# Production Essay

656 Words3 Pages
QUESTION 1: Gas | 8 MMSCFD at 0.6 SG | Oil | 1800 BPD at 40°API | Pressure | 800 psia | Temperature | 75°F | Droplet size | 120 micron removal | Retention time | 2 min | ft3 Ppc = 756.8 – 131.0g – 3.6 2g Tpc = 169.2 + 349.5g - 74.0g For specific 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 Data Book, Gas Processor Suppliers Association, Tulsa) g= SG.P.T.29 lb lbmoleP.TZ .lb mole 379 SCF g= 0.6800535(29)14.75350.88(379) =2.84 lbg dm=120 micron  cp (GPSA Fig 16.26) Assume CD = 0.34 : Determine Vt : Vt = 0.0119 [((51.5- 2.84) / 2.84) x (120 / 0.34)] ^ (1/2) = 0.549 Re = ( 0.0049 x 2.84 x 120 x 0.549) / 0.013 = 219.5 Determine Cd: Cd = (24 / 219.5) + (3 / (219.5^(1/2))) + 0.34 = 0.64 Repating steps using Cd = 0.64 : Determine 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 using 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) /