Drilling hydraulics · field and metric units

Annular velocity and annular capacity formulas

Annular velocity is the speed at which drilling fluid travels up the space between the drill string and the hole or casing. In oilfield units, annular velocity (ft/min) = 24.5 × flow rate (gpm) ÷ (Dh² − Dp²), and annular capacity (bbl/ft) = (Dh² − Dp²) ÷ 1029.4, with the hole diameter Dh and the pipe outside diameter Dp in inches. At 600 gpm in a 12¼ in. hole around 5 in. drill pipe, annular velocity is 117.5 ft/min. This page sets out both formulas in field and metric units, shows where the constants come from, and works through annular volume, bottoms-up strokes and the mistakes that most often spoil the answer.

Reference material only. Hydraulics and hole-cleaning decisions must follow your company's approved procedures and the drilling programme for the well. DrillX calculators are advisory.

1. What the annulus is and why its size matters

The annulus is the ring-shaped space between the outside of the drill string and the wall of the hole or the inside of the casing. Two numbers describe it. Annular capacity is the volume it holds per foot of length, and it is what you need for annular volume, bottoms-up strokes and kick height. Annular velocity is how fast the mud moves up it at a given pump rate, and it is what carries cuttings out of the hole and what drives the friction behind equivalent circulating density (ECD). Both depend on the same term: the hole diameter squared minus the pipe outside diameter squared.

2. The formulas

Annular capacity

Capacity (bbl/ft) = (Dh² − Dp²) ÷ 1029.4

Annular volume

Volume (bbl) = capacity (bbl/ft) × length (ft)

Annular velocity, from gpm

AV (ft/min) = 24.5 × Q (gpm) ÷ (Dh² − Dp²)

Annular velocity, from bbl/min

AV (ft/min) = pump output (bbl/min) ÷ capacity (bbl/ft)

Dh is the hole diameter in open hole, or the casing inside diameter in a cased section. Dp is the outside diameter of the drill pipe, drill collars or tubing. Both are in inches. The two velocity formulas are the same calculation: 1 bbl is 42 US gallons, and 1029.4 ÷ 42 = 24.5. The same relationships in other unit sets:

To getDiameters inFormula
Capacity, bbl/ftin.(Dh² − Dp²) ÷ 1029.4
Capacity, gal/ftin.(Dh² − Dp²) ÷ 24.51
Capacity, m³/mmm(Dh² − Dp²) ÷ 1,273,240
Velocity, ft/minin.24.5 × Q (gpm) ÷ (Dh² − Dp²)
Velocity, ft/minin.1029.4 × Q (bbl/min) ÷ (Dh² − Dp²)
Velocity, m/minmm1,273.24 × Q (L/min) ÷ (Dh² − Dp²)

3. Where 1029.4 and 24.5 come from

Both constants are cylinder geometry and nothing else. One foot of a cylinder with a diameter of D inches has a volume of π ÷ 4 × D² × 12 = 9.4248 × D² cubic inches. A US oilfield barrel is 42 gallons of 231 cubic inches each, which is 9,702 cubic inches. So the capacity in barrels per foot is D² × 9.4248 ÷ 9,702 = D² ÷ 1029.4, and in gallons per foot it is D² × 9.4248 ÷ 231 = D² ÷ 24.51, usually rounded to 24.5. An annulus is a large cylinder with a smaller one taken out, which is why the formulas use Dh² − Dp². In metric units the same reasoning gives 4 ÷ π × 1,000,000 = 1,273,240 for diameters in millimetres and capacity in m³/m.

4. Worked example: annular velocity around pipe and collars

InputValue
Hole size12¼ in. (12.25)
Drill pipe OD5 in.
Drill collar OD8 in.
Flow rate600 gpm
ResultWorkingAnswer
Dh² − Dp², drill pipe12.25² − 5² = 150.0625 − 25125.06
Annular capacity, drill pipe125.06 ÷ 1029.40.1215 bbl/ft
Annular velocity, drill pipe24.5 × 600 ÷ 125.06117.5 ft/min
Dh² − Dp², collars12.25² − 8² = 150.0625 − 6486.06
Annular velocity, collars24.5 × 600 ÷ 86.06170.8 ft/min

The pump rate is the same everywhere, but the mud moves about 45% faster past the collars than past the drill pipe because the gap is smaller. The lowest velocity in a well is always in the widest annulus, which is usually the top of the hole or the riser, and that is where hole cleaning has to be checked. To turn the formula round and find the flow rate for a target velocity, use Q = AV × (Dh² − Dp²) ÷ 24.5: holding 120 ft/min around the 5 in. pipe here needs 120 × 125.06 ÷ 24.5 = 613 gpm.

5. Worked example: annular volume and bottoms-up

Take an 8½ in. hole at 10,000 ft with 9,400 ft of 5 in. drill pipe above 600 ft of 6½ in. drill collars, and a pump delivering 0.100 bbl/stroke at 100 strokes per minute. To keep the arithmetic short the whole annulus is treated as 8.5 in.; on a real well the cased section uses the casing inside diameter.

