Pe 2322

Drilling and Completion Systems Module 5: Fluids Pressure Control

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Module 5: Fluids Pressure Control Specific Gravity Hydrostatic Pressure Calculation Class Activity: Hydrostatic Pressure Examples Pilot Testing Procedures Class Activity: Pilot Testing Procedure Example Desired Viscosity Weight or Density Control Unit

Lesson 1: Functions of Drilling Fluids Lesson 1: Objectives Functions of Drilling Fluids Negative Functions of Drilling Mud Physical Properties of Drilling Mud Classification of Muds Based on Liquid Phase Pressurized Mud Balance Marsh Funnel Rotational Viscometer Rotational-Viscometer Geometry Assignment 5.1: Module 5 Self Study Review Assignment 5.1: Read Fundamentals of Drilling Engineering pp. 87-98

Lesson 3: Buoyancy and Hook Loads Lesson 3: Objectives Hook Loads Buoyancy: Example of Archimedes Principle Hook Load and Buoyancy Calculation Examples Casing Loads Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory

Lesson 2: Drilling Fluid Properties Lesson 2: Objectives

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Lesson 1: Functions of Drilling Fluids

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Lesson 1: Functions of Drilling Fluids Learning Objectives In this lesson we will: Describe the essential functions a properly designed and maintained drilling fluid performs during well construction List the properties important to the function of removing cuttings of the drilling mud Define how the control of oil, gas or water formation pressure is accomplished by a hydrostatic pressure

Define how gel helps reduce the power costs of cutting Define how suspending solids is accomplished Define how the deposit of cuttings in the mud pit or mud ditch is accomplished List the negative functions of drilling mud List the physical properties of drilling mud List the classification of muds based on liquid phase

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Functions of Drilling Fluids A properly designed and maintained drilling fluid performs essential functions during well construction such as: Transporting cuttings to the surface Preventing well-control issues and wellbore stability Minimizing formation damage Cooling and lubricating the drillstring Providing information about the formation

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Remove Cuttings Properties important to the function of removing cuttings of the drilling mud are: Density = rho (ρ) Viscosity = mu (µ) Annular Velocity = Va Type of Flow, Size, Shape and Density of the Cuttings

Suspension of the Cuttings and/or the Gelling Properties

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Prevent Caving This important property helps us by: Controlling the hydrostatic head Consolidating loose or clay type formations by surrounding the particles and holding them in the fluid and gelled mud. Controlling water intrusion prevents clays from swelling and sloughing into the hole, mud cake and filtration properties.

Source: http://petroleumsupport.com/mechanical-sticking-mechanism-of-stuck-pipe/

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Controlling Oil, Gas and Water Formation Pressures The control of oil gas or water formation pressure is accomplished by: A hydrostatic pressure in this consideration we are worried about:  Loss circulation  Gas cut mud  The formations being drilled

Source: http://www.dcmudcleaningequipment.com/How-Does-Gas-Cut-Affect-MudDensity.html

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Cooling and Lubricating the Drillstring Lubrication of the drill pipe, the hole wall, the casing and the mud pumps is accomplished by: The gel, due to clay content Reducing the power cost to increase the drilling speed

Source: https://www.rigzone.com/training/insight.asp?insight_id=291&c_id=24

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Suspending Solids Suspending of solids is accomplished by: Gel strength thixotropic properties The holding of cuttings when static Returning to fluid state when circulation is restored

Source: http://trenchlessonline.com/index/webapp-stories-action/id.2489/archive.yes/Issue.201304-01/title.gel-strengths-for-horizontal-vs.-vertical-drilling

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Deposit of Cuttings The deposit of cuttings in the mud pit or mud ditch is accomplished by: A careful balance between gel strength and viscosity Considering velocities as an important factor The use of the shale shaker and other separation devices at the surface

Source: http://www.ptarmiganservices.com/news/bak ken-solids-control-and-recycling/

Source: http://indonesiabentonite.blogspot.com/2013/10/ bentonite-drilling-fluid.html

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Negative Functions of Drilling Mud Some of the negative functions - which we don’t want the drilling mud to do, are: Deposit of thick mud cake (reduces the diameter of hole; worsens swabbing and further caving) Fluid loss, allowing a harmful amount of water into the formation

 Causes swelling;  Disintegration of the shales and clays;  And may reduce the permeability to hydrocarbons (oil and gas).

Source: http://servicepompa.blogspot.com/p/kendalakendala-teknis.html

Source: http://inibumi.blogspot.com/2011/02/in vasion-drilling-process.html

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Physical Properties of Drilling Mud Physical properties of drilling mud: Density Viscosity Filtration properties such as water loss and mud cake The yield point

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Classification of Muds Based on Liquid Phase Freshwater Natural or Native Nitrate Phosphate Organic colloidal Alkaline (pH > 10) Calcium Lime Gypsum

Saltwater Saturated salt Emulsion Freshwater, oil in water emulsion Saltwater, oil in water emulsion

Oil-based Note: Muds are listed in order of expense from low to high

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Pressurized Mud Balance

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Marsh Funnel

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Rotational Viscometer

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Viscometer, Rev/Min

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Problem Solving Class Activity In pairs, solve the following problem: At 200 rev/min, what is the shear stress?

