PETE 411
Well Drilling
Lesson 10
Drilling Hydraulics (cont’d)
1
10. Drilling Hydraulics (cont’d)
Effect of Buoyancy on Buckling
The Concept of Stability Force
Stability Analysis
Mass Balance
Energy Balance
Flow Through Nozzles
Hydraulic Horsepower
Hydraulic Impact Force
2
READ:
ADE, Ch. 4 to p. 135
HW #5:
ADE # 4.3, 4.4, 4.5, 4.6
due September 27, 2002
3
Fh
Buckling
of
Tubulars
Fh - Fb
l
Neutral Point
l
Slender pipe
suspended
in wellbore
Partially
buckled
slender
pipe
Neutral Point
Fb
4
Buckling of Tubulars
• Long slender columns, like DP,
l
Neutral
Point
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?
Neutral
Point
Fb
5
What is NEUTRAL POINT?
• One definition of NEUTRAL
POINT is the point above
which there is no tendency
towards buckling
l
• Resistance to buckling is
indicated, in part, by:
Neutral
Point
The Moment of Inertia
Neutral
Point
I


d
64
4
n
d4
  in 
4
6
Consider the
following:
19.5 #/ft drillpipe
Depth = 10,000 ft.
Mud wt. = 15 #/gal.
DPHYD = 0.052 (MW) (Depth)
= 0.052 * 15 * 10,000
DPHYD = 7,800 psi
Axial tensile stress in pipe at bottom
= - 7,800 psi
What is the axial force at bottom?
7
What is the axial force at bottom?
Cross-sectional area of pipe
= (19.5 / 490) * (144/1) = 5.73 in2
Axial compressive force = pA
lbf
2
 7,800

5
.
73
in
2
in
= 44,700 lbf.
Can this cause the pipe to buckle?
8
Axial Tension:
FT = W1 - F2
FT
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!
9
F2
Stability Force:
Ai
FS = Aipi - AO pO
FS = (Ai - AO) p
(if pi = pO)
At surface, FS = - 5.73 * 0 = 0
At 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!!
10
Compression
44,770
Tension
150,306
0
FS
FT
Zero Axial Stress
Neutral Point
Depth of Zero Axial Stress Point =
150,306
= 7,708
11 ft
19.5
Length
of
Drill
Collars
Neutral Point
Neutral Point
12
Length of Drill Collars
In Air:
In Liquid:
In Liquid
with S.F.:
(e.g., S.F =1.3)
LDC
FBIT

WDC
LDC 
LDC
FBIT

f 

WDC 
1


s 


 lbf 


 lbf / ft 
 lbf 


 lbf / ft 
FBIT * S.F.


f
WDC 
1  
s





13
State of stress in pipe at the neutral point?14
At the Neutral Point:
The axial stress is equal to the average
of the radial and tangential stresses.
r  t
Z 
2
15
Stability Force:
0
FT
FS
FT
FS = Ai Pi - Ao Po
If FS > axial tension then
the pipe may buckle.
If FS < axial tension then
the pipe will NOT buckle.
16
At the neutral point:
FS = axial load
To locate the neutral point:
Plot FS vs. depth on
“axial load (FT ) vs. depth plot”
The neutral point is located where the
lines intersect.
17
NOTE:
If pi = po = p,
then Fs =
or,


d
4

2
o
 di
2
p
AS
Fs = - AS p
18
Axial Load with
FBIT = 68,000 lbf
19
Stability
Analysis with
FBIT = 68,00020 lbf
Nonstatic Well Conditions
FLUID FLOW
Physical Laws
Rheological Models
Equations of State
21
Physical Laws
Conservation of mass
Conservation of energy
Conservation of momentum
22
Rheological Models
 Newtonian
 Bingham Plastic
 Power – Law
 API Power-Law
23
Equations of State
Incompressible fluid
Slightly compressible fluid
Ideal gas
Real gas
24
Average Fluid Velocity
Pipe Flow
Annular Flow
q
v 
2
2.448 d
q
v 
2
2
2.448 d 2  d1


WHERE
v = average velocity, ft/s
q = flow rate, gal/min
d = internal diameter of pipe, in.
d2 = internal diameter of outer pipe or borehole, in.
d1 =external diameter of inner pipe, in.
25
26
Law of Conservation of Energy
States that as a fluid flows
from point 1 to point 2:
 E 2  E1    p2V2 
p1V1

1 2
2
 g  D2  D1  
v 2  v1
2
W Q
In the wellbore, in many cases
{


Q = 0 (heat)
 = constant27
In practical field units
this equation simplifies to:
p2  p1  0.052 
D2  D1 


8.074 *10  4  v22  v12  DPp  Dp f
where
p1 and p2

v1 and v2
Dpp
Dpf
D1 and D2
are pressures in psi
is density in lbm/gal.
are velocities in ft/sec.
is pressure added by pump
between points 1 and 2 in psi
is frictional pressure loss in psi
are depths in ft.
28
Determine the pressure at the
bottom of the drill collars, if
Dp f  1,400 psi

 400 gal/min.
 12 lbm/gal.
D2
 10,000 ft.
(bottom of drill collars)
D1
0
(mud pits)
q
IDDC  2.5 in.
Dp p  3,000 psi
29
Velocity in drill collars
q
v2 
2
2.448 d
(gal/min)
2
(in )
400
v2 
 26 .14 ft/sec
2
2.448 * (2.5)
Velocity in mud pits, v1  0
30
p 2  p1  0.052  (D2  D1 )
 8.074 * 10
-4
 ( v  v )  DPp  DPf
2
2
2
1
p 2  0  0.052 * 12 (10,000 - 0)
- 8.074 * 10 -4 * 12 (26.14 2  0 2 )  3,000  1,400
 0  6,240  6.6  3,000  1,400
Pressure at bottom of drill collars = 7,833 psig
NOTE: KE in collars
0
May be ignored in many cases31
p2  p1  0.052  ( D2  D1 )
 8.074 *10
-4
 (v  v )  DPp  DPf
2
2
2
1
32
Fluid Flow Through Nozzle
Assume:
D2  D1
DPp  0
v1
0
v 2  vn
DPf  0
4
p 2  p1  8.074 * 10 v
and
vn 
2
n
Dp
4
8.074 * 10 
33
If
DPf  0
Equation may be written as
vn  c d
Dp
4
8.074 * 10 
c d  0.95
This accounts for all the losses in the nozzle.
Example:
vn  0.95
1,000
 305 ft/sec
4
8.074 * 10 * 12
34
35
For multiple nozzles in //
Vn is the same for each nozzle
even if the dn varies!
This follows since Dp is the same
across each nozzle.
vn  c d
Dp
8.074 * 10 4 
Δp bit 
q
& vn 
3.117 A t
8.311 * 10 -5  q2
2
d
C A
2
t
36
Hydraulic Horsepower
of pump putting out 400 gpm at 3,000 psi = ?
Power  rate of doing work
 F* s/t
 Dp * A 
q
A
PH  qDp
In field units:
qDp
HHP 
1714
400 * 3,000

 700hp
1714
37
What is Hydraulic Impact Force
developed by bit?
Consider:
CD  0.95
q  400 gal/min
  12 lb/gal
Δp n  1,169 psi
38
Impact = rate of change of momentum
 q vn
Dmv   m 
Fj 
   Dv 
Dt
32.17 * 60
 Dt 
CD  0.95
Fj  0.01823 c d q  Dp
q  400 gal/min
  12 lb/gal
Δp n  1,169 psi
Fj  0.01823 * 0.95 * 400 12 * 1,169  820 lbf
39