5H cuttings transport slip velocity copy

Petroleum Engineering 405 Drilling Engineering 1 Well Drilling Engineering Lifting Capacity of Drilling Fluids & Particle Slip Velocity Dr DO QUANG KHANH 2  Fluid Velocity in Annulus  Particle Slip[.]

Well Drilling Engineering

Lifting Capacity of Drilling Fluids

& Particle Slip Velocity

Dr DO QUANG KHANH

2

 Fluid Velocity in Annulus

 Particle Slip Velocity

 Particle Reynolds Number

 Friction Coefficient

 Example

 Iterative Solution Method

 Alternative Solution Method

 API RP 13D Method

Read:

Applied Drilling Engineering, Ch 4

HW # ADE 4.55, 4.56

4

Lifting Capacity of Drilling Fluids

Historically, when an operator felt that

the hole was not being cleared of cuttings

at a satisfactory rate, he would:

Increase the circulation rate

Thicken the mud

(increase YP/PV)

Lifting Capacity of Drilling Fluids

More recent analysis shows that:

 Turbulent flow cleans the hole better

 Pipe rotation aids cuttings removal

 With water as drilling fluid, annular

velocities of 100-125 ft/min are generally adequate (vertical wells)

6

Lifting Capacity of Drilling Fluids

 A relatively “flat” velocity

profile is better than a highly pointed one

 Mud properties can be

modified to obtain a flatter profile in laminar flow

e.g., decrease n

Drilled cuttings typically

have a density of about 21 lb/gal

Since the fluid density is

less than 21 lb/gal the cuttings will tend to settle, or ‘slip’ relative

to the drilling mud

VV

V

8

Velocity Profile

The slip velocity can be reduced by

modifying the mud properties such that the velocity profile is flattened:

Increase the ratio (YP/PV)

(yield point/plastic viscosity) or

Decrease the value of n

Plug Flow

 Plug Flow is good for hole

cleaning Plug flow refers

to a “completely” flat velocity profile

 The shear rate is zero

where the velocity profile

is flat

10

Participle Slip Velocity

Newtonian Fluids:

The terminal velocity of a small

spherical particle settling

(slipping) through a Newtonian

fluid under Laminar flow

conditions is given by STOKE’S

s s

d ) (

138 v

Particle Slip Velocity - small particles

Where

cp viscosity,

fluid

in particle, of

diameter d

lbm/gal fluid,

of density

lbm/gal particle,

solid of

density

ft/s velocity,

slip v

s

f s s

= µ

=

= ρ

= ρ

s s

d ) (

138 v

12

Particle Slip Velocity

Stokes’ Law gives acceptable accuracy for a particle Reynolds number < 0.1

For Nre > 0.1 an empirical friction factor

N

What forces act

14

Sphericities for Various Particle Shapes

Shape Sphericity

0.58

20r h

0.87

2r h

0.83

r h

0.59

r/3 h

0.25

r/15 h

Cylinders

0.73

3

* 2

*

0.77

2

*

*

Prism

0.81

Cube

0.85 Octahedron

1.00

particle

Particle Reynolds Number, fig 4.46

) d 104

4 (

Eq

1 f

d 89 1

v

f

s s

Slip Velocity Calculation using Moore’s graph (Fig 4.46)

1 Calculate the flow velocity

2 Determine the fluid n and K values

3 Calculate the appropriate viscosity

(apparent viscosity)

4 Assume a value for the slip velocity

5 Calculate the corresponding

Particle Reynolds number

18

Slip Velocity Calculation (using Moore’s graph)

6 Obtain the corresponding drag coeff., f,

from the plot of f vs Nre

7 Calculate the slip velocity and compare

with the value assumed in step 4 above

8 If the two values are not close enough,

repeat steps 4 through 7 using the calculated Vs as the assumed slip velocity

in step 4

Example

Use (the modified) Moore’s method to

calculate the slip velocity and the net particle velocity under the following assumptions:

Well depth: 8,000 ft Yield point: 4 lbf/100ft2

Drill pipe: 4.5”, 16.6 #/ft Density of Particle: 21 lbm/gal Mud Weight: 9.1 #/gal Particle diameter: 5,000 µm Plastic viscosity: 7 cp Circulation rate: 340 gal/min

Hole size: 7-7/8”

20

Solution - Slip Velociy Problem

1 Calculate the flow velocity

2 Determine the fluid n and K values

11 7

4

300 300

τ

1811

7300

600 300

600

p =θ −θ ⇒ θ = µ +θ = + =

ft/sec 3.325

) 5 4 875

7 ( 448

2

340 )

d d

( 448

2

q v

1

2 2 _

(18/11) log

3.32 log

32

θ

=

7101

0

n =

2 Determine the fluid n and K values - cont’d

Solution - Slip Velociy Problem - cont’d

cp

eq 94

66 K

511

11 )

510 (

511

) 510

22

7101

0 7101

0 1 a

n n

1

_

1 2

a

0208

0

) 7101

0

1 2

( 325

3

5 4 875

7 144

94 66

(4.107) Eq.

0208

0

) n

1 2

( v

d d

144 K

−

−

3 Calculate the appropriate viscosity

Solution - Slip Velociy Problem - cont’d

94.66

7

94

K

cp cp

a

µ µ

663

12

325

32

VV

_

4 Assume a value for the slip velocity

Solution - Slip Velociy Problem - cont’d

5 Calculate the corresponding Particle Reynolds No

17.94

cm 2.54

in m

10

cm m

5000 663)

928(9.1)(1

d v

928 N

4

a

s s f Re

μ ρ

[ ]

24

From graph, f = 2.0

Solution - Slip Velociy Problem - cont’d

6 Obtain the drag coeff., f , from the plot of f vs Nre

1.663

ft/s 0.678

v

f

0.959

1 9.1

21.0 2.0

0.1969 1.89

(4.104d) Eq.

