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 quite invaluable for the educational of recent engineers and petroleum engineering scholars, because it includes nearly a hundred difficulties and their strategies, plus a long bankruptcy giving information important for designing a stimulation therapy. Chapters are integrated containing useful difficulties on reservoir and good issues, rock mechanics, fracturing fluids and proppants, fracture calibration remedies, layout and modeling of propped fractures, evaluate of fracture remedies, layout of matrix remedies, diversion and therapy overview, layout and function of acid fractures and stimulation of horizontal wells. those chapters are classified with letters from A to J to differentiate them from their significant other chapters in Reservoir Stimulation. Equations, figures and tables from the textbook are talked about within the workbook yet usually are not reproduced.
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Additional info for A Practical Companion to Reservoir Stimulation
5-27 for 20140 sand to calculate the stressinduced permeability reduction for 2000 and 4000 psi drawdown. Will the stress on the proppant pack increase or decrease with time? 7. Solution (Ref. Section 5-9) The effective minimum horizontal stress can be‘estimated from Eq. 75 [ = 2303 psi. 7) (5000) (~ooo)] = 6152 psi. 7) ~ (C-30) This is the stress that the proppant will experience without drawdown. Thus, the proppant-pack permeability will be about 300,000 md (from Fig. 5-27). 7) (4000) = 2800 (0= 5100 psi) and result in permeability values equal to 200,000 and 150,000 md, respectively.
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 .
Plot the results on log-log paper. Solution (Ref. 3 and a fluid mechanics textbook) To calculate the friction pressure drop for any type flow, the Reynold's numbers must first be calculated. For a power law fluid this is given by APf = 2 d (C- 16) where the Ap, is the pressure in pounds per square inch. 5 50 (C- 18) and b = While Eq. , and K' is in lbf-sec"'/ft*. 4 - log n' 7 The calculation of the pressure drop is then given by Eq. C-16. A sample calculation is performed with q = 20 BPM. From Eq.