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PETE 411Well Drilling

Lesson 10Drilling Hydraulics (cont’d)

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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

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READ:ADE, Ch. 4 to p. 135

HW #5:

ADE # 4.3, 4.4, 4.5, 4.6

due September 27, 2002

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Buckling of

Tubulars l

l

Slender pipe suspended in wellbore

Partially buckled slender

pipe

Neutral Point

Neutral Point

Fh - Fb

Fh

Fb

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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

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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

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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?

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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?

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73.5800,7 inin

lbf

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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

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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

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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 =

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Length of

Drill Collars

Neutral Point

Neutral Point

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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?

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At the Neutral Point:

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

2

trZ

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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

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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.

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NOTE:

If pi = po = p,

then Fs = pdd io22

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or, Fs = - AS p

AS

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Axial Load with FBIT = 68,000 lbf

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Stability Analysis with

FBIT = 68,000 lbf

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Nonstatic Well Conditions

Physical Laws

Rheological Models

Equations of State

FLUID FLOW

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Physical Laws

Conservation of mass

Conservation of energy

Conservation of momentum

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Rheological Models

Newtonian

Bingham Plastic

Power – Law

API Power-Law

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Equations of State

Incompressible fluid

Slightly compressible fluid

Ideal gas

Real gas

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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

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Law of Conservation of Energy

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

QW

vvDDg

VpVpEE

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2212

112212

2

1

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

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In practical field units this equation simplifies to:

fp pPvv

DDpp

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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

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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)

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Velocity in drill collars

)(in

(gal/min)

d448.2

qv

222

ft/sec 14.26)5.2(*448.2

400v

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Velocity in mud pits, v1 0

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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

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fp PPvv

DDpp

)(10*074.8

)(052.021

22

4-

1212

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0 P

v v0 P

0 vD D

f

n2p

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Fluid Flow Through Nozzle

Assume:

4n

2n

412

10*074.8

pv and

v10*074.8pp

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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

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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

&

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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:

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What is Hydraulic Impact Force

developed by bit?

Consider:

psi 169,1Δp

lb/gal 12

gal/min 400q

95.0C

n

D

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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