PE 2322

Drilling and Completion Systems

Module 5 Fluids Pressure Control

1

2


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 51 Module 5 Self Study Review

Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98

Lesson 2 Drilling Fluid Properties Lesson 2 Objectives

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 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

3

Lesson 1 Functions of Drilling Fluids

4

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

Lesson 1 Functions of Drilling Fluids Learning Objectives

5


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

6

Remove Cuttings

Properties important to the function of removing cuttings of the drilling mud are

Density = rho (ρ)

Viscosity = mu (micro)

Annular Velocity = Va

Type of Flow Size Shape and Density of the Cuttings

Suspension of the Cuttings andor the Gelling Properties

7

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 httppetroleumsupportcommechanical-sticking-mechanism-of-stuck-pipe

8

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 httpwwwdcmudcleaningequipmentcomHow-Does-Gas-Cut-Affect-Mud-Densityhtml

9

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 httpswwwrigzonecomtraininginsightaspinsight_id=291ampc_id=24

10

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

Sourcehttptrenchlessonlinecomindexwebapp-stories-actionid2489archiveyesIssue2013-04-01titlegel-strengths-for-horizontal-vs-vertical-drilling

11

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 httpindonesiabentoniteblogspotcom201310bentonite-drilling-fluidhtml

Source httpwwwptarmiganservicescomnewsbakken-solids-control-and-recycling

12


Some of the negative functions - which we donrsquot 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 httpservicepompablogspotcompkendala-kendala-teknishtml

Source httpinibumiblogspotcom201102invasion-drilling-processhtml

13


Physical properties of drilling mud

Density

Viscosity

Filtration properties such as water loss and mud cake

The yield point

14


Freshwater

Natural or Native

Nitrate

Phosphate

Organic colloidal

Alkaline (pH gt 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

15


16

Marsh Funnel

17


18

Viscometer RevMin

19

Problem Solving Class Activity

In pairs solve the following problem

At 200 revmin what is the shear stress

20


21

What is still unclear

What questions do you have about the topics we have discussed before we move on

Homework



Lesson 1 Wrap Up

22

Lesson 2 Drilling Fluid Properties

23


Calculate specific gravity

Calculate hydrostatic pressure

Demonstrate pilot testing procedures

Calculate weight or density control

Lesson 2 Drilling Fluid Properties Learning Objectives

24

Specific Gravity

The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)

or

SpGr of water = 10 = 10 gmcm3

then

If a fluid weight is 24 gmcm3

SpGr = 24 gmcm3 = 24

Density (ρ) Mass per volume of a material in any units

or

ρ = Mass

Vol

Common units used for drilling fluids

gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl

therefore

Density of fresh water = 1 gmcm3

= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25


Force per unit area exerted by a vertical column of fluid

or

Common units gmfcm2 lbfin2 or lbfft2

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 624 lbf

therefore

Pressure = P = ForceArea = 624 lbf = 624 lbfft2

10 ft2

or

P = 624 lbf = 0433 lbfin2ft = 0433 psift

144 in2

therefore

Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column

Water

26

Hydrostatic Head and Hydrostatic Pressure

27

Hydrostatic PressuremdashOther Fluids

Other fluids

Wtft3 = (624) (SpGr)

Then

Pressure exerted = (624) (SpGr) lbf = lbfin2ft

144 in2

Or

lbfin2ft = (0433) (SpGr)

Or

lbfin2 = (0433 psift) (SpGr) (Height)

28


Example 1

What is the SpGr of a fluid whose density is 78 lbmft3

Solution 1

SpGr = 78 lbmft3 = 125

624 lbmft3

Example 2

What is the density in lbmgal of a fluid whose SpGr is 13

Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3

What is the density in lbmbbl of a fluid whose density is 115 lbgal 13

Solution 3

ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl

Example 4

What is the total weight of 10 bbl of material whose SpGr is 43

Solution 4

Wt = (ρ) (Vol)

lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm

bbl

Class Activity Hydrostatic Pressure Examples (Cont)

