4. wellbore hydraulics, pressure drop calculations

1

PETE 661Drilling Engineering

Lesson 4

Wellbore Hydraulics,

Pressure Drop Calculations

2

Wellbore Hydraulics

• Hydrostatics• Buoyancy• Pipe Tension vs. Depth• Effect of Mud Pressure• Laminar and Turbulent Flow• Pressure Drop Calculations

– Bingham Plastic Model– API Power-Law Model

3

Assignments:

READ: ADE Ch. 4

HW #3: On the Web. - Axial Tension

Due 09-20-04

4

)DD(052.0pp 1ii

n

1ii0

Fig. 4-3. A Complex

Liquid Column

D052.0p

pD052.0p 0

5Fig. 4-4. Viewing the Well as a Manometer (U-Tube)

PPUMP = ?

6

Figure 4.4

})000,10(0.9)000,1(7.16

)700,1(7.12)300(5.8)000,7(5.10{052.00

ppa

psig 00 p

psig 266,1p a

D052.0p

7

Buoyancy Force = weight of fluid displaced (Archimedes, 250 BC)

Figure 4-9. Hydraulic forces acting on a submerged body

8

Effective (buoyed) Weight

s

fe 1WW

Buoyancy Factor

Valid for a solid body or an open-ended pipe!

sf

f

be

W-W

V-W

FWW

We = buoyed weightW = weight in airFb = buoyancy forceV = volume of bodyf = fluid densitys = body density

9

Example

For steel,

immersed in mud,

the buoyancy factor is:

gal/lbm .s 565

)gal/lbm .( f 015

771.05.65

0.1511

s

f

A drillstring weighs 100,000 lbs in air.

Buoyed weight = 100,000 * 0.771 = 77,100 lbs

(= 490 lbm/ft3 )

10

Axial Forces in Drillstring

Fb = bit weightF1 & F1 are pressure forces

11

Simple Example - Empty Wellbore

Drillpipe weight = 19.5 lbf/ft 10,000 ft

OD = 5.000 inID = 4.276 in

22 IDOD4

A

A = 5.265 in2

W = 19.5 lbf/ft * 10,000 ft = 195,000 lbf

AXIAL TENSION, lbf

DE

PT

H,

ft

0 lbf 195,000 lbf

12

Example - 15 lb/gal Mud in Wellbore

Drillpipe weight = 19.5 lbf/ft 10,000 ft

OD = 5.000 inID = 4.276 in

22 IDOD4

A

A = 5.265 in2

W = 195,000 - 41,100 = 153,900 lbf

AXIAL TENSION, lbf

DE

PT

H,

ft

0 195,000 lbf

Pressure at bottom = 0.052 * 15 * 10,000 = 7,800 psiF = P * A= 7,800 * 5.265= 41,100 lbf

153,900- 41,100

13

Axial Tension in Drill String

Example A drill string consists of

10,000 ft of 19.5 #/ft drillpipe and

600 ft of 147 #/ft drill collars

suspended off bottom in 15#/gal mud

(Fb = bit weight = 0).

• What is the axial tension in the

drillstring as a function of depth?

14

Example

Pressure at top of collars

= 0.052 (15) 10,000 = 7,800 psi

Pressure at bottom of collars

= 0.052 (15) 10,600 = 8,268 psi

Cross-sectional area of pipe,

22

2

31 in73.5ft

in144*

ft/lb490

ft/lb5.19A

A1

10,000’

10,600’

15

Cross-sectional area of collars,

22 in2.43144*

490

147A

2

1

537735243 in...

AAarea alDifferenti 2

A2

A1Example – cont’d

16

1. At 10,600 ft. (bottom of drill collars)

Compressive force = p A

= 357,200 lbf

[ axial tension = - 357,200 lbf ]

