Drill Bit Hydraulics

Assumptions 1) Change of pressure due to elevation is negligible. 2) Velocity upstream is negligible compared to nozzles. 3) Pressure due to friction is negligible. 04075.8 2 =−−Δ nB vEP ρ Pressure drop across bit, nozzle velocity BPΔ nv Solving for nozzle velocity

ρ4074.8 −

Δ=

EPv B

n

In the field it has been shown that velocity predicted by this equation is off. So it has been modified,

ρ4074.8 −

Δ=

EPCv B

dn

the recommended valve for Cd is .95. If 3 nozzles are present

3

3

2

2

1

1

Aq

Aq

Aqv === the velocity is equal in all the jets.

321321 AvAvAvqqqq nnn ++=++= That gives us

tn A

qv = In field units t

n Aqv 117.3=

q in gpm, At in inches2, vn in ft/sec solving for the pressure drop

22

25311.8

tdB AC

qEP ρ−=Δ

ρ is #/gal Flow Exponent α It can be deduced that C is a constant αCQPf =

QCPf logloglog α+= So the log log plot of this equation is a straight line with a slope of α. α can found if two Pf and Q are known, this can be achieved by measuring the standpipe or surface pressure for 2 pumping rates. Ps=Pf+PB so by using the above equation P

B

B can be calculated and subtracted from Ps to find Pf.

22

25311.8

tdsf AC

qEPP ρ−−= after finding Pf , α can be found by

⎟⎠⎞⎜

⎝⎛

⎟⎠⎞

⎜⎝⎛

=

1

2

1

2

log

log

QQ

PP

f

f

α

Maximum Drill Bit Hydraulic Horsepower Criterion assumes that optimum hole cleaning is achieved if the hydraulic horsepower across the bit is maximized with respect to the flow rate Q. QPH BHB = Sub in αCQPP sB −=

1+−= αCQPH sHB

Take the first derivative of H with respect to Q set the result to 0.

( ) 01 =+−= αα CQPdQ

gHs

HB

αCQPf = ( ) 01 =−− fs PP α or Sf PP

11+

=α

this is the root that makes HHB a maximum. Hence the optimum bit hydraulics will be achieved if friction pressure loss in the system is maintained at an optimum value of max1

1sfopt PP

+=

α

across the nozzles maxmax 1 sfoptsBopt PPPP

+=−=

αα

Calculate or measure a Pfqa @ some Qa then knowing Pfopt a Qopt can be calculated by

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

fqa

foptaopt P

PantiQQ log1log

α

With Qopt known the PBopt can be rewritten

22

53.8

dtopt

optBopt CA

QEP

ρ−=

solve for Atopt Boptd

opttopt PC

QEA 2

53.8 ρ−=

if all nozzles are the same size 2

4 nopttopt ndA π= n is the number of nozzles

solve for dnopt πn

Ad topt

nopt =

Example: DP 41/2” 20#/ft, Collars 7” 120.3#/ft 1000’ Mud θ300 21, θ600 29, ρ 15.5 #/gal Pump Pmax 5440 psi HHP 1600hp 80% TD 12,000’ Vamin 85 ft/min Bit 8 7/8” 14-14-14 Hole size 9 7/8” Rate data Q1 300 GPM @ Ps1 2966 psi Q2 400 GPM @ Ps2 4883 psi Find α

22

25311.8

tdB AC

QEP ρ=

psiEPB 6.63145099.95.

3005.155311.822

2

1 =⋅⋅

= psiEPB 8.112245099.95.

4005.155311.822

2

2 =⋅⋅

=

psiPf 4.23446.63129661 =−= psiPf 2.37608.112248832 =−=

( )( ) 66.1

300400log

4.23442.3760log

log

log

1

2

1

2

==⎟⎠⎞⎜

⎝⎛

⎟⎠⎞

⎜⎝⎛

=

QQ

PP

f

f

α

Find Qmax and Qmin

( ) gpmQ 403544016008.1714max =⋅= Based on pump

( ) gpmQ 26860

855.4875.9448.2 22min =−= Based on velocity

Optimum friction pressure

psiPP pfopt 20475440166.1

11

1max =⎟

⎠⎞

⎜⎝⎛

+=⎟

⎠⎞

⎜⎝⎛

+=

α

Optimum pressure drop at the bit

psiPPP foptpB 393320475440max =−=−=

Optimum flow rate

gpmantiPP

antiQQfqa

foptaopt 227

23342047log

66.11log300log1log =⎥⎦

⎤⎢⎣⎡=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

α

This is lower than the max and higher than min flow rates. Optimum nozzle area

22

2

2

2

15.339395.

