Bit hydraulics optimisation

Quick refresher

The power of the pumps, $N = p_{\text{SPP}}Q$ (MPa × L/s = kW), goes on friction in the system (the surface lines, the pipe, the drill collars, the annulus) and on speeding up the jets in the nozzles. Only the part that is released at the bit can clean the bottom of the hole: $N_{\text{b}} = \Delta p_{\text{b}}Q$ .
The system losses grow with the flow rate roughly as $Q^m$ . We find the exponent $m$ from the standpipe pressure at two flow rates, subtracting the calculated pressure drop at the bit from each reading. Because of the laminar annulus and the thinning of the mud under shear, $m$ is often clearly smaller than the “turbulent” 1.75–1.86. In the lesson's example it is 1.32.
If the pumps work at the pressure limit $p_{\max}$ , the power at the bit is greatest when the system takes $p_{\max}/(m+1)$ of the pressure and the bit takes $m/(m+1)$ (57% for $m = 1.32$ ). The impact force is greatest when the system takes $2p_{\max}/(m+2)$ : the flow rate is greater and the pressure drop at the bit is smaller.
The optimum is flat: if we move 15–20% away from the optimal flow rate, the power at the bit falls by only a few per cent. So the result is more often decided by the limits (the smallest and the largest flow rate, the pressure and the power of the pumps, the ECD) than by the choice of the criterion.
The required nozzle area is $A = (Q/C_d)\sqrt{500\rho/\Delta p_{\text{b}}}$ (mm², L/s, g/cm³, MPa). We make it up from nozzles of two neighbouring sizes and check the standpipe pressure.
The nozzles stay the same for the whole bit run. The system losses grow with depth, and to stay below $p_{\max}$ the flow rate is gradually lowered. We calculate for the interval of the run and correct the result with pressure readings.

Key relationships

N_{\text{b}} = \Delta p_{\text{b}}\,Q, \qquad F = \rho\,Q\,v_{\text{n}} = C_d\,Q\sqrt{2\rho\,\Delta p_{\text{b}}}

hydraulic power at the bit and jet impact force

m = \dfrac{\ln(\Delta p_{\text{sys}2}/\Delta p_{\text{sys}1})}{\ln(Q_2/Q_1)}

exponent of the system losses from two flow rates

\Delta p_{\text{sys}} = \dfrac{p_{\max}}{m+1}\ \ \text{(power)}, \qquad \Delta p_{\text{sys}} = \dfrac{2\,p_{\max}}{m+2}\ \ \text{(impact force)}

system losses at the optimum when the pumps are at their pressure limit

A = \dfrac{Q}{C_d}\sqrt{\dfrac{500\,\rho}{\Delta p_{\text{b}}}}

total flow area of the nozzles, mm² ( $Q$ in L/s, $\rho$ in g/cm³, $\Delta p_{\text{b}}$ in MPa)

Common pitfalls

Taking the “typical” exponent 1.86 instead of the measured one. With a laminar annulus the calculation gives too low a flow rate and nozzles that are too small.
Forgetting to subtract the pressure drop at the old nozzles from the standpipe pressure before calculating $m$ .
Taking readings at flow rates that are close together. A gauge error of 0.2 MPa changes $m$ noticeably.
Optimising the hydraulics without the limits. A flow rate below what cuttings removal needs, or an ECD above the allowable one, makes any power at the bit worthless.
Keeping the same flow rate until the end of the bit run. The standpipe pressure grows with depth, and the flow rate has to be lowered.

After this lesson you can

explain what the jets do at the bottom of the hole, and how the hydraulic power at the bit, the jet impact force and the jet velocity differ
find the exponent $m$ of the system losses (the pressure losses everywhere except at the bit) from the standpipe pressure (the reading of the gauge on the standpipe, the vertical pipe that feeds mud to the string) at two flow rates, and explain why it is usually below two
derive how to split the pump pressure between the bit and the system, first for the greatest power and then for the greatest impact force
choose the flow rate within the limits of the pumps, the cuttings removal, the flow regime in the annulus and the equivalent circulating density (ECD, the density that the mud effectively has while it circulates), including a graphical method on logarithmic axes
select the nozzles for the required total flow area and check the operating point: the standpipe pressure, the share of the pressure drop at the bit, the specific power and the jet velocity
explain how the result is used on the rig and how it changes towards the end of the bit run

What the jets do at the bottom of the hole

Under the bit the rock breaks into small fragments, and they must be removed before the next tooth or cutter passes over the same place. Two things get in the way. The pressure of the mud in the well is usually higher than the pore pressure (the pressure of the fluid in the rock), and a fresh fragment is pressed to the bottom by the pressure difference until the mud gets into the crack under it. This is called chip hold-down. Also, clayey cuttings stick to the bit (bit balling) and cover the cutting structure. The jets from the nozzles hit the bottom, tear the fragments off and sweep them towards the annulus, and wash the cones (on a roller-cone bit) or the blades (on a bit with fixed cutters). For bits with polycrystalline diamond compact (PDC) cutters, the flow also cools the cutters.

It took time to agree what to count as the “strength” of the jets. There are three measures: the hydraulic power at the bit, which is the energy that the jets bring to the bottom of the hole in a second; the jet impact force, which is the momentum (mass times velocity) that they give to the bottom; and the jet velocity. All three grow with the pressure drop at the bit, but they depend on the flow rate in different ways. Which flow rate turns out to be the best depends on the measure we choose.

Hydraulic power at the bit and specific power

N_{\text{b}} = \Delta p_{\text{b}}\,Q, \qquad N_{\text{sp}} = \frac{N_{\text{b}}}{\pi D^2/4}

where:

$N_{\text{b}}$ hydraulic power that is released in the bit nozzles, W
$\Delta p_{\text{b}}$ pressure drop across the bit nozzles (how it is calculated), Pa
$Q$ flow rate of the pumps, m³/s
$N_{\text{sp}}$ specific hydraulic power: per unit area of the bottom of the hole (hydraulic horsepower per square inch, HSI), W/m²
$D$ bit diameter, m

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

Holds when

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

The pumps give the mud the power $p_{\text{SPP}}Q$ , but only the part released at the bit can clean the bottom of the hole. The rest is the price of making the mud go through the pipe and up the annulus.

