Economics
1 Introduction ... 1
2 Economic Justification...2
2.1 Penetration Rate... 2
2.2 Dilution Rate ... 3
2.2.1 Economic Analysis Calculations... 4
3 Solids Control Economics and Performance Program (SECOP) ... 8
4 Monitoring System Performance ... 9
4.1 API Procedure for Evaluating Total Efficiency of Solids Control Systems (Water-Based Muds) ... 10
5 Summary... 12
FIGURES Fig. 1. Effects of solids content on drilling performance. ... 3
TABLES Table 1 Solids Control Economic Analysis Parameters... 4
1 Introduction
The impact of good solids control can be very significant and can lead to substantial cost savings, but often there is reluctance to invest in solids control for the following reasons:
1. Many of the benefits are indirect and the savings are hard to quantify. 2. Methods to economically justify solids control equipment were not
available.
3. Techniques to measure performance are limited.
4. Disappointing results from ill-chosen or incorrectly-operated equipment.
Although the benefits from good solids control are numerous, the cost savings are not apparent in normal drilling cost accounting. For example, the savings due to reduced trouble costs and improved penetration rate, although substantial benefits, cannot be accurately calculated. Usually the drilling fluid gets most of the credit (or blame) since mud material consumption is easily tracked and the mud properties are the only direct indication of solids control system performance. In a realistic sense, the mud and the solids control equipment are integral parts of one system. One cannot plan the mud without considering the solids control system and vice versa. This does not mean that the benefits of good solids control practices cannot be measured.
2 Economic Justification
2.1 Penetration Rate
The impact of solids control on penetration rate is best depicted by Fig. 1. This has become somewhat of a classic illustration of the benefits of a low solids content mud. For example, a reduction in average solids content from 4.8% (9.0 ppg) to 2.6% (8.7 ppg) results in a 15% reduction in total rig days. Given a 10,000 ft well costing $700,000 excluding mud cost, the estimated savings could reach $100,000. If even half of these savings were realized, it would more than pay for the best solids removal system available.
In soft rock country such as the Gulf Coast, efficient solids removal can reduce the need to control-drill by limiting required dilution rates to manageable levels and reducing operational problems due to overloaded solids removal equipment. The benefits from efficient solids removal, e.g., “low-silt” muds, have been documented for Gulf Coast drilling since the mid-60s when hydrocyclone use was first advocated.
Fig. 1. Effects of solids content on drilling performance.
Note: The benefits of low solids contents are most apparent at less than 5% solids.
2.2 Dilution Rate
Solids removal efficiency directly impacts dilution costs. When dilution water is added to the system, three costs are incurred simultaneously:
1. Dilution water cost.
2. Cost of additives to maintain stable mud properties. 3. Disposal cost.
The savings due to improved penetration rates and reduced trouble time, while real, cannot be reliably predicted as justification for improved solids control equipment. In many cases however, the economic advantages due to reduced dilution and disposal costs are more than enough to justify expenditures for additional equipment. The economic benefits in terms of mud consumption and disposal can be determined through a simple mass balance analysis: Removing a given percentage of drilled solids will result in a certain dilution volume to maintain the desired maximum concentration of drilled solids in the mud. The relevant parameters and their symbols used in the calculations are listed below.
Table 1 Solids Control Economic Analysis Parameters
Vc = Volume of drilled solids generated, bbls Vi = Initial volume in tanks, previous hole/casing, bbls Vf = Final volume in tanks, previous hole/casing, bbls Vd = Volume of addition/dilution fluid required, bbls Vlw = Volume of liquid waste to be disposed, bbls Vsw = Volume of wet solids to be disposed, bbls
Vt = Total volume of solids and liquids to be disposed, bbls
ki = Initial concentration of drilled solids, vol. fraction ks = Maximum volume fraction of drilled solids, vol. fraction X = Drilled solids removed by equipment, vol. fraction Y = Liquid associated with the cuttings, bbl/bbl
D = Hole diameter, in.
