Drilling Questions and Solutions [PDF]

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PETROLEUM ENGINEERING DEPARTMENT PETE-4045 – DRILLING ENGINEERING EXAM # 3 – Fall 2016 (11/18/2016)



Name:



LSU ID #:



General Instructions:  Closed book exam.  No cell phones, I-pods, pocket PC’s, etc. are allowed.  You must show your calculations and underline or box the answer to receive full credit.  Failure to indicate the unit in your calculation will result in losing points.  Use the back side of the previous page or the blank sheets at the end of the exam if you need more space to complete a problem.  Make and state clearly any assumptions that you believe are necessary.  Formula sheet is in the last page of exam.



November / 2016



1) The mud density of a well is being increased from 10 to 12 lbm/gal. If the pump is stopped when the interface between the two muds is at depth of 8,000 ft in the drillstring, what pressure must be held at the surface by the annular blowout preventers to stop the well from flowing? What is the equivalent density in annulus at 4,000 ft after the blowout preventers are closed? Pressure increase at BOP annular = 0.052 * (12 -10) * 8,000 = 832 psig Pressure @ 4,000 ft = 832 + 0.052 * 10 * 4,000 = 2,912 psig Equiv. density @ 4,000 ft = 2,912 / 0.052 / 4,000 = 14 lbm/gal 832 psi 14 lbm/gal 2) Calculate the length of 8” O.D. x 3” I.D 147 lbf/ft drill collars required to safely provide 40,000 lbs WOB if the mud weight is 12.0 ppg. Safety Factor = 30% α = 1 – ρmud / ρsteel = 1 – 12 / 65.5 = 0.817 For safety provide the WOB



433 ft



40,000 lbf:



WOB * 1.3 = 147 * α * LDC LDC = WOB * 1.3 / (147 * α) = 40,000 * 1.3 / (147 *0.817) = 433 ft 3) Given the following conditions: - Flow rate: 500 gpm - Jet bit nozzles total area: 0,45 in2 - Mud weight: 10 lb/gal - Last casing shoe depth: 3,533 ft - Well depth: 10,500 ft - The following frictional pressure losses: Ps = 40 psi ; surface equipment; Pdp= 450 psi; interior of drillpipe; Pdc= 250 psi; interior of drill collars; Padc= 75 psi; annular drill collars - well; Padp= 50 psi; annular drillpipes – well; Pacas= 40 psi; annular casing – drill string. Correlate column A with column B, below: 12345-



Pressure drop across the bit Pressure at casing shoe Pump pressure at surface Bottom hole pressure (annular) Fluid pressure at the shale shakers 



( 4 ) 5,625 psi ( 5 ) 14.7 psi ( 2 ) 1,877 psi ( 1 ) 1,137 psi ( 3 ) 2,042 psi



(∆Pbit) = 8.311x10-5 x ρ x q2 / (Cd2 x At2) = 8.311x10-5 * 10 * 5002 / 0.952 / (0.45)2 = 1,137 psi



4) An operator desiring to increase the density of cement slurry may do so by adding the following material __________________ a) Perlite b) Bentonite c) Barite (primary source for this application) d) Salt 5) In determining the volume of an open hole with the diameter-squared method, a slightly larger figure for diameter should be used to allow for _______________ a) Larger casing ID. b) Hole enlargement c) Errors in calculation d) An extra amount of cement slurry as a safety factor. 6) Pressure should be bled off the casing before ______________ a) The top plug ruptures. b) The cement sets, so the pipe will not bulge. c) The valve in the float collar closes. d) None of the above. 7) Benefits derived from using a bottom plug in the cementing operation include ____________ a) Wiping mud film from inside the casing. b) Reducing slurry contamination. c) Preventing entry of air into slurry. d) All of the above. 8) List three types of cement jobs (applications) that were discussed in class. Cementing casing Setting cement plugs Squeezing cement at casing shoe Repair casing leaks Seal off perforations Combat loss circulation problems, etc.



