Drilling Formulas and Drilling Calculations

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.

Ref book: Formulas and Calculations for drill, production and workover by Norton J. Lapeyrouse

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Retort analysis is the method to extract mud components into oil, water and solid. The retort analysis report shows percentage of each component by volume. We use data from the retort analysis to determine oil water ratio.

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.

Please find the excle sheet used for calculating oil water ratio from retort data.

Ref book: Formulas and Calculations for drill, production and workover by Norton J. Lapeyrouse

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You normally get the oil/water ratio from a retort so you can use these numbers to determine density oil and water mixture by the following formula.

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

Where;

V1 = percentage of oil

V2 = percentage of water

D1 = density of oil

D2 = density of water

DF = final density of mixed fluid between oil and water

Example: If the oil/water (o/w) ratio is 80/20 (80% oil and 20% water), please determine the density of mixed fluid between oil and water from the retort. Oil density = 7.0 ppg. Water density = 8.33 ppg.

V1 = 0.8

V2 = 0.2

D1 = 7.0

D2 = 8.33

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

(0.8)(7.0) + (0.2)(8.33) = (0.8 + 0.2) DF

5.60 + 1.67= 1.0 DF

7.27 = DF

The density of the oil/water mixture = 7.27 ppg

Please the Excel sheet used to calculate the density of oil and water mixture.

Ref book: Formulas and Calculations for drill, production and workover by Norton J. Lapeyrouse

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According to the previous post, you have learnt about Mixing Fluids of Different Densities with Pit Space Limitation. This post you will learn about how to mix different fluid densit withour limitation of mud pit space (Case#2: You don’t have limitation of pit space). For this case, the calculation will be easier because you can directly input value of each variation into the equation to get answer.

First of all, let’s review the formula for mixing fluid density.

(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

Let’s read and understand this example so you will know how to use the formula.

Example 2: No limit is placed on volume:

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.

• 200 bbl of 10.0 ppg mud in pit A

• 300 bbl of 14.0 ppg mud in pitB

Determine the density and volume when the two following muds are mixed together:

Given:

• 200 bbl of 10.0 ppg mud in pit A

• 300 bbl of 14.0 ppg mud in pitB

Solution:

Apply formula of mixing different fluid density.

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

Where;

V1 = 200 bbl

D1 = 10.0 ppg

V2 = 300 bbl

D2 = 14.0 ppg

VF = V1+V2 =500 bbl

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

(200x 10.0) + (300 x 14.0) = 500 x DF

12.4 ppg = DF

You will have total fluid 500 bbl of 12.4 ppg.

Please find the Excel sheet to calculate how to mix different fluid density without pit space limitation.

 
Ref book: Formulas and Calculations for drill, production and workover by Norton J. Lapeyrouse

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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 densiy with limited pit space”

Ref book: Formulas and Calculations for drill, production and workover by Norton J. Lapeyrouse

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You sometimes need to reduce your mud weight in hole. The method to reduce mud weight is dilution. The concept of dilution is to add light weight fluid into heavy fluid and when they mix together whole drilling fluid weight will decrease. The light weight fluid frequely used are fresh water, sea water, base oil, diesel, synthetic oil, etc which are normally your based fluid. You can calculate how much light weitht fluid in barrel required reducing original mud weight to new mud weight with this following formula:

Mud weight reduction by duluting with light weight fluid

Light weight fluid volume in bbl = V1 x (W1 – W2) ÷ (W2 – Dw)

Where;

V1 = Strating volume in bbl

W1 = Start drilling fluid weight in ppg

W2 = Final drilling fluid weight in ppg

Dw = Density of light weight fluid in ppg

Example: Determine the number of barrels of base oil 7.2 ppg (Dw) required to reduce 200 bbl (V1) of 13.8 ppg (W1) to 10.0 ppg (W2):

Light weight fluid volume in bbl = V1 x (W1 – W2) ÷ (W2 – Dw)

Light weight fluid volume in bbl = 200 x (13.8 – 10) ÷ (10 – 7.2)

Light weight fluid volume in bbl = 271.4 bbl

You need to mix 271.4 bbl of base oil 7.2 ppg into 200 bbl of 13.8 ppg in order to achieve final mud weight of 10.0 ppg

Please find the Excel Sheet to calculate mud dilution.

Ref book: Formulas and Calculations for drill, production and workover by Norton J. Lapeyrouse

Note: If you don’t want to miss any post from drillingformulas.com, please consider subscribing to our mail list. So you will not miss our posts.

You know how much volume will be increased due to adding  Hematite into the system; however, you sometimes are limited to total volume due to limit pit volume on the rig so you need to calculate starting volume to achieve the predetermined final volume of desired mud weight. This formula below is used to determine the staring volume of mud with Hematite.