In other unit systems (1)

Oilfield units

N_{\text{b}}\,[\text{hp}] = \frac{\Delta p_{\text{b}}\,[\text{psi}]\;q\,[\text{gpm}]}{1714}, \qquad \text{HSI} = \frac{N_{\text{b}}\,[\text{hp}]}{\pi d^2/4\,[\text{in}^2]}

the power is in British horsepower (hp, 745.7 W, not the metric 735.5 W), the pressure drop in psi, the flow rate in gallons per minute, and the bit diameter in inches.

$1714 = 745.7/(6894.76\cdot 6.309\cdot10^{-5})$ : 1 psi × 1 gpm = 0.435 W. 1 hp/in² = 1.156 W/mm².

Jet impact force

F = \rho\,Q\,v_{\text{n}} = C_d\,Q\sqrt{2\rho\,\Delta p_{\text{b}}}, \qquad v_{\text{n}} = \frac{Q}{A} = C_d\sqrt{\frac{2\,\Delta p_{\text{b}}}{\rho}}

where:

$F$ the force with which the jets press on the bottom, that is, the momentum that they bring in a second, N
$\rho$ density of the mud, kg/m³
$v_{\text{n}}$ velocity of the jets in the nozzles, m/s
$A$ total flow area of the nozzles (TFA), m²
$C_d$ discharge coefficient of the nozzles, about 0.95, —

Units: SI

Holds when

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

The impact force grows as $Q\sqrt{\Delta p_{\text{b}}}$ , and the power as $Q\,\Delta p_{\text{b}}$ . For the same power, the force is greater the greater the flow and the slower the jets.

In other unit systems (2)

Metric units

F\,[\text{kN}] = 0.04472\,C_d\,Q\,[\text{L/s}]\sqrt{\rho\,[\text{g/cm}^3]\;\Delta p_{\text{b}}\,[\text{MPa}]}

the flow rate is in L/s, the density in g/cm³, the pressure drop in MPa, and the force in kN. It is also convenient to use $F\,[\text{kN}] = \rho\,Q\,v_{\text{n}}/1000$ with $v_{\text{n}}$ in m/s.

$0.04472 = \sqrt{2\cdot 10^9}/10^6$ : the conversion of g/cm³ to kg/m³, MPa to Pa, L/s to m³/s and N to kN.

Oilfield units

F\,[\text{lbf}] = \frac{\rho\,q\,v_{\text{n}}}{1930} = 0.01823\,C_d\,q\sqrt{\rho\,\Delta p_{\text{b}}}

the density is in ppg, the flow rate in gpm, the velocity in ft/s, the pressure drop in psi, and the force in pounds-force.

$1930 \approx 60\cdot 32.17$ (minutes to seconds, pound of mass to pound of force); $0.01823 \approx 1/(1930\sqrt{8.074\cdot10^{-4}})$ , where $8.074\cdot10^{-4}\rho v^2$ is the dynamic pressure of the jet in psi.

Worked exampleHow much the bit gets now

Given (assumed data)

Well and mud: from the lesson on pressure losses: hole 215.9 mm, mud 1.30 g/cm³; the bit is at a depth of 2600 m
Flow rate and standpipe pressure (measured): 30 L/s; 12.5 MPa
Nozzles: 5 × 12/32 in, $A = 356.3$ mm²; the pressure drop across them at 30 L/s is 5.11 MPa (calculated)
Allowable pump pressure: 18 MPa

Solution

The pump power is $N = 12.5\cdot 30 = 375$ kW. At the bit, $N_{\text{b}} = 5.11\cdot 30 = 153.3$ kW, which is 41%.
The area of the bottom of the hole is $\frac{\pi}{4}\cdot 215.9^2 = 36{,}610$ mm², and the specific power is $N_{\text{sp}} = 153.3\cdot 1000/36{,}610 = 4.19$ W/mm², or $4.19/1.156 = 3.6$ hp/in².
The velocity of the jets is $v_{\text{n}} = 30\cdot 1000/356.3 = 84.2$ m/s, and the impact force is $F = 1.30\cdot 30\cdot 84.2/1000 = 3.28$ kN.
The pumps allow 18 MPa, but work at 12.5 MPa: 31% of the allowable pressure is not used.

Answer: The pumps give the mud 375 kW, and the bit gets 153.3 kW (41%). The jets leave at 84.2 m/s and press on the bottom with a force of 3.28 kN.

What it means. More than half of the power goes on friction, and almost a third of the allowable pressure is not used. This is a usual starting point: the 5 × 12 nozzles were not chosen for this depth. Next we look at how to choose the flow rate and the nozzles for the next bit, which will drill the interval from 2600 to 3000 m. The data in this example are assumed for illustration.

The pumps and the system losses

The pumps are limited in two ways. The pressure cannot exceed what the fitted liners (the replaceable sleeves in the pump cylinders), valves and surface lines allow, and it must stay below the pressure at which the relief valve (the safety valve) opens, with a margin. This is $p_{\max}$ . The power is limited by the drive: the hydraulic power of the pumps (the pressure times the flow rate) cannot be greater than the power of the engines times the efficiency of the pumps. On a graph of pressure against flow rate, these limits are the horizontal line $p_{\max}$ and the hyperbola $p = N_{\max}/Q$ . They cross at the flow rate $Q_N = N_{\max}/p_{\max}$ . At lower flow rates the pumps run into the pressure limit, and at higher ones into the power limit. Liners of a smaller diameter raise the allowable pressure, but reduce the flow per stroke. How the pumps are built and how their flow rate is calculated is the topic of the lesson on the circulating system of the rig.

The standpipe pressure is the sum of the pressure drop at the bit and the system losses (the parasitic pressure losses). These are the losses in the surface lines, the drill string and the annulus, and also in the downhole motor and the MWD tool (the tool that measures while drilling), if the bottom hole assembly has them (Circulating pressure losses). For given nozzles the pressure drop at the bit grows as $Q^2$ , and the system losses grow more slowly. To choose the nozzles it is convenient to describe the system losses by one power law of the flow rate and to find that law from measurements.

CalculatedThe working window of the pumps and the standpipe pressure in the well of the example, with the bit at 2600 m. The pressure limit is 18 MPa, the power limit is 700 kW (the hyperbola), and the limits of the flow rate taken for the example are 21 L/s at the lowest (for cuttings removal) and 32 L/s at the highest (for the flow regime in the annulus at the drill collars). The old 5 × 12 nozzles give 12.5 MPa at 30 L/s, so almost a third of the pressure is not used. With 4 × 10 + 1 × 11 nozzles the pumps reach the pressure limit at 30.8 L/s. The lower curve is the system losses.

