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# Fluid mechanics for drilling

The three primary functions of a drilling fluid depend on the flow of drilling fluids and the pressures associated with that flow. These functions includes: The transport of cuttings out of the wellbore, prevention of fluid influx, and the maintenance of wellbore stability. If the wellbore pressure exceeds the fracture pressure, fluids will be lost to the formation. If the wellbore pressure falls below the pore pressure, fluids will flow into the wellbore, perhaps causing a blowout. It is clear that accurate wellbore pressure prediction is necessary. To properly engineer a drilling fluid system, it is necessary to be able to predict pressures and flows of fluids in the wellbore. The purpose of this page is to describe the calculations necessary to predict the flow performance of various drilling fluids for the variety of operations used in drilling and completing a well.

## Overview

Drilling fluids range from relatively incompressible fluids, such as water and brines, to very compressible fluids, such as air and foam. Fluid mechanics problems range from the simplicity of a static fluid to the complexity of dynamic surge pressures associated with running pipe or casing into the hole. This page first presents a general overview of one-dimensional (1D) fluid flow so that the common features of all these problems can be studied. The wellbore flow problems listed below are examined in detail, starting from the simplest and progressing to the most complicated. These problems are considered in the following order:

1. Static incompressible fluids
2. Static compressible fluids
3. Circulation of incompressible fluids
4. Circulation of compressible fluids

## Governing equations

A complete fluid mechanics analysis of wellbore flow solves the equations of mass, momentum, and energy for each flow stream and the energy equation for the wellbore and formation. In the usual treatment of these equations, the mass conservation equations are not stated explicitly, and the temperatures are given, rather than calculated from the energy equation. Here, the whole problem is set out, with appropriate assumptions made for special, simplified cases as they are considered.

The flow streams are treated as 1D constant area flow but with recognition of the effects of discontinuous area changes, such as nozzles. The assumption of 1D flow means that the flow variables, such as density, velocity, viscosity, etc., are given as their average values over the cross-sectional area of the flow stream. For instance, for flow in a tube, the frictional pressure drop is formulated in terms of the average velocity, density, and viscosity. The equations of mass and momentum conservation are solved subject to the assumption of steady flow. This assumption is that time variations of all variables are neglected in a time increment. In particular, this means that mass accumulation effects are not considered in the mass balance equation, and that velocity is only a function of position in the momentum balance equation. Fully dynamic momentum equations are considered in a later section.

The balance equations are written in a control volume form. The equations are written as integrals over a specified volume with specified surface areas rather than as partial differential equations at a point. For these flow streams, the volume under consideration has cross-sectional area A and length Δz . The surface areas are the circular or annular cross-sectional areas, A, of the ends and the cylindrical lateral surface area. In the solution of these equations, only the entrance and exit values of the flow variables are calculated. To evaluate the integrals, the variation of variables between the entrance and exit of each space increment may be needed. The usual assumption used in wellbore calculations is that density, velocity, viscosity, and thermal conductivity are constant and equal to entrance conditions through the increment, and that pressure and temperature vary linearly between entrance and exit. Experience has shown these assumptions to be good except for compressible flow. For most cases, the increased accuracy from other interpolation methods does not justify the computation penalty. On the other hand, none of these calculations are so numerically intensive that they cannot be done with more accurate integration methods on a personal computer, and some of these methods are mentioned later in the text.

### Single-phase flow

The balance of mass for single-phase flow is given by ....................(1)

where

ρ = density, kg/m3 ,

= mass flow rate, kg/s,

v = average velocity, m/s,

and

A = area, m2,

where steady flow (time independent flow) has been assumed. By Eq. 1 , the mass flow rate in any flow stream is constant. In other words, the rate of fluid flow into a volume equals the rate of fluid flow out of the volume. Note that this relation does not change with area changes. However for nonsteady-state flow, we find that mass can accumulate in the volume so that flow out does not necessarily equal flow in.

The balance of momentum for single-phase flow has the form ....................(2)

where

P = pressure, Pa;

g = acceleration of gravity, m/s2;

Φ = angle of pipe with vertical;

f = Fanning friction factor;

D hyd = hydraulic diameter, m;

and

Δ z = length of flow increment, m,

where steady flow has again been assumed. The Δv term is called the fluid acceleration and is important only for compressible fluids. The ρg term is the fluid weight term, which is positive for flow downward. The ρv2 term is the fluid friction term. The Fanning friction factor f depends on the:

• Fluid density
• Velocity
• Viscosity
• Fluid type
• Pipe roughness

Appropriate models for f, considering a variety of different fluid types, are considered in detail in the section on rheology. The sign of the friction term is counter to the flow direction (e.g., negative for flow in the positive direction). For area changes, the following relation holds ....................(3)

