1

PETE 411Well Drilling

Lesson 10Drilling Hydraulics (cont’d)

2

10. Drilling Hydraulics (cont’d)

4Effect of Buoyancy on Buckling4The Concept of Stability Force4Stability Analysis4Mass Balance4Energy Balance4Flow Through Nozzles4Hydraulic Horsepower4Hydraulic 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 indd n −= π

7

Consider the following:

19.5 #/ft drillpipeDepth = 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 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 - 44,694

= 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 = 0At 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 Tension44,770 0 150,306

FS FT

ft708,7=5.19306,150

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

lbfWFL

DC

BITDCIn Air:

In Liquid:

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

ρρρρρρρρ−−−−

====

s

fDC

BITDC

W

.F.S*FL1

ρρρρρρρρ−−−−

====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 tr

Zσσσσσσσσσσσσ ++++====

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:

4Plot FS vs. depth on “axial load (FT ) vs. depth plot”

4The neutral point is located where the lines intersect.

18

NOTE:

If pi = po = p,

then Fs = ( )pdd io22

4−− π

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 VelocityPipe Flow Annular Flow

WHEREv = average velocity, ft/sq = flow rate, gal/mind = 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

27

Law of Conservation of Energy

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

(((( )))) (((( ))))(((( )))) (((( ))))

QW

vvDDg

VpVpEE

++++====

−−−−++++−−−−−−−−

−−−−++++−−−−

21

2212

112212

21

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

28

In practical field unitsthis equation simplifies to:

( )

( ) fp pPvv

DDpp

∆−∆+−−

−+=

− 21

22

4

1212

10*074.8

052.0

ρ

ρ

p1 and p2 are pressures in psiρ 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 psiD1 and D2 are depths in ft.

where

29

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)

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

−+−+=

−+−

+=

∆−∆+−ρ−

−ρ+=

Pressure at bottom of drill collars = 7,833 psig

NOTE: KE in collars

May be ignored in many cases

0≈≈≈≈

32fp PPvv

DDpp

∆−∆+−−

−+=

)(10*074.8

)(052.021

22

4-

1212

ρ

ρ

33

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

34

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

36

For multiple nozzles in //Vn is the same for each nozzle

even if the dn varies!This follows since ∆∆∆∆p is the sameacross each nozzle.

tn A117.3

qv =

2t

2d

2-5

bit ACq10*8.311

∆pρ

=

10*074.8

pcv 4dn ρ∆= − &

37

Hydraulic Horsepowerof 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:

38

What is Hydraulic Impact Force

developed by bit?

Consider:

psi 169,1∆plb/gal 12

gal/min 400q95.0C

n

D

==ρ==

39

Impact = rate of change of momentum

(((( ))))60*17.32

vqv

tm

tmvF n

jρρρρ

====∆∆∆∆

∆∆∆∆====

∆∆∆∆∆∆∆∆====

psi 169,1∆plb/gal 12

gal/min 400q95.0C

n

D

==ρ==

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

pqc01823.0F dj ∆∆∆∆ρρρρ====