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:

Core porosity from weights: Pore volume comes from {(saturated weight in air - dry weight in air)}/$\rho_f$. Bulk volume comes from {(saturated weight in air - immersed weight)}/$\rho_f$. Then $\phi = V_p/V_b$.

Answer 1

Correct Answer: 25

Step 1: Calculate pore volume from the absorbed oil. 
When the rock sample becomes oil-saturated, the increase in its weight represents the mass of oil filling the pore spaces.
The mass of oil inside the pores is:

\[ W_{\text{sat,air}} - W_{\text{dry,air}} = 32 - 30 = 2 \text{ g} \]

Given the oil density $\rho_o = 0.88$ g/cc, the pore volume is obtained as:

\[ V_p = \frac{2}{0.88} = 2.2727 \text{ cc} \]

Step 2: Determine bulk volume using buoyancy.
When the saturated sample is immersed in oil, the apparent loss in weight is equal to the weight of the displaced oil.
This weight difference is:

\[ \Delta W = W_{\text{sat,air}} - W_{\text{sat,oil}} = 32 - 24 = 8 \text{ g} \]

Using Archimedes’ principle, the bulk volume of the sample is:

\[ V_b = \frac{8}{0.88} = 9.0909 \text{ cc} \]

Step 3: Evaluate the effective porosity.
Effective porosity is defined as the ratio of pore volume to bulk volume:

\[ \phi_e = \frac{V_p}{V_b} = \frac{2.2727}{9.0909} = 0.25 \]

Expressing this value as a percentage:

\[ \phi_e(\%) = 0.25 \times 100 = 25.00\% \]

Step 4: Final answer.
 

\[ \boxed{25.00\%} \]