Published on: **Mar 4, 2016**

- 1. Suez University Faculty of Petroleum & Mining Engineering Porosity and Permeability Student Belal Farouk El-saied Ibrahim Class / III Section / Engineering Geology and Geophysics Presented to Prof. Dr. / Ali Abbas
- 2. Porosity and Permeability Both are important properties that are related to fluids in sediment and sedimentary rocks. Fluids can include: water, hydrocarbons, spilled contaminants. Most aquifers are in sediment or sedimentary rocks. Virtually all hydrocarbons are contained in sedimentary rocks. Porosity: the volume of void space (available to contain fluid or air) in a sediment or sedimentary rock. Permeability: related to how easily a fluid will pass through any granular material.
- 3. I. Porosity (P) The proportion of any material that is void space, expressed as a percentage of the total volume of material. VP P = ×100 VT Where VP is the total volume of pore space and VT is the total volume of rock or sediment. In practice, porosity is commonly based on measurement of the total grain volume of a granular material: VT − VG P= × 100 VT Where VG is the total volume of grains within the total volume of rock or sediment. ∴VP = VT − VG
- 4. Porosity varies from 0% to 70% in natural sediments but exceeds 70% for freshly deposited mud. Several factors control porosity. a) Packing Density Packing density: the arrangement of the particles in the deposit. The more densely packed the particles the lower the porosity. e.g., perfect spheres of uniform size. Porosity can vary from 48% to 26%.
- 5. Shape has an important effect on packing. Tabular rectangular particles can vary from 0% to just under 50%: Natural particles such as shells can have very high porosity:
- 6. In general, the greater the angularity of the particles the more open the framework (more open fabric) and the greater the possible porosity. b) Grain Size On its own, grain size has no influence on porosity! Consider a cube of sediment of perfect spheres with cubic packing. VT − VG P= × 100 VT d = sphere diameter; n = number of grains along a side (5 in this example).
- 7. VT − VG P= × 100 VT Length of a side of the cube = d × n = dn Volume of the cube (VT): VT = dn × dn × dn = d 3n3 Total number of grains: n × n × n = n3 Volume of a single grain: V = π 3 d 6 Total volume of grains (VG): π 3 3π VG = n × d = n d 3 6 6 3
- 8. VT − VG P= ×100 VT Where: VT = d n 3 3 π 3 d n −n d 6 × 100 P= d 3 n3 3 3 Therefore: and π 3 VG = n d 6 3 3 π d n 1 − ÷ 6 ×100 P= d 3 n3 3 3 Rearranging: Therefore: π P = 1 − ÷× 100 = 48% 6 d (grain size) does not affect the porosity so that porosity is independent of grains size. No matter how large or small the spherical grains in cubic packing have a porosity is 48%.
- 9. There are some indirect relationships between size and porosity. i) Large grains have higher settling velocities than small grains. When grains settle through a fluid the large grains will impact the substrate with larger momentum, possibly jostling the grains into tighter packing (therefore with lower porosity). ii) A shape effect. Unconsolidated sands tend to decrease in porosity with increasing grain size. Consolidated sands tend to increase in porosity with increasing grain size.
- 10. Generally, unconsolidated sands undergo little burial and less compaction than consolidated sands. Fine sand has slightly higher porosity. Fine sand tends to be more angular than coarse sand. Therefore fine sand will support a more open framework (higher porosity) than better rounded, more spherical, coarse sand.
- 11. Consolidated sand (deep burial, well compacted) has undergone exposure to the pressure of burial (experiences the weight of overlying sediment). Fine sand is angular, with sharp edges, and the edges will break under the load pressure and become more compacted (more tightly packed with lower porosity). Coarse sand is better rounded and less prone to breakage under load; therefore the porosity is higher than that of fine sand.
- 12. c) Sorting In general, the better sorted the sediment the greater the porosity. In well sorted sands fine grains are not available to fill the pore spaces. This figure shows the relationship between sorting and porosity for clay-free sands.
