5E an example of pressure drop calculations by api recommended practice

Old Lesson 17B 1 1 Well Drilling Engineering Pressure Drop Calculations API Recommended Practice 13D Dr DO QUANG KHANH 2 Contents The API Power Law Model The Rotational Viscometer A detailed Exampl[.]

Well Drilling Engineering

Pressure Drop Calculations

API Recommended Practice 13D

Dr DO QUANG KHANH

Contents

The API 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

Power Law Model

Fluid Flow in Pipes and Annuli

Fluid Flow in Pipes and Annuli

or RPM(

),

RATE

SHEAR(

n

1

Rotating

Sleeve

Viscometer

Rotating Sleeve Viscometer

VISCOMETER

RPM

3

100

API RP 13D

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

 API RP 13D recommends using only FOUR of

the six usual viscometer 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

Example: Pressure Drop Calculations

in the wellbore shown on the next page,

using the API method

readings are as follows:

PPUMP = ∆PDP + ∆PDC

+ ∆PBIT NOZZLES + ∆PDC/ANN + ∆PDP/ANN + ∆PHYD

Power - Law Constant (n):

Pressure Drop In Drill Pipe

Fluid Consistency Index (K):

Average Bulk Velocity in Pipe (V p ):

OD = 4.5 in

ID = 3.78 in

L = 11,400 ft

737

0 39

65 log

32

3 R

R log 32

600 p

cm

sec

dyne017

2022

,1

65

*11.5022

,1

R11

8 78

3

280

* 408

0 D

Q 408

0

Effective Viscosity in Pipe (µep ):

Pressure Drop In Drill Pipe

Reynolds Number in Pipe (N Rep ):

ep

n 4

1 n

3 D

V

96 K

0

* 4

1 737

0

* 3 78

3

8

*

96 017

2

* 100

737 0 1

737 0

6 53

5 12

* 00 8

* 78 3

* 928 V

D

928 N

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,

b Re

050

93.3737

.0log50

93.3n

log

2690

07

737

0log75

.17

nlog75

1

007126

0616

,6

0759

0N

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

0

LdL

0 78

3

* 81 25

5 12

* 8

* 007126

0 D

81 25

V f dL

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

0 39

65 log

32

3 R

R log 32

3

600 dc

cm

sec

dyne017

2022

,1

65

*11.5022

,1

R11

.

18 5

2

280

* 408

0 D

Q 408

0

Effective Viscosity in Collars ( µec):

Reynolds Number in Collars (NRec):

OD = 6.5 in

ID = 2.5 in

L = 600 ftPressure Drop In Drill Collars

edc

n 4

1 n

3 D

V

96 K

38737

.0

*4

1737

0

*35

.2

28.18

*

96017

.2

*100

737 0 1

737 0

13 21

38

5 12

* 28 18

* 5 2

* 928 V

D

928 N

93.3737

.0log50

93.3n

log

2690

07

737

0log75

.17

nlog75

1

005840

0 870

, 13

0759

0 N

Friction Pressure Gradient (dP/dL)dc :

Friction Pressure Drop in Drill Collars :

OD = 6.5 in

ID = 2.5 in

L = 600 ftPressure Drop In Drill Collars

ft

psi 3780

0 5

2

* 81 25

5 12

* 28 18

* 005840

0 D

81 25

V f dL

dc

2 dc dc dc

0

LdL

Pressure Drop across Nozzles

11

280

* 5 12

*

156 P

+ +

2 2 N

2 1 N

2

Nozzles

D D

D

Q

156 P

+ +

ρ

=

∆

Power-Law Constant (n):

Fluid Consistency Index (K):

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

0 3

20 log

657

0 R

R log 657

0

100 dca

cm

sec

dyne336

62

.170

20

*11.52

.170

R11

5K

=

sec

ft 808

.

3 5

6 5

8

280

* 408

0 D

D

Q 408

0

1

2 2

Effective Viscosity in Annulus ( µea):

Reynolds Number in Annulus (NRea):

55 5413

0

* 3

1 5413

0

* 2 5

6 5 8

808

3

* 144 336

6

* 100

5413 0 1

5413 0

1 20

55

5 12

* 808

3

* 5 6 5 8 928 V

D D

928 N

ea

a 1 2

1 n

1 2

a a

ea

n3

1n

2D

D

V

144K

−

0 600

, 1

24 N

24

f

a Re

psi 05266

0 5

6 5 8 81 25

5 12

* 808

3

* 01500

0 D

D 81 25

V f dL

1 2

2 a a

0

L dL

dP

hole / dc

hole /

Power-Law Constant (n):

Fluid Consistency Index (K):

Average Bulk Velocity in Annulus (V a ):

0 3

20 log

657

0 R

R log 657

0 n

100 dpa

cm

sec

dyne336

62

.170

20

*11.52

.170

R11

5K

=

sec

ft 197

2 5

4 5

8

280

* 408

0 D

D

Q 408

0

1

2 2

Effective Viscosity in Annulus ( µea):

Reynolds Number in Annulus (NRea):

1 n

1 2

a a

ea

n3

1n

2D

D

V

144K

97 5413

0

* 3

1 5413

0

* 2 5

4 5 8

197 2

* 144 336

6

* 100

5413 0 1

5413 0

1 64

97

5 12

* 197

2

* 5 4 5 8 928 V

D D

928 N

ea

a 1 2

0 044

, 1

24 N

24

f

a Re

psi 01343

0 5

4 5 8 81 25

5 12

* 197 2

* 02299

0 D

D 81 25

V f dL

1 2

2 a a a

11

* 01343

0

L dL

dP

hole / dp

hole /

Pressure Drop Calculations

- SUMMARY -

PPUMP = ∆PDP + ∆PDC + ∆PBIT NOZZLES

+ ∆PDC/ANN + ∆PDP/ANN + ∆PHYD

PPUMP = 665 + 227 + 1,026

+ 32 + 153 + 0

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

2,103 psi P CSG

= 0

BHP = 7,985 psig

2103

Hydrostatic Pressures in the Wellbore

Pressures in the Wellbore

2103

Wellbore Pressure Profile

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):

p Re

V f dL

Annular Flow - Turbulent

In 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 N Re > 2,100, then Friction Factor in the Annulus:

b Re

1

=

( 2 1)

2 a a

V f dL

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?)

Critical Circulation Rate

In the Drillpipe/Hole Annulus:

Re

V D D

928 N

ρ

−

=

Optimum Bit Hydraulics

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

v f dp

_ 2

=

n = 1.0

Importance of Pipe Size

or,

25 1

25 0

75 1 _ 75 0 f

d 1800

v dL

∴

75 4

25 0 75

1 75 0 f

d 624 ,

8

q dL

*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