use a cacl2 brine to give 200 psi overbalance. what will it weight in ppg?
Worked Examples
Drilling Exercises
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The post-obit exercises relate to bug and calculations that need to exist solved in the day to solar day activities of an onshore or offshore Driller and Supervisor. To utilize this page as a learning practise go to the API or Metric Formulae Canvas and view the formulae that are used in onshore and offshore drilling applications. To return to this page merely click the number side by side to a formula and you volition be linked to a worked example.
[one] A well is existence drilled to a true vertical depth of 9367 feet (2855 m).
What will exist 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 ten true vertical depth = 10.four ppg 10 0.052 x 9 367 ft = v 066 psi
or
= 1.250 sg x ix.8 x 2 855 thousand = 33 974 kPa
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[2] Whatever mud density can be expressed every bit a gradient which is a charge per unit of change.
If the current mud density beingness used to drill a well is 12.5 ppg (1.500 sg) summate the mud gradient.
Mud Gradient = mud density 10 constant = 12.5 ppg x 0.052 = 0.65 psi/ft
or
= 1.500 sg 10 9.eight = 14.7 kPa/m
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[three] The heaviest mud that tin be placed in a well is called the Maximum Allowable Mud Density (MAMD). This mud density volition produce sufficient hydrostatic pressure to break-down the formation at the casing shoe.
When we turn this mud density into a gradient we phone call it the fracture gradient.
If it was determined from a germination integrity test that the MAMD in a
particular well is 18.5 ppg (2.22 sg), plow this value into a fracture slope.
Fracture Slope = MAMD x constant = 18.5 ppg x 0.052 = 0.962 psi/ft
or
= 2.220 sg x 9.viii = 21.8 kPa/m
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[4] After a boot has been taken and the well shut-in the Shut-in Drill Pipe Pressure level (SIDPP) and the Shut-in Casing Pressure level (SICP) is recorded. A relationship exists between these two values. If the influx that enters the wellbore is lighter than the mud density beingness used then the SICP is greater than the SIDPP. If we know the true vertical acme of the influx then we tin determine the influx'south gradient.
Summate the Influx Slope if the influx true vertical height is 650 ft (198 m) and the electric current mud gradient is 0.52 psi/ft (11.8 kPa/chiliad).
SIDPP = 425 psi (2 933 kPa), SICP = 610 psi (4 209 kPa)
Influx Gradient = Mud Slope - ((SICP - SIDPP) / truthful vertical influx elevation)
= 0.52 psi/ft - ((610 psi - 425 psi) / 650 ft) = 0.24 psi/ft
or
= 11.viii kPa/m - ((4 209 - 2 933) / 198 = v.36 kPa/m
As a "rule of thumb" one tin say that the gradient of water is 0.4 psi/ft (9.8 kPa/chiliad),
oil is 0.2 psi/ft (4.9 kPa/1000) and a gas is 0.i psi/ft (ii.45 kPa/m).
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[5] Any force per unit area being produced by the hydrostatic column or the formation pore force per unit area tin exist expressed as a mud density.
Summate 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 / abiding / TVD = iv 879 psi / 0.052 / 8 580 ft = 10.94 ppg
or
= 33 665 kPa / 9.8 / 2 615 m = i.314 sg
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[six] Any gradient tin be turned back into a density past just dividing the gradient past a abiding.
Express a mud slope of 0.647 psi/ft (14.63 kPa/m) every bit a mud density.
Mud Density = Mud Slope / abiding = 0.647 psi/ft / 0.052 = 12.44 ppg
or
= 14.63 kPa/m / ix.eight = 1.493 sg
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[7] The SIDPP is simply the difference in pressure between the hydrostatic force per unit area in the wellbore and the formation pressure. The formation pressure being the greater of the ii. Because a pressure can exist 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 impale mud density (KMD).
Calculate the impale mud density if the SIDPP is 180 psi (one 242 kPa), the TVD is 5 603 ft
(ane 707 one thousand) 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 = ix.92 ppg
or
= 1 242 kPa / 9.8 / 1 707 m + 1.120 sg = 1.195 sg
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[8] Summate the truthful vertical distance a mud column would need to be to produce a pressure of v 150 psi (35 535 kPa) if the current mud density is ix.five ppg (i.140 sg).
TVD = HSP / constant / Mud Density = 5 150 psi / 0.052 / 9.five ppg = 10 425 ft
or
= 35 535 kPa / 9.8 / 1.140 sg = iii 181 m
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[9] Summate the true vertical distance of a deviated department of hole where the departure angle is 18 degrees and the measured length of the deviated section is three 000 ft (914 thousand).
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.half-dozen m³) and the annular capacity is 0.0291 Bbl/ft (0.01519 chiliad³/m).
