Understanding pressure and pressure

CHAPTER 1

1-1 PRESSURE BASICS Remember to think downhole. The concepts provided in this section cover the foundations for good well control. Understanding pressure and pressure relationships is important if we are to understand well control. By definition, pressure is the force that is exerted or placed on a unit of area, such as pounds per square inch (psi). The pressures that we deal with daily in the oil industry include fluid, formation, friction and mechanical. When certain pressure limits are exceeded, disastrous consequences can result, including blowouts and/or the loss of life. FLUID PRESSURE What is a fluid? A fluid is simply something that is not solid and can flow. Water and oil are obviously fluids. Gas is also a fluid.

1-2 CHAPTER 1 Pressure (Force) What is pressure? Pressure (Force) Fluid Pressure Pressure: 1: the force per unit area that is exerted on a surface. 2: the force that a fluid exerts when it is in some way confined within a vessel. Under extreme temperature and/or pressure almost anything will become fluid. Under some conditions salt or rock become fluid. For our purposes, fluids that we will consider are those normally associated with the oil industry, such as oil, gas, water, muds, packer fluids, brines, completion fluids, etc. Fluids exert pressure. This pressure is the result of the density of the fluid and the height of the fluid column. Density is usually measured in pounds per gallon (ppg) or kilograms per cubic meter (kg/m ³ ). A heavy fluid exerts more pressure because its density is greater. The force or pressure that a fluid exerts at any given point is usually measured in pounds per square inch (psi) or in the metric system, bar. To find out how much pressure a fluid of a given density exerts for each unit of length, we use a pressure gradient. A pressure gradient is normally expressed as the force which the fluid exerts per foot (meter) of depth; it is measured in pounds per square inch per foot (psi/ft) or bars per meter (bar/m). To get the pressure gradient we must convert the fluid s density in pounds per gallon to pounds per square inch per foot (or kilograms per cubic meter, kg/m ³ to bar/m). DENSITY CONVERSION FACTOR The conversion factor used to convert density to pressure gradient in the English system is 0.052. In the metric system, it is 0.0000981. Remember that the definition of pressure gradient is the pressure increase per unit of depth due to its density. For our text, we will use pounds per gallon (ppg) to measure density and feet (ft) for depth measurements in the English system and kilograms per cubic meter (kg/m ³ ) to measure density and meters (m) for depth measurements in the metric system. The way 0.052 is derived is by using a onefoot cube (one foot wide by one foot long by one foot high). It takes about 7.48 gallons to fill the cube with fluid. If the fluid weighs one pound per gallon, and you have 7.48 gallons, then the total weight of the cube is 7.48 pounds, or 7.48 pounds per cubic foot. The weight of one of these square inches, one foot in height, can be found by dividing the total weight of the cube by 144: 7.48 144 = 0.051944 The conversion factor 0.052 is commonly used for oilfield calculations.

PRESSURE BASICS 1-3 PRESSURE GRADIENT To find the pressure gradient of a fluid, multiply the density of the fluid by 0.052; or in metrics, by 0.0000981. Pressure Gradient = Fluid Density Conversion Factor So the pressure gradient of a 10.3 ppg (1234 kg/m ³ ) fluid can be found by multiplying the fluid weight by the conversion factor. Pressure Gradient psi/ft = Fluid Density ppg Conversion Factor = 10.3 ppg 0.052 = 0.5356 psi/ft To calculate pressure at the bottom of a well use true vertical depth. Pressure Gradient bar/m = Fluid Density kg/m ³ Conversion Factor EXAMPLE 1 = 1234 kg/m ³ 0.0000981 = 0.1211 bar/m What is the pressure gradient of a fluid with a density of 12.3 ppg (1474 kg/m ³ )? Pressure Gradient psi/ft = Fluid Density ppg Conversion Factor = 12.3 X 0.052 = 0.6396 psi/ft Pressure Gradient bar/m = Fluid Density kg/m ³ Conversion Factor PROBLEM 1A = 1474 X 0.0000981 = 0.1446 bar/m What is the pressure gradient of a fluid that weighs 9.5 ppg (1138 kg/m ³ )? Pressure Gradient psi/ft = Fluid Density ppg X Conversion Factor Pressure Gradient bar/m = Fluid Density kg/m ³ X Conversion Factor 1' PROBLEM 1B What is the pressure gradient of fresh water which weighs 8.33 ppg (998 kg/m ³ )? If a fluid weighs one pound per gallon, the weight of one square inch in a one foot length is 0.052 lbs. 1' 1'

