Chapter-3

transcript

Drilling Hydraulics

Chapter 3

Effect of Bit Hydraulics on Drilling Rate

DRILLING HYDRAULICS

A good understanding of the drilling hydraulics is required for optimum design

ofthe following tasks:

Bottomhole cleaning while drilling

Drill bit cleaning

Cuttings transport to the surface

DRILLING HYDRAULICS

Friction pressure losses through and around the drillstring, through rig piping and the bit nozzles

Equivalent circulating pressure

Hydraulic energy consumption

Cement and well completion fluid circulation

Surge and Swab Pressure Estimation

Borehole wall erosion

DRILLING HYDRAULICS

DB : Drill Bit

DC : Drill Collar

DP : Drill Pipe

MP : Mud Pump

MT : Mud Tank

SS : Shale Shaker

HYDRAULIC HORSEPOWER REQUIREMENT

Pump Hydraulic Horsepower = Q * Pp /1714

Flow rate, Q (gal/min)

Minimum: Cuttings transportMaximum: Borehole wall erosion (hole wash out)

Pump Pressure (psi) Pp = P

Frictional pressure loss is a function of fluid velocity(flow rate), pipe/wellbore geometry, and drilling fluid rheological characteristics.

DRILLING HYDRAULICS

Mud Hydrostatic Pressure : SBHP = 0.052 m D

Buoyancy Effect : We = Wa (1- m /s)

SBHP = Static bottomhole pressure, lbf/in2

D = Depth, ft

We = Effective weight of the drillstring, lbm

Wa = Air weight of the drillstring, lbm

m = Drilling fluid density, lbm/gal

s = Pipe material density, lbm/gal (65.5 lbm/gal for steel)

DYNAMIC PRESSURE BALANCE

Pump Pressure = PS+ PDP+ PDC+ PB+ PA

PS = Pressure losses through surface installations

PDP = Pressure losses through drill pipes

PDC = Pressure losses through drill collars

PB = Pressure losses through drill bit

PA = Pressure losses through annulus

Circulating Bottomhole Pressure

CBHP (psi) = SBHP + PA

Equivalent Circulating Density

ECD (lbm/gal) = CBHP/(0.052*D)

ECD should always be less than the formation fracture

gradient.

ECD > Fracture Gradient (Lost circulation?)

ECD < Pore Pressure Gradient

Underbalanced drillingKick?

ECD >Pore Pressure Gradient

Overbalanced drillingFormation Damage?Differential pipe sticking?

ECD = Pore Pressure Gradient

Balanced drilling

ESTIMATION OF FRICTION PRESSURE LOSSES

Friction pressure loss is a function of flow rate (flow regime), rheological properties of fluids, and the pipe/wellbore geometry.

There are four recognized flow regimes: Plug Flow Laminar Flow Transitional Flow Turbulent Flow

Plug FlowIf a fluid has a yield strength and its yield strength has a value greater than the shear stress at the wall of a pipe or annulus, then the fluid will be in plug flow.

A suitable criterion is if the value of Reynold’s number is less than 10 and the fluid has a gel strength, then the flow regime is plug.

The other flow regimes are predicted by using Reynold’s number. Depending of the fluid rheology type, however, different criteria need to be used.

Laminar FlowLaminar flow occurs when concentric cylindrical fluid shells (laminae) slide past one another.

The velocity of the shell at the pipe wall is zero, and velocity of the shell at the center of the pipe is maximum.

dv/dr=0

Turbulent Flow

Fluid flow becomes turbulent at very high flow velocities when (high Reynold’s number) fluid particles show a chaotic(random), diffusive motion with three dimensional vorticity(rotational) fluctuations.

No analytical solution for frictional pressure loss estimation is available.

RHEOLOGY OF DRILLING FLUIDS

Newtonian Fluids (Water, High gravity oil)

= Shear stress, dynes/cm2

= Shear rate, 1/sec = Viscosity, poise (dyne-sec/cm2)

Field Units: = lbf/100 ft2

= centipoise

Non-Newtonian Fluids (Drilling Fluids)

Bingham Plastic Model = Y + pPower Law Model = Kn

Yield Power Law Model = Y +Kn

K : Fluid consistency index, lbf-sn/cm2

Y : Yield stress, lbf/100 ft2

p : Plastic viscosity, cp.

n : Flow behavior index

Bingham Plastic

Yield Power Law

Power Law

ESTIMATION OF FRICTIONAL PRESSURE LOSSES

Nomenclature

d: Diameter, in.L: Length, ft.: Viscosity, cp.: Density, lbm/galq: Flow rate, gal/minv : Average fluid velocity, ft/sec

