 R = 48

281 = .1708 x 100 = 17.08%


225 is 15% of what?

P = R x B 225 = .15 x B B = 225 =

.15 To remember formula use following diagr P P

R

P = R x B R = P

B

B = P R

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am

R B

July 2002


Section 3: Fluid Circulation Calculations

This section covers the most commonly used calculations involved with Fluid Hydraulics used by Drill Crews.


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 = large diameter (inside diameter of the hole) d = smaller diameter (outside diameter of the string)

Volumes use:(non annular) 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 inside diameter with 5 drill pipe.


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 = 972.3 x 42 = 40,836.6 gallons


Example: 9 Casing set at 9000 ft. ID = 8.84 8 Open hole to 11,000 ft 5 Drill pipe 19.5 lbs/ft ID = 4.276 600ft of 6 x 2 Drill collars Calculate a) annular volume in bbls, cu ft and gallons

b) 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) x 600 1029 = 21.14 bbls Vol of OH to Pipe annulus = (8.52 - 52) x 1400 1029 = 64.28 bbls Total = 64.28 + 21.14 + 464.83 = 5520.25 bbls


In Gallons = 550.2542 = 23110.5 galls In cubic ft = 550.25 x 5.6146 = 3089.4 cubic ft b) Capacity of drill string = cap. of pipe + cap. of collars = (id) 2 x length (id) 2 x length 1029 1029

(4.2762) 2 (2.52) 2 = 1029 x 10,400 + 1029 x 600 = 184.79 + 3.64 = 188.4 bbls

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) Fill hole after 10 stands of drill pipe have been pulled b) Fill hole after each stand of collars is pulled


c) Total mud required to keep hole full when pulling out d) Quantity of mud displaced running in the hole with extra 300 of 8

x 2 collars

a) Volume of steel in 10 stands of drill pipe

= (52 - 4.2762) x 900 1029 = 0.065 x 900 = 5.87 barrels

b) Volume of steel in 1 stand for drill collars

= (92 - 32) x 90 1029 = .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 x 9400 + .06997 x 600 = 61.1 + 41.98 = 103 barrels


d) Running in the hole

= drill pipe drill collars new drill collar (disp/ft x 9100) + (disp/ft x 600) + (disp/ft x 300) = (.0065 x 9100) + (.06997 x 600) + (82 - 2.752) x 300 1029 = 59.15 + 41.98 + 0.548 x 300 = 101.13 + 16.46 = 117.58 barrels

100 COLLARS

B B L S

50 PIPE I I

50 STANDS 100


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 D2 Volume of Cylinder = 4 x length Volume of 3 cylinders = 3 ( D2 x length) 4 ( = .7854) 4 = 3 (.7854 (62) x 12) = 3 (.7854 x 36 x 12) = 3 x 339.29 = 1017.88 cubic inches


Convert to barrels = 1017.88 1728 x 5.6146 1728 cubic inches = 1 cubic ft 5.6146 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


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:


Pressure inside Kelly = 2950 Pressure inside near bit sub = 2200 Pressure in drill collar annulus = 200 Pressure at flow line = 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. Surface Lines 2. Drill String 3. Drill Bit 4. Annulus

P S I 1 2 3 4


The horsepower required to circulate a known quantity of mud at a certain pressure can be calculated using the formula. P x V Hydraulic Horsepower (HHP) 1714 P = Pump Pressure (psi) V = 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 P x V HHP = 1714 2800 x 350 = 1714 = 572 HHP


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 Example: Pressure at 80 SPM = 3200 psi What is pressure at 70 SPM (70)2 New Pressure = 3200 x (80)2 = 3200 x .8752 = 2450 psi


Changing Mud Weight will affect pump pressure in the following way: New Weight New pump pressure = Old pump pressure x 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 = Old Mud Weight W2 = New Mud Weight


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 15s 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 15s 15 = 15/32 nds of an inch, diameter D2 Area = 2 x 4 = .7854 x (15/32) 2 x 2 Convert fraction to decimal = .7854 x ( .4687)2 x 2 = .1725 sq inches x 2 = .345 sq inches


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 = .3137 square inches Three nozzles = .3137 One nozzle = .10456 sq inches approximately Area = .7854 x D2 Area Solve the equation to get D i.e. D2 = .7854


.10456 D = .7854 = .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 11s or 12s. 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.



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.


