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Louisiana State UniversityPETE 4045HW Assignment 3 - key

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PETROLEUM ENGINEERING 4045 – DRILLING ENGINEERING HOMEWORK ASSIGNMENT # 3 -- FALL 2017 Drilling Fluids 1- A 15-in hole is drilled to a depth of 4,000 ft. The API water loss of the mud is 10 mL. ... Approximately 30% of the lithology is permeable sandstone and the rest is impermeable shale. a. Construct a plot of estimated filtration loss in barrels vs. time in hours (0 to 24 hours) that would occur if the hole where drilled instantaneously. Assume porosity is 25% (answer: 42.4 bbl after 24 hr) 30% of permeable sandstone = 0.30 (4,000 ft) = 1,200 ft Vf = Const. A. √? From API water loss test: Vf = 10 ml = 62.9.10-6 bbl; A= 45 cm2 = 48.44.10-3 ft2 and t = 30 min= 0.5 hr Then: Const = 1.84. 10-3 bbl/ft2/h1/2 For the sandstone area exposed: A = 3.14 (15 in) (1,200) / 12 = 4,712.4 ft2 Then the filtrate volume is: Vf = 8.671. √? (with this equation we construct the plot asked above) After 24 hr we will have: Vf = 8.671. √24 = 42.48 bbl b. Compute the radius of the invaded zone for Part a. in inches after 24 hours. (answer: 9.62 in) Porous volume around the well invaded by the filtrate after 24 hours: Vporous Vporous = (3.14/4).(Dinv2 – Dwell2).L.ϕ = (3.14/4).( Dinv2 – 152).1,200.(0.25) But, Vporous = Vf = 42.48 bbl Then: Dinv = 19.26 in = 9.63 in c. Repeat Part a. assuming a drilling rate of 200 ft/hour. (answer: 31.6 bbl after 24 hr) Here we must consider the area drilled per hour as: A = Dwell . π. R. t / 12 (where R is the drilling rate) Substituting in the integral equation (see textbook page 46; equation (2.7), and integrating, we obtain: Vf = 1.84. 10-3. π. Dwell. R.√?33 / 12 / √3 = 0.834. √?3 For drilling the sandstone @ 200 ft/h drilling rate it is going to take 6 hours, then: Vf (after 6 hours) = 0.834. √63 = 12.16 bbl For the remaining 18 hs we suppose instantaneously drilling, then: Vf = 8.671. (√24 - √6 ) = 21.24 bbl In 24 hours we have: Vf (24 hs) = 12.16 + 21.24 = 33.4 bbl [Show More]

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