Circulating pressure losses

Quick refresher

The standpipe pressure is the sum of the losses in the surface lines, in the drill pipe and the drill collars, at the bit nozzles and in the annulus. The hydrostatic pressures (the weight of the mud columns) inside the string and in the annulus balance each other while the densities are the same. Only the hydrostatic pressure and the annulus losses act on the bottom of the hole.
The laminar losses follow from a balance of forces, $\Delta p/L = 4\tau_w/d$ . Here $\tau_w$ is the shear stress at the wall, and we find it from the rheological model (the formula that describes how the mud flows). For a viscoplastic (Bingham) fluid it is a viscous term plus a yield point term. The yield point (YP) is the part of the stress that does not change with the shear rate. For a power-law fluid, $\tau_w$ is $K$ times the shear rate at the wall, raised to the power $n$ .
The generalised Reynolds number (it compares the forces that stir the flow with the forces that calm it) shows the flow regime. For a power-law fluid, the flow stays laminar up to $3470 - 1370n$ , and turbulent flow starts above $4270 - 1370n$ . In the pipe the flow is usually turbulent, in the annulus usually laminar.
We calculate the turbulent losses with the Fanning equation $2f\rho v^2/d$ . The friction factor $f$ comes from the Blasius formula (for a viscoplastic fluid, with the plastic viscosity) or from the Dodge–Metzner correlation (for a power-law fluid). The losses grow with the flow rate to the power 1.6–1.75 (1.75 for a Newtonian fluid, which has a constant viscosity, and less for a shear-thinning mud, which gets thinner as the shear rate grows), and they depend little on the rheology.
The pressure drop across the bit nozzles, $\rho Q^2/(2C_d^2A^2)$ , does not depend on the rheology and grows as the square of the flow rate. One plugged nozzle out of five makes it more than one and a half times higher.
The ECD at a depth is the mud density plus the annulus losses above this depth, with those losses divided by $g h$ . We compare it with the leak-off pressure below the shoe, leaving a margin for cuttings and for the pressure surges when pipe is run in. The Bingham model usually overestimates the annulus losses.

Key relationships

p_{\text{SPP}} = \Delta p_{\text{surf}} + \Delta p_{\text{dp}} + \Delta p_{\text{dc}} + \Delta p_{\text{b}} + \Delta p_{\text{ann}}

standpipe pressure as the sum of the losses

\dfrac{\Delta p}{L} = \dfrac{2 f \rho v^2}{d}

Fanning equation; in laminar flow in a pipe $f = 16/\text{Re}$

\Delta p_{\text{b}} = \dfrac{\rho Q^2}{2 C_d^2 A^2}

pressure drop across the bit nozzles

\rho_{ECD}(h) = \rho_m + \dfrac{\Delta p_{\text{ann}}(h)}{g\,h}

ECD from the annulus losses above the depth $h$

Common pitfalls

Calculating the flow in the annulus with the Bingham model and the parameters from 600 and 300 rpm. At low shear rates this model overestimates the stress. So it overestimates the ECD, and the pressure surges and swabs when pipe is run in and pulled out even more.
Converting the standpipe pressure to another flow rate with one exponent. At the bit the loss grows as $Q^2$ , in turbulent pipes as a power of 1.6–1.75, and in the laminar annulus much more slowly.
Forgetting the drill collars and the surface lines. Megapascals are also lost in short sections of small diameter.
Comparing the mud density with the leak-off pressure, instead of the ECD at the shoe with a margin for cuttings and pressure surges.
Confusing the Fanning friction factor $f$ with the hydraulic resistance coefficient $\lambda = 4f$ from Russian-language handbooks.

After this lesson you can

list the elements of the circulating system and write the pressure balance at the standpipe
calculate the laminar losses (the losses in smooth flow, in layers) in the pipe and in the annulus for a viscoplastic (Bingham) and a power-law fluid, and find the flow regime from the generalised Reynolds number
calculate the turbulent losses (the losses in swirling flow) with the Fanning equation, using the Blasius or the Dodge–Metzner friction factor
calculate the pressure drop across the bit nozzles from their total flow area
build the ECD along the depth for an annulus of several sections and compare it with the leak-off pressure below the shoe
estimate how each term changes with the flow rate, the density and the rheology of the mud, and calculate the pressure surges (rises) and swabs (falls) when pipe is run in and pulled out

Where the pump pressure goes

The path of the mud is a closed loop. The mud pumps send it through the discharge line and the standpipe (a vertical pipe on the rig's tower, the derrick) into the rotary hose (a flexible hose). From there it goes through the swivel or the top drive (the devices that let the string turn while mud flows into it) into the drill string. The mud passes through the drill pipe, the drill collars and the rest of the bottom hole assembly, leaves in jets from the bit nozzles, and rises up the annulus, first in the open hole and then inside the casing. Through the flowline (a trough) and the shale shakers (vibrating screens that take out the cuttings) it comes back to the tanks at atmospheric pressure. So all the pressure that the pumps made has been used up on the way.

The hydrostatic pressure does not enter this balance. The string and the annulus are the two arms of one U-shaped vessel. While the mud inside and outside has the same density, the two columns balance each other, and the pump only has to overcome friction. If the densities differ, the difference between the columns is added to the standpipe pressure or subtracted from it. This happens when a heavy slug (a small batch of heavier mud) is pumped into the pipe, when there are many cuttings in the annulus, or when the pipe is filled with kill mud (mud heavy enough to stop an inflow from the formation) (Well kill methods).

Standpipe pressure as the sum of the losses

p_{\text{SPP}} = \Delta p_{\text{surf}} + \Delta p_{\text{dp}} + \Delta p_{\text{dc}} + \Delta p_{\text{b}} + \Delta p_{\text{ann}}

where:

$p_{\text{SPP}}$ standpipe pressure (gauge pressure, that is, above atmospheric), the reading of the pressure gauge; at the pump discharge it is a little higher, by the losses in the discharge line, MPa
$\Delta p_{\text{surf}}$ losses in the surface lines after the gauge: the standpipe, the rotary hose, the swivel or the top drive, MPa
$\Delta p_{\text{dp}},\ \Delta p_{\text{dc}}$ friction losses in the drill pipe and in the drill collars (and in other parts of the bottom hole assembly), MPa
$\Delta p_{\text{b}}$ pressure drop across the bit nozzles, MPa
$\Delta p_{\text{ann}}$ friction losses in the annulus from the bottom to the surface, MPa

Units: 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.

Each section “eats up” its share of the pump pressure. The useful work is done first of all by the jets from the bit, which clean the bottom of the hole. The friction losses in the pipe and in the annulus are the price of making the mud go round the loop and carry the cuttings out. Of the whole sum, only $\Delta p_{\text{ann}}$ acts on the formation: the pressure at the bottom while the mud circulates is $\rho g h + \Delta p_{\text{ann}}$ .

CalculatedThe pressure above the hydrostatic pressure along the path of the mud in the well of the example (section “Standpipe pressure”). Inside the string it falls from 13.1 MPa at the top of the string, because of friction in the drill pipe and the drill collars. At the bit it drops suddenly by 5.1 MPa. In the annulus it decreases from 1.8 MPa at the bottom to zero at the surface. Another 0.3 MPa is lost in the surface lines above the string.

Sections of the circulating system

Section: Surface lines
Usual regime: turbulent
What the losses depend on: length and diameter of the lines; usually a few tenths of an MPa

Section: Drill pipe
Usual regime: turbulent
What the losses depend on: flow rate, density, inside diameter; the largest of the “useless” losses

Section: Drill collars and other parts of the BHA
Usual regime: turbulent
What the losses depend on: small inside diameter: a high gradient over a short section

Section: Bit nozzles
Usual regime: jet
What the losses depend on: flow rate, density and the area of the nozzles; the rheology has no effect

Section: Annulus
Usual regime: mostly laminar
What the losses depend on: YP or $K$ of the mud and the width of the gap; it acts on the formation

Section	Usual regime	What the losses depend on
Surface lines	turbulent	length and diameter of the lines; usually a few tenths of an MPa
Drill pipe	turbulent	flow rate, density, inside diameter; the largest of the “useless” losses
Drill collars and other parts of the BHA	turbulent	small inside diameter: a high gradient over a short section
Bit nozzles	jet	flow rate, density and the area of the nozzles; the rheology has no effect
Annulus	mostly laminar	YP or $K$ of the mud and the width of the gap; it acts on the formation

A downhole motor and an MWD tool add their own pressure drops. We take them from the data sheet at the given flow rate.

