Drilling Formulas and Drilling Calculations

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
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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.

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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

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I got an email from Sanan asking me about how to calculate lag time. First of all, you need to know what the lag time is.

Lag time is traveling time interval required for pumping cuttings from each particular depth to surface. It can be expressed in terms of time (minutes) and pump strokes.

The lag time always changes when a well becomes deeper and/or pumping speed change. Two factors, affecting lag time calculation, are annulus volume of drilling fluid in and drilling mud flow rate.

With certain annular volume, the lag time, normally expressed in minutes, can be determined by dividing the annular volume (bbls) by the flow rate (bbl/min).

If there are changes in mud flow rate, the lag time figure will be changed as well. In order compensate for any changes, the lag time is transformed into pump strokes too; therefore, a change in speed of pump will not affect the lag time.

How to Calculate Theoretical Lag Time

There are 3 steps to do in order to calculate lag time as listed below;

1. Calculate pump output
2. Calculate annular volume at certain depth of hole
3. Calculate the theoretical lag time

Example – Determine lag time from bottom to surface with the following information;

Bit depth = 9500’ MD
Pump rate = 300 GPM
Annular volume at 9500’ MD = 250 bbl
Pump details: Triplex pump, 97% efficiency, liner size 6” and stroke length 12”

Solution;

Triplex Pump Output Formula is listed below;

Triplex Pump Output in bbl/stk = efficiency x 0.000243 x (liner diameter in inch) ² X (stroke length in inch)

Triplex Pump Output in bbl/stk = 0.97x 0.000243 x (6) ² X (12)
Triplex pump output = 0.102 bbl/stroke

Pump rate = 300 GPM ÷ 42 = 7.14 bbl / minute

Lag time in minutes = 250 bbl ÷ 7.14 bbl / minute = 35 minutes
Lag time in strokes = 250 bbl ÷ 0.102 bbl/stroke = 2451 strokes

Please find the lag time calculation sheet

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

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 book: Well Control Book

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 book: Well Control Book

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).

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

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 Drilling, Production and Workover, Second Edition

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 Drilling, Production and Workover, Second Edition

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 excel sheet used for calculating oil water ratio from retort data.

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