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PETE 411
Well Drilling
Lesson 12
Laminar Flow - Slot Flow
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Lesson 12 - Laminar Flow - Slot Flow
The Slot Flow Approximation Shear Rate Determination Pressure Drop Calculations
Laminar Flow Turbulent Flow Transition Flow - Critical Velocity
3
Read:Applied Drilling Engineering
Ch.4 to p. 145
Homework #6On the Web
Due Friday, October 4, 2002
4
Representing the Circular Annulus as a Slot
)r(r π W slot of Width
)r(r h slot ofHeight
rr π Wh slot equivalent of Area
12
12
21
22
{ slot approximation is OK if (d1/d2 > 0.3 }
Equal Area and
Height
Simpler Equations
-yet accurate
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Free body diagram for fluid element in a narrow slot
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yWΔLdL
dpp y WpF
ypWF
f22
1
Representing the Annulus as a Slot
F4 = y + yWL = + ddy
y WL
F3 = WL
Consider:- pressure forces- viscous forces
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Representing the Annulus as a Slot
state,steady At
F = maSumming forces along flow:
F = 0
F1
– F2 + F3 – F4 = 0
0LWΔydy
dττ -LWτyWΔL
dL
dp-p -y pW f
,gSimplifyin dp f
dL– d
dy= 0
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Representing the Annulus as a Slot
dy
dvγ
:integrate and variablesSeparate
Model, FluidNewtonian With
dp f
dL– d
dy= 0
dL
dpy 0
f ττ Evaluate 0 at wall where y = 0
γμτ
But,
of
dL
dpy
dy
dvτμτ
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Representing the Annulus as a Slot
dyτy
dL
dp-dvμ 0
f
00f
2
vμ
yτ
dL
dp
2μ
yv
of τ
dL
dpy
dy
dvμ
0 v0,y when 0 vSince 0
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Representing the Annulus as a Slot
μ
hτ-
dL
dp
2μ
h-0 h,y when 0 vSince 0f
2
dL
dp
2
h-τ f
0
2f yhydL
dp
2μ
1v
Hence, substituting for v0 and 0 :
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Representing the Annulus as a Slot
vWdyvdAq
The total flow rate:
h
0
2f dy yhydL
dp
2μ
Wq
dL
dp
12
Whq f
3
g,Integratin
122
12
2 rrh and )r(rπWh But
2f yhydL
dp
2μ
1v
12
Representing the Annulus as a Slot
212
21
22
f )r)(rr(rdL
dp
12μ
πq
)r(rπ
q
A
qv 2
12
2 velocity,averageBut
212
_
f
)r(r
vμ12
dL
dp
In field units,2
12
_
f
)d1000(d
vμ
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 flow representation 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
2221
22
_
ft/s 1.362v_
212
_
f
dd1000
vμ
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%
51.0750 psi 51
)57(1000
)000,10()362.1()15(D
dL
dpΔp
2f
fp
212
_
f
dd1000
vμ
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 is
non-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
rτ fw
w from (Eq. 4.51)
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Determination of Shear Rate
From Eq. 4.54b,
(at the wall)
w
_
w
2w
_
ww
2w
_
f
r
v4μτ
r
v8μ*
2
rτ
r
v8μ
dL
dp
dL
dp
2
rτ fw
w
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Determination of Shear Rate (why?)
Using the Newtonian Model,
Changing to field units,
w
_
w
_
w
r
v4
r
v4μ*
μ
1
μ
τγ
d
v96γ
_
(circular pipe)
sec-1, ft/sec, in
w
_
w r
v4μτ
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Annulus:
From the slot flow approximation,
dL
dp
2
)rr(
dL
dp
2
h f12fw
But, Eq. 4.60 c
212
_
)(
12
rr
v
dL
dp f
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Shear Rate in Annulus
12
_
212
_
12 6
)(
12
2
)(
rr
v
rr
vrrw
12
_
12
_
w
rr
v6
rr
v61
In field units: (annulus)
12
_
dd
v144
Where, ft/secin is v_
inchesin are d and d 21
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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 = 0.3K = 9,400
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Example 4.24
)d2.448(d
qv vel.,Avg. (i) 2
12
2
_
s/ft 803.1v_
22
_
5.4097.8448.2
200v
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Example 4.24
n
n112
_n
f
0.0208n1
2
)d144,000(d
vK
dL
dp ,Press.Drop
0.3
1.3
0.3f
0.02080.31
2
4.5)097144,000(8.
3)9,400(1.80
dL
dp
ftpsi/1,000 77.9psi/ft 0779.0dL
dp
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Example 4.24 cont’d
(ii) Shear rate at pipe wall,
n
12
dd
v48γ
12
_
w
0.3
12
4.58.097
1.803*48γw
1w s128γ
= 75 RPM
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Total Pump Pressure
Pressure loss in surf. equipmentPressure loss in drill pipePressure loss in drill collarsPressure drop across the bit nozzlesPressure loss in the annulus between the drill
collars and the hole wallPressure loss in the annulus between the drill
pipe and the hole wallHydrostatic pressure difference ( varies)
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Total Pump Pressure
)ΔP(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
29
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.
30
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…
31
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
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Turbulent Flow - Newtonian Fluid
The onset of turbulence in pipe flow is characterized by the dimensionless group known as the Reynolds number
dv
N
_
Re
μ
dvρ928N
_
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 μ
in I.D., piped
ft/s velocity,fluid avg. v
lbm/gal density, fluid ρ where_
μ
dvρ928N
_
Re
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