Rig Math

Union Oil of California, dba Unocal 1999

All rights reserved

Basic Rig

Math

2

Basic Rig Math

Knowledge of Basic Math concepts is necessary to understand

increasingly technical drilling technology and practices.

Unocal philosophy is to understand pressures and pressure

changes in the well. The factors that cause the pressure changes

should be calculated to insure that BHP is constant and that the u-

tube is functioning.

3

Well Control With all the emphasis that we place on mathmatics and calculations,

Well Control is still as simple as a playground teeter-totter. As we

continue learning how to calculate BHP, Hydrostatic Pressure,

Gradients, Volumes and Force - Keep in mind this simple picture.

BHP = 5000 psi

0 0

Hydrostatic = 5000 psi Hydrostatic = 5000 psi

4

Rules of Math

Always do multiplication and division before addition and subtraction. For

example, calculate;

When Parentheses ( ) are in an equation they determine what order or

sequence to perform the operations. For example, calculate

2 X 3 + 5 =

2 X 3 =6

6 + 5 = 11

2 X (3 + 5) =

(3 + 5) = 8

2 X 8 = 16

Always perform the function inside the parentheses first.

5

Rules of Math

Without the Rules of Math the equation would give you a different

answer;

When brackets [ ] are in an equation these operations should be

performed after the operations inside the parentheses. For example,

calculate;

7 + [2 X (3 + 5)] =

7 + [2 X 8] =

7 + 16 = 23

7 + 2 X 3 + 5 =

9 X 3 + 5 =

27 + 5 = 32

6

Rules of Math

When Brackets and Parentheses do not appear, carry out

multiplication and/or division in the order they occur: For

example, calculate;

12 6 x 2 =

2 x 2 = 4

7

Worksheet 1

14 + (2 X 7) =

(14 + 2) X 7 =

14 (2 X 7) =

(14 2) X 7 =

14 X (2 + 7) =

(14 X 2) + 7 =

14 - (2 + 7) =

(14 - 2) + 7 =

3 + 8 x 7 =

12 + 4 2 =

28 - 14 7 + 4 =

18 2 + 4 x 3 - 21 =

32 - 16 x 2 + 5 =

10 2 x 5 + 5 =

15 - 3 x 5 1 =

21 + 14 2 x 7 =

3 x 15 + 5 x 9 =

Click for Answers

8

Rules of Math

Exponents: An exponent is a small number to the top right of

another number; example 32. 3 is the base and 2 is the exponent.

To solve this number you would multiply the base times itself as

many times as the exponent says to; example 3 X 3 = 9

In our world we use exponents when we calculate capacity. For

example the capacity of 5 19.5 lb/ft pipe with 4.28 ID would be -

ID2 1029.4 = bbl/ft

4.282 1029.4 =

(4.28 x 4.28) 1029.4 = .01776 bbl/ft

9

Capacity ID2 1029.4 = bbl/ft

4.282 1029.4 =

(4.28 x 4.28) 1029.4 = .01776 bbl/ft

1 ft of pipe

4.28 ID

.01776 bbl of fluid

10

Capacity

To calculate the fluid capacity with a pipe inside of casing (Annular Capacity)

the equation would be; (ID2 - OD2) 1029.4 = bbl/ft

1 ft

of pipe

8.68 ID of Casing (ID2 - OD2) 1029.4 = bbl/ft (8.682 - 52) 1029.4 =

[(8.68 x 8.68) - (5 x 5)] 1029.4 = 50.34 1029.4 = .0489 bbl/ft

.0489 bbl

of fluid

5 OD of pipe

11

Worksheet 2

1 What is the capacity of 1.25 Coiled Tubing with an ID of 1.09?

2 What is the capacity of 6 5/8 Drill Pipe with an ID of 5.965?

3 What is the capacity of 3 1/2 Tubing with an ID of 2.764?

4 What is the annulus capacity of 9 5/8 Casing with an ID of 8.681

with 3 1/2 OD Tubing in the casing?

