Drilling Problems n Drilling Optimization

June 10, 2018 | Author: ngecus | Category: Reynolds Number, Fluid Dynamics, Shear Stress, Pressure, Turbulence



Drilling Problems & Drilling OptimizationDrilling Objectives Drilling Objectives The goals of any drilling venture are safety, minimized cost and a usable completion. Companies may have different ideas ofhow best to attain these goals, and drilling practices may vary according to location, rig type, hole conditions or other factors.But the goals themselves are always the same.  Safety is the primary concern in drilling an oil or gas well. Protection of personnel supersedes all other well objectives, even when it means altering the well plan, incurring unexpected costs or delaying operations. Failure to make safety the top priorityon a rig can and does result in accidents, disabling injuries and deaths. The second priority in drilling is to protect the well and the surrounding environment. We must anticipate potential problems and include provisions in the well plan for minimizing blowout risks or other dangers, and we must continually monitor operations once the rig moves on location. Environmental issues are having an ever-growing impact on all oil and gas operations. In California and other parts of the United States, for example, public opposition--usually based on environmental concerns--has curtailed or even halted drillingand production activities. Those projects that are allowed to proceed are governed by a maze of complex regulations, many ofwhich carry severe penalties for non-compliance. Brown (1992) provides an excellent summary of environmental issues in the petroleum industry. Although his article focuses onU.S. regulations, his basic premise--that environmental management must be an integral part of daily oil and gas operations--applies anywhere. As energy and the environment become matters of public policy throughout the world, the bestthings that engineers or rig supervisors can do are to work closely with their companies' environmental specialists, becomefamiliar with applicable regulations and make sure that their operations are in compliance.  Minimized cost is the basis for most drilling optimization efforts. When looking at this objective, we must consider the total wellcost--it does no good to save money on one procedure if the result is a more expensive well. Effective cost control involves establishing a balance among well objectives. While we should never jeopardize safety in an effort to keep costs down, neither should we spend money unnecessarily. By the same token, while we must provide a usable well, we have to do so within realistic budget constraints. Whether or not an expense is justified is something the engineer must decide for each well. Intermediate casing strings thatmight be necessary in one field, for instance, may be wasted expenditures in another field. Concentrated well- planning andcareful monitoring of rig operations can reduce costs without compromising safety standards or completion requirements.  A usable completion should be the outcome of any drilling operation, whether the well is a producer, an injector or simply a source of information. Even a safely drilled, low-cost well is not entirely successful if it does not meet the needs that led to it being drilled in the first place. As a minimum standard, a well should have: o no irreparable damage to the hole or producing formation ; o a sufficiently large hole diameter for running completion equipment or carrying out other post-drilling activities. We may look at our basic drilling goals from two perspectives: dealing with drilling problems and optimizing normal operations. A drilling problem is any occurrence or condition that stands in the way of well objectives. It could involve anything from weather to transportation delays to blowouts. In this discussion, we concentrate on problems that occur as part of the drilling process itself. A comprehensive, thoroughly researched well plan is our best defense against hole problems. Every aspect of the plan, from pre-spud activities to completion procedures, should be designed to anticipate and control these problems. An effective wellplan requires thorough research and data collection, and a careful analysis of these data. Data sources may include:            bit records mud records mud logs IADC drilling reports ("tour sheets") scout tickets log headers production histories seismic studies well surveys geologic contours databases or service company files The engineer should use all available sources to identify and plan for potential hole problems. Each source may containimportant information not available elsewhere. The preparation and effort that goes into well planning should carry over into daily operations. Rig personnel must continuallymonitor drilling performance for early indications of problems and make sure that the appropriate problem-solving tools,materials and technical expertise are either at the rig or available on short notice. Drilling optimization involves using available resources to minimize overall cost, subject to safety and well completionrequirements. Part of this, of course, entails preventing or successfully solving hole problems. But optimization efforts alsoencompass "normal" operations. The key measure of performance in this area is the cost to drill a given interval. To optimize drilling operations, we must do three things:    Establish criteria for evaluating drilling performance. Identify the variables that affect this performance. Determine how to control these variables to our advantage. Drilling Rig Hydraulics Drilling hydraulics When hole problems occur when we are trying to optimize drilling performance we should look first to the drilling mud and the circulating system. This makes sense when we consider that the muds functions include: · controlling subsurface pressures · stabilizing the wellbore · supporting part of the drill string and casing weight · removing and transporting drilled cuttings · suspending cuttings in the annulus when the pumps are shut down · transmitting hydraulic energy to the bit · cooling and lubricating the bit and drill string · minimizing formation damage · collecting formation data Looking at this list, it is easy to see why most drilling textbooks devote a good deal of discussion to the mud system and rig hydraulics. Wellbore and Formation Pressure Drilling mud exerts a hydrostatic pressure (P) that we can describe using the wellknown relationship (1) where r = fluid density g = acceleration of gravity h = height of fluid column in wellbore 052 becomes 0. the mud is not static. drill string and drill bit. we have to account for these variables.S.) ECD = equivalent circulating density.052 becomes 0. 0. which we control by adjusting the mud weight.S. We can simplify our task by defining equivalent mud weight (EMW) or equivalent circulating density(ECD): (3) (4) where EMW = equivalent mud weight. the well contains fluids of different densities. TVD = True vertical depth. (2) where P = pressure.) TVD = true vertical depth.) Pwell = sum of hydrostatic and applied pressures. This corresponds to the pump pressure minus the pressure losses through the surface equipment. or pore pressure. To determine the actual wellbore pressure. for example. Equation 2 is adequate for determining the pressure of a static. lbm/gal (U. In many instances.00980. Nor is the mud column necessarily homogeneous.S.00980. psi Pann = the frictional pressure loss in the annulus. 