OIL WELL DRILLING
(2003 Course) (412390) (Sem. – II)
Time : 3 Hours
Max. Marks : 100
Instructions to the candidates:
1) Question Nos. 4 and 8 are compulsory. Out of the remaining attempt 2 questions from Section – I and 2 questions from Section – II.
2) Answers to the two sections should be written in separate books.
3) Neat diagrams must be drawn wherever necessary.
4) Figures to the right indicate full marks.
5) Use of logarithmic tables slide rule. Mollier charts, electronic pocket calculator and steam tables is allowed.
6) Assume suitable data, if necessary.
SECTION – I
a) What is well planning? Discuss importance and different parameters considered while making GTO (Geo Technical Order).
b) Find out minimum drilling cost for the following case study.
Data : Rig cost = $ 50,000/ day
Bit cost = $ 40,000/-
Trip time = 6.5 hr.
CASE Bit weight lb Rotory Drilled Rotating
STUDY speed (rpm) footage (ft) (hrs)
Case I 75000 127 149 5.06
Case II 65000 92 192 8.65
Case III 65000 65 218 12.28
a) Make (pendulum assembly) BHA design using following data. 26″ Bit (IADC 1–1–5) 1 × 26″ Integral blade type stabilizer 3 × 9 – ½ Drill-collar, 9 – ½” Bit sub, 2 × cross over 5 × 8″ drill – collar, 1 × 7 – ¾” OD drilling Jar, 18 × 5″OD HWDP Discuss the use of pendulum assembly and bit side force in brief.
b) A drill string consist of 600 ft, 8″ drill-collar & rest is 5″ drill pipe 19.5 PPF. If required Mop is 100,000 lbs and Mud weight is 10 ppg calculate maximum depth of well that can be drilled using new drill pipe pt=501,090lb, steel density 489.5 pcF. B.F. = 0.847.
a) Explain geometrical planning of type I directional well and find measure depth of a well.
slot co – ordinate 15.32 ft N, 5.06 ft E
Target co-ordinate 1650 ft N, 4510 ft E
TVD target = 9800 ft
TVD rop = 1400 ft
Build uprate =1.5 deg. per 100 ft.
b) Discuss any survey method in detail to calculate true vertical depth, north increment and east increment.
a) What is hydraulics? Discuss different pressure losses and hedstrom number? Effect of ECD on bottom hole pressure.
b) Write short note on :
(i) Ton mile calculations.
(ii) MWD Tool.
(iii) Multilateral wells.
SECTION – II
a) Calculate number of Cement sacks required for Lead slurry and tail slurry. Using following data for 13-3/8″ Casing cementation well depth = 915 M, shoe depth = 910 M, float collar depth = 886 m, cement top = 77m, previous casing (20″ inch) shoe depth = 309 m, Hole diameter 19.124″ from 0 to 309 m Hole diameter 17.5″ from 309m to 915m. Casing O.D = 13 – 3/8″ and ID = 12.515 consider 50% excess on lead slurry and 30% excess on tail slurry. Lead slurry cement yield = 2.0630ft3 / sack slurry density = 12.7 ppg, Tail slurry density = 15.8 ppg, Cement yield = 1.1670 ft3 / sack.
b) Discuss use of the following :
(i) Top plug.
(ii) Float collar.
a) Well containing 13-3/8″ casing shoe at 2480M, Mud weight 1.28 gm/cc, collapse pressure at 2480M = 317 kg/cm2, Burst pressure atsurface = 374 kg /cm 2. Use following casing grade P 110, 72 PPF Collapse resistance = 357 kg /cm2Internal yield pressure = 520 kg /cm2Pipe body yield strength = 989 Dan check casing grade with respect to collapse burst & tension.
b) Write casing seat / shoe depth selection procedure in detail.
a) Discuss different types of jack up ratio. Offshore rigs and drilling operations in brief.
b) In a 3000 psi Bop control unit how many ten gallons capacity accumulator bottles with 1000 psi precharge pressure are required when 96.6 gallons of operating fluid is needed including safety factor for all the functions of Bop stack which has 10,000 psi rated working pressure ram preventor with a closing ratio of 7:1.
c) Discuss Blow out preventor accumulator system in detail.
a) Explain wait & weight method of well control in detail.
b) Prepare a kill sheet using following data.
Hole size = 12.25 inches Measured depth = 8762 ft.
True vertical depth = 8462 feet, Mud weight = 11.6 ppg
Drill string volume = 150.81 bbls casing shoe data.
Casing size = 13 –83inch measured depth = 4734 ft True vertical
depth = 4424 ft.
Annulus drill string × Open hole volume = 469.37 bbls
Annulus drill string × casing volume = 605.47 bbls
Mud pump displacement = 0.119 bbl/ strokes slow circulating rate = 360
psi at 30 SPM, The well has been shut in after a kick kick data.
SIDP = 590 psi SICP = 660 psi p.i gain = 12 bbl.
The well will be killed using the wait and weight method at 30 spm.
Answer the following :
i) What is the kill mud weight required to balance the formation?
ii) How many strokes are required to pump kill mud from surface tobit?
iii) How many strokes are required to pump from bit to casing shoe?
iv) What is total annular volume?
v) What is initial circulating pressure?
vi) What is final circulating pressure?
vii) What is the drill pipe pressure reduction per 100 strokes as kill mud is
being pumped to the bit?
viii) How long will it take to circulate kill mud around the well at 30 SPM?
ix) Which pressure kept constant to bring the pump to slow circulation rate?
Give the value.