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 = D 2 (Answer in cubic inches, ft etc) 4 OR D 2 (Answer in barrels per ft) 1029 Capacity of Annular Space = (D2 d2) 4 OR (D2 d2) 1029 Number of strokes required to pump = Volume______ Pump Output/Stroke


Single Stage Job Given 12- diameter hole from surface to 5000 ft Casing O.D. 9- 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 = Annular + Volume of x 40ft Slurry Volume Casing

= (12 2 - 9-2) x 5000 + (92) x 40 1029 = (12.252 9.6252) x 5000 + 81 x 40 1029 1029 = 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 = 1494 sacks 1.06


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 = (id)2 x (5000 40) 1029 = (9)2 x 4960 1029 = 390 bbls at .109 bbls/stk Strokes to bump top plug = 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.


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) Slurry Volume required to bring cement to surface b) Number of sacks required c) 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 + (9.2 - 72)2 x 11250 - 7100

1029 1029 = .0256 x 7100 + .0346 x 4150 = 181.8 + 143.7 = 325.5 bbls


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


Two Stage (Multiple) Cementing: In these jobs we need to calculate: a) Slurry for 1st stage b) Slurry for 2nd stage c) Strokes to bump top plug of 1st stage d) Strokes to bump closing plug of 2nd stage e) 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 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) Slurry volume for 1st stage b) Volume of mud to be pumped between top plug and opening plug c) Slurry column for 2nd stage d) Volume of mud to be pumped behind closing plug of 2nd stage


a) Slurry volume = Annular Capacity x 5000 + Casing Capacity x 50

(12.252) - 9.625 x 5000 (8.7552) x 50 1029 1029 = 279 + 3.7 1st stage Slurry Volume = 282.7 bbls

b) Volume equals capacity between Stage Collar and Float Collar.

Volume = 4950 x (8.7552) 1029 = 368.bbls

c) Slurry Volume for 2nd stage = Annular Capacity x 5000 (12.25 2- 9.6252) x 5000 = 1029 = 279 bbls d) Volume of mud behind closing plug. Casing Volume = 8.7552 x 5000 1029 = 372 bbls


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 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


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 = (12.252) x 500 1029 = 72.9 bbls Second: Calculate height of 8 bbls ahead of water in annulus. Annular Volume = 12.252 52 = .1215 bbls/ft 1029 Ht in Annulus = ____8____ = 65 ft .1215 Third: Calculate barrels of water to fill 65ft in drill pipe Drill Pipe Capacity = 4.2762 = 0.1776 bbls/ft

1029 b) Barrels of water = 65 x .01776 = 1.15bbls


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

.105


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



Section 5: Pressure Control

This section covers a number of the more basic calculations as a preliminary to attending Pressure Control Schools.


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 = 2.5 pounds/square inch

4 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.


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.


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 Depth = 62.4 x 20 = 1248 pounds/sq ft There are 144 square inches in a square foot, therefore: 1248 = 8.67 psi

144

Using oilfield units of pounds per gallon, we must have a conversion factor to get psi values.


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


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) 10,000 ft of 8.5ppg mud b) 7,200 ft of 11.4ppg mud c) 14,280 of 10.7ppg mud

What if we are using Specific Gravity or pounds/cubic foot units ? The following conversion factors are used: Pressure (psi) = Mud weight (S.G.) x Depth (ft) x .433 Pressure (psi) = Mud weight (pcf) x Depth (ft) x .007 Using the same mud weight, it can be seen that pressure will increase with depth.


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 4560 psi 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.


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

6979

6279psi


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 = mud weight + (SIDPP) (Depth x .052) = 10.5 + (700) 11,500 x .052 = 10.5 + (700) (598) = 10.5 + 1.17 Kill Mud Weight = 11.67 ppg In the annulus, the kick fluid has contaminated the hydrostatic head of mud.


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 Gas Pressure = .1psi/ft x 300 = 30psi Total Pressure of mud + gas in annulus = 6115 + 30 = 6145 psi 10.5 ppg Difference = 6978 - 6145 = 833 psi -11200 Therefore, Casing pressure gauge will show 833 psi

833700

11,500


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) SIDPP = 650 psi b) SIDPP = 820 psi c) SIDPP = 300 psi


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) Initial pump pressure b) Pump pressure with heavy mud at bit c) When to adjust choke to get smooth transition between a) and b)

a) 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 = 800 + 700 = 1500 psi b) 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.


