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# PETE 203 DRILLING ENGINEERING Drilling Hydraulics.

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PETE 203 DRILLING ENGINEERING Drilling Hydraulics
Transcript PETE 203

DRILLING ENGINEERING

Drilling Hydraulics Drilling Hydraulics

Energy Balance

Flow Through Nozzles

Hydraulic Horsepower

Hydraulic Impact Force

Rheological Models

Optimum Bit Hydraulics Nonstatic Well Conditions

Physical Laws: Conservation of Mass Conservation of energy Conservation of

momentum Rheological Models

Newtonian Bingham Plastic Power – Law API Power-Law

Equations of State Incompressible fluid Slightly compressible

fluid Ideal gas Real gas Average Fluid Velocity

Pipe Flow Annular Flow

WHERE

v = average velocity, ft/s

q = flow rate, gal/min

d = internal diameter of pipe, in.

d2 = internal diameter of outer pipe or borehole, in.

d1 =external diameter of inner pipe, in.

2448.2 d

qv 2

122448.2 dd

qv  Law of Conservation of Energy

States that as a fluid flows from point 1 to point 2:

QW

vvDDg

VpVpEE

21

2212

112212

2

1

In the wellbore, in many cases Q = 0 (heat) = constant{ In practical field units this equation simplifies to:

fp pPvv

DDpp

21

22

4

1212

10*074.8

052.0

p1 and p2 are pressures in psiis density in lbm/gal.v1 and v2 are velocities in ft/sec.

pp is pressure added by pump between points 1 and 2 in psi

pf is frictional pressure loss in psi

D1 and D2 are depths in ft.

where Determine the pressure at the bottom of the drill collars, if

psi 000,3 p

in. 5.2

0 D

ft. 000,10 D

lbm/gal. 12

gal/min. 400 q

psi 1,400

p

1

2

DC

f

ID

p

(bottom of drill collars)

(mud pits) Velocity in drill collars

)(in

(gal/min)

d448.2

qv

222

ft/sec 14.26)5.2(*448.2

400v

22

Velocity in mud pits, v1 0 400,1000,36.6240,60

400,1000,3)014.26(12*10*8.074-

0)-(10,00012*052.00p

PP)vv(10*074.8

)DD(052.0pp

224-

2

fp21

22

4-

1212

Pressure at bottom of drill collars = 7,833 psig

NOTE: KE in collars

May be ignored in many cases

0 fp PPvv

DDpp

)(10*074.8

)(052.021

22

4-

1212 0 P

v v0 P

0 vD D

f

n2p

112

Fluid Flow Through Nozzle

Assume:

4n

2n

412

10*074.8

pv and

v10*074.8pp If

95.0c 10*074.8

pcv

as writtenbemay Equation

d4dn

0 fP

This accounts for all the losses in the nozzle.

Example: ft/sec 305 12*10*074.8

000,195.0v

4n  For multiple nozzles in //

Vn is the same for each nozzle

even if the dn varies!

This follows since p is the same across each nozzle.

tn A117.3

qv

2t

2d

2-5

bit AC

q10*8.311Δp

10*074.8

pcv

4dn

& Hydraulic Horsepower

HHP of pump putting out 400 gpm at 3,000 psi = ?

Power

pqP

A

qA*p

t/s*F

workdoing of rate

H

hp7001714

000,3*400

1714

pq HHP

In field units: Hydraulic Impact Force

What is the HHP Developed by bit?

Consider:

psi 169,1Δp

lb/gal 12

gal/min 400q

95.0C

n

D Impact = rate of change of momentum

60*17.32

vqv

t

m

t

mvF n

j

psi 169,1Δp

lb/gal 12

gal/min 400q

95.0C

n

D

lbf 820169,1*12400*95.0*01823.0Fj

pqc01823.0F dj Newtonian Fluid Model

Shear stress = viscosity * shear rate

A

F ,

L

VallyExperiment Laminar Flow of Newtonian Fluids

A

F

L

V Newtonian Fluid Model

In a Newtonian fluid the shear stress is directly proportional to the shear rate (in laminar flow):

i.e.,

The constant of proportionality, is the viscosity of the fluid and is independent of shear rate.

sec

12

cm

dyne

. Newtonian Fluid Model

Viscosity may be expressed in poise or centipoise.

poise 0.01 centipoise 1

scm

g1

cm

s-dyne1 poise 1

2

2cm

secdyne

. Shear Stress vs. Shear Rate for a Newtonian Fluid

Slope of line

. Example - Newtonian Fluid Example 4.16

Area of upper plate = 20 cm2

Distance between plates = 1 cm

Force req’d to move upper plate at 10 cm/s = 100 dynes.

What is fluid viscosity? Example 4.16

poise 5.0cm

sdyne5.0

10

52

1-

2

sec 10/1

dynes/cm 20/100

/

/

rate shear

stressshear

LV

AF

cp 50 Bingham Plastic Model Bingham Plastic Model

- if

- if 0

if

yyp

yy

yyp

and y are often expressed in lbf/100 sq.ft Power-Law Model Power-Law Model

n = flow behavior index

K = consistency index

0 if K

0 if K1n

n Rheological Models

1. Newtonian Fluid:

2. Bingham Plastic Fluid:

rate shear

viscosityabsolute

stressshear

*)( py viscosityplastic

point yield

p

y

What if

y 3. Power Law Fluid:

When n = 1, fluid is Newtonian and K = We shall use power-law model(s) to

calculate pressure losses (mostly).

n

)(K

K = consistency index

n = flow behavior index

Rheological Models Velocity Profiles(laminar flow)