30

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 = 1667 lbmft3

3 ft3

ρ = 1667 lbmft3 = 223 lbmgal

748 galft3

ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl

ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6

What pressure will a 94 lbmgal mud exert at a depth of 3500 ft

Solution 6

psi = (SpGr) (0433) (height)

= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)

psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)

834 834

psi = (lbmgal) (0052) (h)

Or

psi = (94) (0052) (3500) = 1711 psi

Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would

weigh 748 pounds per cubic foot The pressure exerted by one foot height of

fluid over the area of the base would be748144 in2=0052 psi


32

Example 7

What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft

Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8

Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure

Solution 8

psi = (SpGr) (0433) (h) where h = vertical height

psi = (10) (0433) (1200-900) = 130 psi


34


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)

35


Another way of looking at it

Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres

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


that 1 blue barrel volume is equivalent

another volume for small units of

mass

A small unit is a gram Scale the

volume for the same proportion of

these units1 gram is 1454 of a pound

New volume is 1454 blue barrels

accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36

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

hellipbut 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

37


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

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

Bentonite added gms Resulting viscosity cP

0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity

The desired viscosity of 20 cP can be read from the curve as shown

or

125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP

or

125 gms350 cmsup3 _ 125 lbs 1 bbl of the system

0

5

10

15

20

25

30

0 5 10 15 20

39


The following relationships are used to calculate mud weighting problems

1Mi + Ma = Mf

2Vi + Va = Vfand

ρ = M M = ρ middotV

Vthen

3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere

Mi = Initial mass Vi = Initial volume ρi = Initial density

Ma = Added mass Va = Added volume ρa = Added density

Mf = Final mass Vf = Final volume ρf = Final density

(Note The above relationships assume no chemical reactions)

40

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 = SpGr lbmgal lbmft3 or lbmbbl

V = cm3 gal ft3 or bbl

41

Class Activity Unit Example 10

What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water

SpGrrsquos water = 1 and clay = 25

Solution 10

Using Example 3) with ρ = SpGr and V = bbl

Assume water = initial and clay = added

then

SpGri bbli + SpGra bbla = SpGrf bblfand

ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then

ρiVi + ρaVa = ρfVf

(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42


What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25

Solution 11

(Using Example 3) with ρ = lbmgal and V = bbl


then

lbmgali Vi + lbmgala Va = lbmgalf Vf

and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl

(SpGr) (350 lbmbbl) 25 times (350 lbmgal)

Vf = (Vi + Va) = (10 +01) = 101 bbl

43

Solution 11 continued

then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)

Note Compare to Example 10 846 = 101 SpGr

834

Class Activity Unit Example 11 (Cont)

44


How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265

Solution 12

Using Equation 3 where ρ = lbmft3 and V = gals


and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46


How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal

Solution 13

Using Equation 3) where ρ = lbmgal and V = bbl

Assume 115 lbmgal mud = initial and water = added

then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then

ρiVi + ρaVa = ρf (Vi + Va)

(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48


How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft

Solution 14

Assume 92 lbmgal = initial and barite = added

and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and

Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs

50


How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24

Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then

ρρiVi + ρaVa = ρfVf

(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)

Vi = 225 bbls (water)

Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52


Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment

Solution 16

Use barite SpGr = 43 as weighting material

Calculate treatment in bbl barite added per 1 bbl of initial system

Assume 915 lbmgal mud = initial and barite = added

and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53

Class Example Unit Example 16 (Cont)