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 = 11.2 ppg Final Circulating Pressure = 11.2

800 x 10 = 896 psi


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 = 1200 psi Final Circulating Pressure = 700 psi 5 Drill Pipe, 4.276 I.D. 600 ft 8 x 3 Collars Pump Output, .2 bbls/strokes Calculate Pump Pressure every 100 strokes. First: Calculate capacity of Drill String in barrels = Drill Pipe Capacity + Drill Collar Capacity 4.2762 32 = 1029 x 9400 + 1029 x 600 = 167 + 5.2 = 172.2 barrels


Second: Calculate number if strokes from the Surface to Bit:

Surface t bit strokes = 172.2 = 861 strokes

.2

Third: 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 x 100

861 = .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.


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 = 700 psi


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. Boyles Law states: If the temperature of a gas is kept constant, then the volume will be inversely proportional to the pressure. This means, if the pressure is reduced by one half, then the volume will double. Boyles Law is expressed: V1 P2 V2 = P1 or V1P1 = P2 V2 V1 = Original Volume V2 = New Volume P1 = Original Pressure P2 = New Pressure


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 = P2 V2 P1 Solve the equation to find V2. V2 = V1 x P1

P2

V2 = 15 x 4500 (10 x 5000 x .052)

= 67500 2600 = 26 barrels


M.A.A.S.P Calculations

M.A.A.S.P is the Mread as the maxim As pressure in thepoints in the syste The weak points ar

a) Casinb) B.O.Pc) Form

Most often the fpressure would cau To find the fractWith the rams clowait, the process straight line will n The pressure at th

Randy Smith Training Soaximum Allowable Annular Surface Pressure, which should be um pressure gauge, before something breaks down.

Annulus builds up, there is a danger of breaking one of the weak m.

e:

g s ation below the casing

ormation below the Shoe is the weakest point. An excess of se the formation to fracture with a resultant loss of mud.

ure point a Leak-off Test is run after drilling out the shoe. sed, a small amount of mud is pumped into the well, after a short is repeated. By plotting volume pumped against Pump Pressure, a ot rise, but level off. This is when the formation is taking mud:

is point is the Leak Off Pressure.

5

4 3

2 1

500 1000 psi

lutions Ltd July 2002

The Leak Off Pressure can then be used to calculate Formation Fracture Pressure. 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.


First: Calculate Formation Fracture pressure and convert to gradient. Frac. Pressure = 1400 + (10 x 6400 x .052) = 4728 psi Frac. Pressure

Convert to gradient: Shoe Depth Gradient = 4728 = .74 psi/ft 6400

Second: Calculate Mud Gradient of mud in the hole Mud Gradient (psi/ft) = 10.5 x .052 x 1ft = .546 psi/ft Third: Apply gradients and Shoe Depth to formula for MAASP MAASP = 6400 x (.74 - .546) = 6400 x .193 = 1235 psi


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 7200 ft Mud Weight 11.5 ppg Leak off Pressure 1200 psi Calculate MAASP if 90% of formation fracture gradient is used. Formation Fracture Pressure = 1200 + (11.5 x 7200 x .052) 5505 psi Formation Fracture Gradient = 5505 = .765 psi/ft 7200 Mud Gradient = 11.5 x .052 = .598 psi/ft Fraction Gradient @ 90% = .765 x .9 = .6885 psi/ft


MAASP = Shoe Depth x (Fracture Gradient - Mud Gradient) = 7200 x ( .6885 - .598) = 7200 x .905 = 651 psi If mud weight was increased then MAASP would decrease.



Section 6: Hoisting Calculations

This section covers the basic theory and calculations behind Lifting Machines; Wire Rope Design factors and Ton Mile accumulations.


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. Weight of Load moved Mechanical advantage = 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 = 4 50 Lifting systems can be categorized into 4 main types:

1. Levers 2. Wheels and axles 3. Inclined Planes 4. Pulleys


Pulleys are used to lift heavy loads vertically.

A load of 500 lbs can be lifted using a 4-line pulley with: 500 = 125 lbs Pull 4 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 Constant

4 .271 6 .1882 8 .1469 10 .1224 12 .1062

14 .0948


Example: Hook load is 280,000 lbs. Blocks are strung with 10 lines. Calculate Fast Line Load tension. Fast Line Load = Weight of Load x Constant = 280,000 x .1224 = 34,272 lbs 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. Nominal Rope Breaking Strength Design Factor = 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 = 55,666 lbs 3 if Fast Load is 55,666 lbs. With 10 lines strung up, what is Hook Load?