0. ft When expressed in SI metric units (with pressure given in kPa). Example 1: Equivalent Circulating Density Determine the Equivalent Circulating Density and the effective wellbore pressure for the following conditions: Mud weight = 12. lbm/gal (U. during cementing operations or mud changeovers. is our primary means of counteracting formation. psi MW = mud weight or density. however. This hydrostatic pressure. circulation and pipe movement can significantly affect wellbore pressure. ft When expressed in SI units (with pressure given in kPa).5 lbm/gal . lbm/gal (U.In field units. homogeneous mud column. Differential pressure (P) is simply the difference between wellbore an dformation pressure: (6) .000 ft Solution: In many instances.500 ft TVD=10. Normal pressure gradient = 0.S. The fluid pressure gradient g typically ranges from 0.300 ft Calculated annular pressure loss = 150 psi Solution: Pore pressure is normally equal to the formation depth times the pressure gradient of the native formation water: (5) where  = formation fluid density. lbm/gal (U.433 psi/ft [9.46 psi/ft Measured depth = 12. which deviate from established area trends.80 kPa/m] for fresh water to 0.TVD =11. Example 2: Determining pore pressure Determine the pore pressure of the following formation.052 becomes 0.54 kPa/m] or higher for brine.)  = fluid pressure gradient. ft When expressed in SI units (with pressure given in kPa). we may encounter abnormal pore pressures. psi/ft TVD = True Vertical Depth. 0.465 psi/ft [10.00980. 600 psi in the other direction -. the mud column must exert at least 4. P is zero.from the well to the formation.When the wellbore and formation pressures are approximately equal. To keep formation fluid from entering the wellbore.1 lbm/gal. Figure 1 Example 3: Balancing formation pressure What mud weight at static conditions is required to balance the formation pressure calculated in Example 2? (Round up to the nearest 0.600 psi is pushing from the formation to the well. We refer to this condition as balanced ( Figure 1 ). Our required mud weight is therefore: .) Solution: This pressure of 4. balanced or underbalanced drilling can result in penetration rates much higher than those we could attain with a weighted mud system. underbalanced conditions. Figure 3 illustrates how penetration rates tend to decrease with increasing differential pressure. wellbore pressure is less than pore pressure). high-strength formations. . Figure 2 Because of favorable rock mechanics effects and the fact that there is no chip holddown caused by high differential pressure. operators may even drill underbalanced (P<0) using air. foam or low-density liquids ( Figure 2 . gas.Balanced drilling with an unweighted mud system is a viable option in areas or depth intervals having well-established pressure trends and little blowout risk. In lowpermeability. . however. wellbore pressure is greater than pore pressure). In the majority of cases. this requires them to handle large volumes of oil and gas at the surface). we have to keep the mud density higher than that of the formation fluid ( Figure 4 .Figure 3 Underbalanced drilling is also used to minimize formation damage in producing zones. It is common practice for operators drilling horizontal wells through this formation to let the wells flow (of course. Overbalanced conditions. The Austin Chalk formation of central Texas is a good example. In addition. . south Texas. since weighted muds allow us to better control formation pressure and maintain hole integrity. it is not uncommon to see P values on the order of several thousand psi. We should not compromise these objectives unnecessarily by using heavier muds than safety requires.0 lbm/gal and the formation pressure is 7. that our drilling objectives also include optimizing penetration rates and reducing expenses.g.Figure 4 That is. Up to a point. parts of the Middle East). we have to drill overbalanced (P>0) to protect against sudden fluid level drops. unintentional mud weight reductions or other circumstances that might decrease wellbore pressure. We need to keep in mind. Example 4: Differential pressure Determine the differential wellbore pressure at 15..600 psi. we need a safeguard against abnormal formation pressures. along with increased potential for formation damage and stuck pipe. however. south Louisiana. this is a fair trade-off.400 ft TVD on a U. In many areas (e. where the mud weight is 14. the North Sea.S. Gulf Coast well. Solution: Disadvantages of overbalanced drilling are lower penetration rates and higher mud costs. 0. and calculate the matrix stress s. thereby creating a blowout risk. resulting in formation damage and induced fractures.535 becomes 12. Determine the pore pressure (based on seismic measurements.10 kPa/m]. Severe fracturing can cause the wellbore fluid level to drop.0 psi/ft [22. 3. Fracture pressure is related to overburden stress. 2. assuming that S=1. developed the following relationship for sedimentary rocks: (8) where frac = fracture gradient. a number of researchers have developed theoretical methods for determining fracture gradients. Exceeding this limit can cause lost circulation.62 kPa/m] for the overburden.62 kPa/m] and the normal stress gradient equals 0. Their procedure for determining the fracture gradient is as follows: 1. Determine the depth (Di) at which s would have a normal value. ft [m] Ki = matrix stress coefficient for the depth at which the value of s would be the normal matrix stress (dimensionless) Matthews and Kelly based this equation on the assumption that the cohesiveness of a rock matrix is generally related to matrix stress. Assume a gradient of 1.535 psi/ft [12. Therefore. psi/ft [kPa/m] D = depth. . which is equal to the rock matrix pressure plus the pore pressure: (7) where S = overburden stress  = matrix stress at depth of interest Ppore = pore pressure at depth of interest Over the years. well log anlaysis or drilling data correlations).Formation fracture gradients define our upper wellbore pressure limit. Matthews and Kelly (1967).0 psi/ft [22. building on the landmark work of Hubbert and Willis (1957). and thus varies only with the degree of compaction.10. (9) When depth is expressed in meters and s is expressed in kPa. determine the fracture gradient just below the casing seat for the following Louisiana Gulf Coast well. basd on data from U. Casing set at 6. Example 5: Determining fracture gradient Using the Matthews and Kelly procedure. Using Figure 5 (Matrix stress coefficients of Matthews and Kelly.4. Figure 5 Gulf Coast Sands).650 ft TVD Formation pressure = 3. determine Ki and calculate frac from Equation 8 (note that Figure 5 is empirically generated from field data).300 psi Solution: .S. Continue pumping small volumes of mud until the formation begins to take fluid. 2. ..5 lbm/gal). below (mud weight = 12. 