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PETE 411
Well Drilling
Lesson 12
Laminar Flow - Slot Flow
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Lesson 12 - Laminar Flow - Slot Flow
4The Slot Flow Approximation4Shear Rate Determination
4Pressure Drop Calculations
4Laminar Flow
4Turbulent Flow4Transition Flow - Critical Velocity
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Read:
Applied Drilling Engineering
Ch.4 to p. 145
Homework #6
On the WebDue Friday, October 4, 2002
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Representing the Circular Annulus as a Slot
)r(r!WslotofWidth
)r(rhslotofHeight
rr!WhslotequivalentofArea
"2
"2
2
"
2
2
+==
==
==
{ slot approximation is OK if (d1/d2 > 0.3 }
Equal
Area
and
Height
Simpler
Equations
-yetaccurate
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yW#LdL
dppyWpF
ypWF
f
22
"
==
=
Representing the Annulus as a Slot
F4 = y +yWL = +ddyy WL
F3 =WL
Consider:
- pressure forces
- viscous forces
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Representing the Annulus as a Slot
state,steadyAt
F = m aSumming forces along flow:
F = 0
F
"
F2 + F3
F4 = 0
0LW#ydy
d$$-LW$yW#L
dL
dp-p-ypW f =
++
,gSimplifyin dp f
dLd
dy= 0
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Representing the Annulus as a Slot
dydv% =
:integrateandvariablesSeparate
Model,FluidNewtonianWith
dp fdL
ddy
= 0
dL
dpy
0
f $$ +=Evaluate 0 at wall where y = 0
= %$
But,
of
dL
dpy
dy
dv$$ +==
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Representing the Annulus as a Slot
+= dy$ydL
dp-dv 0
f
00f
2
v
y$
dL
dp
2
yv +=
of $
dL
dpy
dy
dv +=
0 v0,y when0vSince 0 ===
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Representing the Annulus as a Slot
h$-
dL
dp
2
h-0h,y when0 vSince 0f
2
===
dL
dp
2
h-$ f0 =
( )2f yhydL
dp
2
"v =
Hence, substituting for v0 and 0 :
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Representing the Annulus as a Slot
== vWdyvdAq
The total flow rate:
( )
=
h
0
2f dyyhydL
dp
2
Wq
dL
dp
"2
Whq f
3
=
g,Integratin
"2
2
"
2
2 rrhand)r(r!Wh ==But
( )2f yhydL
dp
2
"v =
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Representing the Annulus as a Slot
2
"2
2
"
2
2
f
)r)(rr(rdL
dp
"2
!
q =
)r(r!
q
A
qv
2"
22 ==velocity,averageBut
2
"2
_
f
)r(r
v"2
dL
dp
=
In field units,2
"2
_
f
)d"000(d
v
dL
dp
=
psi/ft, cp., ft/sec, in
dL
dp
12
Whq f
3
=
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Example 4.22
Compute the frictional pressure loss for a 7 x
5 annulus, 10,000 ft long, using the slot flowrepresentation in the annulus. The flow rate
is 80 gal/min. The viscosity is 15 cp.
Assume the flow pattern is laminar.
7 5 1
6
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Example 4.22
The average velocity in the
annulus,
)52.448(7
80
)d2.448(d
qv
222
"
2
2
_
=
=
ft/s".362v_
=
( )2"2
_
f
dd"000
v
dL
dp
=
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Example 4.22
A somewhat more accurate answer, using an
exact equation for a circular annulus, results
in a value of 50.9792 psi.
Difference = 0.0958 psi i.e., within 0.2%
( )5".0750psi5"
)57("000
)000,"0()362."()"5(
DdL
dp
#p 2f
==
==
fp
( )2
"2
_
f
dd"000
v
dL
dp
=
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Determination of Shear Rate...(why?)
If shear rate in well is known:
1. Fluid can be evaluated in viscometer at
the proper shear rate.
