Drilling Calculations CD Complete Course
Short Description
Drilling Calculation...
Description
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
© Randy Smith Training Solutions Ltd
July 2002
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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July 2002
6
6 x 3 = 18 sq. ins 3
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July 2002
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
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July 2002
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
.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
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
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
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
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
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
=
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
Buoyancy factor
Immersed Collar Weight
Length of collar string
No. of collars required
=
55,000 + 20,000
=
75,000 pounds
=
1 -
=
.836
10.8 ÷ 8.33 7.9
=
105 x .836
=
87.8 pounds/ft
=
75,000 87.8
=
854 ft
=
854 30
=
28.5 or 29 collars
To place these steps into a single formula (assuming a known Buoyancy Factor)
© Randy Smith Training Solutions Ltd
July 2002
No. of 30’ collars
=
Wt of collars required_____ Collar Buoyancy Collar wt/ft x factor x length
Apply this to the above example.
75,000 x .836 x
=
105
=
75,000 2633.4
=
28.5 collars
30
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.
© Randy Smith Training Solutions Ltd
July 2002
Buoyancy factor
Number of collars required
b)
24 collars at 30’ length
=
1-
=
.845
=
60,000 + 30,000 147 x .845 x 30
=
24.1 or 24 collars
=
24 x 30
Length of drill pipe
10.2 ÷ 8.33 7.9
=
720 ft of collars
=
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
=
© Randy Smith Training Solutions Ltd
July 2002
Calculate volume of steel in collars:
Total mud displaced
=
82 - 32 1029.4
=
38.47 barrels
=
60.56
=
99.03 barrels
x
+
720’
38.47
To check our calculation, the drill collars have displaced 38.47 barrels of mud weighing 10.2 ppg. This volume weighs
38.47 x =
Number of collars weighing
42
10.2
16480.5 pounds
16480.5
=
Number of collars in air to make 90,000 lbs =
Total number of collars =
x
16480.5 147 x 30 = 3.7 collars
90,000 147 x 30 = 20.4
20.4 + 3.7
= 24.1 collars The same, as calculated above. © Randy Smith Training Solutions Ltd
July 2002
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
=
60,000 147
=
483 ft
x
.845
This is 80% of length required.
© Randy Smith Training Solutions Ltd
July 2002
If
80% =
483 ft
1%
=
483 80
100% =
483 80
=
x
100
603.75 ft
The number of collars required
Drill pipe
=
603.75 30
=
20.1 or 20 collars
Tension 20 % above
Neutral Point
80 % below Drill collars
© Randy Smith Training Solutions Ltd
Compression
July 2002
Drilling Calculations Course
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.
© Randy Smith Training Solutions Ltd
July 2002
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 ¼, 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. 2.
Volume of pill Volume of mud to spot the pill
First:
Calculate volume of pill required to cover collars
Annular volume around collars
© Randy Smith Training Solutions Ltd
=
(12.252- 82) x 600’ 1029
=
50 barrels
July 2002
Second:
Calculate 25% excess 25% of 50 bbls
Total volume of pill
Third:
=
50
=
.25
x
=
12.5 bbls
+
12.5
=
50
62.5 barrels
Calculate height of pill retained in the string
=
(32) 1029 bbls/ft
=
.0087 bbls/ft
cap. of collars
=
5.25 bbls
Total pill inside string
=
12.5
Pill inside drill pipe
=
12.5
=
7.25 bbls
=
4.2762 1029 bbls/ft
=
.01776 bbs/ft
cap. of collars
Cap. of pipe
© Randy Smith Training Solutions Ltd
-
x
600
5.25
July 2002
Height of pill in drill pipe
=
7.25 0.1776
=
408 ft
Total height of pill in string
=
600’
+
400’
=
1008ft
Fourth:
Calculate volume of mud in remainder of string Length of pipe with mud =
Volume of mud
9000’ - 1008’
=
7992ft
=
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
© Randy Smith Training Solutions Ltd
July 2002
TORQUE Example: 9 inch drill collar with 7 up to 83,000 ft-lbs.
5/8
API regular connection and 3-inch bore is to be made
The Rig Tongs are 4ft long. Calculate the reading to be obtained on the Drillers Torque Gauge calibrated in ftlbs.
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
© Randy Smith Training Solutions Ltd
July 2002
Example: At what value should torque gauge be set if 50,000 ft. lbs is required using 5ft tongs?
