AM0801-Design of Sample Strategies and Sampling Techniques_Widiastuti Setyaningsih

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Widiastuti SetyaningsihEuropean Master in Quality Analytical in Laboratory

Gdansk University of Technology

E-mail [email protected]

Mobile +48889664465 or +6287878121767

AM0801-DESIGN OF SAMPLE STRATEGIES

AND SAMPLING TECHNIQUES

Date : February 17, 2009

Sampling course. Assessment work

1. Hands-on PC seminar (EXCEL work. Use of Gy’s formula and replication experiments)

2. Mass reduction. Biased and unbiased sampling devices.3. Case study 1-D (gas production plant)

1. Case study 0-D (waste container). Describe briefly the sampling procedure to obtain 30 g of representative sample out of 10 tons of heterogeneous material from a waste container. Write

the sequence of SUOs used in the sampling procedure.

Answer :

Sampling Procedure

Objective

to Obtain 30 g of Representative Sample Out of 10 Ton Lot

Definition

Lot = Heterogeneous Material from a Waste Container Increment = Group of heterogeneous material fragment from each waste container

sampling

StepsProcedures

SUO1 Since the lot was huge as 3-Dimensional, we need to reduce the dimension with

put it into a conveyor belt.“Let assume that our conveyor belt transporting (V c ) 60 ton per hour”

SUO6 Do the composite sampling by take the increments from conveyor belt with cross

section samplers to reduce the effect of distributional heterogeneity.

“ Assume that cross section belt velocity (V s ) is 10 cm/s

and sample container width = 10 cm”

The minimum increments weight M i-min as much as

M i-min = 10 000 000 gram x 10 cm = 16666.67 gram

10 x 60 s x 10 cm/s

Vs = 10 cm/s

X = 10 cm

Vc =60 Ton/Hr

mailto:[email protected]

mailto:[email protected]



If the waste material consists of crushed sandstone, it might have C : 1370g/cm3

(ref. http://www.simetric.co.uk/si_materials.htm ); assume d (Average) = 2 cm,So we can calculate Absolute Minimum Sample Weight M s-min , which is tied to

tolerated primary fundamental error (assume S (FSE 1 ) 10%)

M s-min = Cd 3 = 1370 g/cm3 x 23 cm3 = 1096000

S 2 (FSE 1 ) (10/100)2

Therefore, this minimum sample (Q) should be made of :

Q = M s-min = 1096000 = 65.7599 66 increments

M i-min 16666.67

If the sample should be representative of 10 Ton waste material (or 10 minutes

in the conveyor belt), then we shall collect 1 increment every 9 seconds.

Collect the increments into a composite which known as Primary Sample(Considering for the sampling strategies which might applied. It might stratified randomized during 9 seconds). Primary Sample should be taken as much as

S 1 = 10 cm x 2 cm x 2 cm x 1370 g/cm

3

= 54800 gram

SUO4 Particle Size Reduction (Commutation) is needed since the material have constitutionalheterogeneity. It might done with solid waste shredder.

“ Assume that the waste material will be crushed to 0.5 cm before reaching the secondary sampling”

SUO5 Blend the shredded waste material to reduce distributional heterogeneity.

SUO7 Split it into parts with the same size to obtain Representative Mass Reduction.

So, we can have M s-min as belowM s-min = Cd 3 = 1370 g/cm3 x 0.53 cm3 = 17125 gram

S 2 (FSE 2 ) (10/100)2

Sampling Ratio = 17125 gram = 0.3125 or 33.33 %

54800 gram

Allot the primery sample into 3 same part then we will get secondary sample

as much as 54800/3 gram = 18266.67 gram

SUO4 Refined the tertiary sample to make it closer to homogeneous level with grinder.

“ Assuming that we are grinding the secondary sample to 16 mesh (0.10 cm)”

SUO5 Blend again to make it distribute well.

