Drilling waste consists of waste drilling fluid, drilled cuttings with associated drilling fluid, and, to a lesser extent, miscellaneous fluids such as excess cement, spacers, and a variety of other fluids. The amount of drilling waste depends on a number of factors. These include hole size, solidscontrol efficiency, the ability of the drilling fluid to tolerate solids, the ability of the drilling fluid to inhibit degradation or dispersion of drilled cuttings, and the amount of drilling fluid retained on the drilled cuttings.

One simple expression states the amount of wet drilled solids to be discarded as:

*S = ε × HV / Fs*

*where : S = volume of wet drilled solids, in bbl;*

* ε = efficiency of solids control, expressed as a fraction;*

* HV = hole volume, in bbl;*

* Fs = fraction of solids in the discard stream.*

The fraction of the solids in the discard stream varies from a maximum of about 50% to a lower value of about 2530%. There is always some amount of drilling fluid associated with drilled cuttings being discarded.

Solids-control systems, no matter how good, cannot totally separate the drilling fluid from the drilled cuttings. By the same token, rarely can all of the drilled cuttings be separated from the circulating system. This means that, with time, drilled solids will build up in the circulating system.

## Example 1

If 16 m³ of hole volume is drilled, if the solids-control efficiency is 70%, and if the fraction of solids in the discard stream is 0.5 (50%), how much drilled-solids discard is generated?

Sixteen cubic meters of earth (as drilled solids) is removed from the hole, and 70% of that volume is removed from the drilling fluid by the solids-control system. That means that 11.2 m³ of ‘‘dry’’ drilled solids would be removed (16 m³ × 0.7). But, since removing dry material without drilling fluid is impossible, some drilling fluid will be attached. In this case, a fraction of 0.5 indicates that an equal volume of drilling fluid is removed. Thus, two times the dry amount is removed (one part fluid and one part drilled solids). Therefore, the amount of drilled cuttings removed from the system is 22.4 m³ (11.2/0.5).

## Example 2

From a drilling rate of 50 feet per hour in a 121 4-inch hole, dry cuttings are being generated at 7.3 bbl/hr (1.16 m3/hr). Tests indicate that the solids-control system is removing 15.3 bbl/hr of drilled solids with associated drilling fluid. Tests on the discard indicate that the solids fraction in the discard stream is 0.33, or 33%. What is the solids-control efficiency?

The solids-control efficiency can be estimated by using the formula:

*ε = S / ( HV / Fs )*

In this case, hole volume (HV) is 7.3 bbl/hr, solids removal (S) is 15.3 bbl/hr, and the fraction of solids in the discard (Fs) is 0.33. Thus, the efficiency is 70%:

15.3 / (7.3 / 0.33).

Solids buildup in the drilling fluid is the portion of solids not removed and leads to dilution and discard of whole dirty fluid. The solids buildup is calculated by multiplying the hole volume by 1 minus the efficiency. Whole, dirty drilling fluid is discarded when it reaches the limit of drilled-solids tolerance for the fluid type. Typically, 5, 7, and 10% are used in water-based systems. Five percent indicates that the drilling fluid type is very intolerant to solids. If 7% is used, then the system is of about average tolerance. When 10% is used, then the system is reasonably tolerant for water-based fluid. A quick way of estimating the amount of drilling-fluid discard is to divide the solids buildup in the fluid by the drilled-solids tolerance (or the maximum percentage allowable of low gravity solids).

*L = HV × ( 1 – ε ) / T * or * L = DSL / T*

where:

*L = liquid discard, in bbl;*

*HV = hole volume, in bbl;*

*ε = efficiency of solids control, expressed as a fraction;*

*T = tolerance of the fluid system to solids contamination, expressed as a fraction;*

*DSL = solids buildup in the drilling fluid.*

The total waste or discard for water-based drilling-fluid systems is the sum of the waste solids with associated fluid (S) and the liquid discard (L).

As the solids-control efficiency increases, the waste solids increase slightly, but the liquid discard decreases dramatically. For this reason, small improvements in solids-control efficiencies can significantly reduce

the total amount of waste for disposal.

## Example 3

If 16 m³ of hole volume is drilled and the solids-control efficiency is 70%, how much solids buildup in the circulating system will occur?

Since the solids-control efficiency is 70%, the amount of drilled solids not removed must be 30% (1 – ε). Therefore, the amount of drilled solids buildup in the circulating system is 4.8 m³ (16 × 0.3).

## Example 4

If 189 m³ of hole volume is drilled and the solids-control efficiency is 70%, how much drilling fluid will need to be discarded if the tolerance of the system is 7%? In this example, 189 m3 of hole volume represents the approximate volume of 4000 feet of 17½-inch hole, which is a large but reasonable hole section. Since 30% of the hole volume is not removed by the solids-control equipment, 56.7 m3 of cuttings (1890.3) will build up in the system. Since the tolerance of the system is 7%, the amount of fluid discard is 810 m³ (56.7/0.07).

