dc mod 7 hydraulics presentation

Drilling and Completions SystemsModule 7: Hydraulics

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Module 7: HydraulicsLesson 1: Hydraulic PumpsLesson 1: Learning ObjectivesPumpsCirculating System for Typical Rotary Drilling RigFormula for Volumetric Displacement for CylindersClass Activities: CalculationsDuplex TypePump Displacement and Output Flow RateDouble Acting Cylinder Pump Displacement and Output Volume RatesDouble Acting Duplex PumpHorse Power Requirements

Lesson 2: Flow Velocity Lesson 2: Learning ObjectivesClass Activities: CalculationsRequired Flow RateGeometry of the Well BoreFlow Rate of the Mud PumpsViscosityNewtonian Fluid Fanning Friction Factor vs. Reynolds Number

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Lesson 3: Rheology Models (non-Newtonian Models)Lesson 3: Learning ObjectivesLaminar and Turbulent Flow Patterns in Pipe Plastic Fluid Class ActivityThe Hagen-Poiseuille Equation for Laminar FlowOptional: Density and Viscosity Video

Lesson 4: Pressure Drop in PipeLesson 4: Learning ObjectivesFlow in a Cylindrical Annulus Pressure DropΔP Parasitic ComponentsCritical Reynolds Number versus Hedstrom PlotFigure Reynolds Number versus Fanning Friction Factor Bit Nozzles

Module 7: Hydraulics Cont.

Lesson 5: Newtonian Fluids (Possible Homework)Lesson 5: Learning ObjectivesCalculation

Lesson 6 Plastic Fluids Calculation (Possible Homework) Lesson 6: Learning ObjectivesCalculation

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Lesson 1: Hydraulic Pumps

Image source: http://www.botta-equipment.com/content/catalog/mud-pump-and-spares

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In this lesson we will:Distinguish two types of pumpsDefine the circulating system for a typical rotary drilling Identify a working drilling rig circulation system Explain how the Double Acting Duplex Pump worksExplain how mud pumps work and label parts of a pumpCalculate the output flow rateNote: Instructor will go through a series of examples and expect students to work through remaining examples as homework or in class in groups.

Lesson 1: Hydraulic Pumps Learning Objectives

http://www.hddtrenchless.com.au/product/Mud-Pump-Parts-Pistons-and-Liners.cfm

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Two Different Types of Pumps

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Circulating System for Typical Rotary Drilling Rig

http://www.conservation.ca.gov/dog/picture_a_well/PublishingImages/DRILLING-RIGnew.gif

http://www.conservation.ca.gov/dog/picture_a_well/PublishingImages/DRILLING-RIGnew.gif

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Mud Pumps

http://www.pacificoilfield.com/images/mud_pump_working.gif

http://www.pacificoilfield.com/images/mud_pump_working.gif

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How Single Acting, Single Cylinder Pumps Work

Suction Cycle and Discharge Cycle

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Formula for Volumetric Displacement for Single Cylinder

Let D = inside the diameter of the cylinder (or piston diameter), inches S = length of the stroke (or piston movement), inches

Then, Volume displacement per stroke = () D2 s = in3/stroke

4or,Pump displacement = PD (.7854) (in2) (in) (7.48 gal)

(144 in2/ft2) (12 in/ft.) ft3= gal/stroke

If N = strokes/minthen, PD = .7854(D2)(S )(7.48)(N) = gal/min where, PD = gal/min

144 12 D = inS = in N = strokes/min

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https://encrypted-tbn1.gstatic.com/images?q=tbn

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Formula for Volumetric Displacement for Double Cylinder

Let D = inside the diameter of the cylinder (or piston diameter), inchesS = length of the stroke (or piston movement), inches

qout = (PD)(ev) =.7854(D2)(S )(7.48)(N)(2)(ev) = gal/min

144 12Where, q = gal/minD = inS = inN = strokes/min(2) = No. of cylinders

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What is the output flow rate for a 10“ × 20" Single-acting, Single- cylinder pump operating at 40 strokes per minute?

Class Activity: Example 1 Volumetric Displacement

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10"x 20" describes the pump as; D = 10"S = 20"then, PD = q =.7854(102)(20)(7.48)(40) = 272 gal/min

144 12Since the placement of fluid in the cylinder by the piston and the operation of the discharge and suction valves is not 100% efficient due to worn mechanical parts, the output volume rate will be less than the calculated pump displacement. A volumetric efficiency, ev, must be used,or,qoutput = (PD)(ev)

Example 1 Solution

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What is the output flow rate in Example #1 if the volumetric efficiency of the pump is 85%?


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qout = (PD)(ev) = 272(.85) = 231 gal/min

Example 2 Solution

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Using a flow meter, the output flow rate for a 12" x 18" Single-acting, Single cylinder pump operating at 30 strokes per minute was measured at 215 gal/min. Calculate the volumetric efficiency of the pump.


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EV = qout

PDqout = 215 gal/minPD = .7854(122)(18)(7.48)(30) = 264 gal/min

144 12Ev = 215 = 0.814 = 81.4%

264

Example 3 Solution

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Duplex Type

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Pump displacement and output flow rate can be calculated in the same manner as for the single cylinder by multiplying by 2 cylinders, or,qout = (PD)(ev) =.7854(D2)(S ) (7.48)(N)(2)(ev) = gal/min

(144) (12)Where, q = gal/minD = inS = inN = strokes/min(2) = No. of cylinders

Pump Displacement and Output Flow Rate

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Calculate the cylinder ID (or piston diameter) needed in a Single-acting, Duplex pump needed to have an output of 125 gal/min if the stroke length 16" and operates at 25 strokes per minute with a volumetric efficiency of 80%.

Class Activity: Example 4 Calculate the Cylinder ID

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qout = (PD)(ev) = 125 gal/min = .7854(D2)(16 )(7.48)(25)(2)(0.80) = gal/min144 12

D2 = 125 = 572.18

D = 7.58 in

Example 4 Solution

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Double Acting Cylinder

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Pump displacement and output volume rates can be calculated for each side of the piston is the total volume of the cylinder, or,

q1 = (PD1)(ev) = .7854(D2)( S )(N)(7.48)(ev) = gal/min144 12

The PD in the connector rod side of the cylinder is the total volume of the cylinder minus the volume occupied by the connecting rod, or,

q2 = (PD2)(ev) = .7854(D2 - d2)(S )(N)(7.48)(ev) = gal/min144 12

Then,qout = (PD1 + PD2)(ev) = .7854(D2 + D2 - d2)(S )(N)(7.48)(ev) = gal/min

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Pump Displacement and Output Volume Rates

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Calculate the output flow rate for a 7"x14", Double-acting, Single-cylinder pump when operating at 45 strokes/min using a 2" connector rod and a volumetric efficiency of 90%.

Class Activity: Example 5 Calculate Output Flow

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qout = (PD)(ev) = .7854(7² + 7² - 2²)(14 )(45)(7.48)(0.90) = 181 gal/min144 12

Example 5 Solution

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What length of stroke is needed to give 250 gal/min output when using a Double-acting, Single cylinder pump with an 8" cylinder diameter and a 2.25" diameter connector rod when operating at 40 strokes/min and 80% volumetric efficiency?

