Advanced Mud School Part XIII Rheology & Hydraulics Presented By: Jeff Imrie
August 2006
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Rheology • Fluid types – There are two basic types of fluids, Newtonian and non-Newtonian. – Rheological and hydraulic models have been developed to characterize characterize the flow behavior of these two types of fluids.
• Newton Newtonian ian fluids fluids have have a const constant ant visc viscosi osity ty at at a given given temperature and pressure condition. Common Newtonian fluids include: – – – –
Diesel Water Glycerin Clear brines
• Non-Ne Non-Newto wtonian nian fluids fluids have viscosit viscosities ies that that depend depend on measured shear rates for a given temperature and pressure condition. Examples of non-Newtonian fluids include: – Most drilling fluids fluids – Cement
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Rheology • Fluid types – There are two basic types of fluids, Newtonian and non-Newtonian. – Rheological and hydraulic models have been developed to characterize characterize the flow behavior of these two types of fluids.
• Newton Newtonian ian fluids fluids have have a const constant ant visc viscosi osity ty at at a given given temperature and pressure condition. Common Newtonian fluids include: – – – –
Diesel Water Glycerin Clear brines
• Non-Ne Non-Newto wtonian nian fluids fluids have viscosit viscosities ies that that depend depend on measured shear rates for a given temperature and pressure condition. Examples of non-Newtonian fluids include: – Most drilling fluids fluids – Cement
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Rheology heology - Rheologica heologicall mode m odels ls • Rheolo Rheologic gical al mod models els hel help p pred predict ict flu fluid id be behav havior ior across a wide range of shear rates. • Mo Most st dril drilli ling ng flui fluids ds are are non non-N -New ewto toni nian an,, pseudo pse udopla plasti sticc flu fluids ids.. • Th Thee most most imp impor orta tant nt rhe rheol olog ogic ical al mod model elss that that pertain to them are the: – Bingha Bing ham m mode mod el – Powe ow er la l aw mod m ode el – Yield-power law law or mod m odif ified ied powe pow er law
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Rheology - Rheological models
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Rheology - Rheological models • The Bingham model describes laminar flow using the following equation: t = YP + (PV × y) – – – – –
Where t is the measured shear stress in lb/100 ft2 YP is the yield point in lb/100 ft2 PV is the plastic viscosity in cP y is the shear rate in sec-1
• Current API guidelines require the calculation of YP and PV using the following equations: • PV = 600rpm – 300rpm • YP = 300rpm – PV
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Rheology - Rheological models • The power law model describes fluid rheological behavior using the following equation: t = K × yn • This model describes the rheological behavior of polymer-based drilling fluids that do not exhibit yield stress (i.e., viscosified clear brines). • Some fluids viscosified with biopolymers can also be described by power-law behavior.
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Rheology - Rheological models • The general equations for calculating a fluid's flow index and consistency index are: n = log(t2/t1) log(y2/y1) K = t2 y2n • • • • • • • •
t is the calculated shear stress in lb/100 ft2 t2 is the shear stress at higher shear rate (600 rpm) t1 is the shear stress at lower shear rate (300 rpm) n is the flow index y is the shear rate in sec -1 y2 is the higher shear rate (600) y1 is the lower shear rate (300) K is the consistency index
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Rheology - Rheological models
• Because most drilling fluids exhibit yield stress, the Yield-power law [YPL]) model describes the rheological behavior of drilling muds more accurately than any other model. • The YPL model uses the following equation to describe fluid behavior:
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Rheology - Rheological models
t = t0 + (K × y)n – Where – t is the measured shear stress in lb/100 ft 2 – t0 is the fluid's yield stress (shear stress at zero shear 0 rate) in lb/100 ft2 – K is the fluid's consistency index in cP or lb/100 ft2 secn – n is the fluid's flow index – y is the shear rate in sec-1
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Rheology - Rheological models • K and n values in the YPL model are calculated differently than their counterparts in the power law model. • The YPL model reduces to the Bingham model when n = 1 and it reduces to the power law model when t0 = 0. • An obvious advantage the YPL model has over the power law model is that, from a set of data input, only one value for n and K are calculated. – Note: The YPL model requires: • A computer algorithm to obtain solutions. • A minimum of three shear-stress/shear-rate measurements for solution. Model accuracy is improved with additional data input.
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Hydraulics • Once the rheological properties for a fluid have been determined and modeled to predict flow behavior, hydraulics calculations are made to determine what effect this particular fluid will have on system pressures. • The critical pressures are total system pressure (pump pressure), pressure loss across the bit and annular pressure loss (converted to ECD)
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Hydraulics • Many wells are drilled under pressure limitations imposed by the drilling rig and associated equipment. – The pressure ratings of the pump liners and surface equipment and the number of mud pumps available limit the circulating system to a maximum allowable circulating pressure.
