## Increase mud weight by adding barite

Weighting up is the critical operation while drilling so we need to know how much weighting agent such as barite, calcium carbonate, hematite, etc required in order to get proper weight to prevent a well control situation. This mud calculation is separated into several sub topics so you can follow each one easily. For this article, it will demonstrate how to calculate how many sacks of barite needed to increase mud weight in ppg with barite. (average specific gravity of barite – 4.2)

The formula for weight up with barite is listed below:

Sacks of Barite per 100 bbl of mud = 1470 x (W2 – W1) ÷ (35 – W2)

Where; W1 = current mud weight

W2 = new mud weight

Example: Determine the number of sacks of barite required to increase the density of

500 bbl of 10.0 ppg (W1) mud to 13.0 ppg (W2):

Sacks of Barite per 100 bbl of mud = 1470 x (13.0 – 10.0) ÷ (35 – 13.0)

Sacks of Barite per 100 bbl of mud = 201 sacks (round up)

If you have total volume of 500 bbl of mud, the barite required to increase mud weight from 10.0 ppg to 13.0 ppg is 1,003 sacks (201×500/100).

Please find the Excel sheet used for weighting up with Barite.

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.

## Determine height of light weight spot pill to balance formation pressure

When you get differentially stuck, you may consider spotting light weight fluid to reduce force created by differential pressure between mud in wellbore and formation pressure. However, it is imperative to be able to calculate how much light weight fluid added into the well that it will not  accidentally underbalance formation pressure.

Please follow the steps below to determine how height of light weight spot pill in the annulus to balance formation pressure.

a) Determine the difference in pressure gradient in psi/ft between the mud weight and light weight spot fluid:

Difference pressure in pressure gradient in psi/ft = (current mud wt in ppg – light weight spot pill in ppg) x 0.052

b) Determine height in ft of light weight spot fluid that will balance formation pressure in the annulus:

Height ft in vertical = overbalance pressure with current mud weigh in psi ÷ difference in pressure gradient in psi/ft

This height is the maximum allowable height of light weight spot pill in the annulus. If you have higher length of light weight spot pill in the annulus, it may create a well control situation.

Example: Use the following data to determine the height in ft of light weight spot pill that will balance formation pressure in the annulus:

Mud weight = 13.0 ppg

Light weight spot pill = 8.3 ppg

Amount of overbalance = 300 psi

a) Difference in pressure gradient in psi/ft:

Difference pressure in pressure gradient in psi/ft = (13 ppg – 8.3 ppg) x 0.052

Difference pressure in pressure gradient = 0.2444 psi/ft

b) Determine the height in ft of light weight spot liquid that will balance formation pressure in the annulus:

Height = 300 psi ÷ 0.244 psi/ft

Height = 1227 ft

You must ensure that height of light weight pill in the annulus must be less than 1227 ft in order to prevent a wellcontrol situation.

In order to determine light weight pill volume, this simple formula below is used;

volume of pill = annular capacity (bbl/ft) x height of pill (ft)

For instant, the annular capacity is 0.0459 bbl/ft, volume of pill is equal to 56.3 bbl (1,227 x 0.0459).

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.

## Temperature Conversion Formulas

There are several units of temperature used in the drilling industry and you sometimes need to convert one unit into another unit. So you need to know and be able to convert temperature from one unit to another temperature unit. This post demonstrates how to use temperature conversion formulas to convert one temperature unit to another unit.

1 – Convert temperature from °Fahrenheit (F) to °Celsius (C)

°C = ((°F – 32) x 5) ÷ 9

Example: Convert 80 °F to °C:

°C = ((80 – 32) x 5) ÷ 9

°C = 26.7

2 – Convert temperature from ° Celsius (C) to °Fahrenheit

°F = (°C x 9) ÷ 5 + 32

Example: Convert 30 °C to °F:

°F = (30 x 9) ÷ 5 + 32

°F = 86

3 – Convert temperature from ° Celsius (C) to °Kelvin (K)

°K = °C + 273.16

Example: Convert 30 °C to °K:

°K = 30 + 273.16

°K = 303.16

4 – Convert temperature from °Fahrenheit (F) to °Rankine (R)

°R = °F + 459.69

Example: Convert 150 °F to °R:

°R = 150 + 459.69

°R = 609.69

** The 1st and 2nd are the most frequently used in oil field.

Please find the Excel sheet for converting temperature.

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.

## Pressure and Force Relationship and Its Application

Relationship of pressure, force and cross sectional area is one of the most commonly used concepts in oilfield and we will review this concept and its application. Pressure is force divided by cross section area (see an image below).

Pressure = Force ÷ Area

We normally use pressure in many units such as psi (pound per square inch), Pascal, kg/m3, etc.

