Drilling Problems & Drilling Optimization
Drilling 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 well-known relationship
(1)
where r = fluid density
g = acceleration of gravity
In field units,
(2) where P = pressure, psi
MW = mud weight or density, lbm/gal (U.S.) TVD = true vertical depth, ft
When expressed in SI units (with pressure given in kPa), 0.052 becomes 0.00980. This hydrostatic pressure, which we control by adjusting the mud weight, is our primary means of counteracting formation, or pore pressure.
Equation 2 is adequate for determining the pressure of a static, homogeneous mud column. In many instances, however, the mud is not static; circulation and pipe movement can significantly affect wellbore pressure. Nor is the mud column necessarily homogeneous; during cementing operations or mud changeovers, for example, the well contains fluids of different densities. To determine the actual wellbore pressure, we have to account for these variables. We can simplify our task by defining equivalent mud weight (EMW) or equivalent circulating density(ECD):
(3)
(4)
where EMW = equivalent mud weight, lbm/gal (U.S.) ECD = equivalent circulating density, lbm/gal (U.S.)
Pwell = sum of hydrostatic and applied pressures, psi
Pann = the frictional pressure loss in the annulus. This corresponds to the pump pressure minus the pressure losses through the surface equipment, drill string and drill bit.
TVD = True vertical depth, ft
When expressed in SI metric units (with pressure given in kPa), 0.052 becomes 0.00980. Example 1: Equivalent Circulating Density
Determine the Equivalent Circulating Density and the effective wellbore pressure for the following conditions:
TVD =11,300 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.S.) = fluid pressure gradient, psi/ft TVD = True Vertical Depth, ft
When expressed in SI units (with pressure given in kPa), 0.052 becomes 0.00980. The fluid pressure gradient g typically ranges from 0.433 psi/ft [9.80 kPa/m] for fresh water to 0.465 psi/ft [10.54 kPa/m] or higher for brine.
Example 2: Determining pore pressure
Determine the pore pressure of the following formation. Normal pressure gradient = 0.46 psi/ft
Measured depth = 12,500 ft TVD=10,000 ft
Solution:
In many instances, we may encounter abnormal pore pressures, which deviate from established area trends.
Differential pressure (P) is simply the difference between wellbore an dformation pressure:
When the wellbore and formation pressures are approximately equal, P is zero. We refer to this condition as balanced ( Figure 1).
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 0.1 lbm/gal.) Solution:
This pressure of 4,600 psi is pushing from the formation to the well. To keep
formation fluid from entering the wellbore, the mud column must exert at least 4,600 psi in the other direction -- from the well to the formation. Our required mud weight is therefore:
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 low-permeability, high-strength formations, operators may even drill underbalanced (P<0) using air, gas, foam or low-density liquids ( Figure 2 , underbalanced conditions; wellbore pressure is less than pore pressure).
Figure 2
Because of favorable rock mechanics effects and the fact that there is no chip hold-down caused by high differential pressure, balanced or underbalanced drilling can result in penetration rates much higher than those we could attain with a weighted mud system. Figure 3 illustrates how penetration rates tend to decrease with increasing differential pressure.
Figure 3
Underbalanced drilling is also used to minimize formation damage in producing zones. The Austin Chalk formation of central Texas is a good example. It is common practice for operators drilling horizontal wells through this formation to let the wells flow (of course, this requires them to handle large volumes of oil and gas at the surface).
In the majority of cases, however, we have to keep the mud density higher than that of the formation fluid ( Figure 4 , Overbalanced conditions; wellbore pressure is greater than pore pressure).
Figure 4
That is, 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.
In addition, we need a safeguard against abnormal formation pressures. In many areas (e.g., south Louisiana, south Texas, the North Sea, parts of the Middle East), it is not uncommon to see P values on the order of several thousand psi.
Example 4: Differential pressure
Determine the differential wellbore pressure at 15,400 ft TVD on a U.S. Gulf Coast well, where the mud weight is 14.0 lbm/gal and the formation pressure is 7,600 psi. Solution:
Disadvantages of overbalanced drilling are lower penetration rates and higher mud costs, along with increased potential for formation damage and stuck pipe. Up to a point, this is a fair trade-off, since weighted muds allow us to better control
formation pressure and maintain hole integrity. We need to keep in mind, however, that our drilling objectives also include optimizing penetration rates and reducing expenses. We should not compromise these objectives unnecessarily by using heavier muds than safety requires.
Formation fracture gradients define our upper wellbore pressure limit. Exceeding this limit can cause lost circulation, resulting in formation damage and induced fractures. Severe fracturing can cause the wellbore fluid level to drop, thereby creating a blowout risk.
