Drilling Formulas and Drilling Calculations

For coring operation, the ton-miles calculation is also expressed in terms of work in round trip ton-miles. In order to determine the coring ton-miles, you need to figure out ton-miles for one round trip at the depth where coring stopped minus ton-miles for one round trip at the depth where coring began and the multiplied that value by 2.

The formula for calculating ton-mile for coring operation is shown below;

Tc = 2 x (T4 – T3)

Where;

Tc = ton-miles for coring operation
T4 = ton-miles for one round trip at depth where coring operation stopped before coming out of hole
T3 = ton-miles for one round trip at depth where coring get started

Example – Please determine coring ton-mils from 8000 ft to 8050 ft.

Ton-miles @ 8050 ft (end of coring operation) = 200
Ton-miles for trip @ 8000 ft (start of coring operation) = 190

Tc = 2 x (T4 – T3)
Tc = 2 x (200 – 190)
Tc = 20 ton-miles

Referencer book:

Formulas and Calculations for Drilling, Production and Workover, Second Edition

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Drilling or Connection ton-miles is  ton-miles of work in drilling operations. These are the actual ton-miles of work in drilling down the length of a section of drill pipe, usually around +/- 31 ft, plus picking up, connecting, and starting to drill again. In order to figure out connection or drilling ton-miles, it takes 3 times of ton-miles for current round trip minus ton-miles for previous round trip. The formula for calculating drilling ton mile is listed below;

Td = 3 x (T2 – T1)
Where;
Td = Ton-miles for drilling
T2 = Ton-miles for one round trip of last depth before coming out of hole.
T1 = Ton-miles for one round trip of first depth that drilling is started.

Example;
Please determine drilling tome-miles from 8000 ft to 9000 ft.
Ton-miles for trip @ 9000 ft = 230
Ton-miles for trip @ 8000 ft = 195
Td = 3 x (T2 – T1)
Td = 3 x (230 – 195)
Td = 3 x 35
Td = 105 ton-miles
Download the Excel sheet for calculating drilling or connection ton-mile.

Referencer book:

Formulas and Calculations for Drilling, Production and Workover, Second Edition

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All types of ton-mile service should be calculated and recorded in order to obtain a true picture of the total service received from the rotary drilling line. There are several types of ton miles as follows;

1. Round trip ton-miles
2. Drilling or “connection” ton-miles
3. Coring ton-miles
4. Ton-miles setting casing
5. Short-trip ton-miles

For this time, I will show how to calculate round trip ton-mile.

Round Trip Ton-Miles Calculation

The formula for round trip ton-miles is listed below;

RTTM = (Wp x D x (Lp + D) + (2 x D) x (2 x Wb + Wc)) ÷ (5280 x 2000)

where
RTTM = Round Trip Ton-Miles
Wp = buoyed weight of drill pipe in lb/ft
D = hole measured depth in ft
Lp = Average length per stand of drill pipe in ft
Wb = weight of travelling block in lb
Wc = buoyed weight of BHA (drill collar + heavy weight drill pipe + BHA) in mud minus the buoyed weight of the same length of drill pipe in lb
** If you have BHA (mud motor, MWD, etc) and HWDP, you must add those weight into calculation as well not just only drill collar weight. **
2000 = number of pounds in one ton
5280 = number of feet in one mile

Example: Round trip ton-miles

Mud weight = 10.0 ppg
Average length per stand = 94 ft
Drill pipe weight = 13.3 lb/ft
Hole measure depth = 5500 ft
Drill collar length = 120 ft
Drill collar weight = 85 lb/ft
HWDP length = 49 lb/ft
HWDP weight = 450 ft
BHA weight from directional driller = 8,300 lb
BHA length = 94 ft
Travelling block assembly = 95,000 lb

Solution:

a) Buoyancy factor:
BF = (65.5 – 10.0) ÷ 65.5
BF = 0.847

b) Buoyed weight of drill pipe in mud, lb/ft (Wp):
Wp = 13.3 lb/ft x 0.847
Wp = 11.27 lb/ft

c) buoyed weight of BHA (drill collar + heavy weight drill pipe + BHA) in mud minus the buoyed weight of the same length of drill pipe in lb (Wc):

Wc = {[(120x85) + (49x450) + (8300)] x 0.847} – [(120+450+94) x13.3x 0.847]
Wc = 26,866 lb

Round trip ton-miles = [(11.27 x 5500 x (94+ 5500)) + (2 x 5500) x (2 x 95000 + 26,866)] ÷ (5280 x 2000)
RTTM = 258.75 ton-mile

Please find the excel sheet for round trip ton-miles calculation via click this link.

Referencer book:

Formulas and Calculations for Drilling, Production and Workover, Second Edition

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Slug Mud: 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.