Table 1. indicates the amount of waste generated under similar conditions with varying solids-control efficiencies. Hole volume is constantly 6000 barrels (954m³). The solids content in the liquid discard is 50%. Tolerance of the fluid is 5%, meaning that the fluid system is sensitive to buildup of low-gravity solids.

In Table 1, notice that the hole volume is not the amount of waste fluid and cuttings. Yet calculated hole volume is still frequently reported as the total drilling waste. It can also be seen from the table that as the solids-control efficiency increases, the total waste or discard volume decreases. The converse is also true.

Efficiency( % ) |
Solids and associated ( LD ) |
LD |
TW |
Ratio ( TW : HV ) |

30 |
3600 |
84000 |
87600 |
11.4 |

50 |
6000 |
60000 |
66000 |
11 |

70 |
8400 |
36000 |
44400 |
7.6 |

*LD = liquid Discard; TW = Total waste; HV = Hole volume.*

TABLE 1.

Efficiency( % ) |
WS |
FD |
TW |
Ratio ( TW : HV ) |

30 |
3600 |
36000 |
39600 |
6.6 |

50 |
6000 |
30000 |
36000 |
6.0 |

70 |
8400 |
18000 |
264400 |
4.4 |

*WS = waste solids; FD = fluid discharge; TW = Total waste; HV = Hole volume.*

TABLE 2.

Data for total waste in the table are simply compared with the calculated hole volume. The larger the ratio, the lower the solids-control efficiency. In soft rock drilling conditions when water-based fluid is used, this ratio is frequently in the range of 8 or higher.

Table 2. illustrates the amount of waste generated under similar conditions but with solids tolerance at 10% rather than 5%. The total waste in Table 16.2 is much lower than that in Table 16.1 because the drilling-fluid system is much more tolerant to the accumulation of drilled solids. The system is considered more tolerant because the drilling fluid system is more inhibitive, the solids are less dispersive, or the effects

upon the fluid properties can be tolerated.

When an oil-based fluid (or a synthetic fluid with similar inhibitive characteristics) is used, the system is both more tolerant and more inhibitive. Typically, whole fluid is not discarded from an oil-based system, except when fluid is lost or during cleaning, etc. Measured discard rates are on the order of three or four times the hole volume, or less. This indicates that the solids-control efficiencies obtained while drilling with oil-based fluid are high and that the sensitivity of oil-based fluid to solids contamination is high (probably greater than 10%).

This can have significant implications. Modern water-based fluids contain inhibitive additives, fluid loss agents, lubricants, etc., that are used to adjust fluid properties. Many of these additives contain organic materials. When large amounts of the drilling fluid are discarded into the sea, the chemical loading from the organic materials can be high.

Synthetic fluid has potentially higher organic materials per barrel discharged, but less fluid is discharged. Therefore, the resulting chemical loading from organic materials can be less for synthetic-based fluid. All of these factors are used to determine the ultimate fate of the environment in the area of disposal.

## Example 5

If 189 m³ of hole volume is drilled with water-based fluid and the solids control efficiency is 30%, how much total waste will be generated? Assume that the drilling fluid will need to be discarded if the tolerance of the system is 7% and the solid fraction in the discard is 0.5 (50%). What is the ratio of total discard to hole volume?

Use the following formula for solid and associated fluid discard:

*S = ε × HV / Fs*

*where S = volume of wet solids, in bbl; ε = efficiency; HV = hole volume, **in bbl; Fs = solids fraction in the discard. The solids discard is 113.4 m³ **(189 × 0.3/0.5).*

Use the following formula for dirty, whole fluid discard:

*M = HV × (1 – ε ) ⁄ T*

*where M = drilling fluid, in m3, and T = tolerance. Dirty, whole fluid **discard is 1890 cubic meters (1890.7/0.07).*

The total discard is the sum of S and M. Thus, the total discard for this case is 2003 m3. The ratio of total discard to hole volume is 10.6 (2003/189).

## Example 6

If 189 m³ of hole volume (as in example 5) is drilled with oil-based fluid and the solids-control efficiency is 80%, how much drilled-solid waste (S) will be generated? Assume that the fraction of solids in the discard is 0.5. How much solids buildup in the system can be expected? If 400 m3 of drilling fluid exists, then what would the concentration of drilled solids in

the fluid be?

The drilled-solids discard would be 302.4 m³ (1890.8/0.5). The drilled-solids buildup in the system would be 37.8 m³ (189 × 0.2). If this amount is incorporated into 400 m3 of fluid, then the concentration is 9.5%.

Examples 5 and 6 illustrate the difference in waste generation between using water-based fluid and nonaqueous fluid (NAF). Since the waterbased fluid is sensitive to drilled-solids contamination and is not very inhibitive, the total waste amount is high (2003 m³). By using NAF, inhibition of the formation solids is increased and NAF sensitivity to contamination is decreased. The result is less total discard being generated (302 m³). However, the fluid on cuttings being discarded is NAF. And half of the total discard (by volume) is fluid.

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