Pipe: v = q/2.448d2

Annulus: v = q/2.448(d22 - d1

EQUIVALENT DIAMETER CONCEPT

An equivalent diameter is used to extend pipe

flow equations to annular geometry. Use hydraulic radius concept

4221212

2 ddrrrrrrr H

ddrd He 124

Use geometry term in laminar pressure loss

equations derived for pipe flow and concentric

annular flow.

ddddddd

ddddddd e

Use slot approximation

Use Crittendon’s equivalent diameter

ddd e 12816.0

NEWTONIAN FLUIDS

Laminar Flow

Reynold’s Number:

Check for Flow Regime: NRE < 2100

Pressure Losses in Pipe Flow:

Pressure Losses in Annulus:

dddP v

1221000

NEWTONIAN FLUIDS

Turbulent Flow Check for Flow Regime: NRE > 2100 Use the Fanning equation for pressure

losses

Pipe flow Flow in Annulus

ddLvfdP f

vfdPdL

NEWTONIAN FLUIDS

Friction factor, f

Colebrook Equation

fdf N Re

255.1269.0log4

Stanton Chart

BINGHAM PLASTIC FLUIDS

Laminar Flow

Pressure Losses in Pipe Flow:

Pressure Losses in Annulus:

ddLypf

22515002

dddddP ypf

2 2001000 12

Turbulent Flow

Use apparent viscosity in Reynold’s number

criterion developed for Newtonian fluids.

Annulus:

If NRE > 2100 , then the flow is turbulent!

Use the Colebrook equation to determine friction factor.

Determine the frictional pressure loss using the Fanning equation;

vdN RE

ddLvfdP f

vfdPdL

POWER LAW FLUIDS

Reynold’s Number (Pipe Flow)

Reynold’s Number (Annulus Flow)

RE Kvn

0416.0289100

RE Kvn

0208.0 122109000

POWER LAW FLUIDS

Use Fig. 4.34 (Applied Drilling Engineering by Bourgoyne et al.) to determine the critical Reynold’s number.

Following approximate equations can also be used to determine flow regime for power law fluids:

Laminar Flow

NRE < 3470 - 1370n

Turbulent Flow

NRE > 4270 - 1370n

POWER LAW FLUIDS

Laminar Flow Frictional pressure loss in pipes

Frictional Pressure Loss in Annulus

1144000

0416.0

121144000

0208.0

POWER LAW FLUIDS

Turbulent Flow Use the Dodge and Metzner friction factor correlation

(given only for smooth pipes)

Use the Fanning equation to determine frictional pressure loss.

Pipe flow Flow in annulus

f 2.121

Re75.0

395.0log

ddLvfdP f

vfdPdL

FLOW THROUGH JET BITS

10*311.8

At=Total flow area, in2

Cd=Discharge coefficient (0.95)

d= Nozzle diameter, in

vn= Nozzle velocity, ft/sec

Q=Mud flow rate, gal/min

=Mud density, lb/gal

Pb= Bit pressure drop, psi

dddAt2

PRESSURE DROP ACROSS BIT NOZZLES

FLOW THROUGH JET BITS

Bit Hydraulic Horsepower

BHHP= Q Pb /1714

The higher the BHHP, the faster the drilling rate!

Hydraulic (Jet) Impact Force

Fj= Jet impact force, lbf

PCF bdjQ 01823.0

OPTIMIZATION OF DRILLING HYDRAULICS

ObjectiveTo maximize the drilling rate (or to minimize the drilling cost)

Required

• Instantaneous removal of the cuttings from the

rock face, • Effective upward transportation of cuttings to

the surface

Two types of energy sources are brought from the surface to the rock face and

should be applied in an optimal manner:

• Mechanical Energy (WOB, RPM)• Hydraulic Energy (Flow rate, nozzle

Methods of optimal hydraulic design:

• Determine the bit hydraulic horsepower required in order to balance the mechanical energy level

• Maximize arbitrarily selected criterion of estimation, e.g. bit hydraulic horse-power, jet impact force, etc.

Minimum Bit Hydraulic Horsepower vs.WR to Prevent Hydraulic Flounder (After Fullerton)

Example :Is there a proper balance between hydraulic and mechanical energy which has been delivered to therock face ?

Available BHHP : 400 HP Db: 12 ¼”

WOB/ Db : 7000 lbf/in N: 80 RPM

(WOB/Db)*N = 7000*80 = 560X103

From Fullerton Chart Required BHHP = 650 HP < 400 HP

Answer: No

Optimization Criteria

• Maximum Bit Hydraulic Horsepower• Maximum Jet Impact Force• Maximum Nozzle Velocity

Once the objective function for optimal hydraulic program design is selected, the limitations connected with “performance characteristics of the mud pump”

have to be considered.

