Dilution of Mud System to Control Low Gravity Solid

By adding bbl of base fluid required, dilution of mud can help control Low Gravity Solid (LGS) in mud system. This post will demonstrate you how to determine barrels of dilution fluid such as water or base fluid required to achieve the desired low gravity solid.

Formula used to calculate dilution of mud system is listed below;

Vwm = Vm x (Fct – Fcop) ÷ (Fcop)

Where; Vwm = barrels of dilution water or base fluid needed

Vm = total barrels of mud in circulating system

Fct = percent low gravity solids in system

Fcop = percent total low gravity solids desired

Example: Determine how much barrels of base oil to dilute total 2000 bbl of mud in system from total LGS = 7 % to desired LGS of 3.5 %.

Vm = 2000 bbl

Fct = 7%

Fcop = 3.5%

Vwm = 2000 x (7 – 3.5) ÷ 3.5

Vwm = 2000 bbl

In order to dilute total of 2000 bbl of the original mud with 7% LGS down to 3.5% LGS, 2000 bbl of base oil is required to add into the system.

Please find the excel sheet used to calculate how much barrel of base fluid to control Low Gravity Solid (LGS) in mud system.

 

Ref book: Formulas and Calculations for Drilling, Production and Workover, Second Edition

Weight of slug required for desired length of dry pipe with set volume of slug

You can determine how much slug weight required in order to achieve desired length of dry pipe with certain slug volume that you will use.

slug lenght dry pipe

Oilfield Unit

Please follow these steps of calculation below;

Step 1 Determine Length of slug in drill pipe in ft:

Length of slug in drill pipe in ft = slug volume in bbl ÷ drill pipe capacity in bbl/ft

Step 2 Determine hydrostatic pressure required to give desired dry pipe drill pipe:

Hydrostatic Pressure in psi = mud weight in ppg × 0.052 × desired length of dry pipe

Step 3 Determine slug weight needed in ppg:

Slug weight in ppg = (Hydrostatic Prำssure (from step 2) ÷ 0.052 ÷ Length of slug in ft (step1)) + mud weight, ppg, in hole

Example: Determine slug weight required for the following data:

Desired length of dry pipe = 200 ft

Mud weight in hole = 12.0 ppg

Drill pipe capacity = 0.016 bbl/ft

Volume of slug = 20 bbl

Step 1 – Determine Length of slug inside drill pipe in ft:

Slug length = 20 bbl ÷ 0.016 bbl/ft

Slug length = 1250 ft

Step 2 – Determine hydrostatic pressure required to give desired dry pipe drill pipe

Hydrostatic Pressure in psi = 12.0 × 0.052 × 200

Hydrostatic Pressure in psi = 124.8 psi

Step 3 – Determine slug weight needed in ppg:

Slug weight in ppg = (124.8 ÷ 0.052 ÷ 1250) + 12.0

Slug weight in ppg = 13.9 ppg

Metric Unit

Please follow these steps of calculation below;

Step 1 Determine Length of slug in drill pipe in meter (m):

Length of slug in drill pipe in m = slug volume in m³ ÷ drill pipe capacity in m³/m

Step 2 Determine hydrostatic pressure required to give desired dry pipe drill pipe:

Hydrostatic Pressure in psi = mud weight in kg/m³ × 0.00981 × desired length of dry pipe in m

Step 3 Determine slug weight needed in ppg:

Slug weight in ppg = (Hydrostatic Pressure (from step 2) ÷ 0.00981÷ Length of slug in m (step1)) + mud weight, kg/m³, in hole

Example: Determine slug weight required for the following data:

Desired length of dry pipe = 120 m

Mud weight in hole = 1,400 kg/m³

Drill pipe capacity = 0.007824 m³/m

Volume of slug = 4.5 m³

Step 1 – Determine Length of slug inside drill pipe in ft:

Slug length = 4.5 m³÷ 0.007824 m³/m

Slug length = 575 m

Step 2 – Determine hydrostatic pressure required to give desired dry pipe drill pipe

Hydrostatic Pressure in psi = 1,400 × 0.00981 × 120

Hydrostatic Pressure in psi = 1,648 KPa

Step 3 – Determine slug weight needed in ppg:

Slug weight in ppg = (1,648 ÷ 0.00981 ÷ 575) + 1,400

Slug weight in ppg = 1692 kg/m³

Please find the excel sheet used to calculate Weight of slug required for a desired length of dry pipe with a set volume of slug.

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.

Barrels of slug required for desired length of dry pipe

What is slug? Slug: It is heavy mud which is used to push lighter mud weight down before pulling drill pipe out of hole. Slug is used when pipe became wet while pulling out of hole. This article will demonstrate you how to calculate how many barrel of volume slug required for desired light of dry pipe.

Normally, 1.5 to 2 PPG over current mud weight is a rule of thumb to decide how much weight of slug should be. For example, current mud weight is 10 PPG. Slug weight should be about 11.5 to 12 PPG.  Generally, slug is pumped to push mud down approximate 200 ft and slug volume can be calculated by applying a concept of U-tube (See Figure below).

