Drilling fluid rheology and flow regimes

Quick refresher

Shear stress $\tau$ is the friction force between layers of the fluid per unit of area (Pa). Shear rate $\dot\gamma$ is how fast the velocity changes across the flow (s⁻¹). Viscosity is the ratio of the two. A Newtonian fluid (water, for example) has the same viscosity at every shear rate.
Drilling muds get thinner as the shear rate rises, and they do not flow while the stress is small. They are described by the Bingham model ( $\tau = \tau_0 + \eta_p\dot\gamma$ ), the power-law model ( $\tau = K\dot\gamma^n$ ) and the Herschel–Bulkley model ( $\tau = \tau_y + K\dot\gamma^n$ ).
The standard rotational viscometer (the instrument that measures the flow curve of a mud) gives a shear rate of $1.703\,N$ s⁻¹ at $N$ rpm and a stress of $0.511\,\theta$ Pa, where $\theta$ is the reading in degrees on its dial. From the readings at 600 and 300 rpm we get the plastic viscosity (the slope of the Bingham line), $\theta_{600} - \theta_{300}$ (mPa·s), and the flow behaviour index (it shows how strongly the mud thins out under shear), $n = 3.32\log_{10}(\theta_{600}/\theta_{300})$ .
Shear rates in a well differ by orders of magnitude (by factors of ten and more): tens to hundreds of s⁻¹ in the annulus, hundreds to thousands in the pipe, and tens of thousands in the bit nozzles. We take the model parameters from the readings in the range we need.
The Reynolds number decides the flow regime. It compares the forces that stir the flow with the forces that calm it. For a mud we put the effective viscosity into it (the viscosity of a Newtonian fluid that would give the same laminar pressure loss). For the annulus we use the hydraulic diameter, $D - d$ , in place of the pipe diameter. In the pipe the flow is usually turbulent (swirling), in the annulus usually laminar (smooth).
In the laminar flow in the annulus, the pressure loss and the ECD depend strongly on the yield point (YP: in the Bingham model, the part of the stress that does not change with the shear rate). In the turbulent flow in the pipe, they depend on the flow rate, the density and the diameter, and the rheology has little effect.

Key relationships

\tau = \eta\,\dot\gamma

Newtonian fluid

\tau = \tau_0 + \eta_p\,\dot\gamma

Bingham model (a viscoplastic fluid: it does not flow until the stress passes a threshold)

\tau = \tau_y + K\,\dot\gamma^{\,n}

Herschel–Bulkley model; for $\tau_y = 0$ it is the power-law model

\text{Re} = \dfrac{\rho\,v\,d}{\mu_{\text{eff}}}

Reynolds number; in the annulus the hydraulic diameter $D - d$ replaces $d$

\rho_{ECD} = \rho_m + \dfrac{\Delta p_{\text{fr}}}{g\,h}

equivalent circulating density at depth $h$

Common pitfalls

Comparing the plastic viscosity and the yield point measured at different temperatures. When the mud is heated, its viscosity drops noticeably.
Taking the YP as the stress at which the mud starts to flow. It is the extension of a straight line, and the real stress at which the mud starts to flow is usually noticeably smaller.
Using the power-law parameters found from the 600 and 300 rpm readings to calculate the flow in the annulus, where the shear rates are several times smaller.
Calculating the Reynolds number of the mud with the plastic viscosity instead of the effective viscosity, and, in the annulus, with the diameter of the hole instead of the hydraulic diameter.
Confusing the degrees on the viscometer dial with pascals. One degree is about 0.51 Pa.

After this lesson you can

explain what shear stress, shear rate and viscosity are, and how a non-Newtonian fluid differs from a Newtonian one
write down the Newtonian, Bingham, power-law and Herschel–Bulkley models and explain what their parameters mean physically
calculate the parameters of the models from rotational viscometer readings and convert them to SI and oilfield units
estimate the shear rate in the pipe and in the annulus, calculate the Reynolds number with the effective viscosity and find the flow regime
explain how the rheology and the flow regime affect the friction pressure losses and the equivalent circulating density

Shear stress, shear rate and viscosity

When a fluid flows in a pipe, its layers slide over one another. At the wall the fluid does not move. The speed grows towards the middle of the pipe, and every layer rubs against its neighbours. It is like a stack of paper that you push sideways with your palm: the top sheet moves the farthest, the bottom sheet stays in place, and each sheet drags the next one along.

The friction force between neighbouring layers, divided by the area where they touch, is the shear stress $\tau$ . It is measured in pascals (Pa). How quickly the velocity changes across the flow is the shear rate $\dot\gamma = \dd v/\dd y$ , in s⁻¹. It is the difference between the velocities of two layers, divided by the distance between them. The ratio of the two is the dynamic viscosity. It shows how strongly the fluid resists flowing.

Newton's law of viscous friction

\tau = \eta\,\dot\gamma

where:

$\tau$ shear stress, Pa
$\dot\gamma$ shear rate, s⁻¹
$\eta$ dynamic viscosity of the fluid, Pa·s

Units: 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.

