Past Year 2010 Drilling

MUHAMMAD RASHDAN BIN RASHIDIOIL AND GAS ENGINEERING

UiTM

2010 Drilling Engineering Exam Question (UiTM)

1.a

(i) Hosting system function is to raise the drill and casing strings while drilling operation or logging tools while logging the formation.

(ii)

Fd=W+ WEn

+Wn

n = 6

E=0.874

W=250,000 lbs

Fd= Total vertical load

Fd=250,000+ 250,000(0.874 )(6)

+ 250,0006

Fd =339340.2 lbs


1.b

Three systems that requires the most powers on a drilling rig are Hoisting system,Rotary system and Circulating system.

Power optimization could be achieved on a drilling rig by use one system at a time .So,the lowest rating generator can be used on on the drilling rig without compromise the whole operation.

2.a

Hydrostatic pressure :

Max.Overburden Pressure :

Overpressure :

Underpressure :

The maximum principal stress at any depth below the earth’s surface is usually assumed to be the weight of the overburden and is approximately equal to 1 psi/ft.

The pore pressure that is simply equal to the weight of the overlying fluid in the capillary pores of the formation.

Mud pressure in the wellbore is kept slightly higher pressure than pore pressure

Mud pressure is less than the pore pressure


2.b

(i) If the Mud Pressure is too high ,there is a possibilities that the mud may loss into the formation and eventually if it is constantly kept in high pressure,it will fracture the formation and may cause the wall of the well to cave in .

(ii) Overpressure zone allow the engineer identify the best casing setting depth just from mud weight method where the pore pressure,mud weight and fracture pressure are plotted in a pressure-depth plot.from the plot,engineer can clearly see the suitable setting depth for every casing.

2.c (i)

P= ρgh

@5000ft

Pressure gradient =0.565psi/ftPressure @ 5000ft = 0.565 psi/ft x 5000ft = 2825psiOverbalance =250psi

(2825 + 250 )psi = ρ x 0.052 (psi/ft)/ppg x 5000 ft

ρ = 11.827 ppg

@6500ft

Pressure gradient =0.784psi/ftPressure @ 5000ft = 0.784 psi/ft x 6500ft = 5096psiOverbalance =250psi

(5096 + 250 )psi = ρ x 0.052 (psi/ft)/ppg x 6500 ft

ρ = 15.816 ppg

(ii) If mud weight of 11.827 is used at 6500ft ,that means the formation pressure is

higher than the mud hydrostatic pressure,thus the formation fluid will flow into low pressure gradient zone which is the drilled well.This will cause influx where the mud pumped cant flow into the annular as it is filled by formation fluid,the returning mud volume will increase more than the volume of mud pumped.If the influx is flowing uncontrolly blowout will occur.


3.a- Provide WOB and hold the bit - Transmit rotational torque from kelly to the drill bit- provide path for circulating mud to lubricate the bit

3.b

Volume of drillstring

ID2

183.35 x length of pipe = Vp(ft3) ÷ 5.615ft3/bbl = Vp (bbl)

-volume of drillpipe-(OD=5”,ID=4.276”,5000ft)

4.2762

183.35 x 5000ft= 498.614(ft3) ÷ 5.615ft3/bbl = 88.80(bbl)

-Volume of drill collar- (OD =6 ¼” ,ID= 2 1316

“ ,360ft)

21316

2

183.35 x 360ft= 19.1744(ft3) ÷ 5.615ft3/bbl = 19.1744(bbl)

Total Volume of Drill string = 88.80 + 19.1744= 107.9744bbls

Volume of Annular

@ 0-5000ft

Dh2−Dp2

183.35 x length of pipe = Va(ft3) ÷ 5.615ft3/bbl = Va (bbl)

Diameter of hole ,Dh =8” Diameter of drilling pipe,Dp = 5”

82−52

183.35 x 5000 ft = 1063.5397(ft3) ÷ 5.615ft3/bbl = 189.41045(bbl)

@5000ft-5350ft

Diameter of hole ,Dh =8” Diameter of drilling pipe,Dp = 6 ¼ ”

82−614

2

183.35 x 5000 ft = 680.05(ft3) ÷ 5.615ft3/bbl = 121.113(bbl)

Total Volume of annular = 189.41045 + 121.113 = 310.52345 bbls

No.of Pump cycle(stroke)


Strokes = barrels ÷ pump output, bbl/stkStrokes = (107.9744 +310.52345)bbls ÷ 0.172bbl/stkStrokes = 2433.1270stk ≈2433 stk

4.a Drilling fluids are vital for drilling operation as it provides lubrication for the drilling bit to cooling the drilling bit.Besides,Drilling fluid allow the cuttings at the bottom of the well be lifted to the surface as it circulating.The Most important function of this drilling fluid is it equalize the wellbore pressure with the formation pressure to prevent formation fluid influx.

4.b

(i)The most suitable type of mud for this clay rich zone are Oil based mud or synthetic Mud

(ii)

The advantage of Oil based mud is It doesnt react with clay(shale).Clay can absorb water easily from water in the mud.so by using this oil base the water absorb by the clay can be reduced,thus the swelling of the clay can be minimized and the risk of stuck pipe is prevented.Other advantages are thin mud cake produced ,high temperature stability,low formation damages and higher penentration rate.

The disadvanteges of this type of mud are expensive,not environmental friendly and require special logging tool.

5.a

(i)Kick is a condition when formation fluid managed to flow into the wellbore and thus increases the volume of mud returning to the surface as the annular are filled by the formation fluid.besides,it also can be detected by a sudden increases in drilling rate and pressure in standpipe.

(ii)

Kick while drilling Raise Kelly above the rotary table until a tool joint appears

Stop the mud pumps

Close the annular preventer

read shut in drill pipe pressure, annulus pressure and pit gain

Kick while tripping Set the top tool joint on slips


Install a safety valve (open) on top of the string

Close the safety valve and annular preventer

make up the Kelly

open the safety valve

read the shut in pressure and the pit gain

5.b

(i)

Pf = PSIDPP + (0.052 psiftppg

x MW) x D

Pf = 120psi+ (0.052 psiftppg

x 10.2ppg) x 10000 ft

Formation pressure, Pf = 5424psi

Kill Mud weight = (Pf + Overbalanced) ÷ Depth

Kill Mud weight = (5424psi +200psi) ÷ 10000ft

Kill Mud weight = 0.5624 psi/ft ÷ 0.052 psi/ft/ppg = 10.815 ppg

(ii)

Pf = Psicp + (0.052 psiftppg

x MW) x (D-hKI) + (0.052 psiftppg

x ρki x hKI )

Pf = 140psi + (0.052 psiftppg

x MW) x (D-hKI) + (0.052 psiftppg

x ρki x hKI )

hki =

Vpit x1029.4

Dh2−Dc2

hki =

11 x1029.4

8.52−6.252

hki = 341.197 ft

5424psi = 140psi + (0.052 psiftppg

x 10.2) x (10000 – 341.197) + (0.052 psiftppg

x ρki x 341.197)


5424psi – 5263.029 psi = (0.052 psiftppg

x ρki x 341.197)

160.971 ÷ (0.052 x 341.197) = ρki

ρki = 9.07275 ppg

9.07275 ppg > 8.33 ppgSalt water kick.

Past Year 2010 Drilling

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