## 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.

## 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.

## Oilfield Unit

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

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.

## Decrease oil/water ratio

Decrease oil/water ratio: The concept of decrease oil water ratio is to increase water volume in the system without any changes in oil volume to meet new oil water ratio.

How can we decrease oil water ratio to 70/30?

Let’s make it simple so I still use the same information as my previous post. We start with 100 bbl of mud and we have the following information from the retort analysis.

Retort analysis:

% by volume oil = 56

% by volume water = 14

% by volume solids = 30

According to this retort analysis, the oil water ratio is 80/20 (learn how to calculate oil water ratio from a retort analysis) and there are 56 bbl of oil, 14 bbl of water and 30 bbl of solid from 100 bbl of mud.

In order to decrease oil water ratio, water must be added but oil volume remains the same. Therefore, 56 bbl of oil will represent 70% of oil ratio for the new system. We give X equals to the new total liquid volume (combination of oil and water volume).

Then; 70 = (56×100) ÷X

X = 80.0 bbl

Total liquid volume is equal to 80.0 bbl.

Oil volume is still the same but water volume will be added into the system. With this concept, the volume of water will added into the system can be described with the following relationship;

Water added = new total liquid volume – original volume

Water added = 80 – 70 = 10 bbl

If you have the total mud volume of 300 bbl, you will need 30 bbl of water added (10 x 300 ÷ 100) in order to decrease oil water ratio from 80/20 to 70/30

## Increase oil/water ratio

The concept of increase oil water ratio is to increase oil volume in the system without any changes in water to meet new oil water ratio.

How can we increase oil water ratio from 80/20 to 85/15?

Let’s make it simple to understand. We start with 100 bbl of mud and we have the following information from the retort analysis.

Retort analysis:

% by volume oil = 56

% by volume water = 14

% by volume solids = 30

According to this retort analysis (learn how to calculate oil water ratio), the oil water ratio is 80/20 and there are 56 bbl of oil, 14 bbl of water and 30 bbl of solid from 100 bbl of mud. In order to increase oil water ratio, oil must be added and water volume remains the same. Therefore, 14 bbl of water will represent only 15% of water ratio for the new system. We give X equals to the new total liquid volume.

Then; 15 = (14×100)÷X

X = 93.33 bbl

Total new liquid volume is 93.33 bbl

Barrel of base oil added per 100 bbl of mud

Oil added = new total liquid volume – original volume

Oil added = 93.33 – 70 = 23.33 bbl

It means that you need to add oil 23.33 bbl per 100 bbl of original mud without adding any volume of water in order to achieve 85/15 oil water ratio.

Please find the Excel sheet for calculating how to increase oil water  ratio.

## Calculate Oil-Water Ratio from Retort Data

Oil Water Ratio (OWR) is a figure representing the fraction of oil and water in oil based drilling mud.  Generally speaking, it is a ratio between the percent oil in liquid phase and the percent water in liquid phase. In order to determine OWR, volume of oil/water/solid in drilling mud comes from a retort analysis. A sample of oil based mud is controlled burnt in a retort kit at required temperature. When the mud is heated, water and oil will be extracted out and solid is left in the retort kit. The retort analysis report shows percentage of each component by volume so we use data from the retort analysis to determine oil water ratio.

Retort kits, OilfieldPix.com 2017

The formulas below demonstrate how to calculate oil water ratio from retort data.

a) % oil in liquid phase = (% by volume oil x 100) ÷ (% by volume oil + % by volume water)

b) % water in liquid phase = (% by volume water x 100) ÷ (% by volume oil + % by volume water)

c) Result: The oil/water ratio equals to the percent oil in liquid phase and the percent water in liquid phase.

Example: Determine oil water ratio from following information

Data from a retort analysis:

% by volume oil = 56

% by volume water = 14

% by volume solids = 30

Solution:

a) % oil in liquid phase = (56 x 100) ÷ (56+14)

% oil in liquid phase = 80

b) % water in liquid phase = (14 x 100) ÷ (56+14)

% water in liquid phase = 20

c) According to this retort report, the oil/water ratio equals to 80/20.

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.