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PETE 411
Well Drill ing
Lesson 15
Surge and Swab Pressures
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Lesson 15 - Surge and Swab Pressures
Surge and Swab Pressures- Closed Pipe
- Fully Open Pipe
- Pipe with Bit
Example
General Case (complex geometry, etc.) Example
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READ:
APPLIED DRILLING ENGINEERINGChapter 4 (all)
HW #8
ADE #4.46, 4.47due 10 14 02
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Surge Pressure - Closed PipeNewtonian
The velocity profile developed for the slotapproximation is valid for the flow
conditions in the annulus; but the
boundary conditions are different,
because the pipe is moving:
21f
2
dL
dp
2c
yc
yV ++=
V = 0
V = -Vp
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When y = 0, v = - vp ,
When y = h, v = 0,
Substitutingfor c1 and c2:
p2 vc =
p1f
2
v
hc
dL
dp
2
h0 +=
21f
2
c
yc
dL
dp
2
yv ++=
h
v
dL
dp
2
hc
pf1 +=
( )
=
h
y1vyhy
dL
dp
2
1v p
2f
At Drillpipe
Wall
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Velocity profile in the slot
=== vWdyvdAdqq
( )
=
h
y1vyhy
dL
dp
2
1v p
2f
=h
0
2f )dyy(hydL
dp
2
Wq dy)
h
y1(Wv
h
0
p
2
Whv
dL
dp
12
Whq
pf
3
+=
h
0W
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Changing from SLOT to ANNULAR
notation
A = Wh = ( )2
1
2
2 rr!
)rr(
qv
rrh
21
22
12
=
=
2
Whv
dL
dp
12
Whq
pf3
+=Substitute in:
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Or, in field units
( )212
p
f
dd1000
2
vv
dL
dp
+
=
( )212
p
f
rr
2
vv12
dLdp
+
=
or, in field units:
Frictional Pressure Gradient
Same as for pure slot flow if vp = o (Kp = 0.5)
Results in:
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Open
Pipe
Pulling outof Hole
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Surge Pressure - Open Pipe
Pressure at top and bottom is the
same inside and outside the pipe.i.e.,
annulus
f
pipe
f
dL
dp
dL
dp
=
( )( )212
p
a
2
i
pi
dd1000
2
Vv
d1500
vv
+
=+
From Equations
(4.88) and(4.90d):
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ai qq +=tqAlso,
( ) ( )
+
= 21
2
2a
2
ii
2
i
2
1p dd4
!vd
4
!vdd
4
!Vi.e.,
( )
( )( )p2
1
2
2
2
12
4
i
2
12
2
1
4
ia v
dddd46d
dd4d3dv
=
Surge Pressure - Open Pipe
Valid for laminar flow, constant geometry, Newtonian
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Example
Calculate the surge pressures that
result when 4,000 ft of 10 3/4 inch OD(10 inch ID) casing is lowered inside a
12 inch hole at 1 ft/s if the hole is filled
with 9.0 lbm/gal brine with a viscosity
of 2.0 cp. Assume laminar flow.
1. Closed pipe2. Open ended
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( )212
p
a
f
10.751210002
14.0642
)d1000(d2
vv
dLdp
+
=
+
=
ft/s4.064
10.7512
(1)10.75
)d(d
vdv
22
2
2
1
2
2
p
2
1
a =
=
=
1. For Closed Pipe
ft
psi
0.00577dL
dp f
=
psi23.14,0000.00577"# f ==
1d
d
vv
2
1
2
p
a
=
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( )
( )( ) p21222124
212
21
4
a
Vdddd4d6
ddd4d3V
=
sec
ft
0.4865
(1.0))10.75(1210.75)4(126(10)
10.75)(124(10.75)3(10)V 2224
224
a
=
=
2. For Open Pipe,
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2. For Open Pipe,
ft
psi0.00001728
10.75)1000(122
10.48652
)d1000(d
2
VV
dLdp 22
12
p
a
f
=
+
=
+
=
e)(negligiblpsi0.07
4,000*0.00001728"# f
=
=
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2
12
2
p
2
a
2
12p
i
)d4(d
d3vdv6)d(d4v-v
++=
Derivation of Eq. (4.94) contd
From Equation (4.93):
)d(dvdv)d(dv
2
1
2
2a
2
i
22
1p +=Substituting for vi:
2
12
4
p
4
a
2
12
2
p221p
)d4(d
d3vdv6)d(dd4v)d(dv
++=
)d(dv 212
2a +
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[ ]21221224a
42221
212p
)d)(dd4(d6dv
3d)dd(d)d4(dv
+=
+
So,
p21222124
42
12
2
1
a
v)d(d)d-4(d6d
3d)d(d4dv
+
=
i.e.,p2
1
2
2
2
12
4
2
12
2
1
4
a v)d(d)d4(d6d
)d(d4d3dv
=
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Fig. 4.42
Simplified
hydraulicrepresentation
of the lower
part of a
drillstring
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General Solution Method
1. Start at the bottom of the drillstring and
determine the rate of fluid displacement.
