# Worked Examples

**Drilling Exercises**

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The following exercises relate to problems and calculations that need to be solved in the day to day activities of an onshore or offshore Driller and Supervisor. To use this page as a learning exercise go to the API or Metric Formulae Sheet and view the formulae that are used in onshore and offshore drilling applications. To return to this page just click the number next to a formula and you will be linked to a worked example.

[1] A well is being drilled to a true vertical depth of 9367 feet (2855 m).

What will be the hydrostatic pressure (HSP) on bottom if the mud density in the well is currently 10.4 ppg (1.250 sg)

HSP = mud density x constant x true vertical depth = 10.4 ppg x 0.052 x 9 367 ft = 5 066 psi

or

= 1.250 sg x 9.8 x 2 855 m = 33 974 kPa

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[2] Any mud density can be expressed as a gradient which is a rate of change.

If the current mud density being used to drill a well is 12.5 ppg (1.500 sg) calculate the mud gradient.

Mud Gradient = mud density x constant = 12.5 ppg x 0.052 = 0.65 psi/ft

or

= 1.500 sg x 9.8 = 14.7 kPa/m

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[3] The heaviest mud that can be placed in a well is called the Maximum Allowable Mud Density (MAMD). This mud density will produce sufficient hydrostatic pressure to break-down the formation at the casing shoe.

When we turn this mud density into a gradient we call it the fracture gradient.

If it was determined from a formation integrity test that the MAMD in a

particular well is 18.5 ppg (2.22 sg), turn this value into a fracture gradient.

Fracture Gradient = MAMD x constant = 18.5 ppg x 0.052 = 0.962 psi/ft

or

= 2.220 sg x 9.8 = 21.8 kPa/m

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[4] After a kick has been taken and the well shut-in the Shut-in Drill Pipe Pressure (SIDPP) and the Shut-in Casing Pressure (SICP) is recorded. A relationship exists between these two values. If the influx that enters the wellbore is lighter than the mud density being used then the SICP is greater than the SIDPP. If we know the true vertical height of the influx then we can determine the influx's gradient.

Calculate the Influx Gradient if the influx true vertical height is 650 ft (198 m) and the current mud gradient is 0.52 psi/ft (11.8 kPa/m).

SIDPP = 425 psi (2 933 kPa), SICP = 610 psi (4 209 kPa)

Influx Gradient = Mud Gradient - ((SICP - SIDPP) / true vertical influx height)

= 0.52 psi/ft - ((610 psi - 425 psi) / 650 ft) = 0.24 psi/ft

or

= 11.8 kPa/m - ((4 209 - 2 933) / 198 = 5.36 kPa/m

As a "rule of thumb" one can say that the gradient of water is 0.4 psi/ft (9.8 kPa/m),

oil is 0.2 psi/ft (4.9 kPa/m) and a gas is 0.1 psi/ft (2.45 kPa/m).

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[5] Any pressure being produced by the hydrostatic column or the formation pore pressure can be expressed as a mud density.

Calculate the mud density needed to produce a hydrostatic pressure of

4 879 psi (33 665 kPa) over a true vertical distance of 8 580 ft (2 615 m).

Mud Density = HSP / constant / TVD = 4 879 psi / 0.052 / 8 580 ft = 10.94 ppg

or

= 33 665 kPa / 9.8 / 2 615 m = 1.314 sg

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[6] Any gradient can be turned back into a density by simply dividing the gradient by a constant.

Express a mud gradient of 0.647 psi/ft (14.63 kPa/m) as a mud density.

Mud Density = Mud Gradient / constant = 0.647 psi/ft / 0.052 = 12.44 ppg

or

= 14.63 kPa/m / 9.8 = 1.493 sg

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[7] The SIDPP is simply the difference in pressure between the hydrostatic pressure in the wellbore and the formation pressure. The formation pressure being the greater of the two. Because a pressure can be converted into a mud density you can therefore convert the SIDPP into a mud density and add that value onto the current mud density and thus derive the kill mud density (KMD).

Calculate the kill mud density if the SIDPP is 180 psi (1 242 kPa), the TVD is 5 603 ft

(1 707 m) and the current or original mud density (OMD) is 9.3 ppg (1.120 sg).