Eq. 3 simplifies for incompressible flow and is written as ....................(4)

where subscript o indicates upstream properties, and subscript 1 indicates downstream properties. The quantities Cd are the discharge coefficients for the flow through an area change. Exact treatment of the effect of area change on momentum would have Cd equal to 1. However, real flow is not 1D through area changes, so a factor is needed to account for the real three-dimensional (3D) flow effects. Flow into a smaller area results in a reversible pressure drop plus an irreversible pressure drop. Flow into a larger area results in a reversible pressure increase plus an irreversible pressure drop. Thus, the values of Cd are different for flow into a restriction (reduced area) and flow out of a restriction (increased area). The following values of Cd are typical:

Cd = .95 into reduced area.

Cd = 1.00 into increased area.

The basic balance of energy equation for single-phase flow is given by ....................(5)

where

ε = internal energy, J/kg;

Φ = viscous dissipation, W;

Q = heat transferred into volume, W;

Θ = rate of volume energy added, W;

and

This equation is given in the fully transient form because temperature variation with time may be significant and because steady-state temperatures are usually not achieved in typical wellbore hydraulic operations. The viscous dissipation term Φ depends on the fluid friction model. The term Q is usually written as the total heat flux into the control volume. An example of Θ would be the heat of hydration for cement. Eq. 3 can be rewritten in terms of enthalpy h. ....................(6)

and by choosing pressure and temperature as independent variables can be further rewritten as ....................(7)

Cp = = heat capacity at constant pressure, J/kg-K.

β = = coefficient of thermal expansion, 1/K

T = absolute temperature, °K.

At area changes, the following relation holds ....................(8)

## Static wellbore pressure solutions

Static wellbore pressure solutions are the easiest to determine and are the most suitable for hand calculation. Because velocity is zero and no time dependent effects are present, we need only consider Eq. 2 with velocity terms deleted. ....................(9)

Temperatures are assumed to be static (often the undisturbed geothermal temperature) and known functions of measured depth.

### Constant density

The simplest version of Eq. 9 is the case of an incompressible fluid with constant density ρ. ....................(10)

where ΔZ is the change in true vertical depth (TVD) (i.e., hydrostatic head). For constant slope Φ, ΔZ equals cos Φ Δz. For a slightly compressible fluid, such as water, Eq. 9 could be used for small ΔZ increments where temperature and pressure values do not vary greatly.

### Compressible gas

To show a somewhat more complicated static pressure solution, consider the density equation for an ideal gas: where T is absolute temperature, and R is a constant. For an ideal gas, density has an explicit dependence on pressure and temperature. The solution to Eq. 9 for a well with constant slope Φ is ....................(11)

where the initial condition for P is Po . T(z) is a given absolute temperature distribution, and z is the measured depth. For constant T, we see that the pressure of an ideal gas increases exponentially with depth, while an incompressible fluid pressure increases linearly with depth.

## Flowing wellbore pressure solutions

The next level of complexity in hydraulic calculations is the steady flow of the wellbore fluids. One part of this complexity is the calculation of the Fanning friction factor, f. This subject is postponed to the section on rheology (See Fluid Rheology), and f is assumed to be known in the following calculations.

### Constant density

The simplest version of Eq. 2 for flowing fluids is the case of an incompressible fluid with constant density ρ and rheological properties, and a constant-slope wellbore. ....................(12)

In this case, we have evaluated the Fanning friction factor from the appropriate equation in the rheological section. Note that Δv is zero through the mass conservation equation.

### Linearly varying density

The next simplest solution has a linearly varying density along z. Conservation of mass requires that ....................(13)

where the o subscript indicates initial values, and no subscript indicates final values. With Eq. 13 , we can calculate Δv in terms of Δρ . ....................(14)

Assuming a linear variation in density, constant wellbore angle, and constant Fanning friction factor, the pressure drop equation gives ....................(15)

Eq. 15 may be used directly to calculate ΔP if the final density is insensitive to pressure. Otherwise, this equation must be solved numerically for the pressure. For instance, the density terms could be linearized with respect to P and Newton ’ s method used to converge to the final pressure.

### Compressible fluid

The flow of a compressible fluid can often produce results that seem counter to intuition. For example, consider the steady flow of air in a constant area duct. As with all fluids, there is a pressure loss because of friction, and the pressure decreases continuously from the entrance of the duct to the exit. Unlike the flow of incompressible fluids, the fluid velocity increases from the entrance of the duct to the exit. How could friction make the fluid speed up?

Two facts account for this acceleration:

• First, the gas pressure is proportional to the density (as in the ideal gas law P = ρRT). As the pressure of the gas decreases, the density must decrease also.
• Second, because the mass flow through the duct is constant, the product of density and velocity is constant. Thus, as the density decreases with the pressure, the velocity must increase to maintain the mass flow.