Fracture pressure is related to overburden stress, 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, a number of researchers have developed theoretical methods for determining fracture gradients. Matthews and Kelly (1967), building on the landmark work of Hubbert and Willis (1957), developed the following relationship for
sedimentary rocks:
(8)
where frac = fracture gradient, psi/ft [kPa/m] D = depth, 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, and thus varies only with the degree of compaction. Their procedure for determining the fracture gradient is as follows:
1. Determine the pore pressure (based on seismic measurements, well log anlaysis or drilling data correlations).
2. Assume a gradient of 1.0 psi/ft [22.62 kPa/m] for the overburden, and calculate the matrix stress s.
3. Determine the depth (Di) at which s would have a normal value, assuming that S=1.0 psi/ft [22.62 kPa/m] and the normal stress gradient equals 0.535 psi/ft [12.10 kPa/m]. Therefore,
(9)
4. Using Figure 5 (Matrix stress coefficients of Matthews and Kelly, basd on data from U.S.
Figure 5
Gulf Coast Sands), determine Ki and calculate frac fromEquation 8 (note that
Figure 5 is empirically generated from field data). Example 5: Determining fracture gradient
Using the Matthews and Kelly procedure, determine the fracture gradient just below the casing seat for the following Louisiana Gulf Coast well.
Casing set at 6,650 ft TVD Formation pressure = 3,300 psi Solution:
In the field, we can estimate the minimum fracture gradient at each new casing point by performing a leak-off test as follows:
1. Close the blowout preventer and apply pressure down the drill pipe in small increments, using a low-volume pump.
2. Continue pumping small volumes of mud until the formation begins to take fluid, or until the pressure reaches a pre-set test limit.
3. Plot pressure versus the pumped volume to determine the initial fracture pressure.
Example 6: Leak-off test-- calculating fracture gradient (after Adams, 1985):
Determine the fracture gradient at a well's casing point (6,750 ft. TVD), given the information in Table 1., below (mud weight = 12.5 lbm/gal).
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.
Figure 6
The fracture pressure is equal to this applied pressure plus the hydrostatic wellbore pressure:
Volume pumped, bbl Pressure, psi
0 0 1 4 1.5 100 2.0 190 2.5 280 3.0 370 3.5 460 4.0 550 4.5 640 5.0 730 5.5 820 6.0 850 6.5 880
Table 1: Results of leak-off test. Buoyancy
An object immersed in fluid is subject to an upward-acting buoyant force, which equals the weight of the fluid the object displaces (i.e., the product of the fluid volume and density). The net effect of buoyant force is that an object submerged in liquid weighs less than it would in air.
In practical terms, this means that a string of pipe in a mud-filled wellbore weighs less than it would if suspended in air, by an amount that corresponds to the weight of the displaced mud. It is important to account for buoyancy effects when calculating hook loads, bit weights, drill collar requirements, casing specifications, rig capacity and other weight-related parameters.
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, and for steel pipe is equal to
(2)
where MW is the mud weight and steel is the density of steel, which equals 65.5 lbm/gal [7860 kg/m3]. Table 2.2 lists buoyancy factors for mud weights ranging from 8.33 to 16.0 pps.
Example 1: Buoyancy effects
Determine the maximum weight-on-bit provided by 450 ft of 7 3/4-inch o.d., 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.)
· 9.5 lbm/gal · 16.0 lbm/gal Solution:
lbm/gal Factor 8.33 [1000] 0.873 9.0 [1078] 0.862 9.5 [1138] 0.855 10.0 [1198] 0.847 10.5 [1258] 0.840 11.0 [1318] 0.832 11.5 [1378] 0.824 12.0 [1438] 0.817 12.5 [1498] 0.809 13.0 [1558] 0.801 13.5 [1618] 0.794 14.0 [1678] 0.786 14.5 [1738] 0.778
15.0 [1798] 0.771
15.5 [1858] 0.763
16.0 [1918] 0.756
Table 1. Buoyancy Factors for Various Mud Weights (Based on Steel Density of 65.5 lbm/gal [7860 kg/m3])
Rheological Models
Rheology, a term often used in drilling engineering, is the study of the flow or deformation of matter. 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 ).
Figure 1
If we apply a force to the upper plate while holding the lower plate stationary, the upper plate attains a constant velocity (v), which depends on the applied force (F), the distance between the plates, the surface area (A) and the fluid viscosity ():
(1)
(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, Bingham Plastic and Power Law models. (A fourth
relationship, the Three Parameter model, combines elements of the Bingham Plastic and Power Law models.)
Newtonian fluids exhibit direct proportionality between shear stress and shear rate ( Figure 2 ):
Figure 2
(4)
The proportionality constant () is the fluid viscosity. Although this model has often been used to describe drilling fluid behavior, it is an oversimplification; most drilling fluids do not exhibit this straight-line relationship.
A common method for describing the shear stress/shear rate relationship is the Bingham-Plastic model ( Figure 3 ):
Figure 3
(5) where
p = plastic viscosity (PV), cp
y = yield stress or yield point (YP), lbf /100 ft 2 (1.0 cp = 0.001 Pa-s; 1 lbf /100 ft 2
= 0.4788 Pa)
Note that a certain initial pressure is required to initiate fluid movement, and that the mud's circulating pressure depends on the initial yield point and the plastic viscosity. 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, respectively, and PV is given in cp.
