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.

2 PB  8.075E  4vn  0

PB Pressure drop across bit, vn nozzle velocity

Solving for nozzle velocity

P v  B n 8.074E  4

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

P v  C B n d 8.074E  4

the recommended valve for Cd is .95.

If 3 nozzles are present q q q v  1  2  3 the velocity is equal in all the jets. A1 A2 A3

q  q1  q2  q3  vn A1  vn A2  vn A3

That gives us

q q vn  In field units vn  At 3.117At

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

8.311E  5q 2 PB  2 2 Cd At

 is #/gal

Flow Exponent  It can be deduced that

 Pf  CQ C is a constant

log Pf  logC   logQ

So the log log plot of this equation is a straight line with a slope of .  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 PB can be calculated and subtracted from Ps to find Pf.

8.311E  5q 2 Pf  Ps  2 2 after finding Pf ,  can be found by Cd At

 P  log f 2  Pf 1     Q  log 2   Q1  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.

H HB  PBQ

 Sub in PB  Ps  CQ  1 H HB  Ps  CQ

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

gH HB  P   1CQ  0 dQ s

 1 Pf  CQ Ps   1Pf  0 or Pf  PS  1 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

1 P  P fopt  1 s max

across the nozzles

 P  P  P  P Bopt s max fopt  1 s max

Calculate or measure a Pfqa @ some Qa then knowing Pfopt a Qopt can be calculated by  1  P  Q  Q anti log log fopt  opt a   P    fqa  With Qopt known the PBopt can be rewritten

8.3E  5Qopt PBopt  2 2 AtoptCd

8.3E  5Qopt solve for Atopt Atopt  2 Cd PBopt

 2 if all nozzles are the same size Atopt  nd nopt n is the number of nozzles 4

Atopt solve for dnopt d  nopt n

Example: DP 41/2” 20#/ft, Collars 7” 120.3#/ft 1000’

Mud 300 21,  29,  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 @ Pf1 2966 psi Q2 400 GPM @ Pf2 4883 psi

Find  8.311E5Q 2 PB  2 2 Cd At

8.311E515.5 3002 8.311E515.5 4002 PB1  2 2  631.6 psi PB2  2 2  1122.8psi .95 .45099 .95 .45099

Pf 1  2966  631.6  2344.4 psi  Pf 2  4883 1122.8  3760.2 psi

 P  log f 2  P  log3760.2     f 1   2344.4  1.66 Q2  log 400 log   300  Q1 

Find Qmax and Qmin

Q  1714 .8 1600  403gpm max  5440 Based on pump

Q  2.448 9.8752  4.52 85  268gpm min   60 Based on velocity

Optimum friction pressure

 1   1  Pfopt   Pp max   5440  2047 psi   1 1.66  1

Optimum pressure drop at the bit

PB  Pp max  Pfopt  5440  2047  3933psi

Optimum flow rate   1  Pfopt   1 2047 Q  Q anti log log   300anti log log  227gpm opt a   P  1.66 2334   fqa   

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

Optimum nozzle area

2 2 8.311E  5  Qopt 8.311E  515.5 227 2 Atopt  2  2  .15in Cd PBopt .95 3393

For 3 equal sized jets

d  2 .15  .25in nopt 3

Homework 10/31/08 The rate in gpm to have turbulent flow in the annulus of the previous homework. What will the friction drop be in the drill string and annulus if the fluid is water.

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

Fj  BQ PB B  .01823Cd  Q in gpm  in #/gal

 PB  Ps  Pf  Ps  CQ

 Fj  BQ Ps  CQ

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

H P  p max s Q substituting

H F  BQ p max  CQ  B H Q  CQ 2 j Q p max

Differentiate and set to 0

 1 dFj .5BH p max  (  2)CQ    0 dQ  2 H p maxQ  CQ

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

Solve for the optimum friction pressure

1 P  P fopt   2 sopt then solve for the optimum bit pressure

 1 P  P  P  P Bopt sopt fopt   2 sopt

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.

 Pj  BQ Ps max  CQ

Differentiate and set to 0

 1 dFj .5BP  (  2)CQ   s max  0 dQ  2 Ps maxQ  CQ

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

2 P  P fopt   2 s max Gives

 P  P  P  P Bopt s max fopt   2 s max

Example Same data as Hydraulic example

So =1.66 Qmax=4.3 gpm Qmin=268 gpm At 12,000 feet the pump pressure is the limiting factor.

2 2 5440 P  P   2975psi fopt   2 s max 1.66  2

PBopt  Ps max  Pfopt  5440  2975  2465psi

  1  Pfopt   1 2975 Q  Q anti log log   300anti log log  347gpm opt a   P  1.66 2334   fqa   

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

2 2 8.311E  5  Qopt 8.311E  515.5347 2 Atopt  2  2  .26in Cd PBopt .95  2465

d  2 .26  .332in  10.6 / 32" nopt 3

3 - 11 jets have an area of .27in2.

Section 4.13 in text, pages 156, 157