2275.155311.85311.8inE

PCQE

ABoptd

opttopt =

⋅⋅−=

⋅⋅−=

ρ

For 3 equal sized jets

indnopt 25.315.2 == π

The maximum jet impact force criterion assumes that the bottom-hole cleaning is achieved by maximizing the jet impact force with respect to the flow rate. The impact force at the bottom of the hole can be derived form Newton’s second law of motion Bj PBQF = ρdCB 01823.= Q in gpm ρ in #/gal αCQPPPP sfsB −=−=

αCQPBQF sj −= limitations

1) maximum pump horsepower 2) maximum surface pressure

For the shallow portion of the well Pf is small and the flow rate requirement is large the impact force is limited only by the pump horsepower, therefore, the allowable surface pressure, expressed as

Q

HP p

smax=

substituting

2max

max −−=−= αα CQQHBCQQ

HBQF p

pj

Differentiate and set to 0

[ ]

0)2(5.

2max

1max =

−

+−=

+

+

α

αα

CQQH

CQHBdQdF

p

pj

For a valid solution the numerator must be equal to zero.

Solve for the optimum friction pressure soptfopt PP

21+

=α

then solve for the optimum bit pressure soptfoptsoptBopt PPPP

21

++

=−=αα

In the deeper sections of the well the friction pressure loss increases, while the flow rate requirement decreases. Therefore the impact force will limited by the maximum allowed pump pressure, Psmax. αCQPBQP sj −= max Differentiate and set to 0

[ ]0

)2(5.2

max

1max =

−

+−=

+

+

α

αα

CQQP

CQPBdQdF

s

sj

For a valid solution the numerator must be equal to zero. max2

2sfopt PP

+=

α

Gives maxmax 2 sfoptsBopt PPPP

+=−=

αα

Example Same data as Hydraulic example So α=1.66 Qmax=4.3 gpm Qmin=268 gpm At 12,000 feet the pump pressure is the limiting factor. psiPP sfopt 2975

266.154402

22

max =+

⋅=

+=

α

psiPPP foptsBopt 246529755440max =−=−=

gpmantiPP

antiQQfqa

foptaopt 347

23342975log

66.11log300log1log =⎥⎦

⎤⎢⎣⎡=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

α

It is bounded by the min and max flow rates, so

22

2

2

2

26.246595.

3475.155311.85311.8inE

PCQE

ABoptd

opttopt =

⋅⋅⋅−

=⋅⋅−

=ρ

"32/6.10332.3

26.2 === indnopt π 3 - 11 jets have an area of .27in2. Section 4.13 in text, pages 156, 157

Cuttings Lifting

Rock weights about 21 ppg, so it will fall in any fluid that has a lower density. The rate that the cutting fall in the drilling fluid is the slip velocity. To maintain good hole cleaning the velocity of the drilling fluid has to be greater than the slip velocity of the cuttings. The slip velocity depends on the difference in densities, viscosity of the fluid and the size of the cuttings.

( ) 5.

4.113⎥⎥⎦

⎤

⎢⎢⎣

⎡ −=

fD

fpps C

dv

ρρρ

the diameter of the cuttings inches pd

ρp the density of the cuttings 21 ppg CD Drag coefficient Particle Reynolds number

μρ ps

p

dvR

47.15=

which gives

PD R

C 40=

Substituting in the first equation

( )μ

ρρ fpps

dv

−=

24980

For values of Rp greater than 1 which means laminar flow around the particle the drag coefficient can be found using 5.

22

pD R

C =

So the slip velocity equation becomes

( )

333.333.

667.175μρ

ρρ

f

fpps

dv

−=

Designing the hydraulic system

1) Break the well down into sections, hole size, drilling fluid changes and depth etc. Design the drilling fluids for each.

2) Calculate the maximum pump rate using the pump specifications.

3) Calculate the friction loss in the pipe and annulus for each section using 2 flow rates. From this calculate the flow exponent α for that section.

4) Using this flow exponent optimize the bit hydraulics. 5) When drilling confirm your plan by finding the α by

measuring the friction pressure at 2 pump rates. 6) Find the annular velocity at the optimal rate and compare it

to the slip velocity, verify that this rate will clean the hole. 7) Calculate the pressures and horsepower required to pump

the optimal rate for the bit and verify the equipment can handle it.

8) Trail and error may be required to find the optimal rate and jet sizes.