The Bingham Plastic model thus depends on a linear relationship (the difference between the two viscometer readings) to describe drilling fluid flow. Although widely used, its application to turbulent flow conditions, where fluid behavior deviates from linear relationships, is limited.
The Power Law model is a non-linear relationship, which is more descriptive of real drilling fluid behavior ( Figure 4 ):
Figure 4
(8)
where (9)
n = flow behavior index = 3.32 x log( 600/300) (10)
While it is more realistic than the Bingham Plastic model, the Power Law model is also more complex.
Flow Regimes
The flow regimes most commonly encountered in drilling are laminar, turbulent and transitional.
In laminar flow, fluid behaves as a series of parallel layers moving at uniform or near-uniform velocities. There is no large-scale movement of fluid particles between layers. The fluid layers nearest the center of the pipe or annulus generally move faster than the layers adjacent to the hole wall ( Figure 1, Laminar flow profile in drill pipe/hole annulus).
Figure 1
Turbulent flow is characterized by velocity fluctuations among the fluid stream particles, both parallel and axial to the mean flow stream. These fluctuations break down the boundaries between the fluid layers, resulting in a chaotic flow pattern ( Figure 2 , Turbulent flow in drill pipe and at bit face).
Figure 2
Transitional flow exhibits characteristics of both laminar and turbulent regimes. It describes the often hard-to-define region where flow is neither wholly laminar or wholly turbulent.
(An additional term, plug flow, describes the low-velocity, sub-laminar condition of a fluid moving as a homogeneous, relatively undisturbed body.)
We usually prefer to see laminar flow in the annulus to move cuttings up the hole and prevent erosion. Turbulent flow, on the other hand, is more desirable at the bottom of the hole, because it promotes cleaning and cuttings removal.
While they are conceptually easy to visualize, flow regimes may be difficult to
identify. Not only does fluid behavior vary within the circulating system, but we often find that more than one flow regime exists at the same point in the system. For example, while the main flow stream in an annulus may exhibit laminar behavior, the adjacent fluid at the pipe boundary may be in turbulent flow.
The most common method for determining a fluid's flow regime is to calculate its Reynolds number. In its simplest form (i.e., for the case of a Newtonian, non-elastic liquid inside pipe) we may define the Reynolds number as
(1)
where = fluid density, kg/m3
(2) = inside diameter of pipe, m
= fluid viscosity, Pa-s
For annular flow, the Reynolds number becomes
(3) where d2 = hole or casing diameter, m
d1 = outside diameter of pipe, m
(4)
The term 0.816(d2-d1) inEquation3 is the equivalent circular diameter of a slot
representation of the annulus (Bourgoyne et al, 1986).
In field units (where is given in lbm/gal, v in ft/s, q in gal/min, d, d1 and d2 in inches and in cp), Equations 1 through 4 become:
(Pipe flow):
(5)
(6) (Annular flow):
(7)
(8)
As a general guideline, Reynolds numbers of less than 2,100 indicate laminar flow, while Reynolds numbers greater than 4,000 indicate turbulent flow. Between these values, flow is considered transitional.
In field applications, we identify flow regimes by determining the critical Reynolds number or critical velocity (vc) at which flow changes from laminar to turbulent. For
example, we may calculate vc and compare it to the actual fluid velocity (v). If vc < v, flow is laminar, while if vc >v, it is turbulent. Because the transition between
laminar and turbulent flow may not be clear-cut, it often becomes necessary to calculate a range of critical velocities to determine the flow regime.
Example 1: Flow regimes
A 10.5 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. Determine the flow regime inside 4.5 inch o.d., 16.60 lb/ft drill pipe (3.826 inch inside diameter), and in the drill pipe/hole annulus. For this example, assume that the mud behaves as a Newtonian fluid. Solution:
Drill pipe:
NRe > 4,000 Turbulent flow
NRe < 2,100 Laminar flow
Unfortunately, determining flow regimes is seldom this straightforward. Laminar flow has been observed under controlled conditions for Reynolds numbers as low as 1200 and as high as 40,000 (Bourgoyne et al., 1986), although we do not usually
encounter such extremes in drilling operations. 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, cp PV = plastic viscosity, cp YP = yield point, lbf /100 ft2
d = inside pipe diameter, inches d2 = hole or casing diameter, inches d1= outside pipe diameter, inches v = average velocity, ft/s
When Equations 9 and 10 are expressed in SI units, 6.66 becomes 0.1669, and 5 becomes 0.1253.
(Pipe flow) (11)
(Annular flow) (12)
Using these equations, the criterion for turbulence is the same as for Newtonian fluids, with laminar flow occurring below a Reynolds number of 2100.
Power Law Fluids
For power law fluids, 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, lbm/gal (U.S.)
v = fluid velocity, ft/s
d2 and d1 are expressed in inches
When expressed in SI units, Equations 13 and 14 become
(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. 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, the critical Reynolds number is
(16)
System Pressure Loss
For mud to flow through the circulating system, it must overcome frictional forces between the fluid layers, solid particles and pipe or borehole walls. 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, mud weight, viscosity, pipe diameter and hole diameter. The maximum amount of friction loss that we can overcome is limited to the working pressure of our mud pumps.
