Drilling Engineering
Drilling Engineering - PE 311
Laminar Flow in Pipes and Annuli
Newtonian Fluids
Prepared by: Tan Nguyen
Drilling Engineering
Frictional Pressure Drop in Pipes and Annuli
Under flowing conditions
In the annulus: Pwf = DPf(a) + rgTVD
(1)
In the drillpipe: Pp – Pwf = DPf(dp) + DPb – rgTVD
Pwf = Pp - DPf(dp) – DPb + rgTVD
From (1) and (2) give
Pp = DPf(dp) + DPf(a) + DPb
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(2)
Drilling Engineering
Frictional Pressure Drop in Pipes and Annuli
When attempting to quantify the pressure losses in side the drillstring and in the annulus it is
worth considering the following matrix:
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Drilling Engineering
Momentum Equation
Assumptions:
1. The drillstring is placed concentrically in the casing or open hole
2. The drillstring is not being rotated
3. Sections of open hole are circular in shape and of known diameter
4. Incompressible drilling fluid
5. Isothermal flow
Momentum equation:



dV
r
  F  P  r g    
dt
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Drilling Engineering
Force Analysis
d ( r )
dp
r
dr
dL
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Drilling Engineering
Force Analysis
p1 (2rDr )  p2 (2rDr )  (r  Dr ) r Dr [2DL]  r r (2DL)
d ( r )
dp
r
dr
dL
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Drilling Engineering
Force Analysis
d ( r )
dp
r
dr
dL
 d ( r )   r

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dp
dr
dL
r dp C1

2 dL r
Drilling Engineering
Pipe Flow – Newtonian Fluids
r dp f C1


2 dL
r
B.C. r = 0 -->  = 0: then C1 = 0

r dp
2 dL
For Newtonian fluids   mg  m du/dr
B.C. r = R --> u = 0: then
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Drilling Engineering
Pipe Flow – Newtonian Fluids
R
R
0
0
Q   udA   u 2rdr
R
Q
0


1 dP 2
r  R 2 2rdr
4m dL
2 dP
Q
r 2  R 2 rdr

4m dL 0
R


2 dP  R 4 R 4 


Q

4m dL  4
2 
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Q
 dP 4
R
8m dL
Q
 dP 4
R  uR 2
8m dL
u
1 dP 2
R
8m dL
Drilling Engineering
Pipe Flow – Newtonian Fluids
Maximum velocity:
Average fluid velocity in pipe u = umax / 2
From this equation, the pressure drop can be expressed as:
In field unit:
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dP
mu

dL 1,500d 2
Drilling Engineering
Pipe Flow – Newtonian Fluids
From equation:

r dp
2 dL
Combining with the definition of Fanning friction factor:
f 
w
1 2
ru
2
Pressure drop can be calculated by using Fanning friction factor:
Field unit:
This equation can be used to calculate pressure under laminar or turbulent conditions
Prepared by: Tan Nguyen
Drilling Engineering
Pipe Flow – Newtonian Fluids
From this equation, the pressure drop can be expressed as:
Combining this equation and equation
gives
where
This equation is used to calculate the Fanning friction factor when the flow is laminar. If Re <
2,100 then the flow is under laminar conditions
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Drilling Engineering
Pipe Flow – Newtonian Fluids
Prepared by: Tan Nguyen
Drilling Engineering
Pipe Flow – Newtonian Fluids
Combining these two equations  
r dp
and
2 dL
The relationship between shear stress and shear rate can be express as
Where
is called the nominal Newtonian shear rate. For Newtonian fluid, we can used
this equation to calculate the shear rate of fluid as a function of velocity.
Prepared by: Tan Nguyen
Drilling Engineering
Pipe Flow – Newtonian Fluids
Determine whether a fluid with a viscosity of 20 cp and a density of 10 ppg
flowing in a 5" 19.5 lb/ft (I.D. = 4.276") drillpipe at 400 gpm is in laminar or
turbulent flow. What is the maximum flowrate to ensure that the fluid is in laminar
flow ? Calculate the frictional pressure loss in the drillpipe in two cases:
Q = 400 GPM
Q = 40 GPM
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Drilling Engineering
Pipe Flow – Newtonian Fluids
Velocity:
Reynolds number:
NRe = 17725 : The fluid is under turbulent flow.
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Drilling Engineering
Pipe Flow – Newtonian Fluids
Prepared by: Tan Nguyen
Drilling Engineering
Pipe Flow – Newtonian Fluids
With turbulent flow and assuming smooth pipe, the Fanning friction factor:
f 
0.0791
0.0791

