Schedule:Tuesdays: 08:15 10:00 P11 Exercises 1 hour Fridays: 12:15 14:00 P11 TPG4220 Drilling Fluid, 2014 Teacher: Pl Skalle pal.skalle@ntnu.no Stud. ass.: Thomas Sahl thomas.a.sahl@gmail.com Stud. ass.: NN nn@stud.ntnu.noText book: 1. BookBoone: Drilling Fluid2. ADE from SPE; Recommended:Chapter 2, minus 2.1.4. and page 46 left column 2.1.6, 7, 8, 9 2.3.6, 8, 9, 10 2.3.13.1 and 3 + page 77 2.5.11 and 12Chapter 3Chapter 4, minus 4.1.2 + page 117 fight column + page 118 4.4.5 and 9 4.11.3 + page 151 + 2 4.12 and 13 4.15.1 and 3 + page 172 and 175 - 183Exercise:1. Solve the exercise and compare with existing solution2. Write a short report, and for each of the exercises make these statements:Was the exercise text clear enough? If not make an improved suggestion.Was existing solution correct? If not make a clear text of how to solve it correctly3. Make a sketch of totally two completely new exercises, includingMotivation for this suggestionTaskExpected results

Ch. 1IntroductionCh. 2CirculationCh. 3Mud propertiesCh. 4HydraulicsCh. 5Important tasks

Remove cuttings from under the bit; drill fastTransport cuttings out of wellECDStable wellbore

Mud types Tasks:

1.Non dispersed. Spud mudsa2.Dispersed b, c3.Calcium treated. Gyp systems Lime systemsd4.Polymerb5.Low solidsa6.Saturated saltd Saturated salt systems Saltwater systems Brackish or seawater systems.7.Workover / Completion d8.Oil/syntheticd Invert emulsion muds Synthetic (esters, poly alpha olefins, paraffins)9.Air, mist, foam and gasa

Ch. 1. Introduction1930 1950 1970 1990 2010 2030 2050Non-con GasGasNGLsPolar oilDeep waterHeavyRegular

1. IntroductionStatus 2003

Ch. 2. Mud circulation

SPP

2. Mud circulation The pump

2. Mud Circulation Pump characteristics

2. Mud circulation Pump characteristics EmaxPmaxpMax, recommendedqMax, recommended1 600 Hp====EMax, recommended=1 440 Hp0.0487 m3/s = 2 922 l/min383 barE = p * qE = p * q (W)E = p * qE = p (psi) * q (GPM) / 1714 (Hp)

Liner sizein5 1/25 3/466 1/46 1/26 3/477 1/4Discharge pressurepsibar5555383.05085350.646703224305296.83980274.43690254.43430236.53200220.6Pump flow rate GPMm3/s4440.02154860.0307529.0334574.0362621.0392669.0422720.0454772.0487PowerHp1439.01441.81441.31441.71442.01440.31440.81441.3

2. Mud circulation Cleaning of mud 1. Settling2. Dilution3. Mechanical separation4. Chemical treatment

2. Mud circulation Cleaning of mud BARITEOutOutOutUnweighted

WeightedColloids

8.02802463220010.02502108200020.016086380050.009027925060.0072230205100.0045140125120.0037117105200.003074 68325.00184440

Ch. 3. Fluid properties Mud:FluidColloidal phaseInert phase1. Rheology controlCh. 32. Particle control (cleaning)Ch. 53. Density controlCh. 74. Filtration controlCh. 8tasksClay chemistryPolymer chemistryRheologyAdditives

3. Fluid properties 3.1. Clay chemistry Sandstone reservoir

3. Fluid properties 3.1. Clay chemistry Montmorilonite crystal

1 m = 10 m1 = 10 mdH2O = 3 -6-10 0.3 m

9 12 1 m200 stillsheared3. Fluid properties 3.1. Clay chemistry Montmorilonite crystal

Flokkulated suspension

Solid surface

Electrostatic potential

Repulsive force

Resultant force

Attractive force

Bulk liquid

Electrostatic layer

0

Dispersed suspension

More salt, less repulsion

Destilled waterSaturated with NaCl exchangeswelling40

20

03. Fluid properties 3.1. Clay chemistry Montmorilonite crystal

OH

O

O

Si

Si

Si

Mg

Si

O

O

Na

Al

Si

Si

+

Na

+

OH

OH

10 % NaCl10 % CaCl2H2OFlocculate: High colloid concentration High electrolyt concentration High TDispersed: Dispersants Increase pH with NaOH or KOH Precipitate cationsDispersedFlocculated

AggregatedAggre./ floccl.Aggregation: Collapse of diffusion layer -> parallell plates 20 apartSalt CaCl2 NaClGypsum CaSO2 *2H2OLimestone CaCO3Cement Ca(OH) (lime)3. Fluid properties 3.1. Clay chemistry Montmorilonite crystal. Viscosity and Filtration control