SUO7 Split it into parts with the same size to obtain Representative Mass Reduction.So, we can have M s-min as below

M s-min = Cd 3 = 1370 g/cm3 x 0.103 cm3 = 137 gram

S 2 (FSE 2 ) (10/100)2

Sampling Ratio = 137 gram = 0.00749 or 0.75 %

18266.67 gram

http://www.simetric.co.uk/si_materials.htm

http://www.simetric.co.uk/si_materials.htm



Since the sample was very fine and mixed well. We can split into the desired

sample size for the analysis (or weigh it as much as 30 gram)

Keep the rest in a good storage location (for retain sample).

RESUME

LOT

10 Ton, Heterogeneous Material from a Waste Container

SUO1-Dimensional Reduction

conveyor belt transporting (V c ) 60 ton per hour

SUO6-Composite Sampling

collect 66 increments; 1 increment in range 9 seconds

Primary Sample : 54800 gram

SUO4-Particle Size Reduction

crushed to 0.5 cm

SUO5-Blend/Mix

SUO7-Representative Mass Reduction

Allot the 1 st sample into 18266.67 gram for each 2nd sample

SUO4-Particle Size Reduction

grinding into 16 mesh (0.10 cm)

SUO7- Representative Mass Reduction

SUO7- Representative Mass Reduction

Weighing as much as 30 g for 3rd sample for analysis needs

Store the Retain Sample



2. Hands-on PC seminar (EXCEL work. Use of Gy’s formula and replication experiments)

1. Given a lot of C = 540 g/cm3, and lot mass ML = 100 Kg,

a) calculate the fundamental sampling error if the particle size, d = 0,1 cm, and the sample

mass, MS = 50 g. How big would be the absolute error if the lot grade (aL) is 0,05%?

Answer :

(Reference : M H Ramsey and S L R Ellison. (2007). Measurement uncertainty

arising from sampling : a guide to methods & approaches. From135.196.210.195/images/ EURACHEM1_tcm18-102815.pdf )

C = 540 g/cm3σ a = ?

d = 0.1 cmMs = 50 g

ML = 100 Kg = 100 000 g

aL = 0.05%

σr 2 = (540 g/cm3)(0.1 cm)3((1/50 g)-(1/100000 g))

σr

2

= 0.0107946σr

= (0.0107946)½

σr = 0.103897064

σa = σr . aL = (0.103897) (0.05) % = 0.519 %

b) calculate the total fundamental sampling error if the particle size, d = 0,1 cm, and we

perform a primary sampling taking sample mass, MS = 1000 g followed by a reduction of particle size, d =0,05 g, and a secondary sampling, taking MS = 50 g.

Answer :

s2 (FSET) = s2 (FSE1) + s2 (FSE2)

C = 540 g/cm3 Ms1 = 1000 gd1 = 0.1 cm Ms2 = 50 g

d2 = 0.05 cm ML = 100 Kg = 100 000 g

s2 (FSET) ??

s2 (FSE1) = (540 g/cm3)(0.1 cm)3((1/1000 g)-(1/100000 g)) = 0.0005346

s2 (FSE2) = (540 g/cm3)(0.05 cm)3((1/50 g)-(1/1000 g)) = 0.0012825s2 (FSET) = 0.000534 + 0.0012825 = 0.0018165

s (FSET) = 0.042620418 = 4.26 %

−= LS

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c) For a lot of the same material and d = 0,1 cm, how much sample mass should be taken to

reach a fundamental sampling error of 2 %? And 1%, 5%?

Answer :

In Gnumeric File (Spreadsheet of Linux)Widiastuti.Setyaningsih_assessment_material >> Sheet 1

d) For a lot of the same material and taking a sample size of 500 g, how small should be the particle size to attain fundamental sampling errors of 1%, 2%, 5%?

(Use the SOLVER tool for parts c) and d))

Answer :

In Gnumeric File (Spreadsheet of Linux)

Widiastuti.Setyaningsih_assessment_material >> Sheet 1

e) For a lot of the same material, plan sample strategies that allow for having a fundamentalsampling error of 2 % (particle size range allowed: 0,01 cm - 0,1 cm; sample mass range

allowed: 20 g – 200 g). Build a contour plot representing CV (%) (z) vs. particle size (x) and

sample mass (y) to detect the possible ranges of particle size and sample mass that can provide the desired error level.

i) Can we work with a sample mass equal to 20 g? Which particle size range should be

used? Answer :


ii) How low the error can get if the particle size is 0,1 cm?