( )p
22
1t vdd4
!
q =
2. Assume a split of this flow stream with a
fraction, fa, going to the annulus, and
(1-fa) going through the inside of the pipe.
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3. Calculate the resulting total frictional
pressure loss in the annulus, using the
established pressure loss calculationprocedures.
4. Calculate the total frictional pressure lossinside the drill string.
General Solution Method
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5. Compare the results from 3 and 4, and if
they are unequal, repeat the abovesteps with a different split between qaand qp.
i.e., repeat with different values of fa, until
the two pressure loss values agreewithin a small margin. The average of
these two values is the surge pressure.
General Solution Method
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NOTE:
4The flow rate along the annulus need not be
constant, it varies whenever the cross-sectional area varies.
4The same holds for the drill string.
4An appropriate average fluid velocity must be
determined for each section. This velocity
is further modified to arrive at aneffective mean velocity.
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Fig. 4.42
Simplified
hydraulicrepresentation
of the lower
part of a
drillstring
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Burkhardt
Has suggested using an effective mean
annular velocity given by:
Where is the average annular velocity
based on qa
Kc is a constant called the mud clingingconstant; it depends on the annular
geometry. (Not related to Power-law K!)
v
pcaae vKvv +=
av
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The value of Kp lies between 0.4 and 0.5for most typical flow conditions, and is
often taken to be 0.45.
Establishing the onset of turbulence under
these conditions is not easy.
The usual procedure is to calculate surge
or swab pressures for both the laminarand the turbulent flow patterns and then
to use the larger value.
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Kc
Kc
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Table 4.8. Summary of Swab Pressure
Calculation for Example 4.35
Variable
fa=(qa/qt)1 0.5 0.75 0.70 0.692
(qp)1, cu ft/s 0.422 0.211 0.251 0.260
(qp)2, cu ft/s 0.265 0.054 0.093 0.103
(qp)3, cu ft/s 0.111 -0.101 -0.061 -0.052
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Table 4.8 Summary of Swab
Pressure Calculation Inside Pipe
Variable
fa=(qa/qt)1 0.5 0.75 0.70 0.692
pBIT, psi 442 115 160 171pDC, psi 104 33 44 46
pDP
, psi 449 273 293 297
Totalpi, psi 995 421 497 514
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Table 4.8 Summary of Swab
Pressure Calculation in Annulus
Variable
fa=(qa/q t)1 0.5 0.75 0.70 0.6920.422 0.633 0.594 0.585
0.012 0.223 0.183 0.174
104 139 128 126
335 405 392 389
Total pa, psi 439 544 520 515
Total pi, psi 995 421 497 514
psip
psip
cuq
cuq
a
a
,
,
ft/s,)(
ft/s,)(
dpa
dca
2
1
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Table 4.8 Summary of Swab Pressure
Calculation for Example 4.35
( )1.000.990.941.39:
514.5
"#"#21
ai+
fa: 0.5 0.75 0.70 0.692
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vp
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VELOCITY
SURGEPRESSURE
ACCELERATION
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Inertial EffectsExample 4.36
Compute the surge pressure due to
inertial effects caused by downward 0.5ft/s2 acceleration of 10,000 ft of 10.75 csg.
with a closed end through a 12.25 borehole
containing 10 lbm/gal.
Ref. ADE, pp. 171-172
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From Equation (4.99)
psi271"p
(10,000)10.7512.25
75))(0.5)(10.0.00162(10"p
dd
da0.00162
dL
dp
a
22
2
a
2
1
2
2
2
1pa
=
=
=
Inertial Effects - Example 4.36
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END of
Lesson 15Surge and Swab