System losses: the power law and the exponent from two flow rates

\Delta p_{\text{sys}} = \Delta p_{\text{sys}1}\left(\frac{Q}{Q_1}\right)^{m}, \qquad m = \frac{\ln(\Delta p_{\text{sys}2}/\Delta p_{\text{sys}1})}{\ln(Q_2/Q_1)}, \qquad \Delta p_{\text{sys}i} = p_{\text{SPP}\,i} - \Delta p_{\text{b}\,i}

where:

$\Delta p_{\text{sys}}$ pressure losses in the system: all the standpipe pressure except the pressure drop at the bit, MPa
$Q_1,\ Q_2$ flow rates at which the standpipe pressure was measured, L/s
$p_{\text{SPP}\,i}$ standpipe pressure at the flow rate $Q_i$ (measured), MPa
$\Delta p_{\text{b}\,i}$ pressure drop across the nozzles at the flow rate $Q_i$ , calculated from their actual sizes, MPa
$m$ exponent of the system losses, —

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

Holds when

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

For the turbulent pipes and surface lines the exponent is about 1.6–1.75. For the laminar annulus it is below one (for a power-law fluid it is the flow behaviour index $n$ , 0.38 in the example). For the MWD tool and the motor with no load it is close to 2. So the greater the share of the annulus, the smaller $m$ is. With a shear-thinning mud (one that gets thinner as the shear rate grows) it is often clearly smaller than the “turbulent” 1.75–1.86.

Worked exampleThe exponent from two readings

Given (assumed data)

Readings before pulling the old bit (2600 m, off bottom): 24 L/s — 8.8 MPa; 32 L/s — 13.9 MPa
Nozzles of the old bit: 5 × 12/32 in, $A = 356.3$ mm², $C_d = 0.95$
Mud: 1.30 g/cm³

Solution

Pressure drop across the old nozzles: $\Delta p_{\text{b}1} = \dfrac{500\cdot 1.30\cdot 24^2}{0.95^2\cdot 356.3^2} = 3.27$ MPa. At 32 L/s it is $(32/24)^2$ times greater, 5.81 MPa.
System losses: $\Delta p_{\text{sys}1} = 8.8 - 3.27 = 5.53$ MPa, $\Delta p_{\text{sys}2} = 13.9 - 5.81 = 8.09$ MPa.
The exponent: $m = \dfrac{\ln(8.09/5.53)}{\ln(32/24)} = \dfrac{0.3804}{0.2877} = 1.32$ . The law of the losses: $\Delta p_{\text{sys}} = 5.53\,(Q/24)^{1.32}$ MPa.
Sensitivity: if the first reading were 0.2 MPa higher (9.0 MPa), we would get $m = 1.20$ . If it were 0.2 MPa lower (8.6 MPa), $m = 1.45$ .

Answer: The system losses grow as the flow rate to the power 1.32:

\Delta p_{\text{sys}} = 5.53\,(Q/24)^{1.32}

MPa.

What it means. The exponent is far from 1.86, the value often quoted for turbulent flow. In this well a fifth of the system losses is in the laminar annulus, and the mud thins out strongly under shear. The section-by-section calculation for the same well (Circulating pressure losses) gives almost the same exponent between 24 and 32 L/s: 1.33. We take readings at flow rates that differ by at least a third, better at three flow rates: a gauge error of only 0.2 MPa shifts $m$ from 1.20 to 1.45. The data in this example are assumed for illustration.

Two criteria for the optimum

Let the pumps work at the pressure limit $p_{\max}$ : in this problem, pressure that is not used is simply lost. Then, at any flow rate, the bit gets $\Delta p_{\text{b}} = p_{\max} - \Delta p_{\text{sys}}(Q)$ , and we choose the nozzles so that exactly this pressure drop is obtained. At a low flow rate the pressure drop at the bit is large, but little mud goes through the nozzles. At a high flow rate there is a lot of mud, but the system losses eat up the pressure. The optimum lies between the extremes.

Optimum by the hydraulic power at the bit

N_{\text{b}} = \left(p_{\max} - c\,Q^{m}\right)Q, \qquad \frac{dN_{\text{b}}}{dQ} = p_{\max} - (m+1)\,c\,Q^{m} = 0 \;\Rightarrow\; \Delta p_{\text{sys}} = \frac{p_{\max}}{m+1}, \quad \Delta p_{\text{b}} = \frac{m}{m+1}\,p_{\max}

where:

$p_{\max}$ the limit pressure of the pumps, at which they work, MPa
$c$ the coefficient of the loss law $\Delta p_{\text{sys}} = cQ^m$ , equal to $\Delta p_{\text{sys}1}/Q_1^m$ , depends on $m$
$m$ exponent of the system losses, —

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

Holds when

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

At the optimum the bit gets the share $m/(m+1)$ of the pressure: 50% for $m = 1$ , 57% for $m = 1.32$ , and about 64% for $m = 1.75$ .

Optimum by the jet impact force

F^2 \propto Q^2\left(p_{\max} - c\,Q^{m}\right), \qquad \frac{d(F^2)}{dQ} = 0 \;\Rightarrow\; \Delta p_{\text{sys}} = \frac{2\,p_{\max}}{m+2}, \quad \Delta p_{\text{b}} = \frac{m}{m+2}\,p_{\max}

where:

$F$ jet impact force, $F = C_dQ\sqrt{2\rho\,\Delta p_{\text{b}}}$ , N
$p_{\max},\ c,\ m$ as in the power criterion, —

Units: any consistent units

Holds when

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

The impact force depends on the pressure drop at the bit only as a square root, so it loses less when the growing system losses take pressure away from the bit. Its optimum lies at a higher flow rate and a smaller pressure drop at the bit: for $m = 1.32$ the bit gets 40% of the pressure instead of 57%.

