12 Laminar Flow Slot Flow

8/12/2019 12 Laminar Flow Slot Flow

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PETE 411

Well Drilling

Lesson 12

Laminar Flow - Slot Flow


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Lesson 12 - Laminar Flow - Slot Flow

4The Slot Flow Approximation4Shear Rate Determination

4Pressure Drop Calculations

4Laminar Flow

4Turbulent Flow4Transition Flow - Critical Velocity


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Read:

Applied Drilling Engineering

Ch.4 to p. 145

Homework #6

On the WebDue Friday, October 4, 2002


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Representing the Circular Annulus as a Slot

)r(r!WslotofWidth

)r(rhslotofHeight

rr!WhslotequivalentofArea

"2

"2

2

"

2

2

+==

==

==

{ slot approximation is OK if (d1/d2 > 0.3 }

Equal

Area

and

Height

Simpler

Equations

-yetaccurate


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5Free body diagram for fluid element in a narrow slot


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yW#LdL

dppyWpF

ypWF

f

22

"

==

=

Representing the Annulus as a Slot

F4 = y +yWL = +ddyy WL

F3 =WL

Consider:

- pressure forces

- viscous forces


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state,steadyAt

F = m aSumming forces along flow:

F = 0

F

"

F2 + F3

F4 = 0

0LW#ydy

d$$-LW$yW#L

dL

dp-p-ypW f =

++

,gSimplifyin dp f

dLd

dy= 0


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dydv% =

:integrateandvariablesSeparate

Model,FluidNewtonianWith

dp fdL

ddy

= 0

dL

dpy

0

f $$ +=Evaluate 0 at wall where y = 0

= %$

But,

of

dL

dpy

dy

dv$$ +==


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+= dy$ydL

dp-dv 0

f

00f

2

v

y$

dL

dp

2

yv +=

of $

dL

dpy

dy

dv +=

0 v0,y when0vSince 0 ===


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h$-

dL

dp

2

h-0h,y when0 vSince 0f

2

===

dL

dp

2

h-$ f0 =

( )2f yhydL

dp

2

"v =

Hence, substituting for v0 and 0 :


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== vWdyvdAq

The total flow rate:

( )

=

h

0

2f dyyhydL

dp

2

Wq

dL

dp

"2

Whq f

3

=

g,Integratin

"2

2

"

2

2 rrhand)r(r!Wh ==But

( )2f yhydL

dp

2

"v =


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2

"2

2

"

2

2

f

)r)(rr(rdL

dp

"2

!

q =

)r(r!

q

A

qv

2"

22 ==velocity,averageBut

2

"2

_

f

)r(r

v"2

dL

dp

=

In field units,2

"2

_

f

)d"000(d

v

dL

dp

=

psi/ft, cp., ft/sec, in

dL

dp

12

Whq f

3

=


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Example 4.22

Compute the frictional pressure loss for a 7 x

5 annulus, 10,000 ft long, using the slot flowrepresentation in the annulus. The flow rate

is 80 gal/min. The viscosity is 15 cp.

Assume the flow pattern is laminar.

7 5 1

6


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Example 4.22

The average velocity in the

annulus,

)52.448(7

80

)d2.448(d

qv

222

"

2

2

_

=

=

ft/s".362v_

=

( )2"2

_

f

dd"000

v

dL

dp

=


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Example 4.22

A somewhat more accurate answer, using an

exact equation for a circular annulus, results

in a value of 50.9792 psi.

Difference = 0.0958 psi i.e., within 0.2%

( )5".0750psi5"

)57("000

)000,"0()362."()"5(

DdL

dp

#p 2f

==

==

fp

( )2

"2

_

f

dd"000

v

dL

dp

=


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Determination of Shear Rate...(why?)

If shear rate in well is known:

1. Fluid can be evaluated in viscometer at

the proper shear rate.

2. Newtonian equations can sometimes

give good accuracy even if fluid isnon-Newtonian.


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Determination of Shear Rate

The maximum value of shear rate will

occur at the pipe walls.