Fluids whose viscosity does not depend on the shear rate are called Newtonian. Examples are water and brines (salt solutions), diesel fuel, and most crude oils without wax. The viscosity of water at 20 °C is about 1 mPa·s.

In other unit systems (1)

Oilfield units

\tau\,[\text{lb/100 ft}^2] = \frac{\eta\,[\text{cP}]\;\dot\gamma\,[\text{s}^{-1}]}{478.8}

the stress is in pounds-force per 100 square feet, the viscosity in centipoise, the shear rate in s⁻¹.

1 lb/100 ft² = 0.4788 Pa and 1 cP = 0.001 Pa·s, so $\tau = 0.001\,\eta\dot\gamma/0.4788 = \eta\dot\gamma/478.8$ .

Drilling muds are non-Newtonian fluids

In a drilling mud, clay particles and long polymer molecules (chain-like molecules added to the mud) form a loose network. At rest the network gets stronger, and the mud turns into a weak gel (a soft, jelly-like mass). When the mud flows, the network breaks, the particles and molecules line up along the flow, and the resistance falls. So the mud has no single viscosity: the faster the shear, the “thinner” the mud. This behaviour is called pseudoplastic, or shear thinning.

Yield stress. While the stress is below a certain value, the mud does not flow. It only deforms elastically: it gives a little and springs back, like jelly. This is how the mud holds cuttings and weighting material (a heavy powder added to make the mud denser) in suspension, so they do not settle.
Thixotropy means that the behaviour of the mud depends on time. At rest the structure is rebuilt gradually, so the gel is stronger after 10 min than after 10 s. It is described by the gel strength, the stress needed to start the mud moving after it has rested (properties of the mud).
Dilatant (shear-thickening) fluids behave the other way round: they get thicker when sheared fast. They are rare among drilling muds. An everyday example is a thick mix of starch and water.

The ratio of the stress to the shear rate at a given point of the flow curve is called the apparent viscosity: $\mu_{\text{a}} = \tau/\dot\gamma$ . For a non-Newtonian fluid this is not a property of the fluid. It is a function of the shear rate. For a typical mud it is many times greater at slow shear than at fast shear. So when we speak about the viscosity of a mud, we must always say at what shear rate it was measured.

Rheological models

A rheological model is a formula $\tau(\dot\gamma)$ with a few parameters that describes the flow curve of the mud roughly. Drilling engineers use three main models.

Bingham model (viscoplastic fluid)

\tau = \tau_0 + \eta_p\,\dot\gamma \quad (\tau > \tau_0)

where:

$\tau_0$ yield point (YP): the intercept, where the straight line cuts the stress axis, Pa
$\eta_p$ plastic viscosity (PV): the slope of the straight line, Pa·s
$\tau,\ \dot\gamma$ stress and shear rate, Pa, s⁻¹

Units: 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.

The two quantities have a clear meaning. The PV grows with the amount of solids and with the viscosity of the base liquid. The YP grows with the forces of attraction between the particles. This makes the model convenient for controlling the mud, but at low shear rates it overestimates the stress.

Power-law model (Ostwald–de Waele)

\tau = K\,\dot\gamma^{\,n}

where:

$K$ consistency index: the stress at a shear rate of 1 s⁻¹, Pa·sⁿ
$n$ flow behaviour index; $n < 1$ means shear thinning, $n = 1$ a Newtonian fluid, and $n > 1$ a dilatant fluid, —

Units: 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.

On a graph with logarithmic axes, the power-law model is a straight line with the slope $n$ . The smaller $n$ is, the more the mud thins out under shear. In polymer muds $n$ can be 0.3–0.6.

Herschel–Bulkley model

\tau = \tau_y + K\,\dot\gamma^{\,n} \quad (\tau > \tau_y)

where:

$\tau_y$ yield stress: the smallest stress at which the mud starts to flow, Pa
$K,\ n$ the consistency and flow behaviour indices, as in the power-law model, Pa·sⁿ, —

Units: SI

Holds when

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

The model combines the yield stress of the Bingham model with the non-linearity of the power-law model. For $\tau_y = 0$ it becomes the power-law model, and for $n = 1$ it becomes the Bingham model. It describes most muds over the whole range of shear rates, so modern methods of hydraulic calculation use it.

