#### Transcript Normal Pore Pressure

DRILLING ENGINEERING Vahid Salimi Textbook Applied Drilling Engineering, by :Adam T. Bourgoyne Jr., Martin E. Chenevert, Keith K. Millheim F.S. Young Jr.,. Contents: pore pressure and fracture pressure drilling hydraulics casing design under balanced drilling directional drilling Chapter 1 pore pressure and fracture pressure Hydrostatic Pressure • Hydrostatic pressure is defined as the pressure exerted by a column of fluid. • The pressure is a function of the average fluid density and the vertical height or depth of the fluid column. • Mathematically, hydrostatic pressure is expressed as: HP (psi) = 0.052 x ρf (ppg) x D (ft) where: HP = hydrostatic pressure ρf = average fluid density D = true vertical depth or height of the column Hydrostatic Pressure(cont’d) Hydrostatic pressures can easily be converted to equivalent mud weights and pressure gradients. Hydrostatic pressure gradient is given by: HG = HP / D … (psi/ft) Example Calculate the hydrostatic pressure for the following wells: • a. mud weight = 9 ppg, hole depth = 10100 ft MD (measured depth), 9900 ft TVD (true vertical depth) • b. mud gradient = 0.468 psi / ft, hole depth = 10100 ft MD (measured depth), 9900 ft TVD (true vertical depth) solution a. HP (psi) = 0.052 x ρf (ppg) x D (ft) = 0.052 x 9 x 9900 = 4632 psi b. Hydrostatic pressure = fluid gradient (psi / ft) x depth (ft)..........psi = 0.468 (psi /ft) x 9900(ft) = 4633 psi Mud Weight (MW) should be kept heavy enough so that hydrostatic head of mud column is higher than formation pressure at any depth. Usually 150 psi Need to know formation pressure in order to determine MW Pf + 150 = 0.052 MW D Pf MW D 150 Formation Pressure, psi Mud Weight, ppg True Vertical Depth, ft Safety, psi Example You are drilling with 7.9 ppg oil base mud. If the formation pressure is predicted 5,000 psi at 9,000 ft true vertical depth, what is the required MW in order to have 150 psi overpressure ? 5,000 + 150 = 0.052 MW 9,000 MW = 11 ppg OVERBURDEN PRESSURE The overburden pressure is defined as the pressure exerted by the total weight of overlying formations above the point of interest. The total weight is the combined weight of both the formation solids (rock matrix) and formation fluids in the pore space. The overburden pressure can therefore be expressed as the hydrostatic pressure exerted by all materials overlying the depth of interest: σov = 0.052 x ρb x D where σov = overburden pressure (psi) ρb = formation bulk density (ppg) D = true vertical depth (ft) OVERBURDEN PRESSURE(cont’d) Overburden gradient under field conditions of varying lithological and pore fluid density is given by: σovg= 0.433[(1 – φ)ρma +(φxρf)] where σovg= overburden gradient, psi/ft φ= porosity expressed as a fraction ρf= formation fluid density ρma= matrix density matrix and fluid densities Substance Sandstone Limestone Dolomite Anhydrite Halite Gypsum Clay Freshwater Seawater Oil Gas Density (gm/cc) 2.65 2.71 2.87 2.98 2.03 2.35 2.7 - 2.8 1.0 1.03 - 1.06 0.6 - 0.7 0.15 To convert densities from gm/cc to gradients in psi/ft use: Gradient (psi/ft) = 0.433 x (gm /cc) To convert from psi/ft to ppg, use: Density (ppg) = gradient (psi/ft) / 0.052 Pore pressure The magnitude of pressure in the pore of formation known as the pore pressure Pore pressure = formation pressure =formation fluid pressure =reservoir pressure =pressure in fluid contained in the pore spaces of the rock Example Determine the pore pressure of a normally pressured formation in the Gulf of Mexico at 9,000’ depth. Solution p = 0.465 psi/ft * 9,000 ft = 4,185 psig Homework: Pore Pressure Profiles The following pore pressure information has been supplied for the well you are about to drill. a. Plot the following pore pressure/depth information on a P-Z diagram : b. Calculate the pore pressure gradients in the formations from surface; to 8000ft; to 8500ft; and to 9500ft. Plot the overburden gradient (1 psi/ft) on the above plot. Determine the mud weight required to drill the hole section: down to 8000ft; down to 8500ft; and down to 9500ft. Assume that 200 psi overbalance on the formation pore pressure is required. c. If the mudweight used to drill down to 8000ft were used to drill into the formation pressures at 8500ft what would be the over/underbalance on the formation pore pressure at this depth? d. Assuming that the correct mudweight is used for drilling at 8500ft but that the fluid level in the annulus dropped to 500 ft below drillfloor, due to inadequate hole fill up during tripping. What would be the effect on bottom hole pressure at 8500ft ? e. What type of fluid is contained in the formations below 