Equations: Drilling fluids and hydraulics

any consistent units: densities in g/cm³ (t/m³), volumes in m³, mass in tonnes

Holds when: The volumes of the mixed materials add up: $\rho_2 (V_1 + V_{\text{wm}}) = \rho_1 V_1 + \rho_{\text{wm}} V_{\text{wm}}$. This is where the formula comes from. The weighting material is dry and does not dissolve. In practice a little water is often added with it to wet the particles, and then slightly more weighting material is needed. No mud is lost and none is diluted during weighting up.

oilfield: $\eta_p$ in mPa·s, $\tau_0$ in lb/100 ft²; the formulas are valid for the standard geometry of the instrument

Holds when: The flow is described by the Bingham model from the two highest rotation speeds. At low shear rates real muds usually deviate from this line. The measurement is made at a stated temperature, for water-based muds usually about 50 °C (120 °F). Viscosity depends strongly on temperature. The formulas are a convention accepted in field practice. They leave out small constants of the instrument, so we compare values obtained in the same way.

any identical units of density

Holds when: The mass of the bubbles is negligible compared with the mass of the liquid and the solids. In the pressurised balance the gas is squeezed to a negligible volume, and the compression of the mud itself can be neglected. Both samples are taken at the same place and at the same time.

$V$ in cm³, $t$ in min; the expression for $C$ is written in consistent SI units (m³, m², Pa, Pa·s, s)

Holds when: It is a static test: the mud above the cake does not move and does not erode it, and the cake only grows. The cake is incompressible, and its composition and permeability are constant. In real cakes the permeability falls as the pressure difference grows. The viscosity of the filtrate and the temperature are constant during the test.

SI (any consistent system)

Holds when: The same as for static filtration: the cake is incompressible and uniform in thickness. The resistance of the paper and the screen is small compared with the resistance of the cake. The thickness is measured after the liquid mud above the cake has been removed. An error in the thickness goes straight into an error in $k_{mc}$.

densities in g/cm³, shares as fractions (or all shares in per cent)

Holds when: The volumes of the components add up. The solids consist of two groups with known densities. If the weighting material is hematite or calcium carbonate, its density is used. The mud is fresh. Otherwise the share of water is replaced by the share of brine (the correction is in the advanced section). There is no air in the sample.

volumes in cm³, $c_{\text{bent}}$ in kg/m³

Holds when: All the active clay of the sample is counted as bentonite with a cation exchange capacity of about 70 meq/100 g. The coefficient comes from this. The organic substances that take up the dye have been oxidised with the peroxide. The dye is 0.01 N: 1 cm³ carries 0.01 meq of the dye cation.

$c_{\text{lime}}$ in kg/m³; $P_m$, $P_f$ in cm³ per cm³

Holds when: All the difference between the alkalinity of the mud and of the filtrate is given by lime (carbonates and other alkaline particles make it too high). The filtrate has the same composition as the water inside the mud.

mg/L; volumes in cm³

Holds when: Silver nitrate is 0.0282 N: 1 cm³ precipitates 1 mg of chloride. For salty muds 0.282 N is used, and the coefficient becomes 10,000. There are no other ions in the filtrate that silver precipitates: bromides, iodides, sulfides.

$\rho_b$ in g/cm³, $c_s$ in g/L

Holds when: The density of the water in the retort is 1.0 g/cm³. There is one salt, and it is known (usually NaCl). For a mixture of salts we need reference data on the density of the brine.

kg/m³ = g/L; capacities in meq/100 g

Holds when: The dye is 0.01 N: $\mathrm{MBC}$ cm³/cm³ means $10\,\mathrm{MBC}$ meq per litre of mud, and 1 kg/m³ of clay with a capacity $E$ takes up $E/100$ meq per litre. Sand and carbonates among the low-gravity solids take up almost no dye. If their share is large, it is included in the $E_d$ of the cuttings. The sample of cuttings is representative of the fine drilled rock in the mud.

the first form is SI; the second has $I$ in mol/L, $\kappa^{-1}$ in nm, water at 25 °C

Holds when: The Gouy–Chapman theory for a flat surface in a dilute solution: the ions are point charges, and the water is a continuous medium. At an ionic strength above about 0.1 mol/L the formula gives only the order of magnitude. A real layer is never thinner than a hydrated ion, so for brines the formula understates the thickness. The constant 0.304 is for water at 25 °C. At 100 °C the layer is thinner by only about 6%, because the permittivity of water falls as the temperature rises.