L = Section length, ft
W = Washout, vol. fraction
rd = Density of dilution fluid, ppg rc = Density of drilled cuttings, ppg
ri = Mud weight at the start of the section, ppg re = Desired mud weight, end of section, ppg
2.2.1 Economic Analysis Calculations
First, the volume of cuttings generated in a given interval must be calculated:
V
c= 0.000971 x D x L x W
2For a given percent of drilled solids removed, X, the required dilution volume is computed by:
(
)
(
)
V
V
V
k
k
V
d c i i s i=
1- k
k
1- X
s s-
+
The following equations may be used to calculate the solids removal efficiency, Xc, and the associated dilution volume required to discharge only wet solids:
(
)
(
)
X
k V
V
k V
V
k Y
c s f c i i c s=
V
c-
+
+
+
1
(
)
(
)
V
d= V
f−
V
i+
X V
c c1
+
Y
The required mud weight (density) of the dilution volume, Vd, is based on the specified starting and ending densities and is calculated by:
(
)
ρ
dρ
eρ
eρ
i cρ
ρ
d c eV
V
X
=
V
V
i d+
−
−
(
1
−
)(
−
)
The total volume of solids and liquid generated in an interval is given by:
V
t= V
i+
V
c+
V
dThe wet solids volume, Vsw, and liquid volume, Vlw, discharged while drilling the interval is computed by:
(
)
V
sw= XV
c1
+
Y
(
)
V
lw= V
t−
V
f+
V
c+
V
swThe remaining circulating volume includes the volume of solids not removed by the solids removal equipment. Since the solids are assumed to be too fine to be removed by the solids control equipment, their volume is counted as liquid volume for disposal purposes.
When the entire circulating system is to be discharged at the end of the interval, the total liquid for disposal is calculated by:
Once the waste volumes are calculated, the total dilution and disposal cost for the interval may be determined by estimating the equipment rental cost and the cost/bbl for addition/dilution and liquid/solids disposal:
1. Solids Control Equipment Cost
- Estimate rental, transport, service, and maintenance (e.g., screens) cost for the interval.
2. Addition/Dilution Cost
- Estimate the cost/bbl by including purchase cost for dilution liquid, trucking, and additive cost.
3. Liquid/Solids Disposal Cost
- Estimate the cost/bbl by including hauling, disposal, treatment, reserve pit construction and reclamation.
Example Calculations
Interval Data:
Vc = Volume of drilled solids generated, bbls
Vi = 360 bbls
Vf = 360 bbls
Vd = Volume of addition/dilution fluid required, bbls Vlw = Volume of liquid waste to be disposed, bbls Vsw = Volume of wet solids to be disposed, bbls
Vt = Total volume of solids and liquids to be disposed, bbls ki = 0 (fresh mud, no drilled solids)
ks = 0.06 (6% maximum drilled solids) X = 0, 0.1, 0.5 (3 cases)
Y = 1.0 (1:1 solids to liquid ratio in wet solids discharge)
D = 12.25 in.
L = 1600 ft
W = 1.10 (10% washout)
rd = Density of dilution/addition fluid, ppg rc = 2.6 x 8.34 = 21.68 ppg
ri = 8.6 ppg initial mud weight re = 9.4 ppg final mud weight Dilution Cost: $5.00/bbl
Liquid Waste Cost: $3.00/bbl Solid Waste Cost: $5.60/bbl
Calculations: 1. Cuttings volume:
V
c= 0.000971 x D x L x W
2(
) (
) ( )
V
c= 0.000971 x 12.25 x 1600 x 1.1 = 256 bbls
22. Dilution volumes for each solids removal efficiency:
(
)
(
)
V
V
V
k
k
V
d c i i s i=
1- k
k
1- X
s s-
+
For X = 0.0(
)
( )
V
d=
1- 0.06
= 3650 bbls
0 06
1 0 256
360
0
0 06
360
.
(
−
)
−
+
.
