Download Advanced Reservoir Management and Engineering by Tarek Ahmed PhD PE, Nathan Meehan PDF

By Tarek Ahmed PhD PE, Nathan Meehan

Content material:
Front-matter

, Pages i,iii
Copyright

, Page iv
Preface

, Page ix
Chapter 1 - good checking out Analysis

, Pages 1-226
Chapter 2 - Water Influx

, Pages 227-279
Chapter three - Unconventional fuel Reservoirs

, Pages 281-432
Chapter four - functionality of Oil Reservoirs

, Pages 433-483
Chapter five - Predicting Oil Reservoir Performance

, Pages 485-539
Chapter 6 - advent to greater Oil Recovery

, Pages 541-585
Chapter 7 - financial Analysis

, Pages 587-649
Chapter eight - monetary Analysis

, Pages 651-660
Chapter nine - Professionalism and Ethics

, Pages 661-682
References

, Pages 683-688
Index

, Pages 689-702

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Extra info for Advanced Reservoir Management and Engineering

Example text

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. Recall that the gas formation volume factor Bg as expressed in bbl/scf is given by: Solution (a) The m(p) method: Step 1. Calculate tD:  0:000264ð50Þð4Þ ð0:2Þð0:0168Þð6:25 3 1024 Þð0:322 Þ 5 279; 365:1 Bg 5 tD 5 ψD 5 0:5½lnðtD Þ 1 0:80907Š 5 0:5½lnð279; 365:1Þ 1 0:80907Š 5 6:6746 m(pwf) by applying   1422Qg T ψD mðpwf Þ 5 mðpi Þ 2 kh !

6 Radial Flow of Compressible Fluids Gas viscosity and density vary significantly with pressure and therefore the assumptions of CHAPTER 1 Eq. , compressible fluids. To develop the proper mathematical function for describing the flow of compressible fluids in the reservoir, the following two additional gas equations must be considered: (1) Gas density equation: ρ5 pM ZRT Combining the above two basic gas equations with Eq. 98) where t 5 time, hours k 5 permeability, md ct 5 total isothermal compressibility, psi21 φ 5 porosity Al-Hussainy et al.

Calculate ψD from Eq. 111): ψD 5 0:5½lnðtD Þ 1 0:80907Š 5 0:5½lnð224; 498:6Þ 1 0:8090Š 5 6:565 Step 2. Calculate m(pwf) by using Eq. 5 3 106, this gives a corresponding value of pwf 5 4367 psi. Second Solution: Pressure-squared Method. 113) The bars over μ and Z represent the values of the gas viscosity and deviation factor as evaluated at the average pressure p. This average pressure is given by: sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 pi2 1 pwf p5 2  ! 114) Combining Eq. 110), gives: 2 pwf 5 pi2 1637Qg T μZ 2 kh !

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