5 What is the annulus capacity of 9 5/8 Casing with an ID of 8.681

with 5 OD Drill Pipe in the casing?

6 What is the annulus capacity of 3 1/2 Tubing with an ID of 2.764

with 1 .25 OD Coiled Tubing in the tubing?

Click for Answers

12

Deviated Well Volumes

TD = 14,000 ft TVD = 12,800 ft

Kick off = 7,500 ft

Given Information:

Drillpipe = 4 OD; 15.7 ppf; 3.24 ID

Casing Size = 6 OD; 20 ppf; 5.352 ID

14.2 ppg Mud

Deviation = 40

6 Shoe

Calculate Drillpipe Capacity: Bbl/ft Total Barrels

Calculate Annulus Capacity: Bbl/ft Total Barrels

Click for Answers

13

Pressure

Pressure is a force that is felt over an area. ( Force Area )

1 lb

Within Unocal, we generally measure pressure in pounds per square inch

14

Pressure

0

1 lb 1 lb 1 lb

1 2 3

The total force felt downward is 3 lbs but is this a pressure?

lb

15

Pressure

1 lb

1 lb

1 lb

0 1 2 3

The force felt downward is still 3 lbs

but it is felt over a total surface area

of 1 square inch. Is this pressure?

Force = 3 lbs = 3 psi

Area 1 sq. in.

lb

1 1

16

0 lb

1 1

1

Pressure

In our industry, when we are measuring pressure it is usually pressure created

with a fluid. We will describe most of these in our Well Control class. For now

lets talk about fluid at rest.

Fluid at rest creates a pressure that we call Hydrostatic Pressure.

hydro (Fluid) static (at rest)

Weight

of

Fluid

Phydrostatic = Fluid Weightppg x .052 x Vertical Height of fluid

17

Pressure

12 X 12 = 144 in2

12

12

12

1 1

1 ft. = .052 gal.

A one cubic foot container will hold 7.5 gallons of fluid.

Because we are measuring our pressure in square inches, we section the base into

square inches.

If I now divide the 7.5 gallons by 144 square inches, we find that a column of fluid

1in X 1in X 1ft tall contains .052 gallons of fluid.

18

Gradient

1 1

1 ft. = .052 gal.

If our fluid density is measured in pounds per gallon you can then multiply the

fluid weight (ppg) by .052 to find the hydrostatic pressure (psi) exerted by one

foot of this fluid. This is called the pressure gradient (G) of the fluid or the

pressure change per foot (psi/ft).

Gradientpsi/ft = Fluid Weightppg x .052 x 1ft

If we fill the .052 gallon container with 10 ppg fluid, what will be the pressure?

10ppg x .052gal/sq. in./ft = Pressureft

10 x .052 = .52 psift

This means that for every foot of mud in the well, the pressure

increases by .52 psi. So, Gradientpsi/ft x TVDft = Pressurehydrostatic

19

Deviated Well

Hydrostatic

TD = 14,000 ft TVD = 12,800 ft

Kick off = 7,500 ft

Given Information:



14.2 ppg Mud

Deviation = 40

6 Shoe

Click for Answers

TVD/TD = 12,800 ft

Calculate the Hydrostaticpsi for

both of these wells.

20

Worksheet 3

Convert Mud Weight to Gradient:

1 8.6 ppg

2 9.6 ppg

3 10.2 ppg

4 12.7 ppg

5 14.0 ppg

6 15.1 ppg

7 16.8 ppg

8 17.2 ppg

Convert Gradient to Mud Weight:

1 .46 psi/ft

2 .52 psi/ft

3 .55 psi/ft

4 .6 psi/ft

5 .64 psi/ft

6 .71 psi/ft

7 .83 psi/ft

8 1.00 psi/ft

Click for Answers

21

U- Tube

If I started filling the

glass tube with a fluid

that weighed 9.6 ppg

where would the fluid go

and what would the

gauge read?

10 ft

9.6ppg x .052 x 10ft = 5

22

U- Tube

If I then put another few

gallons of a 12 ppg fluid

in the tube what would

happen and what would

the gauge read?