For X = 0.1(
)
( )
V
d=
1- 0.06
= 3250 bbls
0 06
1 0 1 256
360
0
0 06
360
.
(
−
. )
−
+
.
For X = 0.5(
)
( )
V
d=
1- 0.06
= 1645 bbls
0 06
1 0 5 256
360
0
0 06
360
.
(
−
. )
−
+
.
3. Dilution density:In this example, the required density will not change with each case. The parameters for X=1 are chosen for illustration purposes.
(
)
ρ
dρ
eρ
eρ
i cρ
ρ
d c eV
V
X
=
V
V
i d+
−
−
(
1
−
)(
−
)
(
)
ρ
d= .4
360
360
= 8.6 ppg
9
9 4
8 6
256
3250
1 0 1 217
9 4
+
.
−
.
−
(
−
. )(
.
−
. )
4. Solids removal efficiency and dilution volume to achieve zero whole-mud discharge while drilling:
(
)
(
)
X
k V
V
k V
V
k Y
c s f c i i c s=
V
c-
+
+
+
1
(
) ( )
(
)
X
c=
256
x 1.0
= 0.81
−
+
+
+
0 06 360
256
0 360
256 1 0 06
.
.
(
)
(
)
V
d= V
f−
V
i+
X V
c c1
+
Y
(
)
( )( )
V
d= 360
−
360
+
0 81 256 1 1
.
+
= 415 bbls
5. Summary of waste disposal volumes:
Total Volume bbls
Wet Solids bbls
Liquid While Drilling bbls Total Liquid bbls X = 0.00 4266 0 3650 4266 X = 0.10 3866 51 3199 3815 X = 0.50 2261 256 1389 2005 X = 0.81 1030 414 0 616
6. Cost estimate for each case, discarding total liquid volume (last column in Step 5): Drilled Solids Removed Equipment Costs Addition/Dilution Costs Disposal Costs Solids Liquids Total Costs 0% $0 $18,250 $0 $12,678 $30,928 10% $100 $16,250 $286 $11,445 $28,081 50% $500 $8225 $1434 $6015 $16,174 81% $5000 $2075 $2318 $1848 $11,241
The example illustrates how an increase in equipment costs to improve solids removal efficiency is justified by the savings in addition/dilution and disposal costs, even without considering savings attributable to higher penetration rates or reduced trouble costs.
3 Solids Control Economics and Performance Program (SECOP)
A natural question arising from the economic analysis exercise is “What equipment will I need to achieve the optimum solids removal efficiency?” It is also apparent that the determination of an economically-optimum solids control system can be a time-consuming, iterative process. The equipment costs to achieve the minimum required dilution volume (commonly called a “closed-loop” mud system) may not be economic in all cases. It may not even be physically possible with available mechanical solids removal
technology. The Solids Control Economic and Performance Analysis Program (SECOP) was developed at APR to assist drilling personnel in the optimum selection of solids control equipment. It is available as an Integrated Drilling Assistance Program for use on the PC.
1. The economics of solids control in terms of potential savings in mud dilution and disposal costs versus the percent drill solids removed. 2. The performance of solids control equipment. It predicts the drill solids
removed by each piece of equipment selected.
3. The loss of weighting material and mud from each piece of equipment for weighted muds and the predicted recovery from barite-recovery centrifuging.
4. The performance for different equipment options to determine the most effective solids control system for drilling a well.
SECOP predicts only the savings in mud and disposal costs. As discussed previously, no model exists to predict additional savings from higher penetration rates and lower trouble costs that result from effective solids control. The program uses models developed as a result of extensive equipment testing at APR to predict individual equipment and total system performance. The overall economics calculations are based on the same equations described above. A complete description of the program is provided in the IDAP reference manual.
The recommended application of SECOP is to match the performance history of the solids control system for an offset well. This can be done by selecting the proper lithology and resulting particle size distribution which matches the mud volumes and costs for the offset well. Once a lithology match has been made, different equipment options may be tried to find the most economically-effective solids control equipment for the proposed well. A successful economic analysis for future wells will depend on determining a representative particle size distribution from the offset well which, in turn, is dependent upon having accurate records of dilution volumes and equipment operation. This emphasizes the importance of accurately metering water additions and equipment performance while drilling. SECOP may then be used to monitor equipment performance and establish representative particle size distributions for future economic analysis and equipment selection.