3.In the field. Plot pressure versus the pumped volume to determine the initial fracture pressure. using a low-volume pump.750 ft. TVD). given the information in Table 1. Solution: Figure 6 (Results of leak-off test from Example 6) indicates that the formation begins to fracture at an applied pressure of about 820 psi. 1985): Determine the fracture gradient at a well's casing point (6. Close the blowout preventer and apply pressure down the drill pipe in small increments. Example 6: Leak-off test-. we can estimate the minimum fracture gradient at each new casing point by performing a leak-off test as follows: 1.calculating fracture gradient (after Adams. or until the pressure reaches a pre-set test limit. 0 2.5 Pressure. psi 0 4 100 190 280 370 460 550 640 730 820 850 880 .5 3.5 4.0 5.5 5.Figure 6 The fracture pressure is equal to this applied pressure plus the hydrostatic wellbore pressure: Volume pumped.0 6.5 2.0 3.0 4.5 6. bbl 0 1 1. . by an amount that corresponds to the weight of the displaced mud. drill collar requirements. and for steel pipe is equal to (2) where MW is the mud weight and steel is the density of steel. The net effect of buoyant force is that an object submerged in liquid weighs less than it would in air. casing specifications.33 to 16.d.. which equals 65. which equals the weight of the fluid the object displaces (i. Buoyancy An object immersed in fluid is subject to an upward-acting buoyant force. rig capacity and other weight-related parameters. this means that a string of pipe in a mud-filled wellbore weighs less than it would if suspended in air. The weight of pipe in a well is equal to its weight in air multiplied by a buoyancy factor (BF): (1) The buoyancy factor is a function of the fluid density.e.Table 1: Results of leak-off test. Table 2. the product of the fluid volume and density). bit weights.2 lists buoyancy factors for mud weights ranging from 8. 144 lb/ft drill collars for both of the following mud weights. (Assume that all of the drill string compression is in the drill collars.5 lbm/gal · 16.) · 9. In practical terms.0 pps.5 lbm/gal [7860 kg/m3]. It is important to account for buoyancy effects when calculating hook loads.0 lbm/gal Solution: MW. [kg/m3] Buoyancy . Example 1: Buoyancy effects Determine the maximum weight-on-bit provided by 450 ft of 7 3/4-inch o. 794 14.0 [1678] 0.840 11.0 [1198] 0.824 12.801 13.862 9.5 [1498] 0.778 .5 [1138] 0.0 [1438] 0.809 13.0 [1558] 0.832 11.5 [1378] 0.873 9.0 [1318] 0.817 12.lbm/gal Factor 8.0 [1078] 0.786 14.5 [1618] 0.33 [1000] 0.855 10.5 [1738] 0.847 10.5 [1258] 0. a term often used in drilling engineering. Buoyancy Factors for Various Mud Weights (Based on Steel Density of 65. .5 [1858] 0. is the study of the flow or deformation of matter. the distance between the plates.0 [1798] 0.763 16. We generally describe flow or deformation in terms of shear stress and shear rate. We may best understand these concepts by considering a fluid located between two parallel plates separated by a distance d ( Figure 1 ). the upper plate attains a constant velocity (v). which depends on the applied force (F).771 15. Figure 1 If we apply a force to the upper plate while holding the lower plate stationary. the surface area (A) and the fluid viscosity (): (1) Shear stress () is equal to the force per unit area.0 [1918] 0.5 lbm/gal [7860 kg/m3]) Rheological Models Rheology.15.756 Table 1. Bingham Plastic and Power Law models. most drilling fluids do not exhibit this straight-line relationship. combines elements of the Bingham Plastic and Power Law models. it is an oversimplification. (A fourth relationship. A common method for describing the shear stress/shear rate relationship is the Bingham-Plastic model ( Figure 3 ): . the Three Parameter model.(2) while shear rate () is equal to the fluid velocity divided by the distance between the plates: (3) These parameters define the flow relationships used to characterize drilling mud behavior: the Newtonian. Although this model has often been used to describe drilling fluid behavior.) Newtonian fluids exhibit direct proportionality between shear stress and shear rate ( Figure 2 ): Figure 2 (4) The proportionality constant () is the fluid viscosity. 0 cp = 0. and that the mud's circulating pressure depends on the initial yield point and the plastic viscosity. respectively. and PV is given in cp.Figure 3 (5) where p = plastic viscosity (PV). where fluid behavior deviates from linear relationships. 1 lbf /100 ft 2 = 0. Although widely used. The Bingham Plastic model thus depends on a linear relationship (the difference between the two viscometer readings) to describe drilling fluid flow. is limited. . its application to turbulent flow conditions. We can determine these quantities using a rotational viscometer and the following equations: (6) (7) where 600and  300 = the viscometer readings at rates of 600 and 300 RPM. lbf /100 ft 2 (1.4788 Pa) Note that a certain initial pressure is required to initiate fluid movement. cp y = yield stress or yield point (YP).001 Pa-s. The Power Law model is a non-linear relationship. the Power Law model is also more complex. which is more descriptive of real drilling fluid behavior ( Figure 4 ): Figure 4 (8) where (9) n = flow behavior index = 3. Laminar flow profile in drill pipe/hole annulus). fluid behaves as a series of parallel layers moving at uniform or nearuniform velocities.32 x log( 600/300) (10) While it is more realistic than the Bingham Plastic model. The fluid layers nearest the center of the pipe or annulus generally move faster than the layers adjacent to the hole wall ( Figure 1 . turbulent and transitional. Flow Regimes The flow regimes most commonly encountered in drilling are laminar. There is no large-scale movement of fluid particles between layers. In laminar flow. . both parallel and axial to the mean flow stream. These fluctuations break down the boundaries between the fluid layers.Figure 1 Turbulent flow is characterized by velocity fluctuations among the fluid stream particles. resulting in a chaotic flow pattern ( Figure 2 . . Turbulent flow in drill pipe and at bit face). . is more desirable at the bottom of the hole. but we often find that more than one flow regime exists at the same point in the system. It describes the often hard-to-define region where flow is neither wholly laminar or wholly turbulent. plug flow. Not only does fluid behavior vary within the circulating system. For example. on the other hand. flow regimes may be difficult to identify. Turbulent flow.) We usually prefer to see laminar flow in the annulus to move cuttings up the hole and prevent erosion. because it promotes cleaning and cuttings removal. relatively undisturbed body. the adjacent fluid at the pipe boundary may be in turbulent flow. sub-laminar condition of a fluid moving as a homogeneous. describes the low-velocity. while the main flow stream in an annulus may exhibit laminar behavior. (An additional term. While they are conceptually easy to visualize.Figure 2 Transitional flow exhibits characteristics of both laminar and turbulent regimes. for the case of a Newtonian.e. m (4) The term 0. the Reynolds number becomes (3) where d2 = hole or casing diameter.816(d2-d1) in Equation 3 is the equivalent circular diameter of a slot representation of the annulus (Bourgoyne et al. Pa-s For annular flow. In field units (where  is given in lbm/gal. 1986).. kg/m3 (2)  = inside diameter of pipe. Equations 1 through 4 become: (Pipe flow): (5) (6) (Annular flow): . non-elastic liquid inside pipe) we may define the Reynolds number as (1) where  = fluid density. d. d1 and d2 in inches and  in cp).The most common method for determining a fluid's flow regime is to calculate its Reynolds number. m d1 = outside diameter of pipe. In its simplest form (i. m  = fluid viscosity. v in ft/s. q in gal/min. flow is considered transitional.100 indicate laminar flow. Determine the flow regime inside 4. Reynolds numbers of less than 2. Example 1: Flow regimes A 10.5 inch o. Because the transition between laminar and turbulent flow may not be clear-cut. it is turbulent.000 Turbulent flow Annulus: . In field applications.d. For example.826 inch inside diameter). If vc < v. while Reynolds numbers greater than 4.. Between these values.