Drill Bit Hydraulics

1 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. PB E 4 v n2 = 0. PB Pressure drop across bit, vn nozzle velocity Solving for nozzle velocity PB. vn =. E 4 . In the field it has been shown that velocity predicted by this equation is off. So it has been modified, PB. vn = C d E 4 . the recommended valve for Cd is .95. If 3 nozzles are present q1 q 2 q3. v= = = the velocity is equal in all the jets. A1 A2 A3. q = q1 + q 2 + q3 = v n A1 + v n A2 + v n A3. That gives us vn = q In field units vn = q A. At t q in gpm, At in inches2, vn in ft/sec solving for the pressure drop 5 q 2. PB =. C d2 At2. is #/gal Flow Exponent . It can be deduced that Pf = CQ C is a constant log Pf = log C + log Q. 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.

2 Ps=Pf+PB so by using B. the above equation PB can be calculated and subtracted from Ps to find Pf. 5 q 2. Pf = Ps after finding Pf , can be found by C d2 At2. P . log f 2 . Pf1 . =. log Q2 . 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 = PB Q. Sub in PB = Ps CQ . H HB = Ps CQ +1. Take the first derivative of H with respect to Q set the result to 0. gH HB. = Ps ( + 1)CQ = 0. dQ. Ps ( 1)Pf = 0. 1. Pf = CQ 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. Pfopt = Ps max +1. across the nozzles . PBopt = Ps max Pfopt = Ps max +1. Calculate or measure a Pfqa @ some Qa then knowing Pfopt a Qopt can be calculated by 1 P . Qopt = Qa anti log log fopt . P . fqa . With Qopt known the PBopt can be rewritten 5 Qopt PBopt = 2.

3 Atopt C d2. 5 Qopt solve for Atopt Atopt =. C d2 PBopt . if all nozzles are the same size Atopt = 2. nd nopt n is the number of nozzles 4. Atopt solve for dnopt d nopt =. n . Example: DP 41/2 20#/ft, Collars 7 #/ft 1000'. Mud 300 21, 600 29, #/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 . 5 Q 2. PB =. Cd2 At2. 5 300 2 5 400 2. PB1 = = psi PB 2 = = psi . 2 . 2. Pf 1 = 2966 = psi Pf 2 = 4883 1122 .8 = 3760 .2 psi P . =. log f 2. P log f1 . =. (. = ). log Q2 log 400. 300. ( ). Q1 . Find Qmax and Qmin (. Qmax = 1714 .8 1600. 5440. ) = 403gpm Based on pump Qmin = ( )85 = 268 gpm Based on velocity 60. Optimum friction pressure 1 1 . Pfopt = Pp max = 5440 = 2047 psi + 1 + 1 . Optimum pressure drop at the bit PB = Pp max Pfopt = 5440 2047 = 3933 psi Optimum flow rate 1 P . Qopt = Qa anti log log fopt = 300anti log 1 log 2047 = 227 gpm P 2334.

4 Fqa . This is lower than the max and higher than min flow rates. Optimum nozzle area 5 Qopt 2. 5 227 2. Atopt = = = .15in 2. C d2 PBopt 2..95 3393. For 3 equal sized jets d nopt = 2 .15 = .25in 3 . 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 F j = BQ PB B = .01823C d Q in gpm in #/gal PB = Ps Pf = Ps CQ . F j = 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 max Ps =. Q. substituting H p max F j = BQ CQ = B H p max Q CQ 2. Q. Differentiate and set to 0. dF j =. [..5 B H p max ( + 2)CQ +1 ]=0. dQ H p max Q CQ + 2. For a valid solution the numerator must be equal to zero.

5 Solve for the optimum friction pressure 1. Pfopt = Psopt +2. then solve for the optimum bit pressure +1. PBopt = Psopt Pfopt = Psopt +2. 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. dF j =. [..5 B Ps max ( + 2)CQ +1 ]=0. dQ Ps max Q CQ + 2. For a valid solution the numerator must be equal to zero. 2. Pfopt = Ps max +2. Gives . PBopt = Ps max Pfopt = Ps max +2. Example Same data as hydraulic example So = Qmax= gpm Qmin=268 gpm At 12,000 feet the pump pressure is the limiting factor. 2 2 5440. Pfopt = Ps max = = 2975 psi +2 + 2. PBopt = Ps max Pfopt = 5440 2975 = 2465 psi 1 Pfopt . Qopt = Qa anti log log = 300anti log 1 log 2975 = 347 gpm P 2334 . fqa . It is bounded by the min and max flow rates, so 5 Qopt 2. 5 347 2. Atopt = = = .26in 2. C d2 PBopt .95 2465.

6 2. d nopt = 2 .26 = .332in = / 32". 3 . 3 - 11 jets have an area of .27in2. Section 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. d p ( p f ) ..5. v s = . C D f . d p the diameter of the cuttings inches p the density of the cuttings 21 ppg CD Drag coefficient Particle Reynolds number v s d p Rp =.. which gives 40. CD =. RP. Substituting in the first equation 4980 d p2 ( p f ). vs =.. For values of Rp greater than 1 which means laminar flow around the particle the drag coefficient can be found using 22. CD =. R .p5. So the slip velocity equation becomes 175d p ( p f )..667. vs =.

7 F333 .333. 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 for that section. 4) Using this flow exponent optimize the bit Hydraulics . 5) When drilling confirm your plan by finding the 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.