This example demonstrates a typical compressible flow characteristic—the interrelationship of pressure and mass flow. In air drilling, high velocities are needed at bottom of hole to remove the cuttings. High velocities result in friction pressure drops in the drillpipe and annulus, so higher standpipe pressures may be needed to keep the air flowing. Higher standpipe pressures result in higher gas densities and, hence, result in lower velocities. Fortunately, most air drilling operations do not result in the vicious circle situation previously described.

The pressure change in a flowing gas is properly a problem in gas dynamics. Gas dynamic analytic solutions are available for two cases of flow with friction: adiabatic and isothermal flow. Neither case is exactly what we need for these calculations, but the reader is directed to a gas dynamics reference for additional depth of understanding of this problem. If we assume a linearly varying density and temperature, we can use the results of the previous section with the addition of a pressure/volume/temperature (PVT) relationship. For most applications, an ideal gas model is sufficiently accurate. With the relation P = ρRT, we can calculate ΔP in terms of density as ....................(16)

Substituting Eq. 16 into Eq. 15, we derive the quadratic equation for density. ....................(17)

If there are two positive roots for the density, the root that gives a subsonic velocity is the correct root. The speed of sound for an ideal gas is ....................(18)

where cp and cv are the heat capacities at constant pressure and volume, respectively.

## General steady flow wellbore pressure solutions

We make only one assumption in this general discussion of wellbore flow modeling, and that is that the temperature distribution is given. To make any other assumption requires a general solution of the energy equation for the wellbore, which is beyond the scope of this chapter. For review, we repeat the balance of mass and momentum for 1D flow along a constant area duct. ....................(19) ....................(20)

At changes of area, the following conditions hold. ....................(21)

and ....................(22)

Eq. 22 simplifies for incompressible flow. ....................(23)

where subscript o indicates inlet properties.

Given that we have a means of calculating the Fanning friction factor, we need a PVT relationship relating pressure, density, and temperature, and what is often available is a pressure function P(ρ,T) depending on density and temperature. When this is substituted into Eq. 20 , with Eq. 19 , we obtain the first-order differential equation in density. ....................(24)

Eq. 24 can be integrated numerically using any of several methods, such as adaptive Runge-Kutta or Bulirsh-Stoer. The reader is referred to numerical analysis books for details of these two methods. Once the new density has been determined, the pressure can be calculated from the PVT relationship, and the velocity can be calculated from Eq. 19. Alternately, we might have density as a function of pressure and temperature: ρ(P,T ). In this case, the differential equation is in terms of pressure. ....................(25)

Again, this is a first-order differential equation but now in terms of pressure. Once the pressure is determined, density is determined from the PVT relationship, and the density together with Eq. 19 gives the velocity.

Flow-area changes may act like nozzles (area reduction) or diffusers (area increases). The actual calculation of flow-property changes is beyond the capability of a 1D flow analysis. Often, we insert coefficients into the 1D equations to account for the complexity, and then we evaluate these coefficients from experiments. One such coefficient is the discharge coefficient shown in Eqs. 22 and 23 . A further comment is needed about the general Eq. 22 . Some assumption must be made about the variation of temperature with pressure within the nozzle before the integral can be evaluated. A typical assumption is that the flow is adiabatic (i.e., negligible heat transfer takes place in the nozzle). For our purposes, adiabatic is equivalent to isentropic, and entropy functions are available for many fluid models. For isentropic flow, ....................(26)

Or, we can derive the change of temperature with pressure. ....................(27)

For an ideal gas, we can solve Eq. 27 to get ....................(28)

where k = cp/cv . Using the ideal gas law, we can eliminate T. ....................(29)

We can now express pressure in terms of density in Eq. 22 and express v in terms of density with Eq. 21. The resulting equation can be solved numerically for density. Then, Eq. 29 can be used to determine the pressure.

## Calculating pressures in a wellbore

Assuming we can calculate ΔP for each constant area section of drillpipe or annulus and can calculate ΔP for nozzles and area changes, we are now ready to evaluate the pressures in a wellbore. A typical wellbore fluid system is illustrated in Fig. 1. Summing all pressure drops give the standpipe pressure. ....................(30)

In this calculation, we assume that the calculations are started from a known pressure value, most conveniently the atmospheric pressure at the exit of the annulus. This choice is particularly suitable if air or foam drilling is being considered because "choked" gas flow almost never occurs. For this choice of "boundary condition," flow calculations proceed backward from the annulus exit to the standpipe pressure. For flow in the annulus, both fluid density and fluid friction increase pressure going down the annulus. Where fluid type changes, the pressure and flow velocity are continuous.

and ....................(31)

Notice that mass flow rate may not be continuous at the interface between two fluids because the densities may be different: ρ1vAρ2vA, where v and A are continuous at the interface. When calculating from the bit to the standpipe, inside the drillstring, fluid density decreases pressure and fluid friction increases pressure. Pressure changes because of internal upsets and tool joints consist of two area changes and a short flow section, as shown in Fig. 2.