Answer :


2. Study the effect of the factors that modify the fundamental sampling error according to Gy’sformula. Make plots in a), b) and c) for a lot where the critical particles are mixed with the

gangue, i.e., β = 1, α = 100, and for a lot where the critical particles are fully integrated in the

lot, i.e., β = 0.03, α = 30.

Sheet: Gy’s formula liberated particle, Gy’s formula integrated particles. a) Effect of particle size. Represent the variation of CV (%) vs. particle size (particle size

range from 0.01 cm to 0.5 cm).

Answer :

In OpenOffice File (Spreadsheet of Linux)

Widiastuti.Setyaningsih_assessment_material

Sheet : Gy's formula liberated particle and Gy's integrated particles

b) Effect of sample mass. Represent the variation of CV (%) vs. sample mass. (sample mass

range: from lot mass to 10-5 times lot mass). You can use a sample mass logarithmic scale

for a better visualization. Answer :

In OpenOffice File (Spreadsheet of Linux)Widiastuti.Setyaningsih_assessment_material




c) Effect of lot grade. Represent CV (%) vs. lot grade. (lot grade range: from 10% to 10-6 %).

You can use a lot grade logarithmic scale for a better visualization.

Answer :

In OpenOffice File (Spreadsheet of Linux)Widiastuti.Setyaningsih_assessment_material


3. Calculate the GEE, TSE, PSE, SSE and TAE errors (%) from the data of the replication

experiment.

Answer :

In OpenOffice File (Spreadsheet of Linux)

Widiastuti.Setyaningsih_assessment_material Sheet : Replication experiment

EXCEL file: lot0D.xls

Sheet: Gy’s SOLVER and contour plot (exercise 1 c), d) and e))

Sheet: Gy’s formula liberated particle, Gy’s formula integrated particles (exercise 2). Sheet: replication experiment (exercise 3).

4. Mass reduction. Biased and unbiased sampling devices. Reading material: L. Petersen, C.K.Dahl and K.H. Esbensen. Chemom. Intell. Lab. Sys. 74(2004)95-114.

a. Which are the main causes of biased mass reduction in the design of riffle splitters?

Answer :

The main causes of biased mass reduction in the design of riffle splitters are :

1. Riffle Splitters Model (Closed riffle splitters provide lower bias than opened model interm of bouncing materials)

2. Feeding Tray Size (To reduce the bias , feeding tray must have exactly the same width

as the rectangular receiving region of splitter)

3. The Chutes Characteristics

• The width of the chutes (To reduce the bias , it must be minimum 5 mm or 3 times

the largest particle diameter and must all have the same size and form)

• The number of the chutes (The larger number of chutes makes the bias smaller

because the device provide better result to split the sample. And to reach the lowest

bias , it have to be an equal number of chutes)

◦ Compromises an even number, but not less than a total of eight (usually between

12 and 20)

◦ The number of the chutes have to be an equal number for the both sides

(adjacent chutes discharging at opposites sides)

• Wall dimension of chutes (it must be thin in relation to the wall-to-wall dimensions

of the chutes themselves to get the low bias ).• The Chutes Slope (to provide material flow down properly to depress the bias )

◦ Range between 45o until 60o

4 Material feeding

• The material that will be split have to spread evenly on a rectangular scoop designed

so as to fit splitter width.

• The scoop open end covers the set of chutes exactly and the scoop edge is placed in

the middle of the chutes



• The material must be fed perpendicularly to the longitudinal axis

• The scoop must be discharged slowly and evenly to allow the materials flow freely

down the chutes.

5 Equipments Cleaning

• It is required to clean and brush both the chutes and scoop to remove the adhered

fine particles to their surface in order to make the materials flow properly

• To minimize the decontamination effort, disposable drying trays are recommended.6 Split Replication

• Implement an even better than an odd number of riffling stage.

