Transcript Petroleum Engineering 405 Drilling Engineering

PETE 411
Well Drilling
Lesson 9
Drilling Hydraulics
- Hydrostatics
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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
sA = FA/A
2
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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WHY?
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
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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
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Well Control
ppore < pmud < pfrac
Fig. 4-2. The Well Fluid System
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Forces Acting on a Fluid Element
F1 =
FWV = specific wt.
of the fluid
pA
dp
D)A
F2 = (p 
dD
F3 = Fw v AD
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Pressures in a fluid column
At equilibrium,
SF=0
0 = F1 + F2 + F3
dp
 F  0  pA  (p  dD D)A  Fw v AD
 dp  FwvdD
(p = rgh)
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Incompressible Fluids
dp  Fw v dD
Integrating,
p  Fw v D  p 0
[p  p0 when D  0]
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Incompressible Fluids
In field units, Fw v
62.4
r

*
144 8.33
 0.433 *
r
8.33
Fwv  0.052 r
1’ x 1’ x 1’
cube
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p0
Incompressible fluids

If p0 = 0
p  0.052 rD  p0
D
p
(usually the case except during
well control or cementing
procedures)

p  0.052 r D
{psig, lbm/gal, ft}
then,
p
r
0.052 D
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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
dp  Fw v dD
dp  0.052 r dD
T = temperature, R
r = density, lbm/gal
M = gas molecular wt.
m = mass of gas
…………… (1)
…………… (2)
m
pV  Z n R T  Z
R T …… (3)
But,
M
m
pM
pM
r 


…… (4)
V
ZRT 80.3 Z T
from (3)
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Compressible Fluids
p = pressure of gas, psia
o
T = temperature, R
V = gas volume, gal
r = density, lbm/gal
Z = gas deviation factor
M = gas molecular wt.
n = moles of gas
m = mass of gas, lbm
R = universal gas constant,
= 80.3
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Compressible Fluids
From Eqs. (2) and (4):
0.052 p M
 dp 
dD
80.3 Z T


p
p0
Integrating,

Assumptions?
dp
M

p
1544 Z T
[ln p]
p
p0

D
D0
dD
M
D

[D] D 0
1544 Z T
M(D  D 0 )
p  p 0 exp [
]
1544 Z T
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Example
M(D  D 0 )
p  p 0 exp [
]
1544 Z T
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?
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Example (i)
M(D  D 0 )
p  p 0 exp [
]
1544 Z T
(i) What is pressure at 10,000 ft?
p10, 000
M(D - D 0 )
 p 0 exp [
]
1544 Z T
16(10,000 - 0)
 1000 exp [
]  1188 psia
1544(1)( 460  140)
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Example cont’d
(ii) What is density at surface?
pM
1000 *16
lbm
r0 

 0.331
80.3 Z T 80.3 *1* 600
gal
(iii) What is density at 10,000 ft?
r10,000
pM
1188 *16
lbm


 0.395
80.3 Z T 80.3 *1* 600
gal
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Example
(iv) What is psurf if p10,000 = 8,000 psia?
p surf  ?
p  p 0 exp [
M(D  D 0 )
]
1544 Z T
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Fig. 4-3.
A Complex
Liquid
Column
p  0.052 r D  p0
p  0.052 r D
n
p  p 0  0.052 r i (Di  Di 1 )
i 1
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Pa = ?
Fig. 4-4. Viewing the Well as a Manometer
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Figure 4.4
pa  p0  0.052 { 10.5(7,000)  8.5(300)  12.7(1,700)
 16.7(1,000)  9.0(10,000) }
p0  0 psig
 p a  1,266 psig
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Buoyancy Force = weight of fluid
displaced (Archimedes, 250 BC)
Figure 4-9. Hydraulic forces acting on a foreign body
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Effective (buoyed) Weight
We  W  Fb
 W - rf V
W
 W - rf
rs


rf 
We  W 
1  r 

s 

Buoyancy Factor
Valid for a solid body or an open-ended pipe!
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Example
For steel,
rs  65.5 lbm / gal
immersed in mud,
( r f  15.0 lbm / gal )
the buoyancy factor is:

rf
1 
rs

 
15.0 
  1 
  0.771
65.5 
 
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
OD = 5.000 in
ID = 4.276 in
A


OD 2  ID 2
4

195,000 lbf
DEPTH, ft
0 lbf
10,000 ft
A = 5.265 in2
AXIAL TENSION, lbf
W = 19.5 lbf/ft * 10,000 ft = 195,000 lbf
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Example - 15 lb/gal Mud in Wellbore
Drillpipe weight = 19.5 lbf/ft
OD = 5.000 in
ID = 4.276 in


A  OD 2  ID 2
4

0
153,900 195,000 lbf
DEPTH, ft
- 41,100
10,000 ft
A = 5.265 in2
AXIAL TENSION, lbf
F=P*A
= 7,800 * 5.265
= 41,100 lbf
Pressure at bottom = 0.052 * 15 * 10,000 = 7,800 psi
W = 195,000 - 41,100 = 153,900 lbf
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Anywhere in the Drill Collars:
Axial Tension = Wt. - Pressure Force - Bit Wt.
FT  W2  F2  Fb  w dc x dc  p 2 A 2 - Fb
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Anywhere in the Drill Pipe:
Axial Tension = Wts. - Pressure Forces - Bit Wt.
At Drill Pipe : { (c) above }
FT  W1  W2  F1  F2  Fb
FT  w dp x dp  W2  p1 (A 2  A1 )  p 2 A 2  Fb
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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?
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A1
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,
10,000’
19.5 lb / ft 144 in2
2
A1 
*
 5.73 in
3
2
490 lb / ft
ft
10,600’
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Example
A1
Cross-sectional area of collars,
147
2
A2 
*144  43.2in
490
A2
Differenti al area  A 2  A1  43.2  5.73  37.5in
2
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Example
4
1. At 10,600 ft. (bottom of drill collars)
Compressive force = pA
lbf
 8,268 2 * 43.2in 2
in
3
2
1
= 357,200 lbf
[ axial tension = - 357,200 lbf ]
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Example
4
2. At 10,000 ft+ (top of collars)
FT = W2 - F2 - Fb
Fb = FBIT = 0
3
2
= 147 lbm/ft * 600 ft - 357,200
1
= 88,200 - 357,200
= -269,000 lbf
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Example
4
3. At 10,000 ft - (bottom of drillpipe)
FT = W1+W2+F1-F2-Fb
= 88,200 + 7800 lbf/in2 * 37.5in2 - 357,200
3
2
1
= 88,200 + 292,500 - 357,200
= + 23,500 lbf
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Example
4
4. At Surface
FT = W1 + W2 + F1 - F2 - Fb
= 19.5 * 10,000 + 23,500
3
2
= 218,500 lbf
1
Also: FT = WAIR * BF = 283,200 * 0.7710
= 218,345 lbf
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Fig. 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
4. At Surface
FT = +23,500 lbf [tension]
FT = +218,500 lbf [tension]
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