Laminar flow in the pipe and in the annulus

In laminar flow the viscous friction at the wall decides the losses. Let us write a balance of forces for a column of mud of length $L$ . The pressure difference times the cross-sectional area is balanced by the shear stress at the wall, $\tau_w$ , times the wetted surface. For a pipe, $\Delta p\cdot\pi d^2/4 = \tau_w\cdot\pi d L$ . For a narrow annulus it is the same with the width of the gap. What is left is to find the $\tau_w$ that the chosen rheological model gives at the given average velocity.

Pressure loss from the shear stress at the wall

\frac{\Delta p}{L} = \frac{4\,\tau_w}{d}\ \ \text{(pipe)}, \qquad \frac{\Delta p}{L} = \frac{4\,\tau_w}{D - d}\ \ \text{(annulus)}

where:

$\Delta p/L$ friction pressure loss per unit length, Pa/m
$\tau_w$ shear stress at the wall, Pa
$d$ 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 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.

The balance of forces does not depend on the properties of the fluid: for the same wall stress, the narrower the channel, the greater the losses. The properties of the mud come in through $\tau_w$ .

For a Newtonian fluid (one with a constant viscosity, such as water) the exact solution is the Hagen–Poiseuille formula, $\Delta p/L = 32\mu v/d^2$ . For a viscoplastic fluid, a term with the yield point (YP) is added to the viscous term. The mud flows only when the stress at the wall exceeds $\tau_0$ , and part of the pressure difference is spent to overcome this threshold.

Laminar losses of a viscoplastic fluid in a pipe

\frac{\Delta p}{L} = \frac{32\,\eta_p\,v}{d^2} + \frac{16\,\tau_0}{3\,d}

where:

$\eta_p,\ \tau_0$ plastic viscosity and YP, Pa·s, Pa
$v$ average velocity in the pipe, $Q$ divided by the cross-sectional area, m/s
$d$ inside diameter of the pipe, m

Units: 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.

The viscous term grows in proportion to the velocity, and the YP term does not depend on the velocity. A similar solution for the annulus, $48\eta_p v/(D-d)^2 + 6\tau_0/(D-d)$ , is derived in the rheology lesson.

In other unit systems (1)

Oilfield units

\frac{\Delta p}{L} = \frac{\eta_p v}{1500\,d^2} + \frac{\tau_0}{225\,d}\ \ \text{(pipe)}, \qquad \frac{\Delta p}{L} = \frac{\eta_p v}{1000\,(D-d)^2} + \frac{\tau_0}{200\,(D-d)}\ \ \text{(annulus)}

$\Delta p/L$ in psi/ft, $\eta_p$ in cP, $v$ in ft/s, $\tau_0$ in lb/100 ft², diameters in inches.

The conversion of units from the SI formulas: $32\to 1/1500$ and $48\to 1/1000$ (with a rounding of 0.3%), $16/3\to 1/225$ and $6\to 1/200$ (exactly).

For a power-law fluid ( $\tau = K\dot\gamma^n$ ) the exact solution also takes one line. A shear-thinning mud (a mud that gets thinner as the shear rate grows) has a greater shear rate at the wall than a Newtonian fluid with the same average velocity: its velocity profile is flatter, and the shearing is concentrated at the wall. The correction factors $(3n+1)/(4n)$ for a pipe and $(2n+1)/(3n)$ for a slot were mentioned in the rheology lesson. Here they enter the loss formulas.

Laminar losses of a power-law fluid in a pipe and in an annulus

\frac{\Delta p}{L} = \frac{4K}{d}\left(\frac{3n+1}{4n}\cdot\frac{8v}{d}\right)^{n}\ \text{(pipe)}, \qquad \frac{\Delta p}{L} = \frac{4K}{D-d}\left(\frac{2n+1}{3n}\cdot\frac{12v}{D-d}\right)^{n}\ \text{(annulus)}

where:

$K,\ n$ the consistency and flow behaviour indices of the power-law model, Pa·sⁿ, —
$v$ average velocity in the pipe or in the annulus, 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

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).

The expression in brackets is the shear rate at the wall, and the expression before the brackets turns the stress into losses. Since $n < 1$ , the losses grow with the velocity more slowly than in proportion: for $n = 0.4$ , doubling the flow rate raises them by only a third.

In other unit systems (1)

Oilfield units

\frac{\Delta p}{L} = \frac{K}{300\,d}\left(\frac{3n+1}{4n}\cdot\frac{96\,v}{d}\right)^{n}, \qquad \frac{\Delta p}{L} = \frac{K}{300\,(D-d)}\left(\frac{2n+1}{3n}\cdot\frac{144\,v}{D-d}\right)^{n}

$\Delta p/L$ in psi/ft, $K$ in lb·sⁿ/100 ft², $v$ in ft/s, diameters in inches.

$96 = 8\cdot 12$ and $144 = 12\cdot 12$ convert inches to feet in the shear rate. $1/300 = 4\cdot 12/(100\cdot 144)$ comes from 1 lb/100 ft² and square inches to psi.

Worked exampleThe annulus at the drill collars: two models, two answers

Given (assumed data)

Hole and drill collars: 215.9 mm and 158.8 mm (gap $D - d = 0.0571$ m)
Flow rate: 30 L/s
Mud (rheology lesson): 1.30 g/cm³; plastic viscosity (PV) 26 mPa·s, YP 9.20 Pa; from 100 and 3 rpm $n = 0.383$ , $K = 1.64$ Pa·sⁿ (more exactly 0.3832 and 1.641)

Solution

Area of the annulus $\frac{\pi}{4}(0.2159^2 - 0.1588^2) = 0.016804$ m², velocity $v = 0.030/0.016804 = 1.785$ m/s.
Power-law model: the shear rate at the wall is $\dfrac{2n+1}{3n}\cdot\dfrac{12v}{D-d} = 1.5365\cdot 375.2 = 576$ s⁻¹; the stress is $\tau_w = 1.641\cdot 576^{0.3832} = 18.75$ Pa; the losses are $4\cdot 18.75/0.0571 = 1314$ Pa/m.
Regime: the apparent viscosity at the wall is $\tau_w/(12v/(D-d)) = 18.75/375.2 = 0.0500$ Pa·s, and $\text{Re} = 1300\cdot 1.785\cdot 0.0571/0.0500 \approx 2650$ . This is below the limit $3470 - 1370\cdot 0.383 = 2945$ , so the flow is laminar.
Bingham model: $\dfrac{48\cdot 0.026\cdot 1.785}{0.0571^2} + \dfrac{6\cdot 9.20}{0.0571} = 683 + 967 = 1650$ Pa/m. The Reynolds number with the effective viscosity is about 2110, right at the limit of 2100.
Over the 150 m of drill collars: 0.197 MPa by the power-law model and 0.248 MPa by the Bingham model, which is 1.26 times more.

Answer: The losses in the annulus at the drill collars are about 1314 Pa/m by the power-law model and 1650 Pa/m by the Bingham model. The flow is laminar, but close to the transition.

What it means. The Bingham model gives about a quarter more loss, because the YP is the extension of a line from high shear rates. At the walls of the annulus the shear rate is only a few hundred s⁻¹, and the real stress there is lower than the line. The power-law model, built from the readings at 100 and 3 rpm, describes exactly this range. The Reynolds number at the drill collars is the highest in the whole annulus: the gap is narrow and the velocity is high. The data in this example are assumed for illustration.

