5B Hydraulics Cont'd

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Well Drilling Engineering

Drilling Hydraulics (cont’d) Dr. DO QUANG KHANH

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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

HW #:

ADE # 4.14, 4.15, 4.16, 4.21

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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 −=

π

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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?

22 73.5800,7 in

inlbf

∗=

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 - 44,700 = 150,300 lbf.

At bottom, FT = 19.5 * 0 - 7,800 (5.73) = - 44,700 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) = - 44,700 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 44,700 0 150,300

FS FT

ft708,7=5.19306,150

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/lbflbf

wFL

DC

BITDCIn Air:

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

ρρ

−=

s

fDC

BITDC

1w

.F.S*FL

ρρ

−=

ft/lbflbf

1w

FL

s

fDC

BITDC

14 State of stress in pipe at the neutral point

σt

σZ

σr

σr σZ

σt

Steel Elemental Volume

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At the Neutral Point: The axial stress is equal to the average

of the radial and tangential stresses.

2 tr

Zσσσ +

=

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

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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

42

i2

o −π

−

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 dqv = ( )2

122448.2 dd

qv−

=

26

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Law of Conservation of Energy

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

( ) ( )( ) ( )

QW

vvDDg

VpVpEE

+=

−+−−

−+−

21

2212

112212

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

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In practical field units this equation simplifies to:

( )

( ) fp pPvv

DDpp

∆−∆+−−

−+=

− 21

22

4

1212

10*074.8

052.0

ρ

ρ

p1 and p2 are pressures in psi ρ is 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 pin. 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

ρ(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 22 ==

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

−+−+=

−+−

+=

∆−∆+−ρ−

−ρ+=

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

ρ

ρ

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0 P

v v0 P

0 vD D

f

n2p

112

≈∆

==∆

≈≈

Fluid Flow Through Nozzle Assume:

ρ∆

=

ρ−=

−

−

4n

2n

412

10*074.8pv and

v10*074.8pp

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If

{ }95.0c 10*074.8

pcv

as writtenbemay Equation

d4dn ≈ρ

∆= −

0≠∆ fP

This accounts for all the losses in the nozzle.

Example: ft/sec 305 12*10*074.8

000,195.0v 4n == −

35

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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 ACq10*8.311

Δpρ

=

10*074.8

pcv 4dn ρ∆

= − &

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

Power

( )pqP

AqA*p

t/s*F workdoing of rate

H ∆=

∆=

==

hp7001714

000,3*4001714

pq HHP ==∆

=

In field units:

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What is Hydraulic Impact Force

developed by bit? Consider:

psi 169,1Δplb/gal 12

gal/min 400q95.0C

n

D

==ρ==

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Impact = rate of change of momentum

( )60*17.32

vqv

tm

tmvF n

jρ

=∆

∆=

∆∆

=

psi 169,1Δplb/gal 12

gal/min 400q95.0C

n

D

==ρ==

lbf 820169,1*12400*95.0*01823.0Fj ==

pqc01823.0F dj ∆ρ=