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

By Ahmed T. McKinney P.D.

ISBN-10: 0750677333

ISBN-13: 9780750677332

The first concentration of this ebook is to provide the fundamental physics of reservoir engineering utilizing the easiest and most simple of mathematical strategies. it is just via having a whole realizing of physics of reservoir engineering that the engineer can desire to unravel complicated reservoir difficulties in a realistic demeanour. The booklet is prepared in order that it may be used as a textbook for senior and graduate scholars or as a reference booklet for training engineers.

Contents: good trying out research Water inflow Unconventional fuel Reservoirs functionality of Oil Reservoirs Predicting Oil Reservoir functionality advent to grease box Economics.

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**Extra resources for Advanced Reservoir Engineering**

**Example text**

When solving for pwf , it might be sufficient to evaluate the gas properties at pi . 13 is repeated below for convenience. 3 ft is producing at a constant flow rate of 2000 Mscf/day under transient flow conditions. The initial reservoir pressure (shut-in pressure) is 4400 psi at 140◦ F. The formation permeability and thickness are 65 md and 15 ft, respectively. The porosity is recorded as 15%. 000 13. 2 × 106 52. 0 × 106 113. 1 × 106 198. 0 × 106 304. 0 × 106 422. 0 × 106 542. 4 × 106 678. 0 × 106 816.

23 0. 15 1. 5 12 × 10−6 60 = 0. 70] kt φµct rw2 φµct r 2 k = 0. , r = rw . 70 can be applied at r = rw to yield: 162. 6Qo Bo µo log pwf = pi − kh where: 1/23 k φµct rw2 pe − pwf 141. , it is dimensionless) and, accordingly, the left-hand side must be dimensionless. Since the lefthand side is dimensionless, and pe − pwf has the units of psi, it follows that the term Qo Bo µo /0. 00708kh has units of pressure. In fact, any pressure difference divided by Qo Bo µo /0. 00708kh is a dimensionless pressure.

80907] ≈ 3000 Pressure = 0. 5 ln 279 365. 1 + 0. 80907 = 6. 6746 Step 3. 1×10 The corresponding value of pwf = 1200 psi. 6 (b) The p2 method: Step 1. 100: ψD = 0. 5[ln(tD ) + 0. 80907] = 0. 5 ln 279 365. 1 + 0. 80907 = 6. 6747 Step 2. 22 Plot of 1/µBg vs. pressure. 22. 108] 5. 99 gives: pwf = pi − pwf = 1195 psi. where: Combining the above two expressions gives: pi 1 2Tpsc m(pi ) − m(pwf ) = dp 5. 107] − 3. 23 (162. 110] or, equivalently, in terms of dimensionless pressure drop: pwf = pi − Third solution: pressure approximation method The second method of approximation to the exact solution of the radial flow of gases is to treat the gas as a pseudo-liquid.

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