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Rotational-Viscometer Geometry

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Lesson 1 Wrap Up What is still unclear? What questions do you have about the topics we have discussed before we move on?

Homework Assignment 5.1: Module 5 Self Study Review Assignment 5.1: Read Fundamentals of Drilling Engineering pp. 87-98

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Lesson 2: Drilling Fluid Properties

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Lesson 2: Drilling Fluid Properties Learning Objectives In this lesson we will: Calculate specific gravity Calculate hydrostatic pressure Demonstrate pilot testing procedures Calculate weight or density control

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Specific Gravity The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh), or,

Sp.Gr. of water = 1.0 = 1.0 gm/cm3 then, If a fluid weight is 2.4 gm/cm3 Sp.Gr. = 2.4 gm/cm3 = 2.4 Density (ρ); Mass per volume of a material in any units; or, ρ = Mass. Vol. Common units used for drilling fluids, gm/cm3 (or Sp.Gr.), lbm/gal, lbm/ft3, ξ lbm/bbl therefore,

Density of fresh water

= 1 gm/cm3 = 8.34 lbm/gal = 62.4 lbm/ft3 = 350 lbm/bbl

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Hydrostatic Pressure Calculation Water Force per unit area exerted by a vertical column of fluid, or, Common units, gmf/cm2, lbf/in2, or lbf/ft2. Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled with water, the force exerted on the base will be 62.4 lbf, therefore, Pressure = P = Force/Area = 62.4 lbf = 62.4 lbf/ft2 1.0 ft2 or, P = 62.4 lbf = 0.433 lbf/in2/ft = 0.433 psi/ft 144 in2 therefore, Water = Sp.Gr. of 1.0 exerts a pressure of 0.433 psi/foot of vertical column.

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Hydrostatic Head and Hydrostatic Pressure

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Hydrostatic Pressure—Other Fluids

Other fluids; Wt./ft3 = (62.4) (Sp.Gr.) Then, Pressure exerted = (62.4) (Sp.Gr.) lbf = lbf/in2/ft 144 in2 Or lbf/in2/ft = (0.433) (Sp.Gr.) Or

lbf/in2 = (0.433 psi/ft) (Sp.Gr.) (Height)

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Class Activity: Hydrostatic Pressure Examples Example #1: What is the Sp.Gr. of a fluid whose density is 78 lbm/ft3? Solution #1: Sp.Gr. = 78 lbm/ft3 62.4 lbm/ft3

= 1.25

Example #2: What is the density in lbm/gal of a fluid whose Sp.Gr. is 1.3? Solution #2: ρ= (1.3) (8.34 lbm/gal) = 10.84 lbm/gal

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Class Activity: Hydrostatic Pressure Examples (Cont.) Example #3: What is the density in lbm/bbl of a fluid whose density is 11.5 lb./gal 1.3? Solution #3: ρ = (11.5 lbm/gal) (42 gal/bbl) = 483.0 lbm/bbl Example #4: What is the total weight of 10 bbl. of material whose Sp.Gr. is 4.3? Solution #4: Wt. = (ρ) (Vol.) lbm = lbm (bbl) = (4.3) (350 lbm/bbl) (10 bbl) = 15,050 lbm. bbl

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Class Activity: Hydrostatic Pressure Examples (Cont.) Example #5: Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs. Solution #5: ρ = 500 lbm = 166.7 lbm/ft3 3 ft3 ρ = 166.7 lbm/ft3 = 22.3 lbm/gal 7.48 gal/ft3 ρ = (166.7 lbm/ft3) (5.615 ft3/bbl) = 963.0 lbm/bbl ρ = (166.7 lbm) (454 gm/lbm) = 2.67 gm/cm3 (ft3) (28320 cm3/ft3) or, Sp.Gr. = 166.7 lbm/ft3 62.4 lbm/ft3

= 2.67

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Class Activity: Hydrostatic Pressure Examples (Cont.) Example # 6: What pressure will a 9.4 lbm/gal mud exert at a depth of 3500 ft.? Solution #6: psi = (Sp.Gr.) (0.433) (height) = ( 9.4 ) ( 0.433) (3500) = 1708 psi 8.34 (Note : Sp.Gr. = lbm/gal 8.34 and, psi = (Sp.Gr.) (0.433) (h) psi = lbm/gal (0.433) (h) = (lbm/gal) ( 0.433 ) (h) 8.34 8.34 psi = (lbm/gal) (0.052) (h)

Or, psi = (9.4) (0.052) (3500) = 1711 psi 1 cubic foot contains 7.48 U.S. gallons; a fluid weighing 1 ppg would weigh 7.48 pounds per cubic foot; The pressure exerted by one foot height of fluid over the area of the base would be:7.48/144 in2=0.052 psi

Note:

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Class Activity: Hydrostatic Pressure Examples (Cont.) Example #7: What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft.? Solution: ρ=

psi

(0.052) (h)

=

3000

= 11.54 lbm/gal

(0.052) (5000)

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Class Activity: Hydrostatic Pressure Examples (Cont.) Example # 8: Point “A” is at an elevation of 1200 ft. And Point “B” is 2 miles east at an elevation of 900 ft. A 2" pipe line is carrying water from “A” to “B”. What is the difference in hydrostatic pressure? Solution # 8: psi = (Sp.Gr.) (0.433) (h), where h = vertical height psi = (1.0) (0.433) (1200-900) = 130 psi.

