13 API Power-Law Model

5/20/2018 13 API Power-Law Model

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PETE 411

Well Drilling

Lesson 13

Pressure Drop CalculationsAPI Recommended Practice 13D

Third Edition, June 1, 1995


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Homework

HW #7. Pressure Drop Calculations

Due Oct. 9, 2002

The API Power Law Model


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Contents

The Power Law Model

The Rotational Viscometer

A detailed Example - Pump Pressure

Pressure Drop in the Drillpipe

Pressure Drop in the Bit Nozzles

Pressure Drop in the Annulus

Wellbore Pressure Profiles


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Power Law ModelK = consistency index

n = flow behaviour index

SHEARSTRESS

psi

= K n

SHEAR RATE, , sec-10


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Fluid Flow in Pipes and Annuli

LOG(PRESSURE)

(psi)

LOG (VELOCITY) (or FLOW RATE)


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Fluid Flow in Pipes and Annuli

LOG

(SHEAR

STRESS)

(psi)

Laminar Flow Turbulent

)secorRPM(),RATESHEAR(LOG 1

n

1


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Rotating

SleeveViscometer


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Rotating Sleeve Viscometer

VISCOMETERRPM

3

100

300

600

(RPM * 1.703)

SHEAR RATEsec -1

5.11

170.3

511

1022

BOB

SLEEVE

ANNULUS

DRILL

STRING

API RP 13D


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API RP 13D, June 1995

for Oil-Well Drill ing Fluids

API RP 13D recommends using only FOUR of

the six usual viscometer readings:

Use 3, 100, 300, 600 RPM Readings.

The 3 and 100 RPM reading are used forpressure drop calculations in the annulus,

where shear rates are, generally, not very high.

The 300 and 600 RPM reading are used forpressure drop calculations inside drillpipe,

where shear rates are, generally, quite high.


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Example: Pressure Drop Calculations

Example Calculate the pump pressure inthe wellbore shown on the next page, using the

API method.

The relevant rotational viscometer readings

are as follows:

R3 = 3 (at 3 RPM)

R100 = 20 (at 100 RPM)

R300 = 39 (at 300 RPM)

R600 = 65 (at 600 RPM)


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PPUMP = PDP + PDC

+ PBIT NOZZLES

+ PDC/ANN + PDP/ANN

+ PHYD

Q = 280 gal/min = 12.5 lb/gal

Pressure Drop

Calculations

PPUMP


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Power-Law Constant (n):

Pressure Drop In Drill Pipe

Fluid Consistency Index (K):

Average Bulk Velocity in Pipe (Vp):

OD = 4.5 in

ID = 3.78 inL = 11,400 ft

737.039

65

log32.3R

R

log32.3n300

600

p =

=

=

2

n

737.0n

600p

cm

secdyne017.2

022,1

65*11.5

022,1

R11.5K

p===

sec

ft00.8

78.3

280*408.0

D

Q408.0V

22p ===


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Effective Viscosity in Pipe (ep):

Pressure Drop In Drill Pipe

Reynolds Number in Pipe (NRep):

OD = 4.5 in

ID = 3.78 inL = 11,400 ft

ppn

p

p

1n

p

pepn4

1n3DV96K100

+

=

cP53737.0*4

1737.0*378.3

8*96017.2*100

737.01737.0

ep =

+

=

616,653

5.12*00.8*78.3*928VD928N

ep

p

Rep==

=


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NOTE: N Re> 2,100, so

Friction Factor in Pipe (fp):

Pressure Drop In Drill Pipe OD = 4.5 in

ID = 3.78 inL = 11,400 ft

So,

b

Re

p

pN

af =

0759.050

93.3737.0log

50

93.3nloga

p=

+=

+=

2690.07

737.0log75.1

7

nlog75.1b

p=

=

=

007126.0616,6

0759.0

N

af

2690.0b

Re

p

p

===


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Friction Pressure Gradient (dP/dL)p :

Pressure Drop In Drill Pipe OD = 4.5 in

ID = 3.78 inL = 11,400 ft

Friction Pressure Drop in Drill Pipe :

400,11*05837.0L

dL

dPP dp

dp

dp =

=

Pdp = 665 psi

ft

psi05837.0

78.3*81.25

5.12*8*007126.0

D81.25

Vf

dL

dP 22

pp

dp

==

=


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Pressure Drop In Drill Collars


Average Bulk Velocity inside Drill Collars (Vdc):

OD = 6.5 in

ID = 2.5 inL = 600 ft

737.039

65

log32.3R

R

log32.3n300

600

dc =

=

=

2

n

737.0n

600dc

cm

secdyne017.2

022,1

65*11.5

022,1

R11.5K

p===

sec

ft28.18

5.2

280*408.0

D

Q408.0V

22dc ===


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Effective Viscosity in Collars(ec):