$c$ in kg/m³, $Y$ in m³/t, densities in g/cm³

Holds when: The yield was determined for the same viscosity and the same water as needed in the mud. In salt or hard water the yield of bentonite falls sharply. The volumes of clay and water add up.

mole ratios: one mole of reagent per mole of calcium

Holds when: The reactions go to completion, and the precipitate does not dissolve. There are no other consumers of the reagent in the mud: magnesium, or an excess of lime beyond the calculation.

$m$ in kg, $\Delta C$ in mg/L, $V_w$ in m³

Holds when: The reaction goes mole for mole, as for calcium with soda ash or bicarbonate. The ion is spread through all the water of the mud, and the reagent is used only on it. The undissolved reserve of the contaminant (the lime of cement, lumps of anhydrite) is counted separately: it goes on dissolving.

ratios of concentrations

Holds when: The limiting case of the theory, for a high surface potential. At a moderate potential the exponent is smaller. Special effects are not taken into account: bridges between particles, exchange of cations and recharging of the edges.

SI; at 25 °C $RT/\bar V_w \approx 137$ MPa, natural logarithm

Holds when: An ideal membrane. In clays in water-based muds its efficiency is small, a few per cent, and the real force is smaller by the same factor. In time the ions penetrate the clay, and the difference of activities evens out.

dimensionless; in per cent, multiply by 100

Holds when: By established practice, the ratio is calculated from the water from the retort, not from the brine. Some programmes set the ratio “base oil — brine”, and then the water is recalculated into brine (the formulas are below). No formation oil or condensate gets into the retort. Otherwise it is counted as base oil.

mixed: chlorides in mg/L, mass of water in mg per litre of mud

Holds when: All the chlorine is in calcium chloride. If rock salt has got into the brine, we need the calcium of the mud too, to separate CaCl₂ from NaCl. All the salt is dissolved. With oversaturation, undissolved crystals also give chlorides, and the salinity comes out too high.

densities in g/cm³, shares as fractions

Holds when: In a litre of brine there are $1000\rho_b(1 - w)$ grams of water. This is the same correction as for salty water-based muds (the retort in a salty mud), written through the mass fraction of salt.

volumes in m³, shares as fractions

Holds when: The volumes add up, and the base oil and the water in the mud are not lost during the treatment. Water is added as brine of the same salinity. Otherwise the water activity changes. For $\Delta V_w$ cubic metres of water we need $\Delta V_w\,w/(1 - w)$ tonnes of salt. The formulas follow from the condition $(V f_o + \Delta V_o)/(V f_o + \Delta V_o + V f_w) = R$ and a similar one for water.

MPa; natural logarithm

Holds when: An ideal membrane. A real one also lets part of the ions pass, and the acting difference is smaller. But in an OBM the membrane is more efficient than in water-based muds. This is the difference of the osmotic pressures from the lesson on water-based muds: $\Pi = -(RT/\bar V_w)\ln a_w$ for each liquid.

kg/m³; in oilfield units $1.295\,P_{om}$ lb/bbl

Holds when: All the alkali of the sample is lime (its equivalent mass is 37.05 g): $0.1$ eq/L $\times\ 37.05$ g/eq $= 3.7$ g/L. The acid is five times stronger than in the test of a water-based mud (there it is 0.02 N and the coefficient is 0.742). This is why the coefficient is five times larger.

pressure in MPa (absolute), gas volumes at standard conditions

Holds when: Solubility is proportional to pressure (Henry's law) and does not depend on temperature. In reality the dependence is not linear, and the saturation pressure for a given base oil and gas is taken from laboratory data. Gas dissolves only in the base oil; the brine and the solids do not take it up. The gas and the mud are in equilibrium. In practice, dissolving takes time and mixing.

natural logarithm

Holds when: A solution of one salt. For a mixture of CaCl₂ and NaCl the activity is calculated by special methods or measured. 25 °C. The activity of concentrated brines depends only weakly on temperature, but the electric hygrometer is calibrated, and samples are compared at one temperature.