10 ft

9.6ppg x .052 x 10ft =

5

Two columns of fluid connected at the bottom that will balance each other

in a static condition.

23

U- Tube

10,000 ft

While drilling a well, we have a u-tube in effect.

The workstring and

the annulus form our u-tube.

The gauge should be

Bottom Hole Pressure.

24

U- Tube Practice

6000 ft

6000 ft TVD

1,500 ft of 13.6 ppg

AIR

4,000 ft of 10.2 ppg

10.2 ppg

Calculate Bottom Hole Pressure

Click for Answers

25

U- Tube Practice

6000 ft TVD


6000 ft

1,000 ft of 10 ppg

5,000 ft of 9.6 ppg

5,500 ft of 10 ppg

500 ft of 6 ppg

Click for Answers

26

6000 ft TVD

Calculate how far the slug has dropped.

6000 ft

6,000 ft of 10.5 ppg

1,200 ft of 12 ppg

Click for Answers U- Tube Practice

27

U- Tube Practice

6000 ft TVD

If there is no balance between the two columns of fluid and

the fluid cannot escape, pressure will be created.

6,000 ft of 12.5 ppg 6,000 ft of 10 ppg fluid

6000 ft

IF:

12.5 x .052 x 6000 =

3900 psi

3900 Then BHP =

IF:

10 x .052 x 6000 =

3120 psi

Then surface gauge pressure =

3900 - 3120 = 780 psi

780

28

6,000 ft of 12.5 ppg 6,000 ft of 10 ppg fluid

6000 ft

IF:

12.5 x .052 x 6000 =

3900 psi

3900 Then BHP =

IF:

10 x .052 x 6000 =

3120 psi

Then surface gauge pressure =

3900 - 3120 = 780 psi

780

BHP = 5000 psi

0 0


BHP = 3900 psi

0 780


Well Control

Remember:

29

U- Tube Practice

6000 ft TVD

6000 ft

Calculate the gauge readings:

6,000 ft of 9.6 PPG

500 ft of 2 ppg gas

5,500 ft of 9.8 ppg

140

Click for Answers

30

U- Tube Practice

6000 ft TVD

6000 ft


2,000 ft of 14 PPG

3,000 ft of 12.1 ppg

3,000 ft of 9.8 ppg

3504

4,000 ft of 9.8 ppg

Click for Answers

31

U- Tube Practice

6000 ft TVD

6000 ft


1,000 ft of 2 ppg gas

5,000 ft of 9.8 ppg

3558

6,000 ft of 9.8 ppg

Click for Answers

32

Force

AreaSquare inches = .785 x Diameter2

Forcelbs = Pressurepsi x Areasquare inches

6 Piston 3 piston

1415 355

Which direction will the piston travel?

Force = 355 x (.785 x 62)

355 x (.785 x 36)

355 x 28.26

Force = 10032 lbs

Force = 1415 x (.785 x 32)

1415 x (.785 x 9)

1415 x 7.065

Force = 9997 lbs

33

Stripping

Force

500

The Force down is the weight of the

Pipe.

As soon as the BOPs are

closed, any pressure below

the element is a force that is

trying to push the pipe out of

the hole.

The BOP element contacting

the pipe creates friction, which

is a force that must be

overcome for the pipe to move

up or down.

34

Force

6500

The workstring weighs 150,000 lbs. The annular is

closed in around the 5 19.5 lb/ft pipe with 6 5/8

tool joints. Are we in a safe condition? Can we

strip to bottom through the annular? (Ignore

friction)

Click for Answers

35

Taking Tests 1. Your well is shut-in with 500 psi on the casing. You cannot read drillpipe

pressure. The casing pressure is increasing from 500 psi to 600 psi. You must

bleed off some fluid to reduce the hydrostatic by the amount of pressure increase.

How many barrels of fluid do you bleed?

