Drilling Formulas and Drilling Calculations » Drilling Engineering Calculations

Drilling Formulas and Calculation Summary Page

DrillingFormulas.Com — Mon, 05 Jul 2010 03:04:14 +0000

I created summary page for drilling formulas and calculations (http://www.drillingformulas.com/drilling-formulas-and-calculation-sheets/). I wish these excel sheets would be helpful for you.

Amount of cuttings produced per foot of hole drilled
Annular Pressure Loss
Annular-Capacity
Annular-velocity
Bulk Density Calculation
Buoyancy Factor Oilfield
Convert Pressure to Equivalent Mud Weight
Convert-Specific-Gravity
Cost Per Foot Calculation
Critical RPM
D Exponent Calculation
D Exponent Corrected Calculation
Decrease Oil Water Ratio
Density of Oil Water Mixture
Depth of Washout
Dilution LGS control
Dilution LGS control – adding mud
Directional Survey Calculation – Angle Averaging Method
Directional Survey Calculation – Radius of Curvature Method
Dogleg Severity Calculation – Radius of Curvature Method
Dogleg Severity Calculation – Tangential Method
Drilling or Connection Ton-Mile
Drill-pipe-pulled-to-lose-certain-hydrostatic-pressure
Equivalent Circulating Density Calculation
Equivalent Circulating Density with engineering formula
Formation Integrity Test
Formation Temperature
Free Point Constant Calculation
Hydraulic Horse Power
Hydrostatic Pressure Calculation
Hydrostatic Pressure Decreases When POOH
Increase mud weight adding Barite
Increase mud weight adding calcium carbonate
Increase mud weight adding hematite
Increase Oil Water Ratio
Internal Capacity
Lag Time Calculation
Leak off test calculation
Light Weight Spot Pill
Loss hydrostatic due to filling water into annulus
Mix Different Fluid Density – Limit Space
Mix Different Fluid Density – Unlimit Space
Mud Volume Increase Due to Adding Barite
Mud Volume Increase Due to Adding Calcium Carbonate
Mud Volume Increase Due to Adding Hematite
Oil Water Ratio from a Retort Analysis
Pipe Displacement Calculation
Pressure and Force Calculation
Pressure Gradient Calculation
Pressure Required to Break Circulation In Annulus
Pressure Required to Break Circulation Inside Drill String
Pump Output (Duplex and Triplex)
Pump Pressure and Pump Stroke Relationship
Reduce Mud Weight by Dilution
Slug Calculation – Barrel of Slug Required
Slug Calculation – barrels of slug required for a desired length of dry pipe
Slug Calculation – Weight of Slug Required
Specific Gravity Calculation
Starting Volume for Weighting Up with Bartie
Starting Volume for Weighting Up with calcium carbonate
Starting Volume for Weighting Up with Hematite
Stuck Pipe Calculation – Not Know Free Point Constant
Stuck Pipe Calculation – Use Table for Free Point Constant
Temperature Conversion Formulas
Ton-Mile (TM) for Coring Operation
Ton-Mile (TM) for Making Short Trip
Ton-Mile (TM) for Round Trip
Ton-Mile (TM) for Setting Casing
Ton-Mile (TM) for Drilling or Connection

Ref book: Formulas and Calculations for Drilling, Production and Workover, Second Edition

Pressure Loss and Equivalent Circulating Density Review – Reverse Circulation

DrillingFormulas.Com — Sun, 23 May 2010 17:28:53 +0000

We still have the same question as the previous post, Pressure Loss and Equivalent Circulating Density Review, but this case we will do reverser circulation, circulating from annulus to tubing , and see how much pressure and equivalent circulating density at bottom hole.

Data Given

Reverse circulate total of 3 time bottom up from annulus to tubing with 12.7 ppg mud at 10,000′MD/10,000 TVD.

Pump pressure = 1000 psi

Annulus friction loss = 50 psi

Inside tubing friction loss = 925 psi

Surface line friction loss = 25 psi

Determine pressure at bottom hole.

We still apply the concept of frictional pressure so reverse circulation is calculated by this following equation:

Pressure at bottom hole (reference at annulus side) = Hydrostatic Pressure + Pressure from pump- Annular Pressure Loss

Note: Hydrostatic pressure and pressure from pump force downward to bottom hole but annulus pressure forces upward direction.