• When sampling for technical purposes, it should be alternated the choice of the

right and left sample

• When sampling for commercial purpose, the split sample must be chosen at random

b. Describe an unbiased procedure/device for representative mass reduction.

Answer :

Unbiased Procedures/Device for representative mass reduction of “Vario Divider” as

describe bellow.

A. Vario Divider Description

▪ The splitting principle which same as riffle splitters and Boerner divider, is that

every second chute contributes to one of two alternative collecting reservoir – or to

more than two, resulting in variable split rations.

▪ The revolving feeding funnel distributes the sample material equally (in time) over a

number of radial chutes (assuming constant rotational speed). It provide equallyreliable mass reduction.

▪ In this journal, they used 1 G/1-4 Model . The specification of this Vario Divider i.e.

• It designed for grains samples, 1 sample, 4 outlets of which 1, 2 or 3 can be

covered to obtain larger dividing ratios

B. Vario Divider Precision and Accuracy

▪ The precision of this device depend on the rotating speed as the influx velocity of thematerial through the feeding funnel and the grain-size contrast of the material.

▪ The accuracy of the dividing depends on the number of subsamples. To obtain the

greatest accuracy, the time taken for dividing must be never be less than 20 seconds.

C. Vario Divider Advantages

▪ It require very little maintenance

▪ The method was very easy and fast to use, so has no user-dependency

▪ It is possible to get a mass reduction ratio as small as 1:100 (dependently particle size of samples)

D. Vario Divider Disadvantages

▪ Splitting the samples in many streps result higher error since every single step is

error generating.

▪ The loss material caused by rebounding of the rape seeds from the sample boxes.

▪ Set-back would appear to be difficulty to cleaning sticky materials and immobility

when needed in fieldwork.



E. Operational Procedures

For the example 4 kg samples should be divided into 2 samples of 1 kg each (e.g. 1 sample and 1 copy sample)

1. The handle (10) is moved to position (4) on the scale

2. Sample trays (7) and (8) are placed in position.

3. Check the hopper is of the correct size then start the divider by

turning the switch on.4. The sample to be divided poured through a hopper (1) It must

be fed perpendicularly to the longitudinal axis to led it downinto a revolving feeder (2). Wait approximately 15 seconds.

With sample less than 4 kg, it is necessary to use an inlet

hopper with a smaller outlet hole. The sample have to divided in at least two steps for a realistic testing of vario dividing

principle.

5. Material continued to a chute opening (11) or to the bottom as left-over material (9). Inorder to make it flow properly, the chutes must be cleaned well and free form sticky

materials.

6. The machine gives two equivalent samples (7) and (8). The number of subsamples can

be calculated by multiplying the number of rotations of revolving feeder (40 rpm) withthe number of outlet. Considering the purpose of the sampling (technical or

commercial) to define which sample should be taken.

(1) Hopper

(2) Revolving Feeder

(3) Motor (4) Chute Closer

(5) Sample Outlets

(6) Sample Outlets

(7) Sample(8) Sample

(9) Left-over material

(10) Chute Opening Handle(11) Chute

E. Summary of The Characteristics for Vario Divider

Characteristics Score Note

Composition + Standard Deviation (); Relative Bias (); Representativeness ()

Mass + Standard Deviation and Relative Bias of the Final Sample Mass ()

Loss + Total Loss of Material ()

Cleaning - Easiness to clean ()

Initializing + Simplicity Initializing Procedures () and Initialing Time ()

User-dependency + User-dependency ()

Time + Operating Time ()

Score 5 Defined in Very Good (Sum=5) Method/Device



c. Describe a biased procedure/device for mass reduction. Identify the source of bias.

Answer :

Biased Procedures for representative mass reduction of “ Alternate Shoveling ” as describe

bellow.

A. Alternate Shoveling Description

▪ The method is based on the principle that all

extracted shovelfuls from the original sample aredeposited sequentially in two alternative heaps asillustrated in the right picture .

▪ The mass is reduced into two samples with almost

equal weight and (hopefully) equal composition.