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Pilot Testing Procedures Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs, therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample, Or, 1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample Therefore, a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel

Note: Often the density of a fluid is referred to as the weight of the fluid. Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity.)

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Pilot Testing Procedures Another way of looking at it:

Mass

Volume

350 pounds mass Manipulations explain how many pounds per barrel of additives needed to change properties of the fluid Experiments that are scaled down so that adding X more pounds to existing 350 pounds is equivalent to adding X more small units of mass to 350 existing small units of mass. A small unit is a gram. Scale the volume for the same proportion of these units:1 gram is 1/454 of a pound

1 blue barrel of water = 158.9873 litres

Experiments that are scaled down so that 1 blue barrel volume is equivalent another volume for small units of mass New volume is 1/454 blue barrels, accordingly: 158.9873 / 454 = 0.350 liters = 350 cubic centimeters

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Pilot Testing Procedure (Cont.) Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity, water loss, gel strengths, etc. ….but it is not normally used to determine the amount of additive needed to obtain the desired density. The amount of additive to obtain the desired density can be calculated, the amount of additive to control other mud properties cannot be calculated.

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Class Activity: Pilot Testing Procedure Example Example # 9: How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP? Solution # 9: Using a 350 cm3 sample of the original mud, the following laboratory data were obtained: Bentonite added, gms

Resulting viscosity, cP

0

5

4

8

6

12

8

18

16

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Plot gms of bentonite added to a 350 cm3 sample vs. Resulting viscosity in cP

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Desired Viscosity The desired viscosity of 20 cP can be read from the curve as shown, or, 12.5 gms of bentonite added to the original sample of 350 cm³ results in a viscosity of 20 cP, or,

12.5 gms/350 cm³ _ 12.5 lbs/ 1 bbl of the system. 30 25 20 15 10 5 0

0

5

10

15

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Weight or Density Control The following relationships are used to calculate mud weighting problems;

1. Mi + Ma = Mf 2. Vi + Va = Vf and,

ρ= M, V

M = ρ ·V

then,

3. ρi · Vi + ρa · Va = ρf · Vf where, Mi = Initial mass Ma = Added mass Mf = Final mass

Vi = Initial volume Va = Added volume Vf = Final volume

ρi = Initial density ρa = Added density ρf = Final density

(Note: The above relationships assume no chemical reactions)

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Unit Any consistent units can be used in Equation 3, if the product of ρ x V is the same in each term, or, M = lbm ρ = Sp.Gr., lbm/gal, lbm/ft3 , or lbm/bbl V = cm3, gal, ft3, or bbl

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Class Activity: Unit Example # 10 What will be the resulting specific gravity if 0.1 bbl of clay is added to 10 bbl of water? Sp.G.r’s; water = 1 and clay = 2.5

Solution #10: Using Example 3) with ρ = Sp.Gr. and V = bbl Assume water = initial and clay = added then, Sp.Gr.i bbli + Sp.Gr.a bbla = Sp.Gr.f bblf and, ρi = 1.0 Vi = 10 bbl ρa = 2.5 Va = 0.1 bbl ρf = ? Vf = (Vi + Va) = (10 + 0.1) = 10.1 bbl then, ρiVi + ρaVa = ρfVf (1.0) (10) + (2.5) (0.1) = ρf(10.1) ρf = (10 + 0.25) = 1.01 Sp.Gr. (10.1)

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Class Activity: Unit Example # 11 What will be the resulting density in lbm/gal if 87.5 lbs of clay is added to 10 bbl of water? Sp.Gr.’s; water = 1.0 and clay = 2.5 Solution #11: (Using Example 3) with ρ = lbm/gal and V = bbl Assume water = initial and clay = added then, lbm/gali Vi + lbm/gala Va = lbm/galf Vf and,

ρi = (Sp.Gr.) (8.34) = (1.0) (8.34) = 8.34 lbm/gal ρa = (Sp.Gr.) (8.34) = (2.5) (8.34) = 20.85 lbm/gal ρf = ? Vi = 10 bbl Va = Ma/ρa = Ma = 87.5 lbm ____ = 0.1bbl (Sp.Gr.) (350 lbm/bbl) 2.5 × (350 lbm/gal) Vf = (Vi + Va) = (10 +0.1) = 10.1 bbl

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Class Activity: Unit Example # 11 (Cont.) Solution #11, continued: then, ρiVi