Kill Mud Density = SIDPP / constant / TVD + OMD = 180 psi / 0.052 / 5 603 ft + 9.3 ppg = 9.92 ppg

or

= 1 242 kPa / 9.8 / 1 707 m + 1.120 sg = 1.195 sg

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[8] Calculate the true vertical distance a mud column would need to be to produce a pressure of 5 150 psi (35 535 kPa) if the current mud density is 9.5 ppg (1.140 sg).

TVD = HSP / constant / Mud Density = 5 150 psi / 0.052 / 9.5 ppg = 10 425 ft

or

= 35 535 kPa / 9.8 / 1.140 sg = 3 181 m

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[9] Calculate the true vertical distance of a deviated section of hole where the deviation angle is 18 degrees and the measured length of the deviated section is 3 000 ft (914 m).

TVD = Cos deviation x measured length of deviated hole = Cos 18 x 3 000 ft = 2 853 ft

or

= Cos 18 x 914 m = 869 m

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[10] Calculate the length of an influx in the annulus if the pit gain was 10 Bbls (1.6 m³) and the annular capacity is 0.0291 Bbl/ft (0.01519 m³/m).

Influx length in annulus = pit gain / annular capacity = 10 Bbls / 0.0291 Bbls/ft = 343.64 ft

or

= 1.6 m³ / 0.01519 m³/m = 105.33 m

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[11] If the influx in the example above was in a deviated section of well then its true vertical height can be found by multiplying the measured length by the cosine of the well deviation.

Calculate the true vertical influx height in an 18 degree wellbore section.

True Vertical Influx Height = Cos Deviation x influx length = Cos 18 x 343.64 ft = 326.82 ft

or

= Cos 18 x 105.33 m = 100.17 m

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[12] Maximum Allowable Mud Density can be determined from a leak-off test. Calculate the MAMD if a leak-off pressure of 580 psi (4 002 kPa) forced a test mud density of 9.0 ppg (1.080 sg) into the shoe area at a true vertical shoe depth of 1 605 ft (489 m).

MAMD = 580 psi / 0.052 / 1 605 ft + 9.0 ppg = 15.94 ppg

or

= 4 002 kPa / 9.8 / 489 m + 1.080 sg = 1.915 sg

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[13] Convert the total pressures needed at the casing shoe to break-down the formation into a MAMD if the total pressure is 1 331 psi (9 177 kPa) and the shoe is at 1 605 ft (489 m).

MAMD = 1 331 psi / 0.052 /1 605 ft = 15.94 ppg

or

= 9 177 kPa / 9.8 /489 m = 1.914 sg

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[14] Calculate the Equivalent Circulating Density (ECD) when the annular friction loss is 200 psi (1380 kPa) at a true vertical distance of 10000 ft (3048 m) with an original mud density of 10 ppg (1.2 sg).

ECD = annular friction loss / constant / TVD + Mud Density = 200 / 0.052 / 10 000 + 10 = 10.38 ppg

or

= 1 380 / 9.8 / 3 048 + 1.2 = 1.25 sg

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[15] The MAMD for a particular well was calculated to be 18.4 ppg (2.200 sg).

The shoe TVD is at 1 890 feet (576 m).

Calculate the Maximum Allowable Annular Surface Pressure if the current mud density is increase to 10.5 ppg (1.260 sg).

MAASP = (MAMD - Current Mud Density) x constant x TVD of Casing Shoe

= (18.4 ppg - 10.5 ppg) x 0.052 x 1 890 ft = 776 psi

or

= (2.200 sg - 1.260 sg) x 9.8 x 576 m = 5 306 kPa

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[16] In the above example what would be the new MAASP if the well kicked and was killed using a mud density of 11.0 ppg (1.320 sg)?

MAASP = (MAMD - Kill Mud Density) x constant x TVD of Shoe

= (18.4 ppg - 11.0 ppg) x 0.052 x 1 890 ft = 727 psi

or

= (2.200 sg - 1.320 sg) x 9.8 x 576 m = 4 967 kPa

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[17] Calculate the formation pressure if the SIDPP is 180 psi

(1 242 kPa) and the HSP is 5 200 psi (35 880 kPa).

Formation Pressure = HSP + SIDPP = 5 200 psi + 180 psi = 5 380 psi

or

= 35 880 kPa + 1 242 kPa = 37 122 kPa

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[18] Calculate the volume of mud needed to circulate down a tubular with a capacity of 0.01776 bbl/ft (0.0093 m³/m) over a measured distance of 6 000 ft (1 829 m).

