ANSWER FOR TEST Course Oil Well Drilling Technology Answer question 1 The main component parts of a rotary drilling rig and their function of each system o Power system supply energy to all of activit[.]
ANSWER FOR TEST Course: Oil Well Drilling Technology Answer question 1: The main component parts of a rotary drilling rig and their function of each system: o Power system: supply energy to all of activity on rig especially the hoisting and circulating systems o Hosting system: provide a means of lowering or raising drillstrings, casing strings, and other subsurface equipment into or out of the hole o Fluid circulating system:remove the rock cuttings from the hole as drilling progresses o Rotary system:use to achieve bit rotation o Well control system: make sure safety drilling progresses, prevent kick o Monitoring system: detect drilling problem quickly, provides information about the formation Answer question 2: The two basic types of rotary drilling bit are drag bits and rolling cutter bits o Drag bits: fixed cutter blades o Rolling cutter bits: rock bits - with cones Different between drag bits and rolling cutter bits: Drag bits Rolling cutter bits Beside the rotation around the Principle of Rotate around the axis of the axis of the bit, it requires the operating bit rotations around the axis of their rolling cones Average life Higher Lower Structure No seal and bearing bearing Materials Diamond Tungsten carbide Main destroying Cutting and abrasion Cutting and pound principle Tooth Insert tooth Milled tooth The first symbol is letters The first symbol is number Notation (D, M, S, T, O) (1,2,3) Exercise 1: Summary: a triplex pump having dl= in, Ls=11 in, strokes operating at 120spm, ∆P=3000 psia Calculate: a) The pump factor in unit of gal/stroke at 96/98/100%(Ev) volumetric efficiency (Fp) b) The flow rate in gal/min (Q) c) The energy expended by each piston during each cycle (A) d) The pump power developed in Hp under the above condition (PH) Answer/Solution: a) The triplex pump factor in unit of gal/stroke at 96% volumetric efficiency (Fp) is: FP = 3𝜋𝜋 𝐿𝐿𝑠𝑠 𝐸𝐸𝑣𝑣 𝑑𝑑𝑙𝑙2 = 3∗3.14 ∗ 11 ∗ 0.96 ∗ 62 = 895.277 𝑖𝑖𝑖𝑖 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 3.876 b) The flow rate in gal/min (Q) is: Q = 120*3.876= 465.079 gal/min c) The energy expended by each piston during each cycle (A) is: A = F*Ls= P*S*Ls=3000* 3.14*62/4*11/12= 77715 lbf.ft/ cycle d) The pump power developed in Hp under the above condition (PH) is: ∆𝑃𝑃𝑃𝑃 3000 ∗ 465.079 𝑃𝑃𝐻𝐻 = = = 814.024 𝐻𝐻𝐻𝐻 1714 1714 𝑔𝑔𝑔𝑔𝑔𝑔 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 atEv=0.98: Fp= 3.956 gal/stroke; Q= 474.768 gal/min;A= 77715 (lbf.ft/ cycle); PH= 814.024 Hp atEv=1: Fp=4.037gal/stroke; Q= 484.457gal/min;A= 77715 (lbf.ft/ cycle); PH= 847.941 Hp Exercises 2: Summary: Ldp= 11500ft, ODdp= 5in, IDdp= 4in Ldc= 500ft, ODdc= 8in, IDdc= 3in Density of drilling mud 11/12/13 lb/gal; density of steel 490 lb/gal Calculate: a) b) c) d) The axial tension and axial stress (0