Reynolds Number in Collars (NRec):

OD = 6.5 in

ID = 2.5 inL = 600 ft


ppn

p

p

1n

ppedc

n41n3

DV96K100

+

=

cP21.38737.0*4

1737.0*35.2

28.18*96017.2*100

737.01737.0

edc =

+

=

870,1321.38

5.12*28.18*5.2*928VD928N

edc

dcRedc

==

=


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OD = 6.5 in

ID = 2.5 inL = 600 ft


NOTE: N Re> 2,100, so

Friction Factor in DC (fdc):b

Re

dc

dcN

af =

So,

0759.050

93.3737.0log

50

93.3nloga dc =

+=

+=

2690.07

737.0log75.1

7

nlog75.1b dc =

=

=

005840.0870,13

0759.0

N

af

2690.0b

Re

dc

dc

===


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Friction Pressure Gradient (dP/dL)dc :

Friction Pressure Drop in Drill Collars :

OD = 6.5 in

ID = 2.5 inL = 600 ft


ft

psi3780.0

5.2*81.25

5.12*28.18*005840.0

D81.25

Vf

dL

dP 2

dc

2dcdc

dc

==

=

600*3780.0L

dL

dPP dc

dc

dc =

=

Pdc = 227 psi


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Pressure Drop across Nozzles

DN1 = 11 32nds (in)

DN2 = 11 32nds (in)DN3 = 12 32nds (in)

( )22222

Nozzles

121111

280*5.12*156P

++=

PNozzles = 1,026 psi

( )223N22N2

1N

2

Nozzles

DDDQ156P++

=


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Pressure Dropin DC/HOLE

Annulus

DHOLE

= 8.5 inODDC = 6.5 inL = 600 ft

Q= 280 gal/min

= 12.5

lb/gal 8.5 in


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Average Bulk Velocity in DC/HOLE Annulus (Va):

DHOLE = 8.5 in

ODDC = 6.5 inL = 600 ft

Pressure Drop

in DC/HOLE Annulus

5413.03

20log657.0R

Rlog657.0n

3

100dca =

=

=

2

n

5413.0n

100dca

cm

secdyne336.6

2.170

20*11.5

2.170

R11.5K

dca===

sec

ft808.3

5.65.8

280*408.0

DD

Q408.0V

222

1

2

2

dca =

=

=


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Effective Viscosity in Annulus (ea):

Reynolds Number in Annulus (NRea):

DHOLE = 8.5 in

ODDC = 6.5 inL = 600 ft

Pressure Drop

in DC/HOLE Annulus

cP20.555413.0*3

15413.0*25.65.8

808.3*144336.6*100

5413.015413.0

ea =

+

=

( ) ( )600,1

20.55

5.12*808.3*5.65.8928VDD928N

ea

a12Rea

=

=

=

aa n

a

a

1n

12

aaea

n31n2

DDV144K100

+

=


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So,

DHOLE = 8.5 in

ODDC = 6.5 inL = 600 ft

Pressure Drop

in DC/HOLE AnnulusNOTE: N Re< 2,100

Friction Factor in Annulus (fa):

01500.0600,1

24

N

24f

aRe

a ===

( ) ( ) ftpsi

05266.05.65.881.25

5.12*808.3*01500.0

DD81.25

Vf

dL

dP 2

12

2aa

a

=

=

=

600*05266.0LdL

dPP hole/dc

hole/dc

hole/dc =

=

Pdc/hole = 31.6 psi


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q= 280

gal/min

= 12.5 lb/gal

Pressure Drop

in DP/HOLE Annulus

DHOLE = 8.5 inODDP = 4.5 in

L = 11,400 ft


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Average Bulk Velocity in Annulus (Va):

Pressure Drop

in DP/HOLE Annulus

DHOLE = 8.5 in

ODDP = 4.5 inL = 11,400 ft

5413.03

20log657.0R

Rlog657.0n

3

100dpa =

=

=

2

n

5413.0n

100dpa

cm

secdyne336.6

2.170

20*11.5

2.170

R11.5K

dpa===

sec

ft197.2

5.45.8

280*408.0

DD

Q408.0V

222

1

2

2

dpa =

=

=


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Effective Viscosity in Annulus (ea):

Reynolds Number in Annulus (NRea):

Pressure Drop

in DP/HOLE Annulus

aa n

a

a

1n

12

aaea

n31n2

DDV144K100

+

=

cP64.975413.0*3

15413.0*25.45.8

197.2*144336.6*100

5413.015413.0

ea =

+

=

( ) ( )044,1

64.97

5.12*197.2*5.45.8928VDD928N

ea

a12Rea

=

=

=


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So, psi

Pressure Drop

in DP/HOLE AnnulusNOTE: N Re< 2,100

Friction Factor in Annulus (fa):

