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1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics

Transcript

1

PETE 411Well Drilling

Lesson 9

Drilling Hydraulics

- Hydrostatics

2

Drilling Hydraulics - Hydrostatics

Hydrostatic Pressure in Liquid ColumnsHydrostatic Pressure in Gas ColumnsHydrostatic Pressure in Complex ColumnsForces on Submerged BodyEffective (buoyed) Weight of Submerged

Body

Axial Tension in Drill String A = FA/A

3

Read:Applied Drilling Engineering, Ch.4

(Drilling Hydraulics) to p. 125

HW #4

ADE #1.18, 1.19, 1.24

Due Monday, Sept 23, 2002

4

Drilling Hydraulics Applications

Calculation of subsurface hydrostatic pressures that may tend to burst or collapse well tubulars or fracture exposed formations

Several aspects of blowout prevention

Displacement of cement slurries and resulting stresses in the drillstring

WHY?

5

Drilling Hydraulics Applications cont’d

Bit nozzle size selection for optimum hydraulics

Surge or swab pressures due to vertical pipe movement

Carrying capacity of drilling fluids

6Fig. 4-2. The Well Fluid System

Well Control ppore < pmud < pfrac

7

Forces Acting on a Fluid Element

A)DdD

dpp(

DAF vw

F1 =

F2 =

F3 =

pA

FWV = specific wt. of the fluid

8

Pressures in a fluid column

At equilibrium, F = 0

DAFA)DdD

dpp(pA0F vw

dDFdp wv

0 = F1 + F2 + F3

(p = gh)

9

Incompressible Fluids

Integrating,

dDFdp vw

0vw pDFp

]0Dwhen pp[ 0

10

Incompressible Fluids

In field units,33.8

*144

4.62 vwF

33.8*433.0

052.0vwF

1’ x 1’ x 1’cube

11

Incompressible fluids

If p0 = 0 (usually the case except during well control or cementing

procedures)

then,

0pD052.0p

ft} lbm/gal, {psig, 052.0 Dp

D 052.0

p

p0

p

D

12

Compressible Fluids

dDFdp vw

dD 052.0dp

…………… (1)

from (3)

…… (3)

…………… (2)

But, T R M

m ZT R n ZpV

T Z3.80

pM

ZRT

pM

V

m …… (4)

p = pressure of gas, psia

V = gas volume, gal

Z = gas deviation factor

n = moles of gas

R = universal gas constant = 80.3

T = temperature, R = density, lbm/galM = gas molecular wt.m = mass of gas

13

Compressible Fluids

p = pressure of gas, psia

V = gas volume, gal

Z = gas deviation factor

n = moles of gas

R = universal gas constant, = 80.3

T = temperature, oR = density, lbm/gal

M = gas molecular wt.

m = mass of gas, lbm

14

Compressible Fluids

D

D

p

p 00

dD T Z1544

M

p

dp

dD T Z80.3

M p 052.0 dp

From Eqs. (2) and (4):

DD

pp 00

[D] T Z1544

M ]p[ln Integrating,

]T Z1544

)DD(M[ exp p p 0

0

Assumptions?

15

Example

Column of Methane (M = 16)

Pressure at surface = 1,000 psia Z=1, T=140 F

(i) What is pressure at 10,000 ft?

(ii) What is density at surface?

(iii) What is density at 10,000 ft?

(iv) What is psurf if p10,000 = 8,000 psia?

]T Z1544

)DD(M[ exp p p 0

0

16

Example (i)

(i) What is pressure at 10,000 ft?

] T Z1544

)D-M(D [ exp pp 0

0000,10

psia 1188 ] )140460)(1(1544

0)-16(10,000 [ exp 1000

]T Z1544

)DD(M[ exp p p 0

0

17

Example cont’d

gal

lbm331.0

600*1*3.80

16*1000

T Z3.80

pM0

(ii) What is density at surface?

gal

lbm395.0

600*1*3.80

16*1188

T Z3.80

pM000,10

(iii) What is density at 10,000 ft?

18

Example

(iv) What is psurf if p10,000 = 8,000 psia?

?psurf

]T Z1544

)DD(M[ exp p p 0

0

19

)DD(052.0pp 1ii

n

1ii0

Fig. 4-3. A Complex

Liquid Column

Dp

pDp

052.0

052.0 0

20Fig. 4-4. Viewing the Well as a Manometer

Pa = ?