Pressure drop across the bit consists of two area changes: into the nozzles and exit from the nozzles into the openhole annular area. Miscellaneous pressure drops are drops through tools, mud motors, floats, or in-pipe chokes. Sometimes, the manufacturer will have this pressure-loss information tabulated; otherwise, you will have to estimate the pressure loss through use of the tool internal dimensions.

If the standpipe pressure is given, then the flow exiting the annulus must be choked back to atmospheric pressure. ....................(32)

## Nomenclature

 a = acoustic velocity, m/s αvs , bvs = constants that include the viscometer dimensions, the spring constant, and all conversion factors A = flow area (see subscripts), m2 c = average concentration of cuttings overall ca = cuttings concentration in annular region co = feed concentration of cuttings cp = cuttings concentration in plug region C = compressibility Cd = discharge coefficients for the flow through an area change, dimensionless CD = drag coefficient, dimensionless Ce = pressure drop correction factor for pipe eccentricity, dimensionless Cp = heat capacity at constant pressure, J/kg-K Cv = heat capacity at constant volume, J/kg-K ds = particle diameter, m dv/dr = velocity gradient, s –1 dv/dt = total derivative of velocity with respect to time, Pa/s Dh = wellbore diameter, m ƒ = Fanning friction factor, dimensionless g = acceleration of gravity, m/s2 G = mass flow rate density of mixture, kg/m3–s Gs = mass flow rate density of solids, kg/m3–s h = specific enthalpy, J/kg h = total friction pressure drop, Pa/m He = Hedstrom number Hs = holdup of solid particles, volume fraction of solids k = absolute pipe roughness, m k = cp/cv K = consistency index for pseudoplastic fluid, Pa-sn Kb = elastic bulk modulus, Pa L = length of viscometer bob, m m = power-law exponent for Herschel-Bulkley fluids ṁ = mass flow rate, kg/s ṁs = mass flow rate of solid, kg/s MT = torque measured by viscometer, N-m n = power law exponent for pseudoplastic fluids pn = pressure in bit nozzle, Pa pr = pressure in bit annular area, Pa P = pressure, Pa patm = atmospheric pressure, Pa Q = heat transferred into volume, W ri = inside radius of annulus, m ro = outside radius of annulus, m R = ideal gas constant, m3 Pa/kg-K S = entropy, J/K t = time, s T = absolute temperature, °K u = radial displacement, m v* = characteristic velocity for turbulent flow calculations, m/s v = average velocity, m/s W = buoyant weight or particle, N Ws = buoyant weight of the cuttings, N x = parameter in settling velocity equation y = parameter in settling velocity equation Ya = parameter in settling velocity equation z = measure depth, ft Z = true vertical depth, ft αc = parameter in Bingham fluid friction factor β = = coefficient of thermal expansion, 1/K γ = shear stress, s–1 γe = equivalent shear stress, s–1 δ = ratio of the average particle volume to its cross-sectional area δr = the distance between centers of the inside and outside pipes, m ΔP = pressure drop, Pa Δt = time increment, s Δv = change in velocity, m/s Δz = length of flow increment, m ε = internal energy, J/kg ζ = measured depth integration variable, m θ = viscometer reading, degrees ϑ = integration variable κ = ratio of radius of inner cylinder to outer cylinder λP = Dplug/Deq , the plug diameter ratio μ = Newtonian viscosity of the fluid, Pa-s μa = apparent viscosity, Pa-s μp = plastic viscosity, centipoise ξ = integration variable corresponding to depth z, m ρ = fluid density, kg/m3 = fluid in-mixture density, kg/m3 ρƒ = fluid density in solid/fluid mixture, kg/m3 ρs = solid density in solid/fluid mixture, kg/m3 = solid in-mixture density, kg/m3 τ = shear stress, Pa τw = wall shear stress, Pa τy = yield point, Pa υƒ = Poisson’ s ratio for the formation Φ = angle of inclination from the vertical Φ = viscous dissipation, W Ψ = sphericity ω = rotor speed, rev/min Ωo = angular velocity of outer cylinder

Subscripts

c = concentric

e = eccentric

n = properties in bit nozzle, surge calculations

o = upstream, initial, or inlet

r = properties in annulus outside bit, surge calculations

Superscripts

- = upstream properties