Value =
50,000 5
=
10,000 ft lbs
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 Ω 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
© Randy Smith Training Solutions Ltd
July 2002
Example: 16.6 lbs/ft drill pipe. Average pull of 50,000 lbs gave average elongation of 12 inches. Calculate Free Point L
=
735,294 x 12 x 16.6 50,000
=
2929 ft
The string is stuck at and possibly below 2929 ft. WEIGHTING UP Mud density increases are common during normal drilling and essential during some kill operations. To raise the Mud weight Barite is added. However how much is needed to raise the weight? If two substances having different densities are mixed, then the new density is a function of the amount and density of the two substances. The relationship is expressed in the following formula.
V1 D1 + V2 D2 = (V1 + V2) DR
V1 V2 D1 D2 DR
= = = = =
volume of original substance volume of second substance density of original substance density of second substance resulting density
© Randy Smith Training Solutions Ltd
July 2002
Converting this into oilfield units we get V1 W1 + V2 DB = (V1 + V2) W2
= = = = =
V1 V2 W1 W2 DB
volume of mud before weighting, bbls volume of barite added, bbls initial mud density, ppg final mud density, ppg density of Barite, ppg. Which is 35.4ppg
We require to find V2, volume of barite to be added, therefore, solving the equation 2 :V1 W1 + V2 DB = (V1 + V2) W2
V2 =
becomes:
V1 (W2 -W1) WB - W2
Using the following figures we can get barrels of Barite to be added W2 W1 WB V1
= = = =
12ppg 10ppg 35.4 100bbls
© Randy Smith Training Solutions Ltd
V2
= V1 (W2 - W1) WB - W2 =
100 (12 - 10) 35.4 - 12
=
200 23.4
=
8.547 barrels of Barite
July 2002
Thinking of Barite volume in barrels is cumbersome. Barite is mostly measured by weight – for instance:
Number of sacks (1 sack weighs 100 lbs) Number of pounds or tons in Barite container This therefore requires further calculation to convert barrels of barite into pounds. 1 barrel of Barite weighs 1490 lbs (if barite weighs 35.4ppg then 1 barrel weighs 35.4 x 42 lbs = 1490) Therefore, 8.547 bbls weighs
8.547 x 1490 = 12,735 lbs
When mixing, Barite container gauge would have to drop 12,735 lbs (or 5.7 tons) or 127 sacks would be used. To make it easier the formula can be modified.
V2
=
V1 (W2 = W1) 35.4 - W2
1.
Modified formula for long tons (2240 lbs = 1 long ton) required per 100 bbls of mud (V1).
modifies to:-
Convert V1 from barrels to tons, to give answer V2 in Tons. V1
=
100 bbls. 1 bbl of barite weighs 1490 lbs
100 bbls of barite weighs 1490 x 100 = 149,000 lbs
© Randy Smith Training Solutions Ltd
July 2002
V1
=
149,000 2240
Apply V1
=
66.5 tons to formula
66.5 Tons
66.4 (W2 - W1) 35.4 – W2
V2 tons
2.
=
Modified formula Pounds Same as above but do not convert into tons. lbs
V2’ pounds
3.
=
100 bbls of barite = 149,000
149,000 (W2 - W1) 35.4 - W2
Modified formula for Sacks 1 sack weighs 100 pounds Convert formula 2) from pounds to sacks. 149,000 lbs = V2, sacks
=
© Randy Smith Training Solutions Ltd
1490 sacks weighing 100lbs each 1490 (W2 – W1) 35.4 - W2
July 2002
The above formulae give the quantity of barite required to increase the weight of 100 barrels of mud. The following example will be used to check the above formulae. Example: Mud weight of 9.5ppg has to be raised to 11.2ppg, using barite. Calculate: a) b) c)
a)
Number of sacks/100 bbls of mud Number of Pounds/100 bbls of mud Number of Tons/100 bbls of mud
V2
=
1490 (11.2 - 9.5) 35.4 - 11.2
=
105 sacks
b)
149,000 (11.2 - 9.5) 35.4 - 11.2 V2
=
10500 Pounds
=
66.5 (11.2 - 9.5) 35.4 - 11.2
=
4.67 Tons
c) V2
© Randy Smith Training Solutions Ltd
July 2002
4.67 tons
=
4.67 x 2240
105 sacks weighing 100 lbs
=
10,460 pounds
=
10,500 pounds
All answers are same give or take a number of pounds due to rounding off figures for useable numbers.
