Sn pn n n normal 100 di 0465 di 2915 di

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Unformatted text preview: 0.465 * Di + 2,915 Di * (1 - 0.465) = 2,915 2,915 Di = = 5,449 ft 0.535 Fracture Gradients 1.11- 19 Example Matthews and Kelly (iv) Find Ki from the plot on the right, for Di = 5,449 ft For a south Texas Gulf Coast well, Ki = 0.685 Fracture Gradients 1.11- 20 Example - Matthews and Kelly (v) Now calculate F: K iσ P F= + D D 0.685 * 2,915 F= + 0.735 11 000 , = 0.9165 psi / ft 0.9165 F= = 17.63 0.052 Fracture Gradients lb / gal 1.11- 21 Depth, Di 5,449 Fracture Gradients 0.685 Ki 1.11- 22 Example Ben Eaton: P S −P γ + F= * D 1− γ D S =? D Fracture Gradients γ =? 1.11- 23 Variable Overburden Stress by Eaton At 11,000 ft S/D = 0.96 psi/ft Fracture Gradients 1.11- 24 Fig. 5-5 At 11,000...
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This note was uploaded on 09/18/2012 for the course ENGINEERIN 102 taught by Professor Clyde during the Spring '12 term at A.T. Still University.

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