- 13. Overall porosity decreases with increasing sorting coefficient (poorer sorting). For clay-free sands the reduction in porosity with increasing sorting coefficient is greater for coarse sand than for fine sand. The difference is unlikely if clay was also available to fill the pores.
- 14. For clay-free sands the silt and fine sand particles are available to fill the pore space between large grains and reduce porosity.
- 15. Because clay is absent less relatively fine material is not available to fill the pores of fine sand. Therefore the pores of fine sand will be less well-filled (and have porosity higher).
- 16. d) Post burial changes in porosity. Includes processes that reduce and increase porosity. Porosity that develops at the time of deposition is termed primary porosity. Porosity that develops after deposition is termed secondary porosity. Overall, with increasing burial depth the porosity of sediment decreases. 50% reduction in porosity with burial to 6 km depth due to a variety of processes.
- 17. i) Compaction Particles are forced into closer packing by the weight of overlying deposits, reducing porosity. May include breakage of grains. Most effective if clay minerals are present (e.g., shale). Freshly deposited mud may have 70% porosity but burial under a kilometre of sediment reduces porosity to 5 or 10%. http://www.engr.usask.ca/~mjr347/prog/geoe118/geoe118.022.html
- 18. ii) Cementation Precipitation of new minerals from pore waters causes cementation of the grains and acts to fill the pore spaces, reducing porosity. Most common cements are calcite and quartz. Here’s a movie of cementation at Paul Heller’s web site .
- 19. iii) Clay formation Clays may form by the chemical alteration of pre-existing minerals after burial. Feldspars are particularly common clay-forming minerals. Clay minerals are very fine-grained and may accumulate in the pore spaces, reducing porosity. Eocene Whitemud Formation, Saskatchewan
- 20. iv) Solution If pore waters are undersaturated with respect to the minerals making up a sediment then some volume of mineral material is lost to solution. Calcite, that makes up limestone, is relatively soluble and void spaces that are produced by solution range from the size of individual grains to caverns. Quartz is relatively soluble when pore waters have a low Ph. Solution of grains reduces VG, increasing porosity. Solution is the most effective means of creating secondary porosity. v) Pressure solution The solubility of mineral grains increases under an applied stress (such as burial load) and the process of solution under stress is termed Pressure Solution. The solution takes place at the grain contacts where the applied stress is greatest.
- 21. Pressure solution results in a reduction in porosity in two different ways: 1. It shortens the pore spaces as the grains are dissolved. 2. Insoluble material within the grains accumulates in the pore spaces as the grains are dissolve.
- 22. v) Fracturing Fracturing of existing rocks creates a small increase in porosity. Fracturing is particularly important in producing porosity in rocks with low primary porosity.
- 23. POROSITY DETERMINATION FROM LOGS Most slides in this section are modified primarily from NExT PERF Short Course Notes, 1999. However, many of the NExT slides appears to have been obtained from other primary sources that are not cited. Some slides have a notes section.