The general procedure for calculating system pressure losses is as follows: 1. Determine the fluid velocity (or Reynolds number) at the point of interest.
2. Calculate the critical velocity (or Reynolds number) to determine whether the fluid is in laminar or turbulent flow.
3. Choose the appropriate pressure loss equation. (The choice depends on which rheological model and flow regime apply to the point of interest).
Note: In the field, determine both the critical and the actual flow velocities. If vc<v the flow is
turbulent, while if vc v, it is laminar. If the critical and actual velocities are approximately equal,
then perform pressure loss calculations for both flow regimes, and use the results that give the larger pressure loss.
We use this information to maximize the pressure loss through the bit and thereby maximize the hydraulic energy at the bottom of the hole. Table 1 and Table 2 summarize the equations for determining pipe and annular pressure losses based on Bingham Plastic flow, 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)
SI Units:
Nomenclature (SI units in brackets):
d = inside diameter of pipe, inches
[m] v = velocity, ft/s [m/s]
L = pipe length, ft [m] vc = critical velocity, ft/s [m/s]
MW = mud weight, lbm/gal [kg/m3] YP = yield point, lbf/100ft2 [N/m2]
PV = plastic viscosity, cp [Pa-s] Pds = pressure loss, psi [kPa] q = flow rate, gal/min [m3/s]
Table 1 Drill string 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) Oilfield Units: SI Units
Pressure Loss for laminar flow (Pds ) (Use for v < vc)
SI Units:
Pressure Loss for turbulent flow (Use for v vc) Oilfield Units: SI Units: Nomenclature:
d1 = inside diameter of pipe, inches
[m] q = flow rate, gal/min [m
3/s]
d2 = casing or hole diameter, inches
[m] v = velocity, ft/s [m/s]
L = pipe length, ft [m] vc = critical velocity, ft/s [m/s] MW = mud weight, lbm/gal [kg/m3] YP = yield point, lbf/100ft2 [N/m2]
PV = plastic viscosity, cp [Pa-s] Pann = pressure loss, 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)
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, inches [m] v = velocity, ft/s [m/s] K = consistency index
(dimensionless) v
c = critical velocity, ft/s [m/s]
L = pipe length, ft [m] q = flow rate, gal/min [m3/s]
MW = mud weight, lbm/gal [kg/m3] YP = yield point, lbf/100ft2 [N/m2]
n = flow behavior index
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)
SI Units:
Nomenclature:
d1 = outside diameter of pipe, inches
[m]
v = velocity, ft/s [m/s] d2 = casing or hole diameter, inches
[m] vc = critical velocity, ft/s [m/s]
K = consistency index
(dimensionless) q = flow rate, gal/min [m
3/s]
L = pipe length, ft [m] YP = yield point, lbf100ft2 [N/m2]
MW = mud weight, lbm/gal [kg/m3] Pann = pressure loss, psi [kPa]
n = flow behavior index (dimensionless)
PV = plastic viscosity, 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. To determine this equivalent length, we simply match our surface equipment specifications to one of the four groups shown in Table 2.7. For example, if a rig has Group 4 surface equipment and uses 5-inch, 19.5 lb/ft drill pipe, we would, for calculation purposes, add 579 feet to the actual drill pipe length.
Component Typical Combinations
Case 1 Case 2 Case 3 Case 4
i.d.,in L,ft i.d.,in L,ft i.d.,in L,ft i.d.,in L,ft
[cm] [m] [cm] [m] [cm] [m] [cm] [m]
Standpipe 3 40 3.5 40 4 45 4 45
[7.62] [12.192] [8.89] [12.192] [10.16] [13.716] [10.16] [13.716]
Drilling 2 45 2.5 55 3 55 3 55
Swivel, 2 4 2.5 5 2.5 5 3 6 washpipe, [5.08] [1.219] [6.35] [1.524] [6.35] [1.524] [7.62] [1.829] gooseneck
Kelly 2.25 40 3.25 40 3.25 40 4 40
[5.715] [12.192] [8.255] [12.192] [8.255] [12.192] [10.16] 40 Drill pipe: Length of surface connections, expressed as equivalent ft [m] of drill pipe
3.5 inch, 13.3 lb/ft 437 [133.2] 161 [49.1]
4.5inch, 16.6 lb/ft 761 [232.0] 479 [146.0] 340 [103.6]
5 inch, 19.5 lb/ft 816 [248.7] 579 [176.5]
Drill bit pressure losses do not result primarily from friction forces, rather, they are due to the acceleration of the drilling fluid through the bit nozzles. We may express the bit pressure drop in psi as
(2) where
q = circulation rate, gal/min (U.S.) MW = mud weight, lbm/gal
Cd = nozzle discharge coefficient (dimensionless) AT = total nozzle area, in2
In the SI system, where Pbit is expressed in kPa, 12,031 becomes 2000.
Although the nozzle discharge coefficient Cd varies according to nozzle type and size, a value of 0.95 is acceptable for most field calculations.
