05 lifting capacity pps

© 2005 PetroSkills LLC, All Rights Reserved

LIFTING CAPACITYLIFTING CAPACITY

The primary function of the The primary function of the drilling fluid is to clean the holedrilling fluid is to clean the hole

2 © 2005 PetroSkills LLC, All Rights Reserved

Lifting CapacityLifting Capacity

Poor hole cleaning may be Poor hole cleaning may be responsible for up to 70% of all responsible for up to 70% of all drilling problemsdrilling problems

Drag forces on a particle will Drag forces on a particle will determine how fast a particle will determine how fast a particle will fall through a fluidfall through a fluid



Gravity will cause the Gravity will cause the particle to fall through particle to fall through the fluidthe fluid

When the drag forces are When the drag forces are equal to the acceleration equal to the acceleration due to gravity, the due to gravity, the particle will reach its particle will reach its terminal velocityterminal velocity



For a spherical particle, the settling For a spherical particle, the settling velocity is represented by the velocity is represented by the following equationfollowing equation

Settling velocity is a function of the Settling velocity is a function of the diameter of the particle, drag diameter of the particle, drag coefficient, viscosity of the fluid and coefficient, viscosity of the fluid and density differencedensity difference

V

d

Cs

p p f

d f

113 4

1 2

.

/



The drag coefficient can be The drag coefficient can be determined from Figure 5-1 if the determined from Figure 5-1 if the particle Reynolds number is particle Reynolds number is knownknown

Unfortunately, the settling Unfortunately, the settling velocity is required to determine velocity is required to determine the Reynolds numberthe Reynolds number

e

sfpp

VdR

46.15



Based on Figure 5-1, an equation Based on Figure 5-1, an equation can be written for portions of the can be written for portions of the graphgraph

For a Reynolds number of 1 or For a Reynolds number of 1 or less, the equation isless, the equation is

pd RC /24



Substituting into equation 5-1 Substituting into equation 5-1 yieldsyields

Equation 3 would be used very Equation 3 would be used very seldom unless the drilling fluid is seldom unless the drilling fluid is extremely thickextremely thick

V

d

s

p p f

e

8 289

2

,



Between a Reynolds number of Between a Reynolds number of 500 and 200,000, the drag 500 and 200,000, the drag coefficient can be assumed to be coefficient can be assumed to be 0.440.44

Equation 5-4 resultsEquation 5-4 results

V

d

s

p p f

f

171

1 2

/



Equation 5-4 is used when Equation 5-4 is used when drilling with thin drilling fluids drilling with thin drilling fluids such as watersuch as water

Between a Reynolds number of 1 Between a Reynolds number of 1 and 500, the drag coefficient is a and 500, the drag coefficient is a curved line and can be estimated curved line and can be estimated by the following equationby the following equation



Substituting into equation 5-1 Substituting into equation 5-1 yieldsyields

6.0

5.18

p

dR

C

V

ds

p P f

e f

346 61 6

0 6 0 4

0 71

..

. .

.



Equation 5-6 is used for most Equation 5-6 is used for most drilling fluidsdrilling fluids

Calculate the settling velocity, Calculate the settling velocity, then check to make sure the then check to make sure the particle Reynolds number particle Reynolds number corresponds to the equation corresponds to the equation usedused



In order to clean the In order to clean the hole, the drilling fluid hole, the drilling fluid velocity must exceed velocity must exceed the settling velocity of the settling velocity of the particle so that the the particle so that the net particle velocity is net particle velocity is up the holeup the hole

V V Vp f s



If the hydraulics are designed If the hydraulics are designed properly, annular velocity is properly, annular velocity is fixedfixed

Settling velocity must be Settling velocity must be changed by manipulating mud changed by manipulating mud properties such as viscosity or properties such as viscosity or mud weightmud weight

Viscosity is the preferred methodViscosity is the preferred method



Example 5-1 shows how Example 5-1 shows how increasing the mud weight can increasing the mud weight can affect hole cleaningaffect hole cleaning



