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QUESTION 1: posterior up | 8 MMSCFD at 0.6 SG | petroleum | 1800 BPD at 40°API | Pressure | 800 psia | Temperature | 75°F | Droplet coat | great deoxycytidine monophosphate micron removal | computer storage time | 2 min | ???????????????????????????????????ft3 Ppc = 756.8 131.0?g 3.6 ?2g Tpc = 169.2 + 349.5?g - 74.0??g For precise gravity = SG = 0.6 Ppc = 676.904 Tpc = 352.26 Ppr = P / Ppc Tpr = T / Tpc Ppr = 1.18 Tpr = 1.52 Z =0.88 (Compressibility factor is from engineering science Data Book, Gas central processing unit Suppliers Association, Tulsa) ?g= SG.P.T.29 lb lb gram moleculeP.TZ .lb mole 379 SCF ?g= 0.6800535(29)14.75350.88(379) =2.84 lbg dm= great hundred micron ??????? cp (GPSA Fig 16.26) Assume CD = 0.34 : pin down Vt : Vt = 0.0119 [((51.5- 2.84) / 2.84) x ( great hundred / 0.34)] ^ (1/2) = 0.549 Re = ( 0.0049 x 2.84 x 120 x 0.549) / 0.013 = 219.5 interpret Cd: Cd = (24 / 219.5) + (3 / (219.5^(1/2))) + 0.34 = 0.64 Repating go apply Cd = 0.64 : place Vt : Vt = 0.0119 [((51.5- 2.84) / 2.84) x (120 / 0.64)] ^ (1/2) = 0.674 Re = ( 0.0049 x 2.84 x 120 x 0.674) / 0.013 = 86.579 Determine Cd: Cd = (24 / 86.579) + (3 / (86.579^(1/2))) + 0.34 = 0.

94 Repating steps utilise Cd = 0.94 : Determine Vt : Vt = 0.0119 [((51.5- 2.84) / 2.84) x (120 / 0.94)] ^ (1/2) = 0.557 Re = ( 0.0049 x 2.84 x 120 x 0.557) / 0.013 = 71.55 Determine Cd: Cd = (24 / 86.579) + (3 / (86.579^(1/2))) + 0.34 = 1.03 Repating steps using Cd = 1.03 : Determine Vt : Vt = 0.0119 [((51.5- 2.84) / 2.84) x (120 / 1.03)] ^ (1/2) = 0.532 Re = ( 0.0049 x 2.84 x 120 x...If you want to contain a full essay, countersink it on our website: Orderessay

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