The following example illustrates how we can apply pressure loss relationships. Example 1
Determine the bit pressure drop for the following well: Total Depth=9,950 ft
Casing: 9 5/8 inch, 43.50 lb/ft (8.755 in. i.d) cemented at 6,500 ft. Open hole: 8 1/2 inch from 6,500 ft to 9,950 ft (T.D.)
Drill pipe: 9500 ft. of 4 1/2 inch, 16.60 lb/ft (3.826 in. i.d.)
Bottomhole assembly: 450 ft. of 6 3/4 inch o.d. x 2 1/4 inch i.d. drill collars and tools Mud properties:
Mud weight (MW)=10.5 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=2,200 psi Surface Equipment:
Standpipe: 45 ft x 4 in. i.d. Hose: 55 ft x 3 in. i.d.
Swivel, washpipe, gooseneck: 5 ft x 2.5 in. i.d. Kelly: 40 ft x 3.25 in. i.d.
Solution:
1. Note first that the surface components correspond to a Case 3 equipment combination (Table 2.7). The pressure loss through these components is equivalent to 479 ft. of 4 1/2 in., 16.6 lb/ft drill pipe. We can therefore combine Psurf with our calculation of Pds. 2. Determine the pressure losses inside the drill string (Pds). This involves separate calculations for the drill pipe and the bottomhole assembly:
a) Drill pipe:
= 605 psi
(Note that we accounted for surface pressure losses by adding in the drill pipe equivalent length of 479 ft)
b) Bottomhole assembly:
= 4.88 ft/s v vc turbulent flow
= 340 psi
Pds=Pdp+Pbha=605+340=945 psi
3. Determine the pressure losses in the annulus. (DPann). This involves separate calculations for the cased-hole section (surface to 6,500 ft), the drill pipe/hole annulus (6,500 - 9,500 ft) and the BHA/hole annulus (9,500 - 9,050 ft). a) Cased-hole annulus (surface-6500 ft):
= 3.47 ft/s v < vc laminar flow
= 73 psi
b) Drill pipe/hole annulus (6,500 ft - 9,500 ft):
Assume laminar flow (critical velocity will be close to that calculated for the cased-hole annulus)
= 38 psi c) BHA/open hole annulus:
= 31 psi
Total annular pressure losses: DPann=73 + 38 + 31 = 142 psi 4. Determine pressure loss at bit:
Pbit=Ppump-(Psurf+Pds)-Pann=2200 - 945 - 142= 1,113 psi
Surge and Swab Pressure
When we run pipe in a well, it forces drilling mud up the annulus and out of the flowline. At the same time, the mud immediately adjacent to the pipe is dragged downhole, as shown in Figure 1 (Annular flow profile resulting from downward ie movement).
Figure 1
The resulting piston effect generates a surge pressure that is added to the
hydrostatic pressure. Excessive surge pressures can increase the wellbore pressure to such a degree as to induce lost circulation.
Conversely, when we pull pipe out of a well, mud flows down the annulus to fill the resulting void. This causes a suction effect, generating a swab pressure that can lower the differential pressure and possibly bring formation fluid into the wellbore.
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, and illustrates the significance of surge and swab effects.
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, which is referred to as the clinging constant, K ( Figure 3 , Mud clinging constant K as a function of annular geometry).
Figure 3
To apply the clinging constant, we need to know the mud velocity in the annulus. For a closed drill string, this is equal to
(1) where vpipe = pipe velocity
d2 = hole diameter
d1 = pipe outside diameter
For open-ended pipe,
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, and determine whether the total wellbore pressure exceeds the fracture gradient. Assume that the casing is effectively "closed" with a float shoe.
Casing point = 6400 ft
Fracture gradient = 0.82 psi/ft
Pipe velocity = -110 ft/min = -1.83 ft/s ("-" denotes downward velocity) Hole diameter = 14 3/4 inches
Mud: 15.0 lbm/gal; PV=37 cp, YP=6 lbf/100 ft2
Solution:
= 2.07 ft/s
d1/d2= 10.75/14.75=0.73; from Figure 3 , K=0.44 (assume laminar flow)
Use annular pressure loss equation for laminar flow:
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.
Hole Cleaning
From the time that rotary drilling came into its own with the dramatic Spindletop discovery of 1901, it was destined to ultimately replace the once dominant cable tool methods. 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).
Cuttings removal and transport are still primary functions of the drilling mud. Additionally, the mud must be able to hold these cuttings in suspension when circulation stops. The mud's effectiveness at removing cuttings significantly affects drilling efficiency.
Drilled cuttings vary in size and density according to formation lithology, differential pressure, 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, back toward the bottom of the hole. A mud's ability to transport cuttings -- that is, its carrying capacity -- is related to the difference between the annular velocity and the slip velocity with which the cuttings fall ( Figure 1, Carrying capacity depends on a mud's annular velocity and the slip velocity with which cuttings fall back to bottom).
Figure 1
Moore (1974), 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). To use these methods, we need to determine the average densities and diameters of drilled solids by visually inspecting representative cuttings, 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, ft/s
dp = particle diameter, inches
f= fluid density, lbm/gal Cdrag = drag coefficient
1.89 = numerical value of conversion constant, which has units of [ft2/(sec2-in)]1/2. In
SI units, 1.89 becomes 3.615.