Class Activity: Example 6 Length of Stroke

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qout = (PD)(ev) = .7854(82 + 82 - 2.252)(S )(40)(7.48)(0.80) = 250 gal/min144 12

S = 250 = 18.7"1.34

Example 6 Solution

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Formula for Volumetric Displacement for Double-Acting, Single-Cylinder

Let D = inside the diameter of the cylinder (or piston diameter), inchesS = length of the stroke (or piston movement), inches

q1 = (PD1)(ev) = .7854(D2)(S )(N)(7.48)(ev) = gal/min144 12

Or,

q2 = (PD2)(ev) = .7854(D2 - d2)(S )(N)(7.48)(ev) = gal/min144 12

Then,qout = (PD1 + PD2)(ev) = .7854(D2 + D2 - d2)(S )(N)(7.48)(ev) = gal/min

144 12

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Double- Acting, Duplex Type

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Therefore, during this one-half of the stroke, the PD is equal to PD in front of the piston + PD in connector rod side of the piston. Since each one-half of the cycle the discharged volumes are equal, the PD for one stroke is,PD = PD of one cylinder x number of cylindersor,qout = (PD)(ev) = .7854(2D2 - d2)(S )(N)(7.48)(2)(ev) = gal/min

144 12Note: Since the two cylinders are driven by connector rods connected to a common crank, 1 stroke is equivalent to 1 revolution of the crank, or, strokes/min = RPM of the crank.

How the Double Acting Duplex Pump Works

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Calculate the output flow rate of a 6"x12" Double-acting, Duplex pump using 1¾" (1.75 in) connector rods when operating at 35 RPM and a volumetric efficiency of 90%.

Class Activity: Example 7 Calculate the Output Flow Rate

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qout = (PD)(ev) = .7854 {(2)(62) - (1.752)} (12)(35)(7.48)(2)(0.90)144 12

qout = 177 gal/min

Example 7 Solution

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At what RPM of the crank should a 8"x16", Double-acting, Duplex pump be operated to give a flow rate output of 400 gal/min when operating 85% volumetric efficiency and using 2" diameter connecting rods?

Class Activity: Example 8 (Concept of RPM)

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qout = (PD)(ev) = 400 gal/min = .7854(82 + 82 - 22)(16)(N)(7.48)(2)(0.85)144 12

N = 400 = 35 RPM11.5

To improve the efficiency of the operating conditions, a liner (or sleeve) can be inserted in the pump cylinders to decrease the inside diameter of the cylinder (also decreases the diameter of the piston).

Example 8 Solution

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Hydraulic horsepower can be calculated from the flow rate and the pressure, or,Hp = q P

1714where, Hp = hydraulic horsepower

q = flow rate, gal/minP = pressure, psi

Therefore,Pump output Hp = Hpout = qoutPp = (PD)(ev)(Pp)

1714 1714Note: Pump output flow rate can be calculated from the required upward annular circulating from the required upward annular circulating velocity and the pump outputpressure will be the pressure required to overcome the pressure drops around the circulating system (inside the drill string, annulus around the drill string, etc.)

Horse Power Requirements

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What diameter liner (inside) and pistons would be needed for the pump in Example #7 to maintain the same operating conditions with an output flow rate of 300 gal/min?

(Ex. 7: Calculate the output flow rate of a 6"x12" Double-acting, Duplex pump using 1 3/4" connector rods when operating at 35 RPM and a volumetric efficiency of 90%.)qout = (PD)(ev) = .7854 {(2)(62) - (1.752)} (12)(35)(7.48)(2)(0.90)

144 12qout = 177 gal/min

Class Activity: Example 9 (Diameter Liner)

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qout = (PD)(ev) = .7854(2D2 - 22)(16)(35)(7.48)(2)(0.85) = 300 gal/min144 12

D2 = 48.34; D = 6.95“

qout = (PD)(ev) = .7854 {(2)(D2) - (1.752)} (12)(35)(7.48)(2)(0.90) = 300 gal/min144 12.

D = 7.75

Example 9 Solution

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Calculate the output horsepower available from a 7“ x 14", Double-acting, duplex pump with 2" connecting rods operating at 40 RPM with a volumetric efficiency of 85% and an output pressure of 1500 psi.

Class Activity: Example 10 (Calculate the Output Horsepower)

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Hpout = q outPp = (PD) (ev)(Pp)1714 1714

= .7854(72+72-22)(14)(40)(7.48)(2)(0.85)(1500)( 1 )144 12 1714

Hpout = 266HpIn the above calculations, the volumetric efficiency is defined as qout

PD

Example 10 Solution

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Therefore, Hppump = HpPD = Hpout

ev

Pumps have a mechanical efficiency, em, also. The input Hp from the engine must have overcome both mechanical and volumetric losses to produce a given output Hp,then,Hpin = HPPD and HpPD = Hpout

em ev

or,Hpin = Hpout

(em)(ev)

Example 10 Solution Cont.

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What engine output horsepower will be needed to operate the pump in Example # 9 and Example #10 if the mechanical efficiency is 80%?

Class Activity: Example 11 (Engine Output)

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HpPD = Hpout = 266 = 313 Hpev (0.85)

and, Hpin = HPPD = 313 = 391Hp

em (0.80)or,Hpin = Hpout = 266 = 391 Hp

(ev)(em)(0.85)(0.80)

Example 11 Solution

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What is still unclear?What questions do you have about the topics we have discussed before we move on?

HomeworkAssignment 7.1: Self Study ReviewAssignment 7.1: Read Fundamentals of Drilling Engineering Section 5.1 Introduction to Drilling, 5.2 Hydrostatic Pressure Calculations (pp. 179 -182); Section 5.2.4 Equivalent Density Concept (pp. 187-189)Assignment 7.1: Problem Solving: Complete Problems 5.1, 5.2, 5.3 on page 296; Show Your Work!

Lesson Wrap Up

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Lesson 2: Flow Velocity

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In this lesson we will:Interpret geometry of the well boreCalculate the required output flow rate of the mud pumpsPerform a demonstration of viscosity using a rheometerDefine the ratio of viscosity of a fluidCalculate types of flowDefine the Fanning EquationRelate the Fanning Friction Factor to the Reynolds NumberApply the Hagen-Poiseuille equation for laminar flowCalculate the equivalent Reynolds Number

Lesson 2: Flow Velocity Learning Objectives

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Since drilling fluid is incompressible, the volume or flow rate will be constant at any point in the circulating system. The flow velocity will vary due to changes in cross-sectional area, or, q = A v (in consistent units)

Required Flow Rate

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Geometry of the Well Bore

From Bourgoyene et al. (1991)

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To determine the required output flow rate of the mud pumps, an experience factor is used to determine the required minimum upward velocity in the annulus necessary for the efficient removal of the cuttings. This "experience" velocity is usually given in ft./min and can be used to calculate the required flow rate,or,q = A v in consistent units or,q = 2.448 (d2

hole- d2OD pipe) v Note: 2.448 = (0.7854)×(1/122)×(60)×(7.48)

Note: 0.7854 = π/4 Where q = gal/min

d = inv = ft./sec

An Experience Factor to Determine the Required Output Flow Rate of the Mud Pumps

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If the bore hole is 8 ¾" and using 4 ½" OD drill pipe, what is the required flow rate if the minimum upward velocity is 195 ft./min.?