Where laminar flow ends

For a Newtonian fluid the limit is a Reynolds number of about 2100. For a viscoplastic mud the rheology lesson gives a simple rule: put the effective viscosity $\eta_p + \tau_0 d/(6v)$ for a pipe, or $\eta_p + \tau_0(D-d)/(8v)$ for an annulus, into the Reynolds number and compare the result with the same limit. A more exact criterion for such fluids takes the ratio of the YP to the plastic viscosity into account, through the Hedström number (see the advanced section). A flow with a noticeable YP stays laminar at higher Reynolds numbers.

For a power-law fluid we use a similar viscosity, the “apparent viscosity at the wall”: the stress at the wall divided by the shear rate that a Newtonian fluid would have there. The result is the generalised Reynolds number.

Generalised Reynolds number of a power-law fluid

\text{Re} = \frac{\rho\,v\,d}{\mu_a}, \qquad \mu_a = K\left(\frac{3n+1}{4n}\right)^{n}\left(\frac{8v}{d}\right)^{n-1}

where:

$\mu_a$ apparent viscosity at the wall: the viscosity of a Newtonian fluid that would give the same laminar losses at the same velocity, Pa·s
$\rho,\ v$ density of the mud and average velocity, kg/m³, m/s
$d$ inside diameter of the pipe; in the annulus, the hydraulic diameter $D - d$ , m

Units: 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).

For $n = 1$ this is the usual Reynolds number with the viscosity $K$ . The more the mud thins out under shear, the smaller $\mu_a$ is in a fast flow, and the greater the Reynolds number at the same velocity.

Regime limits for a power-law fluid

\text{Re}_{\text{lam}} = 3470 - 1370\,n, \qquad \text{Re}_{\text{turb}} = 4270 - 1370\,n

where:

$\text{Re}_{\text{lam}}$ up to this value the flow is laminar, —
$\text{Re}_{\text{turb}}$ above this value the flow is turbulent; between the two limits the regime is transitional, —
$n$ flow behaviour index in the given section, —

Units: 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.

For $n = 1$ we get 2100 and 2900 (for a Newtonian fluid in a pipe the transition region is often taken wider, up to 4000). For $n = 0.4$ we get about 2900 and 3700. The more the mud thins out under shear, the higher the critical value of the generalised Reynolds number.

Turbulent flow: the friction factor

In turbulent flow the eddies carry momentum (the push of the moving fluid) across the channel. The velocity profile becomes flat in the middle. At the wall a thin viscous sublayer is left, where the velocity changes very steeply. There is no theoretical solution, so we express the losses with an experimental friction factor. This is a dimensionless ratio of the stress at the wall to the dynamic pressure of the flow (the pressure that the moving fluid carries with it).

Fanning equation

\frac{\Delta p}{L} = \frac{2\,f\,\rho\,v^2}{d}

where:

$f$ Fanning friction factor, $f = 2\tau_w/(\rho v^2)$ , —
$\rho,\ v$ density of the mud and average velocity, kg/m³, m/s
$d$ inside diameter of the pipe; in the annulus, the hydraulic diameter $D - d$ , m

Units: 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 $D - d$ .

The losses are proportional to the square of the velocity times the friction factor, which slowly decreases as the Reynolds number grows. In Russian-language hydraulics the same thing is written with the hydraulic resistance coefficient $\lambda = 4f$ : $\Delta p/L = \lambda\rho v^2/(2d)$ .

In other unit systems (1)

Oilfield units

\frac{\Delta p}{L} = \frac{f\,\rho\,v^2}{25.8\,d}

$\Delta p/L$ in psi/ft, $\rho$ in ppg, $v$ in ft/s, $d$ in inches.

$1/25.8 = 2\cdot 7.48\cdot 12/(32.17\cdot 144)$ : 7.48 gallons in a cubic foot, 12 inches in a foot, 32.17 for the change from a pound of mass to a pound of force, and 144 square inches in a square foot.

Friction factor of a Newtonian fluid: smooth and rough pipe

f = \frac{0.0791}{\text{Re}^{0.25}}, \qquad \frac{1}{\sqrt f} = -4\log_{10}\left(\frac{\varepsilon}{3.7\,d} + \frac{1.255}{\text{Re}\sqrt f}\right)

where:

$\text{Re}$ Reynolds number; for a viscoplastic mud, with the plastic viscosity, $\rho v d/\eta_p$ , —
$\varepsilon$ absolute roughness of the wall, m

Units: 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.

The roughness matters at high Reynolds numbers. The inside surface of drill pipe is smooth enough, and we usually do not take roughness into account.

For a power-law fluid, experiments gave their own relation, the Dodge–Metzner correlation (a formula fitted to test data). It links the friction factor with the generalised Reynolds number and the index $n$ . For $n = 1$ it agrees with the friction law of a smooth pipe for a Newtonian fluid.

Friction factor of a power-law fluid (Dodge–Metzner)

\frac{1}{\sqrt f} = \frac{4}{n^{0.75}}\log_{10}\left(\text{Re}\,f^{\,1 - n/2}\right) - \frac{0.395}{n^{1.2}}

where:

$f$ Fanning friction factor, —
$\text{Re}$ generalised Reynolds number (the formula above), —
$n$ flow behaviour index; for pipes, from the readings at 600 and 300 rpm, —

Units: 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.

The smaller $n$ is, the smaller the friction factor at the same Reynolds number: a shear-thinning mud in turbulent flow flows more “easily” than a Newtonian fluid.

Turbulent losses in a smooth pipe in terms of the flow rate

\frac{\Delta p}{L} = 0.241\,\frac{\rho^{0.75}\,Q^{1.75}\,\mu^{0.25}}{d^{4.75}}

where:

$Q$ flow rate, m³/s
$\mu$ viscosity; for a mud, the plastic viscosity, Pa·s
$\rho,\ d$ density and inside diameter, kg/m³, m

Units: 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, $2100 < \text{Re} < 10^5$ .

The formula shows how strongly each quantity matters: a flow rate 20% higher gives losses 38% higher; an inside diameter 10% smaller gives losses 65% higher; a density 10% higher gives losses 7.4% higher; a viscosity twice as large gives losses only 19% higher.

In other unit systems (1)

Oilfield units

\frac{\Delta p}{L} = \frac{\rho^{0.75}\,v^{1.75}\,\mu^{0.25}}{1800\,d^{1.25}} = \frac{\rho^{0.75}\,q^{1.75}\,\mu^{0.25}}{8630\,d^{4.75}}

$\Delta p/L$ in psi/ft, $\rho$ in ppg, $v$ in ft/s, $q$ in gallons per minute, $\mu$ in cP, $d$ in inches.

The same Blasius and Fanning formulas in oilfield units (the numbers 25.8 and 928 are related to them).

Worked exampleLosses in the drill pipe

Given (assumed data)

Pipe: inside diameter 108.6 mm, length 2850 m
Flow rate: 30 L/s
Mud: 1.30 g/cm³; from 600 and 300 rpm $n = 0.670$ , $K = 0.345$ Pa·sⁿ; PV 26 mPa·s

Solution

Velocity $v = 0.030/(\frac{\pi}{4}\cdot 0.1086^2) = 3.2387$ m/s; $8v/d = 239$ s⁻¹.
Apparent viscosity $\mu_a = 0.345\cdot 1.123^{0.670}\cdot 239^{-0.330} = 0.0612$ Pa·s; $\text{Re} = \rho v d/\mu_a \approx 7500$ . This is above the turbulent limit $4270 - 1370\cdot 0.670 = 3352$ .
Friction factor by the Dodge–Metzner formula (by trial and error): $f = 0.006478$ . The explicit approximation: $a = 0.07512$ , $b = 0.2749$ , $f \approx a/\text{Re}^{\,b} = 0.00647$ .
Losses: $2\cdot 0.006478\cdot 1300\cdot 3.2387^2/0.1086 = 1627$ Pa/m, and over 2850 m, 4.64 MPa.
By the Bingham model: $\text{Re} = 1300\cdot 3.2387\cdot 0.1086/0.026 = 17{,}586$ , $f = 0.0791/17{,}586^{0.25} = 0.006869$ , losses 1725 Pa/m, and over 2850 m, 4.92 MPa.
If by mistake we used the laminar formula of the power-law model, we would get 538 Pa/m, which is 3.0 times less.