Share of the pump pressure at the bit at the optimum

$m$: 1.0
by power, $m/(m+1)$: 50.0%
by impact force, $m/(m+2)$: 33.3%

$m$: 1.2
by power, $m/(m+1)$: 54.5%
by impact force, $m/(m+2)$: 37.5%

$m$: 1.4
by power, $m/(m+1)$: 58.3%
by impact force, $m/(m+2)$: 41.2%

$m$: 1.6
by power, $m/(m+1)$: 61.5%
by impact force, $m/(m+2)$: 44.4%

$m$: 1.8
by power, $m/(m+1)$: 64.3%
by impact force, $m/(m+2)$: 47.4%

$m$: 2.0
by power, $m/(m+1)$: 66.7%
by impact force, $m/(m+2)$: 50.0%

$m$	by power, $m/(m+1)$	by impact force, $m/(m+2)$
1.0	50.0%	33.3%
1.2	54.5%	37.5%
1.4	58.3%	41.2%
1.6	61.5%	44.4%
1.8	64.3%	47.4%
2.0	66.7%	50.0%

The rest is the system losses. The smaller $m$ is, the less pressure it is best to give to the bit: the system losses grow slowly with the flow rate, and it pays to pump more mud.

There is also a third criterion, the greatest jet velocity. It grows with the pressure drop at the bit, and the greatest pressure drop comes at the smallest flow rate, so this criterion simply presses the flow rate down to the lower limit. It is rarely used, because it sacrifices cuttings removal.

CalculatedThe power at the bit and the jet impact force against the flow rate at a standpipe pressure of 18 MPa, as percentages of their greatest values. The nozzles are chosen anew for each flow rate, and the system losses follow the law

5.53\,(Q/24)^{1.32}

MPa from the example. The power is greatest at 31 L/s, and the impact force at 40 L/s. Both curves are flat: between 25 and 37 L/s the power is less than its greatest value by no more than 5%.

Flow rate limits and the graphical method

The optimum makes sense only if the flow rate we find is allowable. It is limited from below and from above.

The smallest flow rate, from cuttings removal. In the widest part of the annulus the mud must rise clearly faster than the cuttings settle, or the cuttings pile up. In a vertical hole this limit is set from the experience of neighbouring wells and from a calculation of the settling velocity. In inclined and horizontal holes the cuttings lie on the low side of the wall, and larger flow rates and rotation of the string are needed. The downhole motor and the MWD tool have their own lower limits of flow.
The largest flow rate, from the pumps: $Q_N = N_{\max}/p_{\max}$ , and the largest number of strokes for the fitted liners.
The ECD. The annulus losses grow with the flow rate, and the equivalent circulating density at the shoe and opposite weak formations must not come close to the leak-off pressure (Fracture gradient and leak-off testing, the ECD along the hole).
The flow regime in the annulus. In weak rocks that wash out easily, we try to keep the annulus in the open hole laminar (the regime limits). The narrowest place is at the drill collars.
The equipment data sheets: the largest flow of the downhole motor, the MWD tool and the bit.

The graphical method

1
The grid
On a grid with logarithmic scales of flow rate and pressure, draw the vertical lines $Q_{\min}$ and $Q_{\max}$ and the horizontal line $p_{\max}$ .
2
The readings
Before pulling the old bit, measure the standpipe pressure at two, better three, flow rates. Subtract the calculated pressure drops across its nozzles and plot the points of the system losses. The straight line through them is the power law, and its slope is $m$ .
3
The optimum line
Draw the horizontal line $\Delta p_{\text{sys}} = p_{\max}/(m+1)$ (the power criterion) or $2p_{\max}/(m+2)$ (the impact force criterion).
4
The flow rate
The crossing of the loss line with the optimum line gives $Q_{\text{opt}}$ . If it lies to the right of $Q_{\max}$ , we take $Q_{\max}$ , and if it lies to the left of $Q_{\min}$ , we take $Q_{\min}$ . In both cases the bit gets all that is left of $p_{\max}$ .
5
The nozzles and the check
The pressure drop at the bit is $p_{\max} - \Delta p_{\text{sys}}(Q)$ . From it and the flow rate we find the nozzle area and the set of nozzles. Then we check the standpipe pressure with the chosen set, and the ECD.

CalculatedThe graphical method for the well of the example, with the bit at 2600 m. The dots are the system losses from the two readings, and the straight line through them is the law

5.53\,(Q/24)^{1.32}

MPa. The horizontal lines are the optimal system losses by the power criterion (7.76 MPa) and by the impact force criterion (10.84 MPa), and the pressure limit of 18 MPa. The vertical lines are the flow rate limits taken for the example, 21 L/s (cuttings removal) and 32 L/s (laminar annulus at the drill collars), and the pump power limit of 38.9 L/s. The optimum by power, 31 L/s, lies inside the window. The optimum by impact force, 40 L/s, lies outside it, so for this criterion we take 32 L/s.

With depth the loss line rises, and the crossing point moves to the left. So, with the same hole and mud, at small depths the flow rate is usually limited by the upper limit (the pumps or the annulus). At medium depths it equals the optimal flow rate, and at great depths it runs into the cuttings removal limit. Then we hold the flow rate at the lower limit and give the bit what is left of the pressure limit.

Choosing the nozzles

Total flow area of the nozzles for a given flow rate and pressure drop

A = \frac{Q}{C_d}\sqrt{\frac{\rho}{2\,\Delta p_{\text{b}}}}

where:

$A$ total flow area of the nozzles, m²
$Q$ chosen flow rate, m³/s
$\Delta p_{\text{b}}$ pressure drop at the bit that we need to get, Pa
$\rho,\ C_d$ density of the mud and the discharge coefficient, kg/m³, —

Units: SI

Holds when

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

The area is inversely proportional to the square root of the pressure drop: to double the pressure drop at the same flow rate, we reduce the area by a factor of $\sqrt 2$ .

In other unit systems (2)

Metric units

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

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

This is the formula $\Delta p_{\text{b}} = 500\rho Q^2/(C_d^2A^2)$ from the lesson on pressure losses, turned round.

Oilfield units

A\,[\text{in}^2] = \frac{q\,[\text{gpm}]}{C_d}\sqrt{\frac{8.311\cdot10^{-5}\,\rho\,[\text{ppg}]}{\Delta p_{\text{b}}\,[\text{psi}]}}

the flow rate is in gpm, the density in ppg, the pressure drop in psi, and the area in square inches.

This is the formula $\Delta p_{\text{b}} = 8.311\cdot10^{-5}\rho q^2/(C_d^2A^2)$ , turned round.