9) Complete with the correct bit number:



2 4 1 3



10) Casing having an OD of 9.625” and an ID of 8.535” is to be cemented at a depth of 13,300 ft in a 12.25” borehole. A 40 ft shoe joint will be used between the float collar and the guide shoe. It is desired to place 2,000 ft of cement in the annulus. Each sack of Class G cement will be mixed with 4.97 gal of water to which is added 15% salt (by weight of water). A small quantity of dispersant will be blended with the cement, but this additive has no significant effect on the slurry yield or density. Assume an excess factor of 1.3. The salt will be added to the water phase and thus does not blend with the dry cement. i) Cement specific gravity: 3.14 ii) Salt specific gravity: 2.17 Compute: 15.92 Lbm/gal a. the density of the slurry, b. the yield of the slurry, 1.19 ft3/sack c. the number of sacks of cement required Vol. of water = 4.97 gal/sack 702 or 698 sacks Vol. of cement = 94 lbm/sack / (3.14 * 8.33) lbm/gal = 3.59 gal/sack Vol. of salt = 0.15 * mass of water / (2.17 * 8.33 lbm/gal) Mass of water = 4.97 gal/sack * 8.33 lbm/gal = 41.40 lbm/sack Vol. of salt = 0.15 * 41.40 / (2.17 * 8.33) = 0.344 gal/sack Vol. slurry = 4.97 + 3.59 + 0.344 = 8.90 gal/sack Mass of slurry = mass water + mass cement + mass of salt = 41.4 + 94 + 0.15 * 41.40 = 141.61 lbm/sack Density slurry = mass slurry / vol. slurry = 141.61 / 8.90 = 15.92 lbm/gal Yield = 8.90 gal/sack / 7.48 ft3/gal = 1.1898 ft3/sack = 1.19 ft3/sack Number of sacks: N = Vol. slurry total * 1.3 / yield or 1.3 applied only to the open-hole. Vol. slurry total = (12.252 – 9.6252) * 2000 / 1029.4 + 8.5352 * 40 / 1029.4 = 114.4 bbl = 4,805 gal N = 4,805 gal * 1.3 / 8.90 gal/sack = 702 sacks



OR Vol. slurry total = (12.252 – 9.6252) * 2000*1.3 / 1029.4 + 8.5352 * 40 / 1029.4 = 147.9 bbl = 6210 gal N = 6210 gal / 8.90 gal/sack = 698 sacks Cementing time = time mix + time displace = 1087 / 20 + 8.8352 * (13300 – 40) / 1029.4 / 9 = 166.1 min



11) A well is being drilled at a depth of 5,000 ft using water having a density of 8.33 lb/gal and a viscosity of 1 cP as the drilling fluid. The drillpipe has an external diameter of 4.5 in. and an internal diameter 0f 3.826 in. The diameter of the hole is 6.5 in. The drillstring fluid is being circulated at a rate of 500 gal/min. Assume a relative roughness of zero. a. Determine the flow pattern in the drillpipe. Water is a Newtonian fluid  Mean velocity: V = q / (2.448*d2) V = 500 / (2.448*3.8262) = 13.953 ft/s  NRe = 928*ρ*V*d/µ = 928*8.33*13.953*3.826/1 = 412,673 > 2,100  Flow pattern is turbulent b. Determine the frictional pressure loss gradient inside the drillpipe in psi/ft. dpf/ft = ρ0.75 * V1.75 * µ0.25 / (1,800 * d1.25) = 8.330.75 * 13.9531.75 * 10.25 / (1,800 * 3.8261.25) = 0.0513 psi/ft



Turbulent



0.0513 psi/ft



c. Determine the flow pattern in the annular opposite the drillpipe. Turbulent Mean velocity: V = q / 2.448 / (d22 – d12) V = 500 / 2.448 / (6.52 – 4.52) = 9.284 ft/s NRe = 757*ρ*V*(d2 – d1) / µ = 757*8.33*9.284*(6.5 – 4.5) / 1 = 117,086 > 2,100 Flow pattern is turbulent d. Determine the frictional pressure loss gradient in the drill pipe annulus in psi/ft. 