1-4 CHAPTER 1 Hydrostatic pressure: force exerted by a body of fluid at rest; increases directly with the weight and length of the fluid column. TVD VS. MD Once we know how to find pressure exerted per foot, we can calculate the hydrostatic pressure at a given depth. All we have to do is multiply the pressure gradient by the number of feet to that vertical depth. Here we have to learn the distinction between measured depth (MD) and true vertical depth (TVD). In the illustration below you can see that the depth straight down (the way that gravity pulls) for both wells is 10,000 ft (3048 m). Well A has a measured depth of 10,000 ft (3048 m), and a true vertical depth of 10,000 ft (3048 m). Since gravity pulls straight down, along a true vertical (straight down) path, to calculate the pressure at the bottom of the hole we use the 10,000 ft (3048 m) depth. Well B has a measured depth of 11,650 ft (3550.92 m), and its true vertical depth is 10,000 ft (3048 m). Gravity still pulls straight down, not along the path of the well. You would have a vertical depth of 10,000 ft (3048 m) from the surface straight down to where the well ended. Therefore, to calculate the pressure at the bottom of Well B, you have to use the true vertical depth of 10,000 ft (3048 m). The illustration on page 1-5 offers another way of looking at the difference between true vertical depth and measured depth. In this illustration, we have a picture of square blocks, 15 by 10. Count how many blocks the well covers. This represents the measured depth of the well. Now, count the blocks from the bottom of the well, straight up to surface. The numbers of these blocks represent true vertical depth. Well A 10.0 PPG MUD HYDROSTATIC PRESSURE Hydrostatic pressure is the total fluid pressure created by the weight of a column of fluid, acting on any given point in a well. Hydro means water, or fluid, that exerts pressure like water, and static means not moving. So hydrostatic pressure is the pressure created by the density and height of a stationary (not moving) column of fluid. We already know how to calculate a pressure gradient from the fluid s weight. Hydrostatic pressure can be calculated from a pressure gradient to a given point: Hydrostatic Pressure = Pressure Gradient Depth TVD Or, it may be calculated by: Hydrostatic Pressure = Fluid Density Conversion Factor Depth TVD 10,000' True vertical depth vs. measured depth. Well B 10.0 PPG MUD 11, 650' MD

PRESSURE BASICS EXAMPLE 2 1-5 What is the hydrostatic pressure at the bottom of a well which has a fluid density of 9.2 ppg (1102 kg/m³), a MD of 6,750 (2057.4 m) and a TVD of 6,130 (1868.42 m)? Remember, the formula for calculating hydrostatic pressure is: Hydrostatic Pressure psi = Fluid Density ppg Conversion Factor Depth ft, TVD = 9.2 0.052 6,130 = 2,933 psi Hydrostatic Pressure bar = Fluid Density kg/m ³ Conversion Factor Depth m, TVD PROBLEM 2A = 1102 0.0000981 1868.42 = 201.99 bar TVD MD Find the hydrostatic pressure at the bottom of a well with a 9.7 ppg (1162 kg/m³) fluid in it and a MD of 5,570 (1697.74 m) and TVD of 5,420 (1651.02 m). True vertical depth vs. measured depth. Hydrostatic Pressure psi = Fluid Density ppg 0.052 Depth ft, TVD Hydrostatic Pressure bar = Fluid Density kg/m ³ 0.0000981 Depth m, TVD PROBLEM 2B Find the hydrostatic pressure at 4,300 (1310.64 m) TVD, of a well with fluid density of 16.7 ppg (2001 kg/m³). The well has a MD of 14,980 (4565.9 m) and a TVD of 13,700 (4175.76 m). The above equations for fluid gradient and hydrostatic pressure are basic to understanding the fundamentals of pressure in wells. To prevent the well from flowing, fluid pressure in the well must at least equal the formation pressure. Atmospheric pressure at sea level is about 15 psi; the metric equivalent is approximately one bar. GAUGE/ATMOSPHERIC PRESSURE Although a gauge placed at the bottom of a fluid column would read the hydrostatic pressure of that column, it also would read the atmospheric pressure exerted on the column. This pressure varies with weather conditions and elevation and is normally considered 14.7 psi or 15 psi (approximately one bar) at sea level. If a gauge has a unit notation of psig, it includes the atmospheric column above it. If gauge reads in psig, it has been calibrated to subtract the atmospheric column above it.

1-6 CHAPTER 1 U Tube Analogy U Tubing String Higher density fluid U Tubing String Annulus Annulus U-tubing: the tendency of liquids to seek a pressure balance point in an open well. U-TUBE It is often helpful to visualize the well as a U-tube (see above). One column of the tube represents the annulus and the other column represents the pipe in the well. The bottom of the U-tube represents the bottom of the well. In most cases, there are fluids creating hydrostatic pressures in both the pipe and annulus. Atmospheric pressure can be omitted, since it works the same on both columns. If there were 10 ppg (1198 kg/m ³ ) fluid in both the pipe and annulus, hydrostatic pressures would be equal and the fluid would be static on both sides of the tube. However, what will happen if fluid in the annulus is heavier than the fluid in the string? The heavier fluid in the annulus exerting more pressure downward will flow into the string, displacing some of the lighter fluid out of the string, causing a flow at surface. The fluid level will fall in the annulus, equalizing pressures. When there is a difference in the hydrostatic pressures, the fluid will try to reach balance point. This is called U-tubing, and it explains why there is often flow from the pipe when making connections. This is often evident when drilling fast because the effective density in the annulus is increased by cuttings. Another example of U-tubing is when a slug is pumped. The heavier slug is designed to allow tubing to pull dry by falling to a level below the average length of stand pulled. The depth where the slug will fall and the amount of fluid that U-tubes from the well can be calculated using the following equations: Gain In Pits = (Slug Weight Annulus Weight) Volume of Slug Annulus Weight EXAMPLE 3 Distance of Drop = Gain in Pits Pipe Capacity What will be the gain in the pits, and how far will the slug fall if the mud weight is 10 ppg (1198 kg/m ³ ), the pipe s capacity is 0.0178 bbl/ft (0.00929 m ³ /m)? The volume of the slug is 30 bbls (4.77 m ³ ) and weighs 11 ppg (1318 kg/m ³ ).

