Aug 5

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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Jun 28

When you get differentially stuck, you may consider spotting light weight fluid to reduce force created by differential pressure between mud in wellbore and formation pressure. However, you really need to understand and be able to calculate how much light fluid you can put in hole that it will not create well control situation by accidentally underbalance formation pressure. This post demonstrates you how to determine maximum height of light weight spot pill that can be in the annulus.

Firstly, we really need to know how much under balance with current mud weigh is. You may get this information by calculating current hydrostatic pressure minus expected pressure based on pressure profile. Secondly, you must know how much your light weight pill is. Then please follow the steps below to determine how height of light weight spot pill in the annulus should be.

a) Determine the difference in pressure gradient in psi/ft between the mud weight and light weight spot fluid:

Difference pressure in pressure gradient in psi/ft = (current mud wt in ppg – light weight spot pill in ppg) x 0.052

b) Determine height in ft of light weight spot fluid that will balance formation pressure in the annulus:

Height ft in vertical = overbalance pressure with current mud weigh in psi ÷ difference in pressure gradient in psi/ft

This height is the maximum allowable height of light weight spot pill in the annulus. If you have higher length of light weight spot pill in the annulus, it may create well control situation.

Example: Use the following data to determine the height in ft of light weight spot pill that will balance formation pressure in the annulus:

Mud weight = 13.0 ppg

Light weight spot pill = 8.3 ppg

Amount of overbalance = 300 psi

a) Difference in pressure gradient in psi/ft:

Difference pressure in pressure gradient in psi/ft = (13 ppg – 8.3 ppg) x 0.052

Difference pressure in pressure gradient in psi/ft = 0.2444

b) Determine the height in ft of light weight spot liquid that will balance formation pressure in the annulus:

Height = 300 psi ÷ 0.244 psi/ft

Height = 1227 ft

You must ensure than height of light weight pill in the annulus must less than 1227 ft in order to prevent wellcontrol situation.

Please find the Excel sheet to calculate maximum height in annulus of light weight spot pill.

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

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May 15

After learning about capacity calculation, we can apply the capacity calculation to determine how much barrels of cutting produced per foot of hole drilled and total solid generated in pounds. Please read and understand the following formulas:

Use formular#1 and #2 for calculating amount of cutting generated per feet drilled.

Formula#1 for BARRELS of cuttings drilled per foot of hole drilled:

Barrels of cutting per foot drilled = Dh2 x (1 – % porosity) ÷1029.4

Where: Dh is hole diameter.

Example: Determine barrels of cuttings drilled for one foot of 6-1/8 inch hole with 25% (0.25) porosity:

Barrels/footage drilled = 6.1252 x (1 – 0.25) ÷1029.4
Barrels/footage drilled = 0.02733 bbl/footage drilled

Formula#2 for CUBIC FEET of cuttings drilled per foot of hole drilled:

Cubic feet of cutting per foot drilled = Dh2 x 0.7854 x (1 – % porosity) ÷144
Where: Dh is hole diameter.

Example: Determine barrels of cuttings drilled for one foot of 6-1/8 inch hole with 25% (0.25) porosity

Cubic feet/footage drilled = 6.1252 x 0.7854 x (1 – 0.15) ÷144
Cubic feet/footage drilled = 0.153462 cu ft/footage drilled

Moreover, you also apply simeple density and volume relationship to determine total solids generated. Use the following formula to calculate total solid generated.

Wcg = 350 x Ch x L x (l – porosity) x Cutting density

Wcg = solids generated in pounds
Ch = capacity of hole in bbl/ft
L = footage drilled in ft
Cutting density = cutting density in gm/cc

Example: Determine the total pounds of solids generated in drilling 100 ft of  6-1/8 inch hole (0.03644 bbl/ft). Density of cuttings = 2.20 gm/cc. Porosity = 25%:
Wcg = 350 x 0.03644 x 100 x (1 – 0.25) x 2.2
Wcg = 2104.41 pounds

Please find the excel sheet how to calculate how much cuttings drilled per foot of hole drilled and total solids generated

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

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