Lift Cap Drill Fluid

*PETE 411

Drilling EngineeringLesson 16 - Lifting Capacity of Drilling Fluids - - Slip Velocity -

*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

*Read:Applied Drilling Engineering, Ch. 4 - all HW #8: On the Web - due 10-14-02

*Messages from Darla-Jean WeatherfordThe 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.

*Lifting Capacity of Drilling FluidsHistorically, 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)

*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.Density & Velocity

*Velocity ProfileThe 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.

*Participle Slip VelocityNewtonian Fluids:The terminal velocity of a small spherical particle settling (slipping) through a Newtonian fluid under Laminar flow conditions is given by STOKES LAW:

*Particle Slip Velocity - small particlesWhere

*Particle Slip VelocityStokes Law gives acceptable accuracy for a particle Reynolds number < 0.1

For Nre > 0.1 an empirical friction factor may be used.

*What forces act on a settling particle?Non-spherical particles experience relatively higher drag forces

* Sphericities for Various Particle Shapes Shape SphericitySphericity =

surface area of sphere of same volume as particle

surface area of particle

*

*Particle Reynolds Number, fig. 4.46In field units,Based on real cuttings

*Slip Velocity Calculation using Moores 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.

*Slip Velocity Calculation (using Moores 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.

*ExampleUse (the modified) Moores 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 mm

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

Hole size: 7-7/8

*Solution - Slip Velociy Problem1. Calculate the flow velocity2. Determine the fluid n and K values

*2. Determine the fluid n and K values - contdSolution - Slip Velociy Problem - contd(ADE)

*3. Calculate the appropriate viscositySolution - Slip Velociy Problem - contd

*4. Assume a value for the slip velocitySolution - Slip Velociy Problem - contd5. Calculate the corresponding Particle Reynolds No.

*From graph, f = 2.0Solution - Slip Velociy Problem - contd6. Obtain the drag coeff., f, from the plot of f vs. Nre.

*4 (ii) Assume5 (ii) Particle 6 (ii) From graph, 7 (ii)

Subsequent iterations yield 0.56 ft/s and 0.56 ft/s again...Solution - Slip Velocity Problem - contd

*1. Fully Laminar:Slip Velocity - Alternate Method

*2. Intermediate;Slip Velocity - Alternate Method

*3. Fully Turbulent:Slip Velocity - Alternate MethodNOTE: Check NRe

*For the above calculations:Slip Velocity - Alternate MethodNOTE: Check NRe

*Slip Velocity - Alternate Method_2If 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.

*(i) Intermediate:(ii) Fully Turbulent:Example

*Example - contdIntermediate: Vs = 0.545 ft/secFully Turbulent: Vs = 0.781 ft/sec

The correct slip velocity is 0.545 ft/sec { agrees reasonably well with iterative method on p.12 }Range OK

* Slip Velocity - API RP 13DIterative ProcedureCalculate Fluid Properties, n & KCalculate Shear RateCalculate Apparent ViscosityCalculate Slip VelocityExample

*Settling Velocity of Drilled Cuttings in WaterFrom API RP 13Dp.24

*Calculation Procedure1. Calculate ns for the settling particle2. Calculate Ks for the particle3. Assume a value for the slip velocity, Vs4. Calculate the shear rate, gs5. Calculate the corresponding apparent viscosity, mes6. Calculate the slip velocity, Vs7. Use this value of Vs and repeat steps 4-6 until the assumed and calculated slip velocities ~agree

*Slip Velocity - Example

501

Slip Velocity a la APIPressure Drop via API10/1/94

ASSUMPTIONS:

Drill PipeDrill CollarsDC/HOLE AnnulusDP/HOLE AnnulusCritical Flow Rate in DP/HOLE Annulus

R_3 =6

R_100 =41

R_300 =80A.2n_p = 3.32 log(R_600/R_300) =0.6776783424ok0.6776783424okA.2n_a = 0.657 log(R_100/R_3) =0.5483536224ok0.5483536224okA.9N_Re_crit = 2,100

R_600 =128

A.3K_p = 5.11 R_600/1,022^n_p =5.9729815758ok5.9729815758okA.3K_a = 5.11 R_100/170.2^n_a =12.5270843903ok12.5270843903okTRIAL AND ERROR:

Q =400gal/min

RHO =12.5lb/galA.4V_p = 0.408 * Q/D^2 =8.9257508788ok26.112okA.5V_a = 0.408 * Q/(D_2^2 - D_1^2) =0ok3.453968254okQ, gal/minV, ft/secN_Re

DP_OD =5inA.6Mu_ep = 100 K_p (96 V_p/D)^(n_p-1)*((3n+1)/4n)^nA.7Mu_ea = 100 K_a (144 V_a/(D_2 - D_1))^(n_a-1)*((2n+1)/3n)^n2802.1971,044