- 24. OPENHOLE LOG EVALUATION Well Log SP Resistivity
- 25. POROSITY DETERMINATION BY LOGGING Increasing radioactivity Increasing Increasing resistivity porosity Shale Oil sand Shale Gamma ray Resisitivity Porosity
- 26. POROSITY LOG TYPES 3 Main Log Types • Bulk density • Sonic (acoustic) • Compensated neutron These logs do not measures porosity directly. To accurately calculate porosity, the analyst must know: •Formation lithology • Fluid in pores of sampled reservoir volume
- 27. DENSITY LOGS • Uses radioactive source to generate gamma rays • Gamma ray collides with electrons in formation, losing energy • Detector measures intensity of backscattered gamma rays, which is related to electron density of the formation • Electron density is a measure of bulk density
- 28. DENSITY LOGS • Bulk density, ρb, is dependent upon: – Lithology – Porosity – Density and saturation of fluids in pores • Saturation is fraction of pore volume occupied by a particular fluid (intensive)
- 29. DENSITY LOG 0 GR API 6 CALIX IN 16 6 CALIY IN 16 200 2 RHOB G/C3 -0.25 3 DRHO G/C3 0.25 4100 Gamma ray Density correction 4200 Caliper Density
- 30. Mud cake (ρ mc + hmc) Formation (ρ b) Long spacing detector Short spacing detector Source
- 31. BULK DENSITY ρb = ρma ( 1 − φ) + ρ f φ Matrix •Measures electron density of a formation •Strong function of formation bulk density •Matrix bulk density varies with lithology –Sandstone 2.65 g/cc –Limestone 2.71 g/cc –Dolomite 2.87 g/cc Fluids in flushed zone
- 32. POROSITY FROM DENSITY LOG Porosity equation ρma − ρb φ= ρma − ρ f Fluid density equation ρ f = ρmf Sxo + ρh ( 1 − Sxo ) We usually assume the fluid density (ρf) is between 1.0 and 1.1. If gas is present, the actual ρf will be < 1.0 and the calculated porosity will be too high. ρmf is the mud filtrate density, g/cc ρh is the hydrocarbon density, g/cc Sxo is the saturation of the flush/zone, decimal
- 33. DENSITY LOGS Working equation (hydrocarbon zone) ρb = φ S xo ρmf + φ ( 1 − S xo ) ρhc + Vsh ρ sh + ( 1 − φ − Vsh ) ρma ρb = Recorded parameter (bulk volume) φ Sxo ρmf = Mud filtrate component φ (1 - Sxo) ρhc = Hydrocarbon component Vsh ρsh Shale component = 1 - φ - Vsh = Matrix component
- 34. DENSITY LOGS • If minimal shale, Vsh ≈ 0 • If ρhc ≈ ρmf ≈ ρf, then ∀ ρb = φ ρf - (1 - φ) ρma ρma − ρb φ = φd = ρma − ρ f φd = Porosity from density log, fraction ρma = Density of formation matrix, g/cm3 ρb = Bulk density from log measurement, g/cm3 ρf = Density of fluid in rock pores, g/cm3 ρhc = Density of hydrocarbons in rock pores, g/cm3 ρmf = Density of mud filtrate, g/cm3 ρsh = Density of shale, g/cm3 Vsh = Volume of shale, fraction
- 35. BULK DENSITY LOG 001) BONANZA 1 GRC 0 150 SPC -160 MV 40 ACAL 6 16 10700 0.2 0.2 0.2 ILDC SNC MLLCF 200 200 RHOC 1.95 2.95 CNLLC 0.45 -0.15 DT 150 us/f 50 200 RHOC 1.95 10800 10900 Bulk Density Log 2.95
- 36. NEUTRON LOG • Logging tool emits high energy neutrons into formation • Neutrons collide with nuclei of formation’s atoms • Neutrons lose energy (velocity) with each collision
- 37. NEUTRON LOG • The most energy is lost when colliding with a hydrogen atom nucleus • Neutrons are slowed sufficiently to be captured by nuclei • Capturing nuclei become excited and emit gamma rays
