1PETE 411Well Drilling

Lesson 13Pressure Drop Calculations

API Recommended Practice 13DThird Edition, June 1, 1995

2Homework

HW #7. Pressure Drop Calculations

Due Oct. 9, 2002

The API Power Law Model

3Contents

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

4Power Law ModelK = consistency indexn = flow behaviour index

SHEAR STRESS

psi

= K n

SHEAR RATE, , sec-10

5Fluid Flow in Pipes and Annuli

LOG(PRESSURE)

(psi)

LOG (VELOCITY) (or FLOW RATE)

6Fluid Flow in Pipes and Annuli

LOG

(SHEAR STRESS)

(psi)

Laminar Flow Turbulent

)secor RPM ( ), RATE SHEAR (LOG 1

n1

7RotatingSleeve

Viscometer

8Rotating Sleeve Viscometer

VISCOMETERRPM

3100

300600

(RPM * 1.703)

SHEAR RATEsec -1

5.11170.3

5111022

BOB

SLEEVE

ANNULUS

DRILLSTRING

API RP 13D

9API 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

Example Calculate 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

737.03965log32.3

RRlog32.3n

300

600p =

=

=

2

n

737.0n600

p cmsecdyne017.2

022,165*11.5

022,1R11.5K

p===

secft00.8

78.3280*408.0

DQ408.0V 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

pp n

p

p1n

ppep n4

1n3DV96

K100

+

=

cP53737.0*4

1737.0*378.3

8*96017.2*100737.01737.0

ep =

+

=

616,653

5.12*00.8*78.3*928VD928Nep

pRep

==

=

14

NOTE: NRe > 2,100, soFriction 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

93.3737.0log50

93.3nloga p =+=

+=

2690.07

737.0log75.17

nlog75.1b p ==

=

007126.0616,6

0759.0N

af 2690.0bRe

p

p

===

15

Friction Pressure Gradient (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*05837.0LdLdPP dp

dpdp =

=

Pdp = 665 psi

ftpsi05837.0

78.3*81.255.12*8*007126.0

D81.25Vf

dLdP 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

737.03965log32.3

RRlog32.3n

300

600dc =

=

=

2

n

737.0n600

dc cmsecdyne017.2

022,165*11.5

022,1R11.5K

p===

secft28.18

5.2280*408.0

DQ408.0V 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

pp n

p

p1n

ppedc n4

1n3DV96

K100

+

=

cP21.38737.0*4

1737.0*35.2

28.18*96017.2*100737.01737.0

edc =

+

=

870,1321.38

5.12*28.18*5.2*928VD928Nedc

dcRedc

==

=

18

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

Pressure Drop In Drill Collars

NOTE: NRe > 2,100, soFriction Factor in DC (fdc): bRe

dc

dcN

af =

So,

0759.050

93.3737.0log50

93.3nloga dc =+=+=

2690.07

737.0log75.17

nlog75.1b dc ===

005840.0870,130759.0

Naf 2690.0b

Redc

dc

===

19

Friction Pressure Gradient (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

ftpsi3780.0

5.2*81.255.12*28.18*005840.0

D81.25Vf

dLdP 2

dc

2dcdc

dc

==

=

600*3780.0LdLdPP dc

dcdc =

=

Pdc = 227 psi

20

Pressure Drop across Nozzles

DN1 = 11 32nds (in) DN2 = 11 32nds (in) DN3 = 12 32nds (in)

( )22222

Nozzles121111280*5.12*156P

++=

PNozzles = 1,026 psi

( )223N22N21N2

NozzlesDDD

Q156P++

=

21

Pressure Dropin DC/HOLE

Annulus

DHOLE = 8.5 inODDC = 6.5 in L = 600 ft

Q = 280 gal/min

= 12.5 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

20log657.0R

Rlog657.0n3

100dca =

=

=

2

n

5413.0n100

dca cmsecdyne336.6

2.17020*11.5

2.170R11.5K

dca===

secft808.3

5.65.8280*408.0

DDQ408.0V 222

12

2dca =

=

=

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*25.65.8

808.3*144336.6*1005413.015413.0

ea =

+

=

( ) ( ) 600,120.55

5.12*808.3*5.65.8928VDD928Nea

a12Rea

=

=

=

aa n

a

a

1n

12

aaea n3

1n2DDV144K100

+

=

24

So,

DHOLE = 8.5 inODDC = 6.5 in L = 600 ft

Pressure Dropin DC/HOLE Annulus

NOTE: NRe < 2,100Friction Factor in Annulus (fa):

01500.0600,124

N24f

aRea ===

( ) ( ) ftpsi05266.0

5.65.881.255.12*808.3*01500.0

DD81.25Vf

dLdP 2

12

2aa

a

=

=

=

600*05266.0LdLdPP hole/dc

hole/dchole/dc =

=

Pdc/hole = 31.6 psi

25

q = 280 gal/min

= 12.5 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

20log657.0R

Rlog657.0n3

100dpa =

=

=

2

n

5413.0n100

dpa cmsecdyne336.6

2.17020*11.5

2.170R11.5K

dpa===

secft197.2

5.45.8280*408.0

DDQ408.0V 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

1n2DDV144K100

+

=

cP64.975413.0*3

15413.0*25.45.8

197.2*144336.6*1005413.015413.0

ea =

+

=

( ) ( ) 044,164.97

5.12*197.2*5.45.8928VDD928Nea

a12Rea

=

=

=

28

So, psi

Pressure Dropin DP/HOLE Annulus

NOTE: NRe < 2,100Friction Factor in Annulus (fa):

02299.0044,124

N24f

aRea ===

( ) ( ) ftpsi01343.0

5.45.881.255.12*197.2*02299.0

DD81.25Vf

dLdP 2

12

2aa

a

=

=

=

400,11*01343.0LdLdPP 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 = 665 + 227 + 1,026

+ 32 + 153 + 0

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

PANN = 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

01,0002,0003,0004,0005,0006,0007,0008,0009,000

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

01,0002,0003,0004,0005,0006,0007,0008,0009,000

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

14,0000 2,000 4,000 6,000 8,000 10,000

Pressure, psi

Dept

h, f

t

DRILLSTRING

ANNULUS

(Static)

BIT

2103

36

Pipe Flow - LaminarIn 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, thenFriction Factor in Pipe (fp):

pRep N

16f =D81.25

VfdLdP 2pp

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

aa

Naf =Then and

5093.3nloga a +=

7nlog75.1b a=

( )122

aa

a DD81.25Vf

dLdP

=

38

Critical Circulation RateExampleThe 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 1,044300 2.354 1,154350 2.746 1,446400 3.138 1,756450 3.531 2,086452 3.546 2,099

452.1 3.547 2,100

( )ea

a12Re

VDD928Na

=

40

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, so does the frictional pressure drop.

41

42d 8.25vf

dLdp

_2

f =

n = 1.0

43

Importance of Pipe Size

or,

25.1

25.075.1_

75.0f

d1800v

dLdp

=

75.4

25.075.175.0f

d624,8q

dLdp

=

*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 anincrease of frictional pressure drop by about 65% !!

44

pf = 11.41 v 1.75turbulent flow

pf = 9.11 vlaminar flow

Use max. pf value