drill hydraulics
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PETE 411Well Drilling
Lesson 10Drilling Hydraulics (cont’d)
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10. Drilling Hydraulics (cont’d)
Effect of Buoyancy on BucklingThe Concept of Stability ForceStability AnalysisMass BalanceEnergy BalanceFlow Through Nozzles
Hydraulic Horsepower Hydraulic Impact Force
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Bucklingof
Tubulars l
l
Slender pipe
suspendedin wellbore
Partiallybuckledslender
pipe
Neutral Point
Neutral Point
F h - F b
F h
F b
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Buckling of Tubulars
l
NeutralPoint
NeutralPoint
• Long slender columns, like DP,have low resistance tobending and tend to fail by
buckling if...• Force at bottom (F b) causes
neutral point to move up• What is the effect of buoyancy
on buckling?• What is NEUTRAL POINT?
F b
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What is NEUTRAL POINT?
l
NeutralPoint
NeutralPoint
• One definition of NEUTRALPOINT is the point abovewhich there is no tendencytowards buckling
• Resistance to buckling isindicated, in part, by:
The Moment of Inertia
444
64I ind d n
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Consider thefollowing :
19.5 #/ft drillpipeDepth = 10,000 ft.Mud wt. = 15 #/gal.
P HYD = 0.052 (MW) (Depth)
= 0.052 * 15 * 10,000 P HYD = 7,800 psi
Axial tensile stress in pipe at bottom
= - 7,800 psi
What is the axial force at bottom?
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What is the axial force at bottom?
Cross-sectional area of pipe= (19.5 / 490) * (144/1) = 5.73 in 2
Axial compressive force = pA
= 44,700 lbf.
Can this cause the pipe to buckle?
22 73.5800,7 in
inlbf
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Axial Tension:
FT = W1 - F2
FT = w x - P 2 (AO - Ai )
At surface, FT = 19.5 * 10,000 - 7,800 (5.73)
= 195,000 - 44,694
= 150,306 lbf.
At bottom, FT = 19.5 * 0 - 7,800 (5.73)
= - 44,694 lbf
Same as before!
FT
F2
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Stability Force:
FS = A ip i - A O p O
F S = (A i - A O) p (if p i = p O)
At surface, FS = - 5.73 * 0 = 0
At bottom, FS = (-5.73) (7,800) = - 44,694 lbs
THE NEUTRAL POINT is where F S = F T
Therefore, Neutral point is at bottom!PIPE WILL NOT BUCKLE!!
i
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LengthofDrill
Collars
Neutral Point
Neutral Point
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Length of Drill Collars
ft /lbf
lbf WF
LDC
BITDCIn Air:
In Liquid:
In Liquidwith S.F.:(e.g., S.F =1.3)
s
f DC
BITDC
W
.F.S*FL
1
ft /lbf lbf
W
FL
s
f DC
BITDC
1
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14State of stress in pipe at the neutral point?
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At the Neutral Point:
The axial stress is equal to the averageof the radial and tangential stresses.
2
t r Z
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Stability Force:
F S = A i P i - A o P o
If F S > axial tension thenthe pipe may buckle.
If F S < axial tension thenthe pipe will NOT buckle.
F S
F T
0 F T
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At the neutral point:
FS = axial load
To locate the neutral point:
Plot F S vs. depth on“axial load ( FT ) vs. depth plot”
The neutral point is located where thelines intersect.
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NOTE:
If p i = p o = p,
then F s = pd d io
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4
or, F s = - A S p
AS
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Axial Load with F BIT = 68,000 lbf
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StabilityAnalysis with
FBIT = 68,000 lbf
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Nonstatic Well Conditions
Physical Laws
Rheological Models
Equations of State
FLUID FLOW
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Physical Laws
Conservation of mass
Conservation of energy
Conservation of momentum
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Rheological Models
Newtonian
Bingham PlasticPower – Law
API Power-Law
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Equations of State
Incompressible fluid
Slightly compressible fluid
Ideal gas
Real gas
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Law of Conservation of Energy
States that as a fluid flowsfrom point 1 to point 2:
QW
vv D D g
V pV p E E
2
1
2
212
112212
2
1
In the wellbore, in many cases Q = 0 (heat) = constant{
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In practical field units this equation simplifies to:
f p p P vv
D D p p
21
22
4
1212
10*074.8
052.0
p 1 and p 2 are pressures in psi is density in lbm/gal.v 1 and v 2 are velocities in ft/sec. p p is pressure added by pump
between points 1 and 2 in psi p f is frictional pressure loss in psi
D1 and D2 are depths in ft.
where
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Determine the pressure at thebottom of the drill collars, if
psi 000,3 pin. 5.2
0 D
ft. 000,10 D
lbm/gal. 12
gal/min. 400 q
psi 1,400
p
1
2
DC
f
ID
p
(bottom of drill collars)
(mud pits)
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Velocity in drill collars
)(in
(gal/min) d448.2
qv 222
ft/se14.26)5.2(*448.2
400v 22
Velocity in mud pits, v 1 0
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400,1000,36.6240,60
400,1000,3)014.26(12*10*8.074-
0)-(10,00012*052.00p
PP)vv(10*074.8
)DD(052.0pp
224-
2
f p21
22
4-
1212
Pressure at bottom of drill collars = 7,833 psig
NOTE: KE in collars
May be ignored in many cases
0
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f p P P vv D D p p
)(10*074.8 )(052.0
21
22
4-
1212
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If
95.0c 10*074.8
pcv
as writtenbemayEquation
d4dn
0 f P
This accounts for all the losses in the nozzle.
Example: ft/sec 30512*10*074.8
000,195.0v 4n
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For multiple nozzles in //
Vn is the same for each nozzleeven if the d n varies!
This follows since p is the same across each nozzle.
tn A117.3
qv
2t
2d
2-5
bit AC
q10*8.311 Δp
10*074.8
pcv 4dn &
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What is Hydraulic Impact Force
developed by bit?
Consider:
psi169,1 Δplb/gal12
gal/min400q
95.0C
n
D