Which model to use, and when

Model: Newton
Parameters: $\eta$
Where it works well: water, brines, oils
Weak point: does not describe muds that have a structure

Model: Bingham
Parameters: $\eta_p$ , $\tau_0$
Where it works well: controlling clay-based muds; estimates at high shear rates
Weak point: overestimates the stress at low shear rates

Model: Power-law
Parameters: $K$ , $n$
Where it works well: polymer muds, flow in pipes
Weak point: underestimates the stress at low shear rates

Model: Herschel–Bulkley
Parameters: $\tau_y$ , $K$ , $n$
Where it works well: most muds over the whole range
Weak point: needs readings at several speeds

Model	Parameters	Where it works well	Weak point
Newton	$\eta$	water, brines, oils	does not describe muds that have a structure
Bingham	$\eta_p$ , $\tau_0$	controlling clay-based muds; estimates at high shear rates	overestimates the stress at low shear rates
Power-law	$K$ , $n$	polymer muds, flow in pipes	underestimates the stress at low shear rates
Herschel–Bulkley	$\tau_y$ , $K$ , $n$	most muds over the whole range	needs readings at several speeds

Model parameters from viscometer readings

We measure the flow curve of a mud with a rotational viscometer that has coaxial cylinders, one inside the other (how the instrument is built). The outer sleeve turns at a set speed. The mud in the narrow gap between the cylinders drags the inner cylinder, which hangs on a spring. The angle through which the spring twists is the reading $\theta$ , in degrees on the dial, and it is proportional to the shear stress. In the standard instrument, the sizes of the sleeve and the cylinder and the stiffness of the spring are chosen on purpose. Because of this, the conversion to physical quantities is very simple.

Shear rate and shear stress in the standard viscometer

\dot\gamma = 1.703\,N, \qquad \tau = 0.511\,\theta

where:

$\dot\gamma$ shear rate at the inner cylinder, s⁻¹
$N$ rotational speed of the sleeve, rpm
$\tau$ shear stress, Pa
$\theta$ reading of the instrument, degrees on the dial

Units: $\dot\gamma$ 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.

600 and 300 rpm correspond to shear rates of 1022 and 511 s⁻¹, and 3 rpm to 5.1 s⁻¹. For a Newtonian fluid the reading at 300 rpm is numerically equal to the viscosity in mPa·s: $0.511\,\theta/511 = 0.001\,\theta$ Pa·s.

In other unit systems (1)

Oilfield units

\tau\,[\text{lb/100 ft}^2] = 1.067\,\theta

the stress is in lb/100 ft², and $\theta$ is in degrees on the dial.

$1.067 = 0.511/0.4788$ .

The Bingham model. The straight line through the 600 and 300 rpm points has a slope of $0.511(\theta_{600} - \theta_{300})/511$ Pa·s. This is exactly $\theta_{600} - \theta_{300}$ in mPa·s, and it is the plastic viscosity. The intercept on the stress axis is $0.511(\theta_{300} - \eta_p)$ Pa. In field practice the yield point is simply written as $\theta_{300} - \eta_p$ , and the result is given in pounds per 100 square feet. This convention makes the stress about 6% too small, so in SI calculations it is easier to multiply the degrees by 0.511 at once.

Power-law parameters from two readings

n = \frac{\log_{10}(\theta_{600}/\theta_{300})}{\log_{10} 2} \approx 3.32\,\log_{10}\frac{\theta_{600}}{\theta_{300}}, \qquad K = \frac{0.511\,\theta_{300}}{511^{\,n}}

where:

$n$ flow behaviour index, —
$K$ consistency index, Pa·sⁿ
$\theta_{600},\ \theta_{300}$ readings at 600 and 300 rpm, degrees
$511$ shear rate at 300 rpm, s⁻¹

Units: $K$ 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.

$n$ is the slope of the flow curve on logarithmic axes between two points. $K$ is the stress that this straight line would give at a shear rate of 1 s⁻¹. If the parameters from the “pipe” pair of readings and from the “annulus” pair differ a lot, the power-law model describes the mud poorly.

The Herschel–Bulkley model. We estimate the yield stress from the readings at the lowest speeds. The so-called low-shear yield point is $2\theta_3 - \theta_6$ degrees, that is $\tau_y \approx 0.511\,(2\theta_3 - \theta_6)$ Pa. It is the value at zero shear rate of the straight line through the 6 and 3 rpm points. Then we subtract $\tau_y$ (in degrees) from the readings at 600 and 300 rpm, and find $n$ and $K$ in the same way as for the power-law model.

Worked exampleThree models from the same readings

Given (assumed data)

Readings at 600 / 300 / 200 / 100 rpm: 70 / 44 / 34 / 23
Readings at 6 / 3 rpm: 7 / 6
Mud: water-based, 1.30 g/cm³; measured at 50 °C

