PETE 411 Drilling Engineering Lesson 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

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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)

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

Density & Velocity 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.

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

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 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:

Particle Slip Velocity - small particles Where

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.

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

Sphericities for Various Particle Shapes Shape Sphericity Sphericity = surface area of sphere of same volume as particle surface area of particle

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

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.

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.

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 mm Plastic viscosity: 7 cp Circulation rate: 340 gal/min Hole size: 7-7/8”

Solution - Slip Velociy Problem 1. Calculate the flow velocity 2. Determine the fluid n and K values

Solution - Slip Velociy Problem - cont’d 2. Determine the fluid n and K values - cont’d (ADE)

Solution - Slip Velociy Problem - cont’d 3. Calculate the appropriate viscosity

Solution - Slip Velociy Problem - cont’d 4. Assume a value for the slip velocity 5. Calculate the corresponding Particle Reynolds No.

Solution - Slip Velociy Problem - cont’d 6. Obtain the drag coeff., f, from the plot of f vs. Nre. From graph, f = 2.0

Solution - Slip Velocity Problem - cont’d 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…...

Slip Velocity - Alternate Method 1. Fully Laminar:

Slip Velocity - Alternate Method 2. Intermediate;

Slip Velocity - Alternate Method 3. Fully Turbulent: NOTE: Check NRe

For the above calculations: Slip Velocity - Alternate Method For the above calculations: NOTE: Check NRe

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.

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

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 } Range OK

Slip Velocity - API RP 13D Iterative Procedure Calculate Fluid Properties, n & K Calculate Shear Rate Calculate Apparent Viscosity Calculate Slip Velocity Example

Settling Velocity of Drilled Cuttings in Water From API RP 13D p.24

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, gs 5. Calculate the corresponding apparent viscosity, mes 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”

Slip Velocity - Example

Slip Velocity - Example 1. Calculate ns for the settling particle 2. Calculate Ks for the particle

Slip Velocity - Example 3. Assume a value for the slip velocity, Vs Assume Vs = 1 ft/sec 4. Calculate the shear rate, gs

Slip Velocity - Example 5. Calculate the corresp. apparent viscosity: 6. Calculate the slip velocity, Vs

Slip Velocity - Example 6. Calculate the slip velocity, Vs If y = 0.80, then: Vs = 0.8078 ft/sec Repeat steps 4-6

Slip Velocity - Example Second Iteration - using Vs = 0.8078 ft/sec 4. Shear rate: gs = 19.386 sec-1 5. Apparent viscosity: mes = 162.65 cp 6. Slip velocity: Vs = 0.7854 ft/sec Third Iteration - using Vs = 0.7854 ft/sec 4. Shear rate: gs = 18.849 sec-1 5. Apparent viscosity: mes = 164.75 cp 6. Slip velocity: Vs = 0.7823 ft/sec

Slip Velocity - Example Fourth Iteration - using Vs = 0.7823 ft/sec 4. Shear rate: gs = 18.776 sec-1 5. Apparent viscosity: mes = 165.04 cp 6. Slip velocity: Vs = 0.7819 ft/sec Slip Velocity, Vs = 0.7819 ft/sec { Vs = 1.0, 0.808, 0.782, 0.782 ft/sec }

Transport Ratio

Transport Ratio 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%

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.]

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.

Lesson 16 - Lifting Capacity of Drilling Fluids - - Slip Velocity - The End Lesson 16 - Lifting Capacity of Drilling Fluids - - Slip Velocity -