Drilling Engineering
Drilling Engineering – PE 311
Turbulent Flow in Pipes and Annuli
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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
Turbulent Flow in Pipes – Newtonian Fluids
Introduction
Laminar Flow: In this type of flow, layers of fluid move in streamlines. There is no microscopic or
macroscopic intermixing of the layers. Laminar flow systems are generally represented
graphically by streamlines.
Turbulent Flow: In turbulent flow, there is an irregular random movement of fluid in transverse
direction to the main flow. This irregular, fluctuating motion can be regarded as superimposed on
the mean motion of the fluid.
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Drilling Engineering
Turbulent Flow in Pipes – Newtonian Fluids
Definition of Reynolds Number
Reynolds number, Re, is a dimensionless number that gives a measure of the ratio of inertial
forces to viscous forces. Reynolds number is used to characterize different flow regimes, such as
laminar or turbulent flow. Laminar occurs at low Reynolds number, where viscous forces are
dominant, and is characterized by smooth, constant fluid motion; turbulent flow occurs at high
Reynolds number and is dominated by inertial forces, which tend to produce chaotic eddies,
vortices and other flow instabilities.
_
For pipe
In field unit:
_
Re 
928 ρ u d
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μ
Re 
ud

Drilling Engineering
Turbulent Flow in Pipes – Newtonian Fluids
Determination of Laminar/Turbulent Flow
_
Re
928 ρ u d
μ
where ρ  fluid densit y,lbm/gal
u  avg. fluid velocit y,ft /s
d  pipe I.D.,in
μ  viscosit yof fluid, cp.
If
Re < 2,100
Laminar flow
Re = 2,100 – 4,000
Transition flow
Re > 4,000
Turbulent
Note that this critical Reynolds number is correct only for Newtonian fluids
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Drilling Engineering
Turbulent Flow in Pipes – Newtonian Fluids
Determination of Friction Factor - Laminar Flow
Relationship between shear stress and friction factor:
Pipe flow under laminar conditions:
Therefore,
Newtonian fluids flow in pipe under laminar flow conditions:
Hence,
This equation will be used to calculate the friction factor of Newtonian fluids flow in pipe under
laminar flow conditions.
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Drilling Engineering
Turbulent Flow in Pipes – Newtonian Fluids
Determination of Friction Factor - Turbulent Flow
For turbulent flow, the friction factor can be calculated by using Colebrook correlation.

1
1.255
 4 log 0.269e / d 

f
N Re f





Where e is the absolute roughness. e/d is the relative roughness.
For smooth pipe, the relative roughness e/d < 0.0004, the following equations can be used to
calculate the friction factor in turbulent flow

Re = 2,100 – 100,000:
1
 4 log N Re
f
Blasius approximation:
Re = 2,100 – 100,000:
f 
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0.0791
0.25
N Re

f  0.395
Drilling Engineering
Turbulent Flow in Pipes – Newtonian Fluids
Determination of Friction Factor
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Drilling Engineering
Turbulent Flow in Pipes – Newtonian Fluids
Determination of Friction Factor
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Drilling Engineering
Turbulent Flow in Pipes – Newtonian Fluids
Determination of the Frictional Pressure Loss
From equation:
, an equation of dp/dL can be expressed as
In field unit:
. This equation can be used to calculate the frictional pressure drop
gradient for Newtonian and non-Newtonian fluids.
u 2
0.0791
dp
25.8d

Combining this equation and the Blasius approximation gives
dL  928 vd 0.25


  


dp  0.75u1.75  0.25  0.75 q1.75  0.25
dL

1800d
0.25

8624d 4.75
Note that the Moody friction factor is four times higher than the Fanning friction factor.
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Drilling Engineering
Turbulent Flow in Pipes – Newtonian Fluids
Example
Example: Determine the frictional pressure drop in 10000ft of 4.5-in commercial steel drillpipe
having an internal diameter of 3.826in. If a 20 cp Newtonian fluid having a density of 9 lbm/gal is
pumped through the drillpipe at a rate of 400 gal/min
Solution:
Mean velocity:
u
q
400