Class Activity: Example 12 (Calculate Required Flow Rate)

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q = A v Where, q = ft3/sec

A = ft2

v = ft./secThen Note: π = 0.7854

A = π (8.75)2 - π (4.5)2 = 0.3071ft2 44 12 4 12

v = 195 ft./min = 3.25 ft./sec60 sec/min

And,q = (0.3071) (3.25) = 1 ft3/sec = 449 gal/minOr,q = 2.448 (d2

hole - d2OD pipe) v

q = 2.448 (8.752 - 4.52) (195) = 448 gal/min60

Example 12 Solution

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Viscosity

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Viscosity of a Fluid

Viscosity of a fluid is defined as the ratio of the shearing stress to the rate of shear,or,Force / Velocity [Pressure/Rate of velocity change]Area Distance

then,

(lbf / ft./sec)(gc) = lbm/ft.-sec [Basic Engineering Units]ft2 ft.

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Laminar flow occurs when all individual particles in the fluid flow in a straight line parallel to the axis of the conductor. Under certain conditions (velocity, viscosity, density, and diameter of the conductor), Turbulent flow occurs when the particles flow in a random manner.The Reynolds Number, NR, relationship is used to determine the type of flow under given conditions,Or,NR = dρv = (diameter, ft) * (density, lbm/ft3) * (velocity, ft/sec)

μ viscosity, lbm/ft-secWhere, NR = dimensionless numberd = diameter, ft.v = velocity, ft./sec

μ = viscosity, lbm/ft.-secρ = density, lbm/ft3

Or,NR = (928) d ρ v (field units)

μWhere, NR = dimensionless numberd = diameter, inρ = density, lbm/galv = velocity, ft./secμ = viscosity,cp

Types of Flow

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Through experiment, laminar flow exists when the value of the Reynolds Number is less than 2000 and turbulent flow exists when the value of the Reynolds Number greater than 2000,Or,

NR < 2000; laminar flowNR > 2000; turbulent flow

Note: In some cases, NR < 400; laminar flow

In transition flow, between

NR > 4000; turbulent flow

Types of Flow (Cont.)

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The Hagen-Poiseuille equation states the relationship between pressure drop due to friction and other flow factors for a Newtonian fluid under laminar flow conditions in a straight, circular pipe,or, Δpf = 32 μ L v (Basic units)

gc d2

Or, Δpf = μ L v (Field units)

1500 d2

Where, Δ pf = psi L = ft. d = inμ = cp v = ft./sec

Newtonian Fluid - Laminar Flow - In Pipe

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What is the minimum diameter of a pipe needed to insure laminar flow in a pipe carrying a Newtonian fluid (μ = 20 cp & Sp. Gr. = 0.80) at a rate of 40 gal/min?

Class Activity: Example 13 (Minimum Diameter of a Pipe)

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To insure laminar flow, NR = 2000 then,NR = 2000 = 928 d ρ v;

μ

ρ = (0.80)(8.34) = 6.67 lbm/gal

v = q = 40 . 2.448(d)2 2.448(d)2

μ = 20 cpThen, 2000 = (928)(d)(6.67) ( 40 ) = (928)(6.67)(40)

20 2.448 d2 (20)(2.448)(d)d = 5053 = 2.53"

2000

Example 13 Solution

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Calculate the pressure drop in 10 miles of the pipe in Example #13 if the 2.53" line is replaced with a 6“ ID line.

Class Activity: Example 14 Calculate the Pressure Drop

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NR = 928 d ρ v Where, d = 6"μ ρ = 6.67lbm / gal μ = 20 cp

v = 40 = 0.45 ft./sec2.448d2

Then,NR = (928)(6)(6.67)(0.45) = 836 ; 836<2000, \laminar flow

20Or, Δpf = μ L v = (20)(10)(5280)(0.45) = 9 psi

1500 d2 (1500) (6)2

Example 14 Solution

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The Fanning Equation states the relationship between pressure drop due to friction and other flow factors for a Newtonian fluid under turbulent flow conditions in a straight, circular pipe, Or, Δpf = 2 f ρ L v2 (Basic units) where, f = Fanning Friction Factor

gc d f = dimensionless Or,Δpf = f ρ L v2 (Field units) where, Δpf = psi

25.8 d f = dimensionless ρ = lbm/gal

L = ft.v = ft./secd = in

Pipe, Turbulent Flow

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Fanning Friction Factor vs. Reynolds Number

Text Book p. 254

Fanning Friction Factor Curves

I. Flow inside glass tubesII. Flow inside steel pipeIII. Annular flow between steel pipeIV. Annular flow between open hole and steel pipe

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Class Activity: Example 15 (Calculate the Pressure Drop)Calculate the pressure drop in psi in 1500' of 4"ID pipe carrying oil (Sp.Gr. = 0.825 & viscosity = 12 cp) at a rate of 1550 gal/min. (The oil is a Newtonian fluid; Newtonian fluid is only described with viscosity and does not have a yield point)

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Example 15 Solution

NR = 928 d ρ v ;μ

ρ = (0.825) (8.34) = 6.88 lbm/gald = 4"

v = q = 1550 = 39.5 ft./sec2.448 (d)2 2.448 (4)2

μ = 12 cpThen,NR = (928)(6.88)(4)(39.5) = 84,064 > 2000, turbulent flow

12Then,Δpf =f ρ L v2 ; f = 0.0055 (from Curve II, f-Curves)

25.8(d) ρ= 6.88 lbm/galL = 1500 ft.v = 39.5 ft./secd = 4 inThen,Δpf = (0.0055) (6.88) (1500) (39.5)2 = 858psi

(25.8) (4)

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Class Activity: Example 16 (Calculate the Pressure Drop)

Calculate the pressure drop in Example #15 if the pipe diameter is doubled.

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Example 16 Solution

d = 8"ρ = 6.88 lbm/galv = 1550 = 9.88 ft./sec

2.448(8)2

μ = 12 cpThen,NR = (928)(6.88)(8)(9.88) = 42,053 > 2000, turbulent flow

12f = .0062 (from Curve II, f-Curves)Then,Δpf = (.0062)(6.88)(1500)(9.88)2 = 30 psi

(25.8)(8)

Note: For the same flow rate, doubling the diameter of the pipe decreases the pressure drop from 858 psi to 30 psi or 97%.

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Calculate the pressure drop in psi/100' of a 10" ID line carrying a 9 lbm/gal, 12 cPNewtonian fluid at the rate of 53 gal/min.

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Example 17 Solution

NR = 928 ρ d v ; ρ = 9 lbm/galμ d = 10"

v = 53 = 0.216 ft./sec2.448 (10)2

μ = 12 cpThen,NR = (928)(9)(10)(0.216) = 1503 ; 1503 < 2000, therefore laminar flow

(12)

Δpf = μ L v = (12) (100) (0.216) = 0.0017 psi/100'1500 d2 1500(10)2

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Calculate the pressure drop/100' for the line in Example #17 if the flow rate is doubled.

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Example 18 Solution

NR = 928 ρ d v ; ρ= 9 lbm/galμ d = 10"

v = 106 = 0.433 ft./sec2.448 (10)2

μ = 12 cpThen,NR = (928)(9)(100)(0.433)2 = 3014 ; 3014 > 2000, turbulent flow

(12)Δpf = f ρ L v2 ; f = 0.0115 (equation or curve II)

25.8 dΔpf = (0.0115)(9)(100)(0.4333)2 = 0.0075 psi/100'

(25.8)(10)

Note: Doubling the flow rate in Examples #16 & #17 changed the type of flow from laminar to turbulent and increased the pressure drop by 4+ times.