▪ This is an example of correct an incorrect shovel

design

Incorrect Design Correct Design

Round shape : material at the top of a

flattened sample has more chance to be

part of an increment than the material at

the bottom

Square Shape : all material has the same

chance to be part of the increment

B. Alternate Shoveling Precision and Accuracy

▪ The precision of this device depend on the consistently of the samplers while taking

the increments.

▪ The accuracy of the dividing depends on the number of scoops. The larger number of scoops, the better reduction. Increasing the number of increments should

minimize the effect of the grouping and segregation error.

C. Alternate Shoveling Advantages

▪ More reliable and accurate than coning and quartering

▪ The method is quick and cheap in operation

D. Alternate Shoveling Disadvantages

▪ The equality of the final samples will be highly dependent on the nature of the lot.

▪ Some sample loss was observed due to the practical handling of shovelfuls

▪ Different Operators will have unequal quality of the final reduced samples.

▪ Each scoop tends to select particles adjacent to each other, maintaining much of the

naturally occurring grouping and segregation error.

▪ The analyst must balance the extra time required for small scoop sizes to achieve a

lower grouping and segregation error with being able to accomplish all the

shoveling steps in the time available for sample processing.

▪ The procedure may need to be repeated until the sample is reduced to the mass

required for chemical or physical analysis.



E. Operational Procedures

▪ Extract the material to be sampled one shovelful at a time and place it in twoalternating distinct heaps.

▪ All shovelful should be approximately have the same size and each heap should

consist of the same number of shovelfuls.

▪ One heap should contain only odd increments and the other only even increments.

▪ The increments should be the same size and the minimum number of increments should be around nine (9) for each pile.

▪ One pile is then randomly chosen and subsequently shoveled into two new smaller

piles and so forth.

▪ Sample selected at random to preserve the sampling equity.

E. Summary of The Characteristics for Alternate Shoveling

Characteristics Score Note

Composition - Standard Deviation (); Relative Bias (); Representativeness ()

Mass - Standard Deviation and Relative Bias of the Final Sample Mass ()

Loss 0 Total Loss of Material (can not be defined)

Cleaning + Easiness to clean ()

Initializing + Simplicity Initializing Procedures () and Initialing Time ()

User-dependency - User-dependency ()

Time - Operating Time ()

Score -2 Defined in Poor (Sum < 2) Method/Device

5. Case study 1-D (gas production plant). In a biogas production plant, the formation of the

major product (methane, CH4) and some contaminant by-products, such as hydrogen sulphide

(H2S), must be controlled. To do so, the yield of these two compounds is determined daily as a

part of a systematic quality assurance program. Since the complete process of biogas production takes around two-three weeks, a daily control is considered to be sufficient.

Previous studies showed that the yield of CH4 and H2S is not correlated at all and, therefore, an

independent variographic analysis of each compound should be carried out. The results arerepresented in Figure 1 and 2.

Figure 1. Variogram and error map for H2S.



a) Describe the H2S variograms. Try to estimate the nugget effect , sill and range

Answer :

The Variogram of H 2S describe bellow.

A. Origin Data of Variogram

▪ Variogram data consists of H 2S concentration in produced biogas.

▪ The data have taken in 3 months (January to March, 2005) to obtain 90 samples. By this

means the concentration of H 2S was recorded daily as primer data. B. Variogram Type

▪ The type of the variogram is cyclic.

▪ The graph is cyclic in every 7 days.

▪ The cyclic period is related to the premix capacity per week, which will be restricted during weekend when there is lag of industrial waste being mixed in bioreactors that can

change the profile of raw material of biogas. The amount of Industrial waste which

consist of iron chloride will reduce the H 2S content and makes the concentration of H 2S

drop centered on the weekend.

C. Variogram Characteristics

▪ Nugget Effect ,

• The value is around 0.005• By this means, the H 2S concentration

is still relatively heterogeneous.

▪ Sill

• The estimated of overall sill is around

0.075; or to different ± 0.070 between

the overall sill and the nugget effect.

▪ Range

The distance at which variogram reaches the sill (Range) is ± 14 days

Figure 2. Variogram and error map for CH4.

a) Describe the CH4 variograms. Try to estimate the nugget effect , sill and range in each case.