Influx length in annulus = pit gain / annular chapters = 10 Bbls / 0.0291 Bbls/ft = 343.64 ft
or
= 1.6 grand³ / 0.01519 g³/m = 105.33 m
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[eleven] If the influx in the instance above was in a deviated department of well and then its true vertical height tin be found past multiplying the measured length by the cosine of the well difference.
Summate the truthful vertical influx height in an 18 degree wellbore department.
True Vertical Influx Tiptop = Cos Divergence x influx length = Cos eighteen 10 343.64 ft = 326.82 ft
or
= Cos eighteen 10 105.33 g = 100.17 thousand
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[12] Maximum Allowable Mud Density can be determined from a leak-off exam. Summate the MAMD if a leak-off pressure of 580 psi (4 002 kPa) forced a test mud density of ix.0 ppg (i.080 sg) into the shoe expanse at a true vertical shoe depth of 1 605 ft (489 grand).
MAMD = 580 psi / 0.052 / i 605 ft + 9.0 ppg = 15.94 ppg
or
= 4 002 kPa / 9.8 / 489 m + one.080 sg = i.915 sg
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[13] Convert the total pressures needed at the casing shoe to suspension-downward the formation into a MAMD if the full pressure is 1 331 psi (nine 177 kPa) and the shoe is at 1 605 ft (489 k).
MAMD = ane 331 psi / 0.052 /1 605 ft = fifteen.94 ppg
or
= ix 177 kPa / 9.8 /489 thou = ane.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 altitude of 10000 ft (3048 m) with an original mud density of 10 ppg (i.2 sg).
ECD = annular friction loss / constant / TVD + Mud Density = 200 / 0.052 / 10 000 + ten = 10.38 ppg
or
= 1 380 / 9.8 / 3 048 + 1.2 = 1.25 sg
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[15] The MAMD for a detail well was calculated to exist xviii.4 ppg (2.200 sg).
The shoe TVD is at ane 890 anxiety (576 m).
Calculate the Maximum Allowable Annular Surface Pressure level if the current mud density is increment to ten.5 ppg (1.260 sg).
MAASP = (MAMD - Current Mud Density) x constant x TVD of Casing Shoe
= (xviii.4 ppg - 10.5 ppg) x 0.052 x i 890 ft = 776 psi
or
= (2.200 sg - 1.260 sg) ten 9.viii x 576 m = five 306 kPa
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[16] In the higher up example what would be the new MAASP if the well kicked and was killed using a mud density of eleven.0 ppg (ane.320 sg)?
MAASP = (MAMD - Impale Mud Density) 10 abiding 10 TVD of Shoe
= (18.iv ppg - 11.0 ppg) x 0.052 x ane 890 ft = 727 psi
or
= (2.200 sg - one.320 sg) x 9.viii x 576 m = 4 967 kPa
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[17] Calculate the formation force per unit area 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 = v 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 g³/m) over a measured distance of half-dozen 000 ft (1 829 m).
Volume = Length x Capacity = 6 000 ft x 0.01776 Bbls/ft = 106.56 Bbls
or
= 1 829 one thousand 10 0.0093 m³/m = 17 grand³
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[nineteen] Calculate the capacity of a tubular or hole that has an inside bore (ID) of three inches.
Pipe or Pigsty capacity = ID² / 1029.iv = 0.0087 Bbls/ft
Calculate the length of a 2.4 grand³ slug in a tubular with a capacity of 0.0093 m³/m.
Length = volume / capacity = ii.4 m³ / 0.0093 grand³/yard = 258 thousand
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[20] Calculate the annular chapters between an 8.5 ins (0.216 m) wellbore ID and a 5 ins (0.127 grand) OD drill pipe.
Annular Chapters = (ID² - OD²) / 1029.4 = (eight.5² - 5² ) / 1029.4 = 0.0459 Bbls/ft
or
Annular Chapters = ((ID² - OD²) / 4) x 3.14159
= ((0.216² - 0.127² ) / 4) ten iii.14159
= 0.02397 m³/g
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[21] Calculate the displacement of a tubular with an OD of viii in (0.203 thousand) and an ID of 3 in (0.076 m).
Pipe Displacement = (OD² - ID²) / 1029.four = (viii² - 3²) / 1029.4 = 0.0534 Bbls/ft
or
Piping Deportation = ((OD² - ID²) / 4) x 3.14159 = ((0.203² - 0.076²) / 4) x 3.14159 = 0.0278 m³/1000
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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 = x.71 Bbls/minute
or
= 0.0189 thousand³/stroke x 90 = 1.7 m³/minute
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[23] Calculate the annular velocity around v inch (127 mm) drill pipe in an eight.five inch ID (216 mm) wellbore with a pump output of ten.71 Bbls/minute (1.vii yard³/minute) and annular capacity of 0.0459 Bbls/ft (0.0239 m³/m).