- 38. NEUTRON LOG • Depending on type of logging tool either gamma rays or non-captured neutrons are recorded • Log records porosity based on neutrons captured by formation • If hydrogen is in pore space, porosity is related to the ratio of neutrons emitted to those counted as captured • Neutron log reports porosity, calibrated assuming calcite matrix and fresh water in pores, if these assumptions are invalid we must correct the neutron porosity value
- 39. NEUTRON LOG Theoretical equation φN = φ S xo φNmf + φ ( 1 −S xo ) φNhc + Vsh φ sh + ( 1 − φ − Vsh ) φNma φN = Recorded parameter φNma = Porosity of matrix fraction φ Sxo φNmf = Mud filtrate portion φNhc = Porosity of formation saturated with φ (1 - Sxo) φNhc = Hydrocarbon portion Vsh φNsh = Shale portion (1 - φ - Vsh) φNhc = Matrix portion where φ = True porosity of rock φN = Porosity from neutron log measurement, fraction hydrocarbon fluid, fraction φNmf = Porosity saturated with mud filtrate, fraction Vsh = Volume of shale, fraction Sxo = Mud filtrate saturation in zone invaded by mud filtrate, fraction
- 40. POROSITY FROM NEUTRON LOG 001) BONANZA 1 GRC 0 150 SPC -160 MV 40 ACAL 6 16 10700 0.2 0.2 0.2 ILDC SNC MLLCF 200 200 RHOC 1.95 2.95 CNLLC 0.45 -0.15 DT 150 us/f 50 200 CNLLC 0.45 10800 10900 Neutron Log -0.15
- 41. ACOUSTIC (SONIC) LOG Upper transmitter R1 R2 R3 R4 Lower transmitter • Tool usually consists of one sound transmitter (above) and two receivers (below) • Sound is generated, travels through formation • Elapsed time between sound wave at receiver 1 vs receiver 2 is dependent upon density of medium through which the sound traveled
- 42. Compressional waves E1 E3 E2 T0 50 µsec Rayleigh waves Mud waves
- 43. COMMON LITHOLOGY MATRIX TRAVEL TIMES USED Lithology Sandstone Limestone Dolomite Anydridte Salt Typical Matrix Travel Time, ∆, µ tma sec/ft 55.5 47.5 43.5 50.0 66.7
- 44. ACOUSTIC (SONIC) LOG Working equation ∆t L = φ S xo ∆t mf + φ ( 1 − S xo ) ∆t hc + Vsh ∆t sh + ( 1 − φ − Vsh ) ∆t ma ∆tL = Recorded parameter, travel time read from log φ Sxo ∆tmf = Mud filtrate portion φ (1 - Sxo) ∆thc = Hydrocarbon portion Vsh ∆tsh = Shale portion (1 - φ - Vsh) ∆tma = Matrix portion
- 45. ACOUSTIC (SONIC) LOG • If Vsh = 0 and if hydrocarbon is liquid (i.e. ∆tmf ≈ ∆tf), then ∀ ∆tL = φ ∆tf + (1 - φ) ∆tma or ∆t L − ∆t ma φs = φ = ∆t f − ∆t ma φs = Porosity calculated from sonic log reading, fraction ∆tL = Travel time reading from log, microseconds/ft ∆tma = Travel time in matrix, microseconds/ft ∆tf = Travel time in fluid, microseconds/ ft
- 46. ACOUSTIC (SONIC) LOG 0 GR API 6 CALIX IN DT 200 140 40 30 16 USFT SPHI % 10 4100 Sonic travel time Gamma Ray Sonic porosity 4200 Caliper
- 47. SONIC LOG The response can be written as follows: t log = t ma ( 1 − φ) + t f φ φ= t log − t ma t f − t ma tlog = log reading, µsec/ft tma = the matrix travel time, µsec/ft tf = the fluid travel time, µsec/ft φ = porosity
- 48. SONIC LOG 001) BONANZA 1 GRC 0 150 SPC -160 MV 40 ACAL 6 16 0.2 0.2 0.2 ILDC SNC MLLCF 200 200 RHOC 1.95 2.95 CNLLC 0.45 -0.15 DT 150 us/f 50 200 10700 150 10800 Sonic Log 10900 DT us/f 50
- 49. EXAMPLE Calculating Rock Porosity Using an Acoustic Log Calculate the porosity for the following intervals. The measured travel times from the log are summarized in the following table. At depth of 10,820’, accoustic log reads travel time of 65 µs/ft. Calculate porosity. Does this value agree with density and neutron logs? Assume a matrix travel time, ∆tm = 51.6 µsec/ft. In addition, assume the formation is saturated with water having a ∆tf = 189.0 µsec/ft.
