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Calculate Oil-Water Ratio from Retort Data

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Oil Water Ratio (OWR) is a figure representing the fraction of oil and water in oil based drilling mud.  Generally speaking, it is a ratio between the percent oil in liquid phase and the percent water in liquid phase. In order to determine OWR, volume of oil/water/solid in drilling mud comes from a retort analysis. A sample of oil based mud is controlled burnt in a retort kit at required temperature. When the mud is heated, water and oil will be extracted out and solid is left in the retort kit. The retort analysis report shows percentage of each component by volume so we use data from the retort analysis to determine oil water ratio.

Retort kits, OilfieldPix.com 2017

Retort kits, OilfieldPix.com 2017

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.

Ref books: Lapeyrouse, N.J., 2002. Formulas and calculations for drilling, production and workover, Boston: Gulf Professional publishing.

Bourgoyne, A.J.T., Chenevert , M.E. & Millheim, K.K., 1986. SPE Textbook Series, Volume 2: Applied Drilling Engineering, Society of Petroleum Engineers.

Mitchell, R.F., Miska, S. & Aadny, B.S., 2011. Fundamentals of drilling engineering, Richardson, TX: Society of Petroleum Engineers.

Determine the density of oil and water mixture

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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 Drilling, Production and Workover, Second Edition

Mixing Fluids of Different Densities without Pit Space Limitation

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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 final mud weight and the final volume when mixing 10.0 ppg and 14.0 ppg drilling mud. There is no limitation regarding the pit space.

Mud in mud pits :

200 bbl of 10.0 ppg mud in pit A

300 bbl of 14.0 ppg mud in pit B

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 Drilling, Production and Workover, Second Edition

Mixing Fluids of Different Densities with Pit Space Limitation

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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 density with limited pit space”

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

Reduce mud weight by dilution

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You sometimes need to reduce your mud weight in hole. The method to reduce mud weight is dilution. The concept of dilution is to add light weight fluid into heavy fluid and when they mix together whole drilling fluid weight will decrease. The light weight fluid frequently used are fresh water, sea water, base oil, diesel, synthetic oil, etc which are normally your based fluid.

dilution

You can calculate how much light weight fluid in barrel required reducing original mud weight to new mud weight with this following formula:

Mud weight reduction by diluting with light weight fluid

Light weight fluid volume in bbl = V1 x (W1 – W2) ÷ (W2 – Dw)

Where;

V1 = Intial volume in bbl

W1 = Initial drilling fluid weight in ppg

W2 = Final drilling fluid weight in ppg

Dw = Density of light weight fluid in ppg

Example: Determine the number of barrels of base oil 7.2 ppg (Dw) required to reduce 200 bbl (V1) of 13.8 ppg (W1) to 10.0 ppg (W2):

Light weight fluid volume in bbl = V1 x (W1 – W2) ÷ (W2 – Dw)

Light weight fluid volume in bbl = 200 x (13.8 – 10) ÷ (10 – 7.2)

Light weight fluid volume in bbl = 271.4 bbl

You need to mix 271.4 bbl of base oil 7.2 ppg into 200 bbl of 13.8 ppg in order to achieve final mud weight of 10.0 ppg

Please find the Excel Sheet to calculate mud dilution.

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