Given:Given:Average cutting size 0.5 inch

diameter spheres (13 mm)Mud Viscosity = 50 cpSpecific gravity of cuttings = 2.52 (or

21 ppg) (2520 kg/m3)Case I: Mud Weight = 10.0 ppg

(1200 kg/m3)Case II: Mud Weight = 12.0 ppg

(1440 kg/m3)



Determine:Determine:The slip velocity of the cuttings in

both mud weights

Case I:Case I: Using equation 5-6Using equation 5-6

m/min) (25.91 fpm 01.851050

10215.06.346

71.0

4.06.0

6.1

sV

71.0

4.06.0

6.1

6.346

fe

fPps

dV



Case II:Case II:

Calculate the Reynolds Number Calculate the Reynolds Number to make sure you are using the to make sure you are using the correct equation (should be correct equation (should be between 1 and 500 for eqn 5-6)between 1 and 500 for eqn 5-6)

m/min) (21.32 fpm 00.701250

12215.06.346

71.0

4.06.0

6.1

sV



Both are between 1 and 500Both are between 1 and 500

e

sfpp

VdR

46.15

131

50

01.85105.046.15pR

130

50

00.70125.046.15pR



Symptoms of sloughing or hole Symptoms of sloughing or hole cleaningcleaningDrag or tight hole on trips or

connectionsHigh torque levelsFill after trips or while making

connectionsDifficulty getting logs to bottom and/or

difficulty in running casing



A sloughing problem is a hole A sloughing problem is a hole cleaning problemcleaning problem

If a hole sloughs, the cleaning If a hole sloughs, the cleaning capacity of the well must be capacity of the well must be increasedincreased



Example 5‑2 shows how the Example 5‑2 shows how the lifting capacity of the mud can be lifting capacity of the mud can be enhanced by changing the enhanced by changing the viscosity of the mudviscosity of the mudWhile developing deep gas While developing deep gas reserves in the Wind River Basin, reserves in the Wind River Basin, Wyoming, it is necessary to Wyoming, it is necessary to penetrate approximately 3,000 penetrate approximately 3,000 feet of Waltman shalefeet of Waltman shale



The Waltman is reported to be The Waltman is reported to be both water sensitive and both water sensitive and abnormally pressuredabnormally pressured

The shale was normally drilled The shale was normally drilled with a 14.0 ppg mud (1680 kg/mwith a 14.0 ppg mud (1680 kg/m33))

However, there was no indication However, there was no indication that the shale was over that the shale was over pressuredpressured



Example 5-2 shows how to Example 5-2 shows how to increase lifting capacity with increase lifting capacity with viscosityviscosity

Want a 9.0 ppg (1080 kg/mWant a 9.0 ppg (1080 kg/m33) mud ) mud to have the same lifting capacity to have the same lifting capacity as the 14.0 ppg (1680 kg/mas the 14.0 ppg (1680 kg/m33) mud ) mud in the Waltman shalein the Waltman shale



Find the particle size where the Find the particle size where the slip velocity equals the annular slip velocity equals the annular velocityvelocity

71.0

4.06.0

6.1

6.346

fe

fppsf

dVV

63.0

4.06.04.1

6.346

fp

fes

p

V

d



First must calculate the power First must calculate the power law constants for the mudlaw constants for the mud