8500ft. Normal Pore Pressure Pressure of a column of water extending from the formation to the surface The magnitude of normal pore pressure varies with the concentration of dissolved salts, type of fluid, gases present and temperature gradient. =0.433 psi/ft for fresh water =0.465 psi/ft for seawater Subnormal Formation Pressure Subnormal pore pressure is defined as any formation pressure that is less than the corresponding fluid hydrostatic pressure at a given depth. Subnormal formation pressure can cause lost circulation of water as the drilling fluid. ABNORMAL PORE PRESSURE Abnormal pore pressure is defined as any pore pressure that is greater than the hydrostatic pressure of the formation water occupying the pore space. Abnormal pressure is sometimes called overpressure or geopressure. Abnormal pressure can be thought of as being made up of a normal hydrostatic component plus an extra amount of pressure. This excess pressure is the reason why surface control equipment (e.g. BOPs) are required when drilling oil and gas wells. ABNORMAL PORE PRESSURE(cont’d) Abnormal formation pressure can cause a kick with water as the drilling fluid. Normal and Abnormal Pore Pressure Normal Pressure Gradients West Texas: 0.433 psi/ft Gulf Coast: 0.465 psi/ft Depth, ft Abnormal Pressure Gradients 10,000’ Pore Pressure, psig Pore Pressure vs. Depth 0 0.433 psi/ft 0.465 psi/ft 8.33 lb/gal 9.0 lb/gal 5,000 Normal 10,000 Abormal 15,000 20,000 8 9 10 11 12 13 14 15 Pore Pressure Equivalent, lb/gal 16 { Density of mud required to control this pore pressure } Fracture Gradient Pore Pressure Gradient Transition zone The upper portion of the region of abnormal pressure is called the transition zone Causes Of Abnormal Pore Pressure Compaction Effects Diagenetic Effects Differentional Density Effects Fluid Migration Effects Diagenetic Effects With increasing pressure and temperature, sediments undergo a process of chemical and physical changes collectively known as diagenesis. Diagenesis is the alteration of sediments and their constituent minerals during post depositional compaction. Diagenetic processes include the formation of new minerals, recrystallisation and lithification. Diagenesis may result in volume changes and water generation which if occurring in a seabed environment may lead to both abnormal or sub-normal pore pressure. Clay Diagenesis Clay Diagenesis (Conversion of Smectite to Illite) If the water released in this process cannot escape during compaction, then the pore fluid will support an increased portion of the overburden and will thus be abnormally pressured. Diagenesis of Sulphate Formations Anhydrite (CaSO4) is diagenetically formed from the dehydration of gypsum (CaSO4.2H2O). During the process large volumes of water are released and this is accompanied by a 30-40% reduction in formation volume HIGH PRESSURE NORMAL PRESSURE 7. Abnormal Pressure 661. Drilling Engineering Slide 33 Homework When crossing faults it is possible to go from normal pressure to abnormally high pressure in a short interval. 7. Abnormal Pressure 661. Drilling Engineering Slide 37 Well “A” found only Normal Pressure ... 7. Abnormal Pressure 661. Drilling Engineering Slide 38 Methods of estimating pore pressure Direct measurement It is possible only when the formation has been drilled It is expensive Indirect measurement The main parameter is the variation of porosity with depth (porosity dependent parameter) If pore pressure is normal, porosity-dependent parameter (x) have an easily recognized trend because of the decreased porosity with increased depth of burial and compaction. A departure from the normal pressure trend signals a probable transition zone. Detection of the depth at which this departure occurs is critical because casing must be set in the well before excessively pressured permeable zones can be drilled safely. Prediction and Detection of Abnormal Pressure Zones 1. Before drilling Shallow seismic surveys Deep seismic surveys Comparison with nearby wells 7. Abnormal Pressure 661. Drilling Engineering Slide 40 Prediction and Detection of Abnormal Pressure Zones 2. While drilling Drilling rate, gas in mud, etc. etc. D - Exponent DC - Exponent MWD - LWD Density of shale (cuttings) 7. Abnormal Pressure 661. Drilling Engineering Slide 41 Prediction and Detection of Abnormal Pressure Zones 3. After drilling Resistivity log Conductivity log Sonic log Density log 7. Abnormal Pressure 661. Drilling Engineering Slide 42 Compaction Theory of Abnormal Pressure During deposition, sediments are compacted by the overburden load and are subjected to greater temperatures with increasing burial depth. Porosity