 11.16 ft / s
2.488d 2 2.488* 3.8262
_
Reynolds number:
Re 
928 ρ u d
μ

928* 9 *11.16* 3.826
 17831
20
Since Re > 2,100, the flow is under turbulent flow conditions.
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Drilling Engineering
Turbulent Flow in Pipes – Newtonian Fluids
Example
From table 4.5, the absolute roughness for commercial steel pipe is e = 0.000013 inches.
The relative roughness e/d = 0.000013/3.826 = 0.0000034 < 0.0004 --> smooth pipe

1
1.255
 4 log 0.269e / d 

f
N re f





Solve this equation for the Fanning friction factor: f = 0.00666
Thus the frictional pressure loss can be obtained by
2
dp
f v
0.00666* 9 *11.162
p f 
D
D
*10000 756psi
dL
25.8d
25.8 * 3.826
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Drilling Engineering
Turbulent Flow in Pipes – Newtonian Fluids
Example
Using Blasius approximation
v
f 
0.0791
0.25 ,the equation
N Re
2
0.0791
dp
25.8d

dL  928 vd  0.25


  


1.75
dp  0.75 v  0.25

 0.0777 psi / ft
0.25
dL
1800d
Pressure drop: DP = dp/dL x D = (0.0777)(10,000) = 777 psi
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becomes
Drilling Engineering
Turbulent Flow in Pipes – Newtonian Fluids
Example
Using Blasius approximation
v
f 
0.0791
0.25 ,the equation
N Re
2
0.0791
dp
25.8d

dL  928 vd  0.25


  


1.75
dp  0.75 v  0.25

 0.0777 psi / ft
0.25
dL
1800d
Pressure drop: DP = dp/dL x D = (0.0777)(10,000) = 777 psi
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becomes
Drilling Engineering
Turbulent Flow in Pipes/Annuli – NonNewtonian Fluids
Equivalent Diameter for Annular Geometry – Hydraulic Diameter Method
Hydraulic diameter is defined as:
Equivalent diameter by using hydraulic diameter method:
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Drilling Engineering
Turbulent Flow in Pipes/Annuli – nonNewtonian Fluids
Equivalent Diameter for Annular Geometry – From Momentum Equation
From the momentum equation, frictional pressure drop for Newtonian fluid in the annulus is
dp f
dL

u


2
2
d  d1 

1,500d 22  d12  2
d2 
ln

d1 

For pipe flow, d1 --> 0 then
dp f
dL

u
1,500d 2
Comparing these two equations, the equivalent diameter an annulus can be obtained
d 22  d12
de  d  d 
lnd 2 d1 
2
2
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2
1
Drilling Engineering
Turbulent Flow in Pipes/Annuli – nonNewtonian Fluids
Equivalent Diameter for Annular Geometry – Narrow Slot Approximation
From the narrow slot approximation, frictional pressure drop for Newtonian fluid in the annulus is
For pipe flow
dp f
dL

u
1,500d 2
Comparing these two equations, the equivalent diameter an annulus can be obtained
d e  0.816d 2  d1 
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Drilling Engineering
Turbulent Flow in Pipes/Annuli – nonNewtonian Fluids
Equivalent Diameter for Annular Geometry – Crittendon Correlation
A fourth expression for the equivalent diameter of an annulus was developed by Crittendon.
When using Crittendon correlation, a fictitious average velocity also must be used in describing
the flow system.
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Drilling Engineering
Turbulent Flow in Pipes/Annuli – nonNewtonian Fluids
Bingham Plastic Model
1. Obtain apparent viscosity by combining the frictional pressure loss in
pipe (or annulus) for both Newtonian and Bingham Plastic models
or
a   p 
6.66 y d
a   p 
v
a v
1500d 2

( Pipe)
5 y (d 2  d1 )
v
( Annulus)
2. Use apparent viscosity to determine Reynolds Number
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pv
1500d 2

y
225d
Drilling Engineering
Turbulent Flow in Pipes/Annuli – nonNewtonian Fluids
Bingham Plastic Model
Another way to determine the flow regime (critical Reynolds number) is to use the Hedstrom number
N HE 
yd 2
 p2
In field unit
N HE 
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37100  y d 2
 p2
Drilling Engineering
Turbulent Flow in Pipes/Annuli – nonNewtonian Fluids
Bingham Plastic Model
Turbulent
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Drilling Engineering
Turbulent Flow in Pipes/Annuli – nonNewtonian Fluids
Power Law Model
Apparent Viscosity for use in the Reynolds Number is obtained by comparing the laminar flow
equations for Newtonian and Power Law fluids
Pipe flow:
a v
n
Kv
 3 1/ n 



