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