ResultWorkingAnswer
Capacity, hole × drill pipe(8.5² − 5²) ÷ 1029.4 = 47.25 ÷ 1029.40.0459 bbl/ft
Capacity, hole × collars(8.5² − 6.5²) ÷ 1029.4 = 30.00 ÷ 1029.40.0291 bbl/ft
Volume around drill pipe0.0459 × 9,400431.5 bbl
Volume around collars0.0291 × 60017.5 bbl
Total annular volume431.5 + 17.5449 bbl
Bottoms-up strokes449 ÷ 0.1004,490 strokes
Pump rate0.100 × 10010 bbl/min (420 gpm)
Bottoms-up time449 ÷ 10about 45 minutes
Annular velocity, drill pipe24.5 × 420 ÷ 47.25217.8 ft/min

The volumes above use the unrounded capacities (0.04590 and 0.02914 bbl/ft). Bottoms-up strokes are the figure a driller watches when circulating out gas, a sweep or a sample, and the same annular volumes appear on a kill sheet as strokes from bit to shoe and bit to surface.

6. Metric example

For a 215.9 mm hole (8½ in.) around 127 mm drill pipe (5 in.) at 1,700 L/min: Dh² − Dp² = 46,612.81 − 16,129 = 30,483.81. Annular capacity is 30,483.81 ÷ 1,273,240 = 0.02394 m³/m, or 23.94 litres per metre. Annular velocity is 1,273.24 × 1,700 ÷ 30,483.81 = 71.0 m/min, which is about 233 ft/min. The same well in field units is 449 gpm, and 24.5 × 449 ÷ 47.25 gives the same answer to within rounding.

7. Annular velocity and hole cleaning

Annular velocity has to beat the rate at which cuttings settle through the mud. If it is too low, cuttings build up in the annulus; if it is too high opposite open hole, it can erode the wellbore wall and it raises annular friction and ECD. There is no single correct number. SPE PetroWiki's summary of the research notes that about 200 ft/min was once the usual minimum in practice, that later work suggested 50 ft/min should give satisfactory transport in a vertical well with a typical drilling mud, and that deviated wells are a different problem: cuttings beds were seen to form at inclinations above about 35° from vertical, and raising the annular velocity by increasing the flow rate reduced the bed height significantly. Velocity is one input alongside mud rheology, hole angle, pipe rotation and rate of penetration, so treat a calculated figure as a check on the hydraulics programme, not a replacement for it.

8. Mistakes to avoid

How DrillX helps

DrillX includes an annular capacity and velocity calculator alongside hydrostatic pressure, ECD, annular pressure loss, MAASP from leak-off test, triplex pump output, buoyancy and string weight, and a Driller's and Wait & Weight kill sheet. Values update as you type, in field units, and in English or Arabic. DrillX is in final testing; email info@drillx.app for a demo or early access, or see what DrillX does. For the mud inside the string and the volume its steel displaces on trips, see drill pipe capacity and displacement.

Questions

What is the formula for annular velocity?

Annular velocity (ft/min) equals 24.5 multiplied by the flow rate in gallons per minute, divided by the hole diameter squared minus the pipe outside diameter squared, with both diameters in inches. At 600 gpm in a 12.25 in. hole around 5 in. drill pipe, annular velocity is 24.5 × 600 ÷ (12.25² − 5²) = 117.5 ft/min.

What is the formula for annular capacity?

Annular capacity (bbl/ft) equals the hole or casing inside diameter squared minus the pipe outside diameter squared, divided by 1029.4, with both diameters in inches. An 8.5 in. hole around 5 in. drill pipe holds (8.5² − 5²) ÷ 1029.4 = 0.0459 bbl per foot. Multiply by the length of the section in feet to get annular volume in barrels.

Where do the constants 1029.4 and 24.5 come from?

Both come from the geometry of a cylinder. One foot of a cylinder with a diameter of D inches holds 9.4248 × D² cubic inches. A US oilfield barrel is 9,702 cubic inches and a US gallon is 231 cubic inches, so 9,702 ÷ 9.4248 = 1029.4 and 231 ÷ 9.4248 = 24.51, usually rounded to 24.5.

How do you calculate bottoms-up strokes and time?

Add up the annular volume of every section from the bit to surface, then divide by the pump output. Bottoms-up strokes = annular volume (bbl) ÷ pump output (bbl/stroke), and bottoms-up time in minutes = annular volume (bbl) ÷ pump rate (bbl/min). An annulus of 449 bbl with a pump delivering 0.100 bbl/stroke at 100 strokes per minute needs about 4,490 strokes, or 45 minutes.

What is the annular velocity formula in metric units?

Annular velocity (m/min) equals 1,273.24 multiplied by the flow rate in litres per minute, divided by the hole diameter squared minus the pipe outside diameter squared, with both diameters in millimetres. Annular capacity (m³/m) equals the same difference of squares divided by 1,273,240. At 1,700 L/min in a 215.9 mm hole around 127 mm pipe, annular velocity is about 71 m/min.

What annular velocity is needed for hole cleaning?

There is no single figure. The velocity needed depends on cuttings size and density, mud weight and rheology, hole angle and pipe rotation. SPE PetroWiki notes that about 200 ft/min was once the usual minimum in practice, that later work suggested 50 ft/min should give satisfactory transport in a vertical well with a typical drilling mud, and that deviated wells are harder to clean because cuttings beds form. Use the hydraulics programme for the well, not a rule of thumb.

Sources and further reading