Attraktive van der Waahl3. Fluid properties 3.1. Clay chemistry Montmorilonite crystal

Flokkulated suspension

Solid surface

Electrostatic potential

Repulsive force

Resultant force

Attractive force

Bulk liquid

Electrostatic layer

0

Dispersed suspension

More salt, less repulsion

3. Fluid properties 3.4. Clay chemistry CEC

Type of 0charge: + - Anionic, behave much like clay:Type ofnatural , CMC < 80 C (break up at high T)Polymers:synthetic, PHPA > 250 C003. Fluid properties 3.2. Polymer Chemistry

5111022300600102030170100t = F/A = shear stress (Pa)g = v/y = shear rate (s-1)gRPMMSpeedShear rateReadingRPM s-1 Q

102228300 51114100 170 53. Fluid properties 3.3. Rheology MeasuringFA

RPM g q t t rpm s-1 - lb/100ft2 Pa

600 1022 140 148.4 71.05300 511 98 103.9 49.75200 39.60100 170.3 27.3960306 8.143 5 13 13.8 6.613 10 5 23 11.683 10 5 48 24.37tOFU = Q * 1.06tSI = tOFU * 0.47883. Fluid properties 3.3. Rheology Measuring

Shear Stress (Pa)

Shear Rate (s -1)

Shear Rate (s -1)

Shear Stress (Pa)

Shear Stress (Pa)

Shear Rate (s -1)

Shear Rate (s -1)

Shear Stress (Pa)

Newtonian model: Bingham plastic model:

Power law model:Herschel & Bulkley model: 1 point

2 points

2 points

3 points3. Fluid properties 3.3. Rheology 4 Models

50

050

0

Newtonian modelOne constant is sufficient to describe this model. Bingham - 2 data points - oil fieldBingham - 2 data points - SI Power law - 2 data point - SI PaHerschel Bulkley - 3 data points - oil field Herschel Bulkley - 3 data points - SI IterationCh. 3.3. Rheology Models fitted to 2 data points50

0600102271.05300 51149.753 5 6.61Models fitted to 3 data points

Ch. 3.3. Rheology Models fit to more than 3 points y x . 71.05102249.75 511....Step 1Step 2:Tool Menu Data Analysis RegressionH&B requires non-linear regression:

is a measure of mechanical frictionis a measure of inter-colloidal forcesCh. 3.3. Rheology 2-point Bingham modelTime dependency300300600trThixothropic RheophecticShear thinning150 300

3. Drilling fluid rheology 3.4. Filtration control In sandstone and in the labLab testing of filter cakeReal formation Cuttings(100 2000 mm)

Sulfonated asphaltene(50 400 mm)

Barite(10 -74 mm)

PAC / Clay (0.1 2 mm)1. Spurt loss

2. Initial filter is formed3. Filter cake is formeddporesdpVol %

0 - 4

0 - 2

5 25

1 - 5

Emulsion droplet-size distribution vs. shear during mixing

2. Fluid loss control are polymers

Low-M and low-charge to make higher viscosity in filtrate (only!)Viscosity controlFluid loss controlFriction reducertmud 2mud 1filtrate 2

filtrate 13. Fluid properties 3.5. Additives 1. Viscosity control

Flocculant

Deflocculant

Base fluid + 0.1 % HECm-eff 15 20Dp 700200v(r)C2nH4n+2On+1M > 20 000Polyetylene oxid (PEG)3. Fluid properties 3.5. Additives 3. Friction reducer

Ch. 4. Hydraulic frictionObjective:

Find total loss -> chose pumpFind nozzle p-loss for cleaningMonitor and control ECDNecessary knowledge:

Circulation systemch 2Fluid rheologych 3Fluid mechanicsch 4Plan:

1. Fluid mechanics /Head loss 2. Laminar pressure losspipeannuluseffective viscosityexample3. Turbulent pressure loss4. Singularity loss

Continuity equation: Microscopic

General formCartesian coordinatesCylindrical coordinatesContinuity equation: Macroscopic 4. Hydraulic friction 4.1. Fluid mechanics Concervation equationsView 1: Microscopic. Langrangian observe a point following the flowView 2: Macroscopic. Eularian use a control volumein out

Momentum equation: Microscopic. Navier-StokeGeneral formCartesian coordinates(only x-component)Cylindrical coordinates(only the r-component) For steady state laminar flow, this expression reduces toF is the resultant shear force acting on the system, M the total mass flow The steady state, one dimensional pipe flow the macroscopic expression reduces to: .Momentum equation: Macroscopic 4. Hydraulic friction 4.1. Fluid mechanics Concervation equations