Example: Mud weight is to be raised from 9.6 to 10.8 ppg. Total mud in system = 1420 bbls Barite is stored in pressurized tanks calibrated in Long Tons (1 Ton = 2240 lbs) Present Tank Reading
=
125.8 Tons
Calculate the number of tons of Barite required and Final Tank
Tons barite/100 bbls
=
66.5 (10.8 - 9.6) 35.4 - 10.8
Tons barite for system
=
3.24 x 14.20
=
461 Tons
=
125.8 - 46.1
=
79.7 Tons
Final Tank Reading
© Randy Smith Training Solutions Ltd
July 2002
Adding all this barite will increase total mud volume.
The new mud volume can be calculated using the formula:-
bbls increase/100 bbls of mud =
100 (W2 - W1) 35.4 - W2
Weights in ppg
For watering down the mud, use the following to get barrels of water required.
bbls of water
=
(W1 - W2) x original volume of mud W2 - 8.34
Weights in ppg
© Randy Smith Training Solutions Ltd
July 2002
Drilling Calculations Course
COURSE CONSOLIDATION EXERCISES
The following questions and answers have been compiled as supplemental exercises to be completed after each relevant section, to help consolidate your learning experience
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises SECTION 1 1.
Convert 486 inches to yards ft and inches.
2.
Convert 486 square inches into square ft.
3.
How many US gallons would fill a tank with a capacity of 450 cubic ft?
4.
Convert 10.3 ppg into P.C.F.
5.
What would be the equivalent in P.S.I. of 15 tons resting on a square 2ft by 2ft
6.
How many pounds force would be exerted on a hatch 2ft by 1.5ft if the pressure behind it was 3 P.S.I.
7.
Convert 240 US gallons per minute flow into litres per minute flow.
8.
Convert 1000kg into pounds
9.
How many pounds difference between 1000kg and one long ton?
10.
Convert 90oF to degrees rankin.
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises SECTION 1 : ANSWERS 1.
13 yds, 1 ft 6 inches
2.
3 square ft 54 square inches
3.
3366.4 US gallons
4.
77.05 P.C.F.
5.
15 x 2240 24 x 24
6.
24 x 18 x 3 = 1,512 lbs force
7.
908.4 ltr/min
8.
2,204 lbs
9.
2240 – 2204 = 36 lbs
10.
90 + 460 = 550°R
=
© Randy Smith Training Solutions Ltd
58.33 PSI
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises SECTION 2
1.
Express the following fractions as decimals. a) 5/8
2.
b) 11/16
c) 7/9
d) 23/32
e) 5/24
Round off the following to 2 places of decimal. a) .6356
b) .7945
c).7987
d) .8429
e) .6464
3.
Calculate the circumference of a circle with a diameter of 6 inches.
4.
Calculate the area of a circle with a diameter of 6 inches.
5.
Calculate the annular area between a 13 inch inside diameter pipe and a 5 inch outside diameter pipe.
6.
Calculate the square roots of. a) 69
7a.
b) 138
c) 276
d) 552
Calculate the capacity of a tank in US BBLS with the following dimensions. 15’ long by 6’ wide by 8’ deep
7b.
What volume would be in the tank if the liquid height was 1 foot?
7c.
How much volume has been added to the tank if during drilling operation the lever rose by 1 foot 5 inches?
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises SECTION 2 (Cont) 8. What would be the capacity in BBLS of a sand trap?
Sand Trap SAND TRAP
8’ deep 10’ long by 10’ breath and is triangular in shape.
9a. What percentage of 186 does 42 make up? 9b. What percentage of 93 does 56 make up? 9c. What percentage of 56 does 60 make? 10.What would be the volume of a 1,500 foot annulus between 5 inch pipe and 17 ½ inch hole?