Solution

Shear rates: $1.703\cdot 600 = 1022$ s⁻¹, $1.703\cdot 300 = 511$ s⁻¹, $1.703\cdot 3 = 5.11$ s⁻¹. Stresses: $0.511\cdot 70 = 35.8$ Pa, $0.511\cdot 44 = 22.48$ Pa, $0.511\cdot 6 = 3.07$ Pa.
Bingham model: $\eta_p = 70 - 44 = 26$ mPa·s; $\tau_0 = 0.511\cdot(44 - 26) = 9.20$ Pa. In the field notation the YP is $44 - 26 = 18$ lb/100 ft².
Power-law model from 600 and 300 rpm: $n = 3.32\log_{10}(70/44) = 0.670$ ; $K = 22.48/511^{0.670} = 0.345$ Pa·sⁿ. From 100 and 3 rpm: $n = 0.657\log_{10}(23/6) = 0.383$ , $K = 1.64$ Pa·sⁿ.
Herschel–Bulkley model: $\tau_y = 0.511\cdot(2\cdot 6 - 7) = 2.555$ Pa; $n = 3.32\log_{10}\dfrac{70 - 5}{44 - 5} = 0.737$ ; $K = 0.511\cdot 39/511^{0.737} = 0.201$ Pa·sⁿ.
Check at 3 rpm, where 3.07 Pa was measured: the Bingham model gives $9.20 + 0.026\cdot 5.11 = 9.33$ Pa, the power-law model (from 600 and 300 rpm) gives $0.345\cdot 5.11^{0.670} = 1.03$ Pa, and the Herschel–Bulkley model gives $2.555 + 0.201\cdot 5.11^{0.737} = 3.22$ Pa.
Apparent viscosity: at 600 rpm $35.8/1022 = 35$ mPa·s, and at 3 rpm $3.07/5.11 = 600$ mPa·s.

Answer: PV 26 mPa·s and YP 9.20 Pa (18 lb/100 ft² in the field notation); power-law model:

n

= 0.670,

K

= 0.345 Pa·sⁿ; Herschel–Bulkley model:

\tau_y

= 2.555 Pa,

n

= 0.737,

K

= 0.201 Pa·sⁿ.

What it means. At high shear rates the three models agree, because they were all built from the 600 and 300 rpm points. At low shear rates, as in the annulus and when cuttings settle, the Bingham model gives a stress about three times too high, and the power-law model one about three times too low. The Herschel–Bulkley model almost matches the measurement. The apparent viscosity of the mud at slow shear is 17 times greater than at fast shear. The data in this example are assumed for illustration.

CalculatedThe flow curve of the mud from the example, on logarithmic axes. The dots are the viscometer readings, converted to shear rate and shear stress. The lines are the Bingham model, the power-law model from 600 and 300 rpm, and the Herschel–Bulkley model. Above a shear rate of 300 s⁻¹ the models almost agree. At 5 s⁻¹ they differ by a factor of nine.

What shear rates occur in the well

To choose which readings to use for the model parameters, we must know at what shear rate the mud flows in each part of the well. For an estimate we use the shear rate at the wall for a Newtonian fluid with the same average velocity.

Shear rate at the wall in a pipe and in an annulus (estimate)

\dot\gamma_{\text{pipe}} = \frac{8\,v}{d}, \qquad \dot\gamma_{\text{ann}} = \frac{12\,v}{D - d}

where:

$\dot\gamma_{\text{pipe}},\ \dot\gamma_{\text{ann}}$ shear rate at the wall of the pipe and in the annulus, s⁻¹
$v$ average velocity of the flow: the flow rate divided by the cross-sectional area, m/s
$d$ inside diameter of the pipe; in the annulus, the outside diameter of the pipes, m
$D$ diameter of the hole, or inside diameter of the casing, m

Units: 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.

The shear rate grows with the velocity of the flow and falls with the size of the channel. So it is low in the wide annulus and high in the narrow pipes and bit nozzles.

Typical shear rates

Where the mud flows: At rest, settling of cuttings, gel
Shear rate, s⁻¹: less than 5
Viscometer speed, rpm: 3 and gel strength readings

Where the mud flows: Annulus
Shear rate, s⁻¹: 20–300
Viscometer speed, rpm: 10–180

Where the mud flows: Drill pipe and drill collars (thick-walled heavy pipes at the bottom of the string)
Shear rate, s⁻¹: 200–1500
Viscometer speed, rpm: 100–600 and higher

Where the mud flows: Bit nozzles
Shear rate, s⁻¹: 10⁴–10⁵
Viscometer speed, rpm: beyond the range of the instrument

Where the mud flows	Shear rate, s⁻¹	Viscometer speed, rpm
At rest, settling of cuttings, gel	less than 5	3 and gel strength readings
Annulus	20–300	10–180
Drill pipe and drill collars (thick-walled heavy pipes at the bottom of the string)	200–1500	100–600 and higher
Bit nozzles	10⁴–10⁵	beyond the range of the instrument

These are rough values for usual flow rates and sizes. The viscometer speed is the one at which the shear rate in the instrument is the same.

Laminar and turbulent flow

When the flow is slow, the layers of fluid slide over each other without mixing. This is the laminar regime. When the flow is fast, eddies (small whirls) appear. The fluid particles move at random across the flow. The velocity profile (how the speed changes across the channel) becomes flat in the middle of the channel and very steep near the wall. This is the turbulent regime. In the turbulent regime the friction pressure losses are much greater, and they grow almost in proportion to the square of the velocity.