Pressure in the well at 10,000’ = 1000 + (0.052×10,000×12.7) – 50 = 7554 psi

OR you can referrence to the tubing side as well.

Pressure at bottom hole (reference at tubing) = Hydrostatic Pressure + Annular Pressure Loss in tubing + surface line pressure loss
Note: All pressure force downward to bottom hole so all pressure term must be sum together.

Pressure at bottom hole (reference at tubing) = (0.052×10,000×12.7) + 925+25 = 7554 psi.

Determine Equivalent Circulating Density at bottom hole.

ECD = Current mud weight in PPG + (annular pressure loss /(0.052xTVD)) = Total Pressure at Bottom Hole /(0.052xTVD)

ECD =7554 / (0.052 x 10,000) = 14.53 PPG.

The point that I want you to think of between this example and the previous example, Pressure Loss and Equivalent Circulating Density Review, is about the different of bottom hole pressure and ECD between forward circulation and reverse circulation.
You will see that reverse circulation results in a lot of pressure at bottom hole. Hence, you must keep in mind this concept and try to figure out how much pressure at bottom hole should be for both forward circulation and reverse circulation. Otherwise, you can accidentally break wellbore due to high ECD.

Ref book: Formulas and Calculations for Drilling, Production and Workover, Second Edition

Pressure Loss and Equivalent Circulating Density Review

DrillingFormulas.Com — Thu, 20 May 2010 02:58:34 +0000

This example that I got from my junior member is very simple but it helps you a lot to understand about how to determine pressure loss during normal circulation.

Information given is listed below;

Circulate at 3 bottom up through open end tubing (Down tubing and up annulus) with 12.7 ppg mud.

Pump pressure = 1000 psi

Annulus friction loss = 50 psi

Inside tubing friction loss = 925 psi

Surface line friction loss = 25 psi

Calculate the pressure in the well at 10,000’ (tubing tail). What would ECD at 10,000’ TVD be?

The concept of calculation that you should know : total pressure at bottom = pumping pressure + hydrostatic pressure – pressure loss in the opposite way of fluid flowing.

Then,

If I reference to tubing site, I will get the equation like this.

Pressure at bottom hole= Hydrostatic Pressure at bottom hole + Pressure from pump- Pressure Loss in surface line – Pressure loss in tubing

Pressure in the well at 10,000’ = 0.052×12.7×10000 + 1000 – 25 – 925 = 6654 psi

If I reference to annulus site, I will get the equation like this.

Pressure at bottom hole= Hydrostatic Pressure at bottom hole +Annular pressure loss

Note: Hydrostatic pressure and annular pressure loss force downward.

Pressure in the well at 10,000’ = 0.052×12.7×10000+ 50 = 6654 psi

Note: It doesn’t matter which site of u-tube you refer to the bottom hole pressure is still the same.
ECD (Equivalent Circulating Density) is calculated by this following equation:

ECD = Current mud weight in PPG + (annular pressure loss /(0.052xTVD))

ECD = 12.7 + (50/(0.052 x 10000))

ECD = 12.8 ppg

Ref book: Formulas and Calculations for Drilling, Production and Workover, Second Edition

How much pressure will we see at cement head in case of float shoe fail?

DrillingFormulas.Com — Sun, 16 May 2010 18:19:30 +0000

I still have the simple but interesting question about hydrostatic pressure which you can apply this calculation into drilling/cementing operation. The question I got about how much pressure we will see at cement head in case of float shoe fail.

Given situation as shown in figure above. This situation is in vertical well.

1) Inside 16” shoe, from 0’ to 1975’ = 9.3 ppg mud

From 1975’ to 2000’ = 16.0 ppg mud

2) Outside 16” shoe, From 0’ to 1500’ = 11.6 ppg cement (lead)

From 1500’ to 2000’ = 16.0 ppg cement (tail)

Given the conditions above and assuming the cement is still liquid, how much pressure will we see at cement head in case of float shoe fail?

** Note : you need to understand how to calculate hydrostatic pressure in order to fully understand this question **

Using U-tube concept: Bottom hole pressure both sides are the same.

Let’s work out at annulus side which is heavier due to cement in the annulus

Pressure at bottom hole in annulus = hydrostatic pressure of lead cement + hydrostatic pressure of tail cement

Pressure at bottom hole in annulus = 0.052×11.6×1500 + 0.052x16x(2000-1500) =1320.8 psi

Since, hydrostatic pressure in the annulus is more than hydrostatic pressure in 16″ casing; therefore, there will be pressure in the cement head in order to balance u-tube.