Answer: The losses in the drill pipe are about 4.64 MPa (4.92 MPa by the Bingham model). The flow is turbulent.

What it means. In turbulent flow the two models differ by only 6%, because the rheology enters the calculation only weakly, through the friction factor. But a mistake in the flow regime changes the answer threefold. Compare this with the annulus, where it is the other way round: the regime is laminar, and the choice of the model changes the result by a quarter. The data in this example are assumed for illustration.

Pressure drop across the bit

In the bit nozzles the mud speeds up from a few metres per second to 60–150 m/s. The pressure turns into the kinetic energy of the jet (the energy of its motion), and the energy of the jet is spent at the bottom of the hole, where it tears the cuttings off and carries them away. The nozzle is short, and the friction in it is small compared with the speeding up. So the pressure drop across the bit does not depend on the viscosity or the YP, but only on the density, the flow rate and the area. The discharge coefficient $C_d$ accounts for the unavoidable losses at the entrance and in the narrowing. It is usually 0.95 (up to 0.98 for well-streamlined nozzles).

Pressure drop across the bit nozzles

\Delta p_{\text{b}} = \frac{\rho\,Q^2}{2\,C_d^2\,A^2}, \qquad A = \sum_i \frac{\pi d_i^2}{4}, \qquad v_{\text{n}} = \frac{Q}{A}

where:

$\Delta p_{\text{b}}$ pressure drop across the bit, Pa
$Q$ flow rate, m³/s
$A$ total flow area of the nozzles (TFA), m²
$d_i$ diameters of the nozzles, m
$C_d$ discharge coefficient, —
$v_{\text{n}}$ velocity of the jet in the nozzles, m/s

Units: 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.

The pressure drop grows as the square of the flow rate and is inversely proportional to the square of the area. Nozzles that are 10% smaller in diameter give a pressure drop that is 52% greater.

In other unit systems (2)

Metric units

\Delta p_{\text{b}}\,[\text{MPa}] = \frac{500\,\rho\,[\text{g/cm}^3]\;Q^2\,[\text{L/s}]}{C_d^2\;A^2\,[\text{mm}^2]}

the density is in g/cm³, the flow rate in L/s, and the area in mm².

$500 = 10^3\cdot 10^{-6}/(2\cdot 10^{-12})\cdot 10^{-6}$ : g/cm³ to kg/m³, L/s to m³/s, mm² to m², Pa to MPa.

Oilfield units

\Delta p_{\text{b}}\,[\text{psi}] = \frac{8.311\cdot 10^{-5}\,\rho\,[\text{ppg}]\;q^2\,[\text{gpm}]}{C_d^2\;A^2\,[\text{in}^2]}

the density is in ppg, the flow rate in gallons per minute, and the area in square inches.

$8.311\cdot10^{-5} = 8.074\cdot10^{-4}/3.117^2$ : $8.074\cdot10^{-4}\rho v^2$ is the dynamic pressure in psi for a velocity in ft/s, and $v = q/(3.117A)$ .

By tradition the nozzle diameters are given in 32nds of an inch: “5 × 12” means five nozzles of 12/32 in (9.525 mm). The nozzles of one bit can be different, and then their areas are added. The velocity in the nozzles and the pressure drop are the same for all of them, and the flow through each nozzle is proportional to its area.

Worked exampleA bit with five nozzles, and one plugged nozzle

Given (assumed data)

Nozzles: 5 × 12/32 in, $C_d = 0.95$
Flow rate and mud: 30 L/s, 1.30 g/cm³

Solution

The diameter of a nozzle is $12/32\cdot 25.4 = 9.525$ mm, the area of one is $\frac{\pi}{4}\cdot 9.525^2 = 71.26$ mm², and the total area is $A = 5\cdot 71.26 = 356.3$ mm² (0.552 in²).
The velocity of the jets is $v_{\text{n}} = 0.030/(356.3\cdot 10^{-6}) = 84.2$ m/s.
The pressure drop is $\Delta p_{\text{b}} = \dfrac{500\cdot 1.30\cdot 30^2}{0.95^2\cdot 356.3^2} = 5.11$ MPa.
Check in oilfield units: $q = 30\cdot 15.85 = 475.5$ gpm, $\rho = 1.30\cdot 8.345 = 10.85$ ppg; $\Delta p = 8.311\cdot10^{-5}\cdot 10.85\cdot 475.5^2/(0.95^2\cdot 0.552^2) = 741$ psi, or 5.11 MPa.
One nozzle is plugged: $A = 4\cdot 71.26 = 285.0$ mm², the pressure drop is $5.11\cdot(5/4)^2 = 7.98$ MPa, and the standpipe pressure rises by 2.87 MPa at the same flow rate.

Answer: The pressure drop at the bit is about 5.11 MPa at a jet velocity of 84.2 m/s. If one nozzle plugs, it rises to 7.98 MPa.

What it means. A sudden rise of the standpipe pressure at a constant flow rate is a sign of a plugged nozzle, or of balling of the bottom hole assembly (sticky clay packed round it). A gradual fall is a sign of a washout in the drill string (the mud has eaten a hole through the wall of a pipe, and part of the flow leaks out). A sudden fall is a sign that a nozzle has been lost. A fall of pressure can also be an indirect sign of a kick (an inflow from the formation; see Well control: primary barrier and kick indicators). So we record the standpipe pressure at a known flow rate and keep comparing it with the calculation. The data in this example are assumed for illustration.

Standpipe pressure: calculating the whole system

Now let us put everything together for the well of the rheology lesson, this time with the drill collars, the surface lines and the bit. The calculation goes section by section, and the same steps are repeated in each section.

Order of the calculation

1
Geometry
Divide the path of the mud into sections with constant diameters: the surface lines, the drill pipe, the drill collars, the nozzles, and the sections of the annulus (at the drill collars, at the pipe in the open hole, in the casing).
2
Velocities
For each section, find the average velocity $v = Q/A$ .
3
Regime
Find the Reynolds number with the effective viscosity and compare it with the limits. Take the model parameters from the range of shear rates of the section.
4
Gradients
For laminar flow, use the exact solutions for a pipe and a slot. For turbulent flow, use the Fanning equation with a suitable $f$ . In the transition region, take the larger of the two.
5
Sum
Add the losses to the pressure drop at the bit, and compare the result with the rated pressure of the pumps and with the gauge on the standpipe.

The well and the mud of the example

Element: Surface lines
Dimensions: standpipe 101.6 mm; hose and top drive passage 76.2 mm
Length: 30 m; 20 m; 8 m

Element: Drill pipe
Dimensions: 127 × 108.6 mm
Length: 2850 m

Element: Drill collars
Dimensions: 158.8 × 71.4 mm
Length: 150 m

Element: Bit
Dimensions: 215.9 mm, nozzles 5 × 12/32 in
Length: —

Element: Casing
Dimensions: inside diameter 222.4 mm
Length: down to 2300 m

Element: Open hole
Dimensions: 215.9 mm (the nominal size of the bit)
Length: 2300–3000 m

Element	Dimensions	Length
Surface lines	standpipe 101.6 mm; hose and top drive passage 76.2 mm	30 m; 20 m; 8 m
Drill pipe	127 × 108.6 mm	2850 m
Drill collars	158.8 × 71.4 mm	150 m
Bit	215.9 mm, nozzles 5 × 12/32 in	—
Casing	inside diameter 222.4 mm	down to 2300 m
Open hole	215.9 mm (the nominal size of the bit)	2300–3000 m

The well is vertical. The mud is 1.30 g/cm³, with the viscometer readings from the rheology lesson, and the flow rate is 30 L/s. The data are assumed for illustration.