(7) (8) As a general guideline. we identify flow regimes by determining the critical Reynolds number or critical velocity (vc) at which flow changes from laminar to turbulent. flow is laminar. Solution: Drill pipe: NRe > 4. assume that the mud behaves as a Newtonian fluid.60 lb/ft drill pipe (3. we may calculate vc and compare it to the actual fluid velocity (v). and in the drill pipe/hole annulus. 16. it often becomes necessary to calculate a range of critical velocities to determine the flow regime. For this example. while if vc >v.000 indicate turbulent flow.5 lbm/gal drilling mud with a viscosity of 30 cp is circulating at 250 gallons per minute in an 8 3/4 inch diameter wellbore. 1669.1253. lbf /100 ft2 d = inside pipe diameter. 1986). ft/s When Equations 9 and 10 are expressed in SI units. cp YP = yield point.000 (Bourgoyne et al.66 becomes 0. inches d2 = hole or casing diameter. determining flow regimes is seldom this straightforward.. Bingham Plastic Fluids We can modify Equations 5 and 7 for Bingham-Plastic fluids by defining an apparent viscosity to account for the yield point and plastic viscosity: (Pipe flow) (9) (Annular flow) (10) where a = apparent viscosity. inches v = average velocity. inches d1 = outside pipe diameter.100 Laminar flow Unfortunately.NRe < 2. although we do not usually encounter such extremes in drilling operations. cp PV = plastic viscosity. and 5 becomes 0. 6. Thus. . Laminar flow has been observed under controlled conditions for Reynolds numbers as low as 1200 and as high as 40. for Bingham-Plastic behavior. Equations 13 and 14 become (13 SI) .(Pipe flow) (11) (Annular flow) (12) Using these equations. lbm/gal (U.) v = fluid velocity. ft/s d2 and d1 are expressed in inches When expressed in SI units. with laminar flow occurring below a Reynolds number of 2100. we may use correlations developed by Dodge and Metzner (1959): (Pipe flow): (13) (Annular flow): (14) where n = flow behavior index (dimensionless) K = consistency index (dimensionless) r = fluid density. the criterion for turbulence is the same as for Newtonian fluids.S. Power Law Fluids For power law fluids. The mud pump or standpipe pressure corresponds to the sum of these forces: (1) where Ppump = mud pump pressure Psurf = pressure loss through surface equipment Pds = pressure loss through drill string Pbit = pressure loss through bit Pann = pressure loss in annulus Pressure losses are functions of circulation rate. pipe diameter and hole diameter. Determine the fluid velocity (or Reynolds number) at the point of interest. (The choice depends on which rheological model and flow regime apply to the point of interest). 3. mud weight. the critical Reynolds number is (16) System Pressure Loss For mud to flow through the circulating system. A simple equation for estimating the critical Reynolds number at the upper limit of laminar flow is (15) For the region between transitional and turbulent flow. viscosity.(14 SI) The turbulence criterion for power law fluids is based on a critical Reynolds number ( Nrec ). which depends on the value of the flow behavior index. The general procedure for calculating system pressure losses is as follows: 1. solid particles and pipe or borehole walls. Choose the appropriate pressure loss equation. Calculate the critical velocity (or Reynolds number) to determine whether the fluid is in laminar or turbulent flow. it must overcome frictional forces between the fluid layers. . 2. The maximum amount of friction loss that we can overcome is limited to the working pressure of our mud pumps. Table 1 and Table 2 summarize the equations for determining pipe and annular pressure losses based on Bingham Plastic flow. it is laminar. while if vc v. If v c<v the flow is turbulent. then perform pressure loss calculations for both flow regimes. while Table 3 and Table 4 list the pressure loss equations for Power Law fluids. Fluid velocity (v) Oilfield Units: SI Units: Critical velocity (vc) Oilfield Units: SI Units Pressure Loss for laminar flow (Pds ) (Use for v < vc) Oilfield Units: SI Units: Pressure Loss for turbulent flow (Use for v vc) Oilfield Units: . and use the results that give the larger pressure loss.Note: In the field. If the critical and actual velocities are approximately equal. We use this information to maximize the pressure loss through the bit and thereby maximize the hydraulic energy at the bottom of the hole. determine both the critical and the actual flow velocities. 1985) Fluid velocity (v) Oilfield Units: SI Units: Critical velocity (vc) Oilfield Units: SI Units Pressure Loss for laminar flow (Pds ) (Use for v < vc) Oilfield Units: . inches [m] v = velocity. psi [kPa] q = flow rate. lbm/gal [kg/m3] YP = yield point. ft/s [m/s] MW = mud weight. lbf/100ft2 [N/m2] PV = plastic viscosity. cp [Pa-s] Pds = pressure loss. ft/s [m/s] L = pipe length. and Adams .SI Units: Nomenclature (SI units in brackets): d = inside diameter of pipe. gal/min [m3/s] Table 1 Drill string pressure loss equations for a Bingham Plastic fluid (After Bourgoyne et.1986. al . ft [m] vc = critical velocity. ft [m] vc = critical velocity. inches [m] v = velocity. cp [Pa-s] Pann = pressure loss. gal/min [m3/s] d2 = casing or hole diameter.SI Units: Pressure Loss for turbulent flow (Use for v vc) Oilfield Units: SI Units: Nomenclature: d1 = inside diameter of pipe. ft/s [m/s] L = pipe length. psi [kPa] Table 2 Annular pressure loss equations for a Bingham Plastic fluid [After Bourgoyne et al (1986) and Adams (1985)] Fluid velocity (v) Oilfield Units: SI Units: Critical velocity (vc) . inches [m] q = flow rate. ft/s [m/s] MW = mud weight. lbm/gal [kg/m3] YP = yield point. lbf/100ft2 [N/m2] PV = plastic viscosity. lbm/gal [kg/m3] YP = yield point. gal/min [m3/s] MW = mud weight. ft [m] q = flow rate. inches [m] v = velocity. psi [kPa] . ft/s [m/s] K = consistency index (dimensionless) vc = critical velocity. ft/s [m/s] L = pipe length.Oilfield Units: SI Units: Pressure Loss for laminar flow (Pds ) (Use for v < vc) Oilfield Units: SI Units: Pressure Loss for turbulent flow (Use for v vc) Oilfield Units: SI Units: Nomenclature: d = inside diameter of pipe. lbf/100ft2 [N/m2] n = flow behavior index (dimensionless) DPds = pressure loss. PV = plastic viscosity. cp [Pa-s] Table 3 Drill string pressure loss equations for a Power