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Newtonian Fluid - Laminar Flow - In Annulus

d2 = inside diameter of the outside conductor

d1 = outside diameter of the inside conductor

Therefore,de = 4 rh = (4)(/4)(d2

2 - d2i) = (d2 - di)

π(d2 + di)The actual velocity, va, in an annulus is,va = q ft./sec

2.448 (d22 - d2

i)

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Laminar Flow in Pipe vs. Annulus

(a) (b)

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Type of Flow in an Annulus

Calculate the equivalent Reynolds Number using the equivalent diameter and the actual velocity.

Or,Nre = 757×(d 2 - di) ρ va; Nr < 2000, laminar flow

μNre =757 × ρ de va Nr > 2000, turbulent flow

μIf laminar, use the Hagen-Poiseuille equation modified for annular flow to calculate pressure drop,

Or,Δpf = μ L va = μ L va (Field units)

1000(d 2 - di)2 1000(de)2

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Calculate the pressure drop in the annulus per 1000' of 5"OD pipe suspended in a 15" hole when a Newtonian fluid (Sp.Gr. = 0.92 & viscosity = 8 cP) is flowing at a rate of 100 gal/min.

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Example 19 Solution

NR = 757 ρ de va ; ρ = (0.92)(8.34) = 7.67 lbm/galμ de = (d2 - di) = (15 - 5) = 10"

va = q = 100 __ = 0.204 ft./sec.2.448(d2

2 - di2) 2.448(152 - 52)

μ = 8 cP

Then,NR = (757)(7.67)(10)(0.204) = 1815; 1815 < 2000, Laminar flow

( 8 ) Then,

Δpf = μ L va = (8)(1000)(0.204) = 0.016 psi/1000'1000 d2

e 1000(10)2

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Newtonian Fluid - Turbulent Flow - In Annulus

The Fanning equation for turbulent flow applies only to a straight, circular pipe and cannot be used directly if the cross-sectional area of flow is an annulus. To use this equation for an annulus, the annular cross-sectional area must be expressed as an equivalent cross-sectional area of pipe which will have the same pressure drop per length at the same flow rate. This expression is the diameter of a pipe which will have the same pressure drop-flow rate relation as the equivalent cross section of the annulus,

Or,

de = (d2 – d1)

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Calculate the equivalent Reynolds Number using the equivalent diameter and the actual velocity,Or,Nre = 757 de ρ vₐ ; NR < 2000, laminar flow

μ NR > 2000, turbulent flow

If turbulent flow, use the Fanning equation to calculate pressure drop,Or,Δpf = f ρ L vₐ2 = (field units)(Determine f from NR and f-curves)

21.1 de

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With 3000' of 2“ nom. tubing (OD 2.375" & ID = 1.995") hung in 5", 18 lb./ft. (ID = 4.276") casing, salt water (Sp.Gr. = 1.05 & viscosity = 5 cP) is being pumped down the tubing and back up the annulus at a rate of 400 gal/min. Calculate the pressure drop in the annulus.

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Example 20 Solution

Nre = 757 ρ de va ;ρ = (1.05)(8.34) = 8.76 lbm/galμ de = (d2 - di) = (4.276 - 2.375) = 1.901“

va = q = 400 = 12.91 ft./sec.2.448(d2

2 - di2) 2.448(4.2762 - 2.3752)

μ = 5cP

Then,

Nre = (757)(8.76)(1.901)(12.91) = 39 902; turbulent flow5

f = 0.0070 (from Curve III, f-Curves)

Then,

Δpf = f ρ L v2a = (0.0070)(8.76)(3000)(12.91)2 = 626 psi

21.1 de (21.1)(1.901)

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HomeworkAssignment 7.1: Self Study ReviewAssignment 7.2: Read Fundamentals of Drilling Engineering Section 5.2.6 Effect of Well Deviation (pp. 192-194); Section 5.3 Steady Flow of Drilling Fluids (pp. 194 – 206)Assignment 7.2: Problem Solving, Complete Problems 5.10, 5.11, 5.12 on page 296 - 297; Show Your Work!

Lesson Wrap Up

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Lesson 3: Rheology Models (non-Newtonian Models)

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In this lesson we will:Interpret laminar and turbulent flow patterns in pipeSolve the Hagen-Poiseuille equation for laminar flowsDetermine where viscosity appears in the Fanning equation for turbulent flow Calculate the pressure drop in the annulus

Lesson 3: Rheology Models Learning Objectives

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Laminar and Turbulent Flow Patterns in Pipe

Textbook p. 246

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Plastic Fluid - Laminar Flow - In Pipe (Will discuss)Referring to the plastic fluid viscosity curve, the equation for the straight line portion of the curve is,

Δpf = m v + YBAnd the Hagen-Poiseuille equation for laminar flow is,Δpf = 32 μ v L

gc d2

Or,Slope = m = Δpf = 32 μ L

v gc d2

Then,Δpf = ( 32 μ L)(v) + YB

gc d2

And,YB expressed in equivalent pressure terms = 4 YB L

dThen,

Δpf = 32 μp L v + 4 YB L (Basic units)gc d2 d

Or,Δpf = μp L v + YB L (Field units)

1500 d2 300 d

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Plastic Fluid - Laminar Flow - In Pipe (Cont.)

The Reynolds Number equation and the Hagen-Poiseuille equation for laminar flow apply only to a Newtonian fluid flowing in a straight, circular pipe. If these equations are to be used for a plastic fluid, an equivalent viscosity, μe, which is the viscosity a plastic fluid would have if it were a Newtonian fluid, must be used,And,

μe = 5 YB d + μp ( Field Units )v

Since μe is an equivalent Newtonian viscosity, it can be used in the Reynolds Number equation,

Or,

NR = 928 dρ v ; NR < 2000, laminar flow5 YB d + μp NR > 2000, turbulent flow

v

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Plastic Fluid - Laminar Flow - In Pipe (Cont.)Setting Nr = 2000 and solving for velocity yields a critical velocity, vc, and an actual velocity below which is laminar flow and an actual velocity above which is turbulent flow,

Or, vact < vc , laminar flowvact > vc , turbulent flow

Then,NR = 2000 = 928 dρ v

This leads to a quadratic equation for v solved as:Or,

vc = 1.08 μp + 1.08 (μ2p + 9.3 ρd2 YB)0.5

ρdNote: (Field Units) - is the minus portion discarded because a negative velocity would be meaningless

And,vact = q (Field units)

2.448 d2

5 YB d + μpv

86

Class Activity: Example 21 (Pressure Drop; non-Newtonian fluid)

What is the pressure drop in 4000' of 5 ½, 17 lb./ft. casing (ID = 4.892") when carrying 9.2 lbm/gal drilling mud (Plastic viscosity = μp = 20 cP & Bingham Yield point = YB = 25 lbf/100 ft²) at a rate of 150 gal/min?

87

Example 21 Solution

vc = (1.08)(20) + {1.08 ( (20)2 + (9.3)(9.2)(4.892)2(25)}0.5 = 5.94 ft./sec.(9.2)(4.892)

And,vact = 150 = 2.56 ft./sec

2.448(4.892)2

Therefore,2.63 < 5.94 or v act < vc ; laminar flowThen,ΔPF = (20)(4000)(2.56) + (25)(4000) = 5.706 + 68.14

1500 (4.892)2 300(4.892)ΔPF = 73.8 psi

88

Class Activity: Example 22 (Maximum Flow Rate)

What is the maximum flow rate allowable through 1000' of 3" ID line carrying a 10 lbm/gal plastic fluid (μ = 30 cP & Yt = 15 lbf/100 ft²) to insure laminar flow?