Worked exampleStandpipe pressure section by section

Given (assumed data)

Well, BHA and mud: the table above
Model: power-law: in the pipes $n = 0.670$ , $K = 0.345$ Pa·sⁿ; in the annulus $n = 0.383$ , $K = 1.64$ Pa·sⁿ

Solution

Surface lines: the velocities are 3.70 m/s in the standpipe and 6.58 m/s in the hose, and the flow is turbulent (Re about 8500 and 15,100). The losses are $0.065 + 0.156 + 0.062 = 0.28$ MPa.
Drill pipe (the example above): 4.64 MPa. Drill collars: velocity 7.49 m/s, $\text{Re} \approx 17{,}200$ , gradient 10,418 Pa/m, and over 150 m, 1.56 MPa.
Bit: 5.11 MPa.
Annulus, with laminar flow everywhere: at the drill collars $1314\cdot 150 = 0.20$ MPa; at the pipe in the open hole $622\cdot 550 = 0.34$ MPa; in the casing $545\cdot 2300 = 1.25$ MPa; in all, 1.79 MPa.
Standpipe pressure: $0.28 + 4.64 + 1.56 + 5.11 + 1.79 = 13.38$ MPa (1941 psi). The shares are: the string with the surface lines 48%, the bit 38%, the annulus 13%.
Hydraulic power: $13.38\cdot 30 = 401$ kW.
By the Bingham model (the plastic viscosity in the pipes, the simplified laminar formulas in the annulus) we get 14.64 MPa, of which 2.39 MPa is in the annulus.

Answer: The standpipe pressure is about 13.38 MPa. The drill string eats up almost half of it, the bit gets 38%, and the annulus 13%.

What it means. For the pumps, the annulus is a small item of expense, but only this item presses on the formation: the 1.79 MPa is what creates the ECD, which we look at in the next section. The drill pipe is the main “useless” loss. Pipe with a larger inside diameter would reduce it most of all (the losses fall very fast as the diameter grows: they are inversely proportional to almost the fifth power of the diameter). If the gauge on the standpipe shows a noticeably different number, we look for the reasons in the real well: the tool joints of the drill pipe (the thick threaded ends), washouts in the hole (places where the hole is wider than the bit made it), the temperature at the bottom, cuttings, wear of the nozzles, and the accuracy of the flow meter. The data in this example are assumed for illustration.

Calculation section by section: power-law model, 30 L/s

Section: standpipe
$v$ , m/s: 3.70
Re: 8500
Regime: turbulent
$\Delta p$ , MPa: 0.07

Section: rotary hose
$v$ , m/s: 6.58
Re: 15,100
Regime: turbulent
$\Delta p$ , MPa: 0.16

Section: top drive passage
$v$ , m/s: 6.58
Re: 15,100
Regime: turbulent
$\Delta p$ , MPa: 0.06

Section: drill pipe
$v$ , m/s: 3.24
Re: 7500
Regime: turbulent
$\Delta p$ , MPa: 4.64

Section: drill collars
$v$ , m/s: 7.49
Re: 17,200
Regime: turbulent
$\Delta p$ , MPa: 1.56

Section: bit nozzles
$v$ , m/s: 84
Re: —
Regime: —
$\Delta p$ , MPa: 5.11

Section: annulus: drill collars in the open hole
$v$ , m/s: 1.79
Re: 2700
Regime: laminar
$\Delta p$ , MPa: 0.20

Section: annulus: drill pipe in the open hole
$v$ , m/s: 1.25
Re: 1800
Regime: laminar
$\Delta p$ , MPa: 0.34

Section: annulus: drill pipe in the casing
$v$ , m/s: 1.15
Re: 1600
Regime: laminar
$\Delta p$ , MPa: 1.25

Section	$v$ , m/s	Re	Regime	$\Delta p$ , MPa
standpipe	3.70	8500	turbulent	0.07
rotary hose	6.58	15,100	turbulent	0.16
top drive passage	6.58	15,100	turbulent	0.06
drill pipe	3.24	7500	turbulent	4.64
drill collars	7.49	17,200	turbulent	1.56
bit nozzles	84	—	—	5.11
annulus: drill collars in the open hole	1.79	2700	laminar	0.20
annulus: drill pipe in the open hole	1.25	1800	laminar	0.34
annulus: drill pipe in the casing	1.15	1600	laminar	1.25

For the nozzles, the jet velocity is given instead of the velocity of the flow. The Reynolds number is the generalised one, and in the annulus it uses the hydraulic diameter. The limits of laminar flow are 2552 in the pipes and 2945 in the annulus.

CalculatedThe calculated components of the standpipe pressure against the flow rate for the well of the example (power-law model, logarithmic scales). The slope of a line is the exponent: 2 for the bit, about 1.6 for the string, and about 0.3–0.4 for the laminar annulus. At low flow rates the annulus is a noticeable share of the pressure, and at high flow rates it is hardly noticeable.

The ECD along the hole and the leak-off pressure

The annulus losses build up from the top downwards: at a depth $h$ , the losses of all the sections of the annulus above this depth are added to the hydrostatic pressure. While the hole is cased with one string and the gap is constant, the gradient of the losses is constant, and the ECD does not change with depth in the casing. Below the shoe it grows, and it grows fastest at the drill collars, where the gap is narrow. The dangerous points are the shoe, where the rock is usually the weakest (allowable pressures), and any formation in the open hole with a low leak-off pressure.

ECD for an annulus of several sections

\rho_{ECD}(h) = \rho_m + \frac{1}{g\,h}\sum_i \left(\frac{\Delta p}{L}\right)_i \Delta L_i

where:

$\rho_{ECD}(h)$ equivalent circulating density at depth $h$ , kg/m³
$(\Delta p/L)_i$ gradient of the losses in the $i$ -th section of the annulus, Pa/m
$\Delta L_i$ length of the $i$ -th section above the depth $h$ (for the section that $h$ passes through, only its upper part), m
$\rho_m,\ g,\ h$ density of the mud, acceleration of gravity, true vertical depth, kg/m³, m/s², m

Units: 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).

The ECD is an average weighted by depth. A section below the shoe has no effect at all on the ECD at the shoe, and at the bottom its losses are divided by the whole depth, so a short section with a large gradient raises the ECD there only a little.

CalculatedThe calculated ECD along the depth for the well of the example at 30 L/s: by the power-law model (solid line) and by the Bingham model (dashed). In the casing the ECD is constant, and below the shoe it grows. The Bingham model gives 0.02 g/cm³ more. The dots are the leak-off pressure below the shoe (1.42 g/cm³) and the pore pressure at the bottom (1.24 g/cm³).

Worked exampleECD at the shoe and at the bottom, cuttings and the flow rate

Given (assumed data)

Well and mud: the example “Standpipe pressure section by section”
Leak-off pressure below the shoe (2300 m): 1.42 g/cm³ from the test
Pore pressure at the bottom: 1.24 g/cm³
Rate of penetration (how fast the bit drills): 25 m/h; cuttings of density 2.6 g/cm³

Solution

Above the shoe there is only the section inside the casing, with a constant gradient 545 Pa/m. A column of mud of density 1 g/cm³ gives 9810 Pa/m, so in the casing at any depth $\rho_{ECD} = 1.30 + 545/9810 = 1.356$ g/cm³. The margin to the leak-off pressure is 0.064 g/cm³.
At the bottom there are all the annulus losses: $1.30 + \dfrac{1.79}{0.00981\cdot 3000} = 1.361$ g/cm³. The overbalance (the amount by which the mud pressure exceeds the pore pressure) while circulating is 0.121 g/cm³, instead of 0.06 g/cm³ without circulation.
By the Bingham model: at the shoe 1.375 g/cm³ (margin 0.045 g/cm³), at the bottom 1.381 g/cm³.
Cuttings (we neglect the porosity of the rock, which makes their volume slightly too high): the volume of rock drilled per second is $\dfrac{25}{3600}\cdot 0.0366 = 0.000254$ m³/s. Without slip (the cuttings move as fast as the mud) their share in the annulus is $0.000254/(0.030 + 0.000254) = 0.0084$ , and the added weight is $0.0084\cdot(2.6 - 1.30) = 0.011$ g/cm³. If the cuttings rise half as fast as the mud, the share is 0.0167 and the added weight is 0.022 g/cm³. The ECD at the shoe is then 1.38 g/cm³.
A flow rate of 36 L/s instead of 30: the ECD at the shoe is 1.360 g/cm³ (+0.004), and the standpipe pressure is 18.0 MPa instead of 13.4.