DP_ID =4.276in3002.3541,154

L_DP =15,000ft= 100*2.017 *.....116.7649417802ok69.4888569292ok= 100*2.07 *.....0ok152.5465823962ok3502.7461,446

4003.1381,756

A.8N_Re = 928 D V_p RHO/Mu_ep =3791.6477158251ok10,897okA.9N_Re = 928 (D_2 - D_1) V_a RHO/Mu_ea =0ok919.2674716692ok4503.5312,086

(i)=0.67767834244603.6092,154

A.10a = (log n_p + 3.93)/50 =0.0752ok0.0752A.11f_a = 24/N_Re0ok0.0261077442ok4523.5462,099

452.13.5472,100

(ii)=5.9729815758dyne secn/cm2b = (1.75 - log n_p)/7 =0.2741394848no!0.2741394848no!A.13P/L = f V^2 RHO/(25.81 * (D_2 - D_1)) =0ok0.0430982857ok

f_p = a/N_Re,p^b =0.007857no!0.0058822898no!P_a = 0.08149 * 600 =0ok646.4742860019ok

(iii)=8.9257508788ft/sec

A.12P/L = f V^2 RHO/(25.81 D) =0.0708943836no!0.7769779441no!

DP1,063

(iv)=116.7649417802cpP_p = 0.058379 * 11,400 =1063.4157544384no!466.186766458no!DC466

Bit2,0943,624

DC/Hole0.0

(v)=3791.6477158251NozzlesDP/Hole6460.0

DP_Nozzles =2094.0159467368

TOTAL0.00.0

=0.0752204721A.14ok

=0.2741394848

(vi)=0.0078566607

=0.0708943836psi/ft

(vii)=1,063.42psi

DC_ID =2.5in

L_DC =600ft

A.2n_p = 3.32 log(R_600/R_300) =0.7365378487ok0.7365378487okA.2n_a = 0.657 log(R_100/R_3) =0.5413080428ok0.5413080428ok

D_hole =8.5in

A.3K_p = 510 R_300/511^n_p =201.2705715622ok201.3302552312okA.3K_a = 5.11 R_100/170.2^n_a =6.3359909041ok6.3359909041ok

D_n1 =1132nds

D_n2 =1132ndsA.4V_p = Q/(2.448*D^2) =8.0050310019ok18.3006535948okA.5V_a = 0.408 * Q/(D_2^2 - D_1^2) =3.808ok2.1969230769ok

D_n3 =1232nds

A.6Mu_ep = 100 K_p (96 V_p/D)^(n_p-1)*((3n+1)/4n)^nA.7Mu_ea = 100 K_a (144 V_a/(D_2 - D_1))^(n_a-1)*((2n+1)/3n)^n

= 100*2.017 *.....ok3813.1427770144ok= 100*2.07 *.....55.2070902898ok97.6445906036ok

T_kop =7000ft

BUR =10deg/100 ftA.8N_Re = 928 D V_p RHO/Mu_ep =6634.2141597758ok13,904okA.9N_Re = 928 (D_2 - D_1) V_a RHO/Mu_ea =1600.2582193027ok1043.9618840035ok

L_hor =3000ft

A.10A.11f_a = 24/N_Re0.0149975796ok0.0229893451ok

R_1 =572.9577951308ftA.13P/L = f V^2 RHO/(25.81 * (D_2 - D_1)) =0.0406300823ok0.0081311509ok

f_p =0.00713844650.0057743472

Length of coiled tubing = T_kop + R*Pi/2 + L_horP_a = 0.08149 * 600 =24.3780493909ok92.6951203442ok

=10900ft

A.12P/L = f V^2 RHO/(25.81 D) =0.0586311280.3747886166

DP668

P_p = 0.05873 * 11,400 =668.3948595762224.8731699537DC225

Pressure Drop via APIBit2,1012,995

DC/Hole24

ASSUMPTIONS:NozzlesDP/Hole93117

A_tot =0.2960582921

R_3 =3TOTAL3,1123,112

R_100 =20P_n = 8.311*10^-5 * rho * q^2/2101.2687831312

R_300 =39

R_600 =65Trial f(1/f)^0.5f calc.Trial f(1/f)^0.5f calc.

Q =280gal/min0.0019.12279924330.0120155530.00110.740.0086702756

RHO =12.5lb/gal0.01201555312.55455708770.00634449720.008670275613.720.0053117585DC/HOLE Annulus3/100DP/HOLE Annulus

0.006344497211.67306441280.00733888760.005311758513.040.005876841

DP_OD =4.5in0.007338887611.87403842720.00709256130.00587684113.180.0057530936

DP_ID =3.78in0.007092561311.82691322720.00714919560.005753093613.150.0057788166A.2n_a = 0.657 log(R_100/R_3) =0.5483536224ok0.5483536224

L_DP =11,400ft0.007149195611.83789135360.00713594180.005778816613.160.0057734101

0.007135941811.83533002340.00713903080.005773410113.160.0057745438A.3K_a = 5.11 R_100/170.2^n_a =1250.2569548092ok1250.2569548092

