PETROLEUM ENGINEERING 411

PETROLEUM ENGINEERING 411

Lesson 2A

Drilling Systems

Drilling Rigs

Drilling a Well

2A. Rigs, Drilling a Well PETE 411 Well Drilling Slide 2 of 53

Homework

Read ADE to p. 21

Learn the Definitions in Lesson 2B

ADE # 1.1, 1.2, 1.3 (on the internet)due Monday, September 8,

2003


Drilling Team Drilling Rigs Rig Power System Hoisting System Circulating System . . .

Rotary Drilling

CHAPTER 1 (ADE)


The Rotary System The Well Control System Well-Monitoring System Special Marine Equipment Drilling Cost Analysis

Examples

Rotary Drilling - cont’d


From the Houston Chronicle, August 2001, 2002 and 2003

The Rig Count


From the Houston Chronicle, Sunday, September 1, 2002


From the Houston Chronicle, Sunday, August 31, 2003


Noble Drilling’s

Cecil Forbes

A Jack-Up Rig


Sonat’s George

Washington

A Semi-Submersible

Rig


Zapata’s Trader

A Drillship


Deep Water Development Options

SUBSEA COMPLETIONS

FIXED PLATFORMS

COMPLIANT TOWERS

FLOATING PRODUCTION

SYSTEMS

TENSION LEG

PLATFORMS



TENSION LEG PLATFORM


Shell’s Auger

Tension Leg

Platform


Shell’s Bullwinkle

World’s tallest offshore structure

1,353’ water depth

Production began in 1989

45,000 b/d80MM scf/d


BOP STACK

GUIDE BASEMUDLINE

MARINERISER

GUIDE LINES

SLIP JOINTTo TENSIONERS


Fig. 1.3 - Typical drilling rig organization


Fig. 1.4

The rotary drilling process


Fig. 1.5

Classification of rotary drilling

rigs


Fig. 1.13 Engine power output

P = F . V

Power = Force * Velocity


TABLE 1.1 - HEATING VALUE OF VARIOUS FUELS

Fuel Type

Density (lbm/gal)

Heating Value (Btu/lbm)

diesel gasoline butane methane

7.2 6.6 4.7 ---

19,000 20,000 21,000 24,000


Example 1.1. A diesel engine gives an output torque of 1,740 ft-lbf at an engine speed of 1,200 rpm. If the fuel consumption rate was 31.5 gal/hr, what is the output power and overall efficiency of the engine?

Solution: The angular velocity, , is given by

= 2 (1,200) = 7,539.8 rad/min.)

The power output can be computed using Eq.1.1

hp5.397/hplbf/min-ft 33,000

lbf/min-ft (1,740) 7,539.8T P


Since the fuel type is diesel, the density is 7.2lbm/gal and the heating value H is 19,000 Btu/lbm (Table 1.1). Thus, the fuel consumption rate w f is:

wf = 3.78 lbm/min.

The total heat energy consumed by the engine is given by Eq. 1.2:

minutes 60

hour 1 lbm/gal) (7.2gal/hr 31.5 w f


Qi = w f H

Thus, the overall efficiency of the engine at 1,200rpm given by Eq. 1.3 is

lbf/min/hp-ft 33,000

lbf/Btu-ft 779lbm19,000Btu/lbm/min 3.78iQ

Efficiency = (Power Out / Power in)

23.4%or 0.2341695.4

397.5

it Q

PE


Drilling a Well

Steps in Drilling a Well Duties of Drilling Engineer Making a Connection Making a Trip Rig Selection Criteria Derrick Loading Definitions (Lesson 2B) (separate)


Steps to Drill A Gas/Oil Well

1. Complete or obtain seismic, log, scouting information or other data.

2. Lease the land or obtain concession.3. Calculate reserves or estimate from

best data available. 4. If reserve estimates show payout,

proceed with well.5. Obtain permits from conservation/

national authority.


Steps to Drill a Well - cont’d

6. Prepare drilling and completion program.

7. Ask for bids on footage, day work, or combination from selected drilling contractors based on drilling program.

8. If necessary, modify program to fit selected contractor equipment.



9. Construct road, location/platforms and other marine equipment necessary for

access to site. 10. Gather all personnel concerned for

meeting prior to commencing drilling(pre-spud meeting)

11. If necessary, further modify program.12. Drill well.



13. Move off contractor if workover unit is to complete the well.

14. Complete well.15. Install surface facilities.16. Analysis of operations with

concerned personnel.