Energy equation: Macroscopic(for incompressible flow)Energy equation: Microscopic First bracket-term is called useful or total head. is called the hydraulic head while the velocity head.hvpp specific density.4. Hydraulic friction 4.1. Fluid mechanics Concervation equations

Find average velocity expressed by the other parameters presentedv1 = 1.4 m/sdA = 2prdrRe = rvd / mL = ID * 0.06 Re L = ID * 4.4 Re **1/6 4. Hydraulic friction 4.1. Fluid mechanics Refreshingv0 = vmaxv0 = 2vzv0= 2 vzv = 1.4 . 2 = 2.8 m/s

Entrance length

Singularity loss

EGL (friction less)

EGL (friction)

,4. Hydraulic friction 4.1. Fluid mechanics Head lossTotal head: D velocity head: r . v2 = Total head minus friction

= -vy = vz = 0 (vx = v)1. State2. Constitutive 3. Continuity4. MomentumSeparating, integrating (from r = 0 to r (dv/dx =0), separating, integrating now from r = r to R (v(r) = 0)and taking leading to total flow after integration4. Hydraulic friction 4.2. Laminar flow Pipe. Newtonian. MicroscopicHagen Poiseulle

p1A1 - p2A2 F = mg . sin QDpuniversal = 4/d * tw * Dx(p1 - p2 ) Apressure = F = t Ashear4. Hydraulic friction 4.2. Laminar flow Pipe. Newtonian. Microscopic

4. Hydraulic friction 4.2. Laminar flow Pipe. Bingham

1. Exact solution of momentum eqn.2. Simplified solution of momentum eqn. Slot flowdhydr = douter - dinner4. Hydraulic friction 4.2. Laminar flow; friction loss derivation Annular. Newtonian modle

Ri

Ro

R

r

b

2h

x

y

2 h

b

do

di

Step 1Step 2 use DpNewtonian instead of DpBingham tgm plm eff4. Hydraulic friction 4.2. Laminar flow: Two useful techniques: a. Use Newtonian model, meff, on non-Newtonian fluids

tw = 8v/d . m = m . gStep 1: Find shear rate Step 2: Plot rheology and find tw, pipeStep 3: Find Dp from DpuniversalStep 0: Check if laminar flow4. Hydraulic friction 4.2. Laminar flow: Two useful techniques: b. Use Universal model, independent of rheological modelDpuniversal = Dp Newtonianq = 1000 l/min = 0.0167d = 0.1v= q/A =g = 8v/ d

Shear Stress (Pa)

Shear Rate (s -1)

Shear Rate (s -1)

Shear Stress (Pa)

4. Hydraulic friction 4.2. Laminar flow

Hydraulic friction formulas

q = 4000 Step 1: Turn data into SIStep 2: Find rheology constantsStep 3: Check type of flowStep 4: Find Alternative; use universal 4. Hydraulic friction 4.2. Laminar flow; friction loss derivation Flow example

SymbolRPMqtwtwUnitrpm(-)s-1lb/100 ft2PaData600300200100631409878541613102251134017010.25.1148.4103.982.757.217.0 13.871.0549.7539.6027.398.146.613 103 1023485.15.124.450.911.6824.37

200 m

d = 26 = 0.673 m

200 m

pump

bit

q = 4000 lpm (two pumps)

d = 9 = 0.229 m

DC

Shear Stress (Pa)

Shear Rate (s -1)

Shear Rate (s -1)

Shear Stress (Pa)

Re = dvr / mefffF = shear stress / kinetic pressureF refers to FanningM refers to Moodyhf = friction head > Darcy-Weisback eqn.HGLEGL4. Hydraulic friction 4.3 Turbulent flow Introduction

800

2100

Blasius

Moore

Metzner & Reed

fF= 0.046 * Re-0.2fF = a * Regeneral -ba = f(n)b = f(n)c1 = 0.026 (e/d)0.25 + 0.133 (e/d)c2 = 22 (e/d)0.44 c3 = 1.62 (e/d)0.3416/Re4. Hydraulic friction 4.3 Turbulent flow

Models curve fitColebrook

FANNING FRICTION FACTOR

Re general

Moody Friction Factor

Transition range

Laminar flow

Fully turbulent flow

Smooth

Re

Roughness (mm)

losses (in bits, TJ, MWD, PDM, stabs etc)Dpq4. Hydraulic friction

4.3 Singularity losses

Models

Ch. 5. Cleaning below the bitHydraulic MechanicalTwo ways of finding the optimum:

Necessary cleaning energyMaximize a design criteria:

Rebit ROPIFBHHP1. Cuttings removalCuttings removalBoundary conditions of modelLiner-wise optimizationWell-wise optimization

