PETE 411
Well Drilling
Lesson 15
Surge and Swab Pressures
1
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
2
READ:
APPLIED DRILLING ENGINEERING
Chapter 4 (all)
HW #8
ADE #4.46, 4.47
due 10 –14 – 02
3
v ae  v a  K c v p
4
Surge Pressure - Closed Pipe
Newtonian
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:
2
y dp f
y
V 
 c1  c2
2 dL

V=0
V = -Vp
5
At Drillpipe
Wall
y 2 dp f
y
v
 c1  c 2
2μ dL
μ
When y = 0, v = - vp ,
 c2  vp
h 2 dp f
h
 c1  v p
When y = h, v = 0,  0 
2μ dL
μ
v pμ
h dp f
 c1 

2 dL
h
Substituting
for c1 and c2:


1 dp f
 y
2
v
hy  y  v p 1  
2μ dL
 h
6
Velocity profile in the slot


1 dp f
 y
2
v
hy  y  v p 1  
2μ dL
 h
h
0
W
q   dq   vdA   vWdy
h

y
 v p W (1  )dy
h
0
3
v p Wh
Wh dp f
q 

12μ dL
2
7
Changing from SLOT to ANNULAR
notation

A = Wh = π r  r
2
2
2
1

h  r2  r1
v
q
2
( r2

2
r1 )
3
Substitute in:
v p Wh
Wh dp f
q

12μ dL
2
8
Frictional Pressure Gradient
Results in:
vp 


12μ  v 
2 
dp f


2
dL
r2  r1 
Or, in field units
v

p 
or, in field units:

 v 


2
dp f



2
dL
1000 d 2  d1 
Same as for pure slot flow if vp = o
(Kp = 0.5)
9
How do we evaluate v ?
For closed pipe,
flow rate in annulus = pipe displacement rate:
qa  qp
 π d1
 2
2 
v a  d 2  d1   v p 
 4
4


vp
v 
2
 d2 
   1
 d1 


2




vp
d1
d2
10
Open
Pipe
Pulling out
of Hole
11
Surge Pressure - Open Pipe
Pressure at top and bottom is the
same inside and outside the pipe.
i.e.,
 dp f 
 dp f 

 

 dL  pipe  dL annulus
Vp 
From Equations


μ v a 
(4.88) and
μ vi  v p
2 



(4.90d):
2
2
1500 d i
1000d 2  d1 


12
Surge Pressure - Open Pipe
q t  qi  qa
Also,
i.e.,




π 2
π 2
π 2
2
2 
Vp
d1  d i  v i  d i   v a 
d 2  d1 
4
4

4


3d  4d d 2  d1 
 va  
2
4
2
2
  6d i  4d 2  d1  d 2  d1
4
i
2
1
2


vp


Valid for laminar flow, constant geometry, Newtonian
13
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
14
 va 
1. For Closed Pipe
2
1
d vp
vp
2
 d2 

  1
 d1 
2
10.75 (1)
va  2
 2
 4.064 ft/s
2
2
(d 2  d1 ) 12  10.75
vp 

1




μ v a  
2 4.064  
2
dp f
2




dL 1000(d 2  d1 ) 100012  10.752
dp f
psi
 0.00577
dL
ft
ΔΡ f  0.00577  4,000  23.1 psi
15
2. For Open Pipe,
2
4
2

3d  4d1 d 2  d1 
Va  
2 2
4
2
  6d  4d 2  d1  d 2  d1


 Vp


 3(10)  4(10.75) (12  10.75)

Va  
(1.0)
4
2
2
2 
  6(10)  4(12  10.75) (12  10.75 ) 
ft
  0.4865
sec
4
2
2
16
2. For Open Pipe,
Vp 

1




μ Va  
2  0.4865  
2 
dp f
2




2
dL 1000(d 2  d1 ) 1000(12  10.75) 2
psi
 0.00001728
ft
ΔΡ f  0.00001728 * 4,000
 0.07 psi
(negligibl e)
17
Derivation of Equation (4.94)
From Equation (4.92):
vp 