 0.00685
0.25
0.25
Re
17725
dP
fru 2 0.00685  10  8.937 2


 0.0496 psig / ft
dL 25.8d
25.8  4.276
Wrong calculation !!!!!!! Laminar flow !!!!!!!
dP
mu
20  8.9
5



8
.
2

10
psi / ft
2
2
dL 1,500d
1,500  4.276
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Drilling Engineering
Pipe Flow – Newtonian Fluids
If Q = 40 GPM, the fluid velocity is: u = 0.89 ft/s
The Reynolds number: Re = 1,772  Laminar flow
Frictional pressure loss:
dP
mu
20  0.89


 0.00065 psi / ft
2
2
dL 1,500d
1,500  4.276
f = 16 / Re = 0.009
dP
fru 2 0.009  10  0.89 2


 0.00065 psig / ft
dL 25.8d
25.8  4.276
Prepared by: Tan Nguyen
Drilling Engineering
Annular Flow – Newtonian Fluids
From the general equation for fluid flow:  
For Newtonian fluids:
  mg   m
dv r dp C1


dr 2 dL r
dv r dp C1


dr 2 dL r
r dp C1
 dv  (
 )dr
2m dL mr
r dp C1

dv

(

 2m dL  mr )dr
m
r 2 dp f C1
v
 ln r  C2
4m dL
m
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r dp C1

2 dL r
Drilling Engineering
Annular Flow – Newtonian Fluids
r 2 dp f C1
v
 ln r  C2
4m dL
m
We need two boundary conditions to find C1 and C2
B.C. 1: r = r1 --> u = 0
r2
B.C. 2: r = r2 --> u = 0
r1

r2 
ln
dp
1
f  2
2
2
2
r 
u
r

r

r

r
2
2
1

4 m dL 
ln r2 

r1 


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 

Drilling Engineering
Annular Flow – Newtonian Fluids
Q   v(2r )dr
Flow rate can be calculated as:

r2 
ln
1 dp f  2 2
2
2
r  2rdr
Q   v(2r )dr  
r

r

r

r
2
2
1

r
4 m dL 
2
ln

r1 



2
2 2
dp
r2  r1  

f  4
4
Q
r

r

2
1

r2
8m dL 
ln

r1 


 

Average velocity can be expressed as


Q   r22  r12 u
SI Unit
Pressure drop:
dp f
dL
Prepared by: Tan Nguyen

8m u


2
2
 2 2 r2  r1 
r2  r1  r 
ln 2 

r1 

Field Unit

mu


2
2
d  d1 

1,500d 22  d12  2
d2 
ln

d1 

Drilling Engineering
Summary - Newtonian Fluids
Field unit:
Note: equivalent diameter for an annular section: de= 0.816 (d2 – d1)
Prepared by: Tan Nguyen
Drilling Engineering
Annular Flow – Newtonian Fluids
Example: A 9-lbm/gal Newtonian fluid having a viscosity of 15 cp is being circulated in a 10,000-ft
well containing a 7-in. ID casing and a 5-in OD drillsring at a rate of 80 gal/min. compute the static
and circulating bottomhole pressure by assuming that a laminar flow pattern exists.
Solution:
Static pressure: P = 0.052 r D = 0.052 x 9 x 10,000 = 4,680 psig
Average velocity:
Prepared by: Tan Nguyen
Drilling Engineering
Annular Flow – Newtonian Fluids
Frictional pressure loss gradient
dp f
dL

mu


2
2
d

d

1 
1,500d 22  d12  2

d
2
ln

d1 

Or:
Circulating bottom hole pressure:
P = 4,680 + 0.0051 x 10,000 = 4,731 Psig
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
15(1.362)

7 5
1,5007 2  52 

ln 7
5

2
2




 0.0051 psig / ft
Drilling Engineering
Annular Flow – Newtonian Fluids
Narrow Slot Approximation
Since the exact solution is so complicated, a narrow
rectangular slot approximation is used to arrive at
solutions still very useful for practical drilling
engineering applications. We represent the annulus
as a slot which has the same area and the same
height with the annulus. This approximation is good if
D1 / D2 > 0.3
An annular geometry can be represented by a
rectangular slot with the height h and width w as
given below

Area of equivalent slot  Wh  π r2  r1
2
2
Height of slot  h  (r2  r1 )