The Reynolds number shows which regime sets in. It is the ratio of the inertia forces, which stir the flow, to the viscous forces, which calm it down. For a Newtonian fluid in a round pipe the flow is laminar when Re is below about 2100, and fully turbulent when Re is above about 4000. Between these values there is a transition region.

Reynolds number

\text{Re} = \frac{\rho\,v\,d_{\text{h}}}{\mu}

where:

$\rho$ density of the fluid, kg/m³
$v$ average velocity of the flow, m/s
$d_{\text{h}}$ hydraulic diameter: the inside diameter of a pipe; for an annulus, $D - d$ , m
$\mu$ viscosity; for a mud, the effective viscosity $\mu_{\text{eff}}$ , Pa·s

Units: 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.

High velocity, high density and a large channel make turbulence more likely. Viscosity works against it.

In other unit systems (1)

Oilfield units

\text{Re} = \frac{928\,\rho\,[\text{ppg}]\; v\,[\text{ft/s}]\; d\,[\text{in}]}{\mu\,[\text{cP}]}

the density is in ppg (pounds per gallon), the velocity in ft/s, the diameter in inches, and the viscosity in centipoise.

$928 = 119.83\cdot 0.3048\cdot 0.0254/0.001$ : it converts ppg to kg/m³, ft/s to m/s, inches to metres and cP to Pa·s.

A mud has no single viscosity, so we put the effective viscosity into the Reynolds number. This is the viscosity of a Newtonian fluid that, at the same average velocity in the same channel, would give the same pressure losses in laminar flow. For the Bingham model it is easy to find.

Effective viscosity of a viscoplastic mud

\mu_{\text{eff}} = \eta_p + \frac{\tau_0\,d}{6\,v}\ \ \text{(pipe)}, \qquad \mu_{\text{eff}} = \eta_p + \frac{\tau_0\,(D - d)}{8\,v}\ \ \text{(annulus)}

where:

$\mu_{\text{eff}}$ effective viscosity, Pa·s
$\eta_p,\ \tau_0$ plastic viscosity and yield point, Pa·s, Pa
$v$ average velocity of the flow, m/s
$d,\ D$ inside diameter of the pipe; in the annulus, the outside diameter of the pipes and the diameter of the hole, m

Units: 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.

The contribution of the YP is greater the slower the flow and the wider the channel. In a slow flow in the annulus the effective viscosity of a mud can be several times greater than its plastic viscosity.

Worked exampleFlow regime in the pipe and in the annulus

Given (assumed data)

Mud: 1.30 g/cm³; PV 26 mPa·s, YP 9.20 Pa (from the previous example)
Pump flow rate: 30 L/s
Drill pipe: outside diameter 127 mm, inside diameter 108.6 mm
Open hole: 215.9 mm

Solution

Pipe: $v = 0.030/(\frac{\pi}{4}\cdot 0.1086^2) = 3.24$ m/s; the shear rate at the wall is $8\cdot 3.24/0.1086 = 239$ s⁻¹ (about 140 rpm on the viscometer).
Effective viscosity in the pipe: $0.026 + \dfrac{9.20\cdot 0.1086}{6\cdot 3.24} = 0.0774$ Pa·s; $\text{Re} = \dfrac{1300\cdot 3.24\cdot 0.1086}{0.0774} = 5906$ . The flow is turbulent.
Annulus: $v = 0.030/(\frac{\pi}{4}(0.2159^2 - 0.127^2)) = 1.25$ m/s; the shear rate is $12\cdot 1.25/0.0889 = 169$ s⁻¹ (about 99 rpm).
Effective viscosity in the annulus: $0.026 + \dfrac{9.20\cdot 0.0889}{8\cdot 1.25} = 0.1076$ Pa·s; $\text{Re} = \dfrac{1300\cdot 1.25\cdot 0.0889}{0.1076} = 1346$ . The flow is laminar. If we had used the plastic viscosity, we would have got $\text{Re} = 5570$ and a wrong conclusion that the flow is turbulent.
Check in oilfield units for the pipe: $1.30\cdot 8.345 = 10.85$ ppg, $3.24/0.3048 = 10.63$ ft/s, $108.6/25.4 = 4.276$ in, $\mu = 77.4$ cP; $\text{Re} = 928\cdot 10.85\cdot 10.63\cdot 4.276/77.4 = 5909$ .
At what flow rate does the flow in the annulus come close to the limit $\text{Re} = 2100$ ? Solving the equation for the velocity, we get 1.62 m/s, that is 39 L/s.

Answer: In the pipe the flow is turbulent (Re ≈ 5906), and in the annulus it is laminar (Re ≈ 1346). The flow in the annulus stays laminar up to about 39 L/s.

What it means. This is a typical picture: in the pipe the flow is fast and the channel is narrow, and in the annulus the flow is slow and the channel is wide. In the pipe the losses hardly depend on the YP, and in the annulus they depend on it first of all. The shear rate in the annulus corresponds to about 100 rpm, so the parameters found from the low viscometer speeds suit the annulus calculations better. The data in this example are assumed for illustration.

