By Ahmed T. McKinney P.D.
The first concentration of this e-book is to give the elemental physics of reservoir engineering utilizing the easiest and most simple of mathematical suggestions. it's only via having a whole figuring out of physics of reservoir engineering that the engineer can desire to resolve complicated reservoir difficulties in a pragmatic 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 e-book for working towards engineers.
Contents: good checking out research Water inflow Unconventional fuel Reservoirs functionality of Oil Reservoirs Predicting Oil Reservoir functionality advent to grease box Economics.
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Additional resources for Advanced Reservoir Engineering
5%. It should be pointed out that for x > 10. 9, Ei(−x) can be considered zero for reservoir engineering calculations. 10 An oil well is producing at a constant flow rate of 300 STB/day under unsteady-state flow conditions. The reservoir has the following rock and fluid properties: Bo = 1. 25 bbl/STB, µo = 1. 5 cp, ct = 12 × 10−6 psi−1 ko = 60 md, h = 15 ft, φ = 15%, rw = 0. 25, 5, 10, 50, 100, 500, 1000, 1500, 2000, and 2500 ft, for 1 hour. Plot the results as: (a) pressure versus the logarithm of radius; (b) pressure versus radius.
23 or: pwf = a + m log(t) The above equation indicates that a plot of pwf vs. 25 The second form of solution to the diffusivity equation is called the dimensionless pressure drop solution and is discussed below. φµct r 2 k 162. 23 log kh φµct rw2 = 4000− k = permeability, md t = time, hours ct = total compressibility, psi−1 t > 9. 48 × 104 0. 71 can be used anytime during the transient flow period to estimate the bottom-hole pressure. Step 2. 71: − 3. 23 − 3. 23 0. 15 1. 5 12 × 10−6 60 = 0. 70] kt φµct rw2 φµct r 2 k = 0.
000264 65 1. 5 0. 15 0. 02831 3 × 10−4 0. 32 = 224 498. 6 Step 2. Calculate Bg at pi : Bg = 0. 00504 = 0. 00504 Zi T pi 0. 896 600 = 0. 0006158 bbl/scf 4400 pD = 0. 5[ln(tD ) + 0. 80907] = 0. 5 ln 224 498. 6 + 0. 80907 = 6. 565 Step 4. 0006158 65 15 The “constant” referred to in the above equation can be obtained from a simple material balance using the definition of the compressibility, assuming no free gas production, thus: −1 dV c= V dp Rearranging: cV dp = −dV Step 3. 7 Pseudosteady state In the unsteady-state flow cases discussed previously, it was assumed that a well is located in a very large reservoir and producing at a constant flow rate.
Advanced Reservoir Engineering by Ahmed T. McKinney P.D.