Law fluid [After Bourgoyne et al (1986) and Adams (1985)] Fluid velocity (v) Oilfield Units: SI Units: Critical velocity (vc) Oilfield Units: SI Units: Pressure Loss for laminar flow (Pds ) (Use for v < vc) Oilfield Units: SI Units: Pressure Loss for turbulent flow (Use for v vc) Oilfield Units: . .35] [16. we simply match our surface equipment specifications to one of the four groups shown in Table 2.in L.7. ft [m] YP = yield point. cp [Pa-s] Table 4 Annular pressure loss equations for a Power Law fluid [After Bourgoyne et al (1986) and Adams (1985)] We can find the pressure loss through surface equipment (DPsurf) by treating it as an equivalent length of drill pipe.192] [8.d.5 55 3 55 3 55 hose [5. lbm/gal [kg/m3] Pann = pressure loss.16] [13..5 40 4 45 4 45 [7.ft i.716] [6.d.192] [10.in L.62] [12.ft [cm] [m] [cm] [m] [cm] [m] [cm] [m] 3 40 3.ft i.d. if a rig has Group 4 surface equipment and uses 5-inch. inches [m] v = velocity.SI Units: Nomenclature: d1 = outside diameter of pipe.ft i. inches [m] vc = critical velocity. we would.d.5 lb/ft drill pipe.716] [10.. gal/min [m3/s] L = pipe length.716] Drilling 2 45 2.in L. psi [kPa] n = flow behavior index (dimensionless) PV = plastic viscosity. To determine this equivalent length.. Component Typical Combinations Case 1 Case 2 Case 3 Case 4 i. ft/s [m/s] d2 = casing or hole diameter.62] [16.in L.764] [7. ft/s [m/s] K = consistency index (dimensionless) q = flow rate.08] [13.89] [12.764] Standpipe . for calculation purposes.62] [16.16] [13. 19.764] [7. lbf100ft2 [N/m2] MW = mud weight. add 579 feet to the actual drill pipe length. For example. 25 40 4 40 [5.031 becomes 2000.6 lb/ft 437 [133.500 ft to 9.5 lb/ft 479 [146. Example 1 Determine the bit pressure drop for the following well: Total Depth=9.715] [12.5 inch.829] 2.3 lb/ft 4.) MW = mud weight.D. [5.192] [8.5inch.d) cemented at 6. We may express the bit pressure drop in psi as (2) where q = circulation rate. where Pbit is expressed in kPa. 19.25 40 3.0] 340 [103.755 in. 12. i.16] 40 gooseneck Kelly Drill pipe: Length of surface connections. expressed as equivalent ft [m] of drill pipe 3.95 is acceptable for most field calculations. they are due to the acceleration of the drilling fluid through the bit nozzles.25 40 3.Swivel. lbm/gal Cd = nozzle discharge coefficient (dimensionless) AT = total nozzle area.08] [1.255] [12. rather.50 lb/ft (8.192] [10.1] 761 [232.35] [1. The following example illustrates how we can apply pressure loss relationships.524] [7.524] [6.2] 161 [49.0] 5 inch. in2 In the SI system.500 ft.5 5 2. Although the nozzle discharge coefficient Cd varies according to nozzle type and size.219] [6. a value of 0. Open hole: 8 1/2 inch from 6. 16.950 ft (T.192] [8. 2 4 2. 13.255] [12.5 5 3 6 washpipe.35] [1.5] Drill bit pressure losses do not result primarily from friction forces.7] 579 [176.S.6] 816 [248. gal/min (U.62] [1. 43.950 ft Casing: 9 5/8 inch.) Drill string: . of 4 1/2 inch. gooseneck: 5 ft x 2. i. Kelly: 40 ft x 3. 16.826 in.7).d. We can therefore combine Psurf with our calculation of Pds. Determine the pressure losses inside the drill string (Pds). Note first that the surface components correspond to a Case 3 equipment combination (Table 2. i. 16.) Bottomhole assembly: 450 ft. Hose: 55 ft x 3 in. of 4 1/2 in. This involves separate calculations for the drill pipe and the bottomhole assembly: a) Drill pipe: =3.d.60 lb/ft (3. washpipe.. Swivel. The pressure loss through these components is equivalent to 479 ft. drill collars and tools Mud properties: Mud weight (MW)=10.Drill pipe: 9500 ft. 2.d. Solution: 1.d.d.96 ft/s v vc turbulent flow .200 psi Surface Equipment: Standpipe: 45 ft x 4 in. i.5 lbm/gal Plastic viscosity (PV)=35 cp Yield point (YP)=6 lbf/100 ft2 Assume Bingham Plastic fluid Circulating rate=300 gal/min Pump pressure=2.d. of 6 3/4 inch o.6 lb/ft drill pipe.d. x 2 1/4 inch i.25 in. i. i.5 in. 500 ft) and the BHA/hole annulus (9.= 605 psi (Note that we accounted for surface pressure losses by adding in the drill pipe equivalent length of 479 ft) b) Bottomhole assembly: = 4. This involves separate calculations for the cased-hole section (surface to 6.500 ft).9.9. Determine the pressure losses in the annulus. the drill pipe/hole annulus (6.500 .050 ft). (DPann).500 . a) Cased-hole annulus (surface-6500 ft): .88 ft/s v vc turbulent flow = 340 psi Pds=Pdp+Pbha=605+340=945 psi 3. 28 ft/s v < vc  laminar flow .9.500 ft .500 ft): Assume laminar flow (critical velocity will be close to that calculated for the casedhole annulus) = 38 psi c) BHA/open hole annulus: = 5.= 3.47 ft/s v < vc laminar flow = 73 psi b) Drill pipe/hole annulus (6. . Excessive surge pressures can increase the wellbore pressure to such a degree as to induce lost circulation. mud flows down the annulus to fill the resulting void.945 .= 31 psi Total annular pressure losses: DPann=73 + 38 + 31 = 142 psi 4. This causes a suction effect. Conversely. when we pull pipe out of a well. Figure 1 The resulting piston effect generates a surge pressure that is added to the hydrostatic pressure. the mud immediately adjacent to the pipe is dragged downhole. as shown in Figure 1 (Annular flow profile resulting from downward ie movement).142= 1. Determine pressure loss at bit: Pbit=Ppump-(Psurf+Pds)-Pann=2200 .113 psi Surge and Swab Pressure When we run pipe in a well. it forces drilling mud up the annulus and out of the flowline. generating a swab pressure that can lower the differential pressure and possibly bring formation fluid into the wellbore. At the same time. . Figure 2 Calculating surge and swab pressures can be a complex undertaking. depending on the pipe configuration and hole geometry. Burkhardt (1961) developed a relationship between well geometry and the effect of the mud being dragged by the pipe. Mud clinging constant K as a function of annular geometry). which is referred to as the clinging constant.Figure 2 (Typical pressure surge pattern for a joint of casing lowered into a wellbore) shows the pressure fluctuations that resulted from lowering casing into a mud-filled wellbore. K ( Figure 3 . and illustrates the significance of surge and swab effects. we need to know the mud velocity in the annulus. this is equal to (1) where vpipe = pipe velocity d2 = hole diameter d1 = pipe outside diameter For open-ended pipe. For a closed drill string.Figure 3 To apply the clinging constant. (2) . 