3. Drilling Fluids

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3.4 Oil Based Mud Calculations

Density of oil/water mixture being used

(V₁ × D₁) + (V₂ × D₂) = (V₁ + V₂) × D_F

Example: If the oil/water (o/w) ratio is 75/25 (75% oil, V₁, and 25% water V₂), the following material balance is set up:

NOTE: The weight of diesel oil, D1 = 7.0 ppg
The weight of water, D2 = 8.33 ppg

(0.75) × (7.0) + (0.25) × (8.33) = (0.75 + 0.25) D_F
5.25 + 2.0825 = 1.0 D_F
7.33 = D_F
Therefore: The density of the oil/water mixture = 7.33 ppg

Starting volume of liquid (oil plus water) required to prepare a desired volume of mud

$SV = \frac{35 - W_{2}}{35 - D_{1}} \times DV$

where
SV = starting volume, bbl
W₁ = initial density of oil/water mixture, ppg
W₂ = desired density, ppg
DV = desired volume, bbl

Example: W1 = 7.33 ppg (o/w ratio — 75/25) W2 = 16.0 ppg Dv = 100 bbl
Solution:
$SV = \frac{35 - 16}{35 - 7.33} \times 100$
SV = 68.7 bbl

Oil/water ratio from retort data

Obtain the percent-by-volume oil and percent-by-volume water from retort analysis or mud still analysis. From the data obtained, the oil/water ratio is calculated as follows:

$% oil in liquid phase = \frac{% by vol oil}{% by vol oil + % by vol water} \times 100$

$% water in liquid phase = \frac{% by vol water}{% by vol oil + % by vol water} \times 100$

Result: The oil/water ratio is reported as the percent oil and the percent water.

Example: Retort analysis: % by volume oil = 51
% by volume water = 17
% by volume solids = 32

Solution:
a. % oil in liquid phase = 51 × 100 ÷ (51 + 17)
% oil in liquid phase = 75
b. % water in liquid phase = 17 × 100 ÷ (51 + 17)
% water in liquid phase = 25
c. Result: Therefore, the oil/water ratio is reported as 75/25: 75% oil and 25% water.

Changing oil/water ratio

NOTE: If the oil/water ratio is to be increased, add oil; if it is to be decreased, add water.

Retort analysis: % by volume oil = 51
% by volume water = 17
% by volume solids = 32

The oil/water ratio is 75/25.

Example 1: Increase the oil/water ratio to 80/20:

In 100 bbl of this mud, there are 68 bbl of liquid (oil plus water). To increase the oil/water ratio, add oil. The total liquid volume will be increased by the volume of the oil added, but the water volume will not change. The 17 bbl of water now in the mud represents 25% of the liquid volume, but it will represent only 20% of the new liquid volume.

Therefore: let x = final liquid volume
then, 0.20 x = 17
x = 17 ÷ 0.20
x = 85 bbl
The new liquid volume = 85 bbl

Barrels of oil to be added:
Oil, bbl = new liquid vol − original liquid vol
Oil, bbl = 85 − 68
Oil = 17 bbl oil per 100 bbl of mud

Check the calculations. If the calculated amount of liquid is added, what will be the resulting oil/water ratio?

$% oil in liquid phase = \frac{original vol oil + new vol oil}{original liquid oil + new oil added} \times 100$

$% oil in liquid phase = \frac{51 + 17}{68 + 17} \times 100$
% oil in liquid phase = 80
% water would then be: 100 − 80 = 20
Therefore: The new oil/water ratio would be 80/20.

Example 2: Change the oil/water ratio to 70/30:

As in Example 1, there are 68 bbl of liquid in 100 bbl of this mud. In this case, however, water will be added and the volume of oil will remain constant. The 51 bbl of oil represents 75% of the original liquid volume and 70% of the final volume:

Therefore: let x = final liquid volume
then, 0.70 x = 51
x = 51 ÷ 0.70
x = 73 bbl
The new liquid volume = 73 bbl

Barrels of water to be added:
Water, bbl = new liquid vol − original liquid vol
Water, bbl = 73 − 68
Water = 5 bbl of water per 100 bbl of mud

Check the calculations. If the calculated amount of water is added, what will be the resulting oil/water ratio?

$% water in liquid phase = \frac{17 + 5}{68 + 5} \times 100$
% water in liquid = 30
% oil in liquid phase = 100 − 30 = 70
Therefore: The new oil/water ratio would be 70/30.

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