PETE 203
DRILLING ENGINEERING
Drilling Hydraulics
Drilling Hydraulics
Energy Balance
Flow Through Nozzles
Hydraulic Horsepower
Hydraulic Impact Force
Rheological Models
Optimum Bit Hydraulics
Nonstatic Well Conditions
Physical Laws:
Conservation of Mass
Conservation of energy
Conservation of momentum
Rheological Models
Newtonian
Bingham Plastic
Power – Law
API Power-Law
Equations of State
Incompressible fluid
Slightly compressible fluid
Ideal gas
Real gas
Average Fluid Velocity
Pipe Flow Annular Flow
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.
2448.2 d
qv
2
1
2
2448.2 dd
qv
Law of Conservation of Energy
States that as a fluid flows
from point 1 to point 2:
QW
vvDDg
VpVpEE
2
1
2
212
112212
2
1
In the wellbore, in many cases Q = 0 (heat)
r = constant {
In practical field units this equation simplifies to:
fp pPvv
DDpp
2
1
2
2
4
1212
10*074.8
052.0
r
r
p1 and p2 are pressures in psi r is density in lbm/gal. v1 and v2 are velocities in ft/sec. pp is pressure added by pump between points 1 and 2 in psi pf is frictional pressure loss in psi D1 and D2 are depths in ft.
where
Determine the pressure at the bottom of the drill collars, if
psi 000,3 p
in. 5.2
0 D
ft. 000,10 D
lbm/gal. 12
gal/min. 400 q
psi 1,400
p
1
2
DC
f
ID
p
r
(bottom of drill collars)
(mud pits)
Velocity in drill collars
)(in
(gal/min)
d448.2
qv
222
ft/sec 14.26)5.2(*448.2
400v
22
Velocity in mud pits, v1 0
400,1000,36.6240,60
400,1000,3)014.26(12*10*8.074-
0)-(10,00012*052.00p
PP)vv(10*074.8
)DD(052.0pp
224-
2
fp
2
1
2
2
4-
1212
r
r
Pressure at bottom of drill collars = 7,833 psig
NOTE: KE in collars
May be ignored in many cases
0
fp PPvv
DDpp
)(10*074.8
)(052.0
2
1
2
2
4-
1212
r
r
0 P
v v0 P
0 vD D
f
n2p
112
Fluid Flow Through Nozzle
Assume:
r
r
4n
2
n
4
12
10*074.8
pv and
v10*074.8pp
If
95.0c 10*074.8
pcv
as writtenbemay Equation
d4dn r
0 fP
This accounts for all the losses in the nozzle.
Example: ft/sec 305 12*10*074.8
000,195.0v
4n
For multiple nozzles in //
Vn is the same for each nozzle
even if the dn varies!
This follows since p is the same across
each nozzle.
t
nA117.3
qv
2
t
2
d
2-5
bitAC
q10*8.311Δp
r
10*074.8
pcv
4dnr
&
Hydraulic Horsepower
HHP of pump putting out 400 gpm at 3,000 psi = ?
Power
pqP
A
qA*p
t/s*F
workdoing of rate
H
hp7001714
000,3*400
1714
pq HHP
In field units:
Hydraulic Impact Force
What is the HHP Developed by bit?
Consider:
psi 169,1Δp
lb/gal 12
gal/min 400q
95.0C
n
D
r
Impact = rate of change of momentum
60*17.32
vqv
t
m
t
mvF
n
j
r
psi 169,1Δp
lb/gal 12
gal/min 400q
95.0C
n
D
r
lbf 820169,1*12400*95.0*01823.0Fj
pqc01823.0F dj r
Newtonian Fluid Model
Shear stress = viscosity * shear rate
A
F ,
L
VallyExperiment
Laminar Flow of Newtonian Fluids
A
F
L
V
Newtonian Fluid Model
In a Newtonian fluid the shear stress is directly
proportional to the shear rate (in laminar flow):
i.e.,
The constant of proportionality, is the
viscosity of the fluid and is independent of
shear rate.
sec
12
cm
dyne
.
Newtonian Fluid Model
Viscosity may be expressed in poise or centipoise.
poise 0.01 centipoise 1
scm
g1
cm
s-dyne1 poise 1
2
2cm
secdyne
.
Shear Stress vs. Shear Rate for a Newtonian Fluid
Slope of line
.
Example - Newtonian Fluid
Example 4.16
Area of upper plate = 20 cm2
Distance between plates = 1 cm
Force req’d to move upper plate at 10 cm/s
= 100 dynes.
What is fluid viscosity?
Example 4.16
poise 5.0cm
sdyne5.0
10
52
1-
2
sec 10/1
dynes/cm 20/100
/
/
rate shear
stressshear
LV
AF
cp 50
Bingham Plastic Model
Bingham Plastic Model
- if
- if 0
if
yyp
yy
yyp
and y are often expressed in lbf/100 sq.ft
Power-Law Model
Power-Law Model
n = flow behavior index
K = consistency index
0 if K
0 if K
1n
n
Rheological Models
1. Newtonian Fluid:
2. Bingham Plastic Fluid:
rate shear
viscosityabsolute
stressshear
*)( py viscosityplastic
point yield
p
y
What if y 0
3. Power Law Fluid:
When n = 1, fluid is Newtonian and K =
We shall use power-law model(s) to
calculate pressure losses (mostly).
n
)(K
K = consistency index
n = flow behavior index
Rheological Models
Velocity Profiles (laminar flow)
Fig. 4-26. Velocity profiles for laminar flow:
(a) pipe flow and (b) annular flow
“It looks like concentric rings of fluid
telescoping down the pipe at different velocities”
3D View of Laminar Flow in a pipe
- Newtonian Fluid
Summary of Laminar Flow Equations for Pipes and Annuli
Fig 4.33: Critical Reynolds number for
Bingham plastic fluids.