Friction pressure losses and the ECD

The pumps spend pressure on friction in the surface lines, in the drill pipe, in the bit nozzles and in the annulus. Only one part of it acts on the formation: the losses in the annulus above the depth we look at. They add to the hydrostatic pressure (the pressure of the mud column at rest), so the bottom of the hole “feels” a mud heavier than it really is. This apparent density is called the equivalent circulating density (ECD). We compare it with the leak-off pressure (the pressure at which the rock starts to take mud) below the casing shoe (allowable pressures).

The flow in the annulus is usually laminar. For a Bingham mud we can estimate the pressure losses with a simple formula, a simplified solution for the flow in a narrow slot. The detailed methods of calculating the losses in the pipe, the annulus and the bit nozzles, for different models and regimes, are in the lesson Circulating pressure losses.

Pressure loss in the annulus in laminar flow of a viscoplastic mud

\frac{\Delta p_{\text{fr}}}{L} = \frac{48\,\eta_p\,v}{(D - d)^2} + \frac{6\,\tau_0}{D - d}

where:

$\Delta p_{\text{fr}}/L$ friction pressure loss per unit length of the annulus, Pa/m
$\eta_p,\ \tau_0$ plastic viscosity and yield point, Pa·s, Pa
$v$ average velocity in the annulus, m/s
$D,\ d$ diameter of the hole (inside diameter of the casing) and outside diameter of the pipes, m

Units: 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.

The first term is the viscous friction, and it grows with the velocity. The second term is the contribution of the YP. It does not depend on the velocity, but it grows as the gap gets smaller (it is inversely proportional to the gap). With usual gaps and moderate velocities in the annulus, it is the YP that often gives most of the losses, and so most of the increase of the ECD. When the gap narrows, the viscous term grows faster.

Equivalent circulating density

\rho_{ECD} = \rho_m + \frac{\Delta p_{\text{fr}}}{g\,h}

where:

$\rho_{ECD}$ equivalent circulating density at depth $h$ , kg/m³
$\rho_m$ density of the mud, kg/m³
$\Delta p_{\text{fr}}$ friction pressure loss in the annulus from this depth to the surface, Pa
$h$ true vertical depth, m

Units: SI. In metric units $g\,h$ is replaced by $0.00981\,h$ (g/cm³, MPa, m)

Holds when

The cuttings in the annulus are not counted. They make the column heavier.
The losses are taken along the length of the hole, and the hydrostatic pressure along the vertical.

The ECD is the pressure at the bottom while the mud circulates, expressed as a density. When the pumps stop, it falls to the hydrostatic pressure.

Worked exampleHow much of the ECD comes from the yield point

Given (assumed data)

Mud: 1.30 g/cm³; PV 26 mPa·s, YP 9.20 Pa
Flow rate: 30 L/s
Well (vertical): 3000 m; casing with an inside diameter of 222.4 mm down to 2300 m, below it a 215.9 mm hole
Drill pipe: 127 mm down to the bottom (the heavy drill collars are ignored for simplicity)

Solution

Open hole: $v = 1.25$ m/s; $\dfrac{48\cdot 0.026\cdot 1.25}{0.0889^2} + \dfrac{6\cdot 9.20}{0.0889} = 198 + 621 = 819$ Pa/m.
In the casing: the gap is $0.2224 - 0.127 = 0.0954$ m, $v = 1.15$ m/s; $157 + 579 = 736$ Pa/m. The regime is laminar: Re ≈ 1167.
Losses in the annulus: $(736\cdot 2300 + 819\cdot 700)\cdot 10^{-6} = 2.27$ MPa, and the YP accounts for 78% of this.
ECD at the bottom: $1.30 + \dfrac{2.27}{0.00981\cdot 3000} = 1.377$ g/cm³; at the shoe (2300 m): $1.30 + \dfrac{1.69}{0.00981\cdot 2300} = 1.375$ g/cm³.
If we halve the YP with the same plastic viscosity, the losses fall to 1.38 MPa, and the ECD at the bottom to 1.347 g/cm³.

Answer: The ECD at the bottom is about 1.377 g/cm³ for a mud density of 1.30 g/cm³. The YP gives 78% of the losses in the annulus.

What it means. The increase of 0.077 g/cm³ is 2.27 MPa at the bottom, and almost four fifths of it is made by the YP. So when the window between the pore pressure (the pressure of the fluid in the rock) and the leak-off pressure is narrow, we lower the ECD first of all through the YP and the flow rate. The plastic viscosity has a smaller effect here. But the YP is needed to carry cuttings out, and a mud that is too “thin” cleans the hole badly. So we set the YP as a compromise. The data in this example are assumed for illustration.