Nozzles are made in steps of 1/32 in (0.79 mm), and the bit has as many ports for them as its design provides: usually three on a roller-cone bit, and from three to eight or more on a PDC bit. We make up the required area from nozzles of two neighbouring sizes, which gets closer to the calculated value. Then we check at what flow rate the pumps will reach the pressure limit with the chosen set. If the area is slightly smaller than the calculated one, we have to lower the flow rate a little. If it is larger, we can raise the flow rate a little, as far as the other limits allow. Very small nozzles are easily plugged by cuttings and by bridging agents (particles added to the mud to seal the rock). If the programme includes pumping lost-circulation material (fillers against the loss of mud into the rock), the smallest nozzle size is agreed in advance. A port can be plugged, and nozzles of different diameters can be arranged so that the jets sweep the cuttings towards the junk slots of the bit (the channels for the cuttings).

Worked exampleFlow rate and nozzles for the bit run from 2600 to 3000 m

Given (assumed data)

System losses (the previous example): $\Delta p_{\text{sys}} = 5.53\,(Q/24)^{1.32}$ MPa
Pumps: pressure limit 18 MPa, hydraulic power up to 700 kW
Flow rate limits (taken for the example): not less than 21 L/s, taking 0.8 m/s as the lowest velocity of the mud in the annulus between the casing and the pipe; not more than 32 L/s, taking it that the annulus at the drill collars must stay laminar, because the rocks of the open hole are weak
New bit: 215.9 mm, PDC, five ports; mud 1.30 g/cm³, $C_d = 0.95$ ; the criterion is the greatest power at the bit

Solution

The optimum: $\Delta p_{\text{sys}} = 18/(1 + 1.32) = 7.76$ MPa; the flow rate is $Q = 24\cdot(7.76/5.53)^{1/1.32} = 31.0$ L/s, inside the window; the pump power is $18\cdot 31.0 = 558$ kW, less than 700 kW.
The pressure drop at the bit is $18 - 7.76 = 10.24$ MPa (57% of the pressure), and the nozzle area is $A = \dfrac{31.0}{0.95}\sqrt{\dfrac{500\cdot 1.30}{10.24}} = 260$ mm².
For five nozzles that is 52.0 mm² each, a diameter of $\sqrt{4\cdot 52.0/\pi} = 8.14$ mm, or 10.3/32 in. The nearest set is four nozzles of 10/32 and one of 11/32: $4\cdot 49.48 + 59.87 = 257.8$ mm².
Check at 30.8 L/s: the system losses are $5.53\cdot(30.8/24)^{1.32} = 7.69$ MPa, the pressure drop at the bit is $\dfrac{500\cdot 1.30\cdot 30.8^2}{0.95^2\cdot 257.8^2} = 10.28$ MPa, and the standpipe pressure is 17.97 MPa, right at the limit. At 31.0 L/s it would be 18.17 MPa. So we hold the flow rate at 30.8 L/s.
The operating point: $N_{\text{b}} = 10.28\cdot 30.8 = 316.6$ kW, the specific power is 8.65 W/mm² (7.5 hp/in²), $v_{\text{n}} = 30.8\cdot 1000/257.8 = 119.5$ m/s, $F = 1.30\cdot 30.8\cdot 119.5/1000 = 4.78$ kN. The ECD at the shoe is 1.356 g/cm³, against a leak-off pressure of 1.42 g/cm³.
For comparison, with the “typical” exponent 1.86: $\Delta p_{\text{sys}} = 18/2.86 = 6.29$ MPa, $Q = 24\cdot(6.29/5.53)^{1/1.86} = 25.7$ L/s, and the nozzles are about 202 mm², five of 9/32. With them the flow rate can be raised only to 25.7 L/s, and the power at the bit would be 304 kW.

Answer: A flow rate of about 30.8 L/s and nozzles of 4 × 10/32 + 1 × 11/32 (257.8 mm²): the bit gets 57% of the pressure and 316.6 kW, twice as much as with the old nozzles.

What it means. The pumps now use all the allowable pressure, and the power at the bit has doubled at almost the same flow rate. The jets have become faster (119.5 m/s instead of 84), and the impact force is one and a half times greater. The ECD has hardly changed, because the annulus is laminar and the flow rate is the same. A calculation with the “typical” $m = 1.86$ would have given almost the same power (4% less), but a flow rate only 5 L/s above the lower limit and small nozzles that plug more easily. An error in $m$ affects the flow rate and the nozzles more than the power. The data in this example are assumed for illustration.

Nozzle sets for the bit run: each at the largest allowed flow rate

Set, 1/32 in ( $A$ , mm²): 5 × 12 (356.3)
$Q$ , L/s: 32.0
$\Delta p_{\text{b}}$ , MPa (share): 5.81 (42%)
$N_{\text{b}}$ , kW: 186
$F$ , kN: 3.74

Set, 1/32 in ( $A$ , mm²): 5 × 11 (299.4)
$Q$ , L/s: 32.0
$\Delta p_{\text{b}}$ , MPa (share): 8.23 (50%)
$N_{\text{b}}$ , kW: 263
$F$ , kN: 4.45

Set, 1/32 in ( $A$ , mm²): 3 × 10 + 2 × 11 (268.2)
$Q$ , L/s: 31.6
$\Delta p_{\text{b}}$ , MPa (share): 10.00 (56%)
$N_{\text{b}}$ , kW: 316
$F$ , kN: 4.84

Set, 1/32 in ( $A$ , mm²): 4 × 10 + 1 × 11 (257.8)
$Q$ , L/s: 30.8
$\Delta p_{\text{b}}$ , MPa (share): 10.28 (57%)
$N_{\text{b}}$ , kW: 317
$F$ , kN: 4.78

Set, 1/32 in ( $A$ , mm²): 5 × 10 (247.4)
$Q$ , L/s: 29.9
$\Delta p_{\text{b}}$ , MPa (share): 10.52 (59%)
$N_{\text{b}}$ , kW: 315
$F$ , kN: 4.70

Set, 1/32 in ( $A$ , mm²): 5 × 9 (200.4)
$Q$ , L/s: 25.7
$\Delta p_{\text{b}}$ , MPa (share): 11.84 (66%)
$N_{\text{b}}$ , kW: 304
$F$ , kN: 4.28

Set, 1/32 in ( $A$ , mm²)	$Q$ , L/s	$\Delta p_{\text{b}}$ , MPa (share)	$N_{\text{b}}$ , kW	$F$ , kN
5 × 12 (356.3)	32.0	5.81 (42%)	186	3.74
5 × 11 (299.4)	32.0	8.23 (50%)	263	4.45
3 × 10 + 2 × 11 (268.2)	31.6	10.00 (56%)	316	4.84
4 × 10 + 1 × 11 (257.8)	30.8	10.28 (57%)	317	4.78
5 × 10 (247.4)	29.9	10.52 (59%)	315	4.70
5 × 9 (200.4)	25.7	11.84 (66%)	304	4.28

The bit is at 2600 m, and the system losses follow the law $5.53\,(Q/24)^{1.32}$ MPa. The flow rate of the first two sets is limited by the laminar regime at the drill collars (32 L/s), and that of the others by the pressure of 18 MPa. In the wide middle of the table the power hardly changes, so the choice between neighbouring sets is decided by the lower limit of the flow rate and by the risk of plugging small nozzles.

