Rig Math

© Union Oil of California, dba Unocal 1999All rights reserved

Basic RigMath

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 ControlWith 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. Forexample, calculate;

When Parentheses ( ) are in an equation they determine what order orsequence to perform the operations. For example, calculate

2 X 3 + 5 = 2 X 3 =6 6 + 5 = 11

2 X (3 + 5) = (3 + 5) = 82 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 differentanswer;

When brackets [ ] are in an equation these operations should beperformed 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/ft4.282 1029.4 =(4.28 x 4.28) 1029.4 = .01776 bbl/ft

9

CapacityID2 1029.4 = bbl/ft4.282 1029.4 =(4.28 x 4.28) 1029.4 = .01776 bbl/ft

1 ft of pipe

4.28 “ ID

.01776 bbl of fluid

10

CapacityTo 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 bblof 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 ftTVD = 12,800 ft

Kick off = 7,500 ft

Given Information:Drillpipe = 4” OD; 15.7 ppf; 3.24” IDCasing Size = 6” OD; 20 ppf; 5.352” ID14.2 ppg MudDeviation = 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 lb1 lb1 lb

123

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

lb

15

Pressure

1 lb

1 lb

1 lb

0123

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 psiArea 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 createdwith 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)

Weightof

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

19

Deviated Well Hydrostatic

TD = 14,000 ftTVD = 12,800 ft

Kick off = 7,500 ft


6” Shoe

Click for Answers

TVD/TD = 12,800 ft

Calculate the Hydrostaticpsi for both of these wells.

20

Worksheet 3Convert 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- TubePractice

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- TubePractice

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

266000 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- TubePractice

27

U- TubePractice

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

3900Then 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

3900Then 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- TubePractice

6000 ft TVD6000 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- TubePractice

6000 ft TVD6000 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- TubePractice

6000 ft TVD6000 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

1415355

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 thePipe.

As soon as the BOP’s are closed, any pressure below the element is a force that is trying to push the pipe out of the hole.

The BOP element contactingthe 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 Tests1. 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 givenin the question?

36

Taking Tests


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 thefluid 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


MWppg = 500 572

39

Equation Triangle

Pressurepsi

MWppg X .052 X TVDft


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 & AnswersEquations & 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 ChangeFor Mud Weight Change- New Pump Pressurepsi = Original Pressurepsi x (New Mud Weightppg Old Mud Weightppg)

For Pump Speed ChangeFor 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/ft5 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/ft6 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 ftTVD = 12,800 ft

Kick off = 7,500 ft


6” Shoe

Calculate Drillpipe Capacity: Bbl/ft Total Barrels3.242 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 ftTVD = 12,800 ft

Kick off = 7,500 ft


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 3Convert 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


In the slides used to calculate hole volume andBHP, what angle was the well deviation?

Return to slides

51

U- TubePractice

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- TubePractice

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 + 1563016

Return to slides

53

U- TubePractice

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- TubePractice

6000 ft TVD6000 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 + 522995

9.6 x .052 x 6000 =2995 psi

0

Return to slides

55

U- TubePractice

6000 ft TVD6000 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- TubePractice

6000 ft TVD6000 ft


1,000 ft of 2 ppg gas

5,000 ft of 9.8 ppg

3558

6,000 ft of 9.8 ppg9.8 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 sideBoth sides6,000 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 150,000 < 223,925 We cannot strip the tool joint through the annular.through the annular.

150,000 > 127,563 We are in a safe condition150,000 > 127,563 We are in a safe condition

59

Taking Tests


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 thefluid 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

• 15.5 ppg Water Based Fluid• 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/ftVolume 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

•4” 15.5 lb/ft drillpipe•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 bbl50 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

• 15.5 ppg Water Based Fluid• 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.

?

•4” 15.5 lb/ft drillpipe•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 PressureHydrostatic in DP = 15.5 x .052 x 14,300 = 11,526 psi

11,626 - 11, 526 = 100 psi SIDPP

Rig Math

Documents

basic rig

dry pipe lengthft

convert mud

fluid column

calculate

ignore friction

gauge read

measure pressure