+

ρaVa

=

ρfVf

(8.34) (10) + (20.85) (0.1) = ρf (10.1) ρf = (83.4 + 2.085) = 8.46 lbm/gal (10.1) Note: Compare to Example #10, 8.46 = 1.01 Sp.Gr. 8.34

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Class Activity: Unit Example # 12 How many lbs. of clay must be added to 6000 gals. of water to produce a final density of 65 lbm/ft3? Sp.Gr.’s; water = 1.0 and clay = 2.65

Solution #12: Using Equation 3 where ρ = lbm/ft3 and V = gals Assume water = initial and clay = added and,

ρi = 62.4 lbm/ft3 ρa = (Sp.Gr.) (62.4) = (2.65) (62.4) = 165.4 lbm/ft3 ρf = 65 lbm/ft3

Vi = 6000 gal Va = Ma/ρa = ? Vf = (Vi + Va) = (6000 + Va) 44

Class Activity: Unit Example # 12 (Cont.)

Solution # 12, continued: then, ρiVi

+

ρaVa

=

ρfVf

(62.4) (6000) + (165.4) (Va) = (65) (6000+ Va) (165.4 - 65) (Va) = (65 - 62.4) (6000) Va = 155.4 gal then, Ma = ρaVa = (Sp.Gr.) (8.34) (155.4 gal) = (2.65) (8.34) (155.4) Ma = 3,435 lbs.

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Class Activity: Unit Example # 13 How many bbls of water must be added to an existing system of 400 bbls of 11.5 lbm/gal mud to reduce the density to 10.5 lbm/gal?

Solution #13: Using Equation 3) where ρ = lbm/gal and V = bbl Assume 11.5 lbm/gal mud = initial and water = added then, ρi = 11.5 lbm/gal ρa = 8.34 lbm/gal ρf = 10.5 lbm/gal

Vi = 400 bbl Va = ? Vf = (Vi + Va) = (400 + Va) 46

Class Activity: Unit Example # 13 (Cont.)

Solution # 13, continued: then, ρiVi

+ ρaVa

= ρf (Vi + Va)

(11.5) (400) + (8.34) (Va) = (10.5) (400 + Va) (8.34 - 10.5) (Va) = (10.5 - 11.5) (400)

Va = (0.463) (400) = 185 bbls

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Class Activity: Unit Example # 14 How many lbs. of barite must be added to a 300 bbl, 9.2 lbm/gal system to control a formation pressure of 2550 psi at a depth of 5000 ft.?

Solution # 14: Assume 9.2 lbm/gal = initial and barite = added and, ρi = 9.2 lbm/gal ρa = (Sp.Gr.) (8.34) = (4.2) (8.34) = 35 lbm/gal ρf =

Psi (0.052) (h)

=

2550 (0.052) (5000)

Vi = 300 bbl Va = Ma/ ρa = ? Vf = (Vi + Va) = (300 + Va) 48

= 9.81 lbm/gal

Class Activity: Unit Example # 14 (Cont.)

Solution # 14, continued: then, ρiVi + ρaVa = ρfVf

(9.2) (300) + (35) (Va) = (9.81) (300 + Va) (35 - 9.81) Va = (9.81 - 9.2) (300) Va = 7.26 bbl and, Ma = ρaVa = (Sp.Gr.) (350) (Va) = (4.2) (350) (7.26) = 10,672 lbs.

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Class Activity: Unit Example # 15 How many bbls of water and lbs of clay are needed to make 250 bbls of 9.5 lbm/gal mud? Sp.Gr.’s; water = 1.0 and clay = 2.4.

Solution # 15: Assume water = initial and clay = added and,

ρi = 8.34 lbm/gal ρa = (2.4) (8.34) = 20 lbm/gal ρf = 9.5 lbm/gal

Vi = ? Va = (Vf - Vi) = (250 -Vi) Vf = 250 bbl

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Class Activity: Unit Example # 15 (Cont.)

Solution #15, continued: then, ρρiVi + ρaVa = ρfVf (8.34) (Vi) + (20) (250 - Vi) = (9.5) (250) (8.34 - 20) Vi = (9.5 - 20) (250)

Vi = 225 bbls (water) Va = (250 - Vi) = (250 - 225) = 25 bbls Ma = ρaVa = (2.4) (350) (25) = 21,000 lbs (clay)

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Class Activity: Unit Example # 16 Drilling at 4400 ft. with 9.15 lbm/gal mud. A pressure of 3000 psi is expected at a depth of 5600 ft. Calculate the required mud treatment.

Solution # 16: Use barite, Sp.Gr. = 4.3, as weighting material Calculate treatment in bbl barite added per 1 bbl of initial system Assume 9.15 lbm/gal mud = initial and barite = added and, ρi = 9.15 lbm/gal ρa = (4.3) (8.34) = 35.86 lbm/gal ρf = 3000 = 10.32 lbm/gal (0.0519) (5600) Vi = 1bbl Va = ? Vf = (Vi + Va) = (1 + Va) 52

Class Example: Unit Example # 16 (Cont.)