7365.060

100log32.3log32.3

300

600

n

(2908) 6072.0511

60

511 7365.0300

nk



Calculate the viscosity at the Calculate the viscosity at the annular velocity of 98 fpm (29.9 annular velocity of 98 fpm (29.9 m/min)m/min)

v

DDk

n

n

DDph

n

phe

200

3

124.2

cp127

98

525.126072.0200

7365.03

17365.02

525.12

984.27365.0

e



Calculate maximum particle size Calculate maximum particle size and check the Reynolds Numberand check the Reynolds Number

mm) (29.5 inch 16.11421

14x127x6.346

9863.0

4.06.04.1

pd

194

127

981416.146.1546.15

e

sfpp

VdR



Calculate the viscosity of the 9.0 Calculate the viscosity of the 9.0 ppg mud (1080 kg/mppg mud (1080 kg/m33))

cp417

96.346

98

92116.1

6.346

67.1

4.04.1

6.1

67.1

4.04.1

6.1

fs

fppe

V

d

38

417

98916.145.15pR


Lifting CapacityLifting CapacityClass problemClass problemHole size = 8 ¾” (222 mm), Dp = 4 ½” (114

mm)MW = 9.8 ppg (1180 kg/m3), Q = 275 gpm

(1.04 m3/min)Particle diameter = 0.5” and 1” (13 and 25

mm)100 rpm reading = 2250 rpm reading = 17

Calculate the slip velocity for both Calculate the slip velocity for both particlesparticles



v

DDk

n

n

DD

v ph

n

phe

)(200

3

124.2

1

2log32.3

n

n

i

i

drdv

k

22

5.24

ph DD

Qv

Wdr

dv 7.1



AnswersAnswersn = 0.3718k = 3.2597 (1.5614)v = 119.6 fpm (36.5 m/min)Equivalent thickness μe = 131 cp

0.5” particle = 57 fpm (17.4 m/min)1.0” particle = 126 fpm (38.4 m/min)



Calculate Calculate nn

1

2log32.3

n

3718.017

22log32.3

n



Calculate Calculate kkCalculate the shear rate at 50 rpm

Calculate k

85507.17.1

Wdr

dv

(1.5614) 2597.385

173718.0

n

i

i

drdv

k



Calculate the annular velocityCalculate the annular velocity

22

5.24

ph DD

Qv

m/min 36.5 or fpm 6.119

5.475.8

2755.2422

v



Calculate the viscosity at an Calculate the viscosity at an annular velocity of 119.6 fpm annular velocity of 119.6 fpm (36.5 m/min)(36.5 m/min)

v

DDk

n

n

DD

v ph

n

phe

)(200

3

124.2

6.119

5.475.82597.3200

3718.03

13718.02

5.475.8

6.1194.23718.0

e

cp 131e



Calculate the slip velocity for the Calculate the slip velocity for the 0.5” diameter particle (13 mm)0.5” diameter particle (13 mm)

71.0

4.06.0

6.1

6.346

fe

fPps

dV

m/min) (17.4 fpm 578.9131

8.9215.06.346

71.0

4.06.0

6.1

sV



Calculate the slip velocity for the Calculate the slip velocity for the 1” diameter particle (25 mm)1” diameter particle (25 mm)

71.0

4.06.0

6.1

6.346

fe

fPps

dV

m/min) (38.4 fpm 1268.9131

8.92116.346

71.0

4.06.0

6.1

sV



Check the particle Reynolds Check the particle Reynolds numbernumber

e

sfpp

VdR

46.15

33

131

578.95.046.15pR

146

131

1268.9146.15pR



What happens if the hole is What happens if the hole is washed out to 12 ¼” (311 mm)?washed out to 12 ¼” (311 mm)?The annular velocity will be reduced

in the washout

22

5.24

ph DD

Qv

m/min 15.8 or fpm 52

5.425.12

2755.2422

v



Calculate the viscosity at an Calculate the viscosity at an annular velocity of 52 fpm (15.8 annular velocity of 52 fpm (15.8 m/min)m/min)

v

DDk

n

n

DD

v ph

n

phe

)(200

3

124.2

52

)5.425.12(2597.3200

3718.03

13718.02

5.425.12

524.23718.0

e

cp 322e



Calculate the slip velocity for the Calculate the slip velocity for the 0.5” diameter particle (13 mm)0.5” diameter particle (13 mm)