Fig. 4-26. Velocity profiles for laminar flow: (a) pipe flow and (b) annular flow “It looks like concentric rings of fluid telescoping down the pipe at different velocities”

3D View of Laminar Flow in a pipe - Newtonian Fluid Summary of Laminar Flow Equations for Pipes and Annuli   Fig 4.33: Fig 4.33: Critical Reynolds number for Critical Reynolds number for Bingham plastic fluids.Bingham plastic fluids. Fig 4.34: Fig 4.34: Fraction Factors for Power-law Fraction Factors for Power-law fluid model.fluid model.  Total Pump Pressure

Pressure loss in surf. equipment

Pressure loss in drill pipe

Pressure loss in drill collars

Pressure drop across the bit nozzles

Pressure loss in the annulus between the drill collars and the hole wall

Pressure loss in the annulus between the drill pipe and the hole wall

Hydrostatic pressure difference ( varies) Total Pump Pressure

PUMP SC DP DC

B DCA DPA HYD

P P P P

P P P ( ΔP ) Types of Flow

Laminar Flow

Flow pattern is linear (no radial flow)

Velocity at wall is ZERO

Produces minimal hole erosion Types of Flow - Laminar

Mud properties strongly affect pressure losses

Is preferred flow type for annulus (in vertical wells)

Laminar flow is sometimes referred to as sheet flow, or layered flow:

* As the flow velocity increases, the flow type changes from laminar to turbulent. Types of Flow

Turbulent Flow

Flow pattern is random (flow in all directions)

Tends to produce hole erosion

Results in higher pressure losses (takes more energy)

Provides excellent hole cleaning…but… Types of flow

Mud properties have little effect on pressure losses

Is the usual flow type inside the drill pipe and collars

Thin laminar boundary layer at the wall

Turbulent flow, cont’d

Fig. 4-30. Laminar and turbulent flow patterns in a circular pipe: (a) laminar flow, (b) transition between laminar and turbulent flow and (c) turbulent flow Turbulent Flow - Newtonian Fluid

The onset of turbulence in pipe flow is characterized by the dimensionless group known as the Reynolds number

dv

N

_

Re

μ

dvρ928N

_

Re In field units, Turbulent Flow - Newtonian Fluid

We often assume that fluid flow is

turbulent if Nre > 2,100

cp. fluid, ofviscosity μ

in I.D., piped

ft/s velocity,fluid avg. v

lbm/gal density, fluid ρ where_

μ

dvρ928N

_

Re PPUMP = PDP + PDC

+ PBIT NOZZLES

+ PDC/ANN + PDP/ANN

+ PHYD

Q = 280 gal/min

= 12.5 lb/gal

Pressure Drop Calculations PPUMP "Friction" Pressures

0

500

1,000

1,500

2,000

2,500

0 5,000 10,000 15,000 20,000 25,000

Distance from Standpipe, ft

"Fri

ctio

n" P

ress

ure,

psi

DRILLPIPE

DRILL COLLARS

BIT NOZZLES

ANNULUS

2103 Optimum Bit Hydraulics

Maximum impact force?

Both these items increase when the circulation rate increases.

However, when the circulation rate increases, so does the frictional pressure drop. Jet Bit Nozzle Size Selection

Nozzle Size Selection for Optimum

Bit Hydraulics:

Max. Nozzle Velocity

Max. Bit Hydraulic Horsepower

Max. Jet Impact Force Jet Bit Nozzle Size Selection

Proper bottom-hole cleaning Will eliminate excessive regrinding of

drilled solids, and

Will result in improved penetration rates

Bottom-hole cleaning efficiency

Is achieved through proper selection of bit nozzle sizes Jet Bit Nozzle Size Selection- Optimization -

Through nozzle size selection, optimization may be based on maximizing one of the following:

Bit Nozzle Velocity

Bit Hydraulic Horsepower

Jet impact force• There is no general agreement on which of these three parameters should be maximized. Maximum Nozzle Velocity

From Eq. (4.31)

i.e.

so the bit pressure drop should be maximized in order to obtain the maximum nozzle velocity

4b

dn 10*074.8

PCv

bn Pv Maximum Nozzle Velocity

This (maximization) will be achieved when the surface pressure is maximized and the frictional pressure loss everywhere is minimized, i.e., when the flow rate is minimized.

pressure. surface allowable maximum the and

rate ncirculatio minimum the at

satisfied, are above 2&1 whenmaximized is vn Maximum Bit Hydraulic Horsepower

The hydraulic horsepower at the bit is maximized when is maximized.q) p( bit

dpumpbit ppp

where may be called the parasitic pressure loss in the system (friction).

dp

bitdpump ppp Maximum Bit Hydraulic Horsepower

. turbulentis flow theif

In general, wherem

d cqp 2m0

The parasitic pressure loss in the system, Maximum Bit Hydraulic Horsepower

0)1(p when 0

17141714

pump

1

mHbit

mpumpbit

Hbit

qmcdq

dP

cqqpqpP

dpumpbit ppp md cqp Maximum Bit Hydraulic Horsepower

when maximum is

1

1p when .,.

)1(p when .,.

d

pump

Hbit

pump

d

P

pm

ei

pmei

pumpd pm

p

1

1

0)1(p pump mqmc Maximum Jet Impact Force

The jet impact force is given by Eq. 4.37:

)(c0.01823

01823.0

d dpump

bitdj

ppq

pqcF Maximum Jet Impact Force

But parasitic pressure drop,

2201823.0

mdpdj

md

qcqpcF

cqp

)(c0.01823 d dpumpj ppqF Maximum Jet Impact Force

Upon differentiating, setting the first derivative to zero, and solving the resulting quadratic equation, it may be seen that the impact force is maximized when,

pd p2m

2p

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