PETE 411 Well Drilling

transcript

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

APPLIED DRILLING ENGINEERING Chapter 4 (all)

HW #8ADE #4.46, 4.47

due 10 –14 – 02

pcaae vKvv

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:

V = -Vp

When y = 0, v = - vp ,

When y = h, v = 0,

Substituting

for c1 and c2:

y1vyhy

At Drillpipe Wall

Velocity profile in the slot

vWdyvdAdqq

y1vyhy

Whq pf

Changing from SLOT to ANNULAR notation

A = Wh = 21

22 rrπ

Whq pf

Substitute in:

Or, in field units

dd1000

vv12μ

or, in field units:

Frictional Pressure Gradient

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

Results in:

How do we evaluate v ?

For closed pipe,

flow rate in annulus = pipe displacement rate:

dπvdd

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

dd1000

d 1500

From Equations (4.88) and

(4.90d):

ai qq tq Also,

21p dd

πVi.e.,

a vdddd46d

dd4d3dv

Surge Pressure - Open Pipe

Valid for laminar flow, constant geometry, Newtonian

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

10.75121000

4.0642

)d1000(d

ft/s 4.06410.7512

(1)10.75

1. For Closed Pipe

psi 0.00577

psi 23.14,0000.00577ΔΡf

a Vdddd4d6

ddd4d3V

ft 0.4865

(1.0))10.75(1210.75)4(126(10)

10.75)(124(10.75)3(10)V

2. For Open Pipe,

psi 0.00001728

10.75)1000(1221

0.48652

)d1000(d

e)(negligibl psi 0.07

4,000*0.00001728ΔΡf

Derivation of Equation (4.94)

From Equation (4.92):

)d1000(d

)vvμ(

i )d4(d

d3vdv6)d(d4v- v

Derivation of Eq. (4.94) cont’d

From Equation (4.93):

)d(dvdv)d(dv 21

Substituting for vi:

1p )d4(d

d3vdv6)d(dd4v)d(dv

)d(dv 21

)d)(dd4(d6dv

3d)dd(d)d4(dv

a v)d(d)d-4(d6d

3d)d(d4dv

i.e., p21

a v)d(d)d4(d6d

)d(d4d3dv

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.

Fig. 4.42Simplified hydraulic

representation of the lower

part of a drillstring

General Solution Method

1. Start at the bottom of the drillstring and

determine the rate of fluid displacement.

1t vdd4

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.

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.

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.

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.

Fig. 4.42Simplified hydraulic

representation of the lower

part of a drillstring

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!)

p caae vKvv

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.

For very small values of , K = 0.45 is not a good

approximation

Fig. 4.41 - Mud clinging constant, Kc, for computing surge-and-swab pressure.

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

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

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

ft/s ,)(

Table 4.8 Summary of Swab Pressure Calculation for Example 4.35

1.00 0.99 0.94 1.39 :

ΔΡΔΡ21

fa: 0.5 0.75 0.70 0.692

VELOCITY

SURGE PRESSURE

ACCELERATION

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

From Equation (4.99)

psi 271Δp

(10,000)10.7512.25

75))(0.5)(10.0.00162(10Δp

da 0.00162

Inertial Effects - Example 4.36

END of Lesson 15

Surge and Swab

PETE 411 Well Drilling

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