Solution # 16, continued: then, ρiVi + ρaVa = ρfVf

(9.15) (1) + (35.86)Va = (10.32) (1+ Va) (35.86 - 10.32) Va = (10.32 - 9.15) (1) Va = 0.046 bbl Ma = ρaVa = (4.3) (350) (0.046) = 69.23 lb/bbl Note: When additives are added to increase or decrease mud density, other mud properties must be checked to insure they are within operating limits.

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Lesson 2 Wrap Up What is still unclear? What questions do you have about the topics we have discussed before we move on? Homework Assignment 5.1: Module 5 Self Study Review Assignment 5.2: Read Fundamentals of Drilling Engineering pp. 98 - 119

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Lesson 3: Buoyancy and Hook Loads

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Lesson 3: Buoyancy and Hook Loads Learning Objectives

In this lesson we will: List three methods of calculating hook load Describe buoyancy as an example of the Archimedes Principle Define basic hook loads

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Hook Loads

Three methods of calculating hook load Displacement Buoyancy Factor Hydrostatic Pressure

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Basic Hook Loads

The basic hook loads which must be known are: Weight of casing string, dead weight or suspended in fluid Weight of drill string, dead weight or suspended in fluid Weight of drill string less weight on the bit Weight with pipe or tools stuck in the hole Hole friction, pipe or tools in contact with the hole Weight with applied pump pressures

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Buoyancy: Example of Archimedes Principle

The net force of the fluid on the cylinder is the buoyant force FB

Fup > Fdown because the pressure is greater at the bottom. Hence the fluid exerts a net upward force.

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Archimedes’ Principle Archimedes’ Principle The buoyant force is equal to the weight of the displaced water

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Buoyancy Factor Mud Density, ppg

Mud Density, lb/ft3

Buoyancy Factor (BF) = (65.5 – mud density, ppg) ÷ 65.5

Buoyancy Factor (BF) = (490 – mud density, lb/ft3) ÷ 490

Example: Determine the buoyancy factor for a 13.0 ppg fluid:

Example: Determine the buoyancy factor for a 97.24 lb/ft3 fluid

BF = (65.5 – 13.0) ÷ 65.5 BF = 0.8015

BF = (490 – 97.24) ÷ 490 BF = 0.8015

Note: 65.5 ppg is the density of steel

Note: 490 is the density of steel

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How to Use the Buoyancy Factor Buoyed Weight: The air weight of drilling string x the buoyancy factor = to actual weight in mud For example, determine the string weight in 13.0 ppg mud. Air weight of string is 350000 lbf. The buoyancy factor for a 13.0 ppg fluid: BF = (65.5 – 13.0) ÷ 65.5 BF = 0.8015 The buoyed weight of drill string in 13.0 ppg mud = 350 x 0.8015 = 280000 lbf.

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The Buoyant Force The buoyant force can be expressed as: a.

a. The buoyant force will be equal to the weight of the displaced fluid

b.

b. The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe

c.

c. The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor.

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Hook Load and Buoyancy Calculation: Example # 1

Example #1: Calculate the effective weight of 1000 ft. of 96.21 lb/ft, 6 in. diameter solid steel rod suspended in water. Solution: Using (a) – weight of displaced fluid: Dead weight = (1000) (96.21) = 96210 lbf Volume of displaced fluid = 0.7854 (6)2 (1000) = 196.35 ft3 144 Weight of displaced fluid = (196.3) (62.4) = 12.252 lbf Effective weight = 96210 - 12252 = 83958 lbf.

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Hook Load and Buoyancy Calculation: Example # 1 (Cont.)

Using (b) - hydrostatic pressure; Dead weight = (1000) (96.21) = 96210 lbf Hydrostatic pressure = (62.4/144) (1000) = 433.3 psi Area of exposed bottom = (0.7854) (6)2= 28.27 in2 Buoyant force = (433.3) (28.27) = 12249 Effective weight = 96210 - 12249 = 83961 lbf

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Hook Load and Buoyancy Calculation: Example # 1 (Cont.) Using (c) - Buoyancy factor: Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid, or, BF = m/ft3 in air - m/ft3 of fluid m/ft3 BF = 490 lb/ft3 - 62.4 lb/ft3 = 0.8727 490 lb/ft3 Dead weight = (1000) (96.21) = 96210 lbf then, Effective weight = (Dead weight) (BF) = 96210 * 0.8727 = 83960 lbf

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Hook Load and Buoyancy Calculation: Example # 2

What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lb/gal mud? Solution: (a) Density of oil field steel = 490 lb/ft3 Density of water = 8.34 lb/gal = 62.4 lb/ft3 BF = 490 lb/ft3 - 62.4 lb/ft3 = 0.8727 490 lb/ft3 (b) BF = 490 lb/ft3 - (10 lb/gal) (7.48 gal/ft3) = 0.8473 490 lb/ft3