is reduced as water is forced out. Hydrostatic equilibrium within the compacted layers is retained as long as the expelled water is free to escape If water cannot escape, abnormal pressures occur Compaction Theory In Porous formation the overburden stress is supported by rock matrix stress and pore pressure Bulk Density = ρm (1-Ф) + ρf Ф The average porosity in sediments ,generally decreases with increasing depth - due to the increasing overburden This results in an increasing bulk density with increasing depth, and increasing rock strength Average Porosity Ф = ρm - ρb / ρm – ρf Plot Ф Vs. Depth on similog graph. Example Calculate the overburden stress at a depth of 7,200 ft in the Santa Barbara Channel. Assume φo = 0.37 ρma = 2.6 gm/cc kφ = 0.0001609 ft-1 ρf = 1.044 gm/cc Solution Homeworks Homeworks (cont’d) Homeworks (cont’d) Pore pressure prediction methods Measure the porosity indicator (e.g.density) in normally pressured, clean shales to establish a normal trend line. When the indicator suggests porosity values that are higher than the trend, then abnormal pressures are suspected to be present. The magnitude of the deviation from the normal trend line is used to quantify the abnormal pressure. Equivalent Matrix Stress Method Example Estimate the pore pressure at 10,200’ if the equivalent depth is 9,100’. The normal pore pressure gradient is 0.433 psi/ft. The overburden gradient is 1.0 psi/ft. ASSUME: At 9,100’, pne = 0.433 * 9,100 = 3,940 psig At 9,100’, σobe = 1.00 * 9,100 = 9,100 psig At 10,200’, σob = 1.00*10,200 = 10,200 psig Solution Approach 2: Empirical correlation More accurate Need numerous data Uses (Xo-Xn) or (Xo/Xn) to predict the magnitude of the abnormal pressure Prediction of pore pressure by seismic data Homework Pore pressure indications while drilling Drilling rate (ROP) gas in mud Pit level Flowline temperature Rate Of Penetration(ROP) Drill bits break the rock by a combination of several processes including: Compression (weight-on-bit), shearing (rpm) and sometimes jetting action of the drilling fluid. The speed of drilling is described as the rate of penetration (ROP) and is measured in ft/hr. The rate of penetration is affected by numerous parameters namely: Weight On Bit (WOB) Revolutions Per Minute (RPM) bit type bit wear hydraulic efficiency degree of overbalance drilling fluid properties hydrostatic pressure and hole size Formation properties TABLE 2.8 - Note, that many factors can influence the drilling rate, and some of these factors are outside the control of the operator. 75 Effect of bit weight and hydraulics on penetration rate Inadequate hydraulics or excessive imbedding of the bit teeth in the rock 0 76 Drilling rate increases more or less linearly with increasing bit weight. A significant deviation from this trend may be caused by poor bottom hole cleaning Effect of Differential Pressure on Drilling Rate Differential pressure is the difference between wellbore pressure and pore fluid pressure Decrease can be due to: • The chip hold down effect • The effect of wellbore pressure on rock strength 77 If all parameters affecting ROP are held constant whilst drilling a uniform shale sequence then ROP should decrease with depth. This is due to the natural increased compaction with depth reflecting a decrease in porosity and increased shale density and increased shale (compressive) strength. When entering an abnormally pressured shale, the drill bit sees a shale section which is undercompacted. The increased porosity and decreased density of the undercompacted section results in the formation becoming more ‘drillable’ as there is less rock matrix to remove. Consequently ROP increases, assuming all drilling parameters were kept constant. Drilling underbalanced can further increase the drilling rate. 79 Drilling Rate as a Pore Pressure Predictor Penetration rate depends on a number of different parameters. R = K(P1)a1 (P2)a2 (P3)a3… (Pn)an Modified d-exponent 81 log R 60N d 12W log 6 10 d b R ft /hr N rpm d d exponent W Bit W eight, lbf d b Bit Diamet er,in Or, in its most used form: The D Exponent basically attempts to correct the ROP for changes in RPM, weight on bit and hole size. The D exponent increases linearly with depth, reflecting increased rock strength with depth. For abnormally pressured shales, the D exponent deviates from the normal trend and actually decreases with depth. 82 dc-exponent Mud weight also affects R An adjustment to d may be made: dc = d (rn /rc) where dc = exponent corrected for mud density rn = normal pore pressure gradient rc = effective mud density in use 83 d-exponent The d-exponent normalizes R for any variations in W, db and N Under normal compaction, R should decrease with depth. This would cause d to increase with depth. Any deviation from the trend could be caused by abnormal pressure. 