In drilling operation, we mostly use circular area so area can be calculated by this formula;

## Area = π x (radius)2 or π x (diameter)2÷ 4

Where  π= 22/7 = 3.143, so we can write a formula above in easy way

## Area = 3.143 x (radius)2 or 0.7857 x (diameter)2

Pressure calculation based on the relationship above is shown below;

## Pressure = force ÷ (3.143 x (radius)2) or force ÷ (0.7857 x (diameter)2)

Example : For this example, we will use the oilfield unit so force is in lb, diameter is in square inch (in2), and diameter is in inch.

Let’s try to apply pressure and force relationship in drilling operation. We plan to bullhead well and we still have drill string in the hole.

Drill string weight in the air = 45,000 lb
Mud weight in hole = 12.0 ppg
Bit size = 8.5”
Drill pipe size = 5″

What is the maximum pressure at surface you can apply before drilling string will be hydraulically pushed out due to bull heading pressure?

Solution

Buoyancy factor = (65.5 – 12.0) ÷ 65.5 = 0.817

Buoyed weight of drill string = 45,000 x 0.817 = 36,765 lb

Area = 0.7857 x (diameter)2= 0.7857 x (8.5)2= 56.77 square inch

Pressure = 36,765 lb ÷ 56.77 square inch= 647 psi.

In order to perform safe bullheading operation with drill string in hole, you need to apply bullheading pressure less than 647 psi on surface.

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.

## Loss of Hydrostatic Pressure When Filling Hole with Water or Lighter Mud

In case of totally lost return, the annulus must be fully filled with fluid, normally water, as fast as we can. Water filled in annulus causes loss of hydrostatic pressure in the wellbore. This article demonstrates how to determine hydrostatic pressure reduction when water or other light fluid is used to fully fill the hole.

Note: This calculation is based on if the float is in the drill string. Therefore, there is no way that fluid from annulus can come into the string.

There are two main concepts, annular capacity and hydrostatic pressure, applied to determine loss of hydrostatic pressure.

## Oilfield Unit

Height of feet of water in annulus

Height of water added  (ft)= water added in bbl ÷ annular capacity in bbl/ft

Bottomhole (BHP) pressure reduction

BHP decrease in psi = (current mud weight in ppg – weight of water in ppg) × 0.052 × (ft of water added)

Note: In order to calculate bottom hole pressure reduction, we assume the column of water in annulus is true vertical depth (TVD). This calculation may not be accurate if the well has high angle so you need to determine the actual TVD from directional survey data. However, this method will be the worst case scenario of pressure reduction.

Equivalent Mud Weight at TD

EMW in ppg = current mud weight in ppg – (BHP decrease in psi ÷ 0.052 ÷ TVD ft of hole)

Example: Determine bottom hole pressure loss and equivalent mud weight at TD due to filling up water into annulus.

Mud weight = 13.0 ppg
Water added = 140 bbl required to fill annulus
Weight of water = 8.6 ppg **
Annular capacity = 0.1422 bbl/ft
Hole TVD = 6,000 ft

Number of feet of water in annulus

Feet of water in annulus = 140 bbl ÷ 0.1422 bbl/ft

Feet = 984.5 ft

Bottomhole (BHP) pressure reduction

BHP reduction = (13.0 ppg – 8.6 ppg) × 0.052 × 984.5 ft

BHP reduction = 225.3 psi

Equivalent mud weight at TD

EMW in ppg = 13.0 – (225.3 psi ÷ (0.052× 6,000 ft))

EMW = 12.3 ppg

## Metric Unit

Height of water in annulus

Height of water added  (m) = water added in m3 ÷ annular capacity in m3/m

Bottomhole (BHP) pressure reduction

BHP decrease in psi = (current mud weight in kg/m3 – weight of water in kg/m3) × 0.00981 × (m of water added)

Equivalent Mud Weight at TD

EMW in ppg = current mud weight in kg/m3 – (BHP decrease in KPa ÷ 0.00981 ÷ TVD m of hole)

Example: Determine bottom hole pressure loss and equivalent mud weight at TD due to filling up water into annulus.

Mud weight = 1550 kg/m3
Weight of water = 1030 kg/m3
Annular capacity = 0.07417 m/m
Hole TVD = 2000 m

Height of water in annulus

Height of water in annulus= 20 m3 ÷ 0.07417 m3/m

Height of water in annulus = 270 m

Bottomhole (BHP) Pressure Reduction

BHP reduction = (1550 kg/m3 – 1030 kg/m3) × 0.00981 × 270 m

BHP reduction = 1,375 KPa

Equivalent mud weight at TD

EMW   = 1550 – (1375 ÷ (0.00981× 2,000 m))

EMW = 1480 kg/m3

Please find the Excel sheet for calculating how much pressure loss due to lost return

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.