Volume of slug required for required length of dry pipe can be calculated by this following equations:

Oilfield Unit

Step 1: Determine hydrostatic pressure required to give desired drop inside drill pipe:

Hydrostatic Pressure in psi = mud weight in ppg × 0.052 × ft of dry pipe

Step 2: Determine difference in pressure gradient between slug weight and mud weight:

Pressure gradient difference in psi/ft = (slug weight in ppg – mud weight in ppg) × 0.052

Step 3: Determine length of slug in drill pipe:

Slug length in ft = Hydrostatic Pressure in psi (in step 1) ÷ Pressure gradient difference in psi/ft (step 2)

Step 4: Slug volume required in barrels:

Slug volume in barrel = Slug length in ft × drill pipe capacity in bbl/ft

Example: Determine the barrels of slug required for the following:

Desired length of dry pipe = 200 ft

Drill pipe capacity = 0.016 bbl/ft

Mud weight = 10.0 ppg

Slug weight = 11.5 ppg

slug

Step 1 Hydrostatic pressure required:

Hydrostatic Pressure in psi = 10.0 ppg × 0.052 × 200 ft

Hydrostatic Pressure in psi = 104 psi

Step 2 differences in pressure gradient between slug weight and mud weight:

Pressure gradient difference in psi/ft = (11.5 ppg – 10.0 ppg) × 0.052

Pressure gradient difference in psi/ft = 0.078 psi/ft

Step 3 length of slug in drill pipe:

Slug length in ft = 104 psi ÷ 0.078 psi/ft

Slug length in ft = 1,333 ft

Step 4 Slug volume required in barrels:

Slug volume required = 1,333 ft × 0.016 bbl/ft

Slug volume required = 21.3 bbl

Metric Unit

Step 1: Determine hydrostatic pressure required to give desired drop inside drill pipe:

Hydrostatic Pressure in kPa = mud weight in kg/m³ × 0.00981 × length of dry pipe in m

Step 2: Determine difference in pressure gradient between slug weight and mud weight:

Pressure gradient difference in kPa/m = (slug weight in kg/m³ – mud weight in kg/m³) × 0.00981

Step 3: Determine length of slug in drill pipe:

Slug length in m = Hydrostatic Pressure in kPa (in step 1) ÷ Pressure gradient difference in kPa/m(step 2)

Step 4: Slug volume required in barrels:

Slug volume in m³ = Slug length in m × drill pipe capacity in m³/m

Example: Determine the barrels of slug required for the following:

Desired length of dry pipe = 120 m

Drill pipe capacity = 0.00782 m³/m

Mud weight = 1,380 kg/m³

Slug weight = 1,500 kg/m³

Step 1 Hydrostatic pressure required:

Hydrostatic Pressure in psi = 1,380 kg/m³ × 0.00981 × 120 m

Hydrostatic Pressure in psi = 1,625 kPa

Step 2 differences in pressure gradient between slug weight and mud weight:

Pressure gradient difference in kPa/m = (1,500 kg/m³ – 1,380 kg/m³ ) × 0.00981

Pressure gradient difference in kPa/m = 1.1772 kPa/m

Step 3 length of slug in drill pipe:

Slug length in ft = 1,625 kPa ÷ 1.1772 kPa/m

Slug length in ft = 1,380 m

Step 4 Slug volume required in barrels:

Slug volume required = 1,380 m × 0.00782 m³/m

Slug volume required = 10.79 m³

Please find the excel sheet used to calculate barrels of slug required for desired length of dry pipe

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.

Mixing Fluids of Different Densities with Pit Space Limitation

You many have different drilling fluid densities in your mud pits so you can have option to mix different fluid densities together in order to get you desired mud weight and desired volume. The concept is to weighted average volume and density of each mud component. The calculation is show as a formula below

(V1 x D1) + (V2 x D2) = VF x DF

Where;

V1 = volume of fluid 1 (bbl, gal, etc.)

D1 = density of fluid 1 (ppg,lb/ft3, etc.)

V2 = volume of fluid 2 (bbl, gal, etc.)

D2 = density of fluid 2 (ppg,lb/ft3, etc.)

VF = volume of final fluid mix

DF = density of final fluid mix

There are 2 cases of calculations as follows:

Case#1: You have limitation of pit space.

Case#2: You don’t have limitation of pit space.

Let’s read and understand how to calculate each case.

Case#1: You have limitation of pit space.

This case you can not apply only formula as show above but you need to add one formula and then you use relationship between 2 equations to sovle your problem.

Determine the volume of 10.0 ppg mud and 14.0 ppg mud required to build 300 bbl of

12.0 ppg mud:

You have following mud in you mud pit.

• 300 bbl of 10.0 ppg mud in pit A

• 300 bbl of 14.0 ppg mud in pitB

Solution:

V1 = volume of 10.0 ppg mud required to mix

V2 = volume of 14.0 ppg mud required to mix

So total volume required: V1 + V2 = 300 bbl -> Equation#1

Apply formula of mixing different fluid density.

(V1 x D1) + (V2 x D2) = VF x DF

Where;

V1 = volume of 10.0 ppg mud required to mix

D1 = 10 ppg

V2 = volume of 14.0 ppg mud required to mix

D2 = 14.0 ppg

VF = 300 bbl

DF = 12.0 ppg

(V1 x10.0) + (V2 x 14.0) = 300 x 12.0 -> Equation#2

From Equation#1: V1 = 300 – V2.

Use this relationship to Equation#2

((300-V2) x 10.0) + (V2 x 14.0) = 3,600

3000 – 10V2 + 14V2 = 3600

4V2 = 600

V2 = 150

So volume of 14.0 ppg mud required to mix = 150 bbl

Let’s go back to Equation#1 and put V2 that you just solve so you will get V1

V1 + V2 = 300 bbl

V1 + 150 = 300

V1 = 150 bbl

In order to get 300 bbl of 12.0 ppg mud, you need to mix 150 bbl of 10.0 ppg with 150 bbl of 14.0 ppg.

The case#2 will be posted on the next post so please continue in next few days.

Please find the excel sheet to calculate “mix different mud density with limited pit space”

Ref book: Formulas and Calculations for Drilling, Production and Workover, Second Edition

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.