82 psi/ft Pipe velocity = -110 ft/min = -1. Casing point = 6400 ft Fracture gradient = 0. PV=37 cp. Assume that the casing is effectively "closed" with a float shoe.0 lbm/gal.The effective annular velocity is equal to (3) Example 1: Surge effects Calculate the surge pressure generated by running a string of 10 3/4 -inch casing under the following conditions. K=0.75=0.75/14.83 ft/s ("-" denotes downward velocity) Hole diameter = 14 3/4 inches Mud: 15.07 ft/s d1/d2= 10. from Figure 3 .73. YP=6 lbf/100 ft2 Solution: = 2.44 (assume laminar flow) Use annular pressure loss equation for laminar flow: Psurge = 67 psi . and determine whether the total wellbore pressure exceeds the fracture gradient. Rotary drillers found out early on that they had a major advantage over cable tool operators: the ability to continuously remove drilled cuttings (as opposed to periodically removing them with a bailer). it was destined to ultimately replace the once dominant cable tool methods.EMW < EMWfra We can substitute the effective velocity into the friction pressure loss equations for a given flow regime to compute the surge (or swab) pressure. Additionally. .that is. Hole Cleaning From the time that rotary drilling came into its own with the dramatic Spindletop discovery of 1901. The mud's effectiveness at removing cuttings significantly affects drilling efficiency. Carrying capacity depends on a mud's annular velocity and the slip velocity with which cuttings fall back to bottom). A mud's ability to transport cuttings -. the mud must be able to hold these cuttings in suspension when circulation stops.is related to the difference between the annular velocity and the slip velocity with which the cuttings fall ( Figure 1 . Drilled cuttings vary in size and density according to formation lithology. Cuttings removal and transport are still primary functions of the drilling mud. its carrying capacity -. back toward the bottom of the hole. the cutting action of the bit and other factors. They are usually heavier than the drilling mud. and therefore tend to slip down through the annulus. differential pressure. Figure 1 Moore (1974). ft/s dp = particle diameter. lbm/gal . we need to determine the average densities and diameters of drilled solids by visually inspecting representative cuttings. To use these methods. Chien (1971) and Walker and Mayes (1975) are among those who have developed correlations for estimating slip velocity (this section presents Moore's correlation by way of example). or by doing a sieve or screen analysis. We can use the following relationship to estimate the slip velocity of a particle suspended in a Newtonian fluid: (1) where vs = slip velocity. inches p = particle density. m/s. 928 reduces to 1. Figure 2 . To determine the drag coefficient.m. Pa-s). In SI units. which has units of [ft2/(sec2-in)]1/2.89 = numerical value of conversion constant. We can then obtain Cdrag from Figure 2 (Particle drag coefficient as a function of the particle Reynolds number).615.f = fluid density. 1. lbm/gal Cdrag = drag coefficient 1. we must first compute the particle Reynolds number (NRep): (2) where µ = fluid viscosity.89 becomes 3. cp In SI units (kg/m3. 90 becomes 0. 82.As we can see. 2.. Cdrag has a constant value of about 1. For (NRep) >2. and (5) In SI units. and Equation 1 reduces to (4) In SI units.7053. 10< (NRep) <100) Cdrag = . given in units of centipoise and defined as follows: (6) where K= consistency index = (7) .5. Moore presents the following modifications of Equation 1 for various ranges of the particle Reynolds number: For (NRep) <1.000 (turbulent flow). The term µa in these equations is the apparent viscosity.87 becomes 0.54 becomes 2. For intermediate particle Reynolds number values (i.3267. this is an iterative process. 1. Cdrag=40/(NRep) and (3) In SI units.e.945. Determine apparent viscosity (µa): .8 K=150 cp equivalent Solution: 1.25 inches and a density of 2.1 lbm/gal). ft/s In SI units.n = flow behavior index = (8) v = fluid velocity. given the following mud and hole data: Hole size=12. Note that this method for determining slip velocity may require several iterations. Equation 6 becomes (6 SI) Moore recommends Equation 5 for use in routine field applications. Example 1: Slip velocity (Moore's method) Calculate the slip velocity of drilled cuttings having an average diameter of 0.25 inches Drill pipe=5 inches MW=10.2 lbm/gal Annular Velocity=60 ft/min (1. (with a expressed in Pa-s).65 g/cc (22.0 ft/s) n (from viscometer readings)=0. 7 ft/s Verification: (NRep) >1. Annular velocity = 1.0-0.1 cp 2.39 ft/s Verification: (NRep) = 10.= 88. Net upward velocity of cuttings = (1.0< (NRe p) <100: vs = 0.39) . Assume an (NRep) range and then check the results obtained: Let (NRe p) <1: =0.0 ft/s.0  first assumption was invalid Assume 1.5 assumption was valid 3. flow behavior index (n) and the consistency index (K). Calculate the maximum bit weight that would be available using 750 ft of 6 1/4 x 2 1/4 inch drill collars (91 lb/ft) in 13. Calculate the plastic viscosity (PV). or 37 ft/min. remembering that the drill pipe must remain in tension. What mud weight is required to balance formation pressure and provide a 400 psi trip margin at a depth of 12. Soln . Exercises 1. yield point (YP).= 0. Soln 3.0 bm /gal mud. given viscometer readings of 34 at 300 RPM and 62 at 600 RPM.45 psi/ft? solution 2.300 ft (TVD) and a fluid pressure gradient of 0.61 ft/s. determine the critical velocity and flow regime for a 12.50 lb/ft (i.=4.5 lbm/gal PV=26 cp YP=9 lbf /100ft2 Soln.276 inches) for the following circulating conditions: Q=350 gallons/minute MW=13. determine the critical velocity and flow regime for a 12. Using the Bingham Plastic model.25 x 7. Using the Bingham-Plastic flow model. calculate the frictional pressure loss in a 10.4.000 foot section of 5-inch.75 inch open hole/drill collar annulus: 5.2 lbm/gal PV=32 cp YP=8 lbf/100ft2 Circulation rate: Q=300 gallons/minute Soln Using the Bingham-Plastic flow model.d.75 inch open hole/drill collar annulus: Mud properties: MW=10.25 x 7. . 19. 75 inch... Trip speed = -90 ft/min = -1.60 lb/ft Drill collars=700 ft.sign denotes downward veleocity) Bingham Plastic flow model Assume laminar flow in the annulus Float valve in drill pipe (i.6. YP=8 lbf /100 ft2 Soln.. Annular velocity opposite drill pipe: .0 lbm /gal mud.e.5 in.5 ft/s (. of 4.500 ft Drill pipe=7800 ft. 16. of 6.75 inch bit at 8. Determine the surge pressure for the following conditions: 8. PV=29 cp. closed-ended drill string) 12. Abnormal Pressure Environments Origins Normally pressured formations are open hydraulic systems. They do not exceed native hydrostatic pressure. They are sealed by structural traps or low-permeability barriers that keep high pressures from dispersing. they have unblocked communication with adjacent formations and.7 = 116 psi. with the surface.5 + 28. Abnormally pressured formations. on the other hand. ultimately. because any pressure buildup rapidly dissipates to neighboring intervals.Equivalent velocity ve = vmud -Kvpipe Annular velocity opposite drill collars: Surge pressures: Drill pipe: Drill Collars: Total surge pressure = 87. are closed systems. The most common circumstances under which abnormal pressures occur are: · undercompaction of sediments . The underlying layers are thus subjected to successively increasing overburden pressure. which prevent fluid from escaping). This compaction increases the rock density and matrix strength. it puts more weight on underlying layers and forces them downward. . enabling the rock to support the added pressure.. a shale layer) slows or prevents fluid from escaping. If an underlying layer is normally pressured.g. if the pore fluid has somewhere to go. however. Compaction can only take place. the pressure increase is transmitted to the trapped formation fluid. As long as the fluid escapes fast enough to compensate for the increased pressure. Abnormal pore pressure due to undercompaction of sediments. The lower sandstone formation is surrounded by lowpermeability shales. Since the rock matrix cannot support the higher overburden pressure. increased overburden causes the rock matrix to compact. thereby decreasing the pore space and forcing fluid into adjacent permeable formations. If a low permeability barrier (e. As each new layer forms.