then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl

Note When additives are added to increase or decrease mud density

other mud properties must be checked to insure they are within operating

limits

54

Lesson 2 Wrap Up



Homework


Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119

55

Lesson 3 Buoyancy and Hook Loads

56


List three methods of calculating hook load

Describe buoyancy as an example of the Archimedes Principle

Define basic hook loads

Lesson 3 Buoyancy and Hook Loads Learning Objectives

57

Hook Loads

Three methods of calculating hook load

Displacement

Buoyancy Factor

Hydrostatic Pressure

58

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

59


The net force of the fluid

on the cylinder is the

buoyant force FB

Fupgt Fdown because the pressure is

greater at the bottom Hence the

fluid exerts a net upward force

60

Archimedesrsquo Principle


The buoyant force is equal

to the weight of the

displaced water

61

Buoyancy Factor

Mud Density ppg Mud Density lbft3

Buoyancy Factor (BF) = (655 ndash mud

density ppg) divide 655


density lbft3) divide 490

Example

Determine the buoyancy factor for a

130 ppg fluid

BF = (655 ndash 130) divide 655

BF = 08015

Note 655 ppg is the density of steel

Example


9724 lbft3 fluid


BF = 08015

Note 490 is the density of steel

62

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 130 ppg mud Air weight of string is 350000 lbf

The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015

The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf

63

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

64

Hook Load and Buoyancy Calculation Example 1

Example 1

Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water

Solution

Using (a) ndash weight of displaced fluid

Dead weight = (1000) (9621) = 96210 lbf

Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3

144

Weight of displaced fluid = (1963) (624) = 12252 lbf

Effective weight = 96210 - 12252 = 83958 lbf

65

Using (b) - hydrostatic pressure

Dead weight = (1000) (9621) = 96210 lbf

Hydrostatic pressure = (624144) (1000) = 4333 psi

Area of exposed bottom = (07854) (6)2= 2827 in2

Buoyant force = (4333) (2827) = 12249


Hook Load and Buoyancy Calculation Example 1 (Cont)

66

Using (c) - Buoyancy factor

Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or

BF = mft3 in air - mft3 of fluid

mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then

Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf


67


What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud

Solution

(a) Density of oil field steel = 490 lbft3

Density of water = 834 lbgal = 624 lbft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473

490 lbft3

68


Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud

Solution 1

gt Using weight (wt) = (5000) (2256) = 112800 lbf

Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =

144

= 2302 ft3

Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf

WI = 112800 - 20663 = 92137 lbf

69



Solution 2

gt Using hyd pressure

Dead wt = 112800 lbf

Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2

Buoyant force =pressurearea= (3120) (663) = 20684 lbf

WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf

Note (748) is gallons per cubic foot

71


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 lbft

(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 lbft)

72

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 frac12 166 lbft grade D drill pipe

Wall thickness = 0337 and ID = 3826 (pipe body)

(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)

The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is

07854 (452 - 38262)(1)(490) = 15 lbft

144


73

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

= lbmft (ft) = ft3

lbmft3

then 166(1) = 00339 ftsup3ft displacement

490

This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method


74


Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)

Solution

2 nom tubing OD = 2375

(H-40) ID = 11995

lbft = 470 lbft

Dead wt = (3750)(470) = 17625 lbf (includes couplings)

Displ volume = 470((3750) = 3597 ft3

490

Wt of displ fluid = (3597) (115)(624) = 2581 lbf

WI = 17625 - 2581 = 15044 lbf

or Dead wt = 17625 lbf

BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf

Note The single quote ( ʹ) means

foot and double quote ( ldquo ) means inches

75


A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)

Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

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 manufacturers 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 23 of the total length of collars should be used to put weight on the bit


77


How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud

Solution

Effective wtft of drill collars suspended in mud

BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78


If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling

Solution

Total effective wt of string = (BF)(dead wt) =

(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79


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 23 of total length

Total length = (217) (23) = 3255

No drill collars = 325530 = 1085 or use 11 drill collars (330)

Since the effective wtft 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

80


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 wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft

WI increase = increase in total eff wt =

(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf

(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)

81

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)

82

Casing Load Example 10

Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud

Solution

Dead wt = (6000)(435) = 261000 lbf

Eff wt = (BF) (Dead wt+)

= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490

Total Eff wt = Pipe eff wt + friction load

= Pipe eff wt + (015)(Pipe eff wt)

WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83

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 (ID is flush or constant) or

Volume displaced by the total metal in the pipe = Wtft(length) = ft3

490

Volume of the ID of the pipe = (7854)(ID)2 = ft3

(144)

Casing Load Example 10 (Cont)