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Hydraulics • It is imperative to optimize drilling fluid hydraulics by controlling the rheological properties of the drilling fluid to avoid reaching this theoretical limit. – This is especially true in extended-reach drilling.
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Hydraulics • Fluids in laminar flow “act” differently than fluids in turbulent flow. – These differences make it necessary to use different equations to determine the pressure losses in laminar and turbulent flow. – Different equations are also required to calculate the pressure losses in the annulus and drillstring because of different geometries.
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Hydraulics • The first step in hydraulics calculations is to determine which stage of flow is occurring in each geometric interval of the well. • The velocity of the fluid in each of these intervals can be determined with the following equations.
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Hydraulics – Bulk Velocity • Average bulk velocity in pipe (Vp): Vp (ft/min) = 24.48 x Q (gpm) D2 (in.)
• Average bulk velocity in annulus: Va (ft/min) =24.48 x Q (gpm) (D22 – D12)(in.) – V = Velocity (ft/min) – Q = Flow ratio (gpm) – D = Diameter (in.)
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Hydraulics – Reynolds Number • The Reynolds number (N Re ) is a dimensionless number that is used to determine whether a fluid is in laminar or turbulent flow. – A Reynolds number less than or equal to 2,100 indicates laminar flow. – A Reynolds number greater than 2,100 indicates turbulent flow. – Earlier API hydraulics bulletins and many hydraulics programs that predate the current API hydraulics bulletin define laminar and turbulent flow differently.
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Hydraulics – Reynolds Number • The general formula for Reynolds number is: NRe = V Dr μ
V = Velocity D = Diameter r = Density μ
= Viscosity
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Hydraulics – Reynolds Number • The Reynolds number for inside the pipe is: NRep = 15.467 x Vp Dr μep
• The Reynolds number for the annulus is: NRea = 15.467 Va (D2-D1)r μea
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Hydraulics – Reynolds Number • D = ID drill pipe or drill collars • D2 = ID hole or casing • D1 = OD drill pipe or drill collars • μep = Effective viscosity (cP) pipe • μea = Effective viscosity (cP) annulus
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Hydraulics – Critical Velocity • The critical velocity is used to describe the velocity where the transition occurs from laminar to turbulent flow. • Flow in the drill pipe is generally turbulent. • The equations for critical velocity in the pipe and in the annulus are listed below.
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Hydraulics ydraulics – Critica ritic al Velocity locit y Formulas ormulas • Critical Pipe Velocity Vcp (ft/min)= ((38727 x K p)/r)(1/(2-n)) x ((1.6/D)x((3n+1)/4n) (n(2-n))
• Critical Pipe Flow rate Qca (gpm)= VcpD2 24.51
• Critical Annular Velocity Vca (ft/min)= ((25818 x K a)/r)(1/(2-n)) x ((2.4/(D2-D1))x((3n+1)/4n) (n(2-n))
• Critical Annular Flow rate Qca (gpm)= Vca(D2-D1)2 24.51
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Hydraulic ydraulics s – Pressure ressure losse loss es • The circulating system of a drilling well is made up of a number of components or intervals, each with a specific pressure drop. • The sum of these interval pressure drops is equal to the total system pressure loss or the measured standpipe pressure.
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Hydraulic ydraulics s – Pressure ressure losse loss es • The total pressure loss for this system can described mathematically as: PTotal = PSurf Equip + PDrillstring + PBit + PAnnulus
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Hydraulics – Pressure losses • Surface pressure losses include losses between the standpipe pressure gauge and the drill pipe. – This includes the standpipe, kelly hose, swivel, and kelly or top drive. – To calculate the pressure loss in the surface connections, use the API pipe formula for pressure loss in the drill pipe.
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Hydraulics – Pressure losses • Top Drive Surface connections – There is no current standard case for top drive units. – The surface connections of most of these units consist of an 86-ft standpipe and 86 ft of hose with either a 3.0- or 3.8-in. ID. In addition, there is an “S” pipe that is different on almost every rig.
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Hydraulics – Pressure losses • Drill String Pressure losses – The pressure loss in the drillstring is equal to the sum of the pressure losses in all of the drillstring intervals, including drill pipe, drill collars, mud motors, MWD/LWD/PWD or any other downhole tools.
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Hydraulics – Pressure losses • Friction Factor – Before calculating the pressure loss, the Fanning friction factor (f p) is calculated next with different equations being used for laminar and turbulent flow. – This friction factor is an indication of the resistance to fluid flow at the pipe wall. – The friction factor in these calculations assumes a similar roughness for all tubulars.