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises SECTION 2 ANSWERS 1. a) .625b) .6875 2. a) .64
c) .7778
b) .79
d) .7188
c) .80
e) .2083
d) .84
e) .65
3. πD or 2π r = 18.85 inches 4. πr2 = π32 = 28.27 sq. inches 5. = .7854 (132 - 52) = 113.1 sq. inches 6. a) 8.3
b) 11.75
c) 16.6
d) 23.49
7a. 720 cubic ft ÷ 5.6146 = 128.2 BBLS 7b. 128.2 ÷ 8 = 16.025 BBLS 7c. 17 x 16.025 12
= 22.7 BBLS
8. ½ x 8 x 10 x 10 = 400 cubic ft ÷ 5.6140 = 71.2 BBLS 9.a) b) c)
42 ÷ 186 x 100 = 22.58% 56 ÷ 93 x 100 = 60.22% 60 ÷ 56 x 100 = 107.1%
11. (17.52 - 52) = .2733 BBL/FT x 1,500 ft = 410 BBLS 1029
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 3 1.
From the following well information calculate the volumes: Hole size 12 ¼” hole Hole depth 12,650’ Casing shoe depth 10,200 Casing size 13 3/8” ID 12.46” 5” pipe ID 4.2” a)
What is the hole volume with no pipe in the hole?
b)
What would be the liquid volume with 5 inch pipe in the hole from top to bottom?
c)
What volume would be in the drill pipe open hole annulus?
d)
What volume would be in the drill pipe casing annulus?
e)
What would be the volume in the drill pipe?
2.
Calculate the pump output per stroke of a triplex pump with a 12” stroke and liner size of 6” at 98% volumetric efficiency.
3.
Calculate the pump output in BBL/STK of a triplex cement pump with 5 inch liners and an 8 inch stroke. Use a volumetric efficiency of 95%.
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 3 (Cont)
4a.
What would be the fluid output per minute of a triplex pump running at 80 strokes per minute, with 6.25” liners and 98% volumetric efficiency?
4b.
What is the annular capacity of 8” drill collars in 12 ¼” hole?
4c.
What would be the annular velocity of the fluid passing round the drill collars?
5.
How long would it take to circulate an annular volume of 950 BBLS using the pump in 4a?
6.
What would be the maximum pressure that could be reached pumping at 400 gallons/minute with a pump of 750 hydraulic horse power?
7.
Determine the new pressure required by increasing the pump rate from 60 to 65 strokes/min. Pump pressure at 60 STK/MIN was 2,650 PSI.
8.
Determine the new pump pressure required to pump a lighter fluid at the same rate. 10.5 PPG mud is being pumped at 80 STKS/MIN at 3,000 PSI. The weight is being reduced to 9.8 PPG.
9.
Calculate the total cross sectional area of three jet nozzles 16/32, 16/32 and 14/32.
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 3 ANSWERS 1a.
1b.
10,200’ of casing x 12.462 1029
=
1538.9 BBLS
2,450’ of open hole x 12.252 = 1029
357.3 BBLS
Total volume of hole
1896.2 BBLS
=
Volume of steel per ft of pipe x length (52 – 4.22) x 12,650 1029
=
90.48 BBL
=
1896.2 – 90.48
Volume of liquid in hole
=
1c.
1805.7 BBLS
Drill pipe open hole annular capacity length (12.252 - 52) x 2,450 1029
© Randy Smith Training Solutions Ltd
=
297.8 BBLS
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 3 ANSWERS (Cont)
1d.
Drill pipe casing and capacity x length (12.422 – 52) x 10,200 = 1,281.3 BBLS 1029
1e.
Drill pipe capacity x length (4.22) x 12,650 = 216.9 BBLS 1029
2.
Volume in BBLS for 1 ft of 6 inch diameter =
62 = .034985 BBLS/Cylinder 1029
For 3 cylinders = .10495 BBLS/STK At 98% volumetric efficiency .10495 x 9.8 = .103 BBLS/STK 100
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 3 ANSWERS (Cont)
3.
8 inches is 2/3 of a foot, therefore the output would be 2/3 or .6666 of that of a 12” stroke 8 x (52) 12 1029
= .01619 BBL/Cylinder
For 3 cylinders = .0486 BBL/Stroke At 95% volumetric efficiency .0486 x
95 = 100
.046
BBL/STK
4a.
(6.252) x 3 x .98 x 80 1029
4b.
(12.252 – 82) = .0836 BBL/STK 1029
4c.
Annular velocity = Pump Output Annnular Vol
= 8.928 BBL/MIN
BBL/MIN BBL/FT
= 8.928 .0836 = 106.7 Ft/Min
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 3 ANSWERS (Cont) 5.
Annular volume Pump output
(BBLS) (BBLS/MIN)
950 8.928 = 106.4 Minutes
6.