μ v a  
μ(v i  v p )
2


2
2
1500d
1000(d 2  d1 )
vp  2

3 v a  d
2

 vi  v p 
2
2(d 2  d1 )
18
Derivation of Eq. (4.94) cont’d
- 4v p (d 2  d1 )  6v a d  3v p d
2
 vi 
2
4(d 2  d1 )
2
2
From Equation (4.93):
v p (d  d )  vi d  va (d  d )
2
1
2
Substituting for vi:
v p (d  d ) 
2
1
2
2
2
2
2
1
 4v p d 2 (d 2  d1 ) 2  6v a d 4  3v p d 4
4(d 2  d1 )
2
 v a (d  d )
2
2
2
1
19
So,

v p 4(d 2  d1 ) (d  d  d )  3d
2

2
1
2
2
 v a 6d  4(d  d )(d 2  d1 )
4
2
2
2
1
2

4

 4d (d 2  d1 )  3d


 v a   4
v
2
2
2  p
 6d  4(d 2 - d1 ) (d 2  d1 ) 
2
1
2
4


3d  4d (d 2  d1 )

i.e., v a  
v
4
2
2
2
  6d  4(d  d ) (d  d )  p
2
1
2
1 

4
2
1
2
20
Surge Pressure - General Case
The slot approximation discussed
earlier is not appropriate if the pipe ID
or OD varies, if the fluid is nonNewtonian, or if the flow is turbulent.
In the general case - an iterative
solution technique may be used.
21
Fig. 4.42
Simplified
hydraulic
representation
of the lower
part of a
drillstring
22
General Solution Method
1. Start at the bottom of the drillstring and
determine the rate of fluid displacement.


π 2
2
qt 
d1  d v p
4
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.
23
General Solution Method
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.
24
General Solution Method
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.
25
NOTE:
The flow rate along the annulus need not be
constant, it varies whenever the crosssectional 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.
26
Fig. 4.42
Simplified
hydraulic
representation
of the lower
part of a
drillstring
27
Burkhardt
Has suggested using an effective mean
annular velocity given by:
v ae  v a  K c v p
Where v a is the average annular velocity
based on qa
v
Kc is a constant called the mud clinging
constant; it depends on the annular
geometry.
(Not related to Power-law K!)
28
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.
29
Kc
Kc
30
Kc
For very small values of a,
KK
c = 0.45 is not a good
approximation
Fig. 4.41 - Mud clinging constant, Kc, for computing surge-and-swab pressure.
31
Table 4.8. Summary of Swab Pressure
Calculation for Example 4.35
Variable
fa=(qa/qt)1
(qp)1, cu ft/s
(qp)2, cu ft/s
(qp)3, cu ft/s
0.5
0.422
0.265
0.111
0.75
0.211
0.054
-0.101
0.70 0.692
0.251 0.260
0.093 0.103
-0.061 -0.052
32
Table 4.8 Summary of Swab
Pressure Calculation Inside Pipe
Variable
fa=(qa/qt)1 ………
DpBIT, psi ………
DpDC, psi ………
DpDP, psi ………
Total Dpi, psi ……
0.5
442
104
449
995
0.75
115
33
273
421
0.70
160
44
293
497
0.692
171
46
297
514
33
Table 4.8 Summary of Swab
Pressure Calculation in Annulus
Variable
fa=(qa/qt)1
Total Dpa, psi
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
439
544 520 515
Total Dpi, psi
995
( qa )1 , cu ft/s
( qa ) 2 , cu ft/s
Dpdca , psi
Dpdpa , psi
421
497
514
34
Table 4.8 Summary of Swab Pressure
Calculation for Example 4.35
fa :
1
ΔΡ i  ΔΡ a 
2
:
514.5
0.5 0.75 0.70 0.692
1.39 0.94 0.99 1.00
35
vp
36
SURGE
PRESSURE
VELOCITY
ACCELERATION
37
Inertial Effects
Example 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
38
Inertial Effects - Example 4.36
From Equation (4.99)
dp a
dL

0.00162  a p d
d d
2
2
2
1
2
1

0.00162(10)(0.5)(10. 75) 2
(10,000)
Δp a 
2
2
12.25  10.75

Δp a  271 psi
39
END of
Lesson 15
Surge and Swab
40