Annular Velocity = pump output / annular chapters = 10.71 Bbls/min / 0.0459 Bbls/ft = 233.33 ft/min
or
= i.7 k³ / 0.0239 m³/m = 71.xiii m/min
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[24] Calculate the Initial Circulating Pressure level (ICP) needed to broadcast a well during a kill operation if the dynamic pressure loss is 300 psi (two 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 scrap when the slow circulation pressure (Dynamic Force per unit area Loss) is 300 psi (two 070 kPa), original mud density is 12.five ppg (1.500 sg) and kill mud density is 13 ppg (1.560 sg)
Final Circulating Pressure = Dynamic Pressure level Loss x Impale Mud Density / Old Mud Density
= 300 psi x 13.0 ppg /12.5 ppg = 312 psi
or
= two 070 kPa ten one.560 sg /i.500 sg = 2 153 kPa
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[26] Whatsoever alter in pump rate will effect in a alter in circulating pressure.
Summate the new circulating pressure if the electric current pump rate of xc spm is
increased to 105 spm and the current circulating pressure is 1 500 psi (10 350 kPa).
New Pump Pressure = electric current pressure 10 (new rate/old rate)²
= 1 500 x (105/90)² = 2 042 psi
or
= ten 350 10 (105/90)² = 14 088 kPa
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[27] Any modify in the mud density will result in a change in circulating force per unit area.
Calculate the new circulating pressure level if the current mud density of 12.5 ppg (1.500 sg) is increased to 13.5 ppg (ane.620 sg) and the current circulating pressure is i 500 psi (10 350 kPa).
New Pump Pressure = current pressure x (new Mud Density / quondam Mud Density)
= 1 500 psi x 13.five ppg / 12.five ppg = 1 620 psi
or
= 10 350 kPa x i.620 sg / 1.500 sg = 11 178 kPa
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[28] Calculate the length of pipage 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 piping displacement is 0.0075 Bbls/ft (0.0039 grand³/m).
Pipe to pull before well starts to flow = (overbalance x (casing chapters - metal displacement)
(Mud Density x abiding 10 metallic deportation)
= (150 psi x (0.0767 Bbls/ft - 0.0075 Bbls/ft)) / (fourteen ppg x 0.052 ten 0.0075 Bbls/ft)
= 1 901 ft
or
= (ane 035 kPa ten (0.04 thousand³/grand - 0.0039 m³/yard)) / (1.680 sg x ix.8 ten 0.0039 m³/m)
= 582 k
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[29] An overbalance is the positive difference between hydrostatic pressure and formation pressure.
Calculate the overbalance generated by 12.v ppg (1.500 sg) mud over a true vertical distance of 12 450 ft (three 795 g) when the pore pressure gradient is 0.572 psi/ft (12.9 kPa/m).
Overbalance = HSP - Germination Pressure
= (Mud Density x constant ten TVD) - (pore pressure level slope x TVD)
= (12.five ppg x 0.052 x 12 450 ft) - (0.572 psi/ft ten 12 450 ft)
= 971 psi
or
= (1.500 sg x 9.8 x 3 795 m) - (12.9 kPa/m x iii 795 grand)
= 6 831 kPa
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[30] Calculate the pressure level drib per human foot when drill pipage 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 k³/1000) and the drill pipage displacement is 0.0075 Bbls/ft (0.0039 m³/m).
Force per unit area drop per pes tripping dry out pipe
= (Mud Density 10 abiding x metal deportation)/(casing capacity - metal displacement)
= (12.5 ppg 10 0.052 10 0.0075 Bbls/ft) / (0.0767 Bbls/ft - 0.0075 Bbls/ft) = 0.07 psi/ft
or
= (1.500 sg x ix.8 ten 0.0039 k³/chiliad) / (0.04 m³/m - 0.0039 g³/m) = one.588 kPa/thousand
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[31] Calculate the force per unit area drop per human foot when drill pipage 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 deportation is 0.0075 bbl/ft (0.0039 m³/m) and the drill piping chapters is 0.01776 bbl/ft (0.0093 m³/m).
Force per unit area drop per foot tripping wet pipe
= (Current Density x constant x (metal deportation + 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 ix.viii x (0.0039 m³/grand + 0.0093 yard³/1000)) / 0.0255 m³/m
= 7.609 kPa/m
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[32] Summate the mud level drib if 1 000 ft (304 m) of drill collars are pulled from the well dry out if metal deportation is 0.0349 Bbls/ft (0.0182 chiliad³/thousand) and the casing capacity is 0.0767 Bbls/ft (0.04 m³/one thousand).