15.5 ppg Water Based Fluid

Casing Shoe at 9488 ft

14,300 ft TVD

9 5/8 Casing 8.681 ID

5 19.5 lb/ft drillpipe

6 1/2 Hole

a) 6 bbls

b) 10 bbls

c) 14 bbls

d) 20 bbls

What answer are we looking for? Barrels

What information are we given

in the question?

36

Taking Tests



14,300 ft TVD

9 5/8 Casing 8.681 ID


6 1/2 Hole

1. 100 psi = ? barrels

a) 6 bbls

b) 10 bbls

c) 14 bbls

d) 20 bbls

First, think about how we measure pressure

in the well?

Hydrostaticpsi = MW x .052 x TVD

We know the mud weight so we can find the

fluid column height.

Psi

Psi/ft =

Like terms cancel each other out.

FT

Once we know how many feet of mud we can

multiply it times the capacity to find out how

many barrels.

Ft x BBL/Ft =

Like terms cancel each other out.

BBL

To work the calculations we need to convert mud weight

to Gradient and find the annulus capacity. Give it a try!

Click for Answers

37

Equation Triangle

Pressurepsi

Pressurepsi =

MWppg

MWppg

X

X

.052

.052

X

X

TVDft

TVDft

If you want to solve for MW or TVD, fill

in the known information and the

equation is written for you.

38

Equation Triangle

Pressurepsi

MWppg X .052 X TVDft

1) SIDPP is 500 psi. Hole TVD is 11,000 ft.

How much MW increase is needed to kill

the well?

_______ppg

500 psi

? 11000 ft

500 psi MWppg =

.052

.052 x 11000 ft

On your calculator you would key in:

.052 x 11000 = 572

500 572 = .87ppg

.87

If you want to solve for MW or TVD, fill in the

known information and the equation is written

for you.

MWppg = 500

572

39

Equation Triangle

Pressurepsi

MWppg X .052 X TVDft

If you want to solve for MW or TVD, fill in the

known information and the equation is written

for you.

1) While pulling out of the hole, using 9.6

ppg fluid, you forgot to fill the hole. If your

overbalance is 100 psi, how far can the

fluid level drop before you are

underbalance?

_______ft

FT =

?

100 psi 100psi

9.6ppg x

9.6ppg .052

.052

FT = 100

.5

On your calculator you would key in:

9.6 x .052 = .5 psi/ft

100 .5 = 200ft

200

Click for Answers

Equations & Answers

41

Formulas

1 Phydrostatic = MWppg x .052 x TVDft

2 MWppg = Pressurepsi .052 TVDft

3 TVDft = Pressurepsi .052 MWppg

4 Gradientpsi/ft = MWppg x .052

5 Gradientpsi/ft = Pressurepsi TVDft

6 MWppg = Gradientpsi/ ft .052

7 Capacitybbl/ft = Hole Diameter2 1029.4

8 Annular Capacitybbl/ft = (Hole diameter2 - Pipe Diameter2) 1029.4

9 Fluid Column Heightft = Volumebbls Capacitybbl/ft

42

1 Displacementbbl/ft = Pipe Weightlbs x .00036

2 Triplex Pump Outputbbl/stk = .000243 x Liner Diameterin2 x Stroke Lengthin x Efficiency%

3 Total Pump Strokes = Volumebbls Pump Outputbbl/stk

4 Kill Weight Mudppg = (SIDPPpsi .052 TVDft) + MWppg

5 Volume of Slugbbls = Mud Weight.ppg x Dry Pipe Lengthft x Pipe Capacitybbl/ft

Slug Weightppg - Mud Weightppg

6 Slug Weightppg = Mud Weightppg + Mud Weight.ppg x Dry Pipe Lengthft x Pipe Capacitybbl/ft

Slug Volumebbls

7 Pit Gain from Slugbbls = Volume of Slugbbls x Slug Weightppg - Mud Weightppg

Mud Weightppg

8 Depth Slug Fallsft = Pit Gain from Slugbbls Pipe Capacitybbl/ft

9 Pump Pressure Correction: For Mud Weight Change-

New Pump Pressurepsi = Original Pressurepsi x (New Mud Weightppg Old Mud Weightppg)