PRESSURE BASICS 1-7 Gain In Pits bbls = (Slug Weight ppg Annulus Weight ppg ) Volume of Slug bbls Annulus Weight ppg = (11 10) 30 10 = 1 30 10 = 3 bbls Distance of Drop ft = Gain In Pits bbls Pipe Capacity bbls/ft = 3 0.0178 = 168.5 ft Porosity is the measurement of void space within a rock, expressed as a percentage. Gain In Pits m ³ = (Slug Weight kg/m ³ Annulus Weight kg/m ³) Volume of Slug m ³ Annulus Weight kg/m ³ = (1318-1198) 4.77 1198 = 120 X 4.77 0.00929 1198 = 0.478 m ³ Distance of Drop m = Gain In Pits m ³ Pipe Capacity m ³ /m PROBLEM 3 = 0.478 0.00929 = 51.45 m What will be the gain in the pits, and how far will the slug fall if the mud weight is 11.6 ppg (1390 kg/m ³ ), the pipe s capacity is 0.00579 bbl/ft (0.00302 m ³ /m)? The volume of the slug is 15 bbls (2.39 m ³ ) and weighs 12.4 ppg (1486 kg/m ³ ). Gain In Pits bbls = (Slug Weight ppg Annulus Weight ppg ) Volume of Slug bbls Annulus Weight ppg Distance of Drop ft = Gain In Pits bbls Pipe Capacity bbls/ft Gain In Pits m ³ = (Slug Weight kg/m ³ Annulus Weight kg/m ³) Volume of Slug m ³ Annulus Weight kg/m ³ Distance of Drop m = Gain In Pits m ³ Pipe Capacity m ³ /m Two important characteristics of reservoir rocks are porosity, tiny openings in rock (far left) and permeability, the connection of these holes which allows fluids to move (near left).

1-8 Fracture pressure is the amount of pressure it takes to permanently deform the rock structure of a formation. FORMATION CHARACTERISTICS Porosity and permeability, along with pressure differences, must be considered if we are to understand well control. A reservoir rock looks solid to the naked eye. A microscopic examination reveals the existence of tiny openings in the rock. These openings are called pores. The porosity of the rock is expressed as a percentage. It is the ratio of void (pore) space to solid volume. Another characteristic of a reservoir rock is that it must be permeable. That is, the pores of the rock must be connected so hydrocarbons can move between them. Otherwise the hydrocarbons remain locked in place and cannot flow into a well. FORMATION PRESSURE Formation pressure is the pressure within the pore spaces of the formation rock. This pressure can be affected by the weight of the overburden (rock layers) above the formation, which exerts pressure on both the grains and pore fluids. Grains are solid or rock material, and pores are spaces between grains. If pore fluids are free to move, or escape, the grains lose some of their support and move closer together. This process is called compaction. Normally pressured formations exert a pressure equal to a vertical column of native fluid from the formation to surface. The pressure gradient of the native fluid usually ranges from 0.433 psi/ft (0.0979 bar/m) to 0.465 psi/ft (0.1052 bar/m), and varies depending on the geologic region. Formations pressured in this range are designated normal, depending on the area. For simplicity, this text will designate a gradient of 0.465 psi/ft (0.1052 bar/m) as normal. In normally pressured formations most of the overburden weight is supported by the grains that make up the rock. When the overburden increases with depth, pore fluids are free to move and the amount of pore space is reduced due to compaction. CHAPTER 1 Abnormally pressured formations exert pressure greater than the hydrostatic pressure (or pressure gradient) of the contained formation fluid. When abnormally pressured formations develop, during the compaction phase, the pore fluid movement is restricted or stopped. The pore fluid pressure increases, generally exceeding 0.465 psi/ft (0.1052 bar/m). The result causes the increasing overburden weight to be partially supported by pore fluid rather than by the rock grains. Such formations may require working fluid densities up to, and sometimes greater than, 20 ppg (2397 kg/m ³ ) to control them. Abnormal pressures may be caused in other ways, including the presence of faults, salt domes, uplifting, and differences in elevation of underground formations. In many regions, hundreds of feet of pre-existing rock layers (overburden) have been stripped off by erosion. At the new, shallower depths this loss from erosion can cause the pressure to become abnormal, above 0.465 psi/ft (0.1052 bar/m), or 8.94 ppg (1072 kg/m ³ ). When a normally pressured formation is raised toward the surface while prevented from losing pore fluid in the process, it will change from normal pressure (at a greater depth) to abnormal pressure (at a shallower depth). When this happens, and then you drill into the formation, mud weights of up to 20 ppg (2397 kg/m ³ ) may be required for control. This process accounts for many of the shallow, abnormally pressured zones in the world. In areas where faulting is present, salt layers or domes are predicted, or excessive geothermal gradients are known, drilling operations may encounter abnormal pressure. An abnormally pressured formation can often be predicted using well history, surface geology, downhole logs or geophysical surveys. Subnormally pressured formations have pressure gradients lower than fresh water, or less than 0.433 psi/ft (0.0979 bar/m). Naturally occurring subnormal pressure can be developed when the overburden has been stripped away, leaving the formation exposed at the surface.