To determine the drag coefficient, we must first compute the particle Reynolds number (NRep):
(2) where µ = fluid viscosity, cp
In SI units (kg/m3, m/s,m, Pa-s), 928 reduces to 1.
We can then obtain Cdrag from Figure 2 (Particle drag coefficient as a function of the particle Reynolds number).
As we can see, this is an iterative process.
Moore presents the following modifications of Equation 1 for various ranges of the particle Reynolds number:
For (NRep) <1, Cdrag=40/(NRep) and
(3) In SI units, 82.87 becomes 0.3267.
For (NRep) >2,000 (turbulent flow), Cdrag has a constant value of about 1.5, and
Equation 1 reduces to
(4) In SI units, 1.54 becomes 2.945.
For intermediate particle Reynolds number values (i.e., 10< (NRep)
<100) Cdrag = , and
(5) In SI units, 2.90 becomes 0.7053.
The term µa in these equations is the apparent viscosity, given in units of centipoise
and defined as follows:
(6)
n = flow behavior index = (8) v = fluid velocity, ft/s
In SI units, (with a expressed in Pa-s), Equation6 becomes
(6 SI)
Moore recommends Equation5 for use in routine field applications. Note that this method for determining slip velocity may require several iterations.
Example 1: Slip velocity (Moore's method)
Calculate the slip velocity of drilled cuttings having an average diameter of 0.25 inches and a density of 2.65 g/cc (22.1 lbm/gal), given the following mud and hole data:
Hole size=12.25 inches Drill pipe=5 inches MW=10.2 lbm/gal
Annular Velocity=60 ft/min (1.0 ft/s) n (from viscometer readings)=0.8 K=150 cp equivalent
Solution:
= 88.1 cp
2. Assume an (NRep) range and then check the results obtained:
Let (NRe p) <1:
=0.7 ft/s Verification:
(NRep) >1.0 first assumption was invalid
Assume 1.0< (NRe p) <100:
vs = 0.39 ft/s
Verification:
(NRep) = 10.5 assumption was valid
3. Annular velocity = 1.0 ft/s;
= 0.61 ft/s, or 37 ft/min.
Exercises
1.
What mud weight is required to balance formation pressure and provide a 400 psi trip margin at a depth of 12,300 ft (TVD) and a fluid pressure gradient of 0.45 psi/ft?solution
2.
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 13.0 bm /gal mud, remembering that the drill pipe must remain in tension.Soln
3.
Calculate the plastic viscosity (PV), yield point (YP), flow behavior index (n) and the consistency index (K), given viscometer readings of 34 at 300 RPM and 62 at 600 RPM.4.
Using the Bingham-Plastic flow model, determine the critical velocity and flow regime for a 12.25 x 7.75 inch open hole/drill collar annulus:Mud properties: MW=10.2 lbm/gal PV=32 cp
YP=8 lbf/100ft2
Circulation rate: Q=300 gallons/minute
Soln
Using the Bingham-Plastic flow model, determine the critical velocity and flow regime for a 12.25 x 7.75 inch open hole/drill collar annulus:
5.
Using the Bingham Plastic model, calculate the frictional pressure loss in a 10,000 foot section of 5-inch, 19.50 lb/ft (i.d.=4.276 inches) for the following circulating conditions:Q=350 gallons/minute MW=13.5 lbm/gal
PV=26 cp YP=9 lbf /100ft2
6.
Determine the surge pressure for the following conditions: 8.75 inch bit at 8,500 ftDrill pipe=7800 ft. of 4.5 in., 16.60 lb/ft Drill collars=700 ft. of 6.75 inch.
Trip speed = -90 ft/min = -1.5 ft/s (- sign denotes downward veleocity) Bingham Plastic flow model
Assume laminar flow in the annulus
Float valve in drill pipe (i.e., closed-ended drill string) 12.0 lbm /gal mud, PV=29 cp, YP=8 lbf /100 ft2
Soln..
Annular velocity opposite drill pipe:
Equivalent velocity ve = vmud -Kvpipe
Annular velocity opposite drill collars:
Surge pressures: Drill pipe: Drill Collars:
Total surge pressure = 87.5 + 28.7 = 116 psi.
Abnormal Pressure Environments
OriginsNormally pressured formations are open hydraulic systems; they have unblocked communication with adjacent formations and, ultimately, with the surface. They do not exceed native hydrostatic pressure, because any pressure buildup rapidly dissipates to neighboring intervals. Abnormally pressured formations, on the other hand, are closed systems. They are sealed by structural traps or low-permeability barriers that keep high pressures from dispersing. The most common circumstances under which abnormal pressures occur are:
· 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. As each new layer forms, it puts more weight on underlying layers and forces them downward. The underlying layers are thus subjected to successively increasing overburden pressure.