For Pump Speed Change-

New Pump Pressurepsi = Original Pressurepsi x (New SPM Old SPM)2

Formulas

43

Worksheet 1 Answers

14 + (2 X 7) = 14 + 14 = 28

(14 + 2) X 7 = 16 X 7 = 112

14 (2 X 7) = 14 14 = 1

(14 2) X 7 = 7 X 7 = 49

14 X (2 + 7) = 14 X 9 = 126

(14 X 2) + 7 = 16 + 7 = 23

14 - (2 + 7) = 14 - 9 = 5

(14 - 2) + 7 = 12 + 7 = 19

3 + 8 x 7 = 3 + 56 = 59

12 + 4 2 = 12 + 2 = 14

28 - 14 7 + 4 = 28 - 2 + 4 =

26 + 4 = 30

18 2 + 4 x 3 - 21 = 9 + 12 - 21 =

21 - 21 = 0

32 - 16 x 2 + 5 = 32 - 32 + 5 =

0 + 5 = 5

10 2 x 5 + 5 = 5 x 5 + 5 =

25 + 5 = 30

15 - 3 x 5 1 = 15 - 15 1 =

15 - 15 = 0

21 + 14 2 x 7 = 21+ 7 x 7 =

21 + 49 = 70

3 x 15 + 5 x 9 = 45 + 45 = 90

44

On the first slide that showed the teeter- totter, what was

BHP equal to?

__________ psi

BHP = 5000 psi

0 0


5000 psi

Return to slides

45

Worksheet 2 Answers

1 What is the capacity of 1.25 Coiled Tubing with an ID of 1.09?

1.092 1029.4 = (1.09 x 1.09) 1029.4 = .0012 bbl/ft

2 What is the capacity of 6 5/8 Drill Pipe with an ID of 5.965?

5.9652 1029.4 = (5.965 x 5.965) 1029.4 = .035 bbl/ft

3 What is the capacity of 3 1/2 Tubing with an ID of 2.764?

2.7642 1029.4 = (2.764 x 2.764) 1029.4 = .0074 bbl/ft

4 What is the annulus capacity of 9 5/8 Casing with an ID of 8.681 with 3

1/2 OD Tubing in the casing?

(8.6812 - 3.52) 1029.4 = [ (8.681 x 8.681) - (3.5 x 3.5)] 1029.4 =

(75.36 - 12.25) 1029.4 = .061 bbl/ft

5 What is the annulus capacity of 9 5/8 Casing with an ID of 8.681 with 5

OD Drill Pipe in the casing?

(8.6812 - 52) 1029.4 = [(8.681 x 8.681) - (5 x5)] 1029.4 =

(75.36 - 25) 1029.4 = .048 bbl/ft

6 What is the annulus capacity of 3 1/2 Tubing with an ID of 2.764 with

1 .25 OD Coiled Tubing in the tubing?

(2.7642 - 1.252) 1029.4 = [(2.764 x 2.764) - (1.25 x 1.25) 1029.4 =

(7.64 - 1.56) 1029.4 = .006 bbl/ft

46

1 cubic ft

7.5 gallons of fluid is equal to what?

Return to slides

47

Deviated Well Volumes

TD = 14,000 ft TVD = 12,800 ft

Kick off = 7,500 ft

Given Information:



14.2 ppg Mud

Deviation = 40

6 Shoe

Calculate Drillpipe Capacity: Bbl/ft Total Barrels

3.242 1029.4 = .0102 Bbl/ft .0102 x 14,000 = 142.8 bbls

Calculate Annulus Capacity: Bbl/ft Total Barrels

(5.3522 - 42) 1029.4 = .0123 Bbl/ft .0123 x 14,000 = 172.2 Bbls

Return to slides

48

Deviated Well

Hydrostatic

TD = 14,000 ft TVD = 12,800 ft

Kick off = 7,500 ft

Given Information:



14.2 ppg Mud

Deviation = 40

6 Shoe

TVD/TD = 12,800 ft

Return to slides

Since Hydrostatic pressure is a function

of gravity, only the Vertical Depth is

used.

