HW Assignment # 1 - 2022 - key Rotary Systems

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Louisiana State University *

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4045

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Mechanical Engineering

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Apr 3, 2024

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1 PETROLEUM ENGINEERING 4045 – DRILLING ENGINEERING HOMEWORK ASSIGNMENT # 1 – SPRING 2022 1) There are three fundamental functions that must be performed during drilling. List these below. a) Break the rock b) Remove the broken rock (cuttings) to clean the hole. c) Maintain the wellbore stable (keep the hole open, prevent kicks or loss of return) 2) We will study six major rig systems in this class. List these six systems. (#) a) Power b) Hoisting c) Circulating d) Rotary e) Well control f) Well monitoring Rig Power and Hoisting Systems 3) A 750-ton (1.5 x 10 6 lb) block system is to be used under the following conditions. Determine the Safety Factor. Casing String Section Length, ft Weight, lb/ft 3,200 47 4,100 53 2,900 43 Mud weight = 15.6 lb/gal Buoyance factor equation: BF = 1 – (mud weight / steel density) Steel density = 65.4 lb/gal Solution: • Weight of casing in air: Tcas Tcas = 3200 x 47 +4100 x 53 + 2900 x 43 = 492,400 lbf

2 • Total weight in mud: Tcm Tcm = 492,400 x 0.761 = 374,716 lbf • Safety Factor: SF SF = 1.5x10 6 / 374,716 ; SF = 4.0 4) A diesel engine was running at an angular velocity of 8000 rad/min at the drilling operations side. The torque output of the engine is 1,612.61 ft-lbf. The driller was trying to pull a drillstring 600,000 lbf. The engine was running for one hour. Calculate: a. The speed of the engine in rpm, b. The engine shaft output in hp, c. The drillstring velocity, d. The distance traveled by the drillstring. a) The speed of the engine can be calculated by using the given equation: ω = 2 π N 2 π N = 8000 N = 8000/2 π = 1273.24 rpm b) The engine shaft output is obtained as follows: P s = ω T = 8,000 rad/min x 1612.61 ft-lb / 33,000 ft-lbf/min/hp = 390.94 hp c) The drillstring velocity can be calculated as follows: v = P s / W = 390.94 hp x 33,000 ft-lbf/min/hp / 600,000 lbf = 21.5017 ft/min d) As the engine was running for one hour, so the total distance traveled by the drillstring within one hour is obtained as follows: W.d / t = W.v or d/t = v and then, d = 21.5017 ft/min x 60 min = 1290.102 ft 5) Assume that you are working on a medium-size drilling rig as a trainee. It has diesel engine-driven mechanical drive for the drawworks and rotary table. You have just drilled through a very sticky section of hole and are unable to lift the drillstring to make a connection because the total weight and drag exceeds the safe working strength of the drillpipe. You can move the pipe with much less drag while rotating. The toolpusher (your boss) wants to know if the engine powering the drawworks has enough power to simultaneously rotate while lifting the drillstring (reaming out of the hole). How much total input power is required for the drawworks and rotary given the following

3 assumptions? Hoisting – String weight - 200,000 lbs (includes the block weight) Drag - 20,000 lbs (if rotating) Speed - 50 feet per minute Efficiency Factor - 84% Rotary -- Torque - 8000 ft-lbs Speed - 80 rpm Mech Efficiency - 90% 1. Power required at Drawwork: P dw P dw = W . v b / E • Total Hook Load: W W = string weight + drag (rotating) W = 200,000 + 20,000 = 220,000 lbf • P dw = (220,000 lbf x 50 ft/min) / (0.84 x 33,000 ft-lbf/min/hp) P dw = 397 hp 2. Power required at Rotary Table: P rotary P rotary = 2. π .N.T / E P rotary = 2 x 3.14 x 80 rpm x 8,000 ft-lbf / (0.90 x 33,000 ft-lbf/min/hp) P rotary = 135 hp 3. Total Input Power required: P input req = P dw + P rotary P input req = 397 + 135 P input req = 532 hp

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4 Circulating System 1) Compute the volumetric output per stroke, the output per minute, and the strokes and minutes to complete displace the annulus and drill string for the following well: Pump (duplex) = 5-in bore (liner) 16-in (stroke length) 2,5-in rod 60 spm 85% efficiency Drillpipe = 11,700 ft = 4 1/2 in, 16.6 lb/ft = 0.01422 bbl/ft capacity Drillcollars = 7” OD x 3” ID = 600 ft Annulus = 9 7/8” well diameter Solution: Fp = 0.09633 bbl/stk. For duplex pump Flow rate = 0.09633 bbl/stk x 60 spm = 5.78 bbl/min Vol DP = 3.826 2 x 11700 / 1029.4 = 166.37 bbl Vol DC = 3 2 x 600 / 1029.4 = 5.25 bbl Vol annulus = ((9.875 2 – 7 2 ) x 600 + (9.875 2 – 4.5 2 ) x 11700) 1029.4 = 906.47 bbl Time to displace the well volume = Total volume/Flow rate=(166.37+5.25+906.47)/5.78=106.52 minutes= 1 hour and 47 min Number of strokes=Total volume / Fp =(166.37+5.25+906.47) / 0.09633 = 11,192 strokes

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