PRESSURE BASICS Casing Cement Cement Test existing formation pressure and resisting rock structure. Loosely compacted formations, such as those found offshore in deep water, can fracture at low gradients. Fracture pressures at any given depth can vary widely because of the geology of the area. 1-9 Formation FORMATION INTEGRITY TESTS Depletion of original pore fluids through evaporation, capillary action and dilution produces hydrostatic gradients below 0.433 psi/ft (0.0979 bar/m). Subnormal pressures may also be induced through depletion of formation fluids. FRACTURE PRESSURE Integrity Test Fracture pressure is the amount of pressure it takes to permanently deform (fail or split) the rock structure of a formation. Overcoming formation pressure is usually not sufficient to cause fracturing. If pore fluid is free to move, a slow rate of entry into the formation will not cause fractures. If pore fluid cannot move out of the way, fracturing and permanent deformation of the formation can occur. Fracture pressure can be expressed as a gradient (psi/ft), a fluid density equivalent (ppg), or by calculated total pressure at the formation (psi). Fracture gradients normally increase with depth due to increasing overburden pressure. Deep, highly compacted formations can require very high fracture pressures to overcome the An accurate evaluation of a casing cement job as well as of the formation is extremely important during the drilling of a well and for subsequent work. The information resulting from Formation Integrity Tests (FIT) is used throughout the life of the well and also for nearby wells. Casing depths, well control options, formation fracture pressures and limiting fluid weights may be based on this information. To determine the strength and integrity of a formation, a Leak Off Test (LOT) or a Formation Integrity Test (FIT) may be performed. Whatever the name, this test is first: a method of checking the cement seal between casing and the formation, and second: determining the pressure and/or fluid weight the test zone below the casing can sustain. Whichever test is performed, some general points should be observed. The fluid in the well should be circulated clean to ensure it is of a known and consistent density. If mud is used for the test, it should be properly conditioned and gel strengths minimized. The pump used should be a high-pressure, low-volume test or cementing pump. Rig pumps can be used if the rig has electric drives on the mud pumps, and they can be slowly rolled over. If the rig pump must be used and the pump cannot be easily controlled at low rates, then the leak-off technique must be modified. It is a good idea to make a graph of the pressure versus time or volume for all leak-off tests as shown in the illustrations on the next page. The information resulting from formation integrity tests is used throughout the life of a well.

1-10 CHAPTER 1 Increments of Volume Generally about 20 Gal (75 Liters) Cumulative Volume Pumped Increments of Pressure PRESSURE Stop Here SURFACE PRESSURE (PSI) Pit Limit B C Shut Pumps Down D Instantaneous Shut-in Pressure Shut-In Time E End of Test PRESSURE Increments of Pressure Weight Stop Here A Slack in System TIME PUMP STROKES TIME Pressure vs time or volume for leak-off tests Jug test: limited formation integrity test, often performed when risk of formation damage is high. LEAK-OFF TEST (LOT) A leak-off test is performed to estimate the maximum pressure or mud weight (fluid density) that the test point can withstand before formation breakdown or fracture occurs. LEAK-OFF TECHNIQUE 1 The well is pressured in increments of 100 psi (6.9 bar) or fluid is pumped into the well in approximately one-half barrel (0.079 m ³ ) increments of volume. After each increase in pressure, the pump is stopped and the pressure is held for about 5 minutes. If the pressure holds, the next increment is tested. If the pressure does not hold, the well is pressured again. The test is completed when the pressure will not hold after several attempts, or if the well will not pressure up any further. LEAK-OFF TECHNIQUE 2 The choke is opened on the manifold and the pump is started at an idle. The choke is closed to increase the pressure in increments of 100 psi (6.9 bar). At each interval of pressure, the fluid volume in the pits is watched until it is certain that no fluid is being lost to the formation. The test is complete at the pressure where fluid is continuously being lost to the formation. Some fluid will be lost at each pressure increase. If this technique is to be used, a small tank should be used so large amounts of fluid are not forced into the formation. Circulating frictional pressure losses which are present in this technique add more unseen pressure on the formation tested, which will give slightly different results (lower fracture pressures) than technique number 1. LIMITED INTEGRITY TEST A limited formation integrity test (limited FIT), also called a jug test, is performed when it is not acceptable to cause the formation to fracture. It may also be used on wells drilled in developed fields. In such cases, operators have good data concerning formation strength and do not expect to approach fracture pressures. In the limited formation integrity test, the wellbore is pressured to a predetermined pressure or fluid weight. If the formation can withstand the applied pressure, the test is called good. Both tests, Limited FIT and LOT, have their good and bad points. In the Limited FIT, the formation is not broken down, however, the maximum pressure before the formation starts to accept fluid is not determined. In the LOT, the pressure where the formation starts accepting fluid is determined, but there is a possibility of fracturing the formation.