14.2 x .052 x 12,800 = 9452 psi

49

Worksheet 3

Convert Mud Weight to Gradient:

1 8.6 ppg

2 9.6 ppg

3 10.2 ppg

4 12.7 ppg

5 14.0 ppg

6 15.1 ppg

7 16.8 ppg

8 17.2 ppg

Convert Gradient to Mud Weight:

1 .46 psi/ft

2 .52 psi/ft

3 .55 psi/ft

4 .6 psi/ft

5 .64 psi/ft

6 .71 psi/ft

7 .83 psi/ft

8 1.00 psi/ft

8.6 ppg x .052 = .447 psi/ft

9.6 ppg x .052 = .499 psi/ft

10.2 ppg x .052 = .53 psi/ft

12.7 ppg x .052 = .66 psi/ft

14.0 ppg x .052 = .728 psi/ft

15.1 ppg x .052 = .785 psi/ft

16.8 ppg x .052 = .874 psi/ft

17.2 ppg x .052 = .894 psi/ft

.46 psi/ft .052 = 8.9 ppg

.52 psi/ft .052 = 10 ppg

.55 psi/ft .052 = 10.6 ppg

.6 psi/ft .052 = 11.5 ppg

.64 psi/ft .052 = 12.3 ppg

.71 psi/ft .052 = 13.7 ppg

.83 psi/ft .052 = 16 ppg

1.00 psi/ft .052 = 19.2 ppg

50

Given Information:



14.2 ppg Mud

Deviation = 40

In the slides used to calculate hole volume and

BHP, what angle was the well deviation?

Return to slides

51

U- Tube Practice

6000 ft

6000 ft TVD

1,500 ft of 13.6 ppg

AIR

4,000 ft of 10.2 ppg

10.2 ppg


10.2 x .052 x 6,000 = = (1060) + (2122) 3182

13.6 x .052 x 1,500 = 1060 psi

10.2 x .052 x 4000 = 2122 psi

Return to slides

52

U- Tube Practice

6000 ft TVD


6000 ft

1,000 ft of 10 ppg

5,000 ft of 9.6 ppg

5,500 ft of 10 ppg

500 ft of 6 ppg

10 x .052 x 1000 = 520

9.6 X .052 x 5,000 = 2496

10 x .052 x 5,500 = 2860

6 x .052 x 500 = 156

520 + 2496 = = 2860 + 156 3016

Return to slides

53

U- Tube Practice

6000 ft TVD

Calculate how far the slug has dropped.

6000 ft

6,000 ft of 10.5 ppg

1,200 ft of 12 ppg

10.5 x .052 x 6000 = 3276 psi

3276

12 x .052 x 1200 = 749 psi

2527 .052 10.5 = 4628 ft

3276 - 749 = 2527 psi

6000 - 4628 - 1200 = 372 ft

Return to slides

54

U- Tube Practice

6000 ft TVD

6000 ft


6,000 FT of 9.6 PPG

500 ft of 2 ppg gas

5,500 ft of 9.8 ppg

140

9.8 x .052 x 5500 =

2803 psi

2 x .052 x 500 =

52 psi

= 140 + 2803 + 52 2995

9.6 x .052 x 6000 =

2995 psi

0

Return to slides

55

U- Tube Practice

6000 ft TVD

6000 ft


2,000 ft of 14 PPG

3,000 ft of 12.1 ppg

3,000 ft of 9.8 ppg

3504

4,000 ft of 9.8 ppg

9.8 x .052 x 3000 =

1529 psi

12.1 x .052 x 3000 =

1888 psi

- 1888-1529 = 87 psi

87

14 x .052 x 2000 =

1456 psi

9.8 x .052 x 4000 =

2038 psi

10 psi = 1456 - 2038 -

10

Return to slides

56

U- Tube Practice

6000 ft TVD

6000 ft


1,000 ft of 2 ppg gas

5,000 ft of 9.8 ppg

3558

6,000 ft of 9.8 ppg

9.8 x .052 x 6000 =

3058 psi

3558

- 3058

500 psi

500

9.8 x .052 x 5000 =

2548 psi

2 x .052 x 1000 =

104 psi

- 104 - 2548 = 906 psi

906

57

In a static situation, with the well

open, if the u-tube is broken will

each side of the u-tube be

effected or only the side that is

broken?