Worked exampleThe same bit run by the impact force criterion

Given (assumed data)

Data: the same as in the previous example
Criterion: the greatest jet impact force

Solution

The optimum: $\Delta p_{\text{sys}} = 2\cdot 18/(1.32 + 2) = 10.84$ MPa, $Q = 24\cdot(10.84/5.53)^{1/1.32} = 40.0$ L/s. This is above 32 L/s (the annulus at the drill collars), and the pumps would need $18\cdot 40.0 = 720$ kW, more than 700.
We take the largest allowable flow rate, 32 L/s: $\Delta p_{\text{sys}} = 5.53\cdot(32/24)^{1.32} = 8.08$ MPa, and the bit is left with $18 - 8.08 = 9.92$ MPa. The area is $\dfrac{32}{0.95}\sqrt{\dfrac{500\cdot 1.30}{9.92}} = 273$ mm².
The nearest set is three nozzles of 10/32 and two of 11/32, 268.2 mm². With it the flow rate can be raised to 31.6 L/s, and above that the pressure would exceed 18 MPa. The impact force is 4.84 kN, and the power at the bit 316 kW.

Answer: By the impact force criterion, about 31.6 L/s and nozzles of 3 × 10/32 + 2 × 11/32. The impact force is 1.2% greater than in the solution by power, and the power at the bit is practically the same.

What it means. The free optimum by the impact force, 40.0 L/s with a pressure drop at the bit of only 7.16 MPa, is not available: both the annulus and the pumps cut it off. After the limits are applied, the two criteria lead to almost the same solution. If the rocks were strong and a laminar regime at the drill collars were not needed, the flow rate would be limited by the pump power (38.9 L/s), and the solutions would differ more. The data in this example are assumed for illustration.

How the result is used on the rig

The hydraulics calculation is part of the hydraulics programme of the drilling. For each interval and bit it states the flow rate, the nozzles, the expected standpipe pressure, the share of the pressure drop at the bit, the specific power, the jet velocity and the ECD. On the rig the programme is corrected with measurements. Before pulling each bit, the standpipe pressure is recorded at two or three flow rates. This gives the law of the losses at this depth with this mud, and the nozzles of the next bit are chosen with it.

The nozzles are not changed during the run, but the system losses grow with depth, because the drill pipe and the annulus get longer. If the pumps were working near the pressure limit, the flow rate has to be lowered little by little, and the power at the bit falls. So the nozzles are chosen for the interval of the run, not for one depth. For a long run we choose for its middle, accepting that at the start the pumps work below the limit.

Worked exampleThe same nozzles at the end of the run

Given (assumed data)

Nozzles: 4 × 10/32 + 1 × 11/32, 257.8 mm² (the example “Flow rate and nozzles for the bit run”)
Bit: at 3000 m, the end of the run
Pressure limit: 18 MPa

Solution

The system losses at 3000 m at the same 30.8 L/s are, by the section-by-section calculation, 8.57 MPa: the drill pipe and the annulus are 400 m longer (at 2600 m, by the law from the readings, it was 7.69 MPa).
With the same flow rate the standpipe pressure would be $8.57 + 10.28 = 18.85$ MPa, above the limit.
We lower the flow rate to 29.9 L/s: the system losses are 8.24 MPa, the pressure drop at the bit is $\dfrac{500\cdot 1.30\cdot 29.9^2}{0.95^2\cdot 257.8^2} = 9.69$ MPa, and together they are 17.9 MPa.
The power at the bit is $9.69\cdot 29.9 = 289.7$ kW, which is 8.5% less than at the start of the run. The bit takes 54% of the pressure. The flow rate is still above the lower limit of 21 L/s.

Answer: Towards the end of the run the flow rate is lowered from 30.8 to 29.9 L/s, and the power at the bit falls by 8.5%.

What it means. The system losses have grown by 12%, but the power at the bit has fallen by only 8.5%. This is the same flatness of the optimum. The section-by-section calculation and the law from the readings differ a little; on the rig such a calculation is corrected with the actual standpipe pressure. If the mud were made heavier towards the end of the run, both the pressure drop at the bit and the system losses would grow, and the flow rate would have to be lowered more. The data in this example are assumed for illustration.

The downhole motor and the MWD tool. Their pressure drops are part of the system losses. The pressure drop across the motor also grows with the weight on the bit, and it is used to judge the load (downhole motors). The readings for the loss law are taken with no weight on the bit.
The actual flow rate. It is usually calculated from the number of pump strokes, but worn valves and liners lower the volumetric efficiency (the share of the calculated volume that is actually delivered). The real flow rate is lower than the calculated one, and the calculated pressure drops at the bit are too high. A flow meter at the outlet of the well helps, and so does comparing the pressure with the calculation.
Wear of the nozzles. An abrasive mud erodes the nozzles, the area grows, and the pressure drop at the bit falls during the run. The standpipe pressure falls smoothly at the same flow rate.
The mud. After the mud is made heavier or treated, the loss law changes, and the old exponent $m$ must not be used.
Pressure control. A sudden rise of the standpipe pressure at a constant flow rate is a sign of a plugged nozzle, a sudden fall is a sign of a lost nozzle, and a gradual fall is a sign of a washout in the drill string (a hole eaten through a pipe wall) (pressure drop at the bit, Well control: primary barrier and kick indicators).