Fig 4.34: Fraction Factors for Power-law
fluid model.
Total Pump Pressure
Pressure loss in surf. equipment
Pressure loss in drill pipe
Pressure loss in drill collars
Pressure drop across the bit nozzles
Pressure loss in the annulus between the drill
collars and the hole wall
Pressure loss in the annulus between the drill
pipe and the hole wall
Hydrostatic pressure difference (r varies)
Total Pump Pressure
PUMP SC DP DC
B DCA DPA HYD
P P P P
P P P ( ΔP )
Types of Flow
Laminar Flow
Flow pattern is linear (no radial flow)
Velocity at wall is ZERO
Produces minimal hole erosion
Types of Flow - Laminar
Mud properties strongly affect
pressure losses
Is preferred flow type for annulus
(in vertical wells)
Laminar flow is sometimes referred to
as sheet flow, or layered flow:
* As the flow velocity increases, the flow type
changes from laminar to turbulent.
Types of Flow
Turbulent Flow
Flow pattern is random (flow in all directions)
Tends to produce hole erosion
Results in higher pressure losses (takes more energy)
Provides excellent hole cleaning…but…
Types of flow
Mud properties have little effect on pressure losses
Is the usual flow type inside the drill pipe and collars
Thin laminar boundary layer at the wall
Turbulent flow, cont’d
Fig. 4-30. Laminar and turbulent flow patterns in a circular pipe: (a) laminar
flow, (b) transition between laminar and turbulent flow and (c) turbulent flow
Turbulent Flow - Newtonian Fluid
The onset of turbulence in pipe flow is
characterized by the dimensionless group
known as the Reynolds number
r dvN
_
Re
μ
dvρ928N
_
Re In field units,
Turbulent Flow - Newtonian Fluid
We often assume that fluid flow is
turbulent if Nre > 2,100
cp. fluid, ofviscosity μ
in I.D., piped
ft/s velocity,fluid avg. v
lbm/gal density, fluid ρ where
_
μ
dvρ928N
_
Re
PPUMP = PDP + PDC
+ PBIT NOZZLES
+ PDC/ANN + PDP/ANN
+ PHYD
Q = 280 gal/min
r = 12.5 lb/gal
Pressure Drop Calculations PPUMP
"Friction" Pressures
0
500
1,000
1,500
2,000
2,500
0 5,000 10,000 15,000 20,000 25,000
Distance from Standpipe, ft
"F
ricti
on
" P
ressu
re, p
si DRILLPIPE
DRILL COLLARS
BIT NOZZLES
ANNULUS
2103
Optimum Bit Hydraulics
Under what conditions do we get the
best hydraulic cleaning at the bit?
Maximum hydraulic horsepower?
Maximum impact force?
Both these items increase when the circulation
rate increases.
However, when the circulation rate increases,
so does the frictional pressure drop.
Jet Bit Nozzle Size Selection
Nozzle Size Selection for Optimum Bit
Hydraulics:
Max. Nozzle Velocity
Max. Bit Hydraulic Horsepower
Max. Jet Impact Force
Jet Bit Nozzle Size Selection
Proper bottom-hole cleaning
Will eliminate excessive regrinding of
drilled solids, and
Will result in improved penetration rates
Bottom-hole cleaning efficiency
Is achieved through proper selection of bit
nozzle sizes
Jet Bit Nozzle Size Selection - Optimization -
Through nozzle size selection, optimization may
be based on maximizing one of the following:
Bit Nozzle Velocity
Bit Hydraulic Horsepower
Jet impact force
• There is no general agreement on which of
these three parameters should be maximized.
Maximum Nozzle Velocity
From Eq. (4.31)
i.e.
so the bit pressure drop should be maximized in order
to obtain the maximum nozzle velocity
r4
bdn
10*074.8
PCv
bn Pv
Maximum Nozzle Velocity
This (maximization) will be achieved when the
surface pressure is maximized and the
frictional pressure loss everywhere is
minimized, i.e., when the flow rate is
minimized.
pressure. surface allowable maximum the and
rate ncirculatio minimum the at
satisfied, are above 2&1 whenmaximized is vn
Maximum Bit Hydraulic Horsepower
The hydraulic horsepower at the bit is
maximized when is maximized. q) p( bit
dpumpbit ppp
where may be called the parasitic pressure
loss in the system (friction). dp
bitdpump ppp
Maximum Bit Hydraulic Horsepower
. turbulentis flow theif
cqpppppp 75.1
dpadcadcdpsd
In general, where m
d cqp 2m0
The parasitic pressure loss in the system,
Maximum Bit Hydraulic Horsepower
0)1(p when 0
17141714
pump
1
mHbit
m
pumpbitHbit
qmcdq
dP
cqqpqpP
dpumpbit ppp m
d cqp
Maximum Bit Hydraulic Horsepower
when maximum is
1
1p when .,.
)1(p when .,.
d
pump
Hbit
pump
d
P
pm
ei
pmei
pumpd pm
p
1
1
0)1(p pump mqmc
Maximum Jet Impact Force
The jet impact force is given by Eq. 4.37:
)(c0.01823
01823.0
d dpump
bitdj
ppq
pqcF
r
r
Maximum Jet Impact Force
But parasitic pressure drop,
2201823.0
m
dpdj
m
d
qcqpcF
cqp
rr
)(c0.01823 d dpumpj ppqF r
Maximum Jet Impact Force
Upon differentiating, setting the first derivative
to zero, and solving the resulting quadratic
equation, it may be seen that the impact force is
maximized when,
pd p2m
2p
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