Answer: The ECD at the shoe is 1.356 g/cm³ (1.375 by the Bingham model), and the margin to the leak-off pressure is 0.064 g/cm³. With cuttings at 25 m/h, the margin shrinks to 0.04 g/cm³.

What it means. A margin of a few hundredths of a g/cm³ is quite usual, and it is easy to “eat it up”: cuttings in fast drilling, a mud that has thickened, a pressure surge when pipe is run in. The flow rate has little effect on the ECD here, because the annulus is laminar and the mud thins out strongly under shear. But it raises the standpipe pressure strongly, so the pumps limit the flow rate, not the formation. In narrow gaps (liners, small-diameter wells) it is the other way round, and the ECD limits the flow rate. The calculation by the Bingham model errs on the safe side. When the window (the gap between the pore pressure and the leak-off pressure) is narrow, we measure the properties of the mud at low shear rates and calculate with the models that describe them. The data in this example are assumed for illustration.

How the losses change with the flow rate and the properties of the mud

The terms of the balance respond differently to the flow rate, the density and the rheology. We need to know this to understand what will change after we make the mud heavier or dilute it, or change the nozzles, and so that we do not recalculate the pressures with one formula.

What each term depends on, and how

Term: Bit
Flow rate $Q$: $Q^2$
Density: $\rho$
Rheology: no effect

Term: Pipes, turbulent
Flow rate $Q$: $Q^{1.6}$ – $Q^{1.75}$
Density: about $\rho^{0.75}$
Rheology: weak: about $\eta_p^{0.25}$

Term: Annulus, laminar (Bingham)
Flow rate $Q$: the viscous part $\propto Q$ ; the YP part does not depend on it
Density: no effect
Rheology: $\eta_p$ and $\tau_0$ directly

Term: Annulus, laminar (power-law)
Flow rate $Q$: $Q^{\,n}$
Density: no effect
Rheology: $K$ directly

Term: Hydrostatic pressure
Flow rate $Q$: no effect
Density: $\rho$
Rheology: no effect

Term	Flow rate $Q$	Density	Rheology
Bit	$Q^2$	$\rho$	no effect
Pipes, turbulent	$Q^{1.6}$ – $Q^{1.75}$	about $\rho^{0.75}$	weak: about $\eta_p^{0.25}$
Annulus, laminar (Bingham)	the viscous part $\propto Q$ ; the YP part does not depend on it	no effect	$\eta_p$ and $\tau_0$ directly
Annulus, laminar (power-law)	$Q^{\,n}$	no effect	$K$ directly
Hydrostatic pressure	no effect	$\rho$	no effect

The diameters matter most for the turbulent losses in the pipes ( $d^{-4.75}$ ) and for the pressure drop at the bit ( $A^{-2}$ ). The laminar losses in the annulus depend on the width of the gap: the viscous part as $(D-d)^{-3}$ at the same flow rate, and the YP part as $(D-d)^{-1}$ .

Because of this, the system losses (the parasitic pressure losses: all the losses except the pressure drop at the bit) grow with the flow rate roughly as $Q^m$ . Here $m$ is a weighted average exponent: about 1.6–1.8 when the turbulent pipes give most of the losses, and clearly smaller when the share of the laminar annulus is large. We find $m$ from two readings of the standpipe pressure at different flow rates, subtracting the calculated pressure drop at the bit from each: $m = \ln(\Delta p_2/\Delta p_1)/\ln(Q_2/Q_1)$ . It is used when choosing the nozzles (Bit hydraulics optimisation).

Worked exampleWhy we measure the pressure at a reduced flow rate, and do not calculate it

Given (assumed data)

Well and mud: the example “Standpipe pressure section by section”: 30 L/s gives 13.38 MPa
Reduced flow rate (for example, for well killing): 10 L/s

Solution

A naive conversion with the exponent 1.8: $13.38\cdot(10/30)^{1.8} = 1.85$ MPa.
Calculation section by section at 10 L/s: the string with the surface lines 1.050 MPa (in the drill pipe the flow becomes laminar: $\text{Re} = 1734$ is below 2552), the bit 0.567 MPa, the annulus 1.177 MPa; in all, 2.794 MPa.
The share of the annulus has grown from 13% to 42%: the laminar losses hardly fall with the flow rate.

Answer: At 10 L/s the standpipe pressure is about 2.8 MPa, and not 1.85 MPa, as the conversion with one exponent would give.

What it means. A mistake by a factor of one and a half would mean a wrong bottomhole pressure during well killing, because we control that pressure by the pressure in the drill pipe. So the pressure at a reduced flow rate (the slow circulating rate pressure, SCR) is measured regularly with the rig pump (Well kill methods). The model of this lesson also simplifies. In the transitional regime the real losses may be different, and this is one more argument for measuring. The data in this example are assumed for illustration.

Running and pulling pipe: pressure surges

The same formulas describe the pressure changes when the string moves without circulation. When the string is run in, it displaces the mud, and the mud must go up the annulus. The pressure below the bit rises (this is called surge). When the string is pulled out, the volume that it leaves is filled with mud from above, and the pressure falls (swabbing, the piston effect of the string; this is called swab). What this means for well control is explained in the lesson Well control: primary barrier and kick indicators. Here we show how to calculate it.

Velocity of the mud in the annulus when the string moves with a closed end

v_{\text{ann}} = v_{\text{pipe}}\left(\frac{d^2}{D^2 - d^2} + K_c\right)

where:

$v_{\text{ann}}$ calculated velocity of the mud in the annulus relative to the wall, m/s
$v_{\text{pipe}}$ speed of movement of the string; we take the highest speed during the movement of one stand (several joints of pipe screwed together), about one and a half times the average, m/s
$d,\ D$ outside diameter of the pipes (drill collars) and the diameter of the hole or the casing, m
$K_c$ clinging constant: part of the mud moves together with the pipe; in laminar flow it is about 0.4–0.5, —

Units: 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.

The first term is the displacement of the volume of the metal, and the second is the mud that the pipe drags along with it. The pressure change is the sum of the losses over all the sections of the annulus from the bit to the surface at these velocities.

Worked exampleSwabbing when pulling pipe out of the well of the example

Given (assumed data)

Well and mud: the example “Standpipe pressure section by section”; the string has a float valve
Highest pulling speed: 0.6 m/s; $K_c = 0.45$
Pore pressure at the bottom: 1.24 g/cm³ (the overbalance without circulation is 0.06 g/cm³)

Solution

The ratios $d^2/(D^2 - d^2)$ : in the casing 0.484, at the pipe in the open hole 0.529, at the drill collars 1.179. The velocities of the mud are 0.560, 0.587 and 0.977 m/s.
The gradients by the power-law model (100 and 3 rpm); the flow is laminar (Re from 496 to 1001): 414.3, 465.1 and 1042.8 Pa/m.
The fall of the pressure at the bottom: $414.3\cdot 2300 + 465.1\cdot 550 + 1042.8\cdot 150 = 1.365$ MPa, that is $1.365/(0.00981\cdot 3000) = 0.046$ g/cm³ in equivalent density.
While the pipe is being pulled, the bottom feels $1.30 - 0.046 = 1.254$ g/cm³. This is above the pore pressure of 1.24 g/cm³, but the margin is only 0.40 MPa.
At twice the speed, 1.2 m/s, the swab is 1.78 MPa, which is more than the whole overbalance 1.77 MPa.
By the Bingham model (the simplified formula) at 0.6 m/s we would get 2.10 MPa, which is more than the whole overbalance.

Answer: At 0.6 m/s the pressure at the bottom falls by 1.365 MPa (0.046 g/cm³). At 1.2 m/s the swab will exceed the overbalance.

What it means. Here the swab depends only weakly on the speed (twice as fast gives about a third more). A mud with $n = 0.38$ behaves almost like a plastic solid. So slow pulling is not the only help against swabbing. A mud with a low YP and low gels helps, and above all so does keeping track of the fill-up volumes (the volume of mud that must be added to the hole as the pipe comes out). The Bingham model greatly overestimates the pressure changes: at low shear rates the true stress is much smaller than the YP. But in the first seconds of the movement, until the gel is broken, the pressure peak really can be large. When pipe is run in, the same change has the opposite sign: it adds to the hydrostatic pressure, and we compare the sum with the leak-off pressure below the shoe. The data in this example are assumed for illustration.

