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4 DESCRIBED AS A SUBSTANCE’S RESISTANCE TO FLOW

5 The frictional drag of one layer of liquid sliding over another layer of liquid divided by the relative velocity of the layers

6 VISCOSITY Drilling Fluid Viscosity Varies With Shear Rate. Funnel Viscosity - Seconds per Quart. Water = 26 seconds +/- ½ second Plastic Viscosity - Centipoise. (one gram /cm-sec) PV =  600 -  300 Apparent Viscosity - Centipoise. (  600÷2) Effective Viscosity - The Measured or Calculated Viscosity at a Given Shear Rate. Low Shear Rate Viscosity - Centipoise.

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8 ... is the velocity gradient of a fluid It is the speed of 1 layer of liquid sliding by another layer of fluid divided by the distance between them. Is dependent on the average velocity of the fluid in the geometry in which it is flowing Higher Shear rate usually cause a greater resistive force of Shear Stress Measured in: 1/sec (reciprocal second)

9 The rate at which one layer is moving past the next layer is the Shear Rate. Shear Rate is therefore a velocity gradient.  (sec -1 ) = Velocity B (ft/sec) - Velocity A (ft/sec) Distance (ft)

10 When a fluid is flowing, a force exists in the fluid which opposes the flow. This force is known as SHEAR STRESS. It is a frictional force which develops as one layer of fluid slides by another.

11 Shear Stress is reported as the pound of force per hundred square feet required to maintain the Shear Rate.  (lb/100ft²) = 1.0678 x  The difference is small, therefore conversion is not used when reported  = viscometer dial reading

12 Internal resistance to an applied stress Measured in: dynescm²

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14 Shear Stress Shear Rate dyne / cm² 1 / sec = = dyneseccm² = Poise

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16 Dial Units RPM 511 x RPM 300 = 1/sec

17 Dial Reading x 1.0678 = lb/100 ft 2

18 Dial Reading x 1.0678 511 x RPM 300 Viscosity = = lb sec/100 ft 2

19 Viscosity = 300 x Dial Reading RPM = Centipoise

20 Thixotropic Behavior

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23 Shear stress is directly proportional to shear rateShear stress is directly proportional to shear rate Yield stress (stress required to initiate flow) is always zeroYield stress (stress required to initiate flow) is always zero Will not suspend cuttings and weight materialsWill not suspend cuttings and weight materials

24 Viscosity is dependent on the shear rate Do not exhibit a linear relationship Shear thinning When in laminar flow they are thinner at high shear rates than at low shear rates. At low shear rates, particles link together, increasing the resistance to flow. At higher shear rates the linking bonds are broken.

25 Example 1 600 DR = 40 300 DR = 20 Example 2 600 DR = 68 300 DR = 40

26 Rheological Models Bingham Plastic Power Law Herschel - Buckley Casson Robertson - Stiff

27 Bingham Plastic Model Describes a fluid in which a finite force is required to initiate flow (Yield Point). Then it exhibits a constant viscosity with increasing shear rate (Plastic Viscosity)

28 Bingham Plastic Model R P M FANN DIAL READING 300600 40 25 10 0 600 = 40 300 = 25 PV = 15 YP = 10 YP = Stress as zero Shear Rate Bingham YP Bingham PV slope

29 Bingham Plastic Model Accurately describes SS/SR relationship of most fluids at high shear rates (> 300 rpm). At low shear rates the initial gel strength and YP should be the same if all fluids were Bingham.

30 Bingham Plastic F = YP + PV F = YP + PV PV=Plastic Viscosity YP=Yield Point YP=Yield Point R=RPM Rotary Speed R=RPM Rotary Speed F=Dial Reading at Speed R F=Dial Reading at Speed R R300

31 Plastic Viscosity Described as that part of resistance to flow caused by mechanical friction Affected by: Solids concentration Size and shape of the solids Viscosity of the fluid phase

32 Plastic Viscosity Bingham Plastic theoretical lowest limit of shear thinning of a fluid A good approximation the bit nozzle viscosity

