1 pete 411 well drilling lesson 9 drilling hydraulics - hydrostatics
TRANSCRIPT
1
PETE 411Well Drilling
Lesson 9
Drilling Hydraulics
- Hydrostatics
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Drilling Hydraulics - Hydrostatics
Hydrostatic Pressure in Liquid ColumnsHydrostatic Pressure in Gas ColumnsHydrostatic Pressure in Complex ColumnsForces on Submerged BodyEffective (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 or collapse well tubulars or fracture exposed formations
Several aspects of blowout prevention
Displacement of cement slurries and resulting stresses in the drillstring
WHY?
5
Drilling Hydraulics Applications cont’d
Bit nozzle size selection for optimum hydraulics
Surge or swab pressures due to vertical pipe movement
Carrying capacity of drilling fluids
6Fig. 4-2. The Well Fluid System
Well Control ppore < pmud < pfrac
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Forces Acting on a Fluid Element
A)DdD
dpp(
DAF vw
F1 =
F2 =
F3 =
pA
FWV = specific wt. of the fluid
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Pressures in a fluid column
At equilibrium, F = 0
DAFA)DdD
dpp(pA0F vw
dDFdp wv
0 = F1 + F2 + F3
(p = gh)
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Incompressible Fluids
Integrating,
dDFdp vw
0vw pDFp
]0Dwhen pp[ 0
10
Incompressible Fluids
In field units,33.8
*144
4.62 vwF
33.8*433.0
052.0vwF
1’ x 1’ x 1’cube
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Incompressible fluids
If p0 = 0 (usually the case except during well control or cementing
procedures)
then,
0pD052.0p
ft} lbm/gal, {psig, 052.0 Dp
D 052.0
p
p0
p
D
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Compressible Fluids
dDFdp vw
dD 052.0dp
…………… (1)
from (3)
…… (3)
…………… (2)
But, T R M
m ZT R n ZpV
T Z3.80
pM
ZRT
pM
V
m …… (4)
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, R = density, lbm/galM = 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 T Z1544
M
p
dp
dD T Z80.3
M p 052.0 dp
From Eqs. (2) and (4):
DD
pp 00
[D] T Z1544
M ]p[ln Integrating,
]T Z1544
)DD(M[ exp p p 0
0
Assumptions?
15
Example
Column of Methane (M = 16)
Pressure at surface = 1,000 psia Z=1, T=140 F
(i) What is pressure at 10,000 ft?
(ii) What is density at surface?
(iii) What is density at 10,000 ft?
(iv) What is psurf if p10,000 = 8,000 psia?
]T Z1544
)DD(M[ exp p p 0
0
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Example (i)
(i) What is pressure at 10,000 ft?
] T Z1544
)D-M(D [ exp pp 0
0000,10
psia 1188 ] )140460)(1(1544
0)-16(10,000 [ exp 1000
]T Z1544
)DD(M[ exp p p 0
0
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Example cont’d
gal
lbm331.0
600*1*3.80
16*1000
T Z3.80
pM0
(ii) What is density at surface?
gal
lbm395.0
600*1*3.80
16*1188
T Z3.80
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?
?psurf
]T Z1544
)DD(M[ exp p p 0
0
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)DD(052.0pp 1ii
n
1ii0
Fig. 4-3. A Complex
Liquid Column
Dp
pDp
052.0
052.0 0
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
psig 00 p
psig 266,1p 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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Effective (buoyed) Weight
s
fe 1WW
Buoyancy Factor
Valid for a solid body or an open-ended pipe!
sf
f
be
W-W
V-W
FWW
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Example
For steel,
immersed in mud,
the buoyancy factor is:
gal/lbm 5.65s
)/ 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 inID = 4.276 in
22 IDOD4
A
A = 5.265 in2
W = 19.5 lbf/ft * 10,000 ft = 195,000 lbf
AXIAL TENSION, lbf
DE
PT
H,
ft
0 lbf 195,000 lbf
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Example - 15 lb/gal Mud in Wellbore
Drillpipe weight = 19.5 lbf/ft 10,000 ft
OD = 5.000 inID = 4.276 in
22 IDOD4
A
A = 5.265 in2
W = 195,000 - 41,100 = 153,900 lbf
AXIAL TENSION, lbf
DE
PT
H,
ft
0 195,000 lbf
Pressure at bottom = 0.052 * 15 * 10,000 = 7,800 psiF = P * A= 7,800 * 5.265= 41,100 lbf
153,900- 41,100
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Anywhere in the Drill Collars:
Axial Tension = Wt. - Pressure Force - Bit Wt.
b22 dcdcb22T F - Ap xwFFWF
29b221212dpdp
b2121T
FAp)AA(pWxw
FFFWWF
} above (c) { :Pipe Drill At
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 of 19.5 #/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 the
drillstring 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,
22
2
31 in73.5ft
in144*
ft/lb490
ft/lb5.19A
A1
10,000’
10,600’
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Example
Cross-sectional area of collars,
22 in2.43144*
490
147A
212 in5.3773.52.43AAarea alDifferenti
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 ]
22
in2.43*in
lbf268,8
4
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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
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1
Fb = FBIT = 0
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Example
3. At 10,000 ft - (bottom of drillpipe)
FT = W1+W2+F1-F2-Fb
= 88,200 + 7800 lbf/in2 * 37.5in2 - 357,200
= 88,200 + 292,500 - 357,200
= + 23,500 lbf
4
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1
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Example
4. At Surface
FT = W1 + W2 + F1 - F2 - Fb
= 19.5 * 10,000 + 23,500
= 218,500 lbf
Also: FT = WAIR * BF = 283,200 * 0.7710
= 218,345 lbf
4
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1
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]