Chuong 5g surge and swab

PETROLEUM ENGINEERING 405 * Well Drilling Engineering Surge and Swab Pressures Dr DO QUANG KHANH * 15 Surge and Swab Pressures Surge and Swab Pressures Closed Pipe Fully Open Pipe Pipe with Bit Exampl[.]

Well Drilling Engineering

Surge and Swab Pressures

Dr DO QUANG KHANH

15 - Surge and Swab Pressures

 Surge and Swab Pressures

- Closed Pipe

- Fully Open Pipe

- Pipe with Bit

 Example

 General Case (complex geometry, etc.)

 Example

Surge Pressure due to Pipe Movement

When a string of pipe is

being lowered into the

wellbore, drilling fluid is

being displaced and forced

out of the wellbore

The pressure required to

force the displaced fluid out

of the wellbore is called the

surge pressure.

Surge Pressure due to Pipe Movement

An excessively high surge pressure can

result in breakdown of a formation

When pipe is being withdrawn a similar

reduction is pressure is experienced This

is called a swab pressure, and may be

high enough to suck fluids into the wellbore, resulting in a kick

swab surge P

P

, v

fixed

p c a

v  

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 1

2

v μ

h c dL

dp 2μ

h



2 1

f

2

c μ

y c dL

dp 2μ

dp 2

v y

hy dL

dp 2μ

Velocity profile in the slot

v y

hy dL

dp 2μ

1

dy

) h

y 1

( W

dp 12μ

Wh

q  3 f  p



h 0

W

Changing from SLOT to ANNULAR

) r

r (

q v

r r

h

2 1

2 2

1 2

dp 12μ

Wh

Substitute in:

Or, in field units

1 2

p

f

d d

1000

2

v v

p

f

rr

2

vv

12μdL

or, in field units:

Frictional Pressure Gradient

Same as for pure slot flow if v = o (K = 0.5)

Results in:

How do we evaluate v ?

For closed pipe,

flow rate in annulus = pipe displacement rate:

d

d4v

2 1 p

2 1

2 2

a 

d1d

vp

1d

Open

Pipe

Pulling out

of Hole

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

p a

2 i

p

i

d d

1000

2

V v

μ d

1500

v v

2 i i

2 i

2 1

4

π v

d 4

π v

d

d 4

2 1 2

4 i

2 1 2

2 1

4 i

d d

d d

4 6d

d d

Surge Pressure - Open Pipe

Valid for laminar flow, constant geometry, Newtonian

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

Derivation of Equation (4.94)

From Equation (4.92):

2 1 2

2

p a

p i

2 1 2

p a

2

p i

) d 2(d

d 2

v v

3 v

v

) d 1000(d

2

v v

μ 1500d

) v v

2 1 2

2 p

2 a

2 1 2

p i

) d 4(d

d 3v d

v 6 )

d (d

4v

v

v d

v )

d (d

v p 12  2  i 2  a 22  12

Substituting for v i :

2 1 2

4 p

4 a

2 1 2

2 p 2

2 1 p

) d 4(d

d 3v d

v 6 )

d (d

d

4v )

d (d

v 2  2



 

1 2

2 1

2 2

4 a

4 2

2

2 1

2 1 2

p

) d )(d

d 4(d

6d v

3d )

d d

(d )

d 4(d

2 2

2 1 2

4

4

2 1 2

2 1

) d (d

) d - 4(d

6d

3d )

d (d

4d v

2 1 2

4

2 1 2

2 1

4

) d (d

) d 4(d

6d

) d (d

4d

3d v

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

General Solution Method

1 Start at the bottom of the drillstring and

determine the rate of fluid displacement

2 Assume a split of this flow stream with a

fraction, f a , going to the annulus, and (1-f a ) going through the inside of the pipe

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

5 Compare the results from 3 and 4, and if

with a different split between q a and q p.

i.e., repeat with different values of f a, until

values is the surge pressure

General Solution Method

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

Has suggested using an effective mean

annular velocity given by:

Where is the average annular velocity

based on q a

Kc is a constant called the mud clinging

constant; it depends on the annular geometry

(Not related to Power-law K!)

v

p c

The value of Kc 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

K c

K c

For very small values of ,

K = 0.45 is not a good

approximation

K c

K c

Table 4.8 Summary of Swab Pressure

Calculation for Example 4.35

Table 4.8 Summary of Swab

Pressure Calculation Inside Pipe

Table 4.8 Summary of Swab Pressure Calculation in Annulus

psi p

cu q

cu q

a

a

, ,

ft/s ,

)

(

ft/s ,





Table 4.8 Summary of Swab Pressure

Calculation for Example 4.35

1.00

0.99

0.94

1.39

: 514.5

vp

SURGE PRESSURE

Inertial Effects Example 4.36

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

From Equation (4.99)

psi271

Δp

(10,000)10.75

12.25

75))(0.5)(10

0.00162(10Δp

dd

da

0.00162

dL

dp

a

2 2

2 a

2 1

2 2

2 1 p a

END of

Lesson