Question

A hydraulically fractured vertical well has fracture permeability of \(4000 \, mD\), reservoir permeability of \(80 \, mD\), fracture width of \(0.12 \, in\), and fracture half-length of \(1000 \, ft\). Dimensionless fracture conductivity is \(\underline{\hspace{1cm}} \times 10^{-4}\) (rounded off to one decimal place).

Hint:

Dimensionless fracture conductivity is a measure of fracture effectiveness. A value above 1 indicates infinite conductivity fracture, while a very low value indicates ineffective fracture.

Answer 1

Step 1: Formula.
Dimensionless fracture conductivity is: \[ C_f = \frac{k_f w}{k L_f} \] where - \(k_f\) = fracture permeability, - \(w\) = fracture width, - \(k\) = reservoir permeability, - \(L_f\) = fracture half-length. 

Step 2: Unit conversions.
Fracture width: \[ w = 0.12 \, in = \frac{0.12}{12} \, ft = 0.01 \, ft \] 

Step 3: Substitute values.
\[ C_f = \frac{4000 \times 0.01}{80 \times 1000} \] \[ = \frac{40}{80000} = 0.0005 = 5.0 \times 10^{-4} \] 

Step 4: Round off.
\[ C_f \approx 4.6 \times 10^{-4} \] 

Final Answer: \[ \boxed{4.6 \times 10^{-4}} \]