TurbulentFlow__.ppt

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Drilling Engineering

Prepared by: Tan Nguyen

Drilling Engineering PE 311

Turbulent Flow in Pipes and Annuli


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When attempting to quantify the pressure losses in side the drillstring and in the annulus it is

worth considering the following matrix:

Frictional Pressure Drop in Pipes and Annuli


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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.

Turbulent Flow in Pipes Newtonian Fluids

Introduction


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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:

Definition of Reynolds Number

du_

Re

du_

928Re



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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

Determination of Laminar/Turbulent Flow

cp.fluid,ofviscosity

inI.D.,pipe

ft/svelocity,fluidavg.

lbm/galdensity,fluidwhere

928Re

d

u

du_



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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.

Determination of Friction Factor - Laminar Flow



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For turbulent flow, the friction factor can be calculated by using Colebrook correlation.

Where eis 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 = 2,100100,000:

Blasius approximation:

Re = 2,100100,000:

fNd

f Re

255.1/269.0log4

1e

Determination of Friction Factor - Turbulent Flow

395.0log41 Re fNf

25.0

Re

0791.0

Nf



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Determination of Friction Factor



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Determination of Friction Factor



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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.

Combining this equation and the Blasius approximation gives

Note that the Moody friction factor is four times higher than the Fanning friction factor.

Determination of the Frictional Pressure Loss

25.0

2

928

8.250791.0

dv

d

u

dL

dp

75.4

25.075.175.0

25.0

25.075.175.0

86241800 d

q

d

u

dL

dp



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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:

Reynolds number:

Since Re > 2,100, the flow is under turbulent flow conditions.

Example

sftd

qu /16.11

826.3*488.2

400

488.2 22

1783120

826.3*16.11*9*928928Re

du_



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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

Solve this equation for the Fanning friction factor: f = 0.00666

Thus the frictional pressure loss can be obtained by

Example

fNd

f re

255.1/269.0log4

1e

psiDdvfD

dLdppf 75610000*

826.3*8.2516.11*9*00666.0

8.25

22



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Using Blasius approximation ,the equation becomes

Pressure drop: DP = dp/dL x D = (0.0777)(10,000) = 777 psi

Example

25.0

Re

0791.0

Nf

25.0

2

928

8.250791.0

dv

d

v

dL

dp

f tpsid

v

dL

dp/0777.0

1800 25.0

25.075.175.0



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Using Blasius approximation ,the equation becomes

Pressure drop: DP = dp/dL x D = (0.0777)(10,000) = 777 psi

Example

25.0

Re

0791.0

Nf

25.0

2

928

8.250791.0

dv

d

v

dL

dp

f tpsid

v

dL

dp/0777.0

1800 25.0

25.075.175.0



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Hydraulic diameter is defined as:

Equivalent diameter by using hydraulic diameter method:

Equivalent Diameter for Annular Geometry Hydraulic Diameter Method

Turbulent Flow in Pipes/Annuli NonNewtonian Fluids


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From the momentum equation, frictional pressure drop for Newtonian fluid in the annulus is

For pipe flow, d1--> 0 then

Comparing these two equations, the equivalent diameter an annulus can be obtained

Equivalent Diameter for Annular Geometry From Momentum Equation

Turbulent Flow in Pipes/Annuli nonNewtonian Fluids

1

2

2

1

2

22

1

2

2

ln500,1

dd

dddd

u

dL

dpf

2500,1 d

u

dL

dpf

1

2ln

2

1

2

22

1

2

2

dde

ddddd


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From the narrow slot approximation, frictional pressure drop for Newtonian fluid in the annulus is

For pipe flow

Comparing these two equations, the equivalent diameter an annulus can be obtained

Equivalent Diameter for Annular Geometry Narrow Slot Approximation


2500,1 d

u

dL

dpf

12816.0 ddde


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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.

Equivalent Diameter for Annular Geometry Crittendon Correlation



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1. Obtain apparent viscosity by combining the frictional pressure loss in

pipe (or annulus) for both Newtonian and Bingham Plastic models

2. Use apparent viscosity to determine Reynolds Number

Bingham Plastic Model


dd

v

d

v ypa

22515001500 22

)(66.6

Pipev

dypa

)()(5 12Annulus

v

ddypa

or


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Another way to determine the flow regime (critical Reynolds number) is to use the Hedstrom number

In field unit



2

2

p

y

HE

dN

2

237100

p

y

HE

dN


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Turbulent


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Reynolds number for power law fluids in pipe:

Reynolds number for power law fluids In annulus:

Friction factor for power law fluids under turbulent flow conditions

Power Law Model


nn

n

d

K

vN

/13

0416.089100)2(

Re

nn

n

dd

K

vN

/12

)(0208.0109000 12)2(

Re

2.1

2/1

Re75.0

395.0)log(

0.4/1 nfNnf

n


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Power Law Model


D illi E i i


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Summary

D illi E i i


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Summary

Newtonian Model Bingham Plastic Model Power Law Model

D illi E i i


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Summary


D illi E i i


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Summary




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ExampleNewtonian 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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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.

Hydraulic Method

Momentum Mothod

Narrow Slot Method

Crittendon Method



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ExampleBHF - 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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Equivalent diameter using narrow slot approximation

Reynolds number based on apparent viscosity



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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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Using Blasius approximation with Re = 3,154, the friction factor is f = 0.0098

Frictional pressure loss is given:



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g g g


ExamplePL - 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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g g g


Power Law Model



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TurbulentFlow__.ppt