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Hook Load and Buoyancy Calculation Example # 3 Calculate the weight indicator reading when 5000 ft. of 5 1/2 in. O.D., 4.67 in. I.D. , 22.56 lb./ft/ (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lb./gal. mud. Solution 1: > Using weight (wt.) = (5000) (22.56) = 112800 lbf. Volume (Vol.) of displaced (displ.) fluid = 0.7854 (5.52 - 4.672) (5000) = 144 = 230.2 ft3 Wt. of displ. fluid = (230.2 ft3) (12 lb/gal) (7.48 gal/ft3) = 20663 lbf W.I. = 112800 - 20663 = 92137 lbf

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Hook Load and Buoyancy Calculation Example # 3 Calculate the weight indicator reading when 5000 ft. of 5 1/2 in. O.D., 4.67 in. I.D. , 22.56 lb./ft/ (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lb./gal. mud. Solution 2: > Using hyd. pressure; Dead wt. = 112800 lbf Ph = (0.052) (12*5000) = 3120 psi Exposed area = (3120) (6.63) = 20684 lbf -- =0.7854*(5.52 - 4.672)= = 6.63 in2 Buoyant force =pressure*area= (3120) (6.63) = 20684 lbf W.I. = 112800 - 20684 = 92116 lbf.

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Hook Load and Buoyancy Calculation: Example # 3 (Cont.)

> Using B.F. Dead wt. = 112800 lbf B.F. = 490 - (12) (7.48) = 0.8168 490 W.I. = (0.8168) (112800) = 92135 lbf Note: (7.48) is gallons per cubic foot

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Hook Load and Buoyancy Calculation: Example # 3 (Cont.)

Displacement Volume: Since there are coupling on tubing, tool joints on drill pipe, collars on casing, etc., the volume and weight of these couplings must be considered. Oil field tubular goods are described by the outside diameter, OD, in inches and fractions of an inch and by the weight per foot, lb/ft. (Note: This is not always true when describing oil well tubing. Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing. When given the nominal diameter of tubing, it is necessary to refer to a handbook to determine the OD, ID, and lb/ft.)

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Hook Load and Buoyancy Calculation: Example # 3 (Cont.)

For the same outside diameter, an increase in the weight per foot will decrease the inside diameter. The weight per foot also includes the weight of the couplings as a distributed weight. Consider 4 ½", 16.6 lb/ft, grade D drill pipe; Wall thickness = 0.337" and ID = 3.826" (pipe body) (Note: ID = OD -2 x wall thickness and that ID's are expressed as inches and decimals of an inch) The weight of this pipe is 16.6 lb/ft which is a distributed weight including tool joints. Using 4 ½" OD, 3.826" ID, and density of steel = 490 lbm/ft3, the weight per foot of the pipe body is: 0.7854 (4.52 - 3.8262)(1)(490) = 15 lb/ft 144 72

Hook Load and Buoyancy Calculation: Example # 3 (Cont.)

Therefore, to calculate the volume displaced by this pipe, the weight per foot including connections must be used, or, Displ. volume = weight per foot (length) density = lbm/ft (ft) = ft3 lbm/ft3 then, 16.6(1) = 0.0339 ft³/ft displacement 490 This displacement volume must be used to calculate the buoyant force when using the wt. of displaced fluid method.

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Hook Load and Buoyancy Calculation: Example # 4 Calculate the W.I. reading when 3750' of 2" nominal tubing is suspended in a hole filled with salt water (Sp.Gr. = 1.15). Solution: 2" nom. tubing; OD = 2.375" (H-40)

ID = 1.1995"

lb/ft = 4.70 lb/ft Dead wt. = (3750)(4.70) = 17625 lbf (includes couplings) Displ. volume = 4.70((3750) = 35.97 ft3 490 Wt. of displ. fluid = (35.97) (1.15)(62.4) = 2581 lbf W.I. = 17625 - 2581 = 15044 lbf or, Dead wt. = 17625 lbf B.F. = 490 - (1.15)(62.4) = .08536 490 W.I. = (0.8536)(17625) = 15045 lbf

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Note: The single quote ( ʹ) means foot and double quote ( “ ) means inches.

Hook Load and Buoyancy Calculation: Example # 5

A drill string consists of 9,000' if 24.7 lb/ft drill pipe and 450' of 7", 109.68 lb/ft drill collars. Hole fluid = 10.5 lb/gal mud. Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud). Solution: Dead wt. = (9000)(24.7) + (450)(109.68) = 271656 lbf B.F. = 490 - (10.5)(7.48) = 0.8397 490 W.I. = (0.8397)(271656) = 228113 lbf

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Hook Load and Buoyancy Calculation: Example # 5 (Cont.) Bit Weight: Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time), weight on the bit (lb.), rotation speed (RPM), bit wear (teeth or bearings), and efficient removal of the cuttings (mud properties and circulation rate). The proper combinations are based on manufacturer's recommendations, experimental data, rules of thumb, and experience. Generally, hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds. The weight on the bit should be applied by the drill collars. (Note: Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression). A general rule for determining the number of drill collars to be used is that approximately 2/3 of the total length of collars should be used to put weight on the bit.