SI (metres and hours)

Holds when: The pore fluid of the rock is not part of the solids balance. All the destroyed rock goes into the mud. Caving of the walls would add more to it, and lost circulation would carry part of it into the formation.

SI

Holds when: A spherical particle with laminar flow around it: the Reynolds number of the particle, $\rho_f v_s d/\mu$, is below about 1. The liquid is Newtonian (its viscosity is constant). For a drilling mud we take the effective viscosity at the shear rate around the particle, and at rest the gel may hold fine particles altogether. The particles do not hinder each other. In a thick suspension, settling is slower.

any identical units

Holds when: The particles are separated by their settling velocity in the Stokes regime, and the working conditions of the device are the same. In a weighted mud a large particle is surrounded by a mixture of liquid and fine solids, not by water, so the estimate is approximate.

any identical units of volume

Holds when: The content of drilled rock is held constant, at the limit $f_{ds}$. The fine particles in the mud carried away with the waste of the equipment are counted as removed. No mud goes into the formation. Lost circulation carries away part of the rock itself, and it is counted separately. The expression for $V_{\text{new}}$ is true while $V_{\text{dil}}$ is larger than the volume of the new hole. Otherwise the new mud is needed first of all to fill it.

any identical units of volume

Holds when: The water is used only to make the mud. Evaporation and leaks are not taken into account. The share of water in the new mud is the same as in the circulating mud. No mud goes into the formation. In case of lost circulation, that volume is counted separately. The mud that was made, together with what is carried away in the waste, is for simplicity counted as dilution mud. This slightly understates $E$.

dimensionless

Holds when: The form was fitted to tests and describes hydrocyclones and centrifuges. The parameters are found by testing a particular device on a particular mud. In the lesson's figure: the screen has $d_{50c} = 65$ µm, $\alpha = 8$, $R_f = 0$; the desander 45 µm, 3 and 0.05; the desilter 18 µm, 3 and 0.05; the centrifuge 5 µm, 3 and 0.02.

SI. Viscosity is often given in mPa·s, which is numerically equal to the centipoise (cP); 1 Pa·s = 1000 mPa·s

Holds when: The flow is laminar: the layers do not mix. The temperature and the pressure are constant. The viscosity depends on them, but for a Newtonian fluid it does not depend on the shear rate.

SI. In reports the PV is given in mPa·s (cP), and the YP in lb/100 ft² or in dPa (decipascals; 1 lb/100 ft² = 0.4788 Pa = 4.788 dPa)

Holds when: At $\tau \le \tau_0$ the mud does not flow. The straight line is drawn through the readings at high shear rates. At low shear rates real muds usually lie below the line.

SI. In oilfield units $K$ is given in lb·sⁿ/100 ft² (1 lb·sⁿ/100 ft² = 0.4788 Pa·sⁿ)

Holds when: There is no yield stress: as $\dot\gamma \to 0$, the stress tends to zero. $K$ and $n$ depend on the range of shear rates from which they were found.

SI

Holds when: The three parameters are found from the readings at several shear rates, including low ones.

$\dot\gamma$ in s⁻¹, $N$ in rpm, $\tau$ in Pa, $\theta$ in degrees

Holds when: The instrument has the standard geometry and the standard spring. With another spring, the stress factor is different. The shear rate is calculated for a Newtonian fluid at the surface of the inner cylinder. For strongly shear-thinning muds it is slightly different. The mud is measured at a stated temperature, usually about 50 °C for water-based muds.

$K$ in Pa·sⁿ, logarithms to base 10

Holds when: The power-law curve passes exactly through the two chosen points. For the annulus, where the shear rates are lower, we take the readings at 100 and 3 rpm: $n \approx 0.657\log_{10}(\theta_{100}/\theta_3)$ and $K = 0.511\,\theta_{100}/170.3^{\,n}$. Here $0.657 = 1/\log_{10}(100/3)$, and 170.3 s⁻¹ is the shear rate at 100 rpm.

SI

Holds when: The flow is the laminar flow of a Newtonian fluid. In a shear-thinning mud the shear rate at the wall is greater, by the factor $(3n + 1)/(4n)$ in a pipe and $(2n + 1)/(3n)$ in an annulus. The annulus is replaced by a flat slot of width $(D - d)/2$. This works well when the gap is small compared with the diameter.

