7.1.1.3 Step 3: Estimate fracture conductivity

Fracture conductivity is estimated using Eqs. (21) and (22):

$$k\_f w\_f = \frac{121.74}{\sqrt{(0.08)(0.45)(17.7 \times 10^{-6})(0.76)}} \left(\frac{(101)(0.45)(1.507)}{(42)(40)}\right)^2 = 290.77 \text{ md-ft}$$

$$k\_f w\_f = \frac{1947.46}{\sqrt{(0.08)(0.45)(17.7 \times 10^{-6})(0.76)}} \left(\frac{(101)(0.45)(1.507)}{(42)(160)}\right)^2 = 290.7 \text{ md-fit.}$$

From Figure 3, tBLRi = 200 hr. A very close value is obtained from Eq. (33):

$$t\_{BLRi} = 1677 \frac{\left(0.08\right) \left(0.45\right) \left(17.7 \times 10^{-6}\right)}{\left(0.76\right)^{3}} \left(290.7\right)^{2} = 205.71 \text{ hr}.$$

which indicates that the calculation of the fracture conductivity is accurate. Notice that instead of estimating tBLRi the fracture conductivity can be found instead to obtain another value of fracture conductivity; then, Eq. (33) can also be expressed as

$$k\_f w\_f = \sqrt{\frac{k^3 t\_{\text{BLRi}}}{1677 \xi \phi \mu c\_t}} = \sqrt{\frac{0.76^3 (205)}{1677 (1) (0.08) (0.45) (17.7 \times 10^{-6})}} = 290.2 \text{ md-ft.}$$

7.1.1.4 Step 4: Half-fractured length and dimensionless fracture conductivity estimation

Find half-fracture length with Eqs. (23) and (24):

$$x\_f = \frac{4.064(101)(1.507)}{(42)(120)} \sqrt{\frac{0.45}{(0.08)(17.7 \times 10^{-6})(0.76)}} = 79 \text{ ft.}$$

$$\mathbf{x}\_f = \sqrt{\frac{kt\_{Lri}}{1207\xi\phi\mu c\_l}} = \sqrt{\frac{(0.76)(10)}{1207(0.08)(0.76)\left(17.7 \times 10^{-6}\right)}} = 76.5 \text{ ft.}$$

Solve for half-fracture length from Eq. (13) and find this:

$$x\_f = \frac{1.92173}{\frac{\varepsilon'}{r\_w} - \frac{3.31739k}{w\_f k\_f}} = \frac{1.92173}{\frac{\varepsilon^{-4.6844}}{0.28} - \frac{3.31739(0.76)}{290.7}} = 79 \text{ ft.} $$

Find the dimensionless fracture conductivity using Eq. (5):

Well Test Analysis for Hydraulically-Fractured Wells DOI: http://dx.doi.org/10.5772/intechopen.80996

$$\mathbf{C}\_{f\mathbf{D}} = \frac{w\_f k\_f}{\mathbf{x}\_f k} = \frac{290.7}{79(0.76)} = 4.8... $$

The above value confirms that the fracture has finite conductivity.