Volume = Length x Capacity = 6 000 ft x 0.01776 Bbls/ft = 106.56 Bbls

or

= 1 829 m x 0.0093 m³/m = 17 m³

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[19] Calculate the capacity of a tubular or hole that has an inside diameter (ID) of 3 inches.

Pipe or Hole capacity = ID² / 1029.4 = 0.0087 Bbls/ft

Calculate the length of a 2.4 m³ slug in a tubular with a capacity of 0.0093 m³/m.

Length = volume / capacity = 2.4 m³ / 0.0093 m³/m = 258 m

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[20] Calculate the annular capacity between an 8.5 ins (0.216 m) wellbore ID and a 5 ins (0.127 m) OD drill pipe.

Annular Capacity = (ID² - OD²) / 1029.4 = (8.5² - 5² ) / 1029.4 = 0.0459 Bbls/ft

or

Annular Capacity = ((ID² - OD²) / 4) x 3.14159

= ((0.216² - 0.127² ) / 4) x 3.14159

= 0.02397 m³/m

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[21] Calculate the displacement of a tubular with an OD of 8 in (0.203 m) and an ID of 3 in (0.076 m).

Pipe Displacement = (OD² - ID²) / 1029.4 = (8² - 3²) / 1029.4 = 0.0534 Bbls/ft

or

Pipe Displacement = ((OD² - ID²) / 4) x 3.14159 = ((0.203² - 0.076²) / 4) x 3.14159 = 0.0278 m³/m

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[22] Calculate the pump output when the pump displacement is 0.119 Bbls/stroke

(0.0189 m³/stroke) and it's pump rate is 90 strokes per minute.

Pump Output = pump displacement x pump rate = 0.119 Bbls/stroke x 90 = 10.71 Bbls/minute

or

= 0.0189 m³/stroke x 90 = 1.7 m³/minute

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[23] Calculate the annular velocity around 5 inch (127 mm) drill pipe in an 8.5 inch ID (216 mm) wellbore with a pump output of 10.71 Bbls/minute (1.7 m³/minute) and annular capacity of 0.0459 Bbls/ft (0.0239 m³/m).

Annular Velocity = pump output / annular capacity = 10.71 Bbls/min / 0.0459 Bbls/ft = 233.33 ft/min

or

= 1.7 m³ / 0.0239 m³/m = 71.13 m/min

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[24] Calculate the Initial Circulating Pressure (ICP) needed to circulate a well during a kill operation if the dynamic pressure loss is 300 psi (2 070 kPa) and the SIDPP is 250 psi (1 729 kPa).

ICP = Dynamic Pressure Loss + SIDPP = 300 + 250 = 550 psi

or

= 2 070 + 1 729 = 3 799 kPa

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[25] Calculate the Final Circulating Pressure (FCP) that must be maintained after kill mud reaches the bit when the slow circulation pressure (Dynamic Pressure Loss) is 300 psi (2 070 kPa), original mud density is 12.5 ppg (1.500 sg) and kill mud density is 13 ppg (1.560 sg)

Final Circulating Pressure = Dynamic Pressure Loss x Kill Mud Density / Old Mud Density

= 300 psi x 13.0 ppg /12.5 ppg = 312 psi

or

= 2 070 kPa x 1.560 sg /1.500 sg = 2 153 kPa

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[26] Any change in pump rate will result in a change in circulating pressure.

Calculate the new circulating pressure if the current pump rate of 90 spm is

increased to 105 spm and the current circulating pressure is 1 500 psi (10 350 kPa).

New Pump Pressure = current pressure x (new rate/old rate)²

= 1 500 x (105/90)² = 2 042 psi

or

= 10 350 x (105/90)² = 14 088 kPa

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[27] Any change in the mud density will result in a change in circulating pressure.

Calculate the new circulating pressure if the current mud density of 12.5 ppg (1.500 sg) is increased to 13.5 ppg (1.620 sg) and the current circulating pressure is 1 500 psi (10 350 kPa).

New Pump Pressure = current pressure x (new Mud Density / old Mud Density)

= 1 500 psi x 13.5 ppg / 12.5 ppg = 1 620 psi

or

= 10 350 kPa x 1.620 sg / 1.500 sg = 11 178 kPa

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[28] Calculate the length of pipe that can be pulled from a well before the hydrostatic pressure overbalance disappears and the well goes under-balanced. Currently the well has an overbalance of 150 psi (1 035 kPa) with a mud density of 14 ppg (1.680 sg). Casing capacity is 0.0767 Bbls/ft

(0.04 m³/m), drill pipe displacement is 0.0075 Bbls/ft (0.0039 m³/m).