We can simply write equation as follows;

Bottom hole pressure = hydrostatic pressure inside 16” casing + surface pressure at cement head

1320.8 = 0.052x16x(2000-1975) + 0.052×9.3×1975 + surface pressure at cement head

Surface pressure at cement head = 1321 – 976 = 345 psi

Ref book: Formulas and Calculations for Drilling, Production and Workover, Second Edition

Round trip ton-miles Calculation

DrillingFormulas.Com — Mon, 01 Mar 2010 22:57:12 +0000

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.
Ref book: Formulas and Calculations for Drilling, Production and Workover, Second Edition

Accumulator Capacity – Usable Volume per Bottle Calculation for Subsea BOP

DrillingFormulas.Com — Mon, 14 Dec 2009 03:37:03 +0000

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: Well Control Book

Ref book: Formulas and Calculations for Drilling, Production and Workover, Second Edition

Accumulator Capacity – Usable Volume per Bottle Calculation (Surface Stack)

DrillingFormulas.Com — Wed, 09 Dec 2009 06:52:51 +0000

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: Well Control Book

Ref book: Formulas and Calculations for Drilling, Production and Workover, Second Edition

Understand Hydrostatic Pressure

DrillingFormulas.Com — Mon, 26 Oct 2009 02:33:46 +0000

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 book: Well Control Book

Corrected D exponent

DrillingFormulas.Com — Fri, 23 Oct 2009 02:13:41 +0000

The original “d” exponent is good for constant mud weight but in reality several drilling operations drill with various mud weights in hole due to weight up. In order to account for mud weight variation, so modification of d exponent, called “corrected d exponent”, has been made to correct for mud weight changes.

The corrected d-exponent is listed below.

dc = log (R ÷ 60N) ÷ log (12W ÷ 1000D) x (MW1 ÷ MW2)

Where;

dc = corrected “d” exponent

R = penetration rate in feet per hour

d = exponent in drilling equation, dimensionless

N = rotary speed in rpm

W = weight on bit in kilo pound

D = bit size in inch

MW1 = initial mud weight in ppg

MW2 = actual mud weight in ppg

Example: Determine the corrected d-exponent from following information.

Rate of penetration (R) = 90 ft/hr

Rotary drilling speed (N) = 110 rpm

Weight on bit (W) = 20 klb

Bit Diameter (D) = 8.5 in

MW1 = 9.0 ppg

MW2 = 12.0 ppg

Solution: dc = log [90÷ (60 x 110)] ÷ log [(12 x 20) ÷ (1000 x 8.5)] x (9.0 ÷ 12.0)

dc = 1.20 x 0.75

dc = 0.9

** Please remember that single d exponent or corrected d exponent valve does not help identify abnormal pressure. The trend of d exponent will help drilling personnel detect high formation pressure zones while drilling.

Please find the excel sheet for calculating the corrected D Exponent

Ref book: Formulas and Calculations for Drilling, Production and Workover, Second Edition

D Exponent Calculation

DrillingFormulas.Com — Tue, 20 Oct 2009 02:34:39 +0000

D exponent is an extrapolation of drilling parameters to get a trend while drilling into over-pressured zones. Usually, mud logger will correct all data, calculate d-exponent and plot the d exponent valve on the curve. The d-exponent can be utilized to detect transition from normal pressure regime to abnormal formation pressure. While drilling, if the change of trend is observed, rig supervisors must be cautious about this situation because this is one of the possible well control indications.

The “d” exponent described from the equation below:

d = log (R ÷ 60N) ÷ log (12W ÷ 1000D)

Where; R = penetration rate in feet per hour

d = exponent in drilling equation, dimensionless

N = rotary speed in rpm

W = weight on bit in kilo pound

D = bit size in inch

** Note: this equation is is valid for constant drilling fluid weight.

Example: Determine the d-exponent from following information.

Rate of penetration (R) = 90 ft/hr

Rotary drilling speed (N) = 110 rpm

Weight on bit (W) = 20 klb

Bit Diameter (D) = 8.5 in.

Solution:

d = log [90÷ (60 x 110)] ÷ log [(12 x 20) ÷ (1000 x 8.5)]

d = 1.20

Please find the Excel Sheet for calculating d-exponent.

Ref book: Formulas and Calculations for Drilling, Production and Workover, Second Edition