Answer :

The Variogram of CH 4 describe bellow.

A. Origin Data of Variogram

▪ Variogram data consists of CH 4 concentration in produced biogas.

▪ The data have taken in 3 months (January to March, 2005) to obtain 90 samples. By this

means the concentration of CH 4 was recorded daily as primer data.

▪ But the data of increments No. 2 and 56 are outliers that fatally influenced the Variogramanalysis.



B. Variogram Type

▪ After remove the significant outliers (increments number 2 and 56), the type of thevariogram appear as cyclic.

▪ The graph is periodically cyclic in 20± 2 days ( ~ 3 weeks)

▪ The cyclic period is related to the transportation of animal manure in which small farm

deliver raw manure weekly while larger farm per 3 weeks.

C. Variogram Characteristics

▪ Nugget Effect ,

• The value is not negligible ( ± 0.001)

• It is mean that the variability of the CH 4concentration is still big since all the

type of waste are heterogeneous

(consists of 60% cow manure, 20% pig

manure, 20% organic industrial food processing waste)

• Pre-mixing with collect and deliver the

the type of manure to plant, will help to bring constant overall raw material composition.

▪ Sill

• The estimated of overall sill is around 0.009; or to different ± 0.008 between the

overall sill and the nugget effect.

▪ Range

• The unclear range display on the variogram. But it may around ~ 6-8 days.

b) Design a sampling strategy valid to control simultaneously the yield of H2S and CH4.

Answer :

Valid Sampling Strategies Design to control simultaneously the yield of H 2S and CH 4(1)It need composite sampling because both of the the nugget effect is not negligible.(2)Use Stratified Random for sampling

◦ The cyclic variogram indicate that there were found heterogeneity of H 2S and CH 4concentration within one period (dissimilar population).

◦ Stratified random sampling is most appropriate when the entire population from which the

sample is taken is heterogeneous.

◦ To obtain representative sample, increment must be taken randomly per stratum (a period which is defined based on H 2S and CH 4 sharing a specific attribute or characteristic).

(3) Sampling Rate, The Sample must be taken below range

◦ For controlling H 2S concentration, the range is ± 14 days. This means, sample may took

per 2 weeks. Since composite sampling is needed, increments should be taken. It isrecommended to take sample within a periodic cycle (variogram cyclic in every 7 days)

◦ Range in CH 4 controlling is ~ 6-8 days. So, the increments should be taken bellow 6 days.

◦ Concerning in efficiency, sampling may done in two purpose (H 2S and CH 4 ). Because of

the range of CH 4 is lower than H 2S (but nearly with the H 2S cyclic), the sampling should

be taken below 1 week with increments. The number of increments may calculated based

on defined target error (usually 10%).

◦ From figure 2. Increments may take bellow the range 7 ( lag 2 ) as much as 2 increments to

achieve Total Sampling Error bellow 10% ( 9.33% of error )



c) Comment on possible options to perform the sampling operation to determine the CH 4 yield

achieving a Total Sampling Error below 15%. Answer :

Possible options to perform the sampling operation to determine the CH 4 yield to achieve TSE

< 15% mentioned bellow.

Option 1.

Number of Increment : 1

Lag J : 2 TSE achievement :13.198 %Comment.

The number of increment is not enough to

get the representative sample because the

nugget effect is not negligible and needscomposite sampling.

Option 6.


Lag J : 10 TSE achievement :8.383%Comment.

This option reduce TSE of option number 5

by take more increments in the same lag, but

still not applicable to control the yield of CH 4 since the sampling done > range.

Option 2.


Lag J : 2 TSE achievement :9.333%

Comment.

It is sufficient to collect increments as much

as 2 increments per 2 days. Moreover if tightened sampling is applied (< 10%) to

control CH 4 , this scheme is useful.

Option 7.



Comment.

Regarding to the sampling rate, increments

must be taken < range (range for CH 4 is ~6-8 days). Although this scheme has sufficient

number of increments, it can not be applied

to control CH 4 yield (lag > Range).