84


Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud

Solution

WI = Dead weight - buoyant force

Dead wt = (8200)(29) = 237800 lbf

Volume of displ fluid = lbft + (7854)(ID)2( length)

490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or

Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing

86


Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume

Solution


Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3

144

Wt of displ fluid = (2191)(95)(748) = 155692 lbf

WI = 237800 - 155692 = 82108 lbf

(Note Compare to Example 11)

87


Calculate the WI reading for Example 13 after the pipe is filled with mud

Solution


WI = Eff wt = (BF)(dead wt)

= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

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

89


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 14 cable

Solution

FLL = HL

(No of supporting lines)(ef)

HL = 254317 lbf

No of supporting lines = 6

ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)

The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load

90


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


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)

With this system the fast line load is less than the maximum recommended load

91

Hook LoadsmdashStuck 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 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension

92


Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string

Solution

Stretch constant = 4545x10-8 inftlb

Feet of free pipe = 1128 in

(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93

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

93

Safety Video 7 Handling Compressed Gas Cylinders in the

Laboratory Learning Objectives

94

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

94


Laboratory

95

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

httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv


Laboratory

96

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 ldquoStandard compressionrdquo

b As a liquid

c Dissolved in a solvent

d All of the above

3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder

a True

b False

96

Safety Video 7 In Class Recap

a True

d All of the above

b False

97

4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo 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

6 Cylinders that contain corrosive gases should not be stored for more than how many months

a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98

7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together

a True

98


b False

99

Lesson 3 Wrap Up



Homework


Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory


100

Credits

Developer

Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University

Contributors

Rui V Sitoe PhD Department of Mechanical Engineering UEM

Victoria Johnson Instructional Designer

2


Lesson 1 Functions of Drilling Fluids Lesson 1 Objectives






Marsh Funnel





Lesson 2 Drilling Fluid Properties Lesson 2 Objectives

Specific Gravity





Desired Viscosity


Unit

Lesson 3 Buoyancy and Hook Loads Lesson 3 Objectives

Hook Loads


Hook Load and Buoyancy Calculation Examples

Casing Loads

Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory

3


4












5








6

Remove Cuttings


Density = rho (ρ)





7

Prevent Caving






8




Loss circulation

Gas cut mud



9






10

Suspending Solids






11

Deposit of Cuttings







12





Causes swelling





13



Density

Viscosity


The yield point

14


Freshwater

Natural or Native

Nitrate

Phosphate

Organic colloidal

Alkaline (pH gt 10)

Calcium

Lime

Gypsum

Saltwater

Saturated salt

Emulsion



Oil-based


15


16

Marsh Funnel

17


18

Viscometer RevMin

19




20


21



Homework



Lesson 1 Wrap Up

22


23







24

Specific Gravity


or


then




or

ρ = Mass

Vol



therefore


= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



3


4












5








6

Remove Cuttings


Density = rho (ρ)





7

Prevent Caving






8




Loss circulation

Gas cut mud



9






10

Suspending Solids






11

Deposit of Cuttings







12





Causes swelling





13



Density

Viscosity


The yield point

14


Freshwater

Natural or Native

Nitrate

Phosphate

Organic colloidal

Alkaline (pH gt 10)

Calcium

Lime

Gypsum

Saltwater

Saturated salt

Emulsion



Oil-based


15


16

Marsh Funnel

17


18

Viscometer RevMin

19




20


21



Homework



Lesson 1 Wrap Up

22


23







24

Specific Gravity


or


then




or

ρ = Mass

Vol



therefore


= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



4












5








6

Remove Cuttings


Density = rho (ρ)





7

Prevent Caving






8




Loss circulation

Gas cut mud



9






10

Suspending Solids






11

Deposit of Cuttings







12





Causes swelling





13



Density

Viscosity


The yield point

14


Freshwater

Natural or Native

Nitrate

Phosphate

Organic colloidal

Alkaline (pH gt 10)

Calcium

Lime

Gypsum

Saltwater

Saturated salt

Emulsion



Oil-based


15


16

Marsh Funnel

17


18

Viscometer RevMin

19




20


21



Homework



Lesson 1 Wrap Up

22


23







24

Specific Gravity


or


then




or

ρ = Mass

Vol



therefore


= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



5








6

Remove Cuttings


Density = rho (ρ)





