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1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics
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Page 1: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

1

PETE 411Well Drilling

Lesson 9

Drilling Hydraulics

- Hydrostatics

Page 2: 1 PETE 411 Well 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

Page 3: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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

Page 4: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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?

Page 5: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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

Page 6: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

6Fig. 4-2. The Well Fluid System

Well Control ppore < pmud < pfrac

Page 7: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

7

Forces Acting on a Fluid Element

A)DdD

dpp(

DAF vw

F1 =

F2 =

F3 =

pA

FWV = specific wt. of the fluid

Page 8: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

8

Pressures in a fluid column

At equilibrium, F = 0

DAFA)DdD

dpp(pA0F vw

dDFdp wv

0 = F1 + F2 + F3

(p = gh)

Page 9: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

9

Incompressible Fluids

Integrating,

dDFdp vw

0vw pDFp

]0Dwhen pp[ 0

Page 10: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

10

Incompressible Fluids

In field units,33.8

*144

4.62 vwF

33.8*433.0

052.0vwF

1’ x 1’ x 1’cube

Page 11: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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

Page 12: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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

Page 13: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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

Page 14: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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?

Page 15: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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

Page 16: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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

Page 17: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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?

Page 18: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

18

Example

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

?psurf

]T Z1544

)DD(M[ exp p p 0

0

Page 19: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

19

)DD(052.0pp 1ii

n

1ii0

Fig. 4-3. A Complex

Liquid Column

Dp

pDp

052.0

052.0 0

Page 20: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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

Pa = ?

Page 21: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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

Page 22: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

22

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

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

Page 23: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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

Page 24: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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

Page 25: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

25

Axial Forces in Drillstring

Fb = bit weight

Page 26: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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

Page 27: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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

Page 28: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

28

Anywhere in the Drill Collars:

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

b22 dcdcb22T F - Ap xwFFWF

Page 29: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

29b221212dpdp

b2121T

FAp)AA(pWxw

FFFWWF

} above (c) { :Pipe Drill At

Anywhere in the Drill Pipe:

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

FT

Page 30: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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?

Page 31: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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’

Page 32: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

32

Example

Cross-sectional area of collars,

22 in2.43144*

490

147A

212 in5.3773.52.43AAarea alDifferenti

A2

A1

Page 33: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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

Page 34: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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

Page 35: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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

Page 36: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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

Page 37: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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

Page 38: 1 PETE 411 Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics.

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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