71.0

4.06.0

6.1

6.346

fe

fPps

dV

m/min) (11.9 fpm 398.9322

8.9215.06.346

71.0

4.06.0

6.1

sV



Calculate the slip velocity for the Calculate the slip velocity for the 1” diameter particle (25 mm)1” diameter particle (25 mm)

71.0

4.06.0

6.1

6.346

fe

fPps

dV

m/min) (26.2 fpm 868.9322

8.92116.346

71.0

4.06.0

6.1

sV



Check the particle Reynolds Check the particle Reynolds numbernumber

e

sfpp

VdR

46.15

9

322

398.95.046.15pR

40

322

868.9146.15pR



The CCI is the Carrying Capacity The CCI is the Carrying Capacity Index for a drilling mudIndex for a drilling mud

There are only three hole There are only three hole cleaning variables that can be cleaning variables that can be controlled at the rigcontrolled at the rigMud weightAnnular VelocityViscosity



The hole cleaning variables that The hole cleaning variables that cannot be controlled on the rig cannot be controlled on the rig are:are:Diameter of particleDensity of particleTo an extent, hole enlargement



From empirical data, it was From empirical data, it was found that hole cleaning was found that hole cleaning was usually adequate when the usually adequate when the product of the mud weight, product of the mud weight, viscosity and annular velocity viscosity and annular velocity was equal to approximately was equal to approximately 400,000400,000The equation determining the The equation determining the CCI is:CCI is:



= Mud weight in ppg

K = drilling fluid viscosity, equivalent cp

= Annular velocity, feet per minute

000,400

vKCCI f

f

v

SI Units

000,000,14

vKCCI f



YPPV

YPPVn

2log32.3

)(511 )1( YPPVK n

)479.0

(511 )1( YPPVK n

SI Units



If the CCI is 1.0 or greater, hole If the CCI is 1.0 or greater, hole cleaning is assumed to be cleaning is assumed to be adequateadequate

The value of The value of KK can also be can also be determined from a chart of yield determined from a chart of yield point and plastic viscositypoint and plastic viscosity



Example 5-3 in the book with the Example 5-3 in the book with the 14 ppg mud (1680 kg/m14 ppg mud (1680 kg/m33), and ), and annular velocity of 98 fpm (29.9 annular velocity of 98 fpm (29.9 m/min)m/min)PV = 100 – 60 = 40YP = 60 – 40 = 20 (10)n = 0.7365

Calculate the K valueCalculate the K value



Calculate the K valueCalculate the K value

)(511 )1( YPPVK n

cpK 310)2040(511 )7365.01(



310



Calculate the CCICalculate the CCI

000,400

vKCCI f

1.1

000,400

9831014CCI



Class problemClass problemHole size = 8 ¾” (222mm)Dp = 4 ½” (114mm)

MW = 9.8 ppg (1180 kg/m3)Q = 275 gpm (1.04 m3/min)Plastic Viscosity = 14Yield Point = 12 (6)

Calculate the Carrying Capacity Calculate the Carrying Capacity Index or CCIIndex or CCI



Calculate the annular velocityCalculate the annular velocity

22

5.24

ph DD

Qv

m/min 36.5 or fpm 6.119

5.475.8

2755.2422

v



Determine Determine KK from the graph from the graph



12

14

276



Or you can calculate the Or you can calculate the K K value value

)(511 )1( YPPVK n

cpK 276)1214(511 )6211.01(

YPPV

YPPVn

2log32.3

6211.0

1214

12142log32.3

n



Calculate the CCICalculate the CCI

May need more viscosity or May need more viscosity or annular velocityannular velocity

000,400

vKCCI f

81.0

000,400

6.1192768.9CCI


Directional WellsDirectional Wells

Hole cleaning in a Hole cleaning in a vertical well is a vertical well is a function offunction ofAnnular velocityParticle diameterMud viscosity, andMud density