· chemical diagenesis · tectonic activity · presence of salt structures · differences in fluid density · fluid migration between zones · presence of artesian systems Sedimentary layers form through rock particle deposition. then the pore volume cannot decrease quickly enough to provide adequate rock matrix strength. resulting in an overpressured condition ( Figure 1 . there is a balance between the overburden and the rock matrix stress. and is thought to be a principle cause of abnormal pressure (McClure. Water that was formerly bound to the montmorillonite is thus released to occupy pore space ( Figure 2 .Figure 1 Chemical diagenesis refers to the chemical alteration of minerals by geologic processes. 1983). . 1967). temperature increases during rock deposition may cause montmorillonite clay to dehydrate and turn into illite (Powers. For example. .Figure 2 Figure 3 and Figure 4 . Chemical diagenesis of montmorillonite to illite). Figure 3 Because free water occupies more space than bound water, it becomes overpressured if held in by a low-permeability seal. Figure 4 Precipitation of minerals from solution is another form of diagenesis. Gypsum (CaSO4.2H2O), for instance, may give up water to become anhydrite (CaSO4). Such evaporites can form extreme low-permeability barriers, allowing pore pressure to build up. Tectonic action can provide both a pressure source and a seal, as illustrated in Figure 5 (Abnormal pressure resulting from tectonic activity). Figure 5 In this case, faulting and uplift have moved a formerly buried formation from an area of high overburden stress to one of lower overburden stress. The fault has placed this permeable formation against an impermeable shale, effectively forming a trap. The result is an abnormal pressure condition. Salt beds are a leading cause of abnormal pressures. They not only form impermeable barriers, but they are plastic and have very little matrix strength. Instead of supporting overburden stress, a salt bed transmits the entire load to underlying formations. Salt domes, because they are plastic and usually less dense than surrounding rocks, can literally flow upward into shale sections, overcompacting shallow formations and creating unusually high pressures. Abnormal pressures can result from fluid density differences in zones having connected permeability. Figure 6 (Abnormal pressure due to fluid density difference) illustrates a situation where the pressure in a gas-producing interval having a significant dip is higher than the native salt water gradient. lower-pressured formation. An Artesian water system may result in abnormal pressures by transmitting hydrostatic pressure from a high elevation to a lower elevation through a continuous. water-bearing zone ( Figure 7 . Abnormal pressure in an Artesian water system).Figure 6 Under certain circumstances. This type of overpressuring cannot be predicted using conventional techniques. channeling through cement or underground blowouts. These circumstances could include leakage across a fault. the fluid that may have been normally pressured at its original depth will be overpressured at the shallower depth. When this occurs. fluid migrating from a relatively deep formation can charge a shallower. . casing leaks. geologists and engineers. and should be part of an integrated. Pressure prediction involves the talents of geophysicists. to quantitative methods that we can use in selecting mud weights. which are performed during the well testing and completion stages. designing casing strings. planning cement jobs and specifying completion requirements. They include: seismic methods well log analysis mud logging measurement-while-drilling (MWD) drilling equations Pressure transient tests. can also provide valuable information for future drilling. Geophysical predictions of abnormal pressure are based on reflection seismic principles. team approach to well planning and operations. These range from qualitative indicators. Most prediction techniques are based on finding the relatively low bulk densities and high pore volumes characteristic of overpressured zones. which serve primarily as warning signs. detect and evaluate overpressured zones.Figure 7 Prediction and Detection There are a number of ways to anticipate. which in the simplest sense involve . Normal pressure trend line for interval transit time. In a normal pressure environment. Interval transit time plot indicating abnormal pressure trend. . we should see interval transit times that are above normal ( Figure 2 . where porosity decreases with depth. decreasing porosity). we should see a corresponding decrease in interval transit time ( Figure 1 . To put it another way. the interval transit time (which is the reciprocal of sonic velocity) decreases. based on seismic data from south Texas Frio trend). base on seismic data survey from US gulf coast). Figure 1 It follows that in an overpressured formation.e. the velocity of sound through a subsurface horizon increases with increasing rock density (i.generating acoustic energy at the surface measuring the interval transit time for this energy to reflect back from a subsurface horizon plotting this transit time as a function of depth With other factors being equal. where porosities are abnormally high.. the objective of log analysis is to detect the high porosities characteristic of abnormally pressured formations. Figure 3 (Generalized shale resistivity plot) illustrates this deviation from the normal. higher conductivity) than rock matrix particles. linear shale resistivity trend. With the exception of measurement-whiledrilling (MWD) data. an increase in the fluid-filled pore volume due to overpressuring should result in a resistivity decrease. sonic. We can do this by obtaining electrical (resistivity/conductivity). well logs are used for "post-drilling" evaluations. As with seismic interpretation. which can be used during the early stages of well planning. is that they provide "pre-spud" data.Figure 2 While this is a qualitative indication of abnormal pressure. A major benefit of seismic techniques. We can also obtain interval transit times from sonic logs. we can arrive at quantitiative presure estimates by analyzing seismic data in more detail.e. . however. Because water has lower electrical resistivity (i.. density and other measurements from open-hole or cased-hole tools. Empirical correlation of reservoir fluid pressure gradients versus ratio of noremal to observed shale resistivities) and the following three-step procedure: 1. ) To estimate formation pressure from resistivity measurements. . only resistivities of clean shales are used in constructing resistivity plots. Hottman and Johnson (1965) recommend determining the ratio between observed and normal rock resistivities. using an empirical correlation ( Figure 4 .Figure 3 (Note that since the abnormal pressures originally developed in shale sections. 1985): Given the resistivity data in Table 1. Determine the pressure gradient at the depth of interest. 