Drilling OperationsField Engineers, Drilling Foremen

A. Well planning prior to SPUDB. Monitor drilling operationsC. After drilling, review drilling results and recommend future improvements

- prepare report.D. General duties.

What are the well requirements? Objectives, safety, cost


Makinga

Connection

Makinga

Trip


Making a mouse hole connection


Making a mouse hole connection - cont’d

Single Added.

Ready to Drill

Moving Kelly to Single in Mousehole

Stabbing the Pipe


Use Elevators

fortripping

Making a trip

Put Kelly in Rathole

Why trip?


Making a trip - cont’d

Tripping one stand at a time

60-90 ft


Criteria for determining depth limitation

Derrick Drawworks Mud Pumps Drillstring Mud System Blowout Preventer Power Plant


T W

• FIG 1-1 Simple Pulley System

T = W

LD = 2W (no friction in sheave)

T W

W W

2W


W = 4 T T = W/4

LD = 6 T = 6 W/4W

n2n

LD

n = numberof lines

W = weight(hook load)

LD = load on derrick

Assuming no friction

• FIG 1-2 Block and Tackle System


Example 1.1(no friction)

The total weight of 9,000 ft of 9 5/8-inch casing for a deep well is determined to be 400,000 lbs. Since this will be the heaviest casing string run, the maximum mast load must be calculated. Assuming that 10 lines run between the crown and the traveling blocks and neglecting buoyancy effects, calculate the maximum load.


Solution:

The tension, T, will be distributed equally between the 10 lines. Therefore,

T = 400,000/10 = 40,000 lbfThe tension in the fast line and dead line will also be 40,000 lbf, so the total load is

40,000 X 12 = 480,000 lbf


Solution, cont.

Example 1.1 demonstrates two additional points.

1. The marginal decrease in mast load decreases with additional lines.

2. The total mast load is always greater than the load being

lifted.


A Rotary Rig Hoisting System


Projection of Drilling Lines on Rig Floor

TOTAL


Load on Derrick(considering friction in sheaves)Derrick Load = Hook Load

+ Fast Line Load + Dead Line Load

Fd = W + Ff + Fs

F WW

En

W

n

E En

EnWd

=

1

E = overall efficiency, e.g., E = en = 0.98n


Example 1.2

A rig must hoist a load of 300,000 lbf. The drawworks can provide an input power to the block and tackle system as high as 500 hp. Eight lines are strung between the crown block and traveling block. Calculate

1. The static tension in the fast line when upward motion is impending,2. the maximum hook horsepower

available,


Example 1.2, cont.

3. the maximum hoisting speed,4. the actual derrick load,5. the maximum equivalent derrick load, and,6. the derrick efficiency factor.

Assume that the rig floor is arranged as shown in Fig. 1.17.


Solution

1. The power efficiency for n = 8 is given as 0.841 in Table 1.2. The tension in the fast line is given by Eq. 1.7.

Tension in the Fast Line,

lbnE

WF 590,44

8*841.0

000,300

( 0.988 = 0.851 )


Solution

2. The maximum hook horsepower available is

Ph = Epi = 0.841(500) = 420.5 hp.


Solution

3. The maximum hoisting speed is given by

vP

Wbh

hp ft - lbf / min

hp

300,000 lbf

= 46.3 ft / min

420 533 000

.,


Solution to 3., cont.

To pull a 90-ft stand would require

t 90

1 9 ft

46.3 ft / min . min.


Solution

4. The actual derrick load is given by Eq.1.8b:

FE En

EnWd

1

=1+0.841 +0.841(8)

0.841(8)(300,000)

= 382,090 lbf.


Solution

5. The maximum equivalent load is given by Eq.1.9:

lbfF

Wn

nF

de

de

000,450

000,300*8

484

n

W

4

WFDLL

Wn4

4nFDLL


Solution

6. The derrick efficiency factor is:

000,450

090,382

F

FE

de

dd

84.9% or 849.0E d

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