dpf / 1,000 ft = ρ0.75 * V1.75 * µ0.25 / 1,396 / (d2 – d1)1.25 = 8.330.75 * 9.2841.75 * 10.25 / 1,396 / (6.5 – 4.5)1.25 = 0.0729 psi/1,000 ft



0.0729 psi/ft



12) The bit currently in use has three 12/32” nozzles. The driller has recorded that when the 10 lbm/gal mud is pumped at a rate of 500 gal/min, a pump pressure of 3,000 psig is observed, and when the pump is slowed to a rate of 250 gal/min, a pump pressure of 800 psig is observed. The pump is rated a 1,000 hp and has an overall efficiency of 0.9. The minimum flow rate to lift the cuttings is 240 gal/min. The maximum allowable surface pressure is 3,000 psig. a) Determine the proper pump operating conditions (qmax) and bit jet nozzles total area (At) for maximum bit hydraulic horsepower for the next bit run, if optimum hydraulics occurs in Interval 1. 514 gpm 0.344 in2 Pressure losses @ two points measured by the driller: ∆pbit 1 = 8.311x10-5 *10*5002 / 0.952 / [3*π*(12/32)2/4]2 = 2,097 psi ∆pd 1 = 3,000 – 2,097 = 903 psi ∆pbit 2 = 8.311x10-5 *10*2502 / 0.952 / [3*π*(12/32)2/4]2 = 525 psi ∆pd 2 = 800 – 525 = 275 psi Determine m m = log (∆pd1/∆pd2) / log (q1/q2) = log (903/275) / log (500/250) = 1.715 Maximum Flow Rate (interval 1); qmax = 1,714 * PHP * Efficiency / Pmax = 1,714 * 1,000 * 0.9 / 3,000 = 514 gpm Constant C for the Parasitic Pressure Loss equation: C = ∆pd 1 / qm = 903 / 5001.715 = 0.02123 ∆pd @ qmax = 0.02123 * 5141.715 = 947 psi Then ∆pbit optimum = 3,000 – 947 = 2,053 psi o At opt = [8.311x10-5 *10*5142 / 0.952 / 2,053]0.5 = 0.344 in2 b) What bit horsepower will be obtained at the conditions selected? 



HPbit = 514 * 2,053 / 1,714 = 616 hp



c) What impact force will be obtained at the conditions selected? 



1,275 lbf



Fj = 0.01823*0.95*514*(10*2053)0.5 = 1,275 lbf



d) What nozzle velocity will be obtained at the conditions selected? 



616 hp



Vnozzle = qopt / At = 514 gal/min *0.3209 min-ft-in2/gal-s / 0.344 in2 = 479 ft/s



479 ft/s



PETROLEUM ENGINEERING 4045 – DRILLING ENGINEERING EXAM #3 – FALL 2016 Formula Sheet



1 bbl = 5.615 ft^3, 1 gal = 231 in^3



pp  pb  pd



Nre =



̅.𝑑 928.ρ.𝑣 μ



𝑁𝐻𝑒 = 𝑑𝑝𝑓 𝑑𝐿



37,100.𝜌.𝜏𝑦. 𝑑 2 𝜇𝑝 2



= 𝐶𝑜𝑛𝑠𝑡 𝑥 𝑞1.75 = 𝐶𝑜𝑛𝑠𝑡 𝑥 𝑞𝑚



𝑑𝑝𝑓 𝑑𝐿



=



𝜌0.75 .𝑉 1.75 𝜇0.25 1,800.𝑑 1.25



=



𝜌0.75 .𝑞1.75 𝜇0.25 8,624.𝑑 1.25



- Maximum bit hydraulic horsepower pp pd  m 1 - Maximum jet impact force 2p p pd  m2 - Maximum pump discharge pressure qmax 



1,714 PHP E 1,714(1600)(0.85)   681gpm pmax 3,423