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PETE 411 Well Drilling. Lesson 13 Pressure Drop Calculations API Recommended Practice 13D Third Edition, June 1, 1995. Homework. HW #7. Pressure Drop Calculations Due Oct. 9, 2002 The API Power Law Model. Contents. The Power Law Model The Rotational Viscometer - PowerPoint PPT Presentation
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1 PETE 411 Well Drilling Lesson 13 Pressure Drop Calculations API Recommended Practice 13D Third Edition, June 1, 1995
Transcript 1

PETE 411Well Drilling

Lesson 13Pressure Drop Calculations

API Recommended Practice 13D

Third Edition, June 1, 1995 2

Homework

HW #7. Pressure Drop Calculations

Due Oct. 9, 2002

The API Power Law Model 3

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 4

Power Law Model

K = consistency index

n = flow behaviour index

SHEAR STRESS

psi

= K n

SHEAR RATE, , sec-1

0 5

Fluid Flow in Pipes and Annuli

LOG(PRESSURE)

(psi)

LOG (VELOCITY) (or FLOW RATE) 6

Fluid Flow in Pipes and Annuli

LOG

(SHEAR STRESS)

(psi)

Laminar Flow Turbulent

)secor RPM ( ), RATE SHEAR (LOG 1

n1 7

RotatingSleeve

Viscometer 8

Rotating Sleeve Viscometer

VISCOMETERRPM

3100

300600

(RPM * 1.703)

SHEAR RATE

sec -1

5.11170.3 511

1022

BOB

SLEEVE

ANNULUS

DRILLSTRING

API RP 13D 9

API RP 13D, June 1995for Oil-Well Drilling 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 for

pressure drop calculations in the annulus, where shear rates are, generally, not very high.

The 300 and 600 RPM reading are used for pressure drop calculations inside drillpipe, where shear rates are, generally, quite high. 10

Example: Pressure Drop Calculations

ExampleCalculate the pump pressure in the 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) 11

PPUMP = PDP + PDC

+ PBIT NOZZLES

+ PDC/ANN + PDP/ANN

+ PHYD

Q = 280 gal/min

= 12.5 lb/gal

Pressure DropCalculations

PPUMP 12

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 in L = 11,400 ft

737.039

65log32.3

R

Rlog32.3n

300

600p

2

n

737.0n600

p 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 13

Effective Viscosity in Pipe (ep):

Pressure Drop In Drill Pipe

Reynolds Number in Pipe (NRep):

OD = 4.5 in ID = 3.78 in L = 11,400 ft

ppn

p

p

1n

ppep n4

1n3

D

V96K100

cP53737.0*4

1737.0*3

78.3

8*96017.2*100

737.01737.0

ep

616,653

5.12*00.8*78.3*928VD928N

ep

pRep 14

NOTE: NRe > 2,100, so

Friction Factor in Pipe (fp):

Pressure Drop In Drill Pipe OD = 4.5 in ID = 3.78 in L = 11,400 ft

So,

bRe

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

2690.0bRe

p

p 15

(dP/dL)p :

Pressure Drop In Drill Pipe OD = 4.5 in ID = 3.78 in L = 11,400 ft

Friction Pressure Drop in Drill

Pipe :400,11*05837.0L

dL

dPP dp

dpdp

Pdp = 665 psi

ft

psi05837.0

78.3*81.25

5.12*8*007126.0

D81.25

Vf

dL

dP 22pp

dp 16

Power-Law Constant (n):

Pressure Drop In Drill Collars

Fluid Consistency Index (K):

Average Bulk Velocity inside Drill Collars (Vdc):

OD = 6.5 in ID = 2.5 in L = 600 ft

737.039

65log32.3

R

Rlog32.3n

300

600dc

2

n

737.0n600

dc 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 17

Effective Viscosity in Collars(ec):

Reynolds Number in Collars (NRec):

OD = 6.5 in ID = 2.5 in L = 600 ft

Pressure Drop In Drill Collars

ppn

p

p

1n

ppedc n4

1n3

D

V96K100

cP21.38737.0*4

1737.0*3

5.2

28.18*96017.2*100

737.01737.0

edc

870,1321.38

5.12*28.18*5.2*928VD928N

edc

dcRedc 18

OD = 6.5 in ID = 2.5 in L = 600 ft

Pressure Drop In Drill Collars

NOTE: NRe > 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

2690.0bRe

dc

dc 19

(dP/dL)dc :

Friction Pressure Drop in Drill

Collars :

OD = 6.5 in ID = 2.5 in L = 600 ft

Pressure Drop In Drill Collars

ft

psi3780.0

5.2*81.25

5.12*28.18*005840.0

D81.25

Vf

dL

dP 2

dc

2dcdc

dc

600*3780.0LdL

dPP dc

dcdc

Pdc = 227 psi 20

Pressure Drop across Nozzles

DN1 = 11 32nds

(in) DN2 = 11

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

2222

2

Nozzles121111

280*5.12*156P

PNozzles = 1,026 psi

223N

22N

2

1N

2

Nozzles

DDD

Q156P 21

Pressure Dropin DC/HOLE

Annulus

DHOLE = 8.5 inODDC = 6.5 in L = 600 ft

Q = gal/min

= lb/gal 8.5 in 22

Power-Law Constant (n):