Terms and symbols

Standpipe pressure: The pressure on the gauge of the discharge line during circulation. It is the sum of the losses in the whole system.
System losses (parasitic pressure loss): The pressure losses in the surface lines, the string and the annulus: everything except the pressure drop at the bit.
Shear stress at the wall, $\tau_w$: The friction force on the wall per unit of area. The laminar losses are $4\tau_w/d$ .
Generalised Reynolds number: The Reynolds number with the apparent viscosity at the wall. For a power-law fluid the laminar limit is $3470 - 1370n$ .
Fanning friction factor, $f$: A dimensionless number, $2\tau_w/(\rho v^2)$ . In laminar flow in a pipe it is $16/\text{Re}$ . The related coefficient is $\lambda = 4f$ .
Hydraulic diameter: Four times the cross-sectional area divided by the wetted perimeter. For an annulus, it is $D - d$ .
Total flow area of the nozzles (TFA): The sum of the cross-sectional areas of all the bit nozzles. We use it to calculate the pressure drop and the velocity of the jets.
Discharge coefficient, $C_d$: A correction to the velocity of the jet for the losses in the nozzle. It is usually 0.95.
ECD: The mud density plus the annulus losses above a given depth, expressed as a density.
Clinging constant, $K_c$: The share of the speed of the string at which the mud moves along with the pipe when pipe is run in or pulled out.

Common mistakes and misconceptions

Wrong

“The thicker the mud, the higher the pump pressure, so let us thin the mud and the pressure problem is solved.”

In fact

The main losses are at the bit and in the turbulent pipes, and they hardly depend on the rheology. Thinning the mud noticeably lowers the annulus losses, that is, the ECD, but it changes the standpipe pressure little.

Wrong

“The annulus can be calculated with the same parameters as the pipes.”

In fact

The shear rates in the annulus are several times smaller. With the parameters from 600 and 300 rpm, the power-law model gives losses that are too low, and the Bingham model gives losses that are too high. For the annulus we use the readings at low speeds, or the Herschel–Bulkley model.

Wrong

“If the flow is turbulent, the YP matters as much as in laminar flow.”

In fact

In turbulent flow the YP hardly affects the losses. They are decided by the density, the flow rate, the diameter and (weakly) the plastic viscosity.

Wrong

“The ECD at the bottom and at the shoe is the same.”

In fact

At the shoe we count only the losses above the shoe. A short section with a narrow gap near the bottom raises the ECD at the bottom, but not at the shoe.

Wrong

“Slow pulling removes swabbing completely.”

In fact

In a shear-thinning mud the losses depend only weakly on the speed, and the gel gives a pressure peak at the start of the movement. Both the properties of the mud and the control of the fill-up volumes matter.

In depth: advanced detail

Exact laminar solutions and equivalent diameters

For a viscoplastic fluid in a pipe, the exact solution (the Buckingham equation) links the average velocity with the stress at the wall: $v = \dfrac{\tau_w d}{8\eta_p}\left(1 - \dfrac{4}{3}x + \dfrac{x^4}{3}\right)$ , where $x = \tau_0/\tau_w$ . If we drop the term $x^4/3$ , we get $\tau_w \approx 8\eta_p v/d + 4\tau_0/3$ and, after multiplying by $4/d$ , the simplified formula of the lesson. The dropped term is positive, so the simplification overestimates the losses, and the more so the closer $\tau_0$ is to $\tau_w$ .

For a slot (the annulus) the general way is this. The stress grows linearly from zero in the middle of the slot to $\tau_w$ at the wall. The model gives the shear rate at each stress, and integrating over the cross-section links the average velocity with $\tau_w$ : $v = \dfrac{h}{\tau_w^2}\displaystyle\int_0^{\tau_w}\tau\,\dot\gamma(\tau)\,\dd\tau$ , where $h = (D-d)/4$ is half the width of the slot. For a Newtonian fluid this gives $48\mu v/(D-d)^2$ , and for a power-law fluid the formula of the lesson. For the Herschel–Bulkley model the integral can be taken in elementary functions, and $\tau_w$ is found by trial and error.

Newtonian fluid in a concentric annulus: the exact solution

\frac{\Delta p}{L} = \frac{32\,\mu\,v}{D^2 + d^2 - \dfrac{D^2 - d^2}{\ln(D/d)}}

where:

$\mu$ viscosity, Pa·s
$v$ average velocity in the annulus, m/s
$D,\ d$ outside and inside diameters of the ring, m

Units: SI; natural logarithm

Holds when

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

Comparing with the slot and with the pipe gives three “equivalent diameters” of the annulus. The hydraulic diameter is $D - d$ . The laminar diameter is the square root of the denominator of this formula, for which the Poiseuille formula gives the exact answer. The slot diameter is $0.816\,(D - d)$ , for which the Poiseuille formula gives the answer for a slot. For a casing of 222.4 mm and pipe of 127 mm these are 95.4, 78.1 and 77.9 mm. The slot is almost exact, and the hydraulic diameter is suitable for the Reynolds number, but not for the laminar losses.

Which model is closer to the truth

Let us compare five ways to calculate the laminar losses for the section of the annulus in the casing (30 L/s, velocity 1.15 m/s). All of them use the readings of the same mud. The Herschel–Bulkley model describes the mud best over the whole range of shear rates (the rheology lesson), so its exact solution can be taken as the reference.

The annulus in the casing: the loss gradient and the ECD at the shoe by different models

Model and method: Bingham, simplified formula (as in the rheology lesson)
Gradient, Pa/m: 736
ECD at the shoe, g/cm³: 1.375

Model and method: Bingham, exact solution for a slot
Gradient, Pa/m: 672
ECD at the shoe, g/cm³: 1.369

Model and method: power-law, from 600 and 300 rpm
Gradient, Pa/m: 447
ECD at the shoe, g/cm³: 1.346

Model and method: power-law, from 100 and 3 rpm
Gradient, Pa/m: 545
ECD at the shoe, g/cm³: 1.356

Model and method: Herschel–Bulkley, exact solution for a slot
Gradient, Pa/m: 507
ECD at the shoe, g/cm³: 1.352

Model and method	Gradient, Pa/m	ECD at the shoe, g/cm³
Bingham, simplified formula (as in the rheology lesson)	736	1.375
Bingham, exact solution for a slot	672	1.369
power-law, from 600 and 300 rpm	447	1.346
power-law, from 100 and 3 rpm	545	1.356
Herschel–Bulkley, exact solution for a slot	507	1.352

The mud is 1.30 g/cm³ with the readings of the rheology lesson. The exact solutions for a slot are found by trial and error on the stress at the wall.

The Bingham model overestimates the losses in the annulus by about a third, even in the exact solution. This is a property of the model itself, not of the simplification of the formula. The power-law model from 600 and 300 rpm underestimates them by 10–15%, and from 100 and 3 rpm it differs from the reference by less than 10%. The difference between the extreme estimates is 0.03 g/cm³ of ECD. For a narrow window this is a lot, so modern methods of hydraulic calculation use the Herschel–Bulkley model with the readings at low speeds, measured if possible at the temperature and pressure of the bottom of the hole.

The Hedström criterion for a viscoplastic fluid

The simple rule of “effective viscosity and a limit of 2100” is only an estimate for a viscoplastic fluid. The core of the flow, which moves like a solid, damps the disturbances. So the laminar regime stays up to Reynolds numbers, calculated with the plastic viscosity, that are much greater than 2100. The dimensionless Hedström number shows more exactly where the transition comes. In Russian-language practice it is traditionally used in calculating the flow of viscoplastic muds.