PRESSURE BASICS FLUID DENSITY/PRESSURE The total pressure applied causes leak off or formation damage. This is usually a combination of the hydrostatic pressure of a fluid plus an additional pressure, such as pump pressure on a leak off test. The applied pressure raises the total pressure against the formation. From test data, calculations estimate the integrity fluid density. This is the total pressure, represented as fluid density, above which leak off or formation damage may occur. This may also be called the maximum allowable mud weight or frac mud weight. The calculations to find the estimated integrity fluid density follow. 1-11 When computing formation integrity values decimals in the results are not rounded up. Estimated Integrity Fluid Density ppg = (Test Pressure psi 0.052 Depth of Test ft, TVD ) + Test Fluid Density ppg Est. Integrity Fluid Density kg/m ³ = (Test Pressure bar 0.0000981 Depth of Test m, TVD ) + Test Fluid Density kg/m ³ Test fluid density is seldom used throughout the entire well. If the fluid density changes, then the surface pressure that may damage the formation should be re-calculated. To find the new estimated integrity pressure with a different density fluid: Est. Integrity Pressure psi = (Est. Integrity Fluid Density ppg Present Fluid Density ppg ) Depth of Test ft, TVD 0.052 Est. Integrity Pressure bar = (Est. Int. Fluid Density kg/m ³ Present Fluid Density kg/m ³) Depth of Test m, TVD 0.0000981 EXAMPLE 4 Solve the following equations for the formation's estimated integrity fluid density (maximum fluid weight without causing formation damage), and the estimated integrity pressure that may cause damage with a different fluid density using the following data. Note: When doing the following exercises, decimals in the answers should not be rounded up. Safety against formation fracture lies in the lower values. The well has a TD of 11,226' (3421.68 m) and a Casing Shoe set at 5,821' (1774.24 m) TVD. The Leak Off Test Pressure was 1,250 psi (86.19 bar), with a Leak Off Test Fluid of 9.6 ppg (1150 kg/m ³ ). The Present Fluid Wt. is 10.1 ppg (1210 kg/m ³ ). First find the Estimated Integrity Fluid Density: Estimated Integrity Fluid Density ppg = (Test Pressure psi 0.052 Depth of Test ft, TVD ) + Test Fluid Density ppg = (1,250 0.052 5,821) + 9.6 = 4.1 + 9.6 = 13.7 ppg. Est. Integrity Fluid Density kg/m ³ = (Test Pressure bar 0.0000981 Depth of Test m, TVD ) + Test Fluid Density kg/m ³ = (86.19 0.0000981 1774.24) + 1150 = 495 + 1150 = 1645 kg/m ³

