09 drilling hydraulics hydrostatics
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PETE 411
Well Drill ing
Lesson 9
Drilling Hydraulics- Hydrostatics
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Drilling Hydraulics - Hydrostatics
Hydrostatic Pressure in Liquid Columns
Hydrostatic Pressure in Gas Columns
Hydrostatic Pressure in Complex Columns
Forces on Submerged Body
Effective (buoyed) Weight of Submerged
Body
Axial Tension in Drill String A = FA/A
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Read:
Applied Drilling Engineering, Ch.4
(Drilling Hydraulics) to p. 125
HW #4
ADE #1.18, 1.19, 1.24
Due Monday, Sept 23, 2002
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Drilling Hydraulics Applications
Calculation of subsurface hydrostatic
pressures that may tend to burst orcollapse well tubulars or fracture exposedformations
Several aspects of blowout prevention
Displacement of cement slurries andresulting stresses in the drillstring
WHY?
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Drilling Hydraulics Applications
contd
Bit nozzle size selection for optimum
hydraulics
Surge or swab pressures due to vertical
pipe movement
Carrying capacity of drilling fluids
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6Fig. 4-2. The Well Fluid System
Well Control ppore < pmud < pfrac
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Incompressible Fluids
Integrating,
dDFdp vw=
0vw pDFp +=
]0Dwhenpp[ 0 ==
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Incompressible Fluids
In field units, 33.8*144
4.62 =vwF
33.8*433.0 =
052.0=vwF
1 x 1 x 1
cube
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Incompressible fluids
If p0 = 0 (usually the case except during
well control or cementingprocedures)
then,
0pD052.0p +=
ft}lbm/gal,{psig,052.0 Dp =
D052.0
p=
p0
p
D
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Compressible
Fluids
dDFdp vw=dD052.0dp =
(1)
from (3)
(3)
(2)
But, TRMmZTRnZpV ==
TZ3.80
pM
ZRT
pM
V
m=== (4)
p = pressure of gas, psia
V = gas volume, gal
Z = gas deviation factorn = moles of gas
R = universal gas constant = 80.3
T = temperature, R
= density, lbm/gal
M = gas molecular wt.m = mass of gas
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Compressible Fluids
p = pressure of gas, psia
V = gas volume, gal
Z = gas deviation factor
n = moles of gas
R = universal gas constant,
= 80.3
T = temperature,oR
= density, lbm/gal
M = gas molecular wt.
m = mass of gas, lbm
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Compressible Fluids
=D
D
p
p 00
dD
TZ1544
M
p
dp
dD
TZ80.3
Mp052.0dp =
From Eqs. (2) and (4):
D
D
p
p 00[D]
TZ1544
M]p[ln =
Integrating,
]
TZ1544
)DD(M[exppp 00
=
Assumptions?
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Example (i)
(i) What is pressure at 10,000 ft?
]TZ1544
)D-M(D[exppp 00000,10 =
psia1188])140460)(1(1544
0)-16(10,000[exp1000 =
+=
]
TZ1544
)DD(M[exppp 00
=
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Example contd
gallbm331.0
600*1*3.8016*1000
TZ3.80pM
0 ===
(ii) What is density at surface?
gal
lbm
395.0600*1*3.80
16*1188
TZ3.80
pM000,10 ===
(iii) What is density at 10,000 ft?
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Example
(iv) What is psurf if p10,000 = 8,000 psia?
?p surf =
]TZ1544
)DD(M[exppp 00
=
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)DD(052.0pp 1ii
n
1i
i0
=
+=
Fig. 4-3.A Complex
Liquid
Column
Dp
pDp
=
+=
052.0
052.00
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20Fig. 4-4. Viewing the Well as a Manometer
Pa = ?
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Figure 4.4
})000,10(0.9)000,1(7.16
)700,1(7.12)300(5.8)000,7(5.10{052.00
+
+++=ppa
psig00 =p
psig266,1p a =
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Buoyancy Force = weight of fluid
displaced (Archimedes, 250 BC)
Figure 4-9. Hydraulic forces acting on a foreign body
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Example
For steel,
immersed in mud,
the buoyancy factor is:
gal/lbm5.65s =
)/0.15( gallbmf =
7710565
01511 .
.
.
s
f =
=
A drillstring weighs 100,000 lbs in air.
Buoyed weight = 100,000 * 0.771 = 77,100 lbs
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Axial Forces in Drillstring
Fb = bit weight
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Simple Example - Empty Wellbore
Drillpipe weight = 19.5 lbf/ft 10,000 ft
OD = 5.000 in
ID = 4.276 in
( )22 IDOD4
A =
A = 5.265 in2
W = 19.5 lbf/ft * 10,000 ft = 195,000 lbf
AXIAL TENSION, lbf
DEP
TH,
ft
0 lbf 195,000 lbf
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29b221212dpdp
b2121T
FAp)AA(pWxw
FFFWWF}above(c){:PipeDrillAt
++=
++=
Anywhere in the Drill Pipe:
Axial Tension = Wts. - Pressure Forces - Bit Wt.
FT
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Axial Tension in Drill String
Example
A drill string consists of 10,000 ft of19.5 #/ft drillpipe and 600 ft of 147 #/ft
drill collars suspended off bottom in
15#/gal mud (Fb = bit weight = 0).
What is the axial tension in thedrillstring as a function of depth?
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Example
Pressure at top of collars = 0.052 (15) 10,000
= 7,800 psi
Pressure at bottom of collars = 0.052 (15) 10,600
= 8,268 psi
Cross-sectional area of pipe,
2
2
2
31in73.5
ftin144*
ft/lb490ft/lb5.19A ==
A1
10,000
10,600
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Example
Cross-sectional area of collars,
2
2 in2.43144*
490
147A ==
2
12 in5.3773.52.43AAareaalDifferenti ===
A2
A1
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Example
1. At 10,600 ft. (bottom of drill collars)
Compressive force = pA
= 357,200 lbf
[ axial tension = - 357,200 lbf]
2
2 in2.43*in
lbf268,8=
4
32
1
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Example
2. At 10,000 ft+ (top of collars)
FT = W2 - F2 - Fb
= 147 lbm/ft * 600 ft - 357,200
= 88,200 - 357,200
= -269,000 lbf
4
32
1
Fb = FBIT= 0
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37Fig. 4-11. Axial tensions as a function of depth for Example 4.9
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Example - Summary
1. At 10,600 ft FT = -357,200 lbf[compression]
2. At 10,000 + ft FT = -269,000 lbf[compression]
3. At 10,000 - ft FT = +23,500 lbf[tension]
4. At Surface FT = +218,500 lbf[tension]
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