10 hydraulics cont'd
TRANSCRIPT
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PETE 411Well Drilling
Lesson 10Drilling Hydraulics (cont’d)
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10. Drilling Hydraulics (cont’d)
4Effect of Buoyancy on Buckling4The Concept of Stability Force4Stability Analysis4Mass Balance4Energy Balance4Flow Through Nozzles4Hydraulic Horsepower4Hydraulic Impact Force
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READ:ADE, Ch. 4 to p. 135
HW #5:ADE # 4.3, 4.4, 4.5, 4.6
due September 27, 2002
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Buckling of
Tubulars l
l
Slender pipe suspended in wellbore
Partially buckled slender
pipe
Neutral Point
Neutral Point
Fh - Fb
Fh
Fb
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Buckling of Tubulars
l
Neutral Point
Neutral Point
• Long slender columns, like DP, have low resistance to bending and tend to fail by buckling if...
• Force at bottom (Fb) causes neutral point to move up
• What is the effect of buoyancy on buckling?
• What is NEUTRAL POINT?
Fb
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What is NEUTRAL POINT?
l
Neutral Point
Neutral Point
• One definition of NEUTRAL POINT is the point above which there is no tendency towards buckling
• Resistance to buckling is indicated, in part, by:
The Moment of Inertia
( ) { }444
64I indd n −= π
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Consider the following:
19.5 #/ft drillpipeDepth = 10,000 ft.Mud wt. = 15 #/gal.
∆∆∆∆PHYD = 0.052 (MW) (Depth)= 0.052 * 15 * 10,000
∆∆∆∆PHYD = 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 in2
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 - P2 (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 = Aipi - AO pO
FS = (Ai - AO) p (if pi = pO)
At surface, FS = - 5.73 * 0 = 0At bottom, FS = (-5.73) (7,800) = - 44,694 lbs
THE NEUTRAL POINT is where FS = FT
Therefore, Neutral point is at bottom!PIPE WILL NOT BUCKLE!!
Ai
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Compression Tension44,770 0 150,306
FS FT
ft708,7=5.19306,150
Zero Axial Stress
Neutral Point
Depth of Zero Axial Stress Point =
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Length of
Drill Collars
Neutral Point
Neutral Point
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Length of Drill Collars
====ft/lbf
lbfWFL
DC
BITDCIn Air:
In Liquid:
In Liquidwith S.F.: (e.g., S.F =1.3)
ρρρρρρρρ−−−−
====
s
fDC
BITDC
W
.F.S*FL1
ρρρρρρρρ−−−−
====ft/lbf
lbf
W
FL
s
fDC
BITDC
1
14State of stress in pipe at the neutral point?
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At the Neutral Point:The axial stress is equal to the average
of the radial and tangential stresses.
2 tr
Zσσσσσσσσσσσσ ++++====
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Stability Force:
FS = Ai Pi - Ao Po
If FS > axial tension then the pipe may buckle.
If FS < axial tension then the pipe will NOT buckle.
FS
FT
0 FT
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At the neutral point:
FS = axial load
To locate the neutral point:
4Plot FS vs. depth on “axial load (FT ) vs. depth plot”
4The neutral point is located where the lines intersect.
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NOTE:
If pi = po = p,
then Fs = ( )pdd io22
4−− π
or, Fs = - AS p
AS
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Axial Load with FBIT = 68,000 lbf
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Stability Analysis 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 Plastic
�Power – Law
�API Power-Law
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Equations of State
�Incompressible fluid
�Slightly compressible fluid
�Ideal gas
�Real gas
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Average Fluid VelocityPipe Flow Annular Flow
WHEREv = average velocity, ft/sq = flow rate, gal/mind = internal diameter of pipe, in.d2 = internal diameter of outer pipe or borehole, in.
d1 =external diameter of inner pipe, in.
2448.2 dqv = ( )2
122448.2 dd
qv−
=
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Law of Conservation of Energy
States that as a fluid flows from point 1 to point 2:
(((( )))) (((( ))))(((( )))) (((( ))))
QW
vvDDg
VpVpEE
++++====
−−−−++++−−−−−−−−
−−−−++++−−−−
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2212
112212
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In the wellbore, in many cases Q = 0 (heat)ρρρρ = constant{
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In practical field unitsthis equation simplifies to:
( )
( ) fp pPvv
DDpp
∆−∆+−−
−+=
− 21
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4
1212
10*074.8
052.0
ρ
ρ
p1 and p2 are pressures in psiρ is density in lbm/gal.v1 and v2 are velocities in ft/sec.∆pp is pressure added by pump
between points 1 and 2 in psi∆pf is frictional pressure loss in psiD1 and D2 are depths in ft.
where
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Determine the pressure at the bottom 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
ρ(bottom of drill collars)
(mud pits)
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Velocity in drill collars
)(in(gal/min)
d448.2 qv 222 ====
ft/sec 14.26)5.2(*448.2
400v 22 ========
Velocity in mud pits, v1 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
fp21
22
4-
1212
−+−+=
−+−
+=
∆−∆+−ρ−
−ρ+=
Pressure at bottom of drill collars = 7,833 psig
NOTE: KE in collars
May be ignored in many cases
0≈≈≈≈
32fp PPvv
DDpp
∆−∆+−−
−+=
)(10*074.8
)(052.021
22
4-
1212
ρ
ρ
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0 P
v v0 P
0 vD D
f
n2p
112
≈∆
==∆
≈≈
Fluid Flow Through Nozzle Assume:
ρ∆=
ρ−=
−
−
4n
2n
412
10*074.8pv and
v10*074.8pp
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If
{ }95.0c 10*074.8
pcv
as writtenbemay Equation
d4dn ≈ρ
∆= −
0≠∆ fP
This accounts for all the losses in the nozzle.
Example: ft/sec 305 12*10*074.8
000,195.0v 4n ======== −−−−
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For multiple nozzles in //Vn is the same for each nozzle
even if the dn varies!This follows since ∆∆∆∆p is the sameacross each nozzle.
tn A117.3
qv =
2t
2d
2-5
bit ACq10*8.311
∆pρ
=
10*074.8
pcv 4dn ρ∆= − &
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Hydraulic Horsepowerof pump putting out 400 gpm at 3,000 psi = ?
Power
( )pqP
AqA*p
t/s*F workdoing of rate
H ∆=
∆=
==
hp7001714
000,3*4001714
pq HHP ==∆=
In field units:
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What is Hydraulic Impact Force
developed by bit?
Consider:
psi 169,1∆plb/gal 12
gal/min 400q95.0C
n
D
==ρ==
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Impact = rate of change of momentum
(((( ))))60*17.32
vqv
tm
tmvF n
jρρρρ
====∆∆∆∆
∆∆∆∆====
∆∆∆∆∆∆∆∆====
psi 169,1∆plb/gal 12
gal/min 400q95.0C
n
D
==ρ==
lbf 820169,1*12400*95.0*01823.0Fj ========
pqc01823.0F dj ∆∆∆∆ρρρρ====