By M. J. Economides

This workbook is a pragmatic better half to the second one version of the textbook Reservoir Stimulation. the 2 books are meant for use jointly. This new quantity can be fairly beneficial for the learning of latest engineers and petroleum engineering scholars, because it includes nearly a hundred difficulties and their options, plus a long bankruptcy giving facts invaluable for designing a stimulation remedy. Chapters are incorporated containing sensible difficulties on reservoir and good issues, rock mechanics, fracturing fluids and proppants, fracture calibration remedies, layout and modeling of propped fractures, review of fracture remedies, layout of matrix remedies, diversion and remedy review, layout and function of acid fractures and stimulation of horizontal wells. those chapters are classified with letters from A to J to tell apart them from their significant other chapters in Reservoir Stimulation. Equations, figures and tables from the textbook are mentioned within the workbook yet aren't reproduced.

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**Example text**

1 but use h, = 140 ft (instead of 70 ft). Use the same well and pressure data as in Tables 7-4 and 7-5. Solution (Ref. 1) The fracture compliance can be calculated from Eq. 1. 8 x IO-’ft/psi. 1. The fracture half-length can then be calculated from Eq. 6). 1. D-9 This Page Intentionally Left Blank E. Design and Modeling of Propped Fractures EXAMPLE E-1 and finally, from Eq. 8-9, Calculation of Transient IPR Curves Calculate transient inflow performance relationships (IPR) for 10 days, 30 days and 365 days using the data given in Table E-I.

Solution (Ref. 5. From Eq. 7-7 (converting appropriate units) and using 2g,, G 3, x, = - Doubling the leakoff coefficient (or, conversely, halving it by its control) has a much more pervasive influence on the created length than the ability to increase the fracture width. For this exercise, doubling the leakoff coefficient to 2x f t l G n would result in a reduction in the fracture length to 1 1 26 ft. If it were an order of magnitude larger, the fracture half-length would be 338 ft. 5) (J60)] = 1587ft Table D-1-Well, Example D-1 .

09 x 1 0 - ~ ~ ~ i - ’ . 5) (10,000) and from Eq. 2-14, To calculate Biot’s poroelastic constant, the total and bulk compressibilities are required. C, Since no information was given on reservoir pressure, then c, = 10-5psi +3x (B-9) Table B-1-Formation data for Example B-2. B-3 PRACTICAL COMPANION T O RESERVOIR STIMULATION EXAMPLE B-3 EXAMPLE B-4 Calculation of Horizontal Stresses Critical Depth for Horizontal vs. Vertical Hydraulic Fractures Calculate the minimum horizontal stress at 10,000 ft depth for a formation with the properties used in Example B-2.