Transcript

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PETE 411

Well Drilling

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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APPLIED DRILLING ENGINEERING Chapter 4 (all)

due 10 –14 – 02

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pcaae vKvv

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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

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When y = 0, v = - vp ,

When y = h, v = 0,

Substituting

for c1 and c2:

p2 vc

p1f

2

hc

dL

dp

h0

21f

2

yc

dL

dp

yv

h

μv

dL

dp

2

hc pf

1

h

y1vyhy

dL

dp

1v p

2f

At Drillpipe Wall

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Velocity profile in the slot

h

y1vyhy

dL

dp

1v p

2f

dy)h

y1(Wv

h

0

p

2

Whv

dL

dp

12μ

Whq pf

3

h0

W

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Changing from SLOT to ANNULAR notation

A = Wh = 21

22 rrπ

)rr(

qv

rrh

21

22

12

2

Whv

dL

dp

12μ

Whq pf

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Substitute in:

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Or, in field units

212

p

f

dd1000

2

vv

dL

dp

212

p

f

rr

2

vv12μ

dL

dp

or, in field units:

Same as for pure slot flow if vp = o (Kp = 0.5)

Results in:

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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

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Open Pipe

Pulling out

of 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

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

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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(

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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

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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.

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Fig. 4.42Simplified hydraulic

representation 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.

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.

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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

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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

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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.

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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

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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

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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

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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 ……… 442 115 160 171

pDC, 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/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

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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

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vp

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VELOCITY

SURGE PRESSURE

ACCELERATION

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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.

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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

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END of Lesson 15

Surge and Swab

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