Fluid Consistency Index (K):

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

DHOLE = 8.5 inODDC = 6.5 in L = 600 ft

Pressure Dropin DC/HOLE Annulus

5413.03

20log657.0

R

Rlog657.0n

3

100dca

2

n

5413.0n100

dca 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

2221

22

dca 23

Effective Viscosity in Annulus (ea):

Reynolds Number in Annulus (NRea):

DHOLE = 8.5 inODDC = 6.5 in L = 600 ft

Pressure Dropin DC/HOLE Annulus

cP20.555413.0*3

15413.0*2

5.65.8

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 n3

1n2

DD

V144K100 24

So,

DHOLE = 8.5 inODDC = 6.5 in L = 600 ft

Pressure Dropin DC/HOLE Annulus

NOTE: NRe < 2,100 Friction Factor in

Annulus (fa): 01500.0600,1

24

N

24f

aRea

ft

psi05266.0

5.65.881.25

5.12*808.3*01500.0

DD81.25

Vf

dL

dP 2

12

2aa

a

600*05266.0LdL

dPP hole/dc

hole/dchole/dc

Pdc/hole = 31.6 psi 25

q = gal/min

= lb/gal

Pressure Dropin DP/HOLE Annulus

DHOLE = 8.5 inODDP = 4.5 in L = 11,400 ft 26

Power-Law Constant (n):

Fluid Consistency Index (K):

Average Bulk Velocity in Annulus (Va):

Pressure Dropin DP/HOLE Annulus

DHOLE = 8.5 inODDP = 4.5 in L = 11,400 ft

5413.03

20log657.0

R

Rlog657.0n

3

100dpa

2

n

5413.0n100

dpa 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

2221

22

dpa 27

Effective Viscosity in Annulus (ea):

Reynolds Number in Annulus (NRea):

Pressure Dropin DP/HOLE Annulus

aa n

a

a

1n

12

aaea n3

1n2

DD

V144K100

cP64.975413.0*3

15413.0*2

5.45.8

197.2*144336.6*100

5413.015413.0

ea

044,1

64.97

5.12*197.2*5.45.8928VDD928N

ea

a12Rea 28

So, psi

Pressure Dropin DP/HOLE Annulus

NOTE: NRe < 2,100 Friction Factor in

Annulus (fa): 02299.0044,1

24

N

24f

aRea

ft

psi01343.0

5.45.881.25

5.12*197.2*02299.0

DD81.25

Vf

dL

dP 2

12

2aa

a

400,11*01343.0LdL

dPP hole/dp

hole/dphole/dp

Pdp/hole = 153.2 psi 29

Pressure DropCalculations

- SUMMARY -

PPUMP = PDP + PDC + PBIT NOZZLES

+ PDC/ANN + PDP/ANN + PHYD

PPUMP = + +

+ + +

PPUMP = psi 30

PPUMP = 1,918 + 185 = 2,103 psi

PHYD = 0

PPUMP = PDS + PANN + PHYD

PDS = PDP + PDC + PBIT NOZZLES

= 665 + 227 + 1,026 = 1,918 psiPANN = PDC/ANN + PDP/ANN

= 32 + 153 = 185

2,103 psi

P = 0 31

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 32

"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

"Fri

ctio

n" P

ress

ure,

psi

DRILLPIPE

DRILL COLLARS

BIT NOZZLES

ANNULUS

2103 33

Hydrostatic Pressures in the Wellbore

0

1,000

2,000

3,000

4,000

5,000

6,000

7,000

8,000

9,000

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

Distance from Standpipe, ft

Hyd

rost

atic

Pre

ssur

e, p

si

BHP

DRILLSTRING ANNULUS 34

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

Distance from Standpipe, ft

Pre

ssur

es,

psi

STATIC

CIRCULATING

2103 35

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

De

pth

, f

t

DRILLSTRING

ANNULUS

(Static)

BIT

2103 36

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: if NRe < 2,100, then

Friction Factor in Pipe (fp):

pRep N

16f

D81.25

Vf

dL

dP2

pp

dp

Then and 37

Annular Flow - TurbulentIn the above example the flow up the annulus

was laminar.

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

NOTE: if NRe > 2,100, then Friction Factor in the Annulus:

bRe

a

aN

af Then and

50

93.3nloga a

7

nlog75.1b a

12

2aa

a DD81.25

Vf

dL

dP 38

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?) 39

Critical Circulation RateIn 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 2,086 452 3.546 2,099

452.1 3.547 2,100

ea

a12Re

VDD928N

a 40

Optimum Bit Hydraulics

maximum hydraulic horsepower? maximum impact force?

Both these items increase when the circulation rate increases.

However, when the circulation rate increases, so does the frictional pressure drop. 41 42d 8.25

vf

dL

dp_2

f

n = 1.0 43

Importance of Pipe Size

or,

25.1

25.075.1_

75.0f

d1800

v

dL

dp

75.4

25.075.175.0f

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% !! 44

pf = 11.41 v 1.75

turbulent flow

pf = 9.11 vlaminar flow

Use max. pf value

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