Option 3.

Number of Increment : 4 Lag J : 2 TSE achievement :6.599%

Comment.

Higher increments take in the same lag will

reduce the TSE. Another concern is theavailability of sampler and sampling

equipments for this increments.

Option 8.

Number of Increment : 8 Lag J : 24 TSE achievement :10.269 %

Comment.

Same with option 7, this scheme is not

suitable to control CH 4 yield in terms of sampling rate (must be < range) although

they reduce the TSE with take more

increments in the same lag.

Option 4.



Comment.

This is the best sampling scheme in terms of

TSE achievement. But the increments that

should be taken are enormous and run out from efficiency if (far away or) not suitable

on the defined target (<<< 15%).

Option 9.


Lag J : 43 TSE achievement :14.838 %

Comment.

This is the most simple and practical one to

have composite samples since only 4

increments per 43 days. But, when it comeinto CH 4 discussion which sampling must be

taken bellow range (~6-8 days), this schemeis not applicable.

Option 5.


Lag J : 10 TSE achievement :11.855 %

Comment.

TSE of this scheme still bellow 15% with

adequate increments. But It is not suitable toapply to control CH 4 since the Lag > Ratio.

Option 10.



Comment.

Not suitable to obtain representative sample

because the Lag is very high (>>> Range)



Conclusion

To obtain the representative sample to control CH 4 yield with TSE target as much as 15%, we haveto concern about the sampling strategy requirements (Number of increments, and sampling rate

(range > lag)). So we have 3 rules which must be obeyed to get the representative sample, i.e.

1. Number of increment > 1 increment

2. Lag < Range (~6-8 days)

3. Total Sampling Error (TSE) < 15 % From mentioned options, it can be defined the most suitable scheme for sampling to control CH 4 yield. Option 2 is recommended as a guide of sampling scheme since it fulfill all of the requirement of 3 rules listed in above. It is quite adequate for the number of increments (2 increments) and no

need many samplers or sampling equipments to take the increments (relatively no need higher cost

than other options which have more increments). The sampling rate also meet the rule which must be done bellow the range. The only disadvantage is it may have tightened sampling scheme since

the TSE relatively quite far from target error, but it can be re-evaluated in the future based on

sampling history. If this sampling scheme success to control CH 4 properly, the lag should be re-calculated in between 2 until ~6-8 days to come closer with TSE 15%.

ANNEX -Useful formula-

C = cfgβ (1- aL/α)2

c = ρc + (1- aL/α) ρmaL/α

(1- aL/α)2

c = ρc + (1- aL/α) ρmaL/α

CV = 100 × s2 FSE For more than one source of error: CV total

= 100× ∑i

si2

EXCEL HINTS

SOLVER tool

To optimize a parameter of Gy’s equation in order to reach a certain error level for FSE, youmay use the SOLVER tool. You should introduce the Gy’s formula in the objective cell and ask for

an optimization varying the suitable parameter (particle size, sample mass,…) in the changing cell.

The goal of the optimization is making the value of the objective cell equal to the desired error level by modifying the parameters in the changing cell. After running the SOLVER tool, the result

of the objective function will get close to the error level desired and the changing cell will adopt the

optimized value.

Objective cell: it should contain the equation to be optimized (the Gy’s formula in this case).When writing the equation, the cell containing the parameter to be modified should be called.

Changing cell: it should contain a number, an initial guess of the parameter to be optimized.

Syntax: Tools –SOLVER

Dialog box: objective cell (mark cell with Gy’s formula)

Objective cell value: equal to s2(FSE) (numerical value)

Changing cells: (mark cell with parameter to be optimized )

ResolveCONTOUR PLOTS

Represents a flat plot of z values (with contour lines) vs. x/y pairs. A table with the z valuesfor the different x/y pairs should be calculated (WARNING!!! values in x and y axes should be

equispaced).

Select the whole table of x,y,z values

Syntax: plot wizard icon – Surface – contour plot

To modify intervals among contour lines, click legend and change scale values (may vary according

to EXCEL version).

−=LS

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