1-12 CHAPTER 1 In formation integrity calculations, it is more conservative to not round up. So, in the previous calculations 4.1 ppg was used instead of 4.13 ppg (495 kg/m ³ instead of 495.19 kg/m ³ ). In this example, the present mud weight is higher than the test mud weight so we must solve for the present estimated integrity pressure. Estimated Integrity Pressure psi If fluid density is changed, surface pressure that may damage the formation must be recalculated. = (Est. Integrity Fluid Density ppg Present Fluid Density ppg ) Depth of Test ft, TVD 0.052 = (13.7 10.1) 5,821 0.052 = 1,089 psi Estimated Integrity Pressure bar = (Est. Int. Fluid Density kg/m ³ Pres. Fluid Density kg/m ³) Depth of Test m, TVD 0.0000981 = (1645 1210) 1774.24 0.0000981 = 75.71 bar PROBLEM 4 What is the estimated integrity fluid density and estimated integrity pressure that may damage the formation for a well with an MD of 12,000 (3657.6 m), TVD of 10,980 (3346.7 m). The Casing Shoe is at 8,672 (2643.23 m) TVD. The Leak Off Test Pressure was 1,575 psi (108.59 bar) with a Leak Off Test Fluid Density of 11.1 ppg (1330 kg/m ³ ), and the Present Fluid Density is 11.6 ppg (1390 kg/m ³ ). First, solve for the estimated integrity fluid density: Estimated Integrity Fluid Density ppg = (Test Pressure psi 0.052 Depth of Test ft, TVD ) + Test Fluid Density ppg Estimated Integrity Fluid Density kg/m ³ = (Test Pressure bar 0.0000981 Depth of Test m, TVD ) + Test Fluid Density kg/m ³ Second, solve for the present estimated integrity pressure: Estimated Integrity Pressure psi = (Est. Integrity Fluid Density ppg Present Fluid Density ppg ) Depth of Test ft, TVD 0.052 Estimated Integrity Pressure bar = (Est. Int. Fluid Density kg/m ³ Pres. Fluid Density kg/m ³) Depth of Test m, TVD 0.0000981 Often a chart is generated to post on the rig floor showing incremental mud weight increases and the estimated integrity pressures with each. To do so, calculate the gain in hydrostatic pressure for an increment of 0.1 ppg (11.98 kg/m³). Hydrostatic Pressure = Fluid Weight Increase Conversion Factor Depth TVD The estimated integrity pressure that may be applied is reduced by the hydrostatic pressure increase gained by each increase in mud weight. A chart beginning with the present mud weight up to integrity fluid density versus integrity pressure can then easily be prepared.

PRESSURE BASICS EXAMPLE 5 Prepare a chart of Estimated Integrity Pressure on surface for mud weights ranging from 10.1 to 11.1 ppg (1222 to 1330 kg/m³). The Casing Shoe depth is 5,821 (1774.24 m) TVD and Estimated Integrity Pressure for present fluid weight 10.1 ppg (1210 kg/m³) is 1,250 psi (86.19 bar). First, find the increase in hydrostatic pressure for 0.1 ppg (11.98 kg/m³): Hydrostatic Pressure psi = Fluid Wt. Inc. ppg 0.052 Depth ft, TVD = 0.1 0.052 5,821 = 30 psi Hydrostatic Pressure bar = Fluid Wt. Inc. kg/m ³ 0.0000981 Depth m, TVD = 11.98 0.0000981 1774.24 = 2.09 bar Based on the gain in hydrostatic pressure, subtract this value from the calculated Estimated Integrity for each corresponding increase in mud weight. Estimated Integrity Pressure on Surface Fluid Estimated Fluid Estimated Density Integrity Press. Density Integrity Press. (ppg) (psi) (kg/m 3 ) (bar) 10.1 1250 1210 86.19 10.2 1220 1222 84.1 10.3 1190 1234 82.01 10.4 1160 1246 79.92 10.5 1130 1258 77.83 10.6 1100 1270 75.74 10.7 1070 1282 73.65 10.8 1040 1294 71.56 10.9 1010 1306 69.47 11.0 980 1318 67.38 11.1 950 1330 65.29 1-13 PROBLEM 5 Prepare a chart of Estimated Integrity Pressure on surface for mud weights ranging from 11.7 to 12.6 ppg (1402 to 1510 kg/m³). The Casing Shoe depth is 8,672 (2,643.23 m) TVD and the Estimated Integrity Pressure for present fluid weight 11.6 ppg (1390 kg/m³) is 1,352 psi (93.22 bar). Estimated Integrity Pressure on Surface Hydrostatic Pressure psi = Fluid Wt. Inc. ppg 0.052 Depth ft, TVD Fluid Estimated Fluid Estimated Density Integrity Press. Density Integrity Press. (ppg) (psi) (kg/m 3 ) (bar) Hydrostatic Pressure bar = Fluid Wt. Inc. kg/m ³ 0.0000981 Next, fill in chart at right. Depth m, TVD Alternate terms such as fracture mud weight, and either MASP (Maximum Allowable Surface Pressure) or MAASP (Maximum Allowable Annular Surface Pressure) are also used for estimated integrity fluid density and estimated integrity pressure. If such terms and information are used as limiting factors and without proper understanding of limiting pressures vs. maintaining control of the well, serious well control complications can result. If this information is used during a well kill operation you must also take kick position, distribution and density into account.