22

in2.43*in

lbf268,8

4

32

1

Example - cont’d

17

Example - cont’d

2. At 10,000 ft+ (top of collars)

FT = W2 - F2 - Fb

= 147 lbm/ft * 600 ft - 357,200

= 88,200 - 357,200

= -269,000 lbf

4

32

1

Fb = FBIT = 0

18

3. At 10,000 ft - (bottom of drillpipe)

FT = W1+W2+F1-F2-Fb

= 88,200 + 7800 lbf/in2 * 37.5in2 - 357,200

= 88,200 + 292,500 - 357,200

= + 23,500 lbf

4

32

1

Example - cont’d

19

4. At Surface

FT = W1 + W2 + F1 - F2 - Fb

= 19.5 * 10,000 + 88,200

+ 292,500 - 357,200 - 0

= 218,500 lbf

Alternatively: FT = WAIR * BF

= 283,200 * 0.7710 = 218,345 lbf

4

32

1

Example - cont’d

20Fig. 4-11. Axial tensions as a function of depth for Example 4.9

21

Example - Summary

1. At 10,600 ft FT = -357,200 lbf [compression]

2. At 10,000 + ft FT = -269,000 lbf [compression]

3. At 10,000 - ft FT = +23,500 lbf [tension]

4. At Surface FT = +218,500 lbf [tension]

22

Axial Load with FBIT = 68,000 lbf

23

24

For multiple nozzles in parallel

Vn is the same for each nozzle even if the dn varies!

This follows since p is the same across each nozzle.

tn A117.3

qv

2

2

t2d

-5

bit AC

q10*8.311Δp

10*074.8

pcv

4dn

&

Cd = 0.95

25

Hydraulic Horsepower

… of pump putting out 400 gpm at 3,000 psi = ?

Power, in field units:

1714

000,3*400 HHP

1714

pq HHP

Hydraulic Horsepower of Pump = 700 hp

26

What is Hydraulic Impact Force

… developed by bit?

If:

psi 169,1Δp

lb/gal 12

gal/min 400q

95.0C

n

D

pqc01823.0F dj

27

Impact = rate of change of momentum

lbf 820169,1*12400*95.0*01823.0F

pqc01823.0F

60*17.32

vqv

t

m

t

mvF

j

dj

nj

28

Laminar Flow

Rheological Models Newtonian Bingham Plastic Power-Law (ADE & API)

Rotational Viscometer

Laminar Flow in Wellbore Fluid Flow in Pipes Fluid Flow in Annuli

29

Laminar Flow of Newtonian Fluids

A

F

L

V

Experimentally:

30

Newtonian Fluid Model

In a Newtonian fluid the shear stress is directly proportional to the shear rate (in laminar flow):

i.e.,

The constant of proportionality, is the viscosity of the fluid and is independent of shear rate.

sec

12

cm

dyne

31

Newtonian Fluid Model

Viscosity may be expressed in poise or centipoise.

poise 0.01 centipoise 1

scm

g1

cm

s-dyne1 poise 1

2

2cm

secdyne

32

Shear Stress vs. Shear Rate for a Newtonian Fluid

Slope of line

.

33

Apparent Viscosity

Apparent viscosity =

is the slope at each shear rate, .,, 321

/

34

Typical Drilling Fluid Vs. Newtonian, Bingham and Power Law Fluids

(Plotted on linear paper)

35

Rheological Models

1. Newtonian Fluid:

2. Bingham Plastic Fluid:

viscosityplastic

point yield

p

y

What if

y

py

rate shear

viscosity absolute

stress shear

36

RotatingSleeve

Viscometer

37

Figure 3.6Rotating Viscometer

Rheometer

We determine rheological properties of drilling fluids in this device

Infinite parallel plates

38

Rheometer (Rotational Viscometer)

Shear Stress = f (Dial Reading)

Shear Rate = f (Sleeve RPM)

Shear Stress = f (Shear Rate)

)(f BOB

sleeve

fluid

Rate Shear the (GAMMA), of value

the on depends Stress Shear the ),TAU(

39

Rheometer - base case

N (RPM) sec-1) 3 5.11 6 10.22 100 170 200 340 300 511 600 1022

RPM * 1.703 = sec-1

40

Example

A rotational viscometer containing a Bingham plastic fluid gives a dial reading of 12 at a rotor speed of 300 RPM and a dial reading of 20 at a rotor speed of 600 RPM

Compute plastic viscosity and yield point

12-20

300600p

cp 8p

= 20 = 12

See Appendix A

41

Example

8-12

p300y

2y ft lbf/100 4

= 20 = 12

(See Appendix A)

42

Gel Strength

43

Gel Strength

= shear stress at which fluid movement begins

• The yield strength, extrapolated from the 300 and 600 RPM readings is not a good representation of the gel strength of the fluid

• Gel strength may be measured by turning the rotor at a low speed and noting the dial reading at which the gel structure is broken

(usually at 3 RPM)

44

Gel Strength

In field units,

In practice, this is often approximated to

06.1g 2ft 100/lbf

2ft 100/lbf

The gel strength is the maximum dial reading when the viscometer is started at 3 rpm.