DC_OD =6.5in0.007139030811.83592740040.00713831020.005774543813.160.0057743059

DC_ID =2.5in0.007138310211.83578806310.00713847820.005774305913.160.0057743558A.5V_a = 0.408 * Q/(D_2^2 - D_1^2) =3.808ok2.1969230769

L_DC =600ft0.007138478211.83582056270.0071384390.005774355813.160.0057743454

0.00713843911.83581298230.00713844820.005774345413.160.0057743476

D_hole =8.5in0.007138448211.83581475040.00713844610.005774347613.160.0057743471A.13P/L =0.1050550964ok0.0265658908

0.007138446111.8358143380.00713844650.005774347113.160.0057743472

D_n1 =1132nds0.007138446511.83581443420.00713844640.005774347213.160.0057743472P_a =63.0330578628ok398.49

D_n2 =1132nds0.007138446411.83581441170.00713844650.005774347213.160.0057743472

D_n3 =1232nds0.007138446511.8358144170.00713844650.005774347213.160.0057743472

0.007138446511.83581441570.00713844650.005774347213.160.0057743472

0.007138446511.8358144160.00713844650.005774347213.160.0057743472

APIDP668

DC225

Bit2,1012,995

DC/Hole63

DP/Hole398462

TOTAL3,4563,456

RED BOOKDP665

withDC227

3/100Bit1,0261,918

DC/Hole32

DP/Hole153185

TOTAL2,1032,103

502

WQ#4 405/502Pressure Drop via API10/1/94

ASSUMPTIONS:

R_3 =6

R_100 =41

R_300 =80

R_600 =128

Q =600gal/min

RHO =12.5lb/gal

DP_OD =5in

DP_ID =4.276in

L_DP =15,000ft

(i)=0.6776783424

(ii)=5.9729815758dyne secn/cm2

(iii)=13.3886263182ft/sec

(iv)=102.4600210144cp

(v)=6481.525970902

=0.0752204721

=0.2741394848

(vi)=0.0067827344

=0.1377086326psi/ft

(vii)=2,065.63psi

Slip Velocity

Slip Velocity a la APIAPI RP 13D - APPENDIX B10/1/94

Third Edition, June 1, 1995

ASSUMPTIONS:

3 RPM ReadingR33lbf/100 ft2=147.4764752715

100 RPM ReadingR10020lbf/100 ft2

Particle Densityrp22.5lb/gal

Mud Densityr12.5lb/gal

Particle Dia. =Dp0.5in

Slip VelocityVs1ft/sec(assumed)

(i)=0.5413OK

(ii)=6.3360dyne secn/cm2OK

First(iii)=24.0sec-1OK

Iteration

(iv)=147.48cpOK

(v)OK

Vs=0.8077546268ft/sec

Second(iii)=19.386111044sec-1OK

Iteration

(iv)=162.6495693985cpOK

(v)

=0.7853643498ft/secOK

Third(iii)=18.8487443946sec-1OK

Iteration

(iv)=164.7603681034cpOK

(v)

Vs=0.7823133957ft/sec

Fourth(iii)=18.7755214972sec-1

Iteration

(iv)=165.0547906367cp

(v)

Vs=0.7818890643ft/sec

More than close enough!

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*Slip Velocity - Example1. Calculate ns for the settling particle2. Calculate Ks for the particle

*Slip Velocity - Example3. Assume a value for the slip velocity, VsAssume Vs = 1 ft/sec 4. Calculate the shear rate, gs

*Slip Velocity - Example5. Calculate the corresp. apparent viscosity:6. Calculate the slip velocity, Vs

*Slip Velocity - Example6. Calculate the slip velocity, VsIf y = 0.80, then:Vs = 0.8078 ft/secRepeat steps 4-6

*Slip Velocity - ExampleVs = 0.8078 ft/sec4. Shear rate: gs = 19.386 sec-15. Apparent viscosity: mes = 162.65 cp 6. Slip velocity: Vs = 0.7854 ft/secSecond Iteration - using4. Shear rate: gs = 18.849 sec-15. Apparent viscosity: mes = 164.75 cp 6. Slip velocity: Vs = 0.7823 ft/secThird Iteration - usingVs = 0.7854 ft/sec

*Slip Velocity - ExampleVs = 0.7823 ft/sec4. Shear rate: gs = 18.776 sec-15. Apparent viscosity: mes = 165.04 cp 6. Slip velocity: Vs = 0.7819 ft/secFourth Iteration - usingSlip Velocity, Vs = 0.7819 ft/sec{ Vs = 1.0, 0.808, 0.782, 0.782 ft/sec }

*Transport Ratio

*Transport RatioA 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 Problems1. 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.]

*Potential Hole-Cleaning Problems3. 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 EndLesson 16 - Lifting Capacity of Drilling Fluids - - Slip Velocity -