1-14 EQUIVALENT MUD WEIGHT From the previous discussions, it should be apparent that any applied pressure raises the total pressure at a given point. If the applied pressure is known, then it can be calculated to an equivalent weight. Alternatively, if a zone must be pressure tested to an equivalent weight, then calculations CHAPTER 1 may be performed to determine the test pressure. The equivalent mud weight (EMW) is also the summation of all pressures (hydrostatic pressure, choke or back pressure, applied pressure, kick pressure, circulating pressure losses, etc.) at a given depth or zone and is expressed as a fluid density. If these pressures are known or can be estimated, the EMW can be calculated as follows: EMW = (Pressure Conversion Factor Depth of Interest TVD ) + Present Fluid Density EXAMPLE 6 What is the EMW for a zone with an MD depth of 3,120 (950.97 m) and TVD of 3,000 (914.4 m) when the well is shut in with 375 psi (25.86 bar) registering on the casing gauge? Present Fluid Density is 8.8 ppg (1055 kg/m ³ ). EMW ppg = (Pressure psi 0.052 Depth of Interest ft, TVD ) + Present Fluid Density ppg = (375 0.052 3,000) + 8.8 = 2.4 + 8.8 = 11.2 ppg EMW kg/m ³ = (Pressure bar 0.0000981 Depth of Int. m, TVD ) + Present Fluid Density kg/m ³ Frictional resistance: the opposition to flow created by a fluid when it flows through a line or other container. PROBLEM 6 = (25.86 0.0000981 914.4) + 1055 = 288 + 1055 = 1343 kg/m ³ What is the EMW for a zone with a MD of 7,320 (2231.14 m) and TVD of 6,985 (2129.03 m) if the estimated choke and friction pressures total 730 psi (50.33 bar)? The Present Fluid Density is 13.8 ppg (1654 kg/m ³ ). EMW ppg = (Pressure psi 0.052 Depth of Interest ft, TVD ) + Present Fluid Density ppg EMW kg/m ³ = (Pressure bar 0.0000981 Depth of Interest m, TVD ) + Present Fluid Density kg/m ³

PRESSURE BASICS 1-15 To determine how much applied pressure is required to test to a pre-determined EMW at a given depth: Test Pressure psi = (EMW ppg Present Fluid Density ppg ) 0.052 Depth Tested ft, TVD Test Pressure bar = (EMW kg/m ³ Present Fluid Density kg/m ³) 0.0000981 Depth Tested m, TVD EXAMPLE 7 How much test pressure should be used to test a formation with a MD of 5,890 (1795.27 m) and a TVD of 5,745 (1751.08 m) to an equivalent fluid density of 13.4 ppg (1606 kg/m ³ )? The Present Fluid Density is 9.1 ppg (1090 kg/m ³ ). Most pressure loss will occur circulating down the string and through restrictions such as jet nozzles. Test Pressure psi = (EMW ppg Present Fluid Density ppg ) 0.052 Depth Tested ft, TVD = (13.4 9.1) 0.052 5,745 = 4.3 0.052 5,745 = 1285 psi Test Pressure bar = (EMW kg/m ³ Present Fluid Density kg/m ³) 0.0000981 Depth Tested m, TVD = (1606 1090) 0.0000981 1751.08 = 88.64 bar PROBLEM 7 How much test pressure should be used to test a formation with a MD of 7,590 (2313.43 m) and a TVD of 7450 (2270.76 m) to an equivalent fluid density of 14.3 ppg (1714 kg/m ³ )? The Present Fluid Density is 8.9 ppg (1067 kg/m ³ ). Test Pressure psi = (EMW ppg Present Fluid Density ppg ) 0.052 Depth Tested ft, TVD Test Pressure bar = (EMW kg/m ³ Present Fluid Density kg/m ³) 0.0000981 Depth Tested m, TVD

1-16 Bottomhole pressure: 1: pressure exerted by a column of fluid in the wellbore. 2: formation pressure at depth of interest. PRESSURE LOSSES/CIRCULATING Friction is the resistance to movement. It takes force, or pressure, to overcome friction to get anything to move. Friction has to be overcome to lift pipe, move fluid, or even to walk. How much friction is present to overcome depends upon many factors. These include density or weight, type and roughness of the surfaces making contact, surface area, thermal and electrical properties of the surfaces and direction and velocity of the objects. The amount of force used to overcome friction is called frictional loss and can be measured in many ways. Torque, drag (amps, foot-pounds, horsepower) and force (psi or bar) to move fluid are a few. Thousands of psi (bar) of pressure can be lost to the well s Standpipe CHAPTER 1 circulating system as fluid is pumped through surface lines, down the string, and up the annulus. The pressure on the pump is actually the amount of friction that must be overcome to move fluid throughout the wellbore at a given flow rate. Most of the pressure loss will occur when circulating down the string and through restrictions such as jet nozzles. Pressure losses also occur in other parts of the circulating system, such as when the choke is used to hold back pressure on the casing side during well killing operations. When fluid finally returns to the pits it is under atmospheric, or almost zero, pressure. When the well is being circulated, bottomhole pressure is increased by the amount of friction overcome in the annulus. When pumps are shut off, wellbore pressure is reduced because no frictional force is being overcome. 3000 2950 Drill Pipe Pump 0 Flowline TANK Casing Circulating Pressure Bit 900

