petroleum engineering 405 drilling engineering · ppt file · web view2004-02-10 · pete 411...

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PETE 411Drilling Engineering

Lesson 16 - Lifting Capacity of Drilling Fluids -

- Slip Velocity -

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Lifting Capacity of Drilling Fluids - Slip Velocity -

Fluid Velocity in Annulus Particle Slip Velocity Particle Reynolds Number Friction Coefficient Example Iterative Solution Method Alternative Solution Method API RP 13D Method

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Read:Applied Drilling Engineering, Ch. 4 - all

HW #8:On the Web - due 10-14-02

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Messages from Darla-Jean Weatherford

The seniors were supposed to have submitted the drafts of their papers for the Student Paper Contest to me last Friday; a few more than half did. Will you please remind the rest that I need those papers to complete their grades for 485?

We also are looking for recruiters for the fairs in Houston, which will be 18 to 22 November this year. If they can go with us any evening or Friday morning, they need to let Larry Piper know soon so we can get t-shirts and transportation (and meals!) arranged.

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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)

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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)

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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

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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.

slipfluidparticle VVV

Density & Velocity

slipV

particleVfluid

_

V

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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

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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.

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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 LAW:

2

sfss

d)(138v

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Particle Slip Velocity - small particles

Where

cp viscosity, fluidin particle, of diameterd

lbm/gal fluid, of densitylbm/gal particle, solid of density

ft/s velocity, slipv

s

f

s

s

2

sfss

d)(138v

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Particle Slip Velocity

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

For Nre > 0.1 an empirical friction factor

may be used.

ssfRe

dv928N

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What forces act on a settling

particle?

Non-spherical particles

experience relatively

higher drag forces

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Sphericities for Various Particle Shapes Shape Sphericity

0.58 20rh 0.87 2rh 0.83 rh 0.59 r/3h 0.25 r/15h

Cylinders0.73 3*2* 0.77 2**

Prism0.81 Cube0.85 Octahedron1.00 Sphere

Sphericity =

surface area of sphere of same

volume as particle

surface area of particle

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Particle Reynolds Number, fig. 4.46

)d104.4.(Eq...........1f

d89.1vf

sss

In field units,

Based on real cuttings

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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.

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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.

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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 m

Plastic viscosity: 7 cp Circulation rate: 340 gal/min

Hole size: 7-7/8”

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Solution - Slip Velociy Problem

1. Calculate the flow velocity

2. Determine the fluid n and K values

1174 300300y pyp

18117 300600300600p p

ft/sec 3.325

)5.4875.7(448.2340

)dd(448.2qv 222

12

2

_

22

(18/11)log3.32log32.3n300

600

7101.0n

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

Solution - Slip Velociy Problem - cont’d

cp.eq 94.66K

51111)510(

511)510(K 7101.0n

300

(ADE)

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7101.0

7101.01

a

nn1

_12

a

0208.0

)7101.012(

325.35.4875.7

14494.66

(4.107) Eq. .......... 0208.0

)n12(

v

dd144K

3. Calculate the appropriate viscosity


cp eq 94.66

7 94.17 p

K

cpcpa

24

sec/ 663.12325.3

2VV ___

s ft

4. Assume a value for the slip velocity


5. Calculate the corresponding Particle Reynolds No.

17.94

cm2.54in

m10cmm5000.663)928(9.1)(1

dv928N

4

a

ssfRe

μμ

μρ

}in 1969.0{d v92.8 154N ssRe

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From graph, f = 2.0


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

1.663 ft/s 0.678vf

0.959 19.1

21.02.0

0.19691.89

(4.104d) Eq. 1f

d1.89v

s

f

sss

ρρ

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4 (ii) Assume

5 (ii) Particle

6 (ii) From graph,

7 (ii)

Subsequent iterations yield 0.56 ft/s and 0.56 ft/s again…...

