Question

A homogeneous sandstone reservoir is under a radial steady state flow. The wellbore radius is 0.1 m. The formation near the wellbore is damaged up to 0.9 m from the sand face. The permeability impairment results in \( k/k_s = 5 \), where \( k \) is the permeability in the undamaged region and \( k_s \) is that of the damaged region. The value of skin factor is (rounded off to two decimal places) ______________.

Hint:

Skin factor increases significantly when permeability of the near-wellbore zone is much lower than the undamaged region.

Answer 1

Correct Answer: 9

For radial steady flow, the skin factor for a damaged zone is given by:
\[ S = \left(\frac{k}{k_s} - 1\right)\ln\left(\frac{r_s}{r_w}\right) \]
Given:
Wellbore radius \( r_w = 0.1 \, \text{m} \)
Damaged radius \( r_s = 0.9 \, \text{m} \)
Permeability ratio \( k/k_s = 5 \)
Substitute values into the formula:
\[ S = (5 - 1)\ln\left(\frac{0.9}{0.1}\right) \]
\[ = 4 \ln(9) \]
\[ \ln(9) = 2.1972 \]
\[ S = 4 \times 2.1972 = 8.7888 \]
Rounded off to two decimal places:
\[ S = 8.79 \]
Final Answer: 8.79