89

Example 22 Solution

vc = (1.08)(30) + {1.08( (30)2 + (9.3)(10)(3)2 (4/3) (15)}0.5 = 5.86 ft./sec(10)(3)

To insure laminar flow, vc = vact = 5.86 ft./sec

Then, vact = 5.86 = q . 2.448(3)2

Or, q = 129.2 gal/min

90

Determine f using the f-Curves and the Reynolds Number, NR, calculated using the actual velocity, vact , and plastic viscosity, μP,

Or,

NR = 928 d vact (Field units)

μP

Then,

Dpf = f L v2act (Field units)

25.8 de

Determine f

91

Plastic Fluid - Turbulent Flow - In Pipe

Determine if the flow is laminar or turbulent by calculating the critical velocity, vc, and comparing to the actual velocity, vact , or,

vact < vc ; laminar flowvact > vc ; turbulent flow

If flow is turbulent, the Fanning Equation must be used to calculate the pressure in a straight, circular pipe.

92


Calculate the pressure drop in 600' of 4“ ID pipe carrying 9.0 lb./gal mud (μP

= 18 cP & YB = 25 lbf/100 ft²) at a rate of 400 gal/min.

93

Example 23 Solution

vc = (1.08)(18) + (1.08){(18)2 + (9.3)(9.0)(4)2(25)}0.5 = 6.06 ft./sec(9.0)(4)

vact = 400 = 10.2 ft./sec2.448(4)2

10.2 > 6.06 or vact > vc ; turbulent flow

Then,NR = (928)(4)(9.0)(10.2) = 18931 ; f = 0.0075

(18)And,

Δpf = (0.0075)(9.0)(600)(10.2)2 = 41 psi(25.8)(4)

94

This expression is the diameter of a circular pipe which will have the same pressure drop-flow rate relation as the equivalent diameter of the annulus, Or, Equivalent diameter, de, is defined as , rh = Cross-sectional area of flow

Wetted perimeter

Where, the wetted perimeter is the total length of surface contacted by the fluid;

The Hagen-Poiseuille Equation for Laminar Flow

95

The Hagen-Poiseuille Equation for Laminar Flow (Cont.)

d² = inside diameter of the outside conductor

d1 = outside diameter of the inside conductor

Therefore,de = 4 rh = (4)(π/4)(d2

² - d2i) = (d² - di)

π(d² + di)The actual velocity, va, in an annulus is,va = q ft./sec

2.448 (d2² - d2

i)

96

The expression for determining the critical velocity, vc, is modified when the flow is in an annulus, or,vc = 1.08μp + 1.08( μ2

p + 6.98 ρd2e YB) 0.5 where, de = (d² - d1)

ρ de

Determine if the flow is laminar or turbulent by calculating the critical velocity, vc, using the equivalent diameter, (d² - d1), and comparing to the actual velocity, vact,

Or,vact < vc ; laminar flowvact > vc ; turbulent flow

If laminar flow , the Hagen-Poiseuille equation modified for annular flow must be used to calculate pressure drop using de, vact, and μp

Δpf = μp L vact + YB L (Field units)

1000 (d² - d1)2 200 (d² - d1)


97


Calculate the pressure drop in the annulus with 3000' of 4 ½" drill pipe suspended in a 7 ⅞" hole. The mud is 10.2 lb./gal (μp = 30 cP & YB = 28 lbf/100 ft²) and the flow rate is that required to give an upward annular velocity of 165 ft./min.

98

Example 24 Solution

de = 7.875 - 4.5 = 3.375"vc = (1.08)(30) + (1.08) (30) 2 + (6.98)(10.2)(3.375)2(28) = 5.76 ft./sec

(10.2)(3.375)

vact = 165 ft./min = 2.75 ft./sec60 sec/min

2.75 < 5.76 or vact < vc ; laminar flow

Then,Δpf = (30)(3000)(2.75) + (28)(3000) = 21.73 + 93.22 = 115 psi

(1000)(3.375)2 (200)(3.375)

99


Determine if the flow is laminar or turbulent by calculating the critical velocity, vc, and comparing to the actual velocity, vact or,vact < vc ; laminar flowvact > vc ; turbulent flow

If flow is turbulent, the Fanning equation must be used to calculate the pressure in a straight, circular pipe.

100

Type of Flow in an Annulus (Cont.)

Note: Viscosity does not appear in the Fanning equation for turbulent flow except in determining the friction factor, f, from the Reynolds Number equation and the f-Curves and has little effect on the pressure drop calculation when turbulent flow exists.

Determine f using the f-curves and the Reynolds Numbers, NR, calculated using the actual velocity, vact, and plastic velocity, μp, and equivalent diameter, de

orNR = 757 de ρvact (Field units)

μpthen,Δpf = fρ L v2

act (Field units)21.1 de

101


Calculate the pressure drop in the annulus for 390' of 6 ⅝" OD drill collars suspended in a 8 ½" hole. The mud is 9.0 lb./gal (μp = 15 cP & Yb = 22 lbf/100 ft²) and the flow rate is 485 gal/min.

102

Example 25 Solution

de = 8.5 - 6.625 = 1.875"vc = (1.08)(15) + (1.08) {(15)2 + (6.98)(9.0)(1.875)2(22)} 0.5 = 5.52 ft./sec

(9.0)(1.875)

Vact = 485 = 6.98 ft./sec(2.448)(8.52 - 6.6252)

6.98 > 5.52 or vact > vc ; turbulent flow

Then,NR = (757)(9.0)(1.875)(6.98) = 5,944; f = 0.0105 (Curve IV, f-Curves)

(15)And,Δpf = (0.0105)(9.0)(390)(6.98)2 = 44.9 psi

(21.1)(1.875)

103

Instructions

Review Video

Complete questions (Handout in SharePoint)

Prepare to discuss in class.

Additional questions regarding this video may be on a quiz or test.This experiment will help you to:

Grasp the rheological models of drilling fluids, such as:Newtonian ModelBingham Plastic ModelPower Law Model

Use instruments, such as mud balance, Fann VG meter (Viscometer), pH scale and Beckman pH meter, Stirrer, and StopwatchKnow something about common chemicals used in drillingGet familiar with flow and chemical parameters, such as:

DensityViscosityGel strengthpH

Density and Viscosity Video (Optional if Lab is not Available or has not been completed in Module 3)

1. Bentonite clay is a gelling material and helps increase the viscosity of water.a. Trueb. False

2. What is the specific gravity of Bentonite clay?a. 2.56b. 2.65c. 2.78d. 2.85

3. Fresh water is also known as a Newtonian Fluid.a. Trueb. False

104

Density and Viscosity Video—Questions and Answers

a. True

b. 2.65

4. What is the reason for using 350 cc of water in lab for conducting such experiments?a. Short way of measuring things in labb. Corresponds to the oil filed unitsc. All of the above

5. What was the density of the Bentonite clay fluid?a. 8.5 lbm/galb. 8.6 lbm/galc. 8.7 lbm/gald. 8.8 lbm/gal

6. What does X denotes in the equation that describes the physical model of a Bingham Plastic Fluid?a. Densityb. Viscosityc. Revolutions per minute (RPM)d. Weight

105

Density and Viscosity Video—Questions and Answers

c. All of the above

c. 8.7 lbm/gal

c. Revolutions per minute (RPM)

106


HomeworkAssignment 7.1: Self Study ReviewAssignment 7.3: Read Fundamentals of Drilling Engineering Section 5.4 Rheological Models of Drilling Fluids (pp. 206-213 only)Assignment 7.3: Problem Solving, Complete Problems 5.20, 5.21, 5.22. 5.23, 5.24, 5.25 on pages 297 - 298; Show Your Work!