1 f

d 1.89

v

s

f

s s

vs =

9 62 678

0

* 7 92

7 2

f =

etc

s / ft 58

0 7

2

959

0

Solution - Slip Velocity Problem - cont’d

f

s s

_ Re

Re

d 82.87 v

; N

40 f

: 3 N

2 Intermediate;

; N

22

f

: 300 N

s

s s

_

)(

)(

d

2.90v

μρ

28

3 Fully Turbulent:

f

f s

s s

Re

ρ

)ρ(ρ

d1.54

v

1.5;

f

:300N

For the above calculations:

d) q.(4.104

E

1 f

d 1.89

v

d v

928 N

f

s

s s

a

s s f Re

μ ρ

Slip Velocity - Alternate Method

NOTE: Check NRe

30

Slip Velocity - Alternate Method_2

If the flow is fully laminar, cuttings transport is not likely to be a problem

(i) Intermediate:

ft/sec 545

.

0 17.94)

* (9.1

9.1) (21

* 0.1969

*

2.90 v

) (

) (

2.90d v

1/3 2/3 s _ 1/3 a f 2/3 f s s s _ = − = − = μ ρ ρ ρ (ii) Fully Turbulent: ft/sec 0.781 9.1) (21 0.1969 1.54 v ρ ) ρ ( ρ d 1.54 v

_

f

f s

s s

_

=

−

=

−

=

Example

32

Example - cont’d

Intermediate: Vs = 0.545 ft/sec

Fully Turbulent: Vs = 0.781 ft/sec

The correct slip velocity is 0.545 ft/sec

{ agrees reasonably well with iterative method on p.12 }

51 94 17 1969 0 * 545 0 * 1 9 * 928 N :

Range OK

Slip Velocity - API RP 13D

Iterative Procedure

Calculate Fluid Properties, n & K

Calculate Shear Rate

Calculate Apparent Viscosity

Calculate Slip Velocity

Example

Calculation Procedure

1 Calculate n s for the settling particle

2 Calculate K s for the particle

3 Assume a value for the slip velocity, V s

4 Calculate the shear rate, γs

5 Calculate the corresponding apparent viscosity, µes

6 Calculate the slip velocity, V s

7 Use this value of V s and repeat steps 4-6 until the

assumed and calculated slip velocities ~“agree”

Slip Velocity - Example

1 Calculate ns for the settling particle

2 Calculate Ks for the particle

2

n 5413

0 S

cm

sec

dyne 336

.

6 2

170

20

* 11

5

5413

0 3

20 log

657

R

R log

657

2 170

R 11 5

38

Slip Velocity - Example

3 Assume a value for the slip velocity, Vs

Assume Vs = 1 ft/sec

4 Calculate the shear rate, γs

p

S S

D

V 12

=

5 0

Slip Velocity - Example

5 Calculate the corresp apparent viscosity:

6 Calculate the slip velocity, Vs

1 n s s

.14724

*336

ρ +

D e

790 ,

920 (

1 D

e 0002403

.

0

V

2 p

p p

03 5 es

03 5 s

40

Slip Velocity - Example

6 Calculate the slip velocity, Vs

ρ +

0 V

2

es

p p

p p

es s

5 12

* 5 0 1 5 12

5 22 5

0

* 465 ,

16

1 5

12

* 5 0

48 147 01344

.

0

V

2 s

Slip Velocity - Example

V s = 0.8078 ft/sec

4 Shear rate: γs = 19.386 sec -1

5 Apparent viscosity: µes = 162.65 cp

6 Slip velocity: V s = 0.7854 ft/sec

Second Iteration - using

4 Shear rate: γs = 18.849 sec -1

5 Apparent viscosity: µes = 164.75 cp

Third Iteration - using V s = 0.7854 ft/sec

6 Slip velocity: V s = 0.7819 ft/sec

Fourth Iteration - using

Slip Velocity, V s = 0.7819 ft/sec

{ V s = 1.0, 0.808, 0.782, 0.782 ft/sec }

Transport Ratio

? Efficiency

Transport

ft/min 120

velocity

Fluid

ft/min 90

velocity

Particle

: Example

100%

* velocity

fluid

velocity

particle Efficiency

Transport

velocity fluid

velocity

particle

Ratio Transport

44

Transport Ratio

%75

%100

*)120/

90(

=

=

efficiency Transport

A transport efficiency of 50% or higher is desirable!

Note: Net particle velocity = fluid velocity - slip velocity

In example, particle slip velocity = 120 - 90 = 30 ft/min

With a fluid velocity of 120 ft/min a minimum particle

velocity of 60 ft/min is required to attain a transport

efficiency of 50%

Potential Hole-Cleaning Problems

1 Hole is enlarged This may result in

reduced fluid velocity which is lower than the slip velocity

2 High downhole temperatures may

adversely affect mud properties downhole

[ We measured these at the surface.]

46

Potential Hole-Cleaning Problems

3 Lost circulation problems may preclude

using thick mud or high circulating

velocity Thick slugs may be the

answer

4 Slow rate of mud thickening - after it has

been sheared (and thinned) through the

bit nozzles, where the shear rate is very

high

The End