Drilling Engineering - PE 311 Laminar Flow in Pipes and Annuli Newtonian Fluids

Drilling Engineering

Prepared by: Tan Nguyen

Drilling Engineering - PE 311

Laminar Flow in Pipes and Annuli

Newtonian Fluids



Under flowing conditions

In the annulus: Pwf = Pf(a) + gTVD (1)

In the drillpipe: Pp – Pwf = Pf(dp) + Pb – gTVD

Pwf = Pp - Pf(dp) – Pb + gTVD (2)

From (1) and (2) give

Pp = Pf(dp) + Pf(a) + Pb

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:

Frictional Pressure Drop in Pipes and Annuli



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

gPFdtVd



Force Analysis

dLdpr

drrd

)(



Force Analysis

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

dLdpr

drrd

)(



Force Analysis

dLdpr

drrd

)(

drdLdprrd )(

rC

dLdpr 1

2



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

For Newtonian fluids dudr

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

Pipe Flow – Newtonian Fluids

rC

dLdpr f 1

2

dLdpr

2




RR

rdruudAQ00

2

R

rdrRrdLdPQ

0

22 241

R

rdrRrdLdPQ

0

22

42

2442 44 RR

dLdPQ

4

8R

dLdPQ

24

8RuR

dLdPQ

2

81 R

dLdPu



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 du

dLdP



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


dLdpr

2

2

21 u

f w



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




Combining these two equations and

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.


dLdpr

2




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




Velocity:

Reynolds number:

NRe = 17725 : The fluid is under turbulent flow.




With turbulent flow and assuming smooth pipe, the Fanning friction factor:

Wrong calculation !!!!!!! Laminar flow !!!!!!!

00685.017725

0791.0Re

0791.025.025.0 f

ftpsigd

ufdLdP /0496.0

276.48.25937.81000685.0

8.25

22

ftpsid

udLdP /102.8

276.4500,19.820

500,15

22




If Q = 40 GPM, the fluid velocity is: u = 0.89 ft/s

The Reynolds number: Re = 1,772 Laminar flow

Frictional pressure loss:

f = 16 / Re = 0.009

ftpsid

udLdP /00065.0

276.4500,189.020

500,1 22

ftpsigd

ufdLdP /00065.0

276.48.2589.010009.0

8.25

22



From the general equation for fluid flow:

For Newtonian fluids:

Annular Flow – Newtonian Fluids

rC

dLdpr 1

2

rC

dLdpr

drdv 1

2

rdr

CdLdprdv

rdr

CdLdprdv

rC

dLdpr

drdv

)2

(

)2

(

2

1

1

1

21

2

ln4

CrC

dLdprv f



We need two boundary conditions to find C1 and C2

B.C. 1: r = r1 --> u = 0

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


21

2

ln4

CrC

dLdprv f

1

2

22

12

222

2ln

ln

41

rr

rr

rrrrdLdp

u f

r1

r2



Flow rate can be calculated as:

Average velocity can be expressed as

Pressure drop:


drrvQ )2(

rdr

rr

rr

rrrrdLdp

drrvQ f

2ln

ln

41)2(

1

2

22

12

222

2

1

2

221

224

14

2ln8

rr

rrrrdLdp

Q f

urrQ 21

22

1

2

21

222

122

1

2

21

222

12

2ln

500,1ln

8

dd

dddd

u

rr

rrrr

udLdp f

SI Unit Field Unit



Field unit:

Note: equivalent diameter for an annular section: de= 0.816 (d2 – d1)

Summary - 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 D = 0.052 x 9 x 10,000 = 4,680 psig

Average velocity:




Frictional pressure loss gradient

Or:

Circulating bottom hole pressure:

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


ftpsig

dd

dddd

udLdp f /0051.0

57ln

5757500,1

)362.1(15

ln500,1

2222

1

2

21

222

122



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

)rr πW )r (r h rr πWh

12

12

21

22

( slot ofh Widt slot ofHeight slot equivalent of Area





P1Wy - P2Wy + yWL - y+yW L = 0

1CydLdp

P1 P2

y

y + y





For Newtonian fluids:

Boundary conditions: y = 0 --> u = 0 and y = h --> v = 0

1CydLdp

dydv

212

21 CyCy

dLdpv

dLdphC

C

2

0

1

2

hyydLdpy

dLdphy

dLdpv 22

21

221





Flow rate:

dLdpWhWdyhyy

dLdpvWdyvdAq

hhh

1221 3

0

2

00

122

12

2 rr h and)r(rπWh

212

21

2212

)r)(rr(rdLdp

μπq f





Average velocity:

Frictional pressure losses gradient

Field unit

vrrvAq 21

22

dLdp)r(r

rr

)r)(rr(rdLdp

μπ

rrq

Aqv f

f

1212 2

122

12

2

212

21

22

21

22

212

12)r(rvμ

dLdp

_

f

2121000 )d(d

vμdLdp

_

f





Determination of shear rate:

Field unit:

2

12

1212

22 )r(rvμrr

dLdph

_

w

1212

126dd

vrr

vww

12

144ddvw

w





Field Unit:

12

12dd

vw

Dv

w8

212

12)r(rvμ

dLdp

_

f

2

8R

vμdLdp

_

f





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’’





Velocity of fluid in the drillpipe:

Velocity of fluid in the annulus:

Pressure drop in the drillpipe:

Pressure drop in the annulus:

sftd

Qudp

dp /79.1276.4448.2

80448.2 22

sftdd

Quann /36.157448.2

80448.2 222

122

ftpsigdu

dLdP

dp

dp

dp

/1079.9276.4150079.115

15004

22

ftpsig

ddu

dLdP ann

ann

/0051.0571000

36.1151000 22

122





Drill bit area:

Pressure drop at the bit:

Total pressure drop:

Shear rate at the wall: sddv

w /19857

)362.1(144144

12

222

39.032/134

34

3 ind

A bt

psigAC

qPtd

b 87.3439.095.0

80910311.810311.822

25

22

25

bitanndpTotal

PdLdP

dLdP

dLdP

psigdLdP

Total

66.9587.3478.6087.34000,100051.01079.9 4

Drilling Engineering - PE 311 Laminar Flow in Pipes and Annuli Newtonian Fluids

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