Lesson Wrap Up

107

Lesson 4: Pressure Drop in Pipe

108

Lesson 4: Pressure Drop in Pipe Learning Objectives

In this lesson we will:Identify the primary type of pressure drops involved in drilling and production operationsCalculate pressure drop in pipe and annuliCompare turbulent report error if disagreement exceeds 0.002Calculate summation pressure drop and cutting transport

109

Flow in a Cylindrical Annulus

Textbook p. 221PT see notes: This is a tau (Font symbol)

110

Depthft.

lbm/gal

μpcp

tylbf/100ft2

n Kequiv. cp

8000 12.50 26.7761 12.48 .6573 339.611

9000 14.56 26.0576 48.63 .2524 7514.073

10000 15.60 35.9729 2.13 .9296 59.396

11000 15.60 53.5043 45.96 .3674 4860.413

12000 16.40 58.1985 32.56 .4924 2057.784

Pressure Drop in Pipe and Annuli—Example Calculation

Given:

Pump: 3500 psi max surface pressure, 1600 hp max input, 0.85 efficiencySurface Equipment: equivalent to 340 ft. of drillpipeDrillpipe: 4.5 inch OD, 16.6 lb./ft., 3.78 inch ID, FH-XHDrill collars: 600 ft., 6.5 inch OD, 2.5 inch IDSurface casing: 3000 ft., 8.835 inch IDBit: 8.5 inchMinimum velocity to lift cuttings = 1 ft./sec

Determine the proper pump operating conditions and bit nozzle sizes for maximum bit horsepower at 1000 ft. increments for an interval of the well between 8000 ft. and intermediate casing at 12 000 ft.

Mud Plan:

111

Mud @ 8000 ft.

112

Pressure Drop Throughout the Circulation System

1

113

• 1 = Surface Equipment• 2 = Inside Drill Pipe • 3 = Inside Drill Collars • 4 = Annulus between Open Hole and Drill Collars • 5 = Annulus between Open Hole and Drill Pipe (may have 0 length) • 6 = Annulus between Cased Hole and Drill Pipe• 7 = Annulus between Cased Hole and Drill Collar (May have 0 length)

Calculate Pressure Drop in Friction in Seven Places

In this picture, Position 5 has 0 length.

In the previous slide, Position 7 has 0 length.

1

114

ΔP Parasitic Components

ΔPFQ = ΔPSE + ΔPDP + ΔPDC + ΔPOH/DC + ΔPOH/DP + ΔPCH/DP + ΔPCH/DC

1 2 3 4 6 *7*5

* Position 5 or Position 7 will have a length of zero

115

“Log of q versus Log of ΔP Parasitic”

116

We will now calculate Δpressure parasitic at qmaxThis is called interval 1This will have 7 steps1. Δpressure through surface equipment, (pipe flow, turbulent)2. Δpressure through drill pipe, (pipe flow, turbulent)3. Δpressure through drill collars, (pipe flow, turbulent)4. Δpressure in annulus between open hole and drill collars, (annular flow, laminar)5. Δpressure in annulus between open hole and drill pipe, (annular flow, laminar)6. Δpressure in annulus between cased hole and drill collars, (annular flow,

laminar)7. Δpressure in annulus between cased hole and drill pipe, (annular flow, laminar)

ΔPressure Parasitic at qmax

117

Critical Reynolds Number versus Hedstrom Plot

118

Determine correct qmax :qmaxpump

= 1714 PHP E/Pmax

qmaxpump = 1714 * 1600 * 0.85 / 3500 = 666.0 gal/min qmaxlaminar:

de = 0.8165 ( d2 - d1 ) Note: at smallest annular cross-section: open hole/collar1.633 inch = 0.8165 * ( 8.5 - 6.5 )NHEAnnulus

= 37100 ρ ty de2 / mp

2

215271 = 37100 * 12.50 * 12.48 * 1.6332 / 26.77612

NReCAnnulus= 5004 from Hedstrom vs. Critical Reynolds Curve

qmaxlaminar= 0.0026385 NREC Annulus mp (d2 + d1) / ρ

424.2 gal/min = 0.0026385 * 5004 * 26.7761 * ( 8.5 + 6.5 ) / 12.50qmax = smallest of qmaxpump or qmax laminar

424.4 gal/min < 666.0 gal/min

Interval 1

119

v = q / 2.448 / d2

12.13 ft./sec = 424.2 / 2.448 / 3.782 Note: drill pipe-equivalent

NReBingham = 928 ρ v d /μp

19864 = 928 * 12.50 * 12.13 * 3.78 / 26.7761

turbulent flow indicatedNHE = 37100 ty ρ d2 / μp

2

115342 = 37100 * 12.50 * 12.48 * 3.782 / 26.77612

NReC= 7000 Note: from Hedstrom vs. Critical Reynolds Curve @ NHe = 115342

turbulent flow indicated

Δp Surface Equipment @ qmax: Position 1

120

flaminar = 16 [ 1 / NRe + ( NHe / NRe2 ) / 7.9 ]

0.001405 = 16 [ 1 / 19864 + ( 115342 / 198642 ) / 7.9 ]fturbulent = 0.006482 from Reynolds vs. Fanning Curve @ NRe = 19864 (See slide 118)

fturbulentDS= 0.057 / NRe

0.2

0.007875 = 0.057 / 198640.2

fturbulentSPE= 0.0791 / NRe

0.25

0.006663 = 0.0791 / 198640.25

compare fturbulent report error if disagreement exceeds 0.002use Reynolds vs. Fanning Curve valuecompare fturbulent with flaminar take largest value0.006482 > 0.001405

Compare f Position 1

121

Figure Reynolds Number versus Fanning Friction Factor

122

Critical Reynolds Numbers to Bingham Plastic Fluids

123

Turbulent Part of the Reynolds Number vs Fanning Friction Factor

124

Δp/DL = 0.03875 f ρ v2 / d0.1222 = 0.03875 * 0.006482 * 12.50 * 12.132 / 3.78Δp = Δp/ DL L41.55 psi = 0.1222 * 340

Calculate ΔPressure Through Surface Equipment (Position 1)

125

v = q / 2.448 / d212.13 ft./sec. = 424.2 / 2.448 / 3.782

NreBingham = 928 ρ v d / μp19864 = 928 * 12.50 * 12.13 * 3.78 / 26.7761turbulent flow indicatedNHE = 37100 ty ρ d2/μp2

115342 = 37100 * 12.50 * 12.48 * 3.782 / 26.77612

NReC = 7000 from Hedstrom vs. Critical Reynolds Curve @ NHe = 115342 turbulent flow indicated

ΔpDrillpipe @ qmax: Position 2

126


0.001405 = 16 [ 1 / 19864 + ( 115342 / 198642 ) / 7.9 ]fturbulent = 0.006482 from Reynolds vs. Fanning Curve @ NRe = 19864f turbulent DS = 0.057 /NRe

