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Q2. a) m w =m M m B V M = m w ρ w + m B ρ B V M = m M m B ρ w + m B ρ B V M = ( m M m B ) ρ B +m B ρ w ρ B ρ w m B = m M ρ B V M ρ B ρ w ρ B ρ w m M =V M ρ M m B = V M ρ M ρ B V M ρ B ρ w ρ B ρ w = V M ρ B ( ρ M ρ w ) ρ B ρ w = 32 × 4200 ( 16001000) 42001000 ¿ 25,200 Kg V w = V M m B ρ B =3225,200 4,200 ¿ 26 m 3 c)
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Answers drilling questions

Feb 06, 2016

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John Murtagh

Answers drilling questions
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Page 1: Answers drilling questions

La Luna

Barinas

Q2. a)

mw=mM−mB

V M=mw

ρw+mB

ρB

V M=mM−mB

ρw+mB

ρB

V M=(mM−mB ) ρB+mB ρw

ρB ρw

mB=mM ρB−V M ρB ρw

ρB−ρw

mM=VM ρM

mB=V M ρM ρB−V M ρB ρw

ρB− ρw=V M ρB (ρM−ρw )

ρB− ρw=

32×4200 (1600−1000 )4200−1000

¿25,200Kg

V w=VM−mB

ρB=32−25,200

4,200 ¿26m3

c)

10 ; 40 ; 59 ; 922.4 ; 2.7 ; 2.87

2 ; 4

Page 2: Answers drilling questions

(i) La Luna

V sh=GR sh−GR zone

GRsh−GRsand

=92−5992−10

=0.4

Barinas

V sh=GR sh−GR zone

GRsh−GRsand

=9 2−4092−10

=0.6

(ii) La Luna

∅=ρma−ρbρma− ρf

=2.87−2.72.87−1

=0.09

Barinas

∅=ρma−ρbρma− ρf

=2.87−2.42.87−1

=0.25

(iii)

Page 3: Answers drilling questions

For La Luna: from the graph at ρ = 2.7 and PET = 4;

Ø = 5%

For Barinas: from the graph at ρ = 2.4 and PET = 2;

Ø = 18%

(iv)

For La Luna: from the graph at ρ = 2.7 and PET = 4;

Lithology: Limestone/Dolomite

For Barinas: from the graph at ρ = 2.4 and PET = 2;

Lithology: Sandstone

Q3. c)(i)

∅=ρma−ρbρma− ρf

=2670−23102670−1000

=0.216

Page 4: Answers drilling questions

∅=∆ t log−∆ tma∆ t f−∆ tma

=2.76×10−4− 1

54721

1642−

15472

=0.219

(ii)

Sw=1.65√ a×Rw

∅m×R t

=1.65√ 1.37×0.04

0.2161.8×27=0.124

Q4. a) W air=L×W Nom=30×19.5¿585 lb

W=585×1.103¿645.3 lb

b) From the table at density of 1438 kgm-3: buoyancy factor = 0.8172

W wet=585×0.8172¿478.1 lb

Drilling 2009/10

Q3. a)

∆ DA=t b×Q p=12×13.8¿165.6 ft

C f (A)=Cb+C r ( tb+t c+tt )

∆ D=

1200+400 (12+0.1+7 )165.6

¿ £53.38 ft−1

CD ( A )=C f ×Df=53.38×6000¿ £320,280

∆ DB=t b×Q p=48×12.6¿604.8 ft

C f (B)=Cb+C r (t b+t c+t t )

∆ D=

2300+400 (48+0.4+7 )604.8

¿ £40.44 ft−1

CD (B )=C f ×D f=40.44×6000¿ £242,640

∆ DC=t b×Q p=72×10.2¿734.4 ft

Page 5: Answers drilling questions

C f (A)=Cb+C r ( tb+t c+tt )

∆ D=

3850+400 (72+0.5+7 )734.4

¿ £48.54 ft−1

CD ( A )=C f ×Df=48.54×6000¿ £291,240

c)