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Hydraulics – Pressure losses • Formulas for friction Factor – If the Reynolds number is less than or equal to 2,100: f p= 16 NRe
p
– If the Reynolds number is greater than or equal to 2,100 f p=((log n + 3.93)/50) NRe
p
((1.75 – log n)/7)
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Hydraulics – Pressure losses • Pipe Interval Pressure loss – Drillstring (including drill collars) intervals are determined by the ID of the pipe. – The length of an interval is the length of pipe that has the same internal diameter. – The following equation is used to calculate the pressure loss for each drillstring interval.
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Hydraulics – Pressure losses • Formula for Pipe Pressure loss • Pp (psi) = fpVp2r x L 92,916D • Vp = Velocity (ft/min) • D = ID pipe (in.) • r = Density (lb/gal) • L = Length (ft)
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Hydraulics – Pressure losses • Pressure loss for motors and tools – If the drillstring contains a downhole motor; an MWD, LWD or PWD tool; a turbine or a thruster, their pressure losses must be included in the system pressure losses when calculating the system’s hydraulics. – These pressure losses can significantly change the pressure available at the bit, as well as bypass flow around the bit.
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Hydraulics – Pressure losses • Pressure loss for motors and tools – The pressure loss through MWD and LWD tools varies widely with mud weight, mud properties, flow rate, tool design, tool size and the data transmission rate. – Some manufacturers publish pressure losses for their tools but these pressure losses can be conservative, because they are usually determined with water .
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Hydraulics – Pressure losses • Pressure loss for motors and tools – The pressure loss across motors and turbines cannot be accurately determined by formula, but, again, this pressure loss data is available from the suppliers.
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Hydraulics – Pressure losses • Pressure loss at the bit – Regular nozzle type bit Pbit =
156rQ2 (D2n1 + D2n2 + D2n3 + …)2
– Diamond type coring bit Pbit =
rQ2 10,858(TFA) 2
– r = Density (lb/gal) – Q = Flow ratio (gpm) – TFA = Total Flow Area (in.2)
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Hydraulics – Pressure losses • Pressure loss in the annulus – The total annular pressure loss is the sum of all of the annular interval pressure losses. – Annular intervals are divided by each change in hydraulic diameter. – A change in drillstring outside diameter and/or a change in casing, liner or open hole inside diameter would result in a hydraulic diameter change. – As with the drillstring pressure loss equations, the friction factor must first be determined before calculating the pressure loss for each annular section.
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Hydraulics – Pressure losses • Pressure loss in the annulus – If the Reynolds number is less than or equal to 2,100: f a= 16 NRe
a
– If the Reynolds number is greater than or equal to 2,100 f a=((log n + 3.93)/50) NRe
a
((1.75 – log n)/7)
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Hydraulics – Pressure losses • Pressure loss in the annulus – The pressure loss for each interval must be calculated separately and added together for the total annular pressure loss.
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Hydraulics – Pressure losses • Pressure loss in the annulus – This equation is used to calculate the individual interval pressure losses. Pa (psi) = f aVa2r
x Lm
92,916D2 - D1) – Va = Velocity (ft/min) – D2 = ID hole or casing (in.) – D1 = OD Drill pipe or collars (in.) – r = Density (lb/gal) – Lm = Length (ft)
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Hydraulics – ECD • Equivalent circulating density (ECD) – The pressure on a formation while circulating is equal to the total annular circulating pressure losses from the point of interest to the bell nipple, plus the hydrostatic pressure of the mud. – This force is expressed as the density of mud that would exert a hydrostatic pressure equivalent to this pressure.
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Hydraulics – ECD • Formula for ECD ECD (lb/gal) =
Pa (psi) 0.052 x TVD (ft)
• Excessive ECD may cause losses by exceeding fracture gradient on a well. • It is important to optimize rheological properties to avoid excessive ECD.
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Hydraulics – Bit Hydraulics • In addition to bit pressure loss, several other hydraulics calculations are used to optimize the drilling performance. • These include hydraulic horsepower, impact force and jet velocity calculations.
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Hydraulics – Bit Hydraulics • Hydraulic Horsepower (hhp) – The recommended hydraulic horsepower (hhp) range for most rock bits is 2.5 to 5.0 Horsepower per Square Inch (HSI) of bit area. – Low hydraulic horsepower at the bit can result in low penetration rates and poor bit performance.
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Hydraulics – Bit Hydraulics • Hydraulic Horsepower (hhp) – The bit hydraulic horsepower cannot exceed the total system hydraulic horsepower. hhpb = QPBit 1,740 – Q = Flow rate (gpm) – PBit = Bit pressure loss (psi)
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Hydraulics – Bit Hydraulics • Hydraulic Horsepower (hhp) – Hydraulic Horsepower per square inch HSI = 1.27 x hhpb Bit Size2 – Bit Size = Bit diameter (in.)