0.2

0.007875 = 0.057 / 198640.2


0.25

0.006663 = 0.0791 / 198640.25

compare fturbulent report error if disagreement exceeds 0.002use Reynolds vs. Fanning Curve valuecompare fturbulent with flaminar take largest value0.006482 > 0.001405


127

Δp/DL = 0.03875 f ρ v2 / d0.1222 = 0.03875 * 0.006482 * 12.50 * 12.132 / 3.78Δp = Δp/DL L904.2 psi = 0.1222 * 7400

Calculate ΔPressure through Drill Pipe (Position 2)

128

v = q / 2.448 / d2

27.74 ft./sec = 424.2 / 2.448 / 2.52

NReBingham= 928 ρ v d /mp

30043 = 928 * 12.50 * 27.74 * 2.5 / 26.7761turbulent flow indicatedNHE = 37100 ty ρ d2 / μp

2

50453 = 37100 * 12.50 * 12.48 * 2.52 / 26.77612

NReC= 5000 from Hedstrom vs. Critical Reynolds Curve @ NHe = 50 453

turbulent flow indicated

ΔpDrill Collars @ qmax: Position 3

129


0.0006457 = 16 [ 1 / 30043 + (50453 / 300432 ) / 7.9 ]fturbulent = 0.006082 from Reynolds vs. Fanning Curve @ NRe = 19 864fturbulentDS

= 0.057 /NRe0.2

0.007250 = 0.057 / 300430.2


0.25

0.006008 = 0.0791 / 300430.25

compare fturbulent report error if disagreement exceeds 0.002use from Reynolds vs. Fanning Curve curve fit valuecompare fturbulent with flaminar take largest value0.006082 > 0.0006457


130

Δp/DL = 0.03875 f ρ v2 / d0.8743 = 0.03875 * 0.006082 * 12.50 * 27.242 / 2.5Δp = Δp/DL L524.6 psi = 0.8743 * 600

Calculate ΔPressure Through Drill Collars (Position 3)

131

v = q / 2.448 / ( d22 - d1

2 )5.776 ft./sec = 424.2 / 2.448 / ( 8.52 - 6.52 )de = 0.8165 ( d2 - d1 )1.633 = 0.8165 * ( 8.5 - 6.5 )NreBingham

= 928 ρ v de /μp

4000 = 928 * 12.50 * 5.776 * 1.633 / 26.7761laminar / turbulent flow indicated -NHE = 37100 ty ρ de

2 / μp2

21527 = 37100 * 12.50 * 12.48 * 1.6332 / 26.77612

NReC= 5004 from Hedstrom vs. Critical Reynolds Curve @ NHe = 21 527

laminar flow indicated

ΔpDrill Collar / Bit @ qmax: Position 4 – OPEN HOLE/DOLLAR

132


0.01010 = 24 [ 1 / 4000 + ( 21527 / 40002 ) / 7.9 ]fturbulent = 0.01002 from Reynolds vs. Fanning Curve fit @ NRe = 4000fturbulentDS

= 0.057 /NRe0.2

0.01085 = 0.057 / 40000.2


0.25

0.009946 = 0.0791 / 40000.25

compare fturbulent report error if disagreement exceeds 0.002use from Reynolds vs. Fanning Curve fit valuecompare fturbulent with flaminar take largest value0.01002 < 0.01010

Friction Position 4 (more discussion)

133

Δp/DL = 0.03875 f ρ v2 / de

0.09815 = 0.03875 * 0.01010 * 12.50 * 5.7762 / 1.663Δp = Δp / DL L58.89 psi = 0.09815 * 600

WHAT IS THE RATIONALE OF USING THE Fanning FACTOR FOR LAMINAR FLOW ???

Pressure Drop Calculation Position 4

134

v = q / 2.448 / ( d22 - d1


= 928 ρ v de /mp

3998 = 928 * 12.50 * 4.839 * 1.907 / 26.7761laminar / turbulent flow indicatedNHE = 37100 ty ρ de

2 / μp2

29357 = 37100 * 12.50 * 12.48 * 1.9072 / 26.77612

NReC= 5400 from Hedstrom vs. Critical Reynolds Curve curve fit @ NHe = 29357


ΔpDrill Collar / Csg @ qmax: Position 7

135


0.01158 = 24 [ 1 / 3998 + ( 29357 / 39982 ) / 7.9 ]fturbulent = 0.009988 from Reynolds vs. Fanning Curve fit value @ NRe = 3998fturbulentDS

= 0.057 /NRe0.2

0.01085 = 0.057 / 39980.2


0.25

0.009946 = 0.0791 / 39980.25


Friction Position 7

136

Δp/DL = 0.03875 f ρ v2 / de

0.06889 = 0.03875 * 0.01158 * 12.50 * 4.839**2 / 1.907Δp = Δp/DL L0.0 psi = 0.06889 * 0

Pressure Position 7

137

v = q / 2.448 / ( d22 - d1


= 928 ρ v de /mp


2 / μp2

86108 = 37100 * 12.50 * 12.48 * 3.2662 / 26.77612

NReC= 6500 from Hedstrom vs. Critical Reynolds Curve fit @ NHe = 86108


ΔpDrillpipe / Bit @ qmax: Position 5

138



= 0.057 /NRe0.2

0.01050 = 0.057 / 47140.2


0.25

0.009546 = 0.0791 / 47140.25


Friction Position 6

139

Δp/DL = 0.03875 f ρ v2 / de

0.01567 = 0.03875 * 0.009514 * 12.50 * 3.3322 / 3.266Δp = (Δp/ DL ) L68.93 psi = 0.01567 * 4400

Pressure Position 6

140

v = q / 2.448 / ( d22 - d1


= 928 ρ v de /μp


2 / μp2

101162 = 37100 * 12.50 * 12.48 * 3.5402 / 26.77612

NReC= 7000 from Hedstrom vs. Critical Reynolds Curve fit @ NHe = 101162


ΔpDrillpipe / Csg @ qmax: Position 6

141



= 0.057 /NRe0.2

0.01056 = 0.057 / 45820.2


0.25

0.009614 = 0.0791 / 45820.25

compare fturbulent report error if disagreement exceeds 0.002use Figure 16 curve fit valuecompare fturbulent with flaminar take largest value0.009988 < 0.01158 (0.009688 < 0.009709)

Friction Position 6

142

Δp/ DL = 0.03875 f ρ v2 / de

0.01194 = 0.03875 * 0.009709 * 12.50 * 2.9982 / 3.540Δp = (Δp/ DL ) L35.81 psi = 0.01194 * 3000

Pressure Position 6

143

Summation of Parasitic Pressure Drops

ΔPFQ = ΔPSE + ΔPDP + ΔPDC + ΔPOH/DC + ΔPOH/DP + ΔPCH/DC + ΔPCH/DP

ΔPFQ = 41.55 + 904.2 + 524.6 + 58.89 + 0.00 + 68.93 + 35.81

1 75 6432

Note: Position 5 or Position 6 will have length of 0.00

144

Bit Nozzles

At each depth a like calculation is made. The results are shown below:

Depth ft.

