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PETE 411 Well Drilling. Lesson 15 Surge and Swab Pressures. 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. READ: APPLIED DRILLING ENGINEERING Chapter 4 (all). - PowerPoint PPT Presentation
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1
PETE 411
Well Drilling
Lesson 15
Surge and Swab Pressures
2
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
3
READ:
APPLIED DRILLING ENGINEERING Chapter 4 (all)
HW #8ADE #4.46, 4.47
due 10 –14 – 02
4
pcaae vKvv
5
Surge Pressure - Closed PipeNewtonian
The velocity profile developed for the slot approximation 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
6
When y = 0, v = - vp ,
When y = h, v = 0,
Substituting
for c1 and c2:
p2 vc
p1f
2
vμ
hc
dL
dp
2μ
h0
21f
2
cμ
yc
dL
dp
2μ
yv
h
μv
dL
dp
2
hc pf
1
h
y1vyhy
dL
dp
2μ
1v p
2f
At Drillpipe Wall
7
Velocity profile in the slot
vWdyvdAdqq
h
y1vyhy
dL
dp
2μ
1v p
2f
dy)h
y1(Wv
h
0
p
2
Whv
dL
dp
12μ
Whq pf
3
h0
W
8
Changing from SLOT to ANNULAR notation
A = Wh = 21
22 rrπ
)rr(
qv
rrh
21
22
12
2
Whv
dL
dp
12μ
Whq pf
3
Substitute in:
9
Or, in field units
212
p
f
dd1000
2
vv
dL
dp
212
p
f
rr
2
vv12μ
dL
dp
or, in field units:
Frictional Pressure Gradient
Same as for pure slot flow if vp = o (Kp = 0.5)
Results in:
10
How do we evaluate v ?
For closed pipe,
flow rate in annulus = pipe displacement rate:
pa qq
4
dπvdd
4v
21
p2
12
2a
d1
d2
vp
1dd
vv 2
1
2
p
11
Open Pipe
Pulling out
of Hole
12
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
2
12
pa
2i
pi
dd1000
2
Vvμ
d 1500
vvμ
From Equations (4.88) and
(4.90d):
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ai qq tq Also,
2
122a
2ii
2i
21p dd
4
πvd
4
πvdd
4
πVi.e.,
p2
122
212
4i
212
21
4i
a vdddd46d
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 pipe
2. Open ended
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212
pa
f
10.75121000
21
4.0642
)d1000(d
2
vvμ
dL
dp
ft/s 4.06410.7512
(1)10.75
)d(d
vdv
22
2
21
22
p21
a
1. For Closed Pipe
ft
psi 0.00577
dL
dpf
psi 23.14,0000.00577ΔΡf
1dd
vv
2
1
2
pa
16
p2
122
212
4
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
psi 0.00001728
10.75)1000(1221
0.48652
)d1000(d
2
VVμ
dL
dp22
12
pa
f
e)(negligibl psi 0.07
4,000*0.00001728ΔΡf
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Derivation of Equation (4.94)
From Equation (4.92):
212
2pa
pi
212
pa
2
pi
)d2(d
d2
vv3
vv
)d1000(d
2
vvμ
1500d
)vvμ(
19
212
2p
2a
212p
i )d4(d
d3vdv6)d(d4v- v
Derivation of Eq. (4.94) cont’d
From Equation (4.93):
)d(dvdv)d(dv 21
22a
2i
221p
Substituting for vi:
212
4p
4a
212
2p22
1p )d4(d
d3vdv6)d(dd4v)d(dv
)d(dv 21
22a
20
2
1221
22
4a
42221
212p
)d)(dd4(d6dv
3d)dd(d)d4(dv
So,
p21
22
212
4
4212
21
a v)d(d)d-4(d6d
3d)d(d4dv
i.e., p21
22
212
4
212
21
4
a v)d(d)d4(d6d
)d(d4d3dv
21
Surge Pressure - General Case
The slot approximation discussed earlier is not appropriate if the pipe ID or OD varies, if the fluid is non-Newtonian, or if the flow is turbulent.
In the general case - an iterative solution technique may be used.
22
Fig. 4.42Simplified hydraulic
representation of the lower
part of a drillstring
23
General Solution Method
1. Start at the bottom of the drillstring and
determine the rate of fluid displacement.
p22
1t vdd4
πq
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.
24
3. Calculate the resulting total frictional pressure loss in the annulus,
using the established pressure loss calculation procedures.
4. Calculate the total frictional pressure loss inside the drill string.
General Solution Method
25
5. Compare the results from 3 and 4, and if they are unequal, repeat the above steps with a different split between qa and qp.
i.e., repeat with different values of fa, until the two pressure loss values agree within a small margin. The average of these two values is the surge pressure.
General Solution Method
26
NOTE:
The flow rate along the annulus need not be constant, it varies whenever the cross-sectional area varies.
The same holds for the drill string. An appropriate average fluid velocity must be
determined for each section. This velocity is further modified to arrive at an
effective mean velocity.
27
Fig. 4.42Simplified hydraulic
representation 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 clinging constant; it depends on the annular geometry.(Not related to Power-law K!)
v
p caae vKvv
av
29
The value of Kp lies between 0.4 and 0.5 for 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 laminar and the turbulent flow patterns and then to use the larger value.
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Kc
Kc
31
For very small values of , K = 0.45 is not a good
approximation
Kc
Fig. 4.41 - Mud clinging constant, Kc, for computing surge-and-swab pressure.
Kc
32
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
33
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 ……… 442 115 160 171
pDC, psi ……… 104 33 44 46
pDP, psi ……… 449 273 293 297
Totalpi, psi …… 995 421 497 514
34
Table 4.8 Summary of Swab Pressure Calculation in Annulus
Variable
fa=(qa/qt)1 0.5 0.75 0.70 0.692
0.422 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
35
Table 4.8 Summary of Swab Pressure Calculation for Example 4.35
1.00 0.99 0.94 1.39 :
514.5
ΔΡΔΡ21
ai
fa: 0.5 0.75 0.70 0.692
36
vp
37
VELOCITY
SURGE PRESSURE
ACCELERATION
38
Inertial EffectsExample 4.36
Compute the surge pressure due to inertial effects caused by downward 0.5 ft/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)
psi 271Δp
(10,000)10.7512.25
75))(0.5)(10.0.00162(10Δp
dd
da 0.00162
dL
dp
a
22
2
a
21
22
21pa
Inertial Effects - Example 4.36
40
END of Lesson 15
Surge and Swab
Recommended