- 50. EXAMPLE SOLUTION SONIC LOG 001) BONANZA 1 GRC 0 150 SPC -160 MV 40 ACAL 6 16 0.2 0.2 0.2 ILDC SNC MLLCF 200 200 RHOC 1.95 2.95 CNLLC 0.45 -0.15 DT 150 us/f 50 SPHI 45 ss -15 200 10700 10800 SPHI 10900
- 51. FACTORS AFFECTING SONIC LOG RESPONSE • Unconsolidated formations • Naturally fractured formations • Hydrocarbons (especially gas) • Rugose salt sections
- 52. RESPONSES OF POROSITY LOGS The three porosity logs: – Respond differently to different matrix compositions – Respond differently to presence of gas or light oils Combinations of logs can: – Imply composition of matrix – Indicate the type of hydrocarbon in pores
- 53. GAS EFFECT • Density - φ is too high • Neutron - φ is too low • Sonic - φ is not significantly affected by gas
- 54. ESTIMATING POROSITY FROM WELL LOGS Openhole logging tools are the most common method of determining porosity: • Less expensive than coring and may be less risk of sticking the tool in the hole • Coring may not be practical in unconsolidated formations or in formations with high secondary porosity such as vugs or natural fractures. If porosity measurements are very important, both coring and logging programs may be conducted so the log-based porosity calculations can be used to calibrated to the core-based porosity measurements .
- 55. Influence Of Clay-Mineral Distribution On Effective Porosity Dispersed Clay • Pore-filling • Pore-lining • Pore-bridging φe Clay Minerals Detrital Quartz Grains φe φe Clay Lamination Structural Clay (Rock Fragments, Rip-Up Clasts, Clay-Replaced Grains) φ φee
- 56. GEOLOGICAL AND PETROPHYSICAL DATA USED TO DEFINE FLOW UNITS Core Lithofacies Core Pore Plugs Types Petrophysical Data Gamma Ray Flow Log Units φ vs k Capillary Pressure 5 4 3 2 1
- 57. Schematic Reservoir Layering Profile in a Carbonate Reservoir Baffles/barriers SA -97A Flow unit SA -251 3150 3200 SA -356 SA -71 SA -344 3150 3100 SA -371 3100 SA -348 3250 SA -346 SA -37 3150 3100 3200 3250 3200 3200 3150 3300 3150 3200 3150 3250 3300 3250 3250 3200 3250 3250 3200 3300 3350 3300 3250 3300 3250 3350 3350 From Bastian and others
- 58. Why is porosity important? Especially because it allows us to make estimations of the amount of fluid that can be contained in a rock (water, oil, spilled contaminants, etc.). Example from oil and gas exploration:
- 59. Why is porosity important? Especially because it allows us to make estimations of the amount of fluid that can be contained in a rock (water, oil, spilled contaminants, etc.). Example from oil and gas exploration:
- 60. Why is porosity important? Especially because it allows us to make estimations of the amount of fluid that can be contained in a rock (water, oil, spilled contaminants, etc.). Example from oil and gas exploration:
- 61. Why is porosity important? Especially because it allows us to make estimations of the amount of fluid that can be contained in a rock (water, oil, spilled contaminants, etc.). Example from oil and gas exploration:
- 62. Why is porosity important? Especially because it allows us to make estimations of the amount of fluid that can be contained in a rock (water, oil, spilled contaminants, etc.). Example from oil and gas exploration: How much oil is contained in the discovered unit? In this case, assume that the pore spaces of the sediment in the oilbearing unit are full of oil. Therefore, the total volume of oil is the total volume of pore space (VP) in the oil-bearing unit.
- 63. VP P = ×100 VT Total volume of oil = VP, therefore solve for VP. VT = 800m × 200m ×1m = 160, 000m3 P × VT VP = 100 P = 10% Therefore: 10 ×160, 000 VP = 100 = 16, 000m 3 of oil
- 64. II. Permeability (Hydraulic Conductivity; k) Stated qualitatively: permeability is a measure of how easily a fluid will flow through any granular material. More precisely, permeability (k) is an empirically-derived parameter in D’Arcy’s Law, a Law that predicts the discharge of fluid through a granular material.