If an underlying layer is normally pressured, increased overburden causes the rock matrix to compact, thereby decreasing the pore space and forcing fluid into adjacent permeable formations. This compaction increases the rock density and matrix
strength, enabling the rock to support the added pressure. As long as the fluid escapes fast enough to compensate for the increased pressure, there is a balance between the overburden and the rock matrix stress.
Compaction can only take place, however, if the pore fluid has somewhere to go. If a low permeability barrier (e.g., a shale layer) slows or prevents fluid from escaping, then the pore volume cannot decrease quickly enough to provide adequate rock matrix strength. Since the rock matrix cannot support the higher overburden pressure, the pressure increase is transmitted to the trapped formation fluid, resulting in an overpressured condition ( Figure 1 , Abnormal pore pressure due to undercompaction of sediments. The lower sandstone formation is surrounded by low-permeability shales, which prevent fluid from escaping).
Figure 1
Chemical diagenesis refers to the chemical alteration of minerals by geologic processes, and is thought to be a principle cause of abnormal pressure (McClure, 1983).
For example, temperature increases during rock deposition may cause
montmorillonite clay to dehydrate and turn into illite (Powers, 1967). Water that was formerly bound to the montmorillonite is thus released to occupy pore space (
Figure 2
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.
Figure 6
Under certain circumstances, fluid migrating from a relatively deep formation can charge a shallower, lower-pressured formation. When this occurs, the fluid that may have been normally pressured at its original depth will be overpressured at the shallower depth. These circumstances could include leakage across a fault, casing leaks, channeling through cement or underground blowouts.
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). This type of overpressuring cannot be predicted using conventional techniques.
Figure 7
Prediction and Detection
There are a number of ways to anticipate, detect and evaluate overpressured zones. These range from qualitative indicators, which serve primarily as warning signs, to quantitative methods that we can use in selecting mud weights, designing casing strings, planning cement jobs and specifying completion requirements.
Pressure prediction involves the talents of geophysicists, geologists and engineers, and should be part of an integrated, team approach to well planning and operations. Most prediction techniques are based on finding the relatively low bulk densities and high pore volumes characteristic of overpressured zones. They include:
seismic methods well log analysis mud logging
measurement-while-drilling (MWD) drilling equations
Pressure transient tests, which are performed during the well testing and completion stages, can also provide valuable information for future drilling.
Geophysical predictions of abnormal pressure are based on reflection seismic principles, which in the simplest sense involve
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, the velocity of sound through a subsurface horizon increases with increasing rock density (i.e., decreasing porosity). To put it another way, the interval transit time (which is the reciprocal of sonic velocity) decreases. In a normal pressure environment, where porosity decreases with depth, we should see a corresponding decrease in interval transit time ( Figure 1, Normal pressure trend line for interval transit time, base on seismic data survey from US gulf coast).
Figure 1
It follows that in an overpressured formation, where porosities are abnormally high, we should see interval transit times that are above normal ( Figure 2, Interval transit time plot indicating abnormal pressure trend, based on seismic data from south Texas Frio trend).
Figure 2
While this is a qualitative indication of abnormal pressure, we can arrive at quantitiative presure estimates by analyzing seismic data in more detail.
We can also obtain interval transit times from sonic logs. A major benefit of seismic techniques, however, is that they provide "pre-spud" data, which can be used during the early stages of well planning.
As with seismic interpretation, the objective of log analysis is to detect the high porosities characteristic of abnormally pressured formations. We can do this by obtaining electrical (resistivity/conductivity), sonic, density and other measurements from open-hole or cased-hole tools. With the exception of measurement-while-drilling (MWD) data, well logs are used for "post-measurement-while-drilling" evaluations.
Because water has lower electrical resistivity (i.e., higher conductivity) than rock matrix particles, an increase in the fluid-filled pore volume due to overpressuring should result in a resistivity decrease. Figure 3 (Generalized shale resistivity plot) illustrates this deviation from the normal, linear shale resistivity trend.
Figure 3
(Note that since the abnormal pressures originally developed in shale sections, only resistivities of clean shales are used in constructing resistivity plots. )
To estimate formation pressure from resistivity measurements, Hottman and Johnson (1965) recommend determining the ratio between observed and normal rock
resistivities. using an empirical correlation ( Figure 4 , Empirical correlation of reservoir fluid pressure gradients versus ratio of noremal to observed shale resistivities) and the following three-step procedure:
Figure 4
Plot the shale resistivity versus depth on semi-log paper to establish a normal resistivity trend.
2. Note the depth at which the plotted points deviate from the established trend. This indicates the top of the overpressured zone.
3. Determine the pressure gradient at the depth of interest.
a. Extrapolate the trend line to determine the "normal" resistivity at that depth. b. Calculate the ratio of the observed resistivity to the extrapolated resistivity.
c. Use Figure 4 to determine the formation pressure.