Both sides One side Both sides

6,000 ft of 9.8 ppg 6,000 ft of 9.8 ppg

Return to slides

58

Force

6500

The workstring weighs 150,000 lbs. The annular is

closed in around the 5 19.5 lb/ft pipe with 6 5/8

tool joints. Are we in a safe condition? Can we

strip to bottom through the annular?(ignore friction)

Return to slides

.785 x 52 =

.785 x (5 x 5) =

.785 x 25 = 19.625

19.625 x 6500 = 127,563 lbs upward force

.785 x 6.6252 =

.785 x (6.625 x 6.625) =

.785 x 43.89 = 34.45

34.45 x 6500 = 223,925 lbs upward force

150,000 < 223,925 We cannot strip the tool joint through the annular.

150,000 > 127,563 We are in a safe condition

59

Taking Tests



14,300 ft TVD

9 5/8 Casing 8.681 ID


6 1/2 Hole

1. 100 psi = ? barrels

a) 6 bbls

b) 10 bbls

c) 14 bbls

d) 20 bbls

First, think about how we measure pressure

in the well?

Hydrostaticpsi = MW x .052 x TVD

We know the mud weight so we can find the

fluid column height.

Psi 100 psi 100

Psi/ft = (15.5 x .052) = .806 = 124 ft

Once we know how many feet of mud we can

multiply it times the capacity to find out how

many barrels.

Ft x BBL/Ft = 124 x [(8.6812 - 52) 1029.4] =

= 124 x .0489 = 6 bbls

a) 6 bbls

Return to slides

60

1417


9 5/8 Casing 8.681 ID

Shoe at 12,600 ft

6 1/2 Hole @14,300 ft TVD

Calculate the height of the gas, Bottom Hole Pressure & SIDPP.

Volume around DC x Open Hole = (6.52 - 4.752) 1029.4 = .019 bbl/ft

Volume around DP x Open Hole = (6.52 - 42) 1029.4 = .026 bbl/ft

?

Volume around DP x CSG = (8.6812 - 42) 1029.4 = .058 bbl/ft

14,300 - 12,600 = 1,700 ft of open hole; 1,700 - 1,000 of DC = 700 ft of DP x OH


1000 ft of 4 3/4 OD Drill Collars

50 bbls gas kick at .12 psi/ft

Volume of Gas around DC = 1000 ft x .019 = 19 bbls

Volume of Gas around DP x OH = 700 x .026 = 18.2 bbl 50 bbls gas - 19 - 18.2 = 12.8 bbls gas in casing

1000

700

18.2 .058 = 221 ft of gas in casing

221

Height of 50 bbls gas = 1000 + 700 + 221 = 1921 ft

61

1417


9 5/8 Casing 8.681 ID

Shoe at 12,600 ft

6 1/2 Hole @14,300 ft TVD

Calculate the height of the gas, Bottom Hole Pressure & SIDPP.

?


1000 ft of 4 3/4 OD Drill Collars

50 bbls gas kick at .12 psi/ft

1000

700

221

1921

14,300 1921 = 12, 379 ft of 15.5 ppg 15.5 x .052 x 12,379 = 9978 psi

1921

1921 ft of gas x .12 = 231 psi

1417 + 9978 + 231 = 11,626 psi Bottom Hole Pressure

Hydrostatic in DP = 15.5 x .052 x 14,300 = 11,526 psi

11,626 - 11, 526 = 100 psi SIDPP

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