g = max,3

45

Velocity Profiles(laminar flow)

Fig. 4-26. Velocity profiles for laminar flow: (a) pipe flow and (b) annular flow

46

“It looks like concentric rings of fluid telescoping down the pipe at different velocities”

3D View of Laminar Flow in a pipe - Newtonian Fluid

47

Table 4.3 - Summary of Equations for Rotational Viscometer

Newtonian Model

Na N

300

Nr

066.52

300a

or

48

Table 4.3 - Summary of Equations for Rotational Viscometer

300

N

or

1pNy 1

rpm 3 atmaxg

Bingham Plastic Model

300600p )(NN

300

or

12 NN12

p

p300y

or

or

49

Example 4.22

Compute the frictional pressure loss for a 7” x 5” annulus, 10,000 ft long, using the slot flow representation 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

50

Example 4.22

The average velocity in the annulus,

)52.448(7

80

)d2.448(d

qv

2221

22

_

ft/s 1.362v_

212

_

f

dd1000

vμ

dL

dp

51

Example 4.22

51.0750 psi 51

)57(1000

)000,10()362.1()15(D

dL

dpΔp

2f

fp

212

_

f

dd1000

vμ

dL

dp

52

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 drill collars and the hole wall

• Pressure loss in the annulus between the drill pipe and the hole wall

• Hydrostatic pressure difference ( varies)

53

Types of flow

Laminar

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

Turbulent

54

Turbulent Flow - Newtonian Fluid

We often assume that fluid flow is

turbulent if Nre > 2100

cp. fluid, ofviscosity μ

in I.D., piped

ft/s velocity,fluid avg. v

lbm/gal density, fluid ρ where_

μ

dvρ928N

_

Re

55

Turbulent Flow - Newtonian Fluid

25.1

25.075.1_

75.0f

d1800

v

dL

dp

Turbulent Flow - Bingham Plastic Fluid

25.1

25.0p

75.1_75.0

f

d1800

v

dL

dp

25.112

25.0p

75.1_75.0

f

dd396,1

v

dL

dp

25.112

25.075.1_

75.0f

dd396,1

v

dL

dp

In Annulus

In Pipe

56

API Power Law Model

K = consistency index

n = flow behaviour index

SHEAR STRESS

psi

= K n

SHEAR RATE, , sec-1

0

API RP 13D

57

Rotating Sleeve Viscometer

VISCOMETERRPM

3100

300600

(RPM * 1.703)

SHEAR RATE

sec -1

5.11170.3 511

1022

BOB

SLEEVE

ANNULUS

DRILLSTRING

58

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)

59

PPUMP = PDP + PDC

+ PBIT NOZZLES

+ PDC/ANN + PDP/ANN

+ PHYD

Q = 280 gal/min

= 12.5 lb/gal

Pressure DropCalculations

PPUMP

60

Power-Law Constant (n):

Pressure Drop In Drill Pipe

Fluid Consistency Index (K):

Average Bulk Velocity in Pipe (V):

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

737.039

65log32.3

R

Rlog32.3n

300

600

2737.0600 sec

017.2022,1

65*11.5

022,1

11.5

cm

dyneRK

n

n

sec

ft00.8

78.3

280*408.0

D

Q408.0V

22

61

Effective Viscosity in Pipe (e):

Pressure Drop In Drill Pipe

Reynolds Number in Pipe (NRe):

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

n1n

e n4

1n3

D

V96K100

cP53737.0*4

1737.0*3

78.3

8*96017.2*100

737.01737.0

e

616,653

5.12*00.8*78.3*928VD928N

eRe

62

NOTE: NRe > 2,100, soFriction Factor in Pipe (f):

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

So,

bReN

af

0759.050

93.3737.0log

50

93.3nloga

2690.07

737.0log75.1

7

nlog75.1b

007126.0616,6

0759.0

N

af

2690.0bRe

63

Friction Pressure Gradient (dP/dL) :

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*05837.0LdL

dPP

Pdp = 665 psi

ft

psi05837.0

78.3*81.25

5.12*8*007126.0

D81.25

Vf

dL

dP 22

64

Power-Law Constant (n):

Pressure Drop In Drill Collars

Fluid Consistency Index (K):

Average Bulk Velocity inside Drill Collars (V):

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

737.039

65log32.3

R

Rlog32.3n

300

600

2

n

737.0n600

cm

secdyne017.2

022,1

65*11.5

022,1

R11.5K

sec

ft28.18

5.2

280*408.0

D

Q408.0V

22

65

Effective Viscosity in Collars(e):

