## Hydrostatic Pressure (HP) Decreases When POOH

When pulling out of hole, volume of steel will be out of hole and mud volume will replace the steel volume.  If we don’t fill hole, hydrostatic pressure will decrease.  This topic shows you how to calculate hydrostatic pressure loss while pulling out of hole without filling the wellbore.  Moreover, there is the Excel sheet for calculating pressure decrease due to pulling out of hole.

## Oilfield Unit

Step 1: Determine Total Volume of Steel Out of Hole

Total Volume of Steel Out of Hole = Length  of pipe pulled out × Pipe Displacement

Where,

Total Volume of Steel Out of Hole in bbl

Length  of pipe pulled out in ft

Pipe Displacement in bbl/ft

Step 2: Determine Hydrostatic Pressure Decrease

Hydrostatic Pressure Decrease = (Total Volume of Steel Out of Hole × 0.052 × mud weight) ÷ (casing capacity – pipe displacement)

Where,

Hydrostatic Pressure Decrease in psi

Total Volume of Steel Out of Hole in bbl

mud weight in ppg

casing capacity in bbl/ft

pipe displacement in bbl/ft

Example: Determine the hydrostatic pressure decrease when pulling pipe out of the hole:

Number of stands pulled = 10
Pipe displacement = 0.0055 bbl/ft
Average length per stand = 91 ft
Casing capacity = 0.0873 bbl/ft
Mud weight = 12.0 ppg

Step 1: Determine of pipe displacement in Barrels = 10 stands × 91 ft/std × 0.0055 bbl/ft displaced

Total Volume of Steel Out of Hole = 5.01 bbl

Step 2: Determine HP, psi decrease = 5.01 barrels × 0.052 × 12.0 ppg ÷ (0.0873 bbl/ft – 0.0055 bbl/ft)

Hydrostatic pressure decrease = 38.2 psi

## Metric Unit

Step 1: Determine Total Volume of Steel Out of Hole

Total Volume of Steel Out of Hole = Length  of pipe pulled out × Pipe Displacement

Where,

Total Volume of Steel Out of Hole in m3

Length  of pipe pulled out in m

Pipe Displacement in m3 /m

Step 2: Determine Hydrostatic Pressure Decrease

Hydrostatic Pressure Decrease = (Total Volume of Steel Out of Hole × 0.00981 × mud weight) ÷ (casing capacity – pipe displacement)

Where,

Hydrostatic Pressure Decrease in KPa

Total Volume of Steel Out of Hole in m3

mud weight in kg/m3

casing capacity in m3 /m

pipe displacement in m3 /m

Example: Determine the hydrostatic pressure decrease when pulling pipe out of the hole:

Number of stands pulled = 10
Pipe displacement = 0.00287 m3 /m
Average length per stand = 30 m
Casing capacity = 0.04554 m3 /m
Mud weight = 1440 kg/m3

Step 1: Determine of pipe displacement in m3 = 10 stands × 30 m/std × 0.00287 m3 /m pipe displacement

Total Volume of Steel Out of Hole 0.86  m3

Step 2: Determine HP, psi decrease = 0.86  m3 × 0.00981 × 1440 kg/m3 ÷ (0.04554 m3 /m- 0.00287 m3 /m)

Hydrostatic pressure decrease = 285 KPa

Please find the Excel sheet for calculating pressure decrease due to pulling out of hole.

Ref books:

Lapeyrouse, N.J., 2002. Formulas and calculations for drilling, production and workover, Boston: Gulf Professional publishing.

Bourgoyne, A.J.T., Chenevert , M.E. & Millheim, K.K., 1986. SPE Textbook Series, Volume 2: Applied Drilling Engineering, Society of Petroleum Engineers.

Mitchell, R.F., Miska, S. & Aadny, B.S., 2011. Fundamentals of drilling engineering, Richardson, TX: Society of Petroleum Engineers.

## Buoyancy Factor Calculation

Buoyancy Factor is the factor that is used to compensate loss of weight due to immersion in drilling fluid.

Before explaining any further, we will explain you about the basic concept of Buoyancy. Buoyancy is the upward force that keeps things afloat. The net upward buoyancy force equals to the amount of the weight of fluid displaced by the body volume. This force will make objects lighter when it immerses in fluid. For example, we feel ourselves lighter when we are in swimming pool because this is the effect of buoyancy.

In drilling operation, we need to know how much weight of string of drill pipe, completion string, etc in drilling fluid. Therefore,  Buoyancy Factor is value that we need to know and be able to calculate this value. Please follow the formulas below to calculate Buoyancy Factor in different mud weight units, ppg and lb/ft3.

## Buoyancy Factor using mud weight in ppg

Buoyancy Factor (BF) = (65.5 – mud weight in ppg) ÷65.5

Note: 65.5 ppg is density of steel.