4900 – 5010 feet

V sh=GR sh−GR zone

GRsh−GRsand

=90−4890−10

=0.525

10 ; 48 ; 69 ; 90

Page 6: Answers drilling questions

5200 – 5250 feet

V sh=GR sh−GR zone

GRsh−GRsand

=9 0−6990−10

=0.263

Q4. a)

μ Water (μ)High Clay 13.7 86.3Low Clay 44 56

ρC=ρw×γ c=1000×2.6¿2600kgm−3

ρmud(Hc)=mHc+mw

mHc

ρc+mw

ρw

= 13.7+86.313.72600

+ 86.31000

¿1092kgm−3

ρmud(Lc)=mLc+mw

mLc

ρc+mw

ρw

= 44+5644

2600+ 56

1000¿1371kgm−3

c) (i)

F=Ro

Rw

= 0.3863.97×10−2 =9.72

(ii)

∅=m√ aF=2√ 0.629.72

=0.253

(iii)

Sw=n√ a× Rw

∅m× Rt

= 1√ 0.62×3.79×10−2

0.2532×3.2=0.115¿11.5%

(iv)

6.4 ; 13.7 ; 44

Page 7: Answers drilling questions

So=1−Sw=1−0.115¿0.885¿88.5 %

Q5. b) (i)

0 2000 4000 6000 8000 100000

1000

2000

3000

4000

5000

6000

7000

8000

9000

10000

PoreOverburden

Pressure (psi)

Dept

h (ft

)

(ii)

PG (8000 ft )=3720−08000−0

=0.465 psi / ft

PG (8500 ft )=6800−08500−0

=0.8 psi / ft

PG (9500 ft )=6900−09500−0

=0.726 psi / ft

(iii)

EMW (8000 ft )= PG0.052

=0.4650.052

=8.94 ppg

Page 8: Answers drilling questions

EMW (8500 ft )= 0.80.052

=15.38 ppg

EMW (9500 ft )=0.7260.052

=13.96 ppg

(iv)P=0.052× EMW ×D=0.052×8.94×8500¿3952 psi

Punder=6800−3952=2848 psi

Q6. (i) Pressure drop in drill pipe

∅ 300=μPlastic+Y P=30+10=40

∅ 600=μPlastic+∅ 300=30+40=70

n=3.32 log(∅ 600∅ 300 )=3.32 log( 70

40 )=0.807

K=5.11×∅ 600

1022n=5.11×70

10220.807=1.333

v=0.408Q

ID2=0.408×300

3.8262=8.36 ft /sec

μe=100K ( 96 vID )

n−1

( 3n+14n )

n

=100×1.333 ( 96×8.363.826 )

0.807−1

( 3×0.807+14×0.807 )

0.807

¿50cP

ℜ=928 IDvρμe

=928×3.826×8.36×1050

=5937 (Turb Flow)

a= logn+3.9350

= log 0.807+3.9350

=0.0767

b=1.75−logn7

=1.75−log 0.8077

=0.263

f= a

ℜb= 0.0767

59370.263=0.007805

( dPdL )= f v2ρ25.81×ID

=0.007805×8.362×1025.81×3.826

=0.05524

Page 9: Answers drilling questions

∆ Pdp=( dPdL )∆L=0.05524×5500=304 psi

Pressure drop in drill collars

v=0.408Q

ID2=0.408×300

2.8132=15.47 ft / sec

μe=100K ( 96 vID )

n−1

( 3n+14n )

n

=100×1.333 ( 96×15.472.813 )

0.807−1

( 3×0.807+14×0.807 )

0.807

¿42cP

ℜ=928 IDvρμe

=928×2.813×15.47×1042

=9615 (Turb Flow)

f= a

ℜb= 0.0767

96150.263=0.006875

( dPdL )= f v2ρ25.81×ID

=0.006875×15.472×1025.81×2.813

=0.2266

∆ Pdc=( dPdL )∆ L=0.2266×500=113 psi

Pressure drop across bit

Assuming Cd = 0.95;

∆ PBit=8.33×10−5 ρQ2

Cd2 A t

2 = 8.33×10−5×10×3002

0.952×( π (132+132+132)4×322 )