4 Monitoring System Performance
The API Recommended Practice 13C contains a field method for evaluating the total efficiency of the drilling fluid processing system in water-based fluids. As with any performance analysis, this procedure depends upon accurate dilution volume information. The API procedure uses the dilution volume over a given interval to compute a dilution factor, DF, which is the volume ratio of actual mud built to mud dilution required to maintain a desired solids concentration with no solids removal equipment. The dilution factor is used to determine the total solids removal efficiency of the system.
This total efficiency can then be used in SECOP to establish a representative particle-size distribution for further analysis and equipment performance predictions.
4.1 API Procedure for Evaluating Total Efficiency of Solids Control
Systems (Water-Based Muds)
1. Over a desired interval length, obtain accurate water additions and retort data.
2. From the retort data, calculate:
- The average drilled solids concentration in the mud, ks.
- The average water fraction in the mud, kw.
3. Calculate the volume of mud built, Vm:
V
V
k
m ww =
4. Calculate the volume of drilled solids, Vc:
V
c=0.000971 x D
2x L x W
5. Calculate the dilution volume required if no solids were removed, Vd:
V
V
k
d cs =
6. Calculate the dilution factor, DF:
DF
V
V
m d=
7. Calculate the total solids removal performance, Et:
Et = (1 - DF) Multiply by 100 to calculate as a percentage. The accuracy of the API procedure depends on a relatively constant solids concentration in the mud, constant surface circulating volume, and consistent averaging techniques over the interval of interest. Regardless, the total solids removal performance should be reported at frequent intervals to facilitate solids control analysis and planning for future wells.
Example Calculation
Interval Data:
Water Added, Vw 1481 bbl
Average Water Fraction, kw 0.9
Interval Length, L 1600 ft
Bit Diameter, D 12.25 in.
Washout, W 10%
Average Drilled Solids Concentration, ks 0.06
Calculations:
1. Calculate the volume of mud built, Vm:
V
V
k
m w w ==
1481
0.9
= 1645 bbls
2. Calculate the volume of drilled solids, Vc:
V
c= 0.000971 x D
2x L x W
= 0.000971 (12.25)
2(1600)(1.1)
= 256 bbls
3. Calculate the dilution volume required if no solids were removed, Vd:
V
d=
V
k
=
256
0.06
= 4267 bbls
c s4. Calculate the dilution factor, DF:
DF
V
V
m d=
=
1645
4267
= 0.386
5. Calculate the total solids removal performance, Et:
(
)
E
t= 1- DF = 1- 0.386 = 0.614
Expressed as a percentage:
5 Summary
· The economic advantages of good solids control practices, while real, are usually difficult to predict in terms of improved penetration rates and reduced trouble time. However, savings in dilution and disposal costs can be predicted and are often ample justification to invest in improved solids control equipment.
· Solids removal efficiency directly impacts the cost of dilution, material consumption and waste disposal. A simple mass balance approach may be used to predict total dilution and waste volumes as a function of solids removal efficiency. Example calculations show how an investment in solids control equipment may be easily justified by the savings realized from reduced addition/dilution and disposal costs. · The solids control economics and performance program “SECOP” may
be used to select the most effective solids control system. This program predicts:
- The savings in mud dilution and disposal costs vs. the percent solids removed.
- The drilled solids removed by each piece of equipment.
- Loss of weighting material and mud from each piece of equipment. - Recovery from barite-recovery centrifuging.
· The program is available as an Integrated Drilling Assistance Program. · The API Recommended Practice 13C contains a field method for
monitoring system performance in the field. This method depends upon accurate dilution volume monitoring to determine total solids removal efficiency. The API procedure and example calculations are presented in this section.
