85
1 PETE 661 Drilling Engineering Lesson 4 Wellbore Hydraulics, Pressure Drop Calculations

4. Wellbore Hydraulics, Pressure Drop Calculations

Embed Size (px)

Citation preview

Page 1: 4. Wellbore Hydraulics, Pressure Drop Calculations

1

PETE 661Drilling Engineering

Lesson 4

Wellbore Hydraulics,

Pressure Drop Calculations

Page 2: 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

Page 3: 4. Wellbore Hydraulics, Pressure Drop Calculations

3

Assignments:

READ: ADE Ch. 4

HW #3: On the Web. - Axial Tension

Due 09-20-04

Page 4: 4. Wellbore Hydraulics, Pressure Drop Calculations

4

)DD(052.0pp 1ii

n

1ii0

Fig. 4-3. A Complex

Liquid Column

D052.0p

pD052.0p 0

Page 5: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

PPUMP = ?

Page 6: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 7: 4. Wellbore Hydraulics, Pressure Drop Calculations

7

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

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

Page 8: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 9: 4. Wellbore Hydraulics, Pressure Drop Calculations

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 )

Page 10: 4. Wellbore Hydraulics, Pressure Drop Calculations

10

Axial Forces in Drillstring

Fb = bit weightF1 & F1 are pressure forces

Page 11: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 12: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 13: 4. Wellbore Hydraulics, Pressure Drop Calculations

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?

Page 14: 4. Wellbore Hydraulics, Pressure Drop Calculations

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’

Page 15: 4. Wellbore Hydraulics, Pressure Drop Calculations

15

Cross-sectional area of collars,

22 in2.43144*

490

147A

2

1

537735243 in...

AAarea alDifferenti 2

A2

A1Example – cont’d

Page 16: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 17: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 18: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 19: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 20: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 21: 4. Wellbore Hydraulics, Pressure Drop Calculations

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]

Page 22: 4. Wellbore Hydraulics, Pressure Drop Calculations

22

Axial Load with FBIT = 68,000 lbf

Page 23: 4. Wellbore Hydraulics, Pressure Drop Calculations

23

Page 24: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 25: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 26: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 27: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 28: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 29: 4. Wellbore Hydraulics, Pressure Drop Calculations

29

Laminar Flow of Newtonian Fluids

A

F

L

V

Experimentally:

Page 30: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 31: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 32: 4. Wellbore Hydraulics, Pressure Drop Calculations

32

Shear Stress vs. Shear Rate for a Newtonian Fluid

Slope of line

.

Page 33: 4. Wellbore Hydraulics, Pressure Drop Calculations

33

Apparent Viscosity

Apparent viscosity =

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

/

Page 34: 4. Wellbore Hydraulics, Pressure Drop Calculations

34

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

(Plotted on linear paper)

Page 35: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 36: 4. Wellbore Hydraulics, Pressure Drop Calculations

36

RotatingSleeve

Viscometer

Page 37: 4. Wellbore Hydraulics, Pressure Drop Calculations

37

Figure 3.6Rotating Viscometer

Rheometer

We determine rheological properties of drilling fluids in this device

Infinite parallel plates

Page 38: 4. Wellbore Hydraulics, Pressure Drop Calculations

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(

Page 39: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 40: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 41: 4. Wellbore Hydraulics, Pressure Drop Calculations

41

Example

8-12

p300y

2y ft lbf/100 4

= 20 = 12

(See Appendix A)

Page 42: 4. Wellbore Hydraulics, Pressure Drop Calculations

42

Gel Strength

Page 43: 4. Wellbore Hydraulics, Pressure Drop Calculations

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)

Page 44: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 45: 4. Wellbore Hydraulics, Pressure Drop Calculations

45

Velocity Profiles(laminar flow)

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

Page 46: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 47: 4. Wellbore Hydraulics, Pressure Drop Calculations

47

Table 4.3 - Summary of Equations for Rotational Viscometer

Newtonian Model

Na N

300

Nr

066.52

300a

or

Page 48: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 49: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 50: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

dL

dp

Page 51: 4. Wellbore Hydraulics, Pressure Drop Calculations

51

Example 4.22

51.0750 psi 51

)57(1000

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

dL

dpΔp

2f

fp

212

_

f

dd1000

dL

dp

Page 52: 4. Wellbore Hydraulics, Pressure Drop Calculations

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)

Page 53: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 54: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 55: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 56: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 57: 4. Wellbore Hydraulics, Pressure Drop Calculations

57

Rotating Sleeve Viscometer

VISCOMETERRPM

3100

300600

(RPM * 1.703)

SHEAR RATE

sec -1

5.11170.3 511

1022

BOB

SLEEVE

ANNULUS

DRILLSTRING

Page 58: 4. Wellbore Hydraulics, Pressure Drop Calculations

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)

Page 59: 4. Wellbore Hydraulics, Pressure Drop Calculations

59

PPUMP = PDP + PDC

+ PBIT NOZZLES

+ PDC/ANN + PDP/ANN

+ PHYD

Q = 280 gal/min

= 12.5 lb/gal

Pressure DropCalculations

PPUMP

Page 60: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 61: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 62: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 63: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 64: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 65: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 66: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 67: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 68: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 69: 4. Wellbore Hydraulics, Pressure Drop Calculations

69

Pressure Dropin DC/HOLE

Annulus

DHOLE = 8.5 inODDC = 6.5 in L = 600 ft

Q = gal/min

= lb/gal 8.5 in

Page 70: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 71: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 72: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 73: 4. Wellbore Hydraulics, Pressure Drop Calculations

73

q = gal/min

= lb/gal

Pressure Dropin DP/HOLE Annulus

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

Page 74: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 75: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 76: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 77: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 78: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 79: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 80: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 81: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 82: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 83: 4. Wellbore Hydraulics, Pressure Drop Calculations

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

Page 84: 4. Wellbore Hydraulics, Pressure Drop Calculations

84

Page 85: 4. Wellbore Hydraulics, Pressure Drop Calculations

85d 8.25

vf

dL

dp_2

n = 1.0