HW Assignment # 1 - 2022 - key Rotary Systems

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Louisiana State University *

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Course

4045

Subject

Mechanical Engineering

Date

Apr 3, 2024

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pdf

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4

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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
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