5b hydraulics cont'd
DESCRIPTION
Hydraulics in drilling wellTRANSCRIPT
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Well Drilling Engineering
Drilling Hydraulics (cont’d) Dr. DO QUANG KHANH
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10. Drilling Hydraulics (cont’d)
Effect of Buoyancy on Buckling The Concept of Stability Force Stability Analysis Mass Balance Energy Balance Flow Through Nozzles Hydraulic Horsepower Hydraulic Impact Force
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READ: ADE, Ch. 4
HW #:
ADE # 4.14, 4.15, 4.16, 4.21
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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 inddn −=
π
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Consider the following:
19.5 #/ft drillpipe Depth = 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 - 44,700 = 150,300 lbf.
At bottom, FT = 19.5 * 0 - 7,800 (5.73) = - 44,700 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 = 0 At bottom, FS = ( - 5.73) (7,800) = - 44,700 lbs
THE NEUTRAL POINT is where FS = FT
Therefore, Neutral point is at bottom! PIPE WILL NOT BUCKLE!!
Ai
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Compression Tension 44,700 0 150,300
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/lbflbf
wFL
DC
BITDCIn Air:
In Liquid: In Liquid with S.F.: (e.g., S.F =1.3)
ρρ
−=
s
fDC
BITDC
1w
.F.S*FL
ρρ
−=
ft/lbflbf
1w
FL
s
fDC
BITDC
14 State of stress in pipe at the neutral point
σt
σZ
σr
σr σZ
σt
Steel Elemental Volume
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At the Neutral Point: The axial stress is equal to the average
of the radial and tangential stresses.
2 tr
Zσσσ +
=
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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:
Plot FS vs. depth on “axial load (FT ) vs. depth plot”
The neutral point is located where the lines intersect.
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NOTE:
If pi = po = p, then Fs = ( )pdd
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i2
o −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 Velocity
Pipe Flow Annular Flow
WHERE v = average velocity, ft/s q = flow rate, gal/min d = 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) ρ = constant {
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In practical field units this 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 psi ρ 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 psi D1 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≈
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fp 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 same
across each nozzle.
tn A117.3
qv =
2t
2d
2-5
bit ACq10*8.311
Δpρ
=
10*074.8
pcv 4dn ρ∆
= − &
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Hydraulic Horsepower of 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 169,1Δplb/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
jρ
=∆
∆=
∆∆
=
psi 169,1Δplb/gal 12
gal/min 400q95.0C
n
D
==ρ==
lbf 820169,1*12400*95.0*01823.0Fj ==
pqc01823.0F dj ∆ρ=