Example: Determine the buoyancy factor for a 13.0 ppg fluid:
BF = (65.5 – 13.0) ÷ 65.5
BF = 0.8015

## Buoyancy Factor using mud weight in  lb/ft3

Buoyancy Factor (BF) = (489 – mud weight in lb/ft3) ÷489

Note: 489 lb/ft3 is density of steel.

Example: Determine the buoyancy factor for a 100 lb/ft3 fluid:
BF = (489 – 100) ÷489
BF = 0.7955

## Buoyancy Factor using mud weight in kg/l

Buoyancy Factor (BF) = (7.85 – mud weight in kg/l) ÷7.85

Note: 7.85 kg/l is density of steel.

Example: Determine the buoyancy factor for a 1.1 kg/l fluid:
BF = (7.85 – 1.1) ÷7.85
BF = 0.860

## Buoyancy Factor Table

This table demonstrates buoyancy factor at different mud density

Buoyancy Factor Table

## How to use the Buoyancy Factor to determine buoyed weight

In order to figure out the actual weight of drilling string in fluid, the air weight of drilling string times the buoyancy factor equal to actual weight in mud, called buoyed weight.

Buoyed weight of drill string = String weight in the air × Buoyancy Factor

Example: Determine the string weight in 13.0 ppg mud. Air weight of string is 350 klb.

The buoyancy factor for a 13.0 ppg fluid:
BF = (65.5 – 13.0) ÷ 65.5
BF = 0.8015

Buoyed weight of drill string = String weight in the air × Buoyancy Factor

The buoyed weight of drill string in 13.0 ppg mud = 350 x 0.8015 = 280.5 Klb.

Ref books:

Lapeyrouse, N.J., 2002. Formulas and calculations for drilling, production and workover, Boston: Gulf Professional publishing.

Bourgoyne, A.J.T., Chenevert , M.E. & Millheim, K.K., 1986. SPE Textbook Series, Volume 2: Applied Drilling Engineering, Society of Petroleum Engineers.

Mitchell, R.F., Miska, S. & Aadny, B.S., 2011. Fundamentals of drilling engineering, Richardson, TX: Society of Petroleum Engineers.

## Amount of cuttings produced per foot of hole and total solid generated

After learning about capacity calculation, we can apply the capacity calculation to determine how much barrels of cutting produced per foot of hole drilled and total solid generated in pounds.

Use formula#1 and #2 for calculating amount of cutting generated per feet drilled.

## Formula#1 for BARRELS of cuttings drilled per foot of hole drilled:

Barrels of cutting per foot drilled = Dh2 x (1 – % porosity) ÷1029.4

Where: Dh is hole diameter in inch.

Example: Determine barrels of cuttings drilled for one foot of 6-1/8 inch hole with 25% (0.25) porosity:

Barrels/footage drilled = 6.1252 x (1 – 0.25) ÷1029.4
Barrels/footage drilled = 0.02733 bbl/footage drilled

## Formula#2 for CUBIC FEET of cuttings drilled per foot of hole drilled:

Cubic feet of cutting per foot drilled = Dh2 x 0.7854 x (1 – % porosity) ÷144

Where: Dh is hole diameter in inch.

Example: Determine barrels of cuttings drilled for one foot of 6-1/8 inch hole with 25% (0.25) porosity

Cubic feet/footage drilled = 6.1252 x 0.7854 x (1 – 0.25) ÷144
Cubic feet/footage drilled = 0.153462 cu ft/footage drilled

Moreover, you also apply sample density and volume relationship to determine total solids generated. Use the following formula to calculate total solid generated.

Wcg = 350 x Ch x L x (l – porosity) x Cutting density

Where;

Wcg = solids generated in pounds
Ch = capacity of hole in bbl/ft
L = footage drilled in ft
Cutting density = cutting density in gm/cc

Example: Determine the total pounds of solids generated in drilling 100 ft of  6-1/8 inch hole (0.03644 bbl/ft).

Density of cuttings = 2.20 gm/cc.

Porosity = 25%:
Wcg = 350 x 0.03644 x 100 x (1 – 0.25) x 2.2
Wcg = 2104.41 pounds

Please find the excel sheet how to calculate how much cuttings drilled per foot of hole drilled and total solids generated

Ref books: Lapeyrouse, N.J., 2002. Formulas and calculations for drilling, production and workover, Boston: Gulf Professional publishing.

Bourgoyne, A.J.T., Chenevert , M.E. & Millheim, K.K., 1986. SPE Textbook Series, Volume 2: Applied Drilling Engineering, Society of Petroleum Engineers.

Mitchell, R.F., Miska, S. & Aadny, B.S., 2011. Fundamentals of drilling engineering, Richardson, TX: Society of Petroleum Engineers.

## Pipe Displacement Calculation

Pipe displacement, normally in bbl/ft or m3/m, is steel volume  per length of steel pipe.  When we either pull out of hole or trip in hole for any kind of pipes such as drill pipe, casing or tubing, you should know how much fluid to displace steel volume.