Reynolds Number in Collars (NRe):

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

Pressure Drop In Drill Collars

n1n

e n4

1n3

D

V96K100

cP21.38737.0*4

1737.0*3

5.2

28.18*96017.2*100

737.01737.0

e

870,1321.38

5.12*28.18*5.2*928VD928N

eRe

66

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

Pressure Drop In Drill Collars

NOTE: NRe > 2,100, soFriction Factor in DC (f): b

ReN

af

So,

0759.050

93.3737.0log

50

93.3nloga

2690.07

737.0log75.1

7

nlog75.1b

005840.0870,13

0759.0

N

af

2690.0bRe

67

Friction Pressure Gradient (dP/dL) :

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 22

600*3780.0LdL

dPP

Pdc = 227 psi

68

Pressure Drop across Nozzles

DN1 = 11 32nds

(in) DN2 = 11

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

2222

2

121111

280*5.12*156P

PNozzles = 1,026 psi

22

3N2

2N

2

1N

2

DDD

Q156P

69

Pressure Dropin DC/HOLE

Annulus

DHOLE = 8.5 inODDC = 6.5 in L = 600 ft

Q = gal/min

= lb/gal 8.5 in

70

Power-Law Constant (n):

Fluid Consistency Index (K):

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

DHOLE = 8.5 inODDC = 6.5 in L = 600 ft

Pressure Dropin DC/HOLE Annulus

5413.03

20log657.0

R

Rlog657.0n

3

100

2

n

5413.0n100

cm

secdyne336.6

2.170

20*11.5

2.170

R11.5K

sec

ft808.3

5.65.8

280*408.0

DD

Q408.0V

2221

22

71

Effective Viscosity in Annulus (e):

Reynolds Number in Annulus (NRe):

DHOLE = 8.5 inODDC = 6.5 in L = 600 ft

cP20.555413.0*3

15413.0*2

5.65.8

808.3*144336.6*100

5413.015413.0

e

600,1

20.55

5.12*808.3*5.65.8928VDD928N

e

12Re

n1n

12e n3

1n2

DD

V144K100

Pressure Dropin DC/HOLE Annulus

72

So,

DHOLE = 8.5 inODDC = 6.5 in L = 600 ft

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

01500.0600,1

24

N

24f

Re

ft

psi05266.0

5.65.881.25

5.12*808.3*01500.0

DD81.25

Vf

dL

dP 2

12

2

600*05266.0LdL

dPP

Pdc/hole = 31.6 psi

Pressure Dropin DC/HOLE Annulus

73

q = gal/min

= lb/gal

Pressure Dropin DP/HOLE Annulus

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

74

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

100

2

n

5413.0n100

cm

secdyne336.6

2.170

20*11.5

2.170

R11.5K

sec

ft197.2

5.45.8

280*408.0

DD

Q408.0V

2221

22

75

Effective Viscosity in Annulus (e):

Reynolds Number in Annulus (NRe):

Pressure Dropin DP/HOLE Annulus

n1n

12e n3

1n2

DD

V144K100

cP64.975413.0*3

15413.0*2

5.45.8

197.2*144336.6*100

5413.015413.0

e

044,1

64.97

5.12*197.2*5.45.8928VDD928N

e

12Re

76

So, psi

Pressure Dropin DP/HOLE Annulus

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

02299.0044,1

24

N

24f

Re

ft

psi01343.0

5.45.881.25

5.12*197.2*02299.0

DD81.25

Vf

dL

dP 2

12

2

400,11*01343.0LdL

dPP

Pdp/hole = 153.2 psi

77

Pressure Drop Calcs.- 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

78

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

79

"Friction" Pressures

0

500

1,000

1,500

2,000

2,500

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

Cumulative Distance from Standpipe, ft

"Fri

ctio

n" P

ress

ure,

psi

DRILLPIPE

DRILL COLLARS

BIT NOZZLES

ANNULUS

80

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

Cumulative Distance from Standpipe, ft

Hyd

rost

atic

Pre

ssur

e, p

si

BHP

DRILLSTRING ANNULUS

81

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

Cumulative Distance from Standpipe, ft

Pre

ssur

es,

psi

STATIC

CIRCULATING

82

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

83

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 (f):

ReN

16f

D81.25

Vf

dL

dP2

Then and

84

85d 8.25

vf

dL

dp_2

n = 1.0