Case Study 6 - Solutions

# Case Study 6 Solutions

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Unformatted text preview: egral function. Currently we have: 16 I will vary the value inside the function until it is equal to 1.4741, making the difference equal zero, as shown in the table below: X (trail) 0.01 0.10 0.15 0.145 0.147 0.14835 ei(X) 4.0379 1.8229 1.4645 1.4937 1.4819 1.4740 1.4741-ei(x) -2.5638 -0.3488 0.0096 -0.0196 -0.0078 0.0001 Therefore from the above trial and error method X = 0.14835. Therefore: Now knowing: 17 Assumption involved in this case study (buildups with a nearby sealing fault): For the infinitely acting region in the plot above (on the far right) pressures are on the straight Horner line. The assumption here is that the reservoir is infinitely acting at t = ∆t and t = t + ∆t. Once a fault is hit, the reservoir is no longer infinitely acting at t + ∆t. Pressure begins to shift from the Horner straight line. After a period of time the pressure will again behave linearly on a shifted Horner straight line, providing the assumptions of infinitely acting a t=∆t and ∆t<<t remain valid. Eventually, providing this is a semi infinite reservoir with one continuous sealing fault, the pressure must return back to the initial pressure. Mathematically, it can be shown these forces itself on to a straight line of doubled slope. Since in reality there are no infinitely reservoirs, the pressure will eventually fall off this line from the influence of faults and reach the average reservoir pressure. 18...
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## This note was uploaded on 01/26/2014 for the course ENG 9111 taught by Professor Drjohanssen during the Spring '12 term at Memorial University.

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