SI. The number is dimensionless

Holds when: The limit of 2100 was found for Newtonian fluids in pipes. For muds that thin out under shear, the transition comes at higher Reynolds numbers, and different methods set it differently. The hydraulic diameter is four times the cross-sectional area divided by the wetted perimeter. For a ring-shaped gap it equals $D - d$. Some methods use $0.816\,(D - d)$ for the annulus.

SI

Holds when: It is found by comparing the simplified formulas for laminar losses for the Bingham model and for a Newtonian fluid. The annulus is represented by a flat slot. The pipe is concentric with the hole and does not rotate.

SI

Holds when: The flow is laminar (we check this with the Reynolds number), the annulus is represented by a flat slot of width $(D - d)/2$, and the pipes are concentric and do not rotate. The solution is simplified (a small term of the exact solution is dropped). It overestimates the losses more, the closer the YP is to the stress at the wall. In the example below the error is about 9%. $L$ is the length along the hole. The real walls of an open hole are uneven. There are washouts (places where the hole is wider than the bit made it), and the pipe lies off-centre, close to the wall. All this changes the result.

SI

Holds when: The gel is uniform along the length and breaks away at the same moment over the whole surface. In reality it breaks gradually, so the estimate is too high.

any consistent pressure units

Holds when: The circulation is steady, and the mud has the same density inside the string and in the annulus. There is no downhole motor (a motor at the bottom of the string, turned by the flow of mud) and no MWD tool (a tool that measures while drilling and sends the data up as pressure pulses in the mud) in the bottom hole assembly. If there are any, take their pressure drops from the data sheets and add them to the sum (Deflection tools and downhole motors). The pressure at the outlet of the annulus is atmospheric. If the mud circulates through a choke (an adjustable valve that holds back pressure), add the pressure at the choke to the sum.

SI

Holds when: The flow is steady, the pipes are concentric with the hole and do not rotate, and the temperature is constant. The annulus is replaced by a flat slot of width $(D - d)/2$. This works well when the gap is small compared with the diameter. The formula holds in any regime, but only in laminar flow can we find $\tau_w$ from the rheological model by theory.

SI

Holds when: The flow is laminar (we check this with the Reynolds number). This is a simplified solution: a term of the exact solution is dropped. It is small while $\tau_0$ is much smaller than $\tau_w$. The simplification overestimates the losses. For $\tau_0 = 0$ the formula becomes the Hagen–Poiseuille formula for a Newtonian fluid.

SI

Holds when: The flow is laminar, the pipes are concentric and do not rotate, and the annulus is a flat slot. $K$ and $n$ are found from viscometer readings in the range of shear rates that occurs in this section: for the pipe, usually from 600 and 300 rpm, and for the annulus, from 100 and 3 rpm (model parameters).

SI. The number is dimensionless

Holds when: In the annulus, $(2n+1)/(3n)$ and $12v/(D-d)$ replace $(3n+1)/(4n)$ and $8v/d$, following the slot solution. Some methods represent the annulus as a pipe of diameter $0.816(D-d)$. Then the Reynolds numbers are 18% smaller and the turbulent losses are greater (see the advanced section).

dimensionless

Holds when: These are empirical limits of one of the common methods, that of the American Petroleum Institute (API). Other methods give different but close values. There is no reliable formula in the transition region. It is safer to take the larger of the laminar and the turbulent values of the losses, and the model of this lesson does so.

SI

Holds when: It is valid for any regime if we put in the right $f$. In laminar flow in a pipe, $f = 16/\text{Re}$. For the annulus the choice of the equivalent diameter changes the result: with $0.816(D-d)$ the turbulent losses are about 29% greater than with $D - d$.

dimensionless; the logarithm is to base 10

Holds when: The first formula (Blasius) is for a smooth pipe at $2100 < \text{Re} < 10^5$. The second (Colebrook) is for a rough pipe, and it is solved by trial and error. For a viscoplastic mud in turbulent flow we put in the plastic viscosity. The YP hardly affects the losses there; this is an experimental result. In Russian-language handbooks the Blasius formula is written as $\lambda = 0.3164/\text{Re}^{0.25}$. This is the same thing multiplied by 4.

dimensionless; the logarithm is to base 10

Holds when: A smooth pipe and fully developed turbulent flow (the generalised Reynolds number is above the turbulent limit). It is solved by trial and error, because $f$ is on both sides. A handy explicit approximation $f \approx a/\text{Re}^{\,b}$ with $a = (\log_{10} n + 3.93)/50$ and $b = (1.75 - \log_{10} n)/7$ differs by a few per cent.