Terms and symbols

Shear stress, $\tau$: The friction force between layers of the fluid per unit of area, in Pa.
Shear rate, $\dot\gamma$: The change of velocity across the flow per unit of distance, in s⁻¹.
Apparent viscosity: The ratio $\tau/\dot\gamma$ at a given point of the flow curve. In a mud it depends on the shear rate.
Effective viscosity: The viscosity of a Newtonian fluid that, in the same channel at the same velocity, gives the same laminar losses. We put it into the Reynolds number.
Plastic viscosity and yield point: The slope and the intercept of the Bingham line. From the viscometer, $\eta_p = \theta_{600} - \theta_{300}$ mPa·s and $\tau_0 = 0.511(\theta_{300} - \eta_p)$ Pa.
Indices $K$ and $n$: The consistency and the flow behaviour in the power-law and Herschel–Bulkley models. $n < 1$ means shear thinning.
Yield stress, $\tau_y$: The smallest stress at which the mud starts to flow. It is smaller than the yield point.
Thixotropy: The gradual strengthening of the structure of a mud at rest, and its breaking when the mud flows.
Reynolds number: The dimensionless ratio of the inertia forces to the viscous forces. It decides the flow regime.
Hydraulic diameter: Four times the cross-sectional area divided by the wetted perimeter. For an annulus, it is $D - d$ .

Common mistakes and misconceptions

Wrong

“The YP is the stress at which the mud starts to flow.”

In fact

The YP is the extension of the straight line through the 600 and 300 rpm readings to zero shear rate. The real stress at which the mud starts to flow is usually noticeably smaller. We estimate it from the readings at 6 and 3 rpm.

Wrong

“The higher the viscosity of the mud, the higher the pressure losses in the whole system.”

In fact

In the turbulent flow in the pipe the viscosity has little effect: what matters is the flow rate, the density and the diameter. The laminar losses in the annulus, that is, the ECD, depend strongly on the rheology, above all on the YP.

Wrong

“The power-law parameters from 600 and 300 rpm are good for any calculation.”

In fact

They describe the mud at shear rates of about 500–1000 s⁻¹. For the annulus, where the shear rates are several times lower, find the parameters from the low viscometer speeds or use the Herschel–Bulkley model.

Wrong

“The Reynolds number of a mud is calculated with the plastic viscosity.”

In fact

We need the effective viscosity, which accounts for the YP. In the lesson's example, the plastic viscosity would have given Re above 5000 in the annulus and a false conclusion that the flow is turbulent. The right value is Re of about 1350.

Wrong

“A viscometer reading of 40 is a stress of 40 Pa.”

In fact

The reading is in degrees on the dial. In the standard instrument one degree corresponds to 0.511 Pa (1.067 lb/100 ft²), so 40 degrees is about 20 Pa.

In depth: advanced detail

Where the factors 1.703 and 0.511 come from

In the standard viscometer the inside radius of the rotating sleeve is $R_1 = 18.42$ mm, and the radius of the suspended cylinder is $R_b = 17.25$ mm. For a Newtonian fluid, the shear rate at the surface of the inner cylinder is $\dot\gamma = 2\omega R_1^2/(R_1^2 - R_b^2)$ , where $\omega = 2\pi N/60$ is the angular velocity. Substituting the numbers gives $\dot\gamma \approx 1.703\,N$ s⁻¹.

The stress at the surface of the cylinder is the torque of the spring divided by the area of the side surface of the cylinder and by its radius. The stiffness of the standard spring is about 386 dyne·cm per degree, and the height of the cylinder is 38 mm with a correction for the ends. As a result, one degree on the dial corresponds to about 5.11 dyne/cm², that is, 0.511 Pa. The numbers are chosen so that at 300 rpm ( $\dot\gamma = 511$ s⁻¹) the reading for a Newtonian fluid equals its viscosity in centipoise. This is where the simple field formulas for the PV and the YP come from.

Pressure needed to restart circulation

After a stop, the mud in the well builds up gel strength. To move it, the pump must overcome the static shear stress along the whole length of the channel. From a balance of forces, the pressure acting on the cross-section against the resistance of the gel on the wetted surface, we get:

Pressure needed to break the gel

\Delta p_{\text{gel}} = \frac{4\,\tau_{\text{gel}}\,L}{d}\ \ \text{(pipe)}, \qquad \Delta p_{\text{gel}} = \frac{4\,\tau_{\text{gel}}\,L}{D - d}\ \ \text{(annulus)}

where:

$\Delta p_{\text{gel}}$ pressure difference needed to move the gel, Pa
$\tau_{\text{gel}}$ static shear stress (gel strength) after a given rest time, Pa
$L$ length of the channel, m
$d,\ D$ diameters of the channel: for a pipe, the inside diameter; for an annulus, the outside diameter of the pipes and the diameter of the hole, m

Units: 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.

In the annulus this pressure difference is added, for a short time, to the pressure at the bottom and at the shoe. For example, a gel of 12 degrees (6.1 Pa) in an annulus 3000 m long with a gap of about 9 cm needs about 0.8 MPa. This is almost 0.03 g/cm³ added to the equivalent mud weight (the pressure written as the density of a mud column). So circulation is restored gradually, and the pipe is often rotated first to break the gel.

