Page 67 - A Practical Companion to Reservoir Stimulation
P. 67

DESIGN AND MODELING  OF PROPPED FRACTURES



            EXAMPLE E-4
                                                                 In the right side of Eq. E- 15, the two parentheses denote the
            Calculation of Fracture Width
                                                                 maximum width  (at the wellbore) and  the PKN geometric
                                                                 factor. The product of the two is the average fracture width.
            Calculate the average fracture width of a 1 000-ft fracture half-   In order to calculate the width with the KGD model, Eq.
            length for a Newtonian fluid. Repeat the calculation for a   8-21 with coherent units is transformed into
            range of fractures from 100 ft to 500 ft for both the PKN and
            the KGD models. Use the data in Table E-3.

            Solution (Ref. 8-2.4)
            The width for the PKN model (which is the appropriate one
            for this problem sincexf” hf) is given by Eq. 8- 19 in coherent   for the common units in Table E-3. Figure E-5 is a graph of
            units. For the common units, listed in Table E-3, Eq. 8-19   the average width for the two models for a range of fracture
            becomes:                                             half-lengths from 100 to 500 ft.
                                                                   As can be seen easily from Eqs. E- 12 and E- 16, control of
                                                                 the fracture width during execution is not accomplished eas-
                                                       (E-12)    ily. To double the width for a given reservoir (i.e., given V, E
                                                                 and hf) at a desired fracture length would require an increase
            with the width, w, in inches.                        of the product qip by a factor of  16. Because the viscosity is
              The elastic shear modulus is given by              associated  with  undesirable  side  effects  such  as  residual
                                                                 proppant-pack damage, increasing the rate is the remaining
                                   E                             means of width control. For this problem, doubling the rate to
                          G=                           (E- 13)
                               2(1 + v)’                         80 BPM (this would have an impact on treating pressure and
                                                                 fracture height migration) would increase the average width
            and therefore, for this problem,
                                                                 by a factor of 2°.25, or 1.2 (i.e., by 20%).
                          3  x  lo6
                    G=              =  1.2 x  lo6.    (E-14)
                         2 (I + 0.25)
            From Eq. E- 12,                                     I  q  =  40BPM                                   I
                                                                   h,  =  100ft
                                             I
                       (40) ( 100) (0.75) ( 1000)
                     [       1.2 x  106
              w  =  0.3                    1 4 [;  0.75)           E   =  3 x  106psi
                =  (0.38) (0.59)  =  0.22in.           (E-  15)   I  ji   =  1oocp                               I
                                                                I  y   =  0.75                                   I

                                                                 Table E-3-Variables  to calculate fracture width for
                                                                 Example E-4.
























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