2=549 psi

Pressure drop in drill collar annulus

Assuming DHole = 8.5 in, Ø100 = 20 and Ø3 = 5;

n=0.657 log(∅ 100∅ 3 )=0.657 log( 20

5 )=0.3956

K=5.11×∅ 100

170.2n= 5.11×20

170.30.3956=13.39

v= 0.408Q

DHole2 −ODdc

2=0.408×300

8.52−6.752=4.57 ft /sec

Page 10: Answers drilling questions

μe=100K ( 144 vDHole−ODdc )

n−1

( 2n+13n )

n

¿100×13.39( 144×4.578.5−6.75 )

0.3956−1

( 2×0.3956+13×0.3956 )

0.3956

¿44 cP

ℜ=928 (DHole−ODdc )vρ

μe=

928× (8.5−6.75 )×4.57×1044

=1687 (Laminar Flow)

f=24ℜ = 24

1687=0.01423

( dPdL )= f v2 ρ25.81× (DHole−ODdc )

=0.01423×4.572×1025.81× (8.5−6.75 )

=0.0658

∆ Pdc /ann=( dPdL )∆ L=0.0658×500=33 psi

Pressure drop in drill pipe annulus

v= 0.408Q

DHole2 −ODdp

2=0.408×300

8.52−4.52=2.35 ft /sec

μe=100K ( 144 vDHole−ODdp )

n−1

(2n+13n )

n

¿100×13.39( 144×2.358.5−4.5 )

0.3956−1

(2×0.3956+13×0.3956 )

0.3956

¿108cP

ℜ=928 (DHole−ODdp ) vρ

μe=

928× (8.5−4.5 )×2.35×10108

=808(Laminar Flow )

f=24ℜ = 24

808=0.0297

( dPdL )= f v2 ρ25.81× (DHole−ODdp )

=0.0297×2.352×1025.81× (8.5−4.5 )

=0.01589

∆ Pdp/ann=( dPdL )∆ L=0.01589×5500=87 psi

Total pressure lost in the system

Page 11: Answers drilling questions

∆ PT=∆PHYD+∆ PDP+∆ PDC+∆PDP /ANN+∆ PDC/ ANN+∆ PBIT¿0+304+113+87+33+549

¿1086 psi

(ii)

PH=∆ PTQ

1714=1086×300

1714=190HP

(iii)

PH=∆ PBitQ

1714=549×300

1714=96HP

(iv)F j=0.01823CdQ√ ρ∆PBit=0.01823×0.95×300√10×549¿385 lbs

Q7. b) (i)

Depth (ft)

ROP (ft/hr)

RPMWOB (1000

lb)

MW (ppg)

d dc

7500 125 120 38 9.51.2317

81.2317

8

7800 66 110 38 9.61.3994

11.3848

4

8000 37 110 37 9.81.5626

11.5147

8

8200 42 110 33 9.91.4735

91.4140

5

8300 41 100 33 101.4528

41.3801

9

8500 34 100 38 10.251.5720

11.4569

9

8600 33 100 40 111.6061

21.3871

8700 32 110 42 111.6701

91.4424

4

Page 12: Answers drilling questions

1.2 1.25 1.3 1.35 1.4 1.45 1.5 1.55 1.6 1.65 1.7 1.75 1.8 1.85 1.9 1.95 26800

7000

7200

7400

7600

7800

8000

8200

8400

8600

8800

dc - Exponent

Dep

th (

ft)

Yes there is an indication of overpressured zone as we can see the dramatic curve change on the graph.

(ii) The depth at the top of the transition zone is 8000 ft.

(iii)

( pD )=( SD )−(( SD )−( PD )n)( dcodcn )

1.2

¿1−(1−0.465 )( 1.38711.86 )

1.2

¿0.624 psi / ft

P=0.624×8600=5366.4 psi

Q8. a)PG pore=9×0.052=0.468 psi/ ft

PGfracture=PG pore+(Pmax

D )=0.468+( 19207000 )=0.742 psi / ft

EMW=PGfracture

0.052=0.742

0.052=14.27 ppg

1.86