Drilling Formulas and Drilling Calculations

Hydrostatic pressure is pressure that exert by density of fluid column. The relationship of hydrostatic pressure is shown in the equation below.

HP (Hydrostatic Pressure) = density x g (gravity acceleration) x h (True Vertical Depth, TVD)

In oilfield term, the formula above is modified so that people can use it easily. The formulas are as follows:

HP = Constant x MW x TVD

HP = 0.052 x MW (ppg) x TVD (ft) ** Most frequent used in the oilfield **

HP = 0.007 x MW (pcf) x TVD (ft)

HP = 0.00981 x MW (kg/m3) x TVD (m)

According to the equation, Hydrostatic Pressure is not a function of hole geometry. Only mud weight and True Vertical Depth (TVD) affect on Hydrostatic Pressure.  For example (a picture below); well A and well B have the same vertical depth. With the same mud density in hole, the bottom hole pressure due to hydrostatic pressure is the same. The only different between Well A and Well B is mud volume.

This concept is basic and very important for many aspects such as well control, balance cementing, u-tube, etc.

This is one of well control series. To be continue 

Ref: Well Control Handbook

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Adding bbl of  drilling fluid can help control low gravity solid (LGS) in mud system. However, this is different from the way to control LGS by adding base fluid as base oil or water because mud that is added into system has some Low Gravity Solid (LGS). Hence, when we calculate it, we need to account for Low Gravity Solid (LGS) of new mud into the calculation as well. This post will demonstrate you how to determine barrels of drilling fluid required to achive the desired Low Gravity Solid (LGS).

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

Vwm = Vm x (Fct – Fcop) ÷ (Fcop – Fca)

Where; Vwm = barrels of dilution water or base fluid

Vm = total barrels of mud in circulating system

Fct = percent low gravity solids in system

Fcop = percent total low gravity solids desired

Fca = percent low gravity solids bentonite and/or chemicals added in mud

Example: Determine how much barrels of oil base mud to diluate total 2000 bbl of mud in system from total LGS = 7 % to desired LGS of 3.5 %. The oil base mud has 2% of bentonite slurry.

Vwm = Vm x (Fct – Fcop) ÷ (Fcop – Fca)

Vwm = 2000 x (7 – 3.5) ÷ (3.5-2)

Vwm = 4667 bbl

In order to dilute total of 2000 bbl of the original mud with 7% LGS down to 3.5% LGS, 4667 bbl of mud that has 2% bentonite is requied to add into the system.

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

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

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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 achive 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: Determin how much barrels of base oil to diluate total 2000 bbl of mud in system from total LGS = 7 % to desired LGS of 3.5 %.

Vm = 200 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 requied 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 drill, production and workover by Norton J. Lapeyrouse

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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. 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 x 0.052 x desired length of dry pipe

Step 3 Determine slug weight needed in ppg:

Slug weight in ppg = (Hydrostatic Prssure (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

Slug length = 1250 ft

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

Hydrostatic Prssure in psi = 12.0 x 0.052 x 200

Hydrostatic Prssure 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 = 13.92 ppg

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

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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.

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:

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

Hydrostatic Pressure in psi = mud weight in ppg x 0.052 x 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) x 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 x 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

Step 1 Hydrostatic pressure required:

Hydrostatic Prssure in psi = 10.0 ppg x 0.052 x 200 ft

Hydrostatic Prssure 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 – 10.5 ppg) x 0.052

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

Step 3 length of slug in drill pipe:

Slug length in ft = 104psi ÷ 0.078

Slug length in ft = 1,333 ft

Step 4 Slug volume required in barrels:

Slug volume required = 1333 ft x 0.016 bbl/ft

Slug volume required = 21.3 bbl

Please find the Excel sheet for calculating barrels of slug required for desired length of dry pipe.

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

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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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