33 Plastic Viscosity Increased by: Hydratable Drill Solids Clays, shales Inert Drill Solids Sand, limestone, etc. Colloidal Matter Starch, CMC

34 Recommended Range of Plastic Viscosity in Various Weight Muds

35 Plastic Viscosity Increased by: Particles breaking thus increasing surface area and more friction Weight material to increase density

36 Area Increase by Breaking of Solids

37 Effect of Particle Size on Viscosity

38 Plastic Viscosity Decreased by: Removal of Solids Shale shaker Desanders, desilters, clay jectors, centrifuges Lowering of gel strength allows larger particles to settle out Dilution of solids with water

39 How Solids Affect Mud Viscosity

40 Yield Point Resistance to flow due to dispersion or attraction between solids Affected by: Type of solids and associated charges Concentration of these solids Dissolved salts

41 Yield Point Increased by: Contaminants Salt, cement, gyp, etc. neutralizes charges of clay particles causing flocculation Clay particles fracturing causing residual forces to be left on particle edges resulting in flocculation

42 Yield Point Increased by: Hydratable drilled clay and shale increasing active solids Insufficient or overtreatment of chemicals Additional solids causing attractive forces to increase due to closeness of solids

43 Yield Point Decreased by: Neutralization of broken bond valences by adsorption of negative ions on edges of particles (tannins, lignins, phosphates, lignosulfonates)

44 Yield Point Decreased by: Contamination by calcium or magnesium removed by precipitation of the ion causing flocculation - soda ash, sodium bicarbonate, phosphates Water can be used, but it lowers mud weight (expensive)

45 Interpretation of VG Values Increasing YP May be accompanied by small or no changes in PV, can be reduced by the addition of chemicals only Increasing PV May be accompanied by small or no changes in YP, can be reduced by dilution or solids control equipment

46 Interpretation of VG Values Simultaneous Large Increases in Both YP and PV Can be controlled by treating for both values

47 Power Law More complicated that BPM Does not assume a linear relationship between SS/SR Try to solve the shortcomings at low shear rates compared to BPM ”The SS increases as a function of the SR mathematically raised to some power”

48 Power Law Model SHEAR RATE SHEAR STRESS n = 1 n = 0.5 n = 0.25 k

49 Power Law F = K (RPM) n F =Shear Stress, dial units RPM =Shear Rate, rotary speed K =Consistency Index n =Power Law Index ”The SS increases as a function of the SR mathematically raised to some power”

50 Power Law Equation

51 n Value Indicates the shear thinning ability of a fluid Influenced by viscosifier concentration As “n” decreases, more shear thinning

52 n Value As the velocity profilebecomes flatter the fluid velocity will be higher over a larger ares of the annulus. This improves hole cleaning. n = 1.0 n = 0.25

53 n Value General Equation n = log(      log(     

54 K General Equation K =       n

55 Power Law Fluids SHEAR RATE SHEAR STRESS n = 1 n = 0.5 n = 0.25

56 Power Law Model F = K(RPM) n

57 n is Affected by: Hydratable solids (clays) Chemicals (thinners, contaminants, etc.)

58 What Causes n to Increase Removal of hydratable solids Addition of chemical thinners

59 What Causes n to Decrease Addition of hydratable solids Chemical contamination (salt, etc.)

60 K Value Consistency Index Indicates a system’s viscosity at low shear rates Influenced by viscosifier and solids concentration

61 n and K Value for Drill Pipe and Annulus 5.11 R 300 511 K p = npnp log(R 600 /R 300 log(1022/511) n p = 3.32 log R 600 R 300 = or 5.11 R 600 1022 npnp

62 n and K Value for Drill Pipe and Annulus 5.11 R 100 511 K a = nana n a = log(R 100 /R 3 ) log(170.2/5.11) = 0.657 log R100 R3 or 5.11 R 3 511 nana

63 K is Affected by: Hydratable solids (clays) Non-hydratable solids (barite) Chemicals

64 What Causes K to Increase Addition of hydratable and non- hydratable solids Chemical contamination (salt, etc.)

65 What Causes K to Decrease Removal of hydratable and non- hydratable solids Addition of chemical thinners

66 Rheological Models

67 Thixotropic Behavior

68 Gel Strength Gel structure develops as a result of charged particles assuming equilibrium positions (positive to negative) with respect to each other Gel strength is a function of time, temperature, concentration and strength of attractive particles

