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1 PETE 411 Well Drilling Lesson 10 Drilling Hydraulics (cont’d)

Transcript

1

PETE 411Well Drilling

Lesson 10Drilling Hydraulics (cont’d)

2

10. Drilling Hydraulics (cont’d)

Effect of Buoyancy on Buckling The Concept of Stability Force Stability Analysis Mass Balance Energy Balance Flow Through Nozzles Hydraulic Horsepower Hydraulic Impact Force

3

READ:ADE, Ch. 4 to p. 135

HW #5:

ADE # 4.3, 4.4, 4.5, 4.6

due September 27, 2002

4

Buckling of

Tubulars l

l

Slender pipe suspended in wellbore

Partially buckled slender

pipe

Neutral Point

Neutral Point

Fh - Fb

Fh

Fb

5

Buckling of Tubulars

l

Neutral Point

Neutral Point

• Long slender columns, like DP, have low resistance to bending and tend to fail by buckling if...

• Force at bottom (Fb) causes neutral point to move up

• What is the effect of buoyancy on buckling?

• What is NEUTRAL POINT?

Fb

6

What is NEUTRAL POINT?

l

Neutral Point

Neutral Point

• One definition of NEUTRAL POINT is the point above which there is no tendency towards buckling

• Resistance to buckling is indicated, in part, by:

The Moment of Inertia

444

64I inddn

7

Consider the following:

19.5 #/ft drillpipe

Depth = 10,000 ft.

Mud wt. = 15 #/gal.

PHYD = 0.052 (MW) (Depth)

= 0.052 * 15 * 10,000

PHYD = 7,800 psi

Axial tensile stress in pipe at bottom

= - 7,800 psi

What is the axial force at bottom?

8

What is the axial force at bottom?

Cross-sectional area of pipe

= (19.5 / 490) * (144/1) = 5.73 in2

Axial compressive force = pA

= 44,700 lbf.

Can this cause the pipe to buckle?

22

73.5800,7 inin

lbf

9

Axial Tension:FT = W1 - F2

FT = w x - P2 (AO - Ai )

At surface, FT = 19.5 * 10,000 - 7,800 (5.73)

= 195,000 - 44,694

= 150,306 lbf.

At bottom, FT = 19.5 * 0 - 7,800 (5.73)

= - 44,694 lbf

Same as before!

FT

F2

10

Stability Force:

FS = Aipi - AO pO

FS = (Ai - AO) p (if pi = pO)

At surface, FS = - 5.73 * 0 = 0

At bottom, FS = (-5.73) (7,800) = - 44,694 lbs

THE NEUTRAL POINT is where FS = FT

Therefore, Neutral point is at bottom!

PIPE WILL NOT BUCKLE!!

Ai

11

Compression Tension 44,770 0 150,306

FS FT

ft708,7=5.19

306,150

Zero Axial Stress

Neutral Point

Depth of Zero Axial Stress Point =

12

Length of

Drill Collars

Neutral Point

Neutral Point

13

Length of Drill Collars

ft/lbf

lbf

W

FL

DC

BITDCIn Air:

In Liquid:

In Liquidwith S.F.: (e.g., S.F =1.3)

s

fDC

BITDC

W

.F.S*FL

1

ft/lbf

lbf

W

FL

s

fDC

BITDC

1

14State of stress in pipe at the neutral point?

15

At the Neutral Point:

The axial stress is equal to the average of the radial and tangential stresses.

2

trZ

16

Stability Force:

FS = Ai Pi - Ao Po

If FS > axial tension then the pipe may buckle.

If FS < axial tension then the pipe will NOT buckle.

FS

FT

0 FT

17

At the neutral point:

FS = axial load

To locate the neutral point:

Plot FS vs. depth on “axial load (FT ) vs. depth plot”

The neutral point is located where the lines intersect.

18

NOTE:

If pi = po = p,

then Fs = pdd io22

4

or, Fs = - AS p

AS

19

Axial Load with FBIT = 68,000 lbf

20

Stability Analysis with

FBIT = 68,000 lbf

21

Nonstatic Well Conditions

Physical Laws

Rheological Models

Equations of State

FLUID FLOW

22

Physical Laws

Conservation of mass

Conservation of energy

Conservation of momentum

23

Rheological Models

Newtonian

Bingham Plastic

Power – Law

API Power-Law

24

Equations of State

Incompressible fluid

Slightly compressible fluid

Ideal gas

Real gas

25

Average Fluid Velocity

Pipe Flow Annular Flow

WHERE

v = average velocity, ft/s

q = flow rate, gal/min

d = internal diameter of pipe, in.

d2 = internal diameter of outer pipe or borehole, in.

d1 =external diameter of inner pipe, in.

2448.2 d

qv 2

122448.2 dd

qv

26

27

Law of Conservation of Energy

States that as a fluid flows from point 1 to point 2:

QW

vvDDg

VpVpEE

21

2212

112212

2

1

In the wellbore, in many cases Q = 0 (heat) = constant{

28

In practical field units this equation simplifies to:

fp pPvv

DDpp

21

22

4

1212

10*074.8

052.0

p1 and p2 are pressures in psiis density in lbm/gal.

v1 and v2 are velocities in ft/sec.

pp is pressure added by pump between points 1 and 2 in psi

pf is frictional pressure loss in psi

D1 and D2 are depths in ft.

where

29

Determine the pressure at the bottom of the drill collars, if

psi 000,3 p

in. 5.2

0 D

ft. 000,10 D

lbm/gal. 12

gal/min. 400 q

psi 1,400

p

1

2

DC

f

ID

p

(bottom of drill collars)

(mud pits)

30

Velocity in drill collars

)(in

(gal/min)

d448.2

qv

222

ft/sec 14.26)5.2(*448.2

400v

22

Velocity in mud pits, v1 0

31

400,1000,36.6240,60

400,1000,3)014.26(12*10*8.074-

0)-(10,00012*052.00p

PP)vv(10*074.8

)DD(052.0pp

224-

2

fp21

22

4-

1212

Pressure at bottom of drill collars = 7,833 psig

NOTE: KE in collars

May be ignored in many cases

0

32

fp PPvv

DDpp

)(10*074.8

)(052.021

22

4-

1212

33

0 P

v v0 P

0 vD D

f

n2p

112

Fluid Flow Through Nozzle

Assume:

4n

2n

412

10*074.8

pv and

v10*074.8pp

34

If

95.0c 10*074.8

pcv

as writtenbemay Equation

d4dn

0 fP

This accounts for all the losses in the nozzle.

Example: ft/sec 305 12*10*074.8

000,195.0v

4n

35

36

For multiple nozzles in //

Vn is the same for each nozzle even if the dn varies!

This follows since p is the same across each nozzle.

tn A117.3

qv

2t

2d

2-5

bit AC

q10*8.311Δp

10*074.8

pcv

4dn

&

37

Hydraulic Horsepower

of pump putting out 400 gpm at 3,000 psi = ?

Power

pqP

A

qA*p

t/s*F

workdoing of rate

H

hp7001714

000,3*400

1714

pq HHP

In field units:

38

What is Hydraulic Impact Force

developed by bit?

Consider:

psi 169,1Δp

lb/gal 12

gal/min 400q

95.0C

n

D

39

Impact = rate of change of momentum

60*17.32

vqv

t

m

t

mvF n

j

psi 169,1Δp

lb/gal 12

gal/min 400q

95.0C

n

D

lbf 820169,1*12400*95.0*01823.0Fj

pqc01823.0F dj

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