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FMEV 2003Assignment 2 Question 1: CALCULATE the porosity of a formation composed of uniform spherical grains of radius 100μm in a cubic packing arrangement. The theoretical maximum porosity for a cubic packed rock made of spherical grains of a uniform size is 0.399 Question 2: CALCULATE the porosity of the formation if it were composed of uniform spherical grains of radius 10μm in a cubic arrangement instead. The theoretical maximum porosity for a cubic packed rock made of spherical grains of a uniform size is 0.4764 Question 3: Comment on your result for question 1 and 2. What is the difference and how can you explain it? As grain size increases past 100 mm, the frictional forces decrease and the porosity decreases until a limit is reached that represents random frictionless packing, which occurs at 0.399 porosity, and is independent of grain size. No further loss of porosity is possible for randomly packed spheres, unless the grains undergo irreversible deformation due to dissolution recrystallization, fracture, or plastic flow, and all such decreases in porosity are termed compaction. Question 4: A core plug 1 inch in diameter and 2 inches long is placed in the chamber of a gas expansion porosimeter whose container is 20cm3 large. The initial and final pressure are 28 and 23 psi respectively. Ignoring the dead volume of the apparatus, what is the porosity of the sample? Note: 1inch = 2.54 cm Vb = π (0.5 x 2.54)2 (2 x 2.54) = 25.74cm3 Using Boyle’s Law: P1 V1 = P2 V2 == 28 x 20 = 23 x Vg Therefore, Vg = 24.34 cm3 Thus, Φ = (Vb – Vg)/Vb == (25.74 - 24.34)/25.74 = 0.05 = 5% Question 5: A dry and clean core sample 1 inch in diameter and 4 inches long weighs 120 grams. Mineral analysis shows that the grains are 80% (by volume) calcite and 20% anhydrite. Estimate the sample's porosity. (ρca= 2.71 g/cm3, ρan= 2.98 g/cm3) Using: ρg= Σ Viρgi == (0.8 x 2.71) + (0.2 x 2.98) = 2.76 g/cm3 & Vg= ms/ ρg == 120/2.76 = 43.5 cm3 Vb = π (0.5 x 2.54)2 (4 x 2.54) = 51.48cm3 Thus, Φ = (Vb – Vg)/Vb == (51.48 – 43.5)/51.48 = 0.16 = 16%

Question 6: Suppose the core sample of Question 5 was obtained from a formation which contains only water (ρw= 1 g/cm3). In one location of the formation, the density log measured a bulk density of 2.48 g/cm3, what is the formation's porosity at that location?

ρlog= φ ρf+ (1 - φ)ρma φ = (ρlog - ρma)/ (ρf - ρma) φ = (2.48 – 2.76) / (1 – 2.76) = 0.159 = 16% Question 7: A dry, cylindrical quartz sandstone core, which is 10 cm long and 5 cm in diameter, weighs 350 grams. Calculate the porosity of the core. Vb= π(2.5)2 x 10 = 196.35 cm3

Φ= Question 8: For the porosity calculated in Question 7, is this the effective porosity of the core? If not, why not? Question 9: A clean and dry core sample weighing 550 g was 100% saturated with a 1.06 specific gravity (γ) brine. The new weight is 493 g. The core sample is 16 cm long and 7 cm in diameter. Calculate the porosity of the rock sample. Vb= π(3.5)2 x 16 = 615.75 cm3 VP = 1/y ( Vwet - Vdry) = (550 – 493)/1.06 = 53.77 cm3

φ = VP / VB = 53.77/615.75 = 0.09 = 9% Question 10. An experiment has been performed to determine the porosity of an irregularly shaped core sample. The clean weigh sample was weighed in air. It was then evacuated and fully saturated with an oil with a density of 0.85 g/cc and then weighed again in air. Afterward, the saturated sample was weighed when it was fully immersed in the oil. Here are the results of the experiment. Weight of dry sample in air = 42.40 g Weight of the saturated sample in air = 45.49 g Weight of the saturated sample immersed in the oil = 28.80 g a. Calculate the porosity of the core. Vg = (42.4 – 28.8)/0.85 = 16 cm3 VO = (45.49-42.4) / 0.85 = 3.64 cm3

VB = (45.49-28.8) – 3.64 = 13.05 cm3

Φ = (VB – Vg) / VB = (13.05 – 3.64) / 13.05 = 0.72 = 72% b. Is there enough information from this experiment to determine the minerology of the sample? If yes, what is it? Please justify your answer with appropriate arguments If the rock is predominantly composed of one mineral, e.g. quartz, then the mass of the clean and dry sample, ms, divided by the density of the mineral, ρg, equals the total volume of the grains Vg= ms/ ρg