1500d 2 144000d (1 n )  0.0416
Kd (1n )  3  1 / n 
a 

(1 n ) 
0
.
0416


96v
n
n
( Pipe)
Annular flow:
a v
n
Kv
 2 1/ n 



1000 (d 2  d1 ) 2 144000 (d 2  d1 ) (1 n )  0.0208 
K (d 2  d1 ) (1n )  2  1 / n 
a 


(1 n )
0
.
0208


144v
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n
( Annulus)
n
Drilling Engineering
Turbulent Flow in Pipes/Annuli – nonNewtonian Fluids
Power Law Model
Reynolds number for power law fluids in pipe:
N Re
89100 v

K
( 2 n )
 0.0416d 


 3 1/ n 
n
Reynolds number for power law fluids In annulus:
N Re
109000 v

K
( 2 n )
 0.0208(d 2  d1 ) 


2  1/ n


n
Friction factor for power law fluids under turbulent flow conditions
1/ f 
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4.0
0.395
1 n / 2
log(
N
f
)

Re
n 0.75
n1.2
Drilling Engineering
Turbulent Flow in Pipes/Annuli – nonNewtonian Fluids
Power Law Model
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Drilling Engineering
Summary
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Summary
Newtonian Model
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Bingham Plastic Model
Power Law Model
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Summary
Newtonian Model
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Bingham Plastic Model
Power Law Model
Drilling Engineering
Summary
Newtonian Model
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Bingham Plastic Model
Power Law Model
Drilling Engineering
Example – Newtonian Fluid in Annulus
Example 1: A 9.0 lbm/gal brine having a viscosity of 1.0 cp is being circulated in a well at a rate
of 200 gal/min. Apply the all the criteria for computing equivalent diameter. Determine the flow
pattern and frictional pressure gradient. The drillpipe has an external diameter of 5.0 in. and the
hole has a diameter of 10 in.
Solution:
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Drilling Engineering
Example – Newtonian Fluid in Annulus
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Drilling Engineering
Example – Newtonian Fluid in Annulus
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Drilling Engineering
Example – Newtonian Fluid in Annulus
Hydraulic Method
Momentum Mothod
Narrow Slot Method
Crittendon Method
Note that the Crittendon correlation is applied for the fourth method. In this
method, we need to calculate the equivalent diameter based on Crittendon
correlation and the fictitious average velocity.
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Drilling Engineering
Example – BHF - Annulus
Example 2: A 10 lbm/gal mud having a plastic viscosity of 40 cp and a yield point of 15 lbf/100ft2
is circulated at a rate of 600 gal/min. Estimate the frictional pressure loss in the annulus opposite
the drill collars if the drill collars are in a 6.5-in hole, have a length of 1,000 ft, and a 4.5 in. OD.
Check for turbulence using both the apparent viscosity concept and the Hedstrom number
approach. Use an narrow slot equivalent diameter to represent the annular geometry.
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Drilling Engineering
Example – BHF - Annulus
Equivalent diameter using narrow slot approximation
Reynolds number based on apparent viscosity
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Drilling Engineering
Example – BHF - Annulus
Reynolds number for a plastic viscosity of 40 cp
Using the graph for Hedstrom number, the critical Reynolds number is 3,300. The flow is turbulent
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Drilling Engineering
Example – BHF - Annulus
Using Blasius approximation with Re = 3,154, the friction factor is f = 0.0098
Frictional pressure loss is given:
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Drilling Engineering
Example – BHF - Annulus
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Example – PL - Annulus
Example 3: A 15.6 lbm/gal cement slurry having a consistency index of 335 eq cp and a flow
behavior index of 0.67 is being pumped at a rate of 672 gal/min between a 9.625-in. hole and a
7.0-in.casing. Determine the frictional pressure loss per 100 ft of slurry. Use the equivalent
diameter based on the narrow slot approximation.
Solution:
The mean velocity:
Reynolds number:
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Drilling Engineering
Turbulent Flow in Pipes/Annuli – nonNewtonian Fluids
Power Law Model
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Example– PL - Annulus
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