For example, when we pull out of hole, a trip sheet must be monitored all time. We must know how much fluid will fill the hole each stand of drill pipe pulled out. If the volume of displacement less than theoretical displacement value, we may have problem due to swabbing formation into wellbore.

This article will demonstrate how to calculate plain pipe displacement with this following formula:

## Oilfield Unit

Pipe Displacement in bbl/ft = (OD2 – ID2 ) ÷ 1029.4

Where;
OD in inch
ID in inch

Where,
OD is out side diameter of pipe in inch.
ID is inside diameter of pipe in inch.

This formula is for plain pipe diplacment such as casing and tubing. It is not for closed-end displacement. It’s not accurate enough for drill pipe because this formula does not account for tool joint displacement therefore you need drill pipe specification sheet for its displacement.

Example: Determine pipe displacement in bbl/ft of 9-5/8” casing 40 ppf, OD = 9.625 in, ID = 8.835 in

Pipe Displacement of 9-5/8” casing 40 ppf in bbl/ft = (9.6252 – 8.8352 ) ÷1029.4
Pipe Displacement of 9-5/8” casing 40 ppf in bbl/ft = 0.01417 bbl/ft

## Metric Unit

Annular capacity in m3/m =  (OD2 – ID2) ÷1,273,240

Where;
OD in mm
ID in mm

Example: Determine pipe displacement in bbl/ft of 9-5/8” casing 40 ppf, OD = 244.475 mm, ID = 244.409 mm

Pipe Displacement of 9-5/8” casing 40 ppf in m3/m = (244.4752 – 244.4092 ) ÷1,273,240
Pipe Displacement of 9-5/8” casing 40 ppf in m3//m = 0.0074

Ref books:

Lapeyrouse, N.J., 2002. Formulas and calculations for drilling, production and workover, Boston: Gulf Professional publishing.

Bourgoyne, A.J.T., Chenevert , M.E. & Millheim, K.K., 1986. SPE Textbook Series, Volume 2: Applied Drilling Engineering, Society of Petroleum Engineers.

Mitchell, R.F., Miska, S. & Aadny, B.S., 2011. Fundamentals of drilling engineering, Richardson, TX: Society of Petroleum Engineers.

## Calculate inner capacity of open hole/inside cylindrical objects (Internal Capacity Factor)

From the previous post, you learn how to calculate annular capacity and this article shows you how to use the same principle to calculate internal capacity factor of  open hole & inside cylindrical objects such as tubular, drill pipe, drill collars, tubing, casing, etc.

There are several formulas to calculate inner capacity depending on unit of inner capacity required. The formula are listed below;

## Calculate Internal Capacity Factor in bbl/ft

Inner Capacity in bbl/ft = (ID)2 ÷1029.4

Where;
Internal Diameter (ID) in inch

Example:

Determine Internal Capacity Factor in bbl/ft of a 6-1/8 in. hole:
Internal Capacity Factor in bbl/ft = 6.1252÷1029.4
Internal Capacity Factor in bbl/ft = 0. 0364 bbl/ft

## Calculate Internal Capacity Factor in gal/ft

Inner Capacity in gal/ft = (ID in.)2 ÷24.51
Where;
Internal Diameter (ID) in inch

Example:

Determine Inner Capacity Factor in gal/ft of 6-1/8 in. hole:
Internal Capacity Factor in gal/ft = 6.1252÷ 24.51
Internal Capacity Factor in gal/ft = 1.53 gal/ft

## Calculate Internal Capacity Factorin cu-meter/meter (m3/m)

Annular capacity in m3/m =  (ID2) ÷1,273,240

Where;
Internal Diameter (ID) in mm

Example:

Hole size (ID) = 155.56  mm.
Internal Capacity Factor in m3/m = (155.562 ) ÷1,273,240
Internal Capacity Factor in m3/m = 0.0190 m3/m

## Calculate Internal Capacity Factorin liter/meter (l/m)

Annular capacity in m3/m =  (ID2) ÷1,273.24

Where;
Internal Diameter (ID) in mm

Example:

Hole size (ID) = 155.56  mm.
Internal Capacity Factor in l/m = (155.562 ) ÷1,273.24
Internal Capacity Factor in l/m = 19.0 l/m

Volume can be determined by this following formula;

## Oilfield Unit

Volume = Internal Capacity Factor x Length

Where;

Volume in bbl

Internal Capacity Factor in bbl/ft

Length in ft

Example:

Internal Capacity Factor =  0. 0364 bbl/ft

Hole Length = 3,000 ft

Volume = 0. 0364 x 3,000 = 109.2 bbl.

## Metric Unit

Volume = Internal Capacity Factor x Length

Where;

Volume in cu.meter

Internal Capacity Factor in cu.meter/meter

Length in meter

Example:

Internal Capacity Factor in m3/m = 0.0190 m3/m

Hole Length = 3,000 m

Volume = 0. 0190 x 3,000 = 57 m3