3. a. Extrapolate the trend line to determine the "normal" resistivity at that depth. c. Note the depth at which the plotted points deviate from the established trend. below.Figure 4 Plot the shale resistivity versus depth on semi-log paper to establish a normal resistivity trend.700 .80 10. -m Depth.600 0.600 ft Resistivity. 2. This indicates the top of the overpressured zone. Calculate the ratio of the observed resistivity to the extrapolated resistivity.000 0. -m Depth.400 0.76 10. Use Figure 4 to determine the formation pressure.54 4. determine: (a) the top of the abnormal pressure zone (b) the formation pressure at a depth of 11. b. Example 1: Abnormal pressure prediction using resistivity data (from Adams. ft Resistivity. ft 0.64 4. 28 11.000 0. .60 5.900 Solution: These data are plotted in Figure 5 .60 7.700 0.27 12.500 0.29 12.30 12.36 11.300 0.76 6400 0.45 11.300 0.000 0.600 0.90 9.74 8.30 11.76 8.0.58 10.700 0.100 0.28 12.900 0.500 0.000 0.000 0.70 7.84 10.500 0.100 0.000 0.900 0.600 0.29 11.70 6.82 9. Shale resistivity overlay).Figure 5 The deviation from the normal pressure trend occurs at about 9. as well as for sonic logs. Overlay plots.1 ppg EMW. The ratio of these resistivities is (1.700 ft. that the curves are correlated with the depth marks and guide lines on the overlays. We need to make sure that overlay scales are consistent with those of the plotted curves. is 1. ( Figure 6 . Figure 6 By shifting overlays horizontally.23/0. or 10.23 -m. Extrapolated "normal" resistivity at 11. and that conductivity curves are not used with resistivity overlays (or vice-versa).39.600 ft. consisting of parallel lines that represent formation pressure in terms of mud weight. Hottman and Johnson's correlation ( Figure 4 ) gives a formation pressure corresponding to about 17.314 psi. while the observed resistivity is 0.28 -m. . are available for resistivity or conductivity curves.28) = 4. we can correlate normal resistivity or conductivity trends with normal pressure lines. Mud logs can give the following indications of abnormal pressure: Paleontology If offset well data indicate that certain fossils are characteristic of an overpressured formation. If the mud weight has not changed. however. such as lithology changes. tsh is greater than it would be in a normally pressured rock matrix (similar to the trend shown in Figure 1 and Figure 2 ). determine reservoir pressure. Operators have for many years used drilled cuttings and mud returns to detect abnormal pressure. that there may be other reasons for higher drilling rates. then the presence of these same fossils may indicate that we have encountered that formation. Mud gas analysis Three general classifications of mud gas are: -. where they experience grinding and re-grinding. the measured parameter is travel time in shale (tsh). We can also use sonic or density logs to plot normal versus observed trends and. Cuttings size increase Increased cuttings sizes may mean that the wellbore differential pressure has decreased. because it takes time for the mud to travel from the bottom of the hole to the shale shaker. Flowline temperature increase Because water does not conduct heat as effectively as clay. it is likely that the formation pressure is increasing. Their main drawback is that they provide information on a "delayed" basis.connection gas the increase in gas concentration due to the swabbing action of the drill string when pulling up to make connections . in conjunction with empirical correlations. Penetration Rate Increase ("drilling break") A steady increase in penetration rate could result from increasing porosity and a consequent lowering of differential pressure. and cuttings are not being held on bottom. Note.The Hottman and Johnson technique assumes that: porosity is the only formation variable affecting pore pressure temperature gradients are constant resistivity measurements are made in water-filled shale water salinity is constant Where formation water salinities vary. In an abnormal pressure environment. Mud returns from an overpressured formation having abnormally high porosity should therefore be hotter than returns from a formation having normal porosity. a water-filled pore space tends to be hotter than its surrounding rock matrix. we can apply techniques such as those developed by Foster and Whalen (1965). They nevertheless are useful pressure indicators because of the amount and variety of information they provide.background gas the total gas concentration in the mud returns during drilling -. For a sonic log. -. Shale density Normally. Perhaps the most well-known drilling model used in the industry is the dc exponent. weight-on-bit. once at the frontiers of technology.. however. revolutions/minute (RPM) Jordan and Shirley (1966) simplified and modified this relationship. therefore. a number of other factors contributing to these increases. the mud logger can detect deviations from this trend. instantaneously transmitted) data. lithology (from gamma ray measurements). For mud resistivity to serve as an indicator. abnormal pressure. rotary speed.. 1980). 1965): (1) where = penetration rate. assigning a value of "1" to the formation drillability constant and replacing Bingham's bit weight constant with a d-exponent: . The models and correlations developed for this purpose are mainly empirical. mud properties) is a basic goal of drilling research. shale density should increase with increasing depth. By measuring the bulk density of the drilled cuttings. have become integral to many drilling operations. including resistivity and formation conductivity. the mud salinity must be measurably different from that of the formation fluid. ft/hour a = formation drillability constant (dimensionless) W = weight-on-bit. There are. These tools. which compromise their reliability as pressure indicators. and range from "rule-of-thumb" field calculations to sophisticated computer simulations. annular temperature.g. and other abnormal pressure indicators. developed from Bingham's drilling rate equation (Bingham. Defining the relationship between penetration rate and various rig parameters (e. They offer the advantage of real time (i. (Adams.e. Mud resistivity and chloride content Changes in these parameters may serve as secondary indicators of increasing porosity and. inches b = bit weight exponent (dimensionless) N = rotary speed.trip gas the increased gas circulated out of the hole after tripping pipe. 1000 lbf dB = bit diameter. Increased gas concentrations have commonly been interpreted as signs of abnormal pressure. Figure 7 . We may plot dc exponents as a function of depth on semi-log paper. a plot of d versus depth typically shows increasing d-exponent values with increasing depth . or even begin to decrease. Typical dc-exponent plot). In abnormally pressured formations.(2) Assuming constant mud weights and normally-pressured formations. and then use overlays to determine formation pressures ( Figure 7 . d should increase at a slower rate. Rehm and McLendon (1971) took into account the effect of increasing mud weight by modifying the d-exponent: (3) where EMWn = Equivalent mud weight of normally pressured formation. with depth.
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