09 Drilling Hydraulics Hydrostatics

8/12/2019 09 Drilling Hydraulics Hydrostatics

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1

PETE 411

Well Drill ing

Lesson 9

Drilling Hydraulics- Hydrostatics


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2

Drilling Hydraulics - Hydrostatics

Hydrostatic Pressure in Liquid Columns

Hydrostatic Pressure in Gas Columns

Hydrostatic Pressure in Complex Columns

Forces on Submerged Body

Effective (buoyed) Weight of Submerged

Body

Axial Tension in Drill String A = FA/A


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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


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4

Drilling Hydraulics Applications

Calculation of subsurface hydrostatic

pressures that may tend to burst orcollapse well tubulars or fracture exposedformations

Several aspects of blowout prevention

Displacement of cement slurries andresulting stresses in the drillstring

WHY?


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5

Drilling Hydraulics Applications

contd

Bit nozzle size selection for optimum

hydraulics

Surge or swab pressures due to vertical

pipe movement

Carrying capacity of drilling fluids


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6Fig. 4-2. The Well Fluid System

Well Control ppore < pmud < pfrac


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9

Incompressible Fluids

Integrating,

dDFdp vw=

0vw pDFp +=

]0Dwhenpp[ 0 ==


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10

Incompressible Fluids

In field units, 33.8*144

4.62 =vwF

33.8*433.0 =

052.0=vwF

1 x 1 x 1

cube


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11

Incompressible fluids

If p0 = 0 (usually the case except during

well control or cementingprocedures)

then,

0pD052.0p +=

ft}lbm/gal,{psig,052.0 Dp =

D052.0

p=

p0

p

D


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12

Compressible

Fluids

dDFdp vw=dD052.0dp =

(1)

from (3)

(3)

(2)

But, TRMmZTRnZpV ==

TZ3.80

pM

ZRT

pM

V

m=== (4)

p = pressure of gas, psia

V = gas volume, gal

Z = gas deviation factorn = moles of gas

R = universal gas constant = 80.3

T = temperature, R

= density, lbm/gal

M = gas molecular wt.m = mass of gas


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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


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14

Compressible Fluids

=D

D

p

p 00

dD

TZ1544

M

p

dp

dD

TZ80.3

Mp052.0dp =

From Eqs. (2) and (4):

D

D

p

p 00[D]

TZ1544

M]p[ln =

Integrating,

]

TZ1544

)DD(M[exppp 00

=

Assumptions?


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Example (i)

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

]TZ1544

)D-M(D[exppp 00000,10 =

psia1188])140460)(1(1544

0)-16(10,000[exp1000 =

+=

]

TZ1544

)DD(M[exppp 00

=


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Example contd

gallbm331.0

600*1*3.8016*1000

TZ3.80pM

0 ===

(ii) What is density at surface?

gal

lbm

395.0600*1*3.80

16*1188

TZ3.80

pM000,10 ===

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


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Example

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

?p surf =

]TZ1544

)DD(M[exppp 00

=


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19

)DD(052.0pp 1ii

n

1i

i0

=

+=

Fig. 4-3.A Complex

Liquid

Column

Dp

pDp

=

+=

052.0

052.00


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

Pa = ?


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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

psig00 =p

psig266,1p a =


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22

Buoyancy Force = weight of fluid

displaced (Archimedes, 250 BC)

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


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Example

For steel,

immersed in mud,

the buoyancy factor is:

gal/lbm5.65s =

)/0.15( gallbmf =

7710565

01511 .

.

.

s

f =

=

A drillstring weighs 100,000 lbs in air.

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


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Axial Forces in Drillstring

Fb = bit weight


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Simple Example - Empty Wellbore

Drillpipe weight = 19.5 lbf/ft 10,000 ft

OD = 5.000 in

ID = 4.276 in

( )22 IDOD4

A =

A = 5.265 in2

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

AXIAL TENSION, lbf

DEP

TH,

ft

0 lbf 195,000 lbf


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29b221212dpdp

b2121T

FAp)AA(pWxw

FFFWWF}above(c){:PipeDrillAt

++=

++=

Anywhere in the Drill Pipe:

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

FT


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Axial Tension in Drill String

Example

A drill string consists of 10,000 ft of19.5 #/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 thedrillstring as a function of depth?


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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,

2

2

2

31in73.5

ftin144*

ft/lb490ft/lb5.19A ==

A1

10,000

10,600


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32

Example

Cross-sectional area of collars,

2

2 in2.43144*

490

147A ==

2

12 in5.3773.52.43AAareaalDifferenti ===

A2

A1


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33

Example

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

Compressive force = pA

= 357,200 lbf

[ axial tension = - 357,200 lbf]

2

2 in2.43*in

lbf268,8=

4

32

1


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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


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


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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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09 Drilling Hydraulics Hydrostatics