Fast Line Load = Weight of Load x Constant


Solve equation to: Fast Line Load

Weight of 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.


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 = Nominal Breaking Strength Design Factor = 167,000 3.5 = 47,714 pounds


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


The minimum Tensile Strength of Grade E 5 pipe is 311,400 pounds. Therefore, this pull cannot be made. Max Overpull = 311,400 - 200,460 = 110,940 pounds 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 ton-mile of work. Example: Drill Collars : 900 ft long weigh 100,000 pounds in mud Drill Pipe: 14,100 ft Long Weigh 250,000 pounds in mud Block and Hook weigh: 45,000 lbs Calculate ton-miles to pull out of hole.

(1 short ton = 2,000 lbs 1 mile = 5280 ft)


First: Calculate ton-miles for drill pipe

Wt of pipe 250,000 = 125 tons 2,000 Distance moved is: 14,100 = 2.67 miles

5,280 Pulling out we have an average of: 125 = 62.5 tons

2 62.5 tons pulled 2.67 miles = 62.5 x 2.67 = 166.8 Ton-Miles

Second: 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 = 50 tons 2,000 Therefore, 50 tons are pulled 2.67 miles = 50 x 2.67 = 133.5 Ton-miles


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


Therefore, 27.5 tons for 5.68 miles = 27.5 x 5.68 = 156.2 Ton-miles Total Ton Miles to pull out of the hole = 304.55 + 156.2 = 460.7 Ton-Miles 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 = 5,280 ft 1 short ton = 2,000 pounds Calculate Ton-Miles for a complete Round Trip? Ton Miles for Drill Pipe pulling out: Weight of pipe = 10,200 x 19.5 = 198,900 x Buoyancy Factor Buoyancy factor = .8328 165,652lbs


Wt in Mud = 198,900 x .8328 = 2,000 pounds = 82.83 Tons 82.38 Average weight 2 = 41.41 Tons 41.41 Tons moving 1,93 Miles = 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 = 48.9 Tons 2,000 48.9 Tons moving 1.93 miles = 48.9 x 1.93 = 94.5 Ton-Miles Ton Miles for Collars to be removed: Average weight = 48.9 Tons = 24.5 Tons 2 Distance pulled = 800 = .15 miles 5,280 24.5 tons moving .15 miles = 3.7 Ton Miles


Ton Miles for Blocks: Travel twice 11,000 = 22,000 = 4.16 miles 5,280 Weight = 40,000 = 20 Tons 2,000 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 For running in hole, is the same again Total Round Trip = 2 x 261.3 = 522.6 Ton Miles



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.


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


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.


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 30drill 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 - 11.5 8.33

7.9 = 1 - .1747 = .825


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 = 541 ft

92.4 Fourth: Divide by collar string by 30 lengths to get the number required = 541 = 18 collars 30


Example: How many 30 drill collars of 105 pounds/ft would be needed for a Bottom Hole Assembly to give 55,000 pounds weight on bit in 10.8 ppg mud, with an excess of 20,000 lbs collar weight? Total collar weight in mud = 55,000 + 20,000 = 75,000 pounds Buoyancy factor = 1 - 10.8 8.33

7.9 = .836 Immersed Collar Weight = 105 x .836 = 87.8 pounds/ft 75,000 Length of collar string = 87.8 = 854 ft 854 No. of collars required = 30 = 28.5 or 29 collars To place these steps into a single formula (assuming a known Buoyancy Factor)


Wt of collars required_____ No. of 30 collars = Collar Buoyancy Collar wt/ft x factor x length Apply this to the above example. 75,000 = 105 x .836 x 30 75,000 = 2633.4 = 28.5 collars Worked Example: 10,000 ft deep hole. Prior to running back in we decided to use 147 pounds/ft collars each 30 ft in length. The required Weight on Bit will be 60,000 lbs with 30,000 lbs excess as Safety Factor.

The mud weight is 10.2 ppg

Drill pipe is 5 OD with 4.276 ID

Drill Collars are 8 OD by 3 ID

Calculate a) Number of joints of collars required b) Expected number if barrels of mud to be displaced from hole.