Starting volume in bbl = VF x (40 – W2) ÷ (40 – W1)

Where; W1 = current mud weigth in ppg

W2 = new mud weight in ppg

VF = final volume in bbl

Example: Determine the barrel of starting volume of 10.0 ppg (W1) mud required to achieve 100 bbl (VF) of 13.0 ppg (W2) mud with Hematite:

Starting volume in bbl = VF x (40 – W2) ÷ (40 – W1)

Starting volume in bbl = 100 x (40 -13.0) ÷ (40 – 10.0)

Starting volume = 90 bbl

In order to achieve 100 bbl of 13.0 ppg mud weight up with Hematite, you must have 90.0 bbl starting volume of 10.0 ppg mud.

Please find the Excel Sheet for calculating starting volume of original mud weight up with Hematite.

Ref book: Formulas and Calculations for drill, production and workover by Norton J. Lapeyrouse

The concept of mud volume increase due to adding Hematite is as same as the mud volume increase by adding barite and calcium carbonate. The formula for calculating volume increment is just different only one factor which is 40 for hematite but 22.5 is usedfor calcium carbonate and 35 is used for barite. Please follow the formula below for determining volume increase because of adding Hematite.

Volume increase per 100 bbl of mud = 100 x (W2 – W1) ÷ (40 – W2)

Where; W1 = current mud weight

W2 = new mud weight

Please read and follow the example below for more understanding.

Example: Determine the volume increase when increasing the density with hematite from 10.0 ppg (W1) to 13.0 ppg (W2):

Volume increase per 100 bbl of mud =100 x (13.0 – 10.0) ÷ (40 – 13.0)

Volume increase per 100 bbl of mud = 11.11 bbl

If you have total volume of 500 bbl of mud, the volume increase due to adding calcium carbonate will be equal to 55.56 bbl (11.11 x 500 ÷ 100).

Please find the Excel sheet used to calculate how much volume increased due to adding Hematite.

Ref book: Formulas and Calculations for drill, production and workover by Norton J. Lapeyrouse

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You have learn how to weight up with Barite and Calcium Carbonate from previous posts. Sometimes, you may need to weight up to extremely high weight therefore you may consider using Hematite , its average specific gravity +/- 4.8, to weight up your mud system. If you are interested in Hematite information, please take a look at Wiki.

The concept of weighting up calculation with Hematite is as same as the concept of weighting up with Barite and Calcium Carbonate. The only one thing change is weight factor used in the equation. Please follow the equation below for calculating how many sacks required to weight up per 100 bbl of original mud.
Sacks of Hematite per 100 bbl of mud = 1680 x (W2 – W1) ÷ (40 – W2)

Where; W1 = current mud weight in ppg

W2 = new mud weight in ppg

Example: Determine the number of sacks of Hematite per l00 bbl required to increase the density from 10.0 ppg (W1) to 13.0 ppg (W2):

Sacks of hematite per 100 bbl of mud = 1680 x (13.0 – 10.0) ÷ (40 – 13.0)

Sacks of hematite per 100 bbl of mud = 186.7 sacks

If you have total volume of 500 bbl of mud, calcium carbonate required to increase mud weight from 10.0 ppg to 13.0 ppg is 933.3 sacks (186.7×500/100).

Please find the Excel used for calculating how many sacks required in case of weighting up with Hematite.

Ref book: Formulas and Calculations for drill, production and workover by Norton J. Lapeyrouse

Note: If you don’t want to miss any post from drillingformulas.com, please consider subscribing to our mail list. So you will not miss our posts.

You know how much volume will be increased due to adding calcium carbonate into the system; however, you sometimes are limited to total volume due to limit pit volume on the rig so you need to calculate starting volume to achieve the predetermined final volume of desired mud weight.
This formula below is used to determine the staring volume of mud (for calcium carbonate system).

Starting volume in bbl  = VF x (22.5 – W2) ÷ (22.5 – W1)

Where; W1 = current mud weight in ppg

W2 = new mud weight in ppg

VF = final volume of mud needed in bbl

Example: Determine the barrel of starting volume of 10.0 ppg (W1) mud required to achieve final volume of 100 bbl (VF) of 13.0 ppg (W2) mud with calcium carbonate:

Starting volume in bbl = VF x (22.5 – W2) ÷ (22.5 – W1)

Starting volume in bbl = 100 x (22.5 – 13.0) ÷ (22.5 – 10.0)

Starting volume = 76 bbl

In order to achieve  final mud volume of 100 bbl of 13.0 ppg mud weight up with calcium carbonate, you must have 76.0 bbl starting volume of 10.0 ppg mud.

Ref book: Formulas and Calculations for drill, production and workover by Norton J. Lapeyrouse

Please find the Excel sheet for calculating Barrel of starting volume of original mud weight required to give a predetermined final volume of desired mud weight with CALCIUM CARBONATE.

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