7

Prevent Caving






8




Loss circulation

Gas cut mud



9






10

Suspending Solids






11

Deposit of Cuttings







12





Causes swelling





13



Density

Viscosity


The yield point

14


Freshwater

Natural or Native

Nitrate

Phosphate

Organic colloidal

Alkaline (pH gt 10)

Calcium

Lime

Gypsum

Saltwater

Saturated salt

Emulsion



Oil-based


15


16

Marsh Funnel

17


18

Viscometer RevMin

19




20


21



Homework



Lesson 1 Wrap Up

22


23







24

Specific Gravity


or


then




or

ρ = Mass

Vol



therefore


= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



6

Remove Cuttings


Density = rho (ρ)





7

Prevent Caving






8




Loss circulation

Gas cut mud



9






10

Suspending Solids






11

Deposit of Cuttings







12





Causes swelling





13



Density

Viscosity


The yield point

14


Freshwater

Natural or Native

Nitrate

Phosphate

Organic colloidal

Alkaline (pH gt 10)

Calcium

Lime

Gypsum

Saltwater

Saturated salt

Emulsion



Oil-based


15


16

Marsh Funnel

17


18

Viscometer RevMin

19




20


21



Homework



Lesson 1 Wrap Up

22


23







24

Specific Gravity


or


then




or

ρ = Mass

Vol



therefore


= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



7

Prevent Caving






8




Loss circulation

Gas cut mud



9






10

Suspending Solids






11

Deposit of Cuttings







12





Causes swelling





13



Density

Viscosity


The yield point

14


Freshwater

Natural or Native

Nitrate

Phosphate

Organic colloidal

Alkaline (pH gt 10)

Calcium

Lime

Gypsum

Saltwater

Saturated salt

Emulsion



Oil-based


15


16

Marsh Funnel

17


18

Viscometer RevMin

19




20


21



Homework



Lesson 1 Wrap Up

22


23







24

Specific Gravity


or


then




or

ρ = Mass

Vol



therefore


= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



8




Loss circulation

Gas cut mud



9






10

Suspending Solids






11

Deposit of Cuttings







12





Causes swelling





13



Density

Viscosity


The yield point

14


Freshwater

Natural or Native

Nitrate

Phosphate

Organic colloidal

Alkaline (pH gt 10)

Calcium

Lime

Gypsum

Saltwater

Saturated salt

Emulsion



Oil-based


15


16

Marsh Funnel

17


18

Viscometer RevMin

19




20


21



Homework



Lesson 1 Wrap Up

22


23







24

Specific Gravity


or


then




or

ρ = Mass

Vol



therefore


= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



9






10

Suspending Solids






11

Deposit of Cuttings







12





Causes swelling





13



Density

Viscosity


The yield point

14


Freshwater

Natural or Native

Nitrate

Phosphate

Organic colloidal

Alkaline (pH gt 10)

Calcium

Lime

Gypsum

Saltwater

Saturated salt

Emulsion



Oil-based


15


16

Marsh Funnel

17


18

Viscometer RevMin

19




20


21



Homework



Lesson 1 Wrap Up

22


23







24

Specific Gravity


or


then




or

ρ = Mass

Vol



therefore


= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



10

Suspending Solids






11

Deposit of Cuttings







12





Causes swelling





13



Density

Viscosity


The yield point

14


Freshwater

Natural or Native

Nitrate

Phosphate

Organic colloidal

Alkaline (pH gt 10)

Calcium

Lime

Gypsum

Saltwater

Saturated salt

Emulsion



Oil-based


15


16

Marsh Funnel

17


18

Viscometer RevMin

19




20


21



Homework



Lesson 1 Wrap Up

22


23







24

Specific Gravity


or


then




or

ρ = Mass

Vol



therefore


= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



11

Deposit of Cuttings







12





Causes swelling





13



Density

Viscosity


The yield point

14


Freshwater

Natural or Native

Nitrate

Phosphate

Organic colloidal

Alkaline (pH gt 10)