Level Drop For POOH Drill Collars (Dry) = (length of collars x metal displacement) / casing capacity
= (ane 000 ft x 0.0349 Bbls/ft) / 0.0767 Bbls/ft = 455 ft
or
= (304 m x 0.0182 m³/1000) / 0.04 m³/chiliad = 138 m
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[33] Calculate the slug driblet inside the drill piping if 844 feet (257 m) of 11 ppg (1.320 sg) slug is pumped into the drill drill pipe on height of x ppg (i.200 sg) original mud.
Slug drib = (Slug density / Mud Density x pill length in pipe) - pill length in pipe
= (11 ppg / ten ppg x 844 ft) - 844 ft = 84.4 ft
or
= (1.320 sg / 1.200 sg x 257 grand) - 257 m = 25.seven m
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[34] Calculate the amount the mud density has to be increment to drop the Slug level inside the drill pipage 150 ft (46 m) if the length of slug pumped inside the pipe is 844 ft (257 m) and the original mud density is x ppg (i.200 sg).
Slug density = Mud Density x ((Slug drop / Slug length in pipe) + 1)
= 10 ppg x ((150 ft / 844 ft) + 1) = xi.77 ppg
or
= 1.200 sg 10 ((46 k / 257 m) + i) = 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 xv Bbls (2.38 m³), slug density is 11 ppg (1.320 sg) and original mud density is 10 ppg (1.200 sg).
Proceeds in Trip Tank = ((Slug Density / Mud Density) - 1) x pill volume pumped
= ((11 ppg / 10 ppg) - 1) 10 15 Bbls = one.v Bbls
or
= ((1.320 sg / one.200 sg) - one) x ii.38 m³ = 0.24 chiliad³
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[36] Calculate the force per unit area drop if a barrel (cubic metre) of 0.52 psi/ft (11.76 kPa/one thousand) mud is bleedfrom a wellbore with no drill-stem in a hole with a capacity of 0.0702 Bbls/ft (0.0366 yard³/m)
PSI Per BBL = mud gradient / hole capacity = 0.52 psi/ft / 0.0702 Bbls/ft = 7.4 psi/bbl
kPa Per 1000³ = mud gradient / hole capacity = 11.76 kPa/m / 0.0366 1000³/m = 321 kPa/1000³
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[37] Summate 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 (nineteen 941 kPa) and the well was shut-in on a 15 Bbl (ii.4 m³) kick.
Volume to Bleed to Maintain Lesser Hole Force per unit area = (increment in pressure level 10 pit proceeds)
( formation pressure - increment in pressure)
= (100 psi 10 15 Bbl) / (two 890 psi - 100 psi) = 0.54 bbl
or
= (690 kPa x ii.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 pipage pressure increases at a rate of 100 psi/hour (690 kPa/hr) in 10 ppg (1.200 sg) mud.
Percolation rate = increase in pressure / (Mud Density x constant)
= 100 psi / (x ppg x 0.052) = 192 ft/60 minutes
or
= 690 kPa / (1.200 sg x 9.8) = 59 grand/hour
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[39] When calculating the volume of a gas at whatever pressure you lot must kickoff calculate a abiding.
Calculate a constant if a 10 Bbl (i.5 grand³) influx of gas was compressed with an applied force of 5 200 psi (35 880 kPa).
Abiding = pressure x volume = five 200 psi x 10 Bbl = 52 000 (no units of measure)
or
= 35 880 kPa x 1.5 thou³ = 53 820 (no units of mensurate)
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[forty] Using the constant derived for a particular situation, you can calculate a new gas volume when the pressure level on the gas changes.
Using information from the example in a higher place, summate the volume of the gas when the strength 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 = iii m³
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[41] Calculate the shut-in casing pressure when the influx gradient is assumed to be 0.one psi/ft (2.26 kPa/m), original mud gradient is 0.52 psi/ft (11.76 kPa/m), truthful 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 truthful vertical influx height) + SIDPP
= ((0.52 psi/ft - 0.1 psi/ft) x 450 ft) + 200 psi = 389 psi
or
= ((11.76 kPa/chiliad - 2.26 kPa/grand) x 137 m) + 1 380 kPa = two 682 kPa
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[42] Calculate the mud density increase, known as a riser margin, if the deviation between the hydrostatic pressure level of mud in the riser {from mud-line to flow line} and the seawater hydrostatic force per unit area is 200 psi (i 380 kPa). The altitude from the mud-line to the well TVD is 3 000 ft (914 1000).
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 / truthful vertical distance mud line to TVD
= 200 psi / 0.052 / three 000 ft = 1.28 ppg
or
= 1 380 kPa / 9.8 / 914 1000 = 0.fifteen sg
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Source: https://www.wellcontrol.com.au/index.php/worked-examples
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