FE 02-16B2 Lag Time Problem

Nội dung text: FE 02-16B2 Lag Time Problem

1. Problem 1: Given A semi-submersible is drilling in 500 ft of water. The air gap (the distance from the rig floor to the sea level) is 60 ft. A 24” riser with an ID of 22” connects the flow line to the seafloor. The following casing joints have been run: – 30” – 800 ft RKB (ID: 29”) – 20” – 2000 ft RKB (ID: 18.75”) – 13-3/8” – 6000 ft RKB (ID: 12.625”)
2. Problem 1: Given • The drilling parameters being used are: – WOB – 10 klbs – RPM – 125 – SPM – 120 – MW – 11.2 ppg – Pump Pressure – 2000 psi
3. Problem 2 • Using the data in Problem 1 calculate the down time in strokes. Assume 5” DP with an ID of 4.276 inches.
4. Solution: 1A. Pump Output Calculations • Output (gals/stroke) = (0.0386 x 12 x (6)2) x 0.2642 x 0.97 • Output (gals/stroke) = 4.273 gals/stroke • Output (bbls/stroke) = 0.102 bbls/stroke
5. Solution: 1C. Lag Time Measured in Strokes Annular volume = (D2 – d2) x 0.000971 x Length of annular section or Annular volume = ((D2 – d2) / 1029.4) x Length of annular section Pump Output (bbls/stroke) = Pump output (gpm) / (spm x 42) Lag time ann. sec (strokes) = Annular volume (bbls) / Pump Output (bbls/stroke)
6. Solution: 1C Total Lag Time (strokes) = (86.03 + 178.71 + 710.21 + 249.7) / 0.102 = 1248.08/0.102 = 12236 strokes
7. Solution: 1C. Total Lag Time (strokes) = (86.03 + 178.71 + 710.21 + 249.7) / 0.102 = 1224.65/0.102 = 12036 strokes
8. Solution: Problem 2 • Drill pipe volume = 4.2762 x 7471 x 0.000971 = 132.64 bbl • HWDP volume = 3.02 x 625 x 0.000971 = 5.46 bbl • Drill collar volume = 2.752 x 404 x 0.000971 = 2.97 bbl • Down Strokes = (132.64 + 5.46 + 2.97 ) / 0.102 = 141.07 / 0.102 = 1386 strokes