Calcium

Lime

Gypsum

Saltwater

Saturated salt

Emulsion



Oil-based


15


16

Marsh Funnel

17


18

Viscometer RevMin

19




20


21



Homework



Lesson 1 Wrap Up

22


23







24

Specific Gravity


or


then




or

ρ = Mass

Vol



therefore


= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



12





Causes swelling





13



Density

Viscosity


The yield point

14


Freshwater

Natural or Native

Nitrate

Phosphate

Organic colloidal

Alkaline (pH gt 10)

Calcium

Lime

Gypsum

Saltwater

Saturated salt

Emulsion



Oil-based


15


16

Marsh Funnel

17


18

Viscometer RevMin

19




20


21



Homework



Lesson 1 Wrap Up

22


23







24

Specific Gravity


or


then




or

ρ = Mass

Vol



therefore


= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



13



Density

Viscosity


The yield point

14


Freshwater

Natural or Native

Nitrate

Phosphate

Organic colloidal

Alkaline (pH gt 10)

Calcium

Lime

Gypsum

Saltwater

Saturated salt

Emulsion



Oil-based


15


16

Marsh Funnel

17


18

Viscometer RevMin

19




20


21



Homework



Lesson 1 Wrap Up

22


23







24

Specific Gravity


or


then




or

ρ = Mass

Vol



therefore


= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



14


Freshwater

Natural or Native

Nitrate

Phosphate

Organic colloidal

Alkaline (pH gt 10)

Calcium

Lime

Gypsum

Saltwater

Saturated salt

Emulsion



Oil-based


15


16

Marsh Funnel

17


18

Viscometer RevMin

19




20


21



Homework



Lesson 1 Wrap Up

22


23







24

Specific Gravity


or


then




or

ρ = Mass

Vol



therefore


= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



15


16

Marsh Funnel

17


18

Viscometer RevMin

19




20


21



Homework



Lesson 1 Wrap Up

22


23







24

Specific Gravity


or


then




or

ρ = Mass

Vol



therefore


= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



16

Marsh Funnel

17


18

Viscometer RevMin

19




20


21



Homework



Lesson 1 Wrap Up

22


23







24

Specific Gravity


or


then




or

ρ = Mass

Vol



therefore


= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



17


18

Viscometer RevMin

19




20


21



Homework



Lesson 1 Wrap Up

22


23







24

Specific Gravity


or


then




or

ρ = Mass

Vol



therefore


= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



18

Viscometer RevMin

19




20


21



Homework



Lesson 1 Wrap Up

22


23







24

Specific Gravity


or


then




or

ρ = Mass

Vol



therefore


= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



19




20


21



Homework



Lesson 1 Wrap Up

22


23







24

Specific Gravity


or


then




or

ρ = Mass

Vol



therefore


= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



20


21



Homework



Lesson 1 Wrap Up

22


23







24

Specific Gravity


or


then




or

ρ = Mass

Vol



therefore


= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



21



Homework



Lesson 1 Wrap Up

22


23







24

Specific Gravity


or


then




or

ρ = Mass

Vol



therefore


= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



22


23







24

Specific Gravity


or


then




or

ρ = Mass

Vol



therefore


= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



23







24

Specific Gravity


or


then




or

ρ = Mass

Vol



therefore


= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



24

Specific Gravity


or


then




or

ρ = Mass

Vol



therefore


= 834 lbmgal

= 624 lbmft3

= 350 lbmbbl

25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



25



or




therefore


10 ft2

or


144 in2

therefore


Water

26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



26


27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



27


Other fluids


Then


144 in2

Or


Or


28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



28


Example 1


Solution 1


624 lbmft3

Example 2


Solution 2

ρ= (13) (834 lbmgal) = 1084 lbmgal

29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



29

Example 3


Solution 3


Example 4


Solution 4

Wt = (ρ) (Vol)


bbl


30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



30

Example 5


Solution 5

ρ = 500 lbm = 1667 lbmft3

3 ft3


748 galft3


ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3

(ft3) (28320 cm3ft3)

or

SpGr = 1667 lbmft3 = 267

624 lbmft3


31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



31

Example 6


Solution 6


= ( 94 ) ( 0433) (3500) = 1708 psi

834

(Note SpGr = lbmgal

834

and

psi = (SpGr) (0433) (h)