Pipe to pull before well starts to flow = (overbalance x (casing capacity - metal displacement)

(Mud Density x constant x metal displacement)

= (150 psi x (0.0767 Bbls/ft - 0.0075 Bbls/ft)) / (14 ppg x 0.052 x 0.0075 Bbls/ft)

= 1 901 ft

or

= (1 035 kPa x (0.04 m³/m - 0.0039 m³/m)) / (1.680 sg x 9.8 x 0.0039 m³/m)

= 582 m

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[29] An overbalance is the positive difference between hydrostatic pressure and formation pressure.

Calculate the overbalance generated by 12.5 ppg (1.500 sg) mud over a true vertical distance of 12 450 ft (3 795 m) when the pore pressure gradient is 0.572 psi/ft (12.9 kPa/m).

Overbalance = HSP - Formation Pressure

= (Mud Density x constant x TVD) - (pore pressure gradient x TVD)

= (12.5 ppg x 0.052 x 12 450 ft) - (0.572 psi/ft x 12 450 ft)

= 971 psi

or

= (1.500 sg x 9.8 x 3 795 m) - (12.9 kPa/m x 3 795 m)

= 6 831 kPa

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[30] Calculate the pressure drop per foot when drill pipe is pulled from the well dry without filling the hole if the current mud density is 12.5 ppg (1.500 sg), the casing capacity is 0.0767 Bbls/ft (0.04 m³/m) and the drill pipe displacement is 0.0075 Bbls/ft (0.0039 m³/m).

Pressure drop per foot tripping dry pipe

= (Mud Density x constant x metal displacement)/(casing capacity - metal displacement)

= (12.5 ppg x 0.052 x 0.0075 Bbls/ft) / (0.0767 Bbls/ft - 0.0075 Bbls/ft) = 0.07 psi/ft

or

= (1.500 sg x 9.8 x 0.0039 m³/m) / (0.04 m³/m - 0.0039 m³/m) = 1.588 kPa/m

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[31] Calculate the pressure drop per foot when drill pipe is pulled from the well wet without filling the hole if the current mud density is 12.5 ppg (1.500 sg), the annular capacity is 0.0489 bbl/ft (0.0255 m³/m), the drill pipe displacement is 0.0075 bbl/ft (0.0039 m³/m) and the drill pipe capacity is 0.01776 bbl/ft (0.0093 m³/m).

Pressure drop per foot tripping wet pipe

= (Current Density x constant x (metal displacement + pipe capacity))/annulus capacity

= (12.5 ppg x 0.052 x (0.0075 Bbls/ft + 0.01776 Bbls/ft)) / 0.0489 Bbls/ft

= 0.3357 psi/ft

or

= (1.500 sg x 9.8 x (0.0039 m³/m + 0.0093 m³/m)) / 0.0255 m³/m

= 7.609 kPa/m

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[32] Calculate the mud level drop if 1 000 ft (304 m) of drill collars are pulled from the well dry if metal displacement is 0.0349 Bbls/ft (0.0182 m³/m) and the casing capacity is 0.0767 Bbls/ft (0.04 m³/m).

Level Drop For POOH Drill Collars (Dry) = (length of collars x metal displacement) / casing capacity

= (1 000 ft x 0.0349 Bbls/ft) / 0.0767 Bbls/ft = 455 ft

or

= (304 m x 0.0182 m³/m) / 0.04 m³/m = 138 m

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[33] Calculate the slug drop inside the drill pipe if 844 feet (257 m) of 11 ppg (1.320 sg) slug is pumped into the drill drill pipe on top of 10 ppg (1.200 sg) original mud.

Slug drop = (Slug density / Mud Density x pill length in pipe) - pill length in pipe

= (11 ppg / 10 ppg x 844 ft) - 844 ft = 84.4 ft

or

= (1.320 sg / 1.200 sg x 257 m) - 257 m = 25.7 m

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[34] Calculate the amount the mud density has to be increase to drop the Slug level inside the drill pipe 150 ft (46 m) if the length of slug pumped inside the pipe is 844 ft (257 m) and the original mud density is 10 ppg (1.200 sg).

