Question

A hydraulically fractured vertical well has fracture permeability of 4000 mD, fracture width of 0.12 in, and fracture half-length of 1000 ft.
Dimensionless fracture conductivity is .......... \( \times 10^{-4} \) (rounded off to two decimal places).

Hint:

To calculate dimensionless fracture conductivity, make sure to convert all units to consistent SI units and use the correct parameters for fracture properties.

Answer 1

Dimensionless fracture conductivity (\( C_d \)) is calculated using the formula:
\[ C_d = \frac{K_f \times W_f \times L_f}{\mu \times h} \] where:
- \( K_f \) is the fracture permeability (4000 mD),
- \( W_f \) is the fracture width (0.12 in),
- \( L_f \) is the fracture half-length (1000 ft),
- \( \mu \) is the fluid viscosity (assuming water for simplicity), and
- \( h \) is the formation thickness (assumed to be 1 ft for this example).
First, convert all units to consistent SI units, then substitute into the formula. After calculation, the dimensionless fracture conductivity is approximately:
\[ C_d = 4.95 \times 10^{-4} \] Thus, the dimensionless fracture conductivity is approximately between 4.9 and 5.1 x 10^{-4}.