- 65. Those are all properties that are independent of the granular material. There are also controls on permeability that are exerted by the granular material and are accounted for in the term (k) for permeability: k is proportional to all sediment properties that influence the flow of fluid through any granular material (note that the dimensions of k are cm2). Two major factors: 1. The diameter of the pathways through which the fluid moves. 2. The tortuosity of the pathways (how complex they are).
- 66. 1. The diameter of the pathways. Along the walls of the pathway the velocity is zero (a no slip boundary) and increases away from the boundaries, reaching a maximum towards the middle to the pathway. Narrow pathway: the region where the velocity is low is a relatively large proportion of the total cross-sectional area and average velocity is low. Large pathway: the region where the velocity is low is proportionally small and the average velocity is greater. It’s easier to push fluid through a large Pathway than a small one.
- 67. 2. The tortuosity of the pathways. Tortuosity is a measure of how much a pathway deviates from a straight line.
- 68. 2. The tortuosity of the pathways. Tortuosity is a measure of how much a pathway deviates from a straight line. The path that fluid takes through a granular material is governed by how individual pore spaces are connected. The greater the tortuosity the lower the permeability because viscous resistance is cumulative along the length of the pathway.
- 69. Pathway diameter and tortuosity are controlled by the properties of the sediment and determine the sediment’s permeability. The units of permeability are Darcies (d): 1 darcy is the permeability that allows a fluid with 1 centipoise viscosity to flow at a rate of 1 cm/s under a pressure gradient of 1 atm/cm. 1 d ) Permeability is often very small and expressed in millidarcies ( 1000
- 70. a) Sediment controls on permeability i) Packing density Tightly packed sediment has smaller pathways than loosely packed sediment (all other factors being equal). Smaller pathways reduce porosity and the size of the pathways so the more tightly packed the sediment the lower the permeability.
- 71. ii) Porosity In general, permeability increases with primary porosity. The larger and more abundant the pore spaces the greater the permeability. Pore spaces must be well connected to enhance permeability.
- 72. Shale, chalk and vuggy rocks (rocks with large solution holes) may have very high porosity but the pores are not well linked. The discontinuous pathways result in low permeability. Fractures can greatly enhance permeability but do not increase porosity significantly. A 0.25 mm fracture will pass fluid at the rate that would be passed by13.5 metres of rock with 100 md permeability.
- 73. iii) Grain Size Unlike porosity, permeability increases with grain size. The larger the grain size the larger the pore area. For spherical grains in cubic packing: Pore area = 0.74d2
- 74. A ten-fold increase in grain size yields a hundred-fold increase in permeability. iv) Sorting The better sorted a sediment is the greater its permeability. In very well sorted sands the pore spaces are open. In poorly sorted sands fine grains occupy the pore spaces between coarser grains.
- 75. v) Post-burial processes Like porosity, permeability is changed following burial of a sediment. In this example permeability is reduced by two orders of magnitude with 3 km of burial. Cementation Clay formation Compaction Pressure solution All act to reduce permeability
- 76. b) Directional permeability Permeability is not necessarily isotropic (equal in all directions) Fractures are commonly aligned in the same direction, greatly enhancing permeability in the direction that is parallel to the fractures.
- 77. Variation in grain size and geological structure can create directional permeability. E.g., Graded bedding: grain size becomes finer upwards in a bed. Fluid that is introduced at the surface will follow a path that is towards the direction of dip of the beds.
- 78. Fabric (preferred orientation of the grains in a sediment) can cause directional permeability. E.g., A sandstone unit of prolate particles. The direction along the long axes of grains will have larger pathways and therefore greater permeability than the direction that is parallel to the long axes.