84 Example • While drilling in a Gulf Coast shale, R = 50 ft/hr W = 20,000 lbf N = 100 RPM ECD = 10.1 ppg (Equivalent Circulating Density) db = 8.5 in • Calculate d and dc 85 Solution 50 log 2.079 60 * 100 d 12 * 20,000 1.554 log 6 10 * 8.5 d 1.34 0.465 dc 1.34 0.052 * 10.1 dc 1.19 86 log R 60N d 12 W log 10 6 d b rn d c d rc Example solution Ratio Method The ratio method is much simpler and does not require values of overburden. To calculate pore pressure, use the following formula: Homework • Using the Eaton Method, calculate the pore pressure at depth 12000 ft given: dcn (from normal trend) = 1.5 d-units dco (from new trend) = 1.1 d-units Overburden gradient = 19 ppg Normal pore pressure in area = 9 ppg Fracture Pressure Prediction of Fracture Gradients Well Planning Theoretical Fracture Gradient Determination Hubbert & Willis Matthews & Kelly Ben Eaton Comparison of Results Experimental Frac. Grad. Determination Leak-off Tests Fracture Gradients 1.11- 101 In-Situ Earth Stresses Example Fracture Gradient Determination 2. Matthews & Kelly: K is P F D D where Ki = matrix stress coefficient s = vertical matrix stress, psi Fracture Gradients 1.11- 106 Example A Texas Gulf Coast well has a pore pressure gradient of 0.735 psi/ft. Well depth = 11,000 ft. Calculate the fracture gradient in units of lb/gal using Matthews & Kelly method Summarize the results in tabular form, showing answers, in units of lb/gal and also in psi/ft. Fracture Gradients 1.11- 107 Example P K is F D D 2. Matthews & Kelly s may be calculated, and K i is determined In this case P and D are known, graphically. (i) First, determine the pore pressure gradient. P 0.735 D Fracture Gradients psi / ft 1.11- 108 (given ) Example - Matthews and Kelly (ii) Next, calculate the matrix stress. S=P+s s=S-P = 1.00 * D - 0.735 * D = 0.265 * D = 0.265 * 11,000 s = 2,915 psi Fracture Gradients 1.11- 109 S overburden, psi s mat rixstress, psi P porepressure, psi D depth, ft Example - Matthews and Kelly (iii) Now determine the depth, D,i where, under normally pressured conditions, the rock matrix stress, s would be 2,915 psi. Sn = Pn + sn n = “normal” 1.00 * Di = 0.465 * Di + 2,915 Di * (1 - 0.465) = 2,915 2,915 Di 5,449 ft 0.535 Example Matthews and Kelly (iv) Find Ki from the plot on the right, for Di = 5,449 ft For a south Texas Gulf Coast well, Ki = 0.685 Fracture Gradients 1.11- 111 Example - Matthews and Kelly (v) Now calculate F: K is P F D D 0.685 * 2,915 F 0.735 11,000 0.9165 psi / ft 0.9165 F 17.63 0.052 lb / gal Leak off Test A test carried out to the point where the formation leaks off Fracture Gradients 1.11- 114 Experimental Determination of Fracture Gradient Example: In a leak-off test below the casing seat at 4,000 ft, leak-off was found to occur when the standpipe pressure was 1,000 psi. MW = 9 lb/gal. What is the fracture gradient? Slide 115 Solution Leak-off pressure = PS + DPHYD = 1,000 + 0.052 * 9 * 4,000 = 2,872 psi PLEAK OFF 2,872 D 4,000 psi ft Fracture gradient = 0.718 psi/ft EMW = ? 13.8 lb/gal Slide 116 Homework While performing a leak off test the surface pressure at leak off was 940 psi. The casing shoe was at a true vertical depth of 5010 ft and a mud weight of 10.2 ppg was used to conduct the test. Calculate: the maximum allowable mud weight at this depth . Homework A leakoff test was carried out just below a 13 3/8" casing shoe at 7000 ft. TVD using 9.0 ppg mud. The results of the tests are shown below. What is the maximum allowable mud weight for the 12 1/4" hole section ? BBLS PUMPED 1 1.5 2 2.5 3 3.5 4 4.5 5 SURFACE PRESSURE 400 670 880 1100 1350 1600 1800 1900 1920 (psi) Equivalent Circulating Density (ECD) When the drilling fluid is circulating through the drillstring, the borehole pressure at the bottom of the annulus will be greater than the hydrostatic pressure of the mud. The extra pressure is due to the frictional pressure required to pump the fluid up the annulus. This frictional pressure must be added to the pressure due to the hydrostatic pressure from the column of mud to get a true representation of the pressure acting against the formation a the bottom of the well. An equivalent circulating density (ECD) can then be calculated from the sum of the hydrostatic and frictional pressure divided by the true vertical depth of the well. Homework If the circulating pressure losses in the annulus of the above well is 300 psi when drilling at 7500ft with 9.5ppg mud, what would be the ECD of the mud at 7500ft.