LaminarFlowPipes&Annuli Newtonian

8/11/2019 LaminarFlowPipes&Annuli Newtonian

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Drilling Engineering

Prepared by: Tan Nguyen

Drilling Engineering - PE 311

Laminar Flow in Pipes and Annuli

Newtonian Fluids


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Under flowing conditions

In the annulus: Pwf= DPf(a) + rgTVD (1)

In the drillpipe: PpPwf= DPf(dp) + DPbrgTVD

Pwf= Pp- DPf(dp) DPb+ rgTVD (2)

From (1) and (2) give

Pp= DPf(dp) + DPf(a) + DPb

Frictional Pressure Drop 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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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:

Momentum Equation

gPFdt

Vdrr


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Force Analysis

dL

dpr

dr

rd

)(


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Force Analysis

)2(]2[)()2()2( 21 LrLrrrrprrp rrr DDDDD D

dL

dpr

dr

rd

)(


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Force Analysis

dLdpr

drrd )(

drdLdp

rrd )(

r

C

dL

dpr 1

2


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B.C. r = 0 --> = 0: then C1= 0

For Newtonian fluids mg m du/dr

B.C. r = R --> u = 0: then

Pipe Flow Newtonian Fluids

r

C

dL

dpr f 1

2

dL

dpr

2


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RR

rdruudAQ00

2

R

rdrRrdL

dPQ

0

22 24

1

m

R

rdrRrdL

dPQ

0

22

4

2

m

2442

44

RRdLdPQ

m

4

8R

dL

dPQ

m

24

8

RuR

dL

dPQ

m

2

8

1R

dL

dPu

m


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Maximum velocity:

Average fluid velocity in pipe u = umax / 2

From this equation, the pressure drop can be expressed as:

In field unit:


2500,1 d

u

dL

dP m


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From equation:

Combining with the definition of Fanning friction factor:

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


dL

dpr

2

2

2

1u

f w

r


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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 u = 0

B.C. 2: r = r2 --> u = 0

Annular Flow Newtonian Fluids

2

1

2

ln4

CrCdLdprv

f

mm

1

2

2

2

1

2

2

22

2

ln

ln

4

1

r

rr

r

rrrrdL

dpu

f

m

r1

r2


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Flow rate can be calculated as:

Average velocity can be expressed as

Pressure drop:


drrvQ )2(

rdr

rrr

r

rrrrdL

dpdrrvQ f

m

2

ln

ln

4

1)2(

1

2

2

2

1

2

2

22

2

1

2

221

224

1

4

2

ln8r

rrrrr

dL

dpQ f

m

urrQ 2

1

2

2

1

2

2

1

2

22

1

2

2

1

2

2

1

2

22

1

2

2

ln500,1

ln

8

dd

dddd

u

rr

rrrr

u

dL

dpf mm

SI Unit Field Unit


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Field unit:

Note: equivalent diameter for an annular section: de= 0.816 (d2d1)

Summary - Newtonian Fluids


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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:


D illi E i i


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Frictional pressure loss gradient

Or:

Circulating bottom hole pressure:

P = 4,680 + 0.0051 x 10,000 = 4,731 Psig


ftpsig

dd

dddd

u

dL

dpf/0051.0

57ln

5757500,1

)362.1(15

ln500,1

22

22

1

2

2

1

2

22

1

2

2

m

D illi E i i


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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


Narrow Slot Approximation

)rrW

)r(rhrrWh

12

12

2

1

2

2

(slotofhWidt

slotofHeightslotequivalentofArea

D illi E i i


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P1WDy - P2WDy + yWDL - y+DyW DL = 0

1CydL

dp

P1 P2

y

y + Dy


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Flow rate:

dL

dpWhWdyhyy

dL

dpvWdyvdAq

hhh

mm 122

1 3

0

2

00

12

2

1

2

2 rrhand)r(rWh

2

12

2

1

2

2

12

)r)(rr(r

dL

dp

q

f



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Average velocity:

Frictional pressure losses gradient

Field unit

vrrvAq 2122

dLdp)r(r

rr

)r)(rr(rdL

dp

rr

q

A

qv

f

f

m 12

12 2

122

1

2

2

2

12

2

1

2

2

2

1

2

2

2

12

12

)r(r

v

dL

dp

_

f

2

121000 )d(d

v

dL

dp_

f



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Determination of shear rate:

Field unit:

2

12

12 12

22 )r(r

vrr

dL

dph_

w

1212126

ddv

rrvww

mg

12

144

dd

vww

m

g


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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



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Velocity of fluid in the drillpipe:

Velocity of fluid in the annulus:

Pressure drop in the drillpipe:

Pressure drop in the annulus:

sftd

Qu

dp

dp /79.1276.4448.2

80

448.2 22

sft

dd

Qu

ann

/36.157448.2

80

448.2 222122

ftpsigd

u

dL

dP

dp

dp

dp

/1079.9276.41500

79.115

1500

4

22

m

ftpsig

dd

u

dL

dP ann

ann

/0051.0571000

36.115

1000 22122

m



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Drill bit area:

Pressure drop at the bit:

Total pressure drop:

Shear rate at the wall:

sdd

vw /198

57

)362.1(144144

12

g

222

39.032/134

34

3 ind

A bt

psig

AC

qP

td

b 87.34

39.095.0

80910311.810311.822

25

22

25

D

r

bit

anndpTotal

PdL

dP

dL

dP

dL

dPD

psigdLdP

Total 66.9587.3478.6087.34000,100051.01079.9

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LaminarFlowPipes&Annuli Newtonian