Effect of temperature and pressure

The viscosity of the base liquid of a mud falls when it is heated, so the plastic viscosity at the bottom of a deep, hot well is lower than in the tanks. The YP and the gels can either fall or rise with heating. In water-based muds at high temperature the clays flocculate (the particles stick together in clumps) more strongly, and some of the polymers break down. In muds with a hydrocarbon base, the viscosity of the base rises noticeably with pressure. So to control the properties we measure the mud at a standard temperature. To calculate the hydraulics of deep wells (the flow rates and the pressure losses) we use high-pressure, high-temperature viscometers, or correction curves for the particular mud (Oil-based muds, Water-based muds and clay chemistry).

Connections to other topics

Drilling fluids: functions and properties — functions and properties of the mud; the PV, YP and gel strength and how they are controlled.
Circulating pressure losses — calculation of the pressure losses in the pipe, the annulus and the bit nozzles for different models and regimes.
Fracture gradient and leak-off testing — why the ECD is compared with the leak-off pressure below the shoe.
Well control: primary barrier and kick indicators — swabbing and pressure surges during trips depend on the rheology.
Field testing of drilling fluids — how measurements are made with the viscometer and other instruments.
Bit hydraulics optimisation — how to split the pump pressure between the circulating system (the pipes and the annulus) and the bit.

Check yourself

1
Which of these fluids behaves as a Newtonian fluid?
1. A.Polymer mud
2. B.Cement slurry
3. C.Fresh water
4. D.Clay-based drilling mud
Show the answer
Correct answer: C. Fresh water
The viscosity of water does not depend on the shear rate. Clay-based muds, polymer muds and cement slurries thin out under shear and have a yield stress. They are non-Newtonian fluids.
2
Viscometer readings: $\theta_{600} = 52$ , $\theta_{300} = 32$ . Find the plastic viscosity, the YP in pascals and the power-law parameters.
Show the answer
$\eta_p = 52 - 32 = 20$ mPa·s; $\tau_0 = 0.511\cdot(32 - 20) \approx 6.1$ Pa (12 lb/100 ft² in the field notation). $n = 3.32\log_{10}(52/32) \approx 0.70$ ; $K = 0.511\cdot 32/511^{0.70} \approx 16.4/78.7 \approx 0.21$ Pa·sⁿ.
3
Why is the YP calculated with the Bingham model usually greater than the stress that the viscometer shows at 3 rpm?
1. A.The YP is the extension of a line drawn through the readings at high shear rates; at low shear rates the real curve lies below this line
2. B.The YP is measured at a different temperature
3. C.At 3 rpm the mud flows in turbulent flow
4. D.The instrument is always wrong at low speeds
Show the answer
Correct answer: A. The YP is the extension of a line drawn through the readings at high shear rates; at low shear rates the real curve lies below this line
The Bingham model is a straight line through the 600 and 300 rpm points. The real flow curve of a mud is not straight. At low shear rates it bends down below the line: as the shear rate falls, the stress falls faster than along the straight line. So the true yield stress is closer to the readings at 6 and 3 rpm.
4
A mud of 1.20 g/cm³, PV 20 mPa·s and YP 6 Pa flows in an annulus at 1.0 m/s with a gap $D - d = 0.10$ m. What is the flow regime? What would we get with the plastic viscosity instead of the effective viscosity?
Show the answer
$\mu_{\text{eff}} = 0.020 + 6\cdot 0.10/(8\cdot 1.0) = 0.095$ Pa·s; $\text{Re} = 1200\cdot 1.0\cdot 0.10/0.095 \approx 1260$ . The flow is laminar. With the plastic viscosity we would get $\text{Re} = 6000$ and a wrong conclusion that the flow is turbulent.
5
In laminar flow in the annulus, the YP of the mud was halved, while the plastic viscosity and the flow rate did not change. What happens to the ECD?
1. A.It falls exactly by half
2. B.It does not change
3. C.It falls noticeably: in laminar flow in the annulus most of the losses are usually made by the YP
4. D.It rises
Show the answer
Correct answer: C. It falls noticeably: in laminar flow in the annulus most of the losses are usually made by the YP
The losses in a laminar annulus are the sum of the viscous term and the YP term. With usual gaps and velocities, the second is usually larger. In the lesson's example, halving the YP reduced the increase of the ECD above the mud density by about 40%. The viscous term stayed the same.
6
Which viscometer speeds are better to use for calculating the flow in the annulus, and why?
Show the answer
The shear rates in the annulus are usually tens to hundreds of s⁻¹, which corresponds to about 10–180 rpm on the instrument. So we find the parameters from the readings at 100, 6 and 3 rpm (or from all the readings in the Herschel–Bulkley model), and not from 600 and 300 rpm, which describe the flow in the pipe.

Updated: 19 October 2026