Pump Performance Characteristics

I Pump Operating Range

Pp = Constant for 1 < Q < Qm

Pp = PD+ PB

II Pump Operating Range

HPp = HPD + HPb = Constant

Mud Pump Performance Characteristics

(Pp)max

q’max

Continental EMSCO F 1600 (Triplex) Pump Performance Characteristics

Pp = PS+ PDP+ PDC+ PB+ PDCA + PDPA

PD = PS+ PDP+ PDC+ PDCA + PDPA

PD= Parasitic pressure losses

Pp = PD+ PB

PD= cQm

m is constant (1.3 < m < 2.1 theoretically 1.75)

c is a function of mud properties and hole geometry

PARASITIC PRESSURE LOSSES vs. FLOW RATE

Log PD

Log QQ2

Slope = m

OBJECTIVE:Determine the optimum combination of flow rate and nozzle sizes (total flow area across the bit): CONSTRAINTS:

• Adequate cuttings transport (Qmin)

• Maximum Available Pump Flow Rate/ Wellbore erosion (Qmax)

PATH OF OPTIMUM HYDRAULICS

Log PD

Slope = m

Qmin Log QQmax

(PD)opt

(Pp)max

MAXIMUM BIT HYDRAULIC HORSEPOWER

Determine the flow rate, Q, at which the bit hydraulic horsepower is maximum.

171417141714

QPPPPm

pDpbcQQQ

MAXIMUM BIT HYDRAULIC HORSEPOWER

• If the intersection occurs in zone 1 Qopt=Qmax

• If the intersection occurs in zone 3 Qopt=Qmin

• If the intersection occurs in zone 2

PPP Db optpopt

dboptt 2

10*311.8

MAXIMUM JET IMPACT FORCE

Determine the flow rate, Q, at which the jet

impact force is maximum.

PPCF pQDdj 01823.0

MAXIMUM JET IMPACT FORCE

• If the intersection occurs in zone 1 Qopt=Qmax

• If the intersection occurs in zone 3 Qopt=Qmin

• If the intersection occurs in zone 2

PPP Db optpopt

dboptt 2

10*311.8

Example :Using maximum bit hydraulic horsepower criterion,determine the optimum nozzle sizes to be used to drill the next depth interval.

Given the following well data:

Drillpipe: 4.5”, 20 lb/ft (ID:3.64”) Drill collar: 7” x 2” 120.3 lb/ft, 1000 ft.Drill Bit: 12 7/8”-tricone, 3-14 nozzles to 12,000 ftNext bit: 8 7/8” (assume hole washed out to 9 7/8” )Pump: 1600 HP- National Duplex-Double acting

(Pp)max : 5440 psi Volumetric Efficiency: 80%Cuttings Slip Velocity: vsl : 25 ft/min

Required Net transport velocity: vt : 60 ft/minLast casing set : 9 7/8” @ 12,000 ft.

Mud Data:Bingham plasticDial reading @ 300 rpm : 21Dial reading @ 600 rpm: 29Density: 15.5 ppg

Field data @ 12,000 ft while using 8 7/8” bit, 3-14 nozzles

Q, gpm

Pp, psi

300 2966

400 4883

Solution :

Total bit nozzle area:

TFA: (3*/4)*(14/32)2 = 0.4509 in2

Bit pressure losses at 300 and 400 gpm:

Pb = 8.311*10-5*15.5*3002/ (0.952*0.45092)=632 psi

Pb = 8.311*10-5*15.5*4002/ (0.952*0.45092)=1123 psi

Parasitic pressure losses at 300 and 400 gpm:

PD @ 300 gpm = 2966-632 = 2334 psi

PD @ 400 gpm = 4883-1123 = 3760 psi

The slope and the intercept of the PD vs.Q line:m = log(3760/2334) / Log (400/300) = 1.657C =2334/3001.657 = 0.18345

The maximum flow rate: Qmax = 1714*E*HPp/(Pp)max

= 1714*0.8*1600/5440 = 403 gpm

The minimum required flow rate:Qmin = 2.448*(Dh

2-Dp2)*vmin = 2.448*(9.8752-4.52)*(85/60)

= 268 gpm

Optimum Parasitic pressure losses using Maximum BHHP Criterion:(PD )opt = 5440/(1+1.657) = 2047 psi

Optimum bit pressure drop: (Pb )opt = 5440-2047 = 3393 psi

Optimum Flow Rate: Qopt = (2047/0.18345)1/1.657 = 277 gpm

Qmax > Qopt > Qmin , Region – 2 in optimum hydraulic path!

Optimum nozzle area:

TFA = (8.311*10-5*15.5* 2772/(0.952*3393))0.5 = 0.1797 in2

For 3 equal-sized nozzles the diameter will be:

dn = (0.1797*4*322/(3*))0.5 = 8.83

Therefore 3 - 9/32 in nozzles need to be installed for the next bit run.

Chapter-3

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