Terms and symbols

Hydraulic power at the bit, $N_{\text{b}}$: The pressure drop at the bit times the flow rate. It is the energy that the jets bring to the bottom of the hole per unit of time.
Specific hydraulic power (HSI): The power at the bit divided by the area of the bottom of the hole. It is given in W/mm² or hp/in² (1 hp/in² ≈ 1.156 W/mm²).
Jet impact force, $F$: The momentum that the jets bring to the bottom of the hole per second: $\rho Q v_{\text{n}}$ .
System losses (parasitic pressure loss), $\Delta p_{\text{sys}}$: All the standpipe pressure except the pressure drop at the bit: the losses in the surface lines, the string, the annulus, the motor and the MWD tool.
Loss exponent, $m$: The power to which the system losses grow with the flow rate. We find it from two or three readings of the standpipe pressure.
Pump working window: The flow rates and pressures allowed by the pressure limit $p_{\max}$ and the power $N_{\max}$ . The corner point is $Q_N = N_{\max}/p_{\max}$ .
Hydraulics programme: The flow rates, nozzles and expected pressures for each interval and bit.
Total flow area of the nozzles, $A$: The sum of the areas of the nozzles. It sets the pressure drop at the bit and the jet velocity at a given flow rate.

Common mistakes and misconceptions

Wrong

“The greater the flow rate, the better the bottom is cleaned.”

In fact

When the pumps work near the pressure limit, a flow rate above the optimum takes pressure away from the bit, and the power of the jets falls. The flow rate is needed to carry cuttings up the annulus. The energy of the jets is set by the nozzles.

Wrong

“The loss exponent is always about 1.86, so there is no need to measure it.”

In fact

This is a guide for the turbulent losses of a low-viscosity fluid. Even in turbulent pipes a shear-thinning mud gives 1.6–1.75, and a laminar annulus lowers $m$ even more, to 1.32 in the example. A calculation with the handbook exponent gives a different flow rate and different nozzles.

Wrong

“The best hydraulics is the greatest pressure drop at the bit.”

In fact

The greatest pressure drop comes at the smallest flow rate. This is the criterion of the greatest jet velocity, and it sacrifices cuttings removal. The optima by power and by impact force are a compromise between the pressure drop and the flow rate.

Wrong

“We chose the nozzles, and we hold the same flow rate for the whole run.”

In fact

The system losses grow with depth. To stay below the pressure limit, the flow rate is lowered. The nozzles are chosen for the interval of the run.

Wrong

“The power of the jets and the impact force are the same thing.”

In fact

At the same power the impact force is inversely proportional to the jet velocity: $F = 2C_d^2N_{\text{b}}/v_{\text{n}}$ . So the two criteria give different flow rates.

In depth: advanced detail

When the pumps run into the power limit

If the optimal flow rate at $p_{\max}$ is greater than $Q_N = N_{\max}/p_{\max}$ , the pumps are at the power limit. Along the hyperbola $p_{\text{SPP}}Q = N_{\max}$ the power at the bit, $N_{\text{b}} = N_{\max} - \Delta p_{\text{sys}}Q = N_{\max} - cQ^{m+1}$ , falls as the flow rate grows. Flow rates below $Q_N$ on this hyperbola need a pressure above $p_{\max}$ . So by the power criterion the best point is the corner one, $Q_N$ and $p_{\max}$ , where the pumps are at the limit of both the pressure and the power.

For the impact force, along the same hyperbola, $F^2 \propto Q^2\Delta p_{\text{b}} = N_{\max}Q - cQ^{m+2}$ , and the maximum is given by the condition $\Delta p_{\text{sys}} = p_{\text{SPP}}/(m+2)$ . It can be met at a pressure below $p_{\max}$ , at small depths where the system losses are small. Then the impact force is a little greater than at the corner point, but the gain is usually small. In practice, in this case too, we work at the corner point or at the largest flow rate allowed by the annulus.

The criterion of the greatest jet velocity always leads to the smallest flow rate: at a constant $p_{\max}$ the pressure drop at the bit, $p_{\max} - cQ^m$ , is the greater the smaller $Q$ is.

A jet in the mud, and the placing of the nozzles

The jet from a nozzle is a submerged jet: it flows out into the mud, draws the mud in, spreads out and slows down. Over a length of about five to six nozzle diameters a core keeps its initial velocity, and farther on the velocity on the axis falls roughly in inverse proportion to the distance. The momentum of the jet is almost conserved, so the impact force is only spread over a larger and larger area, while the kinetic energy (the energy of its motion) is dissipated, that is, turned into heat in the swirling mud. For a 10/32 in nozzle (7.9 mm) the core is about 4–5 cm long. If the bottom is farther away, the total impact force is almost the same, but it is spread over a larger area and tears off fragments less well, and part of the power is lost on the way. So the calculated numbers are upper estimates.

So the design of the bit tries to do three things. It brings the nozzles closer to the bottom (extended nozzles on roller-cone bits). It directs the jets so that they sweep the cuttings towards the slots and the annulus and not against each other. And it washes the cutters that carry the heaviest load. An asymmetric arrangement of nozzles of different diameters, and a plugged port, strengthen the cross flow under the bit. The bit manufacturer makes these decisions, and the hydraulics calculation chooses only the total area and the flow rate.

How accurate is the power law of the losses

The law $\Delta p_{\text{sys}} = cQ^m$ is an approximation: each section has its own exponent, and their contributions change with the flow rate. Let us compare the law from the two readings with the section-by-section calculation for the same well with the bit at 2600 m (the power-law model of the mud, Circulating pressure losses).

System losses: the law from the readings and the section-by-section calculation

$Q$ , L/s: 16
Law, MPa: 3.24
Section by section, MPa: 3.36
Difference: −3.7%

$Q$ , L/s: 20
Law, MPa: 4.35
Section by section, MPa: 4.37
Difference: −0.6%

$Q$ , L/s: 24
Law, MPa: 5.53
Section by section, MPa: 5.50
Difference: 0.6%

$Q$ , L/s: 28
Law, MPa: 6.78
Section by section, MPa: 6.72
Difference: 0.8%

$Q$ , L/s: 32
Law, MPa: 8.08
Section by section, MPa: 8.05
Difference: 0.4%

$Q$ , L/s: 36
Law, MPa: 9.44
Section by section, MPa: 9.49
Difference: −0.4%

$Q$ , L/s: 40
Law, MPa: 10.85
Section by section, MPa: 11.01
Difference: −1.5%

$Q$ , L/s	Law, MPa	Section by section, MPa	Difference
16	3.24	3.36	−3.7%
20	4.35	4.37	−0.6%
24	5.53	5.50	0.6%
28	6.78	6.72	0.8%
32	8.08	8.05	0.4%
36	9.44	9.49	−0.4%
40	10.85	11.01	−1.5%

The law is built from two readings with pressures rounded to 0.1 MPa and an exponent rounded to 0.01, so even at the measured points it does not match the section-by-section calculation exactly. Below about 16.5 L/s the flow in the drill pipe becomes transitional, and below 13.4 L/s it becomes laminar. At 16 L/s the law is already wrong by almost 4%.