HHP = P x V 1714 750 = P x 400 1714 750 x 1714 400
= P
P = 3214 PSI
7.
New Pressure = Old Pressure x (New STKS2) (Old STKS2)
= 2650 x (652) (602) = 3,110 PSI
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 3 ANSWERS (Cont) 8.
New Pressure = Old Pressure x New Mud Old Mud = 3,000 x 9.8 10.5 = 2,800 PSI
9.
Area = .7854 x (Diameter2) 16/32 Jet = .7854 x (16/322) = .19635 sq” Two
= . 3927 sq”
14/32
= .7854 x (14/322) = .1503 sq “
Total area = .3927 + .1503 = .543 sq “
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises SECTION 4 1. a. b. c.
Calculate: Slurry volume with 10% excess Number of Sacks Pump stroke to pump the plug for the following single stage cement job. Hole size 17 ½” Casing size 13 3/8” set at 4,600 FT Casing ID 12.42” Yield 1.05 cubic FT/SACK Pump output .138 BBL/STK Float set 80’ above shoe
2.
Calculate for a balanced cement plug.
a. b. c.
Slurry volume. Volume of water behind the cement. Pump stroke to displace. Hole size 8 ½” Drill pipe size 5” cap .01738 BBL/FT Cement plug height – 500 FT Water ahead of cement 10 BBLS Pump output .103 BBLS/STK Hole depth 10,350’
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 4 ANSWERS 1a.
Annular capacity = (17.52 – 13.3752) = .1238 BBL/FT 1029 4,600’ = .1238 X 4600 = 569.3 BBLS Volume between float and shoe = 80 x (12.422) 1029 = 12 BBLS
Total CMT requirements with no excess = 581 BBLS With excess = 639 BBLS
1b.
Yield is 1.05 cubic FT/SAC 639 BBLS = 639 x 5.6146 SACK = 3588 Cubic Ft Nos of Sacks = 3588 1.05 = 3417 Sacks
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 4 ANSWERS (Cont)
1c.
Nos of STKS To Bump Plug
=
Casing Volume to Float Pump Output PER/STCK
Casing Volume
=
4520 x (12.42) 1029
=
675.4 BBLS
=
675.4
=
4,894 STKS
Nos of STKS
2a.
÷
.138
Volume of slurry to fill 500’ of open hole. = 500 x (8.52) = 35.1 BBLS 1029
2b.
Height of the water placed ahead when in annulus. = 10 BBLS
÷ Annular Cap
= 10 ÷ (8.52 – 52) 1029
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 4 ANSWERS (Cont)
2b (cont) = 10
÷
.0459
= 218’
Volume of water behind = 218’ x DP CAP = 218 x .01738 = 3.78 BBLS 2c.
Height of plug with pipe in the hole. =
35.1 (0.459 + .01738)
= 555’ Add the height of the water in the pipe. 555 + 218 = 773
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises
SECTION 4 ANSWERS (Cont) 2c.(Cont)
Mud will be pumped to a depth of 10,350 – 773 = 9,577 Capacity of 9,577 of DP = 9,577 x .01738 = 166.45 BBLS Nos of STKS = 166.45 ÷ .103 = 1616 STKS
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises SECTION 5
1.
Calculate the hydrostatic pressure exerted by the following columns of fluid. a. b. c.
2a.
Depth 12,000’ (TVD) weight 10 PPG Depth 8,500’ (TVD) weight 15.2 PPG Depth 17,200 (TVD) weight 17.8 PPG
Calculate the mud weight in PPG that would give the following pressures at: i) ii) iii)
5,000 PSI 2,325 PSI 10,950 PSI
at at at
10,000 ft (TVD) 5,000 ft (TVD) 16,450 ft (TVD)
2b.
What would be the increase in mud weight required to exert an additional 350 PSI hydrostatic pressure for the examples in (2a)?
3a.
Calculate the hydrostatic pressure exerted by column of fluid made up of different densities From surface From 2,000’ From 3,000’
3b.
to 2,000’ to 3,000’ to 5,000’
– 9.2 PPG mud – 8.9 PPG – 16.4 PPG
What mud weight would exert this pressure?
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises SECTION 5 (Cont) 4.