Slug density = Mud Density x ((Slug drop / Slug length in pipe) + 1)

= 10 ppg x ((150 ft / 844 ft) + 1) = 11.77 ppg

or

= 1.200 sg x ((46 m / 257 m) + 1) = 1.410 sg

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[35] Calculate the volume of mud that should return into the trip tank after the slug is pumped into the pipe if the slug volume pumped is 15 Bbls (2.38 m³), slug density is 11 ppg (1.320 sg) and original mud density is 10 ppg (1.200 sg).

Gain in Trip Tank = ((Slug Density / Mud Density) - 1) x pill volume pumped

= ((11 ppg / 10 ppg) - 1) x 15 Bbls = 1.5 Bbls

or

= ((1.320 sg / 1.200 sg) - 1) x 2.38 m³ = 0.24 m³

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[36] Calculate the pressure drop if a barrel (cubic metre) of 0.52 psi/ft (11.76 kPa/m) mud is bleedfrom a wellbore with no drill-stem in a hole with a capacity of 0.0702 Bbls/ft (0.0366 m³/m)

PSI Per BBL = mud gradient / hole capacity = 0.52 psi/ft / 0.0702 Bbls/ft = 7.4 psi/bbl

kPa Per m³ = mud gradient / hole capacity = 11.76 kPa/m / 0.0366 m³/m = 321 kPa/m³

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[37] Calculate the volume of fluid that must be bled from the annulus to remove 100 psi (690 kPa) of trapped pressurefrom the well. The formation pressure is 2 890 psi (19 941 kPa) and the well was shut-in on a 15 Bbl (2.4 m³) kick.

Volume to Bleed to Maintain Bottom Hole Pressure = (increase in pressure x pit gain)

( formation pressure - increase in pressure)

= (100 psi x 15 Bbl) / (2 890 psi - 100 psi) = 0.54 bbl

or

= (690 kPa x 2.4 m³) / (19 941 kPa - 690 kPa) = 0.086 m³

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[38] Calculate the percolation rate of a gas influx if the drill pipe pressure increases at a rate of 100 psi/hr (690 kPa/hr) in 10 ppg (1.200 sg) mud.

Percolation rate = increase in pressure / (Mud Density x constant)

= 100 psi / (10 ppg x 0.052) = 192 ft/hr

or

= 690 kPa / (1.200 sg x 9.8) = 59 m/hr

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[39] When calculating the volume of a gas at any pressure you must first calculate a constant.

Calculate a constant if a 10 Bbl (1.5 m³) influx of gas was compressed with an applied force of 5 200 psi (35 880 kPa).

Constant = pressure x volume = 5 200 psi x 10 Bbl = 52 000 (no units of measure)

or

= 35 880 kPa x 1.5 m³ = 53 820 (no units of measure)

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[40] Using the constant derived for a particular situation, you can calculate a new gas volume when the pressure on the gas changes.

Using information from the example above, calculate the volume of the gas when the force applied to it reduces to 2 600 psi (17 940 kPa).

New Volume = constant / new pressure = 52 000 / 2 600 psi = 20 bbl

or

= 53 820 / 17 940 kPa = 3 m³

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[41] Calculate the shut-in casing pressure when the influx gradient is assumed to be 0.1 psi/ft (2.26 kPa/m), original mud gradient is 0.52 psi/ft (11.76 kPa/m), true vertical influx height in the annulus is 450 ft (137 m) and the SIDPP is 200 psi (1 380 kPa).

SICP = ((mud gradient - influx gradient) x true vertical influx height) + SIDPP

= ((0.52 psi/ft - 0.1 psi/ft) x 450 ft) + 200 psi = 389 psi

or

= ((11.76 kPa/m - 2.26 kPa/m) x 137 m) + 1 380 kPa = 2 682 kPa

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[42] Calculate the mud density increase, known as a riser margin, if the difference between the hydrostatic pressure of mud in the riser {from mud-line to flow line} and the seawater hydrostatic pressure is 200 psi (1 380 kPa). The distance from the mud-line to the well TVD is 3 000 ft (914 m).

The riser margin needs to be added to your current mud density so that the riser can be removed in the event of an emergency disconnect of the riser from the subsea BOP stack.

Riser Margin= pressure difference / constant / true vertical distance mud line to TVD

= 200 psi / 0.052 / 3 000 ft = 1.28 ppg

or

= 1 380 kPa / 9.8 / 914 m = 0.15 sg