71.0

4.06.0

6.1

6.346

fe

fPps

dV



If the annular velocity of the drilling If the annular velocity of the drilling fluid exceeds the settling velocity of fluid exceeds the settling velocity of the particle, the particle will be the particle, the particle will be carried out of the holecarried out of the hole

If not, the particle must be ground If not, the particle must be ground smaller until the settling velocity is smaller until the settling velocity is lower than the annular velocitylower than the annular velocity

VVp p = V= Vff – V – Vss



In a directional well, In a directional well, the particle velocity the particle velocity is still a function of is still a function of the velocity of the the velocity of the fluid and settling fluid and settling velocity but they are velocity but they are no longer directly no longer directly opposingopposing

The particle will The particle will seek the low side of seek the low side of the holethe hole



A cuttings bed A cuttings bed will form on the will form on the low side of the low side of the hole unless the hole unless the annular velocity annular velocity is high enough is high enough to erode the to erode the cuttings bedcuttings bed

ShakerWellbore

Cuttings

Mud

Cuttings Bed



After a cuttings bed is formed, the After a cuttings bed is formed, the fluid in the annulus will have to fluid in the annulus will have to erode the cuttings bed in order to erode the cuttings bed in order to carry the cuttings up the hole carry the cuttings up the hole

The bed will continue to grow The bed will continue to grow narrowing the annular space and narrowing the annular space and causing an increase in the annular causing an increase in the annular velocity until the rate of erosion velocity until the rate of erosion equals the rate of depositionequals the rate of deposition



Experiments were conducted in Experiments were conducted in the laboratory to determine how the laboratory to determine how mud viscosity, flow regime and mud viscosity, flow regime and annular velocity affects hole annular velocity affects hole cleaning in a directional wellcleaning in a directional well

Three drilling fluids were used. Three drilling fluids were used. The first was water, which has a The first was water, which has a very low viscosity and is always very low viscosity and is always in turbulent flowin turbulent flow



The second fluid was a lightly The second fluid was a lightly gelled mud with a low viscosity. gelled mud with a low viscosity. The viscosity was low enough so The viscosity was low enough so that the fluid was in turbulent flow that the fluid was in turbulent flow even at lower annular velocitieseven at lower annular velocities

The third fluid was a higher The third fluid was a higher viscosity mud. Even at high flow viscosity mud. Even at high flow rates, the flow was still laminarrates, the flow was still laminar



Water PV = 1 YP = 0, always turbulentMud PV = 3 YP = 2 (1), always

turbulentMud PV = 19 YP = 17 (8), always

laminar

ResultsResults

0° and 10°0° and 10°Wells with inclinations between 0° and

10° behave the same as vertical wells



Increasing annular velocity and viscosity will improve hole cleaning

71.0

4.06.0

6.1

6.346

fe

fPps

dV



0

10

20

30

40

0 20 40 60 80Inclination, degrees

Cu

ttin

gs

Co

nc

en

tra

tio

n

Turb. Water 115'/min Turb. Mud 115'/min Lam. Mud 115'/min

Lam. Mud 172'/min Turb. Mud 229'/min Lam. Mud 229'/min

Laminar Mud PV= 19 YP= 17Turbulent Mud PV= 3 YP= 2Turbulent Water PV= 1 YP= 0



10° to 30° 10° to 30° At velocities

less than 120 fpm (37 m/min), the cuttings will settle to the low side of the hole and slide down the wellbore

ShakerWellbore

Cuttings

Mud



Within a short distance, they will again end up in the higher velocity portions of the annulus and be carried up the hole

ShakerWellbore

Cuttings

Mud



The hole cleaning capacity of the mud at this inclination is not as efficient as vertical wells

At annular velocities above 120 fpm (37 m/min), the cuttings are not able to form a bed on the low side of the hole, but rather are carried up the wellbore along the low side in slugs or dunes



0

10

20

30

40


Cut

tings

Con

cent

ratio

n




At flow rates in excess of 180 fpm (55 m/min), the cuttings are carried smoothly along the low side of the hole