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Widt h of slot  W  π (r2  r1 )

Drilling Engineering
Annular Flow – Newtonian Fluids
Narrow Slot Approximation
P1
y + Dy
y
P1WDy - P2WDy + yWDL - y+DyW DL = 0
dp
 
y  C1
dL
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P2
Drilling Engineering
Annular Flow – Newtonian Fluids
Narrow Slot Approximation
For Newtonian fluids:
  mg   m
v
dv
dp

y  C1
dy
dL
1 dp 2 C1 y
y 
 C2
2m dL
m
Boundary conditions: y = 0 --> u = 0 and y = h --> v = 0
C2  0


h dp
C

1

2 dL
v
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
1 dp 2 h dp
1 dp 2
y 
y
y  hy
2m dL
2m dL
2m dL

Drilling Engineering
Annular Flow – Newtonian Fluids
Narrow Slot Approximation
Flow rate:
h
h
h


1 dp 2
Wh 3 dp
q   vdA   vWdy  
y  hy Wdy  
2
m
dL
12m dL
0
0
0
Wh  π (r2  r1 ) and h  r2  r1
2
2
π dp f 2
2
q
(r2  r1 )(r2  r1 )2
12 μ dL
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Drilling Engineering
Annular Flow – Newtonian Fluids
Narrow Slot Approximation
Average velocity:


q  Av   r22  r12 v
v
q
q


A  r22  r12



π dp f 2
2
(r2  r1 )(r2  r1 )2
(r2  r1 )2 dp f
12 μ dL

 r22  r12
12 m dL
Frictional pressure losses gradient
_
dp f
dL

12 μ v
(r2  r1 )2
Field unit
_
μv

dL
1000(d 2  d1 )2
dp f
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

Drilling Engineering
Annular Flow – Newtonian Fluids
Narrow Slot Approximation
Determination of shear rate:
_
h dp r2  r1  12 μ v
w  

2 dL
2 (r2  r1 )2
w
6v
12v
gw  

m r2  r1  d 2  d1 
Field unit:
gw 
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w
144v

m d 2  d1 
Drilling Engineering
Annular Flow – Newtonian Fluids
Narrow Slot Approximation
Field Unit:
_
dp f
dL

8μ v
R2
gw 
8v
D
_
dp f
dL
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
12 μ v
(r2  r1 )2
gw 
12v
d 2  d1 
Drilling Engineering
Annular Flow – Newtonian Fluids
Narrow Slot Approximation
Example: A 9-lbm/gal Newtonian fluid having a viscosity of 15 cp is being circulated in a
10,000-ft well containing a 7-in. ID casing and a 5-in OD drillsring (ID = 4.276’’) at a rate of 80
gal/min. Compute the frictional pressure loss and the shear rate at the wall in the drillpipe and
in the annulus by using narrow slot approximation method. Assume that the flow is laminar.
Also, calculate the pressure drop at the drill bit which has 3 nozzles: db = 13/32’’
Prepared by: Tan Nguyen
Drilling Engineering
Annular Flow – Newtonian Fluids
Narrow Slot Approximation
Q
80

 1.79 ft / s
2
2.448d dp
2.448  4.276 2
Velocity of fluid in the drillpipe:
u dp 
Velocity of fluid in the annulus:
u ann 
Q
80

 1.36 ft / s
2
2
2
2
2.448d 2  d1  2.448  7  5 
Pressure drop in the drillpipe:
dP
dL

Pressure drop in the annulus:
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dp
m  u dp
2
1500d dp

15  1.79
 9.79  10 4 psig / ft
2
1500  4.276
m  u ann
dP
15  1.36


 0.0051 psig / ft
dL ann 1000 d 22  d12 1000  7  52


Drilling Engineering
Annular Flow – Newtonian Fluids
Narrow Slot Approximation
Drill bit area:
At  3
d b2
4
3

4
13 / 322  0.39
in 2
8.311  10 5 rq 2 8.311  10 5  9  80 2

 34.87 psig
Pressure drop at the bit: DPb 
C d2 At2
0.95 2  0.39 2
Total pressure drop:
dP
dP

dL Total dL


dp
dP
dL
 DPbit
ann

dP
 9.79  10 4  0.0051 10,000  34.87  60.78  34.87  95.66 psig
dL Total
Shear rate at the wall:
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gw 
144v
144(1.362)

 98 1 / s
d 2  d1 
7  5