For circular pipe, at the pipe wall,

dL

dp

2

r

$

fw

w =from (Eq. 4.5")


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Determination of Shear Rate

From Eq. 4.54b,

(at the wall)

w

_

w

2

w

_

ww

2

w

_

f

r

v4$

r

v8*

2

r$

r

v8

dL

dp

=

=

=

dL

dp

2

r$ fww =


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Determination of Shear Rate (why?)

Using the Newtonian Model,

Changing to field units,

w

_

w

_

w

r

v4

r

v4*

"

$% ===

dv96%

_

=

(circular pipe)

sec-", ft/sec, in

w

_

wr

v4$ =


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Annulus:

From the slot flow approximation,

dL

dp

2

)rr(

dL

dp

2

h f"2fw

==

But, Eq. 4.60 c

2

"2

_

)(

"2

rr

v

dL

dp f

=


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Shear Rate in Annulus

"2

_

2

"2

_

"2 6

)(

"2

2

)(

rr

v

rr

vrrw

=

=

"2

_

"2

_

w

rr

v6

rr

v6"

=

==

In field units:(annulus)

"2

_

dd

v"44

=

Where, ft/secinisv_

inchesinaredandd 2"


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Power - Law: Example 4.24

A cement slurry has a flow behavior index of

0.3 and a consistency index of 9,400 eq. cp.

The slurry is being pumped in an 8.097 * 4.5 -inch annulus at 200 gal/min.

(i) Assuming the flow pattern is laminar,

compute the frictional pressure

loss per 1,000 ft of annulus.

(ii) What is the shear rate at the wall?

n = 0.3

K = 9,400


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Example 4.24

)d2.448(d

qvvel.,Avg.(i)

2

"

2

2

_

=

s/ft803."v_

=

( )22_

5.4097.8448.2

200

v =


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Example 4.24

n

n"

"2

_n

f

0.0208n

"2

)d"44,000(d

vK

dL

dp,Press.Drop

+

= +

0.3

".3

0.3

f

0.0208

0.3"2

4.5)097"44,000(8.

3)9,400(".80

dL

dp

+

=

ftpsi/1,00077.9psi/ft0779.0

dL

dp========


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Example 4.24 contd

(ii) Shear rate at pipe wall,

( )

+

=

n

"2

dd

v48%

"2

_

w

+

=

0.3

"2

4.58.097

".803*48% w

"

w s"28%

= = 75 RPM


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Total Pump Pressure

Pressure loss in surf. equipment

Pressure loss in drill pipe

Pressure loss in drill collars

Pressure drop across the bit nozzles

Pressure loss in the annulus between the drillcollars and the hole wall

Pressure loss in the annulus between the drillpipe and the hole wall

Hydrostatic pressure difference ( varies)


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Total Pump Pressure

)#P(PPPPPPP HYDDPADCABDCDPSCPUMP ++++++=

PUMP


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Types of Flow

Laminar Flow

Flow pattern is linear (no radial flow)

Velocity at wall is ZERO

Produces minimal hole erosion


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Types of Flow - Laminar

Mud properties strongly affect

pressure losses

Is preferred flow type for annulus

(in vertical wells)

Laminar flow is sometimes referred toas sheet flow, or layered flow:

* As the flow velocity increases, the flow type

changes from laminar to turbulent.


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Types of Flow

Turbulent Flow

Flow pattern is random (flow in all directions)

Tends to produce hole erosion

Results in higher pressure losses

(takes more energy)

Provides excellent hole cleaningbut


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Types of flow

Mud properties have little effect on pressure losses

Is the usual flow type inside the drill pipe and collars

Thin laminar boundary layer at the wall

Turbulent flow, contd

Fig. 4-30. Laminar and turbulent flow patterns in a circular pipe: (a) laminar

flow, (b) transition between laminar and turbulent flow and (c) turbulent flow


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Turbulent Flow - Newtonian Fluid

The onset of turbulence in pipe flow is

characterized by the dimensionlessgroup known as the Reynolds number

dvN

_

Re =

dv&928N

_

Re =

In field units,


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Turbulent Flow -

Newtonian Fluid

We often assume that fluid flow is

turbulent if Nre > 2,100

cp.fluid,ofviscosity

inI.D.,piped

ft/svelocity,fluidavg.v

lbm/galdensity,fluid&where

_

=

=

=

=

dv&928N

_

Re =

12 Laminar Flow Slot Flow

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