30° to 60°30° to 60°Hole cleaning is the most critical at

inclinations between 30° and 60° with the inclinations between 40° and 50° being the most difficult

A cuttings bed forms at 40° with an annular velocity less than 150 fpm (46 m/min)

At 50°, a bed would form at annular velocities of 180 fpm (55 m/min)



Not only can a cuttings bed form rapidly at these inclinations, but the cuttings slide down the wellbore on the low side of the hole when the pump is turned off

ShakerWellbore

Cuttings

Mud

Cuttings Bed

Slumped Cuttings Bed



In directional wells with inclinations less than 40°, the cuttings will fall to the bottom of the hole

Poor hole cleaning will be evidenced by fill on bottom



In high inclination or horizontal wells, the cuttings will fall to a maximum inclination

Poor hole cleaning will be evidenced by excessive drag while pulling the bottomhole assembly through the section where the cuttings quit falling

While tripping in the hole, bridges will be encountered in this section



6060oo to 90 to 90oo

Above an inclination of 60°, cuttings bed development does not get any worse

A cuttings bed will build up reducing the annular area which increases the annular velocity

As the annular velocity increases, the drilling fluid will erode the bed faster

At some point, an equilibrium will be reached between the deposition and erosion of the cuttings bed



Annular velocityAnnular velocityAnnular velocity is the variable that

will affect hole cleaning the mostIncreasing the viscosity may actually

reduce hole cleaning at lower flow rates

At higher flow rates, viscosity makes less of a difference

0

10

20

30

40


Cu

ttin

gs

Co

nc

en

tra

tio

n






Fluids in turbulent flow have relatively flat velocity profiles; whereas, the laminar velocity profile is much more pointed

In laminar flow, there can be a significant difference between the velocity of the fluid in the center of the annular space as compared to the velocity near the pipe and hole walls

Laminar Turbulent



Pipe movementPipe movementDrill pipe movement is an important hole

cleaning consideration in directional wells Both rotation and reciprocation will

increase the hole cleaning capacity in a directional well

When reciprocating the drill pipe, the annular velocity around the tool joint increases aiding hole cleaning



As an example, if the annular velocity in a 4 1/2 (114mm) by 8 1/2 inch (216mm) annulus is 120 fpm (36.6 m/min), then the annular velocity around 6 1/4 inch (159mm) tool joints would be 208 fpm (63.4 m/min) or a 73 percent increase



Rotation will also aid hole cleaningWhile drilling with a steerable system

in the oriented mode (slide mode), the drag in a horizontal well increased

After the connection was made, rotation was resumed and the drag in the well decreased

In this case, the increased drag was due to a cuttings buildup in the well



As in vertical wells, washouts As in vertical wells, washouts will impair hole cleaningwill impair hole cleaning

The annular velocity in a The annular velocity in a washout will be reduced making washout will be reduced making hole cleaning more difficulthole cleaning more difficult



If the washout is at an inclination of If the washout is at an inclination of 35° to 55°, the cuttings accumulation 35° to 55°, the cuttings accumulation can slide down the hole when the can slide down the hole when the pump is turned offpump is turned offHole cleaning in formations that are Hole cleaning in formations that are sensitive to hole erosion can be sensitive to hole erosion can be difficult difficult The high annular velocities required The high annular velocities required to clean a directional well can to clean a directional well can enlarge the hole causing a reduction enlarge the hole causing a reduction in annular velocity in annular velocity



However, it should be However, it should be remembered that the formation remembered that the formation of a cuttings bed will reduce hole of a cuttings bed will reduce hole size causing an increase in size causing an increase in annular velocity anyway, which annular velocity anyway, which can still lead to erosioncan still lead to erosion

05 lifting capacity pps

Engineering

petroskills llc

spherical particle

drag coefficient

reynolds number ofbetween

particle willdrag forces

results v d s p p f

fluid andcoefficient

fluidthe fluid