– System Hydraulic Horsepower hhpSystem = PTotalQ 1,714 – PTotal = Total system pressure losses (psi) – Q = Flow rate (gpm)
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Hydraulics – Bit Hydraulics • Nozzle Velocity – Although more than one nozzle size may be run in a bit, the nozzle velocity will be the same for all of the nozzles. – Nozzle velocities of 250 to 450 ft/sec are recommended for most bits. – Nozzle velocities in excess of 450 ft/sec may erode the cutting structure of the bit.
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Hydraulics – Bit Hydraulics • Nozzle Velocity Vn (ft/sec) =
417.2 x Q D2n1 + D2n2 + D2n3 + …
– Q = Flow rate (gpm) – Dn = Nozzle diameter (32nds in.)
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Hydraulics – Bit Hydraulics • Percent pressure drop at the bit – It is generally desired to have 50 to 65% of surface pressure used across the bit. %PBit = PBit x 100 PTotal
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Hydraulics – Bit Hydraulics • Hydraulic impact force (IF) IF (lb) = VnQr 1,930 – Vn = Nozzle velocity (ft/sec) – Q = Flow rate (gpm) – r = Density (lb/gal)
– Impact force per inch squared IF (psi) = 1.27 x IF (lb) Bit Size2
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Hydraulics – Surge and Swab Pressures • Swabbing – When the drillstring is picked up to make a connection or trip out of the well, the mud in the annulus must fall to replace the volume of pipe pulled from the well. – The hydrostatic pressure is momentarily reduced while the mud is falling in the annulus.
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Hydraulics – Surge and Swab Pressures • Surging – When the drillstring or casing is lowered or run into the well, mud is displaced from the well. – The frictional pressure losses from the flow of mud in the annulus as it is displaced by the pipe causes pressures in excess of the hydrostatic pressure of the column of mud in the wellbore.
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Hydraulics – Surge and Swab Pressures • Swab and surge pressures are related to the mud’s rheological properties: – The mud’s gel strengths – The speed at which the pipe is pulled from, or run into, the well – The annular dimensions – The length of drillstring in the well
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Hydraulics – Surge and Swab Pressures • The rheological properties affect swab and surge pressures in the same manner as they affect annular pressure losses. • Increases in either the plastic viscosity or the yield point will increase the swab and surge pressures.
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Hydraulics – Surge and Swab Pressures • Since the maximum (not average) swab and surge pressures must be less than the pressures needed to swab the well in or break the formation down, swab and surge pressures must be calculated for the maximum drillstring velocity when tripping. • This is generally calculated as one-and-one-half times the average drillstring velocity. VMaxDrillstring (ft/min per stand) = 1.5 x stand length (ft) x 60 sec/min seconds per stand
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Hydraulics – Surge and Swab Pressures • The annular velocity must be calculated for each annular space. • These annular velocities should be substituted into the API equations for the annular pressure losses for each interval. • The swab and surge pressures are then calculated in the same manner as the ECD.
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Hydraulics – Surge and Swab Pressures AVSwab-Surge (ft/min) = VMaxDrillstring (ft/min) x drillstring displacement (bbl/ft) annular capacity (bbl/ft)
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Hydraulics – Surge and Swab Pressures • The object of calculating swab and surge pressures is to determine safe pulling and running speeds and minimized trip times. • This is done by changing the maximum or minimum time per stand and recalculating the swab and surge pressures until times per stand are found where the swab and surge pressures plus the hydrostatic pressure is approximately equal to the formation pressure and fracture pressure. • This time per stand is only relevant for the present length of drillstring in the well.
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Hydraulics – Surge and Swab Pressures • As pipe is removed from the hole, the drillstring length decreases and the bottom hole assembly will be pulled into large diameter casing. • This will make it possible to pull each stand faster without risk of swabbing in the well. • When tripping in to the well, the length of drillstring will be increasing and the annular spaces will decrease as the BHA is run into smaller diameters. • This will require that the running time per stand be increased to avoid fracturing the formation. • The swab and surge pressures should be calculated at either 500- or 1,000-ft intervals.
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Hydraulics – Slip Velocity • Slip Velocity – Free settling occurs when a single particle falls through a fluid without interference from other particles or container walls
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Hydraulics – Slip Velocity • For Slip velocity we use stokes law VS = gC DS2 (r S - r L) 46.3μ – VS = Slip or settling velocity (ft/sec) – gC = Gravitational constant (ft/sec 2) – DS = Diameter of the solid (ft) – r S = Density of solid (lb/ft 3) – r L = Density of liquid (lb/ft 3) –
μ
= Viscosity of liquid (cP)
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Hydraulics – Slip Velocity • This equation is a mathematical expression of events commonly observed: – The larger the difference between the density of the cutting and the density of the liquid the faster the solid will settle. – The larger the particle is the faster it settles – The lower the liquid’s viscosity (1/μ), the faster the settling rate.