qgal/min

vminaft./sec psi

Δpdpsi

Δpbitpsi

Δptotal Jets32 inch

vbitft./sec

8000 378.68 2.68 1383 2129 3512 11 11 11 436.35

9000 549.87 3.89 2129 1383 3511 15 15 16 325.78

10000 247.91 1.75 951 2542 3493 9 9 9 426.73

11000 199.53 1.41 1615 1903 3518 8 9 9 369.29

12000 239.86 1.69 1643 1871 3514 9 10 10 357.05

145

“Log of q versus Log of ΔP Parasitic”

146

Δpbit = pmaxpump - Δd dopt

2083 psi = 3500 - 1417AT = (0.00008311ρ qopt

2 / Cd2 / Δp bit)0.5

0.2862 inch2 = (0.00008311*12.5*385.02 / 0.952 / 2083 )0.5

nozzle = ( AT 1.333 /p)0.5 * 3211.1557 = ( 0.2862 * 1.333 / p) 0.5 * 32

Bit Nozzle Calculation

147

Depth ft.

qgal/min

voptft./sec psi

Δpdpsi

Δpbitpsi

Δptotal Jets32 inch

vbitft./sec

8000 378.68 2.68 1383 2129 3512 11 11 11 436.35

9000 549.87 3.89 2129 1383 3511 15 15 16 325.78

10000 247.91 1.75 951 2542 3493 9 9 9 426.73

11000 199.53 1.41 1615 1903 3518 8 9 9 369.29

12000 239.86 1.69 1643 1871 3514 9 10 10 357.05

Summary of the Calculation Process for all Depths

148


HomeworkAssignment 7.1 Self Study ReviewAssignment 7.4: Read Fundamentals of Drilling Engineering Section 5.5 Laminar Flow in Pipes and Annuli (pp. 218-240 only); Section 5.6 Turbulent Flow in Pipes and Annuli (pp. 245-260 omit Hershel Bulkley Model), p. 267 only)

Lesson Wrap Up

149

Lesson 5: Newtonian Fluids

150

In this lesson we will:Calculate summation of pressure drop in system and cutting transports for Newtonian Fluids

Lesson 5: Newtonian Fluids Learning Objectives

151

For the following problems : A 8.625" OD 24 lb./ft. 8.097" nominal ID 7.972" drift ID surface casing set and cemented at 2000 ft. TVD & MD;The drilling string, currently at 10,000 ft. (TVD, MD), consists of a 7.875" tri-cone rock bit, 1000 ft. of 4.75" OD 2.25" ID 46.70 lb./ft., drill collars 4.5" OD 16.60 lb./ft. 3.826" ID drill pipe. The surface equipment is equivalent to 500 ft. of drill pipe. The drilling mud is 9.0 lb. /gal, μ = 25 cp Newtonian fluid.

Note: May be assigned as homework and then review answers (next slide) or work on calculation together in class

Newtonian Fluid Calculation

152

1. Continuing with the information from the problems above, the mud pump is a 6” x 10” x 2” single acting triplex, volumetric efficiency 0.95, maximum recommend pump pressure is 2200 psi. (Note 3 strokes per minute = 1 revolution per minute.) The number of strokes per minute to pump at q min is most nearly:a. a. 64b. b. 129c. c. 191d. d. 222

2. The hydraulic horsepower used by the pump at q min and maximum pump pressure is most nearly:a. 285 b. 342 c. 469 d. 496

Newtonian Fluid Calculation (Answers)

153


HomeworkAssignment 7.1 Self Study ReviewAssignment 7.5: Read Fundamentals of Drilling Engineering Section 5.9 Calculating Steady-State Pressures in a Wellbore (pp.267 – 269, omit 5.9.2)Assignment 7.5: Problem Solving, Complete Problems 5.27, 5.28, 5.29. 5.30, 5.31on page 298; Show Your Work!

Lesson Wrap Up

154

Lesson 6: Plastic Fluids

155

In this lesson, students will be able to:Calculate summation of pressure drop in system and cutting transports from Plastic Fluids

Lesson 6: Plastic Fluids Learning Objectives

156

For problems 1 -11: A 9.625" OD 36 lb./ft. 8.921" nominal ID 8.765" drift ID surface casing set and cemented at 3000 ft. TVD & MD; The drilling string, currently at 10,000 ft. (TVD, MD), consists of a 7.875" tri-cone rock bit, 1000 ft. of 5.75" OD 2.25" ID 74.70 lb./ft., 4.5" OD 16.60 lb./ft. 3.826" ID drill pipe. The surface equipment is equivalent to 500 ft. of drill pipe. The drilling mud is 10.0 lb. /gal, μp = 20 cp & Yt = 25 lbf/100 ft2 Bingham Plastic fluid.

Lesson 6: Plastic Fluids Calculation

Note: May be assigned as homework and then review answers (next slide) or work on calculation together in class

157

1. The capacity factor, ft3/ft., of the drill pipe is most nearly:a. 0.0276 b. 0.0798 c. 0.110 d. 0.228 e. 0.324

2. The capacity factor, ft3/ft., of the drill collars are most nearly:a. 0.0276 b. 0.0798 c. 0.180 d. 0.228 e. 0.324

3. The capacity factor, ft3/ft., of the open hole/drill collar annulus is most nearly:a. 0.0276 b. 0.0798 c. 0.158 d. 0.310 e. 0.324

4. The capacity factor, ft3/ft., of the open hole/drill pipe annulus is most nearly:a. 0.0276 b. 0.0798 c. 0.158 d. 0.228 e. 0.324

15

Plastic Fluids Calculation (Answers)

158

5. The capacity factor, ft3/ft., of the cased hole/drill pipe annulus is most nearly:a. 0.0276 b. 0.0798 c. 0.158 d. 0.310 e. 0.324

6. If the minimum velocity in the annulus required to lift cutting is 1 ft./sec, q min (gal/min) is most nearly:a. 71 b. 102 c. 138 d. 145 e. 205 f. 222

7. The pressure drop, lbf/in2, through the surface equipment at q min is most nearly:a. 8 b. 17 c. 68 d. 88 e. 138 f. 193

8. The pressure drop, lbf/in2, through the drill pipe at q min is most nearly:a. 8 b. 17 c. 68 d. 88 e. 138 f. 193

9. The pressure drop, lbf/in2, through the drill collars at q min is most nearly: a. 8 b. 17 c. 68 d. 88 e. 138 f. 193

10. The pressure drop, lbf/in2, in the annulus around the drill collars at q min is most nearly:a. 8 b. 17 c. 68 d. 88 e. 138 f. 237

11. The pressure drop, lbf/in2, in the annulus around the drill pipe at q min is most nearly:a. 8 b. 17 c. 69 d. 88 e. 138 f. 237

Lesson 6: Plastic Fluids Calculation (Answers cont.)

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HomeworkAssignment 7.1 Self Study ReviewAssignment 7.6: Read Fundamentals of Drilling Engineering Section 5.11 Cutting Transport (p.279 – 287, omit 5.11.4, read pp. 289 – 294, omit 5.11.6, read pp. 289 –294, omit 5.11.6Assignment 7.6: Complete Problems 5.32, 5.33, 5.34. 5.35,5.36, 5.37, and 5.40 on page 298 - 299; Show Your Work!

Lesson Wrap Up

160

Credits

DeveloperLloyd R. Heinze, Ph.D., Petroleum Engineering/Texas Tech University

Contributors:Rui V. Sitoe, Ph.D., Department of Mechanical Engineering, UEMVictoria Johnson, Instructional Designer

dc mod 7 hydraulics presentation

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