Assignment #2 Hydrostatic Pressure Due: December 25 Name: Ali kamel abdul-ameer Include additional worksheets with calculations for partial credits in the event your final answer is not correct. 1 Calculate the hydrostatic pressure at the bottom of the fluid column for each case shown. 12 the Calculate the equivalent A well has following profile: density at 7000' depth of the following mixed fluid column. Gravel Sp. gr = 2.68 Porosity 25% 5720 Casing 5720 5720 Total depth 16,400' (vertical well) 16'' Open Hole, Surface casing 13 3/8" OD and 12 3/7" ID of 68.00 lb/ft set at 2000'. 12.5'' Open Hole, Intermediate Casing 9 5/8"OD and ID of 8 2/3", 47 lb/ft, set at 8,458' .8.5'' Open Hole, Drilling liner 7" (6" ID) (Grade N-80) interval between 8,400' to 14,200' Open Hole, Bit diameter 5.5" Drill string: 5" drill pipe 19.50 lb/ft nominal weight, (4.276" ID) surface to 8300'. 3 1/2'' drill pipe 13.30 lb/ft nominal weight between 8,300' to 15,200'. (For the ID on the 3 1/2" drill pipe remember that density of steel is 65.5 lb/gal) Drill collars, 4 3/4" OD, x 2 1/4" ID, 46.70 ppf, 15200' to TD. Surface mud system: 3 mud tanks, each 9' high, 7' wide, 36' long. With the entire drill string out of the hole: Mud level in tanks No. 1 & 2 is 72" of mud and tank No. 3 has 60" of mud (M.W: 11 ppg) For the following 2 π½π½π»π»π»π»π»π»π»π» 3 Calculate the total capacity of the surface mud system in bbl. ππ × ππ × ππππ = ππ × = ππππππππ. ππππ ππππππ ππ. ππππππ Calculate total mud volume in the three tanks in bbls. questions 2 to 9, do calculation by hand using equations from the lectures. ππππ ×ππ×ππππ ππππ ×ππ×ππππ π½π½ππππππ = ππ οΏ½ππππππ.ππππππ οΏ½ + οΏ½ππππππ.ππππππ οΏ½ = ππππππ. ππππ ππππππ 4 Calculate the capacity in bbl/ft for each well section. (ππππ. ππππππ)ππ ππππππ π½π½πΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊ(ππ) = = ππ. ππππ ππππππππ. ππ ππππ (ππ. ππππππππ)ππ ππππππ π½π½πΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊ(ππ) = = ππ. ππππππππ ππππππππ. ππ ππππ (ππ)ππ ππππππ π½π½πΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊ(ππ) = = ππ. ππππππππ ππππππππ. ππ ππππ 5 Calculate total hole volume with the entire drill string out of the hole. οΏ½ππ. ππππππππππ × (ππππππππ)οΏ½ + οΏ½ππππ × (ππππππππππ − ππππππππ)οΏ½ + (ππ. ππππ × (ππππππππππ − ππππππππππ)) π½π½ππππππππ = = ππππππ. ππππ ππππππ ππππππππ. ππ 6 Calculate the displacement of the "5 drill pipe. ππππ − ππ. ππππππππ ππππππ π½π½π π π π π π π π = × ππππππππ = ππ. ππππππππππ ππππππππ. ππ ππππ 7 Calculate the displacement of the complete drill string. π½π½π π π π π π π π π π −ππππππππππππ = [ππππ − (ππ. ππππππ)ππ ]ππππππππ + [(ππ. ππππππ)ππ − (ππ. ππ)ππ ](ππππππππππ − ππππππππ) + [(ππ. ππππ)ππ − (ππ. ππππ)ππ ](ππππππππππ − ππππππππππ) = ππππππ ππππππ ππππππππ. ππ 8 Calculate the total hole volume with the bit on the bottom. π½π½π»π»π»π»π»π»π»π»π»π» = π½π½ππππππππ − π½π½π π π π π π π π π π −ππππππππππππ = ππππππ. ππππ − ππππππ = ππππππ. ππππ ππππππ 9 Calculate the hydrostatic pressure at bottom of hole. π·π·ππ = ππ. ππππππ × ρ × π»π»π»π»π»π» = ππ. ππππππ × ππππ × ππππππππππ = ππππππππ. ππ ππππππ 10 Calculate the pressure at the bottom of each well. 10 Calculate the pressure at the bottom of each well. 4258.8 psi 4212 psi 11 Given the following well design: 735.243 psi a)Calculate total hole volume with the entire drill string in the hole. π½π½ππππππππ = (ππππ. ππππππ)ππ (ππ)ππ (ππ. ππ)ππ − (ππ. ππ)ππ (ππππππππ − ππππππππ) − (ππππππππ) οΏ½ · ππππππππ + οΏ½ = ππππππ. ππππ ππππππ ππππππππ. ππ ππππππππ. ππ ππππππππ. ππ π½π½ππππππππ = (ππππ. ππππππ)ππ (ππ)ππ (ππππππππ − ππππππππ) = ππππππ. ππππππππππ · ππππππππ + ππππππππ. ππ ππππππππ. ππ b)Calculate total hole volume with the entire drill string out of the hole. CSG Specification: (13 3/8" CSG (12.5" ID) and 7" Liner (6" ID)) Drill Pipe Specification: (4.5" DP (3.5" ID)) 12 Calculate the equivalent density at 7000' depth of the following mixed fluid column. 1200ft 2800ft π»π»π»π»π»π» ππππππππππππππππππππππ ππππππππππππππππ ππππ ππππππ ππππππππππππ ππππ βΆ − π·π·ππ = ππ. πππππποΏ½(ππππππππ × ππ) + (ππππππππ × ππππ) + (ππππππππ × ππππ)οΏ½ = ππππππππ. ππ ππππππ πΊπΊπΊπΊ ππππππ ππππππππππππππππππππ π π π π π π π π π π π π π π βΆ − ρ= 4500ft π·π·πͺπͺ ππππππππ. ππ = = ππππ. ππππππ ππππππ ππ. ππππππ × π»π»π»π»π»π» ππ. ππππππ × ππππππππ