Drill Hydraulics

8/18/2019 Drill Hydraulics

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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 BucklingThe Concept of Stability ForceStability AnalysisMass BalanceEnergy BalanceFlow Through Nozzles

Hydraulic Horsepower Hydraulic Impact Force



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Bucklingof

Tubulars l

l

Slender pipe

suspendedin wellbore

Partiallybuckledslender

pipe

Neutral Point

Neutral Point

F h - F b

F h

F b



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Buckling of Tubulars

l

NeutralPoint

NeutralPoint

• Long slender columns, like DP,have low resistance tobending and tend to fail by

buckling if...• Force at bottom (F b) causes

neutral point to move up• What is the effect of buoyancy

on buckling?• What is NEUTRAL POINT?

F b



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What is NEUTRAL POINT?

l

NeutralPoint

NeutralPoint

• One definition of NEUTRALPOINT is the point abovewhich there is no tendencytowards buckling

• Resistance to buckling isindicated, in part, by:

The Moment of Inertia

444

64I ind d n



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Consider thefollowing :

19.5 #/ft drillpipeDepth = 10,000 ft.Mud wt. = 15 #/gal.

P HYD = 0.052 (MW) (Depth)

= 0.052 * 15 * 10,000 P HYD = 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 in 2

Axial compressive force = pA

= 44,700 lbf.

Can this cause the pipe to buckle?

22 73.5800,7 in

inlbf



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Axial Tension:

FT = W1 - F2

FT = w x - P 2 (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 = A ip i - A O p O

F S = (A i - A O) p (if p i = p O)

At surface, FS = - 5.73 * 0 = 0

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

THE NEUTRAL POINT is where F S = F T

Therefore, Neutral point is at bottom!PIPE WILL NOT BUCKLE!!

i



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LengthofDrill

Collars

Neutral Point

Neutral Point



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Length of Drill Collars

ft /lbf

lbf WF

LDC

BITDCIn Air:

In Liquid:

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

s

f DC

BITDC

W

.F.S*FL

1

ft /lbf lbf

W

FL

s

f DC

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 averageof the radial and tangential stresses.

2

t r Z



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Stability Force:

F S = A i P i - A o P o

If F S > axial tension thenthe pipe may buckle.

If F S < axial tension thenthe pipe will NOT buckle.

F S

F T

0 F T



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At the neutral point:

FS = axial load

To locate the neutral point:

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

The neutral point is located where thelines intersect.



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NOTE:

If p i = p o = p,

then F s = pd d io

22

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or, F s = - A S p

AS



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Axial Load with F BIT = 68,000 lbf



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StabilityAnalysis 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 PlasticPower – Law

API Power-Law



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Equations of State

Incompressible fluid

Slightly compressible fluid

Ideal gas

Real gas



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Law of Conservation of Energy

States that as a fluid flowsfrom point 1 to point 2:

QW

vv D D g

V pV p E E

2

1

2

212

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:

f p p P vv

D D p p

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4

1212

10*074.8

052.0

p 1 and p 2 are pressures in psi is density in lbm/gal.v 1 and v 2 are velocities in ft/sec. p p is pressure added by pump

between points 1 and 2 in psi p f is frictional pressure loss in psi

D1 and D2 are depths in ft.

where



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Determine the pressure at thebottom 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

(bottom of drill collars)

(mud pits)



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Velocity in drill collars

)(in

(gal/min) d448.2

qv 222

ft/se14.26)5.2(*448.2

400v 22

Velocity in mud pits, v 1 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

f p21

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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f p P P vv D D p p

)(10*074.8 )(052.0

21

22

4-

1212



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If

95.0c 10*074.8

pcv

as writtenbemayEquation

d4dn

0 f P

This accounts for all the losses in the nozzle.

Example: ft/sec 30512*10*074.8

000,195.0v 4n



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For multiple nozzles in //

Vn is the same for each nozzleeven if the d n varies!

This follows since p is the same across each nozzle.

tn A117.3

qv

2t

2d

2-5

bit AC

q10*8.311 Δp

10*074.8

pcv 4dn &



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What is Hydraulic Impact Force

developed by bit?

Consider:

psi169,1 Δplb/gal12

gal/min400q

95.0C

n

D



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Impact = rate of change of momentum

60*17.32

vqv

tm

tmv

F n j

psi169,1 Δplb/gal12

gal/min400q95.0C

n

D

lbf 820169,1*12400*95.0*01823.0F j

pqc01823.0F d j

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