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70 Gel Strength Assists in performing the functions of: Decreasing the settling rate of cuttings when circulation is interrupted Releasing cuttings at the surface... also entrained gas or air Reducing adverse effects of the mud with the formation

71 Problems Attributed to High Viscosity and Gel Strengths More pump pressure to break circulation Increased pressure losses Lost circulation by pressure surges Swabbing of shale and formation fluids into wellbore Abrasive sand being carried in mud

72 Thixotropy Thixotropy: Due to the clay platelets arranging themselves in positions of free energy. With time a greater force than the YP is required to get the fluid moving again Gel Strength (lbs/100ft 2 ) Yield Point (lbs/100 ft 2 ) Shear Stress Gel Strength Yield Point Decreasing Shear Rate Increasing Shear Rate

73 Effective Viscosity 300 x Dial Reading RPM EV = EV = Effective Viscosity, centipoise

74 100 centipoise (cp) = 1 poise Centipoise is the dimension used to express: Plastic viscosity Apparent viscosity Effective viscosity

75 The dimensions of lb/100 sq ft are used for expressing: Yield point Initial gel 10-minute gel

76 RHEOLOGICAL FLOW REGIMES

77 Flow Regime Determination Reynold’s number Wellbore geometry Fluid properties

78 Reynolds’ Number Function of: Mud weight Hole stability Flow rate Fluid viscosity

79 D=Hydraulic diameter, in V=Velocity, ft/sec MW=Mud weight, lb/gal  =Viscosity, cp R N = 928 DV (MW) 

80 Stage 1: No Flow

81 Stage 2: Plug Flow

82 Stage 3: Transition (Plug to Laminar)

83 Stage 4: Laminar Flow

84 Stage 5: Transition (Laminar to Turbulent)

85 Stage 6: Turbulent Flow

86 Reynold’s Number < 2100Laminar 2100 - 4100Transitional > 4100Turbulent Normally we assume turbulent flow when the Reynolds Number > 2100

87 Type of Flow

88 Solids Causing Change in Flow Profile

89 Hydraulics The study of how fluids exert force and do work

90 Circulating System Shear Rates Shear Rates in the Circulating System SectionShear Rate s-1 Pressure Drop Equivalent of Section Rheometer rpm Drill String 170 -10,000 20 - 45 100 and above Bit 10,000 -100,000 40 - 75 None Annulus 5 - 170 2 - 5 3 - 100 Pits 0 - 3 n/a 0 - 3 SCE* 170 - 1000 n/a 100 - 600 *Solids Control Equipment This demonstrates the importance of low end rheologies in cleaning the annulus.

91 Using data obtained at 600 rpm and 300 rpm, the parameters to be used for inside pipe calculations are: = 3.32 log (R 600 /R 300 ) np =np =np =np = log (R 600 / R 300 ) log (1022/511) 5.11 R 300 511 Kp =Kp =Kp =Kp = npnpnpnp 5.11 R 600 1022 npnpnpnp =

92 Using data obtained at 100 rpm and 3 rpm, the parameters to be used for annulus calculations are: = 0.657 log (R 100 /R 3 ) na =na =na =na = log (R 100 / R 3 ) log (170.2/5.11) 5.11 R 100 170.2 Ka =Ka =Ka =Ka = nananana 5.11 R 3 5.11 nananana =

93 Annular GeometryGeometry

94 Mud Prop Geometries Flow Rate Calculate K value Calculate Bulk Velocity Calculate Effective Viscosity Calculate n value Calculate Reynolds Number If R num  2100If R num  2100 Calculate Laminar Fric Factor Calculate Turbulent Fric Factor Calculate Interval  P Save Interval  P Calculate  P for next interval until last interval calculated After  P Calculated for all intervals Total Interval  Ps Annular: Calculate ECD End Sequence of Pipe and Annular CalculationsCalculations