From 20 to 40 L/s the law differs from the section-by-section calculation by no more than 1.5%. This is enough for choosing nozzles, because the optimum is flat. But the law must not be carried far beyond the range of the readings, still less across a change of the flow regime. So the flow rates of the readings are chosen around the one at which we intend to drill.

Connections to other topics

Circulating pressure losses — all the terms of the standpipe pressure, the pressure drop at the bit and the section-by-section calculation with which the loss law is compared.
Drilling fluid rheology and flow regimes — why the laminar annulus and the thinning of the mud under shear lower the exponent $m$ .
Fracture gradient and leak-off testing — the leak-off pressure below the shoe, which limits the ECD and the flow rate.
Deflection tools and downhole motors — the pressure drop across the downhole motor as part of the system losses.
Drilling fluids: functions and properties — cuttings removal and the requirements for the rheology of the mud.
Solids control in drilling fluids — the same run: what happens to the cuttings at the surface.
Bit performance and cost per metre — how cleaning of the bottom affects the rate of penetration and the cost per metre.
Rig circulating and rotary systems — the mud pumps, their liners and the limits of pressure and power.

Check yourself

1
The exponent of the system losses is $m = 1.5$ , and the pressure limit of the pumps is 20 MPa. What pressure drop at the bit gives the greatest hydraulic power?
1. A.10.0 MPa
2. B.8.0 MPa
3. C.12.0 MPa
4. D.13.3 MPa
Show the answer
Correct answer: C. 12.0 MPa
The optimal system losses are $20/(1.5 + 1) = 8.0$ MPa, and the bit is left with $20 - 8.0 = 12.0$ MPa, which is 60% of the pressure. 8.0 MPa is the system losses, 10 MPa is half of the pressure (the optimum for $m = 1$ ), and 13.3 MPa is two thirds (the optimum for $m = 2$ ).
2
A bit with 3 × 14/32 in nozzles ( $A = 291.0$ mm², $C_d = 0.95$ ), mud 1.20 g/cm³. At 25 L/s the standpipe pressure is 11.8 MPa, and at 35 L/s it is 21.1 MPa. Find the exponent $m$ and the flow rate that is optimal by power, if the pressure limit is 22 MPa.
Show the answer
The pressure drop across the nozzles at 25 L/s is $500\cdot 1.20\cdot 25^2/(0.95^2\cdot 291.0^2) \approx 4.91$ MPa. At 35 L/s it is $(35/25)^2 = 1.96$ times greater, $\approx 9.62$ MPa. The system losses are $11.8 - 4.91 = 6.89$ and $21.1 - 9.62 = 11.48$ MPa; $m = \ln(11.48/6.89)/\ln 1.4 \approx 1.52$ . The optimum: $\Delta p_{\text{sys}} = 22/2.52 \approx 8.73$ MPa, and $Q = 25\cdot(8.73/6.89)^{1/1.52} \approx 29.2$ L/s. The bit is left with about 13.3 MPa, which is 60% of the pressure.
3
Why does the criterion of the greatest impact force give a larger flow rate than the criterion of the greatest power at the bit?
1. A.The impact force does not depend on the flow rate
2. B.The impact force is proportional to the square of the pressure drop at the bit
3. C.The impact force criterion does not take the pressure limit of the pumps into account
4. D.At the same power the impact force is greater for slow jets with a large flow, so it pays to give the system more pressure: $2/(m+2) > 1/(m+1)$
Show the answer
Correct answer: D. At the same power the impact force is greater for slow jets with a large flow, so it pays to give the system more pressure: $2/(m+2) > 1/(m+1)$
$F = 2C_d^2N_{\text{b}}/v_{\text{n}}$ : the impact force grows if the same power comes with a larger flow and a lower velocity. At the optimum by impact force the system gets $2p_{\max}/(m+2)$ , which is more than $p_{\max}/(m+1)$ , and so the flow rate is greater. The force is proportional to the square root of the pressure drop, not to its square.
4
The required total nozzle area is 300 mm², and the bit has six ports. Which set of nozzles of two neighbouring sizes is the closest, and what will the pressure drop at the bit be at 33 L/s with a mud of 1.25 g/cm³ and $C_d = 0.95$ ?
Show the answer
Per port it is 50 mm², and the diameter is $\sqrt{4\cdot 50/\pi} \approx 7.98$ mm, or 10.05/32 in. Six nozzles of 10/32 (49.48 mm² each) give 296.9 mm², which is the closest (five of 10/32 and one of 11/32 give 307.3 mm²). The pressure drop is $500\cdot 1.25\cdot 33^2/(0.95^2\cdot 296.9^2) \approx 8.56$ MPa.
5
The pumps with the chosen nozzles work near the pressure limit. As the hole gets deeper, the standpipe pressure at the same flow rate grows. What do we do?
1. A.Change the nozzles without pulling the bit
2. B.Raise the pressure limit of the pumps to a new value
3. C.Raise the flow rate so that the jets become stronger
4. D.Lower the flow rate to keep the pressure within the limit; the power at the bit falls little by little
Show the answer
Correct answer: D. Lower the flow rate to keep the pressure within the limit; the power at the bit falls little by little
The system losses grow, and the nozzles are fixed for the run. We can keep the pressure within the limit only with a lower flow rate. Because the optimum is flat, the power at the bit falls slowly (in the lesson's example, by 8.5% in 400 m). The flow rate must stay above the lower limit for cuttings removal.
6
Why do we subtract the pressure drop across the nozzles of the old bit from the measured standpipe pressure before finding the exponent $m$ , and why do we try to space the flow rates of the readings widely?
Show the answer
The exponent $m$ describes the system losses. The pressure drop at the bit grows strictly as $Q^2$ , and in the new run it will be different, so it must be excluded, or $m$ comes out closer to 2. The logarithm of the ratio of close flow rates is small, and a small gauge error changes $m$ a lot. In the lesson's example, 0.2 MPa at flow rates of 24 and 32 L/s gives a spread from 1.20 to 1.45. So the flow rates are spaced at least a third apart, and three readings are taken.

Updated: 10 October 2026