PRESSURE BASICS 1-17 Well static Normal circulation Circulation with rotating head Kick circulation PUMP PUMP PUMP PUMP Rotation Head BOP Stack BHP = HP BHP = HP + APL BHP = HP + APL + Rotating Head Back Pressure BHP = HP + APL + Choke Press Since friction adds pressure to the wellbore, it increases the effective weight, or the equivalent circulating density (ECD). The total value is the equivalent of bottomhole pressure with the pump on. If pressure in a permeable formation is closely balanced by ECDs, a well could flow when the pump is turned off. Data obtained from logging while drilling tools (LWD) can be used to get an accurate reading of annular pressure, which may be used to determine ECD. BOTTOMHOLE PRESSURE Pressure is imposed on the walls of the hole. The hydrostatic of the fluid column accounts for most of the pressure, but pressure to move fluid up the annulus also acts on the walls. In larger diameters, this annular pressure is small, rarely exceeding 200 psi (13.79 bar). In smaller diameters it can be 400 psi (27.58 bar) or higher. Backpressure or pressure held on the choke also increases bottomhole pressure, which can be estimated by adding up all the known pressures acting in, or on, the annular (casing) side. Bottomhole pressure can be estimated during the following activities. WELL STATIC If no fluid is moving, the well is static. The bottomhole pressure (BHP) is equal to the hydrostatic pressure (HP) on the annular side. If shut in on a kick, bottomhole pressure is equal to the hydrostatic pressure in the annulus plus the casing (wellhead) pressure. NORMAL CIRCULATION During circulation, the bottomhole pressure is equal to the hydrostatic pressure on the annular side plus the annular pressure loss (APL). ROTATING HEAD During circulating with a rotating head the bottomhole pressure is equal to the hydrostatic pressure on the annular side, plus the annular pressure loss, plus the rotating head backpressure. CIRCULATING A KICK OUT Bottomhole pressure is equal to hydrostatic pressure on the annular side, plus annular pressure loss, plus choke (casing) pressure. (For subsea, add choke line pressure loss.) Hydrostatic pressure is controlled by careful monitoring and control of fluid weight.

1-18 CHAPTER 1 MOVING PIPE SURGE/SWAB Swab Pressure The total pressure acting on the wellbore is affected by pipe movement upwards or downwards. When tripping out swab pressure is created, reducing the pressure on the wellbore. Swabbing occurs because the fluid in the well does not drop as fast as the string is being pulled. This creates a suction force and Sand reduces the pressure below the string. This force can be compared to a plunger in a syringe, with formation fluid being pulled into the wellbore. When lowering the string too fast, surge pressure is created because the fluid does not have a chance to get out of the way. Since liquids do not compress to any appreciable degree, pressure throughout the well can increase and cause leak-off or fracture. Both surge and swab pressures are affected by the rate of pipe movement, clearances between pipe and hole and fluid properties. While it is often impossible to avoid these pressures, they can be minimized by slowing the tripping speed. Calculations are available to estimate maximum trip speed and surge and swab pressures, but these calculations are outside the scope of this manual. Swab Pipe Movement Fluid Properties Using fluid density adjustments for a trip or safety margin requires good judgement. Too large a margin can cause lost circulation. Too small a margin may allow the well to kick. The margin depends on hole size, condition, pipe pulling speed, fluid and formation properties. DIFFERENTIAL PRESSURE The difference between the formation pressure and bottomhole hydrostatic pressure is differential pressure. These are classified as overbalanced, underbalanced and balanced. Swabbing occurs because the fluid in the well does not drop as fast as the string is being pulled. TRIP/SAFETY MARGINS Unless there is an excess of fluid weight to compensate for swabbing, formation fluid can enter the well and a kick can occur. The trip margin is an estimated increase in fluid density prior to a trip to compensate for loss of circulation pressure (ECD). A safety margin also compensates for swabbing pressures as pipe is pulled from the well. OVERBALANCED Overbalanced means the hydrostatic pressure exerted on the bottom of the hole is greater than the formation pressure: HP > FP UNDERBALANCED Underbalanced means the hydrostatic pressure exerted on the bottom of the hole is less than the formation pressure: HP < FP

PRESSURE BASICS BALANCED Balanced means the hydrostatic pressure exerted on the bottom of the hole is equal to the formation pressure: HP = FP Most wells are drilled, and worked, in balanced to overbalanced conditions. If circulating or drilling, friction and cuttings contribute to the effective pressure on bottom. Overbalanced HP > FP SUMMARY There are two main opposing pressures in a well. These are the fluid column hydrostatic pressure and the formation pressure. If one pressure overcomes the other, then a kick or lost circulation may occur. Since hydrostatic pressure is a function of the density of the working fluid in the well, its value may be controlled. By making careful calculations and by manipulating the formula for hydrostatic pressure, it is possible to test cement jobs, to estimate formation integrity, to project maximum mud weights and to control kicking wells. Kicks and blowouts are prevented by people who are able to work quickly and decisively under stress. An important part of the training required for blowout prevention is an understanding of pressure concepts and the ability to perform accurate calculations. t 1-19 Kicks are prevented by people able to work quickly and decisively under stress. Underbalanced HP < FP Differential pressure is the difference between formation pressure and bottomhole hydrostatic pressure. B alanced HP = F P