SI ($\Delta p/L$ in Pa/m)

Holds when: It is found by putting the Blasius formula into the Fanning equation: $0.241 = 2\cdot 0.0791\cdot(4/\pi)^{1.75}$. A smooth pipe, $2100 < \text{Re} < 10^5$.

SI

Holds when: This is the Bernoulli equation (the energy balance of a flowing fluid) for a short narrowing. The velocity before the nozzle is small compared with the velocity of the jet, friction and the difference of heights are neglected, and the mud is incompressible (it does not shrink under pressure). The pressure drop is the same across all the nozzles, so the velocity of the jet is the same in all of them, and the flow divides in proportion to their areas. It does not depend on the rheology, so it is valid for both Newtonian and non-Newtonian fluids.

SI; in metric units $g h \to 0.00981\,h$ (g/cm³, MPa, m)

Holds when: The well is vertical. In an inclined well the losses are taken along the hole, and the hydrostatic pressure and $h$ along the vertical. The cuttings in the annulus are not counted. They make the column heavier (see the example). The circulation is steady. When the pumps are started, the pressure needed to break the gel is added to the losses (pressure to restart circulation).

SI

Holds when: The end of the string is closed (a float valve, a one-way valve, in the bottom hole assembly), so the whole displaced volume goes through the annulus. With an open string part of the mud passes inside it, and the pressure changes are smaller, so the calculation with a closed end is cautious. The flow is steady and laminar. The acceleration and braking of the string and the breaking of the gel at the start of the movement add short peaks to the calculation. The losses are calculated with the laminar formulas for the annulus, with a model that describes the mud at low shear rates.

SI; natural logarithm

Holds when: Laminar flow of a Newtonian fluid in a concentric ring, with the pipe not rotating.

dimensionless

Holds when: A viscoplastic fluid in a round pipe; for an annulus with the hydraulic diameter it is an approximation. As $\text{He} \to 0$ the critical number tends to 2100, as for a Newtonian fluid.

SI. In metric units MPa × L/s = kW, and it is convenient to give the specific power in W/mm² (the same as MW/m²)

Holds when: All the power released in the nozzles is counted as delivered to the bottom. In fact some of it is lost in the jet on the way, so this is an upper estimate. The specific power is a conventional measure for comparing bits of different diameters, not an exact measure of how well the bottom is cleaned.

SI

Holds when: The jet hits the bottom at right angles and spreads over it, losing all its velocity along its own axis. Inclined jets give less to the bottom, so this is an upper estimate. The link between the velocity and the pressure drop is the Bernoulli equation for a short nozzle with the correction $C_d$, as in the lesson on pressure losses.

any consistent units of pressure and flow rate; natural logarithm (the ratio is the same with a base-10 logarithm)

Holds when: The law is empirical. Each section grows with its own exponent, and $m$ is their weighted average near the measured flow rates. Far from them the law is not accurate (see the advanced section). Both readings are taken at the same depth, with the same mud, in steady circulation and with no weight on the bit. The pressure drops across the downhole motor and the MWD tool, if there are any, are part of the system losses (for the motor, with no weight on the bit).

any consistent units: pressure in MPa, flow rate in L/s, power in kW

Holds when: The pumps work at a constant pressure $p_{\max}$, and their power $p_{\max}Q$ does not go beyond the limit (otherwise, see the advanced section). The system losses follow the law $cQ^m$ with a constant $m$. The nozzles can be chosen for any pressure drop we need. The second derivative is negative, so the point we found is a maximum.

any consistent units

Holds when: The same as for the power criterion; the density of the mud and $C_d$ are constant.

SI

Holds when: The formula for the pressure drop at the bit, turned round to give the area. It is valid for any mud, because the viscosity in the nozzle has almost no effect. All the nozzles work at one pressure drop, and the flow divides between them in proportion to their areas.