3. Drilling Fluids

Table of Content

1. Increase Mud Weight
2. Dilution
3. Mixing Fluids of Different Densities
4. Oil Based Mud Calculations
5. Solids Analysis
6. Solids Fractions
7. Dilution of Mud System
8. Displacement - Barrels of Water/Slurry Required
9. Evaluation of Hydrocyclone
10. Evaluation of Centrifuge

3.1 Increase Mud Weight

Mud weight, ppg, increase with barite (average specific gravity of barite - 4.2)

$\text{Barite, sk/100 bbl}=\frac{1470\times (W_2-W_1)}{35-W_2}$

Example: Determine the number of sacks of barite required to increase the density of 100 bbl of 12.0 ppg (W₁) mud to 14.0 ppg (W₂):

$\text{Barite, sk/100 bbl}=\frac{1470\times (14.0-12.0)}{35-14.0}$
Barite = 140 sk/ 100 bbl

Volume increase, bbl, due to mud weight increase with barite

$\text{Volume increase, per 100 bbl}=\frac{100\times (W_2-W_1)}{35-W_2}$

Example: Determine the volume increase when increasing the density from 12.0 ppg (W₁) to 14.0 ppg (W₂):

$\text{Volume increase, per 100 bbl}=\frac{100\times (14.0-12.0)}{35-14.0}$
Volume increase = 9.52 bbl per 100 bbl

Starting volume, bbl, of original mud weight required to give a predetermined final volume of desired mud weight with barite

$\text{Starting volume, bbl}=\frac{V_F\times (35-W_2)}{35-W_1}$

Example: Determine the starting volume, bbl, of 12.0 ppg (W₁) mud required to achieve 100 bbl (V_F) of 14.0 ppg (W₂) mud with barite:

$\text{Starting volume, bbl}=\frac{100\times (35-14.0)}{35-12.0}$
Starting volume = 91.3 bbl

Mud weight increase with calcium carbonate (SG — 2.7)

NOTE: The maximum practical mud weight attainable with calcium carbonate is 14.0 ppg.

$\text{Sacks/ 100 bbl}=\frac{945\times (W_2-W_1)}{22.5-W_2}$

Example: Determine the number of sacks of calcium carbonate/l00 bbl required to increase the density from 12.0 ppg (W₁) to 13.0 ppg (W₂):

$\text{Sacks/ 100 bbl}=\frac{945\times (13.0-12.0)}{22.5-13.0}$
Sacks/ 100 bbl = 99.5

Volume increase, bbl, due to mud weight increase with calcium carbonate

$\text{Volume increase, per 100 bbl}=\frac{100\times (W_2-W_1)}{22.5-W_2}$

Example: Determine the volume increase, bbl/100 bbl, when increasing the density from 12.0 ppg (W₁) to 13.0 ppg (W₂):

$\text{Volume increase, per 100 bbl}=\frac{100\times (13.0-12.0)}{22.5-13.0}$
Volume increase = 10.53 bbl per 100 bbl

Starting volume, bbl, of original mud weight required to give a predetermined final volume of desired mud weight with calcium carbonate

$\text{Starting volume, bbl}=\frac{V_F\times (22.5-W_2)}{22.5-W_1}$

Example: Determine the starting volume, bbl, of 12.0 ppg (W₁) mud required to achieve 100 bbl (V_F) of 13.0 ppg (W₂) mud with calcium carbonate:

$\text{Starting volume, bbl}=\frac{100\times (22.5-13.0)}{22.5-12.0}$
Starting volume = 90.5 bbl

Mud weight increase with hematite (SG — 4.8)

$\text{Hematite, sk/100 bbl}=\frac{1680\times (W_2-W_1)}{40-W_2}$

Example: Determine the hematite, sk/100 bbl, required to increase the density of 100 bbl of 12.0 ppg (W₁) to 14.0 ppg (W₂):

$\text{Hematite, sk/100 bbl}=\frac{1680\times (14.0-12.0)}{40-14.0}$
Hematite = 129.2 sk/100 bbl

Volume increase, bbl, due to mud weight increase with hematite

$\text{Volume increase, per 100 bbl}=\frac{100\times (W_2-W_1)}{40-W_2}$

Example: Determine the volume increase, bbl/100 bbl, when increasing the density from 12.0 ppg (W₁) to 14.0 ppg (W₂):

$\text{Volume increase, per 100 bbl}=\frac{100\times (14.0-12.0)}{40-14.0}$
Volume increase = 7.7 bbl per 100 bbl

Starting volume, bbl, of original mud weight required to give a predetermined final volume of desired mud weight with hematite

$\text{Starting volume, bbl}=\frac{V_F\times (40-W_2)}{40-W_1}$

Example: Determine the starting volume, bbl, of 12.0 ppg (W₁) mud required to achieve 100 bbl (V_F) of 14.0 ppg (W₂) mud with calcium carbonate:

$\text{Starting volume, bbl}=\frac{100\times (40-14.0)}{40-12.0}$
Starting volume = 92.9 bbl

⬆ Table of Content