Normally, slug is pumped to push mud down approximate 200 ft (+/2 stands) and slug volume can be calculated by applying a concept of U-tube (see a figure below)

Volume of slug can be calculated by this following equation:

This equation expresses that the higher slug volume, the deeper of dry in drill pipe is met. As per the above equation, length of dry pipe can be substituted by 200 ft.

In normal practice, slug volume pumped to clean drill pipe is around 15-25 bbl depending on drillpipe size. Moreover, it also depends on situations because sometime mud in annulus side may be heavier than measured MW due to cutting, drilling solid contaminated in mud, hence more slug volume is needed.

Reference: Formulas and Calculations for Drilling, Production and Workover, Second Edition

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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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For subsea applications, hydrostatic pressure exerted by the hydraulic fluid must be accounted for calculation.

In this case, we assume water depth at 1500 ft, therefore hydrostatic pressure exerted by hydraulic fluid (hydraulic fluid pressure gradient = 0.445 psi/ft) = 0.445×1500 = 668 psi. Besides of that, the concept for calculation is as same as surface accumulator. So please take a look about how to calculate usable volume per bottle as following steps.

Step 1 Adjust all pressures for the hydrostatic pressure of the hydraulic fluid:

Pre-charge pressure = 1000 psi + 668 psi = 1668 psi

Minimum system pressure = 1200 psi + 668 psi = 1868 psi

Operating pressure = 3000 psi + 668 psi = 3668 psi

Step 2 Determine hydraulic fluid required to increase pressure from pre-charge pressure to minimum system pressure:

Boyle’s Law for ideal gase: P1 V1 = P2 V2

1668 psi x 10 = 1868 x V2

16,680 ÷1,868 = V2

V2 = 8.93 gal

It means that N2 will be compressed from 10 gal to 8.93 gal in order to reach minimum operating pressure. Therefore, 1.07 gal (10.0 – 8.93 = 1.07 gal) of hydraulic fluid is used for compressing to minimum system pressure.

Step 3 Determine hydraulic required increasing pressure from pre-charge to operating pressure:

P1 V1 = P2 V2

1668 psi x 10 gal = 3668 psi x V2

16,680 ÷ 3668 = V2

V2 = 4.55 gal

It means that N2 will be compressed from 10 gal to 4.55 gal in order to reach operating pressure. Therefore, 5.45 gal (10.0 – 4.55 = 5.45 gal) of hydraulic fluid is used for compressing to operating pressure.

Step 4 Determine usable fluid volume per bottle:

Usable volume per bottle = Total hydraulic fluid/bottle – Dead hydraulic fluid/bottle

Usable volume per bottle = 5.45 – 1.07

Usable volume per bottle = 4.38 gallons

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

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Accumulator (Koomey) is a unit used to hydraulically operate Rams BOP, Annular BOP, HCR and some hydraulic equipment. There are several of high pressure cylinders that store gas (in bladders) and hydraulic fluid or water under pressure for hydraulic activated systems. The primary purpose of this unit is to supply hydraulic power to the BOP stack in order to close/open BOP stack for both normal operational and emergency situation. Stored hydraulic in the system can provide hydraulic power to close BOP’s in well control operation, therefore, kick volume will be minimize. Accumulators should have sufficient volume to close/open all preventers and accumulator pressure must be maintained all time.

This post you will learn how to calculate usable volume per bottle by applying Boyle’s gas law:

Use following information as guideline for calculation:

Volume per bottle = 10 gal

Pre-charge pressure = 1000 psi

Operating pressure = 3000 psi

Minimum system pressure = 1200 psi

Pressure gradient of hydraulic fluid = 0.445 psi/ft

For surface application

Step 1 Determine hydraulic fluid required to increase pressure from pre-charge pressure to minimum:

Boyle’s Law for ideal gase: P1 V1 = P2 V2

P1 V1 = P2 V2

1000 psi x 10 gal = 1200 psi x V2

10,000 ÷ 1200 = V2

V2 = 8.3 gal

It means that N2 will be compressed from 10 gal to 8.3 gal in order to reach minimum operating pressure. Therefore, 1.7 gal (10.0 – 8.3 = 1.7 gal) of hydraulic fluid is used for compressing to minimum system pressure.

Step 2 Determine hydraulic required increasing pressure from pre-charge to operating pressure:

P1 V1 = P2 V2

1000 psi x 10 gals = 3000 psi x V2

10,000 ÷3000 = V2

V2= 3.3 gal

It means that N2 will be compressed from 10 gal to 3.3 gal. Therefore, 6.7 gal (10.0 – 3.3 = 6.7 gal) of hydraulic fluid is used for compressing to operating pressure.

Step 3 Determine usable fluid volume per bottle:

Usable volume per bottle = Hydraulic used to compress fluid to operating pressure – hydraulic volume used to compress fluid to minimum pressure

Usable volume per bottle = 6.7 – 1.7

Usable volume per bottle = 5.0 gallons

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

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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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