678.0vs

9.62678.0*7.92NRe

7.2f

.etc.....s/ft 58.07.2

959.0vs

Solution - Slip Velocity Problem - cont’d

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1. Fully Laminar:

Slip Velocity - Alternate Method

1

fd1.89v

f

sss ρ

ρ fsa

2s

s

_

Re

Re

d82.87v

;N40f

:3N

ρρμ

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2. Intermediate;

;N22f

:300N3

Re

Re

1/3af

2/3fss

s

_

)()(d2.90

v μρ

ρρ


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3. Fully Turbulent:

f

fsss

Re

ρ)ρ(ρd1.54 v

1.5;f

:300N


NOTE: Check NRe

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For the above calculations:

d) q.(4.104.........E 1f

d1.89v

dv928N

f

sss

a

ssfRe

ρρ

μρ


NOTE: Check NRe

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Slip Velocity - Alternate Method_2

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

Method: 1. Calculate slip velocity for Intermediate

mode 2. Calculate slip velocity for Fully Turbulent

Mode. 3. Choose the lower value.

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(i) Intermediate:

ft/sec545.017.94)*(9.1

9.1)(21*0.1969*2.90v

)()(2.90dv

1/3

2/3

s

_

1/3af

2/3fss

s

_

μρρρ

(ii) Fully Turbulent:

ft/sec 0.7819.1

9.1)(210.19691.54v

ρ)ρ(ρd1.54v

s

_

f

fsss

_

Example

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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 }

5194.17

1969.0*545.0*1.9*928N :Check Re

Range OK

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Slip Velocity - API RP 13D

Iterative ProcedureCalculate Fluid Properties, n & KCalculate Shear RateCalculate Apparent ViscosityCalculate Slip VelocityExample

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Settling Velocity of Drilled Cuttings in Water

From API RP 13Dp.24

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Calculation Procedure

1. Calculate ns for the settling particle

2. Calculate Ks for the particle

3. Assume a value for the slip velocity, Vs

4. Calculate the shear rate, s

5. Calculate the corresponding apparent viscosity, es

6. Calculate the slip velocity, Vs

7. Use this value of Vs and repeat steps 4-6 until the assumed and calculated slip velocities ~“agree”

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Slip Velocity - Example

ASSUMPTIONS:

3 RPM Reading R3 3 lbf/100 ft2

100 RPM Reading R100 20 lbf/100 ft2 Particle Density p 22.5 lb/galMud Density 12.5 lb/gal

Particle Dia. = Dp 0.5 in

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1. Calculate ns for the settling particle

2. Calculate Ks for the particle

2

n

5413.0S cmsecdyne336.6

2.17020*11.5K

5413.0320log657.0nS

3

100S R

Rlog657.0n

sn100

S 2.170R11.5K

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3. Assume a value for the slip velocity, Vs

Assume Vs = 1 ft/sec

4. Calculate the shear rate, s

p

SS D

V12 1

S sec0.245.01*12

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5. Calculate the corresp. apparent viscosity:


1nsses

sK100

cp5.14724*336.6*100 15413.0es

1D

1De790,920(1D

e0002403.0V2

es

ppp

03.5

p

es03.5s

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If then:

Vs = 0.8078 ft/sec Repeat steps 4-6

1D

1D465,161D

01344.0V2

es

ppp

p

ess

1

48.1475.12*5.01

5.125.225.0*465,161

5.12*5.048.14701344.0V

2

s

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Slip Velocity - ExampleVs = 0.8078 ft/sec

4. Shear rate: s = 19.386 sec-1

5. Apparent viscosity: es = 162.65 cp 6. Slip velocity: Vs = 0.7854 ft/sec

Second Iteration - using



Third Iteration - using Vs = 0.7854 ft/sec

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Slip Velocity - ExampleVs = 0.7823 ft/sec



Fourth Iteration - using

Slip Velocity, Vs = 0.7819 ft/sec

{ Vs = 1.0, 0.808, 0.782, 0.782 ft/sec }

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Transport Ratio

? Efficiency Transport ft/min 120 velocity Fluid

ft/min 90 velocity Particle :Example

100%*velocity fluid

velocity particleEfficiency Transport

velocity fluidvelocity particle Ratio Transport

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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%

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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.]

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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.

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The End

Lesson 16 - Lifting Capacity of Drilling Fluids -

- Slip Velocity -

petroleum engineering 405 drilling engineering · ppt file · web view2004-02-10 · pete 411...

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