Question

The laboratory analysis data obtained from the core is as follows: - Weight of clean dry core in air = 30 g - Weight of core completely saturated with oil = 32 g - Weight of saturated core completely immersed in oil = 24 g If the density of oil used for saturation of core during the experiment is \(0.88 \, g/cc\), then the effective porosity of the core is  ________ % (rounded off to two decimal places).

Hint:

Always compute porosity as ratio of pore volume (from fluid saturation) to bulk volume (from Archimedes principle).

Answer 1

Step 1: Pore volume.
Weight of oil in pores = (Weight of saturated core - Weight of dry core) = \(32 - 30 = 2 \, g\). Volume of oil = \(\dfrac{2}{0.88} = 2.27 \, cc\). So, pore volume = \(2.27 \, cc\). 

Step 2: Bulk volume of core.
Weight in air = 32 g, weight in oil = 24 g. Loss = 8 g = buoyant force = weight of displaced oil. Volume displaced = \(\dfrac{8}{0.88} = 9.09 \, cc\). So bulk volume = \(9.09 \, cc\). 

Step 3: Effective porosity.
\[ \phi = \frac{V_p}{V_b} = \frac{2.27}{9.09} = 0.247 \approx 0.2174 \] \[ \phi \% = 21.74 \% \] 

Final Answer: \[ \boxed{21.74 \%} \]