Calculate the relevant values and construct a step down chart for pumping kill mud down the drill pipe. Well information: Depth 8,200 Mud weight 10 PPG Shut in drill pipe pressure = 250 PSI Slow circulating pressure = 850 PSI at 30 STK/MIN Strokes required to pump from surface to the bit = 860 STKS a. b. c. d. e.
5.
6a.
Kill mud weight Initial circulating pressure Final circulating pressure Pressure drop per 100 STKS Chart
Calculate in cubic feet the volume a 10 BBL gas kick would occupy on surface. If the original formation pressure was 5,300 PSI and atmosphere pressure is 14.75 PSI.
Calculate the formation strength (fracture pressure) from the following data. Shoe depth 6,200’ (TVD) Mud weight – 9.6 PPG Leak off pressure 1,200 PSI Give answers as a pressure and a pressure gradient.
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises SECTION 5 (Cont)
6b.
Calculate the maximum allowable annular surface pressure for 10 and 12 PPG mud respectively.
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises SECTION 5 ANSWERS
1a. 1b. 1c.
6,240 PSI 6,718 PSI 5,920 PSI
i. ii. iii.
9.61 PPG 8.94 PPG 12.8 PPG
2b.
a) b) c)
10.09 PPG an increase of .48 PPG 10.21 PPG an increase of 1.27 PPG 13.21 PPG an increase of .41 PPG
3a.
2,000 x 9.2 x .052 1,000 x 8.9 x .052 2,000 x 16.4 x .052
3b.
12.02 PPG
4a.
Kill mud wt
=
= = = =
957 463 1706 3,126 PSI
SIDPP Depth x .052
= 10 +
© Randy Smith Training Solutions Ltd
+ Original Mud
250 8200 x 0.52
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises SECTION 5 ANSWERS (Cont)
4a (Cont)
= 10 + .58 = 10.6 PPG
4b.
Initial circulating pressure = Shut in drill pipe pressure + slow circ pressure = 250 + 850 = 1100 PSI
4c.
4d.
Final circ press =
Kill Mud Wt x slow circulating pressure Original Mud Wt
=
10.6 x 850 10
=
901 PSI
Pressure drop per STKS = P.Final circ – P. Initial circ Strokes Surface to Bit = 1100 – 901 850 = .234 PSI/STK
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises SECTION 5 ANSWERS (Cont)
Per 100 STKS = 23.4 PSI (23)
5.
V2 = P1 x V1 P2
STKS
1100 PSI
100 200 300 400 500 600 700 800 850
1077 1054 1031 1008 985 962 939 916 901
P1 = 5,300 PSI V1 = 10 BBLS P2 = 14.75
V2 = 5,300 x 10 14.75 = 3,593 BBLS For cubic ft 3,593 x 5.6146 = 20,174 cubic ft
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises SECTION 5 ANSWERS (Cont)
6.
Fracture Pressure
= Leak off plus hyst to shoe = 1200 + 6200 x 9.6 x 0.52 = 4,295 PSI
Fracture pressure gradient = Fracture Pressure TVD Shoe Depth = 4295 6200 = .693 PSI/FT MAASP = Shoe depth (formation fracture gradient – mud gradient) For 10 PPG = 6,200 (.693 - .52) = 1073 PSI For 12PPG = 6,200 (.693 – 624) = 428 PSI
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises SECTION 6
1.
Calculate the maximum hook load that can be applied with 12 lines 13/8” wire with a breaking strain of 167,000 and a design factor of 3.
2.
With the rig up in question if the string weight was reading 350,000lbs. What would be the loading on the fast line?
3.
Calculate the ton miles for a round trip. With depth 12,500’ 300’ DC weight in mud 12,200’ of DP weight in mud Block weight
© Randy Smith Training Solutions Ltd
= 40,000 lbs = 200,000 lbs = 28,000 lbs
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises SECTION 6 ANSWERS
1.
Max Fast Line load
=
Nominal Breaking Strain Design Factor
=
167,000 3
=
55,660 lbs
=
Fast Line Load Constant
=
55,666 .1062
=
524,161 lbs
Weight of load
=
Fast Line Load Constant
Fast Line Load
=
Weight x Constant
=
350,000 x .1062
Weight of load
2.
=
© Randy Smith Training Solutions Ltd
37,170 lbs
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises SECTION 6 ANSWERS (Cont)
3.