834 834


Or

psi = (94) (0052) (3500) = 1711 psi





32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



32

Example 7


Solution

ρ = psi = 3000 = 1154 lbmgal

(0052) (h) (0052) (5000)


33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



33

Example 8


Solution 8


psi = (10) (0433) (1200-900) = 130 psi


34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



34



Or




35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



35















mass





accordingly

1589873 454 = 0350 liters = 350

cubic centimeters

36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



36





37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



37


Example 9






0 5

4 8

6 12

8 18

16 28

38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



38

Desired Viscosity


or


or


0

5

10

15

20

25

30

0 5 10 15 20

39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



39



1Mi + Ma = Mf

2Vi + Va = Vfand


Vthen






40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



40

Unit


or

M = lbm



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



41




Solution 10



then


ρi = 10 Vi = 10 bbl

ρa = 25 Va = 01 bbl

ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl

then


(10) (10) + (25) (01) = ρf(101)

ρf = (10 + 025) = 101 SpGr

(101)

42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



42



Solution 11



then


and

ρi = (SpGr) (834) = (10) (834) = 834 lbmgal

ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal

ρf =

Vi = 10 bbl

Va = Maρa = Ma = 875 lbm ____ = 01bbl


Vf = (Vi + Va) = (10 +01) = 101 bbl

43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



43


then


(834) (10) + (2085) (01) = ρf (101)

ρf = (834 + 2085) = 846 lbmgal

(101)


834


44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



44



Solution 12



and

ρi = 624 lbmft3

ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3

ρf = 65 lbmft3

Vi = 6000 gal

Va = Maρa =

Vf = (Vi + Va) = (6000 + Va)

45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



45



then


(624) (6000) + (1654) (Va) = (65) (6000+ Va)

(1654 - 65) (Va) = (65 - 624) (6000)

Va = 1554 gal

then

Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)

Ma = 3435 lbs

46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



46



Solution 13



then

ρi = 115 lbmgal

ρa = 834 lbmgal

ρf = 105 lbmgal

Vi = 400 bbl

Va =

Vf = (Vi + Va) = (400 + Va)

47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



47



then


(115) (400) + (834) (Va) = (105) (400 + Va)

(834 - 105) (Va) = (105 - 115) (400)

Va = (0463) (400) = 185 bbls

48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



48



Solution 14


and

ρi = 92 lbmgal

ρa = (SpGr) (834) = (42) (834) = 35 lbmgal

ρf = Psi = 2550 = 981 lbmgal

(0052) (h) (0052) (5000)

Vi = 300 bbl

Va = Ma ρa =

Vf = (Vi + Va) = (300 + Va)

49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



49



then


(92) (300) + (35) (Va) = (981) (300 + Va)

(35 - 981) Va = (981 - 92) (300)

Va = 726 bbl

and


50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



50



Solution 15


and

ρi = 834 lbmgal

ρa = (24) (834) = 20 lbmgal

ρf = 95 lbmgal

Vi =

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

Vf = 250 bbl

51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



51



then


(834) (Vi) + (20) (250 - Vi) = (95) (250)

(834 - 20) Vi = (95 - 20) (250)


Va = (250 - Vi) = (250 - 225) = 25 bbls

Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)

52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



52



Solution 16




and

ρi = 915 lbmgal

ρa = (43) (834) = 3586 lbmgal

ρf = 3000 = 1032 lbmgal

(00519) (5600)

Vi = 1bbl

Va =

Vf = (Vi + Va) = (1 + Va)

53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



53



then


(915) (1) + (3586)Va = (1032) (1+ Va)

(3586 - 1032) Va = (1032 - 915) (1)