Increasing ROP

2. Concerning the friction lossDploss = K1D q m80 90 % of max test pressure5. Cleaning below the bit

5.2. Boundary conditions 1. Concerning the pump

Dploss = K1D q m Step 1: log Dp1 = log K1D + m log q1 log Dp2 = log K1D + m log q2Step 1: Create test data and estimate DpbitStep 3: Estimate DplossStep 4: Estimate m and K15. Cleaning below the bit 5.2. Boundary conditions 2. Concerning hydraulic friction loss

Pump rate Pump pressureEstimated bit pressure loss

vmax < vturb (around drill collars)erosion of weak wall

vmin > qr (cuttings concentration < 0.04)poor hole cleaning5. Cleaning below the bit 5.2. Boundary conditions 3. Conserning annular flow velocityDploss = K1D q m

Dploss = K1D q m Strategy: Find an expression of Ae = A1 +A1 + A15. Cleaning below the bit 5.3. Optimizing ROP. Liner-wise Model development

Dploss = K1D q m1. Boundary conditions:4. Read qopt 2. Find 2 data points 5. Cleaning below the bit 5.3. Optimizing ROP. Liner-wise Graphic solution3. Enter pbit, opt = ppump . m / (m+2)a.b.c. q r and q mlog-log plot

INTERMEDIATE RESULTINPUTOPTIMALOUTPUTLiner was selected on basis of section qr and qturb. Each liner is treated individually, i.e. Liner-wise 5. Cleaning below the bit 5.3. Optimizing ROP. Liner-wise Program + field exampleThis example was from an operation in 1980, and the specific pump liner delivered only 210 bars

This is normally how you og about to determine the optimal hydraulic program:

Step 1. Select the correct q

Comment: The liner is selected for other prioritiesthan to be covering the max flow rate, but rather closer to the optimal flow rate. The intricate selection procedure then becomes much easier (select q-opt.theoretically directly)

Step 2. Calculate de, nozzle

start

end

Q-opt.th < Q-max

D = D-tot

Liner size d-nozzleQ-opt

Q-opt.th< Q-r

no

P-max, m, K-1Q-max, Q-rD-tot, D-DQ-opt.th

yes

D = D-0

no

Q-opt = Q-r

Q-opt = Q-max

Q-opt = Q-opt.th

P-bit= P-max *m/(m+2)

D = D + D-D

yes

yes

no

6 liner pre-selected D = 3 000 m qr = 0.2 m3 /s ppump = 93 . 105 Pa qturb = 0.4 m3 /s ppump = 302 . 105 Pa m = 1.7 K1 = 2.4 . 1065. Cleaning below the bit 5.3. Optimizing ROP. Liner-wise Program + new examplepbit opt = 322 . 1.7 / 3.7 = 148322 - 148 = 1741000

100

100.001 0.01 0.1

p = constE = const5. Cleaning below the bit

5.4. Optimizing complete well Model

p

q

q

bit

loss

m1

m2

q

p

D

D

0

p

I

II

r

q

t

q

Section end depthqr.v q r.hq-turb smallest liner26 1000 0.037 0.042 0.0476.75 6.517.5 2000 0.032 0.038 0.040 6.5 - 612.25 3300 0.022 0.032 0.033 5.758.5 5000 0.016 0.025 0.027 5.5

Typical questions

Which liner should be used, and what is the optimal flow rate?Find graphically the pump schedule in the graph below

When should it be changed to a smaller liner dimension in the 26 section?

What is optimal nozzle size in 4 000 m depth?

Assume K1 = 4.0 * 106, m = 1.7

Principles to find optimal solution graphically:

Select q closest to theoretical optimum

If 1. is not conclusive, then choose smallest possible liner obtain highest ppump

3. Find details like opt. bit size

5. Cleaning below the bit 5.4. Optimizing complete well Example

Sectionend depth qr qturb smallest liner26 1000 0.037 0.0476.75 6.517.5 2000 0.032 0.040 612.25 3300 0.022 0.030 58.5 5000 0.016 0.022 5Which liner should be used?

5. Cleaning below the bit 5.4. Optimizing complete well Solution to the Example

When change to a smaller liner dimension in the 26 section? Find theoretically. K1 = 4.0 * 106, m = 1.6c. Optimal nozzle size in 4000 m depth? Find theoretically. K1 = 4.0 * 106, m = 1.7Sectionend depth qr qturb 26 1000 0.037 0.04717.5 2000 0.032 0.040 12.25 3300 0.022 0.030 8.5 5000 0.016 0.022 D = 690 m21 21 21.00 21.3221 2221.33 21.6621 22 21.67 21.995. Cleaning below the bit 5.4. Optimizing complete well Solution to the Example

Principle:

Find RangeFind linerFind opt. values5. Cleaning below the bit 5.4. Optimizing complete well Program. Summary