Ahmed T. McKinney P.D.'s Advanced Reservoir Engineering PDF

By Ahmed T. McKinney P.D.

ISBN-10: 0750677333

ISBN-13: 9780750677332

The first concentration of this publication is to provide the fundamental physics of reservoir engineering utilizing the easiest and simplest of mathematical concepts. it is just via having an entire figuring out of physics of reservoir engineering that the engineer can wish to resolve complicated reservoir difficulties in a realistic demeanour. The e-book is prepared in order that it may be used as a textbook for senior and graduate scholars or as a reference ebook for working towards engineers.
Contents: good checking out research Water inflow Unconventional gasoline Reservoirs functionality of Oil Reservoirs Predicting Oil Reservoir functionality creation to grease box Economics.

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Example text

97 expresses the bottom-hole real-gas pseudopressure as a function of the transient flow time t. The solution as expressed in terms of m(p) is the recommended mathematical expression for performing gas well pressure analysis due to its applicability in all pressure ranges. The radial gas diffusivity equation can be expressed in a dimensionless form in terms of the dimensionless real-gas pseudopressure drop ψD . 84. 70. That is: ψD = 0. 5[ln(tD ) + 0. 80907] Step 2. 3 ft is producing at a constant flow rate of 2000 Mscf/day under transient flow conditions.

5 × 106 By interpolation at m(pwf ) = 1077. 5 × 106 , this gives a corresponding value of pwf = 4367 psi. 102] µZ The bars over µ and Z represent the values of the gas viscosity and deviation factor as evaluated at the average pressure p. 99, gives: p2wf = p2i − 1637Qg T µZ kh log Solution kt φµi cti rw2 − 3. 104] Step 1. Calculate the dimensionless time tD : tD = = or: p2wf = p2i − 0. 0002637kt φµi cti rw2 0. 15 0. 02831 3 × 10−4 0. 32 = 224 498. 6 Step 2. 105] Equivalently: 0. 0002637 65 1. 5 m(pwf ) = m(pi ) − 1637Qg T µZ kh log 4tD γ 1637 2000 600 65 15 (4)224498.

85] pD kh Step 1. 75: 0. 0002637kt tD = φµct rw2 = 0. 000264 60 1 0. 15 1. 5 12 × 10−6 0. 25 2 = 93 866. 67 Step 2. 80 to calculate the dimensionless pressure drop function: pD = 0. 5[ln(tD ) + 0. 80907] = 0. 5[ln(93 866. 67) + 0. 80907] = 6. 1294 Step 3. 85: 141. 2Qo Bo µo pD p (rw , t) = pi − kh p 0. 25, 1 = 4000 − 141. 2 300 1. 25 1. 5 60 15 × (6. 1294) = 3459 psi This example shows that the solution as given by the pD function technique is identical to that of the Ei function approach. The main difference between the two formulations is that the pD function can only be used to calculate the pressure at radius r when the flow rate Q is constant and known.

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Advanced Reservoir Engineering by Ahmed T. McKinney P.D.


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