2. Newtonian equations can sometimes
give good accuracy even if fluid isnon-Newtonian.
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Determination of Shear Rate
The maximum value of shear rate will
occur at the pipe walls.
For circular pipe, at the pipe wall,
dL
dp
2
r
$
fw
w =from (Eq. 4.5")
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Determination of Shear Rate
From Eq. 4.54b,
(at the wall)
w
_
w
2
w
_
ww
2
w
_
f
r
v4$
r
v8*
2
r$
r
v8
dL
dp
=
=
=
dL
dp
2
r$ fww =
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Determination of Shear Rate (why?)
Using the Newtonian Model,
Changing to field units,
w
_
w
_
w
r
v4
r
v4*
"
$% ===
dv96%
_
=
(circular pipe)
sec-", ft/sec, in
w
_
wr
v4$ =
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Annulus:
From the slot flow approximation,
dL
dp
2
)rr(
dL
dp
2
h f"2fw
==
But, Eq. 4.60 c
2
"2
_
)(
"2
rr
v
dL
dp f
=
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Shear Rate in Annulus
"2
_
2
"2
_
"2 6
)(
"2
2
)(
rr
v
rr
vrrw
=
=
"2
_
"2
_
w
rr
v6
rr
v6"
=
==
In field units:(annulus)
"2
_
dd
v"44
=
Where, ft/secinisv_
inchesinaredandd 2"
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Power - Law: Example 4.24
A cement slurry has a flow behavior index of
0.3 and a consistency index of 9,400 eq. cp.
The slurry is being pumped in an 8.097 * 4.5 -inch annulus at 200 gal/min.
(i) Assuming the flow pattern is laminar,
compute the frictional pressure
loss per 1,000 ft of annulus.
(ii) What is the shear rate at the wall?
n = 0.3
K = 9,400
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Example 4.24
)d2.448(d
qvvel.,Avg.(i)
2
"
2
2
_
=
s/ft803."v_
=
( )22_
5.4097.8448.2
200
v =
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Example 4.24
n
n"
"2
_n
f
0.0208n
"2
)d"44,000(d
vK
dL
dp,Press.Drop
+
= +
0.3
".3
0.3
f
0.0208
0.3"2
4.5)097"44,000(8.
3)9,400(".80
dL
dp
+
=
ftpsi/1,00077.9psi/ft0779.0
dL
dp========
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Example 4.24 contd
(ii) Shear rate at pipe wall,
( )
+
=
n
"2
dd
v48%
"2
_
w
+
=
0.3
"2
4.58.097
".803*48% w
"
w s"28%
= = 75 RPM
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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 drillcollars and the hole wall
Pressure loss in the annulus between the drillpipe and the hole wall
Hydrostatic pressure difference ( varies)
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Total Pump Pressure
)#P(PPPPPPP HYDDPADCABDCDPSCPUMP ++++++=
PUMP
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Types of Flow
Laminar Flow
Flow pattern is linear (no radial flow)
Velocity at wall is ZERO
Produces minimal hole erosion
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Types of Flow - Laminar
Mud properties strongly affect
pressure losses
Is preferred flow type for annulus
(in vertical wells)
Laminar flow is sometimes referred toas sheet flow, or layered flow:
* As the flow velocity increases, the flow type
changes from laminar to turbulent.
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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 cleaningbut
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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, contd
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
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Turbulent Flow - Newtonian Fluid
The onset of turbulence in pipe flow is
characterized by the dimensionlessgroup known as the Reynolds number
dvN
_
Re =
dv&928N
_
Re =
In field units,
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Turbulent Flow -
Newtonian Fluid
We often assume that fluid flow is
turbulent if Nre > 2,100
cp.fluid,ofviscosity
inI.D.,piped
ft/svelocity,fluidavg.v
lbm/galdensity,fluid&where
_
=
=
=
=
dv&928N
_
Re =