Va = 0046 bbl

Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl



limits

54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



54

Lesson 2 Wrap Up



Homework



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



55


56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



56






57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



57

Hook Loads


Displacement

Buoyancy Factor


58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



58

Basic Hook Loads








59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



59




buoyant force FB




60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



60





displaced water

61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



61

Buoyancy Factor






Example


130 ppg fluid


BF = 08015


Example


9724 lbft3 fluid


BF = 08015


62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



62


Buoyed Weight






63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



63

The Buoyant Force





64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



64


Example 1


Solution


Dead weight = (1000) (9621) = 96210 lbf


144



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



65


Dead weight = (1000) (9621) = 96210 lbf



Buoyant force = (4333) (2827) = 12249



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



66




mft3

BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3

Dead weight = (1000) (9621) = 96210 lbf

then



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



67



Solution



BF = 490 lbft3 - 624 lbft3 = 08727

490 lbft3


490 lbft3

68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



68



Solution 1



144

= 2302 ft3


WI = 112800 - 20663 = 92137 lbf

69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



69



Solution 2



Ph = (0052) (125000) = 3120 psi

Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=

= 663 in2


WI = 112800 - 20684 = 92116 lbf

70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



70


gt Using BF


BF = 490 - (12) (748) = 08168

490

WI = (08168) (112800) = 92135 lbf


71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



71


Displacement Volume



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



72






07854 (452 - 38262)(1)(490) = 15 lbft

144


73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



73



density

= lbmft (ft) = ft3

lbmft3


490



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



74



Solution


(H-40) ID = 11995

lbft = 470 lbft


Displ volume = 470((3750) = 3597 ft3

490


WI = 17625 - 2581 = 15044 lbf


BF = 490 - (115)(624) = 08536

490

WI = (08536)(17625) = 15045 lbf



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



75



Solution

Dead wt = (9000)(247) + (450)(10968) = 271656 lbf

BF = 490 - (105)(748) = 08397

490

WI = (08397)(271656) = 228113 lbf

76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



76

Bit Weight






77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



77



Solution


BF = 490 - (96)(748) = 08535

490

Eff wtft = (08535)(1080) = 922 lbft

No of feet = 20000 lbf = 217

922 lbft

78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



78



Solution


(08535)(8500)(20) + (330)(1080) = 175514 lbf

WI = 175514 - 20000 = 155514 lbf

79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



79




Solution



Total length = (217) (23) = 3255




80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



80



Solution




(1707)(9730 - 8830) = 15514 lbf

or

WI = 155514 + 15363 = 170887 lbf


81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



81

Casing Loads





82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



82



Solution

Dead wt = (6000)(435) = 261000 lbf


= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf

490



WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf

83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



83





490


(144)


84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



84



Solution


Dead wt = (8200)(29) = 237800 lbf


490 144

= 29 + (7854)(ID)2 (8200)

490 144

= (0268)(8200) = 2196 ft3

85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



85



WI = 237800 - 156048 = 81752 lbf or


86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



86



Solution



144


WI = 237800 - 155692 = 82108 lbf


87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



87



Solution



= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf

490


88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



88

Hook Loads




89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



89



Solution

FLL = HL


HL = 254317 lbf


ef = 1 - (02)(6) = 088

FLL = 254317 = 48166 lbf

(6)(088)


90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



90



Solution

FLL = HL


HL = 254317


ef = 1 - (02)(8) = 084

FLL = 254317 = 38291 lbf

(8)(084)


91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



91




92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



92



Solution



(4454x10-8 inftlb)(48000 lbf)

= 5171 ft

93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



93








93



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



94





Markings and labels



Leak detection

94


Laboratory

95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



95





Laboratory

96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



96


a True

b False



b As a liquid


d All of the above


a True

b False

96


a True

d All of the above

b False

97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



97


a True

b False


a True


a 3 months

b 6 months

c 9 months

d 12 months

97


a True

b False

b 6 months

98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



98


a True

98


b False

99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



99

Lesson 3 Wrap Up



Homework




100

Credits

Developer


Contributors



100

Credits

Developer


Contributors



PE 2322

Documents

Transcript of PE 2322