15 surge and swab

8/11/2019 15 Surge and Swab

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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.



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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.



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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

15 surge and swab

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