Example 1: Abnormal pressure prediction using resistivity data (from Adams, 1985): Given the resistivity data in Table 1, below, determine:
(a) the top of the abnormal pressure zone
(b) the formation pressure at a depth of 11,600 ft
Resistivity, -m Depth, ft Resistivity, -m Depth, ft
0.54 4,000 0.80 10,400
0.60 5,600 0.58 10,900 0.70 6,000 0.45 11,000 0.76 6400 0.36 11,100 0.60 7,000 0.30 11,300 0.70 7,500 0.28 11,600 0.74 8,000 0.29 11,900 0.76 8,500 0.27 12,300 0.82 9,000 0.28 12,500 0.90 9,700 0.29 12,700 0.84 10,100 0.30 12,900 Solution:
Figure 5
The deviation from the normal pressure trend occurs at about 9,700 ft. Extrapolated "normal" resistivity at 11,600 ft. is 1.23 -m, while the observed resistivity is 0.28 -m. The ratio of these resistivities is (1.23/0.28) = 4.39. Hottman and Johnson's correlation ( Figure 4 ) gives a formation pressure corresponding to about 17.1 ppg EMW, or 10,314 psi.
Overlay plots, consisting of parallel lines that represent formation pressure in terms of mud weight, are available for resistivity or conductivity curves, as well as for sonic logs. ( Figure 6 , Shale resistivity overlay).
Figure 6
By shifting overlays horizontally, we can correlate normal resistivity or conductivity trends with normal pressure lines. We need to make sure that overlay scales are consistent with those of the plotted curves, that the curves are correlated with the depth marks and guide lines on the overlays, and that conductivity curves are not used with resistivity overlays (or vice-versa).
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, we can apply techniques such as those developed by Foster and Whalen (1965).
We can also use sonic or density logs to plot normal versus observed trends and, in conjunction with empirical correlations, determine reservoir pressure. For a sonic log, the measured parameter is travel time in shale (tsh). In an abnormal pressure
environment, tsh is greater than it would be in a normally pressured rock matrix
(similar to the trend shown in Figure 1 and Figure 2 ).
Operators have for many years used drilled cuttings and mud returns to detect abnormal pressure. Their main drawback is that they provide information on a "delayed" basis, because it takes time for the mud to travel from the bottom of the hole to the shale shaker. They nevertheless are useful pressure indicators because of the amount and variety of information they provide.
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, then the presence of these same fossils may indicate that we have encountered that formation.
Penetration Rate Increase ("drilling break") A steady increase in penetration rate could result from increasing porosity and a consequent lowering of differential pressure. Note, however, that there may be other reasons for higher drilling rates, such as lithology changes.
Cuttings size increase Increased cuttings sizes may mean that the wellbore differential pressure has decreased, and cuttings are not being held on bottom, where they experience grinding and re-grinding. If the mud weight has not changed, it is likely that the formation pressure is increasing. Flowline temperature increase Because water does not conduct heat as effectively as clay, a water-filled pore space tends to be hotter than its surrounding rock matrix. Mud returns from an overpressured formation having abnormally high porosity should therefore be hotter than returns from a formation having normal porosity.
Mud gas analysis Three general classifications of mud gas are:
-- background gas the total gas concentration in the mud returns during drilling -- connection gas the increase in gas concentration due to the
-- trip gas the increased gas circulated out of the hole after tripping pipe.
Increased gas concentrations have commonly been interpreted as signs of abnormal pressure. There are, however, a number of other factors contributing to these increases, which compromise their reliability as pressure indicators. (Adams, 1980).
Mud resistivity and chloride content Changes in these parameters may serve as secondary indicators of increasing porosity and, therefore, abnormal pressure. For mud resistivity to serve as an indicator, the mud salinity must be measurably different from that of the formation fluid.
Shale density Normally, shale density should increase with increasing depth. By measuring the bulk density of the drilled cuttings, the mud logger can detect deviations from this trend.
These tools, once at the frontiers of technology, have become integral to many drilling operations. They offer the advantage of real time (i.e., instantaneously transmitted) data, including resistivity and formation conductivity, lithology (from gamma ray measurements), annular temperature, and other abnormal pressure indicators.
Defining the relationship between penetration rate and various rig parameters (e.g., weight-on-bit, rotary speed, mud properties) is a basic goal of drilling research. The models and correlations developed for this purpose are mainly empirical, and range from "rule-of-thumb" field calculations to sophisticated computer simulations. Perhaps the most well-known drilling model used in the industry is the dc exponent, developed from Bingham's drilling rate equation (Bingham, 1965):
(1)
where = penetration rate, ft/hour
a = formation drillability constant (dimensionless) W = weight-on-bit, 1000 lbf
dB = bit diameter, inches
b = bit weight exponent (dimensionless) N = rotary speed, revolutions/minute (RPM)
Jordan and Shirley (1966) simplified and modified this relationship, assigning a value of "1" to the formation drillability constant and replacing Bingham's bit weight constant with a d-exponent:
(2)
Assuming constant mud weights and normally-pressured formations, a plot of d versus depth typically shows increasing d-exponent values with increasing depth . In abnormally pressured formations, d should increase at a slower rate, or even begin to decrease, with depth.
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.
We may plot dcexponents as a function of depth on semi-log paper, and then use overlays to determine formation pressures ( Figure 7 , Typical dc-exponent plot).