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Hook Load and Buoyancy Calculation: Example # 6

How many feet of 6 3/4", 108.0 lb/ft drill collars would be needed to put 20000 lb. weight on the bit when drilling in 9.6 lb/gal mud? Solution: Effective wt./ft of drill collars suspended in mud; B.F. = 490 - (9.6)(7.48) = 0.8535 490 Eff. wt./ft = (0.8535)(108.0) = 92.2 lb/ft No. of feet = 20000 lbf 92.2 lb/ft

= 217'

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Hook Load and Buoyancy Calculation: Example # 7

If there were 330' of drill collars and 8500' if 4 1/2", 20 lb/ft drill pipe in Example # 6, what would be the weight indicator reading while drilling? Solution: Total effective wt. of string = (B.F.)(dead wt.) = (0.8535){(8500)(20) + (330)(108.0)} = 175514 lbf W.I. = 175514 - 20000 = 155514 lbf

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Hook Load and Buoyancy Calculation: Example # 8 Approximately how many drill collars (total) would be needed in Example # 7? (1 drill collar = 30'). Solution: 217' needed for 20000 lbf bit weight By general rule, this is 2/3 of total length Total length = (217) /(2/3) = 325.5' No. drill collars = 325.5/30 = 10.85, or use 11 drill collars (330') Since the effective wt./ft. of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth, the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases. In Example # 6, 217' of drill collars will be needed to put 20000 lbf on the bit at any depth.

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Hook Load and Buoyancy Calculation: Example # 9 What will be the increase in weight indicator reading when increasing drilling depth to 9730' from Example # 8? Solution: Total effective wt. at (8500 +330) = 155514 lbf Eff. wt./ft of drill pipe = (B.F.)(wt./ft.) = (0.8535)(20) = 17.07 lb/ft

W.I. increase = increase in total eff. wt. = (17.07)(9730 - 8830) = 15514 lbf or, W.I. = 155514 + 15,363 = 170887 lbf (Note: W.I. reading increases 17.07 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit.)

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Casing Loads Usually, the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe). The weight per foot of casing is higher than most other strings. Casing allows small clearances between the outside diameter of the casing and the hole, therefore, additional loads due to friction may be added when the casing is hoisted. Frictional loads must be estimated and are usually between 10-25% of the total effective weight (depending on hole condition).

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Casing Load: Example # 10 Calculate the weight indicator reading when hoisting 6,000' of 9 5/8", 43.5 lb/ft casing if the estimated frictional load is 15%. Hole and pipe are filled with 10 lb./gal mud.

Solution: Dead wt. = (6000)(43.5) = 261000 lbf Eff. wt. = (B.F.) (Dead wt.+) = (490 - (10)(7.48))(261000) = (0.8473)(261000) = 221145 lbf 490 Total Eff. wt. = Pipe eff. wt. + friction load = Pipe eff. wt. + (0.15)(Pipe eff. wt.) W.I. = 1.15(Pipe eff. wt.) = 1.15(221145) = 254317 lbf

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Casing Load Example # 10 (Cont.) Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing. This reduces the load on the hoisting system considerably while running in the hole. The total effective weight of the string decreases due to the buoyant force being increased. The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars, therefore, the volume of displaced fluid will be, Volume displaced by the total metal in the pipe (I.D. is flush or constant) or, Volume displaced by the total metal in the pipe = Wt./ft(length) = ft3

490 Volume of the I.D. of the pipe = (.7854)(I.D.)2 = ft3 (144)

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Hook Load and Buoyancy Calculation: Example # 11 Calculate the weight indicator reading when reaching landing depth of 8200' using 7", 29 lb/ft, (I.D. = 6.184") if the pipe is run empty. Hole fluid is 9.5 lb/gal mud.

Solution: W.I. = Dead weight - buoyant force Dead wt. = (8200)(29) = 237800 lbf Volume of displ. fluid = {lb/ft + (.7854)(I.D.)2}( length) 490 144 = {29 + (.7854)(I.D.)2}} (8200) 490 144 = (0.268)(8200) = 2196 ft3

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Hook Load and Buoyancy Calculation: Example # 11 (Cont.)

Wt. of displ. fluid = (2196) (9.5)(7.48) = 156048 lbf W.I. = 237800 - 156048 = 81752 lbf or, Since additional volume occupied by the casing collars is small, using the O.D. of the casing shows very small error , therefore, the casing O.D. is usually used to calculate the displacement volume of casing.

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Hook Load and Buoyancy Calculation: Example # 12 Calculate the W.I. reading in Example # 11 using the casing O.D. to determine the displacement volume.