Buoyancy factor = 1 - 10.2 8.33 7.9 = .845 60,000 + 30,000 Number of collars required = 147 x .845 x 30 = 24.1 or 24 collars b) 24 collars at 30 length = 24 x 30 = 720 ft of collars Length of drill pipe = 10,000 - 720 = 9280 ft Calculate volume of steel in drill pipe: = 52 - 4.2762 bbls/ft

1029.4 = .0065 bbls/ft = .0065 x 9280

= 60.32 bbls


Calculate volume of steel in collars: = 82 - 32 x 720

1029.4 = 38.47 barrels Total mud displaced = 60.56 + 38.47 = 99.03 barrels To check our calculation, the drill collars have displaced 38.47 barrels of mud weighing 10.2 ppg. This volume weighs 38.47 x 42 x 10.2 = 16480.5 pounds Number of collars weighing 16480.5 = 16480.5

147 x 30 = 3.7 collars Number of collars in air to make 90,000 lbs = 90,000

147 x 30 = 20.4 Total number of collars = 20.4 + 3.7 = 24.1 collars The same, as calculated above.


When calculating collar length required, the term NEUTRAL POINT is commonly used. This is the point at which 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. 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 measured from the bit up. Example: 30ft, 147 pound/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 = .845 Length of collars to give 60,000 lbs = 60,000 x .845

147 = 483 ft This is 80% of length required.


If 80% = 483 ft

1% =

100% = = The number

Drill pipe Drill collars

Randy Smith Training So

483 80

483 80 x 100

603.75 ft

603.75 of collars required = 30

= 20.1 or 20 collars

Tension

Neutral Point

Compression

80 % below

20 % above

lutions Ltd July 2002


Section 8: Miscellaneous Calculations

This section covers the calculations used during rig work encounters with Spotting Pills, applying torque, stuck pipe and weighting up of mud.


Miscellaneous Calculations The following calculations are commonly used around the rig.

Spotting Pills Torque Stuck Pipe Weighting up

SPOTTING PILLS: TD 9000, Hole Diameter 12 , Mud Weight 11ppg 5 Drill Pipe ID 4.276 600 of 8 x 3 Drill collars Make up a pill to cover collars plus 25% extra. Pump 1 barrell/20 mins of the extra once the pill has been placed in the drill collar annulus. Calculate: 1. Volume of pill

2. Volume of mud to spot the pill First: Calculate volume of pill required to cover collars

Annular volume around collars = (12.252- 82) x 600 1029 = 50 barrels


Second: Calculate 25% excess 25% of 50 bbls = .25 x 50 = 12.5 bbls Total volume of pill = 50 + 12.5 = 62.5 barrels Third: Calculate height of pill retained in the string (32) cap. of collars = 1029 bbls/ft = .0087 bbls/ft x 600 cap. of collars = 5.25 bbls Total pill inside string = 12.5 Pill inside drill pipe = 12.5 - 5.25 = 7.25 bbls Cap. of pipe = 4.2762 1029 bbls/ft = .01776 bbs/ft


Height of pill in drill pipe = 7.25 = 408 ft 0.1776 Total height of pill in string = 600 + 400 = 1008ft Fourth: Calculate volume of mud in remainder of string Length of pipe with mud = 9000 - 1008 = 7992ft Volume of mud = Drill pipe capacity/ft x 7992 = 0.1776 x 7992 = 142 bbls This value has been calculated for the string only. An addition of the mud inside the surface lines must be made. i.e. Surface line volume = 5 bbls Total mud pumped to follow pill = 142 + 5 = 147 bbls


TORQUE Example: 9 inch drill collar with 7 5/8 API regular connection and 3-inch bore is to be made up to 83,000 ft-lbs. The Rig Tongs are 4ft long. Calculate the reading to be obtained on the Drillers Torque Gauge calibrated in ft-lbs. Explanation: The value of 83, 000 ft-lbs requires a tong 1 foot long to give 83, 000 lbs of pull. If the tong is 4 feet long, the extra leverage will reduce this value by 4. Therefore, Pull required = Ft-lbs

Tong length, ft = 83000 4 = 20,750 ft.lbs Drillers gauge marker should be set at 20,750


Example: At what value should torque gauge be set if 50,000 ft. lbs is required using 5ft tongs? Value = 50,000 = 10,000 ft lbs 5 STUCK PIPE Being able to calculate the depth at which the string is stuck is invaluable when spotting freeing pills. There are a number of ways to calculate Free Points. Section of the I.A.D.C. Drilling Manual covers some of the techniques. One common formula is: L = 735,294 x E x W P L = length of free pipe, feet E = average elongation, inches W = weight of pipe, lbs/ft P = average pull in pounds 735,294 = a Constant


Example: 16.6 lbs/ft drill pipe. Average pull of 50,000 lbs gave average elongation of 12 inches.

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