02299.0044,1

24

N

24f

aRe

a ===

( ) ( ) ftpsi

01343.05.45.881.25

5.12*197.2*02299.0

DD81.25

Vf

dL

dP 2

12

2

aa

a

=

=

=

400,11*01343.0LdL

dP

P hole/dphole/dp

hole/dp =

=

Pdp/hole = 153.2 psi


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Pressure Drop

Calculations- SUMMARY -

PPUMP = PDP + PDC + PBIT NOZZLES

+ PDC/ANN + PDP/ANN + PHYD

PPUMP = 665 + 227 + 1,026

+ 32 + 153 + 0

PPUMP = 1,9181,9181,9181,918++++185185185185 ==== 2,1032,1032,1032,103 psi


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PPUMP = 1,918 + 185

= 2,103 psi

PHYD = 0

PPUMP = PDS + PANN + PHYD

PDS = PDP + PDC + PBIT NOZZLES

= 665 + 227 + 1,026 = 1,918 psi

PANN = PDC/ANN + PDP/ANN

= 32 + 153 = 185

2,103 psi

P

=

0


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BHP = 185 + 7,800

What is the BHP?

BHP = PFRICTION/ANN + PHYD/ANN

BHP = PDC/ANN + PDP/ANN

+ 0.052 * 12.5 * 12,000

= 32 + 153 + 7,800 = 7,985 psig

2,103 psi

P

=

0

BHP= 7,985 psig


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" Friction" Pressures

0

500

1,000

1,500

2,000

2,500

0 5,000 10,000 15,000 20,000 25,000

Distance from Standpipe, ft

"Friction"

Pressure

,psi DRILLPIPE

DRILL COLLARS

BIT NOZZLES

ANNULUS

2103


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Hydrostatic Pressures in the Wellbore

0

1,000

2,000

3,000

4,000

5,000

6,0007,000

8,000

9,000

0 5,000 10,000 15,000 20,000 25,000


Hydrostati

c

Pressure,

psi BHP

DRILLSTRING ANNULUS


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Pressures in the Wellbore

0

1,000

2,000

3,000

4,000

5,000

6,000

7,000

8,000

9,000

10,000

0 5,000 10,000 15,000 20,000 25,000


Pressures,

psi

STATIC

CIRCULATING

2103


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Wellbore Pressure Profile

0

2,000

4,000

6,000

8,000

10,000

12,000

14,000

0 2,000 4,000 6,000 8,000 10,000

Pressure, psi

D

epth,

ft

DRILLSTRING

ANNULUS

(Static)

BIT

2103


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Pipe Flow - Laminar

In the above example the flow down the

drillpipe was turbulent.

Under conditions of very high viscosity,

the flow may very well be laminar.

NOTE: i f N Re< 2,100, t hen

Friction Factor in Pipe (fp):

pRe

pN

16f =

D81.25

Vf

dL

dP2

pp

dp

=

Then and


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Annular Flow - Turbulent

In the above example the flow up the annulus

was laminar.

Under conditions of low viscosity and/or highflow rate, the flow may very well be turbulent.

NOTE: i f N Re> 2,100, t henFriction Factor in the Annulus:

b

Re

a

aN

af =Then and

50

93.3nloga a

+=

7

nlog75.1b a

=

( )12

2

aa

a DD81.25

Vf

dL

dP

=


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Critical Circulation Rate

Example

The above fluid is flowing in the annulus

between a 4.5 OD string of drill pipe

and an 8.5 in hole.

The fluid density is 12.5 lb/gal.

What is the minimum circulation rate that

will ensure turbulent flow?(why is this of interest?)


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Critical Circulation Rate

In the Drillpipe/Hole Annulus:

Q, gal/min V, ft/sec Nre

280 2.197 1,044

300 2.354 1,154

350 2.746 1,446

400 3.138 1,756

450 3.531 2,086452 3.546 2,099

452.1 3.547 2,100

( )

ea

a12Re

VDD928N

a

=


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Optimum Bit Hydraulics

Under what conditions do we get the best

hydraulic cleaning at the bit? maximum hydraulic horsepower?

maximum impact force?

Both these items increase when the circulation

rate increases.

However, when the circulation rate increases, sodoes the frictional pressure drop.


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42d8.25

vf

dL

dp

_2

f =

n = 1.0


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Importance of Pipe Size

or,

25.1

25.0

75.1_75.0

f

d1800

v

dL

dp =

75.4

25.075.175.0

f

d624,8

q

dL

dp

=

*Note that a small change in the pipe diameter results in

large change in the pressure drop! (q = const.)

Eq. 4.66e

Decreasing the pipe ID 10% from 5.0 to 4.5 would result in an

increase of frictional pressure drop by about 65% !!


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pf= 11.41 v1.75

turbulent flow

pf= 9.11 v

laminar flow

Use max. pfvalue

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