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
(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)
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
∅=∆ 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
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
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
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
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
∆ 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
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
∆ 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
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
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