PETROLEUM ENGINEERING 661 Lesson 2 Drilling Systems Drilling Rigs Drilling a Well Definitions Slide 2 of 49 Homework Read ADE to p. 37 (all of Ch. 1) Learn the Definitions in Lesson 2B ADE # 1.1, 1.2, 1.3, 1.6 due Monday, September 8, 2003 Slide 3 of 49 Drilling Team Drilling Rigs Rig Power System Hoisting System Circulating System . . . Rotary Drilling CHAPTER 1 (ADE) Slide 4 of 49 The Rotary System The Well Control System Well-Monitoring System Special Marine Equipment Drilling Cost Analysis Examples Rotary Drilling - contd Slide 5 of 49 Slide 6 of 49 Slide 7 of 49 Slide 8 of 49 Composite Rig Count Baker Hughes Rig Count50070090011001300150017001900210023002500Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov DecMonthAverage Monthly Rigs Drilling2004 U.S.2004 World2003 U.S.2003 World2002 U.S.2002 World2001 U.S.2001 World2000 U.S.2000 WorldSlide 9 of 49 Noble Drillings Cecil Forbes A Jack-Up Rig Slide 10 of 49 Sonats George Washington A Semi-Submersible Rig Slide 11 of 49 Zapatas Trader A Drillship Slide 12 of 49 Slide 13 of 49 TENSION LEG PLATFORM Slide 14 of 49 Shells Bullwinkle Worlds tallest offshore structure 1,353 water depth Production began in 1989 45,000 b/d 80MM scf/d Slide 15 of 49 Fig. 1.4 The rotary drilling process Slide 16 of 49 Fig. 1.5 Classification of rotary drilling rigs Slide 17 of 49 Fig. 1.13 Engine power output P = F . V Power = Force * Velocity Slide 18 of 49 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 24,000Slide 19 of 49 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, e , is given by e = 2t (1,200) = 7,539.8 rad/min.) The power output can be computed using Eq.1.1 ( ) hp 5 . 397/hp lbf/min - ft 33,000lbf/min - ft (1,740) 7,539.8T P = = =eSlide 20 of 49 Since the fuel type is diesel, the density is 7.2 lbm/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 60hour 1 lbm/gal) (7.2 gal/hr 31.5 wfSlide 21 of 49 Qi = w f H Thus, the overall efficiency of the engine at 1,200 rpm given by Eq. 1.3 is ( ) ( )lbf/min/hp - ft 33,000 lbf/Btu - ft 779 lbm 19,000Btu/ lbm/min 3.78=iQEfficiency = (Power Out / Power in) 23.4% or 0.2341695.4 397.5= = =itQPESlide 22 of 49 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) Copies of ADE # 1.1, 1.2 and 1.3 Slide 23 of 49 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. Slide 24 of 49 Steps to Drill a Well - contd 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. Slide 25 of 49 Steps to Drill a Well - contd 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. Slide 26 of 49 Steps to Drill a Well - contd 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. Slide 27 of 49 Drilling Operations Field Engineers, Drilling Foremen A. Well planning prior to SPUD B. Monitor drilling operations C. After drilling, review drilling results and recommend future improvements - prepare report. D. General duties. What are the well requirements? Objectives, safety, cost Slide 28 of 49 Making a Connection Making a Trip Slide 29 of 49 Making a mouse hole connection Slide 30 of 49 Making a mouse hole connection - contd Single Added. Ready to Drill Moving Kelly to Single in Mousehole Stabbing the Pipe Slide 31 of 49 Use Elevators for tripping Making a trip Put Kelly in Rathole Why trip? Slide 32 of 49 Making a trip - contd Tripping one stand at a time 60-90 ft Slide 33 of 49 Criteria for determining depth limitation Derrick Drawworks Mud Pumps Drillstring Mud System Blowout Preventer Power Plant Slide 34 of 49 T W FIG 1-1 Simple Pulley System T W T = W Derrick Load = LD = 2W (assumes no friction in sheave) Slide 35 of 49 W = 4 T T = W/4 LD = 6 T = 6 W/4 Wn2 nLD |.|

\| +=Assuming no friction FIG 1-2 Block and Tackle System Why n + 2? n = number of lines, Crown block To Travelling block W = weight (hook load) LD = load on derrick Slide 36 of 49 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. Slide 37 of 49 Solution: The tension, T, will be distributed equally between the 10 lines. Therefore, T = 400,000/10 = 40,000 lbf The 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 Slide 38 of 49 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. Slide 39 of 49 A Rotary Rig Hoisting System Slide 40 of 49 Projection of Drilling Lines on Rig Floor TOTAL E = efficiency = Ph/Pi = W/(n Ff ) or Ff = W/(nE) (1.7) Slide 41 of 49 Load on Derrick (considering friction in sheaves) Derrick Load = Hook Load + Fast Line Load + Dead Line Load Fd = W + Ff + Fs (F WWEnWnE EnEnWd = + + + +|\

|.| =1E = overall efficiency: E = en e.g., if individual sheave efficiency = 0.98 and n = 8, then E = 0.851 Slide 42 of 49 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, Slide 43 of 49 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. Slide 44 of 49 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. lbn EWF 590 , 448 * 841 . 0000 , 300= = =( alternatively, E = 0.988 = 0.851 ) Slide 45 of 49 Solution 2. The maximum hook horsepower available is Ph = E-pi = 0.841(500) = 420.5 hp. ---Slide 46 of 49 Solution 3. The maximum hoisting speed is given by vPWbh==|\

|.| hp ft - lbf / minhp300,000 lbf = 46.3 ft / min420 533 000.,Slide 47 of 49 Solution to 3., cont. To pull a 90-ft stand would require t901 9ft46.3 ft / min . min.Slide 48 of 49 Solution 4. The actual derrick load is given by Eq.1.8b: FE EnEnWd = + +|\

|.||\

|.|1 =1+0.841+0.841(8)0.841(8)(300,000) = 382,090 lbf.Slide 49 of 49 Solution 5. The maximum equivalent load is given by Eq.1.9: lbf FWnnFdede000 , 450000 , 300 *84 8 4=|.|

\| +=|.|

\| +=Slide 50 of 49 Solution 6. The derrick efficiency factor is: 000 , 450090 , 382FFEdedd = =84.9% or 849 . 0 E d =