The Hedström number and the critical Reynolds number (Hanks' form)

\text{He} = \frac{\rho\,\tau_0\,d^2}{\eta_p^2}, \qquad \frac{x}{(1 - x)^3} = \frac{\text{He}}{16{,}800}, \qquad \text{Re}_{\text{cr}} = \frac{\text{He}}{8x}\left(1 - \frac{4x}{3} + \frac{x^4}{3}\right)

where:

$\text{He}$ the Hedström number, —
$x$ the ratio $\tau_0/\tau_w$ at the moment of the transition, —
$\text{Re}_{\text{cr}}$ the critical Reynolds number $\rho v d/\eta_p$ (with the plastic viscosity), —
$d$ diameter of the pipe; for the annulus, approximately the hydraulic diameter, m

Units: 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.

The greater the YP in comparison with the viscosity, and the wider the channel, the later turbulence sets in.

Worked exampleDrill pipe and the annulus at the drill collars by the Hedström criterion

Given (assumed data)

Mud: 1.30 g/cm³, PV 26 mPa·s, YP 9.20 Pa; 30 L/s
Channels: pipe 108.6 mm; the annulus at the drill collars with a gap of 57.1 mm

Solution

Pipe: $\text{He} = 1300\cdot 9.20\cdot 0.1086^2/0.026^2 = 208{,}662$ ; $x = 0.6299$ ; $\text{Re}_{\text{cr}} \approx 8800$ . The Reynolds number with the plastic viscosity, 17,586, is greater, so the flow is turbulent.
Annulus at the drill collars: $\text{He} = 1300\cdot 9.20\cdot 0.0571^2/0.026^2 = 57{,}684$ , $x = 0.4807$ , $\text{Re}_{\text{cr}} \approx 5700$ . The Reynolds number, 5097, is smaller, so the flow is laminar, although the simple rule put it on the limit (about 2110).

Answer: In the pipe the flow is turbulent (17,586 > 8800), and in the annulus at the drill collars it is laminar (5097 < 5700).

What it means. The Hedström criterion agrees with the conclusion of the power-law model (laminar flow at the drill collars) and removes the doubt that the simple rule left. The data in this example are assumed for illustration.

What the calculation leaves out, and where this shifts the result

Eccentricity (the string is off-centre in the hole). The string usually lies against the wall, especially in an inclined hole. In the wide part of the ring the mud flows more easily, and the laminar losses in an eccentric annulus are smaller than in a concentric one, so the calculation overestimates them (by tens of per cent when the pipe lies on the wall). But in the narrow part the mud almost stands still, and cuttings pile up there.
Rotation of the string. It changes the flow in the annulus. In narrow gaps and at high rotary speeds the losses and the ECD usually grow. A calculation without rotation does not account for this.
Tool joints. Local narrowings inside and outside the string add losses. A calculation for a smooth pipe underestimates them, especially in the annulus with small gaps.
Temperature and pressure. The plastic viscosity at a hot bottom is lower than when measured at 50 °C. A calculation based on such measurements usually overestimates the viscous part of the losses (the YP may also grow when heated). The viscosity of a hydrocarbon base grows with pressure (Oil-based muds), so the opposite error is possible in such muds.
Hole washouts and cuttings. A washed-out hole is wider than the nominal size, so the losses are smaller than calculated. Cuttings, and cuttings beds (layers of cuttings lying on the low side of the hole) in inclined holes, narrow the cross-section and make the column heavier, so the ECD is higher than calculated.
Drag reduction by polymers. In turbulent flow, the long molecules of polymers damp the eddies near the wall, and the losses in the pipes can be noticeably lower than in the correlations for a smooth pipe.

Connections to other topics

Drilling fluid rheology and flow regimes — the flow models, the parameters from the viscometer, the regimes and the simplified formula for the losses in the annulus.
Fracture gradient and leak-off testing — the leak-off pressure below the shoe, with which the ECD is compared.
Well control: primary barrier and kick indicators — swabbing, stopping the pumps and gas cutting of the mud as the causes of kicks.
Well kill methods — the pressure at a reduced flow rate and the circulating pressures during killing.
Bit hydraulics optimisation — how to split the pump pressure between the bit and the rest of the circulating system.
Deflection tools and downhole motors — the pressure drop across the downhole motor and what it says about the load.
Field testing of drilling fluids — how the rheology of the mud, from which the losses are calculated, is measured.
Oil-based muds — the rheology and the density of muds with a hydrocarbon base at the pressure and the temperature of the bottom of the hole.

Check yourself

1
The flow rate was raised from 25 to 30 L/s; the mud and the nozzles are the same. By how many times does the pressure drop across the bit grow?
1. A.About 1.44 times
2. B.It does not change: the nozzles decide the pressure drop
3. C.About 1.38 times
4. D.1.2 times
Show the answer
Correct answer: A. About 1.44 times
The pressure drop across the nozzles is proportional to the square of the flow rate: $(30/25)^2 = 1.44$ . The turbulent losses in the pipes would grow by about 1.34–1.38 times (the exponent is 1.6–1.75), and the laminar losses in the annulus by much less.
2
A mud of 1.40 g/cm³ with a plastic viscosity of 30 mPa·s and a YP of 7 Pa flows at an average velocity of 0.8 m/s in a pipe with an inside diameter of 70 mm. Check the regime and find the losses per metre.
Show the answer
The effective viscosity is $0.030 + 7\cdot 0.07/(6\cdot 0.8) \approx 0.132$ Pa·s; $\text{Re} = 1400\cdot 0.8\cdot 0.07/0.132 \approx 590$ , so the flow is laminar. The losses are $32\cdot 0.030\cdot 0.8/0.07^2 + 16\cdot 7/(3\cdot 0.07) \approx 157 + 533 \approx 690$ Pa/m. Almost four fifths came from the YP term: in a slow flow it is the main one.
3
In which section of the circulating system is the flow of the mud most often laminar?
1. A.In the annulus opposite the drill pipe
2. B.In the bit nozzles
3. C.In the drill pipe
4. D.In the surface lines
Show the answer
Correct answer: A. In the annulus opposite the drill pipe
In the annulus the velocity is low, and the apparent viscosity of a shear-thinning mud at a low shear rate is high, so the generalised Reynolds number is below the limit. In the pipes and the surface lines the velocities are several times higher, and the flow is turbulent. In the nozzles there is a jet.
4
The annulus: 2500 m in the casing with a loss gradient of 500 Pa/m, and below it 400 m of open hole with a gradient of 900 Pa/m. The mud is 1.20 g/cm³. Find the ECD at the shoe (2500 m) and at the bottom (2900 m).
Show the answer
At the shoe: $1.20 + 500\cdot 2500\cdot 10^{-6}/(0.00981\cdot 2500) = 1.20 + 1.25/24.5 \approx 1.251$ g/cm³. At the bottom: the losses are $1.25 + 0.36 = 1.61$ MPa, and $1.20 + 1.61/(0.00981\cdot 2900) \approx 1.257$ g/cm³. The section below the shoe has no effect on the ECD at the shoe, and at the bottom the ECD is only 0.006 g/cm³ greater: the losses of this section are divided by the whole depth.
5
At a constant flow rate the standpipe pressure suddenly rose by 2.5 MPa. What is most likely?
1. A.A washout in a drill pipe
2. B.The mud was diluted with water
3. C.A bit nozzle has been lost
4. D.One of the bit nozzles is plugged, or the bottom hole assembly has balled up
Show the answer
Correct answer: D. One of the bit nozzles is plugged, or the bottom hole assembly has balled up
A smaller area of the nozzles or of the annulus raises the pressure. A washout in a pipe and the loss of a nozzle, on the contrary, open an additional path for the mud and lower the pressure. Diluting the mud also lowers it.
6
Why can we not get the standpipe pressure at a reduced flow rate by multiplying the pressure while drilling by $(Q_2/Q_1)^2$ ?
Show the answer
Only the pressure drop at the bit grows as a square. The turbulent losses in the pipes grow to the power 1.6–1.75, and at a low flow rate the flow in them may become laminar. The laminar losses in the annulus hardly depend on the flow rate. So at a low flow rate the share of the annulus grows, and the conversion gives a pressure that is too low. We measure the pressure at a reduced flow rate.

Updated: 8 October 2026