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: Find pore volume from oil uptake.
Oil mass in pores \( = W_{\text{sat,air}} - W_{\text{dry,air}} = 32 - 30 = 2 \) g.
Given oil density \( \rho_o = 0.88 \) g/cc, pore volume is: \[ V_p = \frac{2}{0.88} = 2.2727 \text{ cc} \] Step 2: Find bulk volume using Archimedes' principle.
Loss of weight on immersion in oil: \[ \Delta W = W_{\text{sat,air}} - W_{\text{sat,oil}} = 32 - 24 = 8 \text{ g} \] This equals the weight of displaced oil: \[ \Delta W = \rho_o \, V_b \Rightarrow V_b = \frac{8}{0.88} = 9.0909 \text{ cc} \] Step 3: Compute effective porosity.
\[ \phi_e = \frac{V_p}{V_b} = \frac{2.2727}{9.0909} = 0.25 \] \[ \phi_e(\%) = 0.25 \times 100 = 25.00 \] Step 4: Conclusion.
\[ 25.00\% \]