For a round trip Drill pipe Weight of the pipe =
Distance moved
Ton Miles
200,000 2,000
=
12,200’
=
12,200 5,280
=
100 x 2.31
=
231 Ton Miles
=
100 Tons
=
2.31 Miles
Drill Collars Move 12,200 before they reach surface. =
2.31 Miles
Ton Miles
=
40,000 x 2.31 2,000
=
46.2 Ton Miles
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises SECTION 6 ANSWERS (Cont)
R/Trip 300’ Weighing 40,000 40,000 2,000 =
x
300 5,280
1.14 Ton Miles
Round trip ton miles for the blocks. Weight
=
28,000 2,000
=
14 Tons
Distance
=
12,500 x 4 5,280
=
9.47 Mile
Ton Miles
=
14 x 9.47
=
132.6 Ton Miles
Total round trip miles = DP + DC + Block = 231 + 46.2 + 1.14 + 132.6 = 411 Ton Miles
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises SECTION 7
1.
Calculate the buoyancy factor for: a. b. c.
2a.
12 PPG 14 PPG 16 PPG
How many 30’ drill collars would be required if 60% of the available collar weight is 20,000 lbs? 8” drill collars in 11.8 PPG mud. Weight = 146 lbs/ft
2b.
Where in the drill calculations would be the neutral point if 18,000 lbs was being applied to the bit?
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises SECTION 7 ANSWERS
1a. 1b. 1c.
.818 .787 .757
2a.
B/Factor for 11.8 PPG = 1 – (11.8
÷ 8.33) 7.9
= .821
60%
= 20,000
100% = 20,000 60
x
100
= 33,333 lbs Buoyed
Buoyed Wt Buoyancy Factor
=
Dry Wt
33.333 .821
=
40,600
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises SECTION 7 ANSWERS (Cont)
Length
=
40,600 146 278’ of DC
Drill Calcs
Run
2b.
=
278 30
=
9.26
9
Length required to make up 18,000 lbs =
18,000 Buoyed Wt of one Ft of DC
=
18,000 .821 x 146
=
150’
Neutral point is 150’ above the bit.
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises SECTION 8
1a.
What volume of pill is required to fill the drill collar annulus and leave 20% of that volume in the pipe? Hole size 8 ½” Depth 12,200’ Drill Collars 6 1/4” OD 2 ¾ ID Drill Collar Length 360’
1b.
How many strokes would be required to spot the pill then displace it out of the string? Drill Pipe 5” Cap .01738 BBLS/FT Pump Output .102 BBL/STK
2.
Calculate the line pull to apply the following torques using an effective tong length of 3.5 ft. 26,500 ft lbs 64,000 ft lbs 92,000 ft lbs
3.
What would be the resulting density if 200 BBLS of 15.2 PPG mud was mixed with 150 BBLS of 12.6 PPG?
4.
How much barite would be required to increase the mud weight in a system of 950 BBLS from 11 PPG to 11.6 PPG?
© Randy Smith Training Solutions Ltd
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises SECTION 8 ANSWERS
1a.
=
8.52 – 6.252 1029
=
.0323 BBL/FT
=
36
=
11.6 BBLS
Plus 20%
=
13.9 BBLS
String Capacity
=
360
Plus
=
11,840
=
2.65 BBLS
=
208.4 BBLS
DC/OH Annular Cap
Total DC/OH Annular Cap
1b.
x
.0323
x (2.752) 1029 x
.01738 +
205.78
Volume to leave 20% in string
© Randy Smith Training Solutions Ltd
=
208.4
–
=
206.1 BBLS
2.3
July 2002
DRILLING CALCULATIONS COURSE
Section End Consolidation Exercises SECTION 8 ANSWERS (Cont) 1b (Cont) No of STKS
To spot pill
=
206.1 .102
=
2,020 STKS
Plus 23 STKS to displace 20% 2.
7,571 lbs Pull 18,285 lbs Pull 26,285 lbs Pull
3.
(200 x 15.2) + (150 x 12.6) = 350 D D
4.
V2
=
(200 x 15.2) + (150 x 12.6) 350
=
14.08 PPG
=
V1 (W2 – W1) WB – W2
=
950 (11.6 – 11) 35.4 – 11.6
=
23.95 BBLS of Barite 1 BBL Weight 1490 lbs
=
WT of Barite
© Randy Smith Training Solutions Ltd
=
23.95 x 1490
=
35,684 lbs
July 2002
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