Solution: Dead wt. = 237800 lbf Volume of fluid displaced = (.7854)(7)2(8200) = 2191 ft3 144 Wt. of displ. fluid = (2191)(9.5)(7.48) = 155692 lbf W.I. = 237800 - 155692 = 82108 lbf (Note: Compare to Example #11)

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Hook Load and Buoyancy Calculation Example # 13 Calculate the W.I. reading for Example #13 after the pipe is filled with mud. Solution: Dead wt. = 237800 lbf W.I. = Eff. wt. = (B.F.)(dead wt.) = {490 - (9.5)(7.48)} (237800) = (.08550) (237800) = 203319 lbf

490 (Note: Compare to Example #12)

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Hook Loads Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system. The casing block system uses more cables thru the traveling block to increase the mechanical advantage, therefore, decreasing the load per line.

Note: Increasing the number of lines and sheave wheels increases friction losses, but the decrease in load per line is greatly reduced.

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Hook Load and Buoyancy Calculation: Example # 14 In Section-2, the hoisting system is 6 lines thru the traveling block. Using this system, calculate the load in the fast line for Example #11. Assume 2% friction per working line (average) and a 1 1/4" cable.

Solution: FLL = HL (No. of supporting lines)(ef) HL = 254317 lbf No. of supporting lines = 6 ef = 1 - (.02)(6) = 0.88 FLL = 254317 = 48166 lbf (6)(0.88) The recommended maximum load for a 1 1/4" cable is approximately 40000 lbf, therefore, the load imposed by the casing is in excess of the recommended load. 89

Hook Load and Buoyancy Calculation: Example # 15 The system can be restrung to use 8 lines through the traveling block. Assuming the same friction losses, calculate the load in the fast line.

Solution: FLL = HL No. of supporting lines (ef) HL = 254317 No. of supporting lines = 8 ef = 1 - (0.2)(8) = 0.84 FLL = 254317 = 38291 lbf (8)(0.84) With this system, the fast line load is less than the maximum recommended load. 90

Hook Loads—Stuck Pipe When pipe is stuck in the hole, the depth at which it is stuck must be determined before any recovery procedure can be used. The depth at which the string is stuck, or Free Point, can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks. Note: Stretch constants for 4 1/2", 16.6 lb/ft Grade D drill pipe is 9.722x10-8 in/ft/lb and for 7", 35 lb/ft, J-55 casing it is 4.545x10-8 in/ft/lb. Each foot of free pipe will stretch this amount for each pound of tension.

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Hook Load and Buoyancy Calculation: Example # 16 Calculate the depth ( Free Point) at which 7000' (total) of 7", 35 lb/ft, J-55 casing is stuck is at a stretch of 11.28" is measured under a tension of 48,000 lb above the total effective weight of the string. Solution: Stretch constant = 4.545x10-8 in/ft.lb Feet of free pipe =

11.28 in

(4.454x10-8 in/ft/lb)(48000 lbf) = 5171 ft.

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Safety Video 7: Handling Compressed Gas Cylinders in the Laboratory Learning Objectives For this video, there are several important objectives: Describe the various methods that are used to compress gases

List the hazards associated with compressed gases and compressed gas cylinders Demonstrate proper storage of compressed gas cylinders Define the safe handling techniques that should be used when working with compressed gas cylinders

Determine what types of fittings and connections are used for most cylinders Test for leaks within a compressed gas system

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Safety Video 7: Handling Compressed Gas Cylinders in the Laboratory The video covers the following topics: Four major ways to compress gases

Hazards of compressed gases Proper storage procedures Markings and labels Handling cylinders safely Connections and fittings Leak detection

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Safety Video 7: Handling Compressed Gas Cylinders in the Laboratory Students, please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory. Complete the assignment as homework. Safety video questions will also be on the Module quizzes. To open the video, hold the control key down and click the link embedded in the assignment. http://enterprise.coe.ttu.edu/LabSafety/Handling Compressed Gas Cylinders.wmv

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Safety Video 7: In Class Recap 1. The proper way to move a compressed gas cylinder is by using a hand

truck, preferably a four-wheeled one? a. True b. False 2. Which of the following are ways to store pressurized gases? a. “Standard compression”

b. As a liquid c. Dissolved in a solvent

d. All Allofofthe theabove above d. 3. “Pressure Relief Devices” (PRD’s) control the speed at which gas comes out of the cylinder? a. True b. False b. False

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Safety Video 7: In Class Recap 4.

The purpose of a regulator is to decrease the “delivery pressure” of compressed gases to a usable and safe level? a. True b. False

5. A good way to tell what type of gas in a cylinder is by the color the cylinder is painted? a. True b. False 6. Cylinders that contain corrosive gases should not be stored for more than how many months? a. 3 months b. 6 6 months months b. c. 9 months d. 12 months

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Safety Video 7: In Class Recap 7. Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together? a. True b. False

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Lesson 3 Wrap Up What is still unclear? What questions do you have about the topics we have discussed before we move on?

Homework Assignment 5.1: Module 5 Self Study Review Assignment 5.3: Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Assignment 5.3: Read Fundamentals of Drilling Engineering pp. 119 - 133

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Credits

Developer Lloyd R. Heinze, Ph.D., Petroleum Engineering/Texas Tech University

Contributors: Rui V. Sitoe, Ph.D., Department of Mechanical Engineering, UEM

Victoria Johnson, Instructional Designer

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