Drilling Formulas and Drilling Calculations

Understand Boyle’s Gas Law

Boyle’s law states that at constant temperature, the absolute pressure and the volume of a gas are inversely proportional in case of constant temperature within a closed system. It may sound pretty hard to understand what it is.

Well, we can describe the statement above into simple mathematics as following formula:

Boyle’s Gas Law: P x V = constant

Or express Boyle’s law in another term: P1 x V1 = P2 x V2

Where; P = Pressure and V = Volume

It sounds easy a little bit to understand.

Let’s apply Boyle’s law into our drilling business

Calculate the volume of gas you will have on the surface, 14.7 psi for atmospheric pressure, when 1 bbl of gas kick is circulated out from reservoir where has formation pressure of 3,000 psi.

Boyle’s Gas Law: P1 x V1 = P2 x V2

P1= 3000 psi (reservoir pressure)

V1 = 1 bbl (volume at bottom hole)

P2 = 14.7 psi (atmosphere pressure)

V2 = ? (volume at surface)

P1 x V1 = P2 x V2

3000 x 1 = 14.7 x V2

V2 = 204 bbl

Ref: Well Control Books

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After learning about U-tube concept, let’s get a example in order to understand clearly about physical meaning of U-tube. This is very important concept so you need to clear about it.

Mud weight inside drill pipe is 9.8 PPG is all the way to bit and mud weight in the annulus is 9.2 PPG all the way to surface. Hole depth is 10,000’MD/8500’TVD. The well is shut in and drill pipe pressure is equal to 0 psi. Determine casing pressure.

According to U-tube concept, both sides (casing and drill pipe) have the same bottom hole pressure so we can write the equation to describe the U-tube concept as shown below;

SP (casing) + HP (casing) = BHP = SP (drill pipe) + HP (drill pipe)

At drill pipe side: BHP = 0 psi (Drill pipe Pressure) + 0.052×9.8×8,500 (Hydrostatic Pressure at drill pipe side) = 4,331 psi

At casing side: BHP = 4,331 psi = (Casing Pressure) + 0.052×9.2×8,500 (Hydrostatic Pressure at casing)

With this relationship (SP (casing) + HP (casing) = BHP = SP (drill pipe) + HP (drill pipe) ),we can solve casing pressure.

4331 = Casing Pressure + 4066

Casing Pressure = 4331 – 4066 = 265 psi

Ref: Well Control Book

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This post will show you Lean about bottom hole pressure relationship because this concept is very important for well control concept.
The bottom hole pressure is sum of all the pressure acting on the bottom hole. We can describe the statement before as the following equation;

Bottom Hole Pressure (BHP) = Surface Pressure (SP) + Hydrostatic Pressure (HP)

The image below demonstrates the relationship of bottom hole pressure.

Note: BHP created by hydrostatic column of drilling fluid is the primary well control in drilling.

Looking more into details,

If BHP is more than FP (formation pressure), this situation is called “Overbalance”.

If BHP is equal to FP (formation pressure), this situation is called “Balance”.

If BHP is less than FP (formation pressure), this situation is called “Underbalance”.

For more understanding, please follow this example below demonstrating the relationship of BHP, SP and HP.

Bottom Hole Pressure (BHP) = Surface Pressure (SP) + Hydrostatic Pressure (HP)

We assume that formation pressure is normal pressure gradient of water gradient (0.465 psi/ft) so formation pressure at 8000’ TVD = 8000 ft x 0.465 psi/ft = 3720 psi. Click here to learn how to calculate hydrostatic pressure in oilfield.

The first case: Hydrostatic column is water which is equal to formation pressure gradient so SP is equal to 0 psi

The second case: BHP is still be water gradient but fluid column is oil (0.35 psi/ft) which is lower density than water gradient (0.465 psi/ft). Therefore, in order to balance BHP, we need Surface Pressure (SP) of 920 psi (SP = 3720 – (0.35 x 8000)).

The third case: BHP is still be water gradient but fluid column is gas (0.1 psi/ft) which is even lower density than water gradient (0.465 psi/ft). Therefore, in order to balance BHP, we need Surface Pressure (SP) of 2,920 psi (SP = 3720 – (0.1 x 8000)).

According to the example, Surface Pressure (SP) will compensate the lack of hydrostatic pressure (HP) in order to balance formation pressure (FP).

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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

Please find the Excel sheet used for decreasing oil water ratio.
Ref book: Formulas and Calculations for drill, production and workover by Norton J. Lapeyrouse

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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

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