21

Figure 4.4

})000,10(0.9)000,1(7.16

)700,1(7.12)300(5.8)000,7(5.10{052.00

ppa

psig 00 p

psig 266,1p a

22

Buoyancy Force = weight of fluid displaced (Archimedes, 250 BC)

Figure 4-9. Hydraulic forces acting on a foreign body

23

Effective (buoyed) Weight

s

fe 1WW

Buoyancy Factor

Valid for a solid body or an open-ended pipe!

sf

f

be

W-W

V-W

FWW

24

Example

For steel,

immersed in mud,

the buoyancy factor is:

gal/lbm 5.65s

)/ 0.15( gallbmf

7710565

01511 .

.

.

s

f

A drillstring weighs 100,000 lbs in air.

Buoyed weight = 100,000 * 0.771 = 77,100 lbs

25

Axial Forces in Drillstring

Fb = bit weight

26

Simple Example - Empty Wellbore

Drillpipe weight = 19.5 lbf/ft 10,000 ft

OD = 5.000 inID = 4.276 in

22 IDOD4

A

A = 5.265 in2

W = 19.5 lbf/ft * 10,000 ft = 195,000 lbf

AXIAL TENSION, lbf

DE

PT

H,

ft

0 lbf 195,000 lbf

27

Example - 15 lb/gal Mud in Wellbore

Drillpipe weight = 19.5 lbf/ft 10,000 ft

OD = 5.000 inID = 4.276 in

22 IDOD4

A

A = 5.265 in2

W = 195,000 - 41,100 = 153,900 lbf

AXIAL TENSION, lbf

DE

PT

H,

ft

0 195,000 lbf

Pressure at bottom = 0.052 * 15 * 10,000 = 7,800 psiF = P * A= 7,800 * 5.265= 41,100 lbf

153,900- 41,100

28

Anywhere in the Drill Collars:

Axial Tension = Wt. - Pressure Force - Bit Wt.

b22 dcdcb22T F - Ap xwFFWF

29b221212dpdp

b2121T

FAp)AA(pWxw

FFFWWF

} above (c) { :Pipe Drill At

Anywhere in the Drill Pipe:

Axial Tension = Wts. - Pressure Forces - Bit Wt.

FT

30

Axial Tension in Drill String

Example A drill string consists of 10,000 ft of 19.5 #/ft drillpipe and 600 ft of 147 #/ft drill collars suspended off bottom in 15#/gal mud (Fb = bit weight = 0).

What is the axial tension in the

drillstring as a function of depth?

31

Example

Pressure at top of collars = 0.052 (15) 10,000

= 7,800 psi

Pressure at bottom of collars = 0.052 (15) 10,600

= 8,268 psi

Cross-sectional area of pipe,

22

2

31 in73.5ft

in144*

ft/lb490

ft/lb5.19A

A1

10,000’

10,600’

32

Example

Cross-sectional area of collars,

22 in2.43144*

490

147A

212 in5.3773.52.43AAarea alDifferenti

A2

A1

33

Example

1. At 10,600 ft. (bottom of drill collars)

Compressive force = pA

= 357,200 lbf

[ axial tension = - 357,200 lbf ]

22

in2.43*in

lbf268,8

4

32

1

34

Example

2. At 10,000 ft+ (top of collars)

FT = W2 - F2 - Fb

= 147 lbm/ft * 600 ft - 357,200

= 88,200 - 357,200

= -269,000 lbf

4

32

1

Fb = FBIT = 0

35

Example

3. At 10,000 ft - (bottom of drillpipe)

FT = W1+W2+F1-F2-Fb

= 88,200 + 7800 lbf/in2 * 37.5in2 - 357,200

= 88,200 + 292,500 - 357,200

= + 23,500 lbf

4

32

1

36

Example

4. At Surface

FT = W1 + W2 + F1 - F2 - Fb

= 19.5 * 10,000 + 23,500

= 218,500 lbf

Also: FT = WAIR * BF = 283,200 * 0.7710

= 218,345 lbf

4

32

1

37Fig. 4-11. Axial tensions as a function of depth for Example 4.9

38

Example - Summary

1. At 10,600 ft FT = -357,200 lbf [compression]

2. At 10,000 + ft FT = -269,000 lbf [compression]

3. At 10,000 - ft FT = +23,500 lbf [tension]

4. At Surface FT = +218,500 lbf [tension]

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