Question

A 1.0” diameter by 1.5” long core was cleaned. The measured dry and saturated densities are 2.160 gm/cc and 2.311 gm/cc. The measured porosity is 15.85%. In the process, the following fluid volumes were removed: 1.84 ml of water and 1.22 ml of oil. The oil has a measured density of 0.77 gm/cc and based on salinity measurements, the water corresponds to a brine having a density of 1.05 gm/cc. Calculate fluid saturations and determine the lithology.

Answer 1

Given :- Page 1 Diameter of core = "=0.0254 m. radius of core = 0.5" = 0.0125 m. Length of core - 1.5" = 0.0381 m. Dry density - 2.160 gefce Saturated Density = 2.31 Porosity - 15.85 1. Volume of water - 1.84 me. Volume of oil = 1,22 me t er von (Soil) Density of oil - 0.779 m/l. (Swater) Density of water = 1:05 gm se, Fluid saturutions - ? Lithology = ? Soin ON Fluid saturation - Total volume of fluid phase. e Pore volume. Total volume of fluid phase volume of water volume of - Oil, = 1.84 + 1.22 = 3.06 me = 3.06 x 100 m3 Total volume of fluid phase volume can following be calculated by using Pore formula Porosity Pore volume (up) Totul volume (vt) Xloo (Pt

15.85 - - Page - Pore volume (up) Total volume (va) xoo Total volume - TN2L = nx (0125)? x 0. 0381 Total volume = 1.8702 x 15 m² put this value in ear ② 15.85 = pore Volume (Ve) 1.8702 x 105 & 100 no lo enter Pore volume = 2.9642 x 100 m² From ear Awid saturactions = Saturations- Total volume of fluid phase olan pore volume. 3.06 x 100 m² 2.9642 x 150 m² Fluid saturations - Fluid saturations - 1.0323 XX