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Enhancing the Adsorption Performance of Carbon Nanotubes with A Multistep Functionalization

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Enhancing the Adsorption

Performance of Carbon Nanotubes with A Multistep Functionalization Method: Optimization of Reactive Blue 19 Removal Through Response Surface Methodology

DEWI AYU LASTARI 5003201004

DWI TIAS APRILLIA 5003201030 FARICHA

GHINA NUFASA 5003201015

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● In the textile industry, The Company usually has a reactive textile color that comes of synthetic origin and a complex aromatic structure that is difficult to unravel.

Untreated disposal of waste that was contaminated by reactive equipment can harm humans/aquatic. Reactive Blue 19 (RB19) dye is commonly used in the textile and leather industry and may be mutagenic and toxic because of the presence of electrophilic vinyl sulfone groups. (McCallum et al., 2000; Siddique et al., 2011). To reduce some effects of Reactive Blue 19, it can be adsorbed with carbon nanotubes that have a multistep functionalization method.

● Then, how to design the experiment to enhance the adsorption capacity of Reactive Blue 19 (RB19) onto multi-walled carbon nanotubes?

● Main aim of this study is to enhance the adsorption capacity of Reactive Blue 19 (RB19) onto multi-walled carbon nanotubes with a response surface methodology.

Intro Duction

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Methodo Logy

Data ANALYSIS

Response surface methodology, or RSM, is a collection of mathematical and statistical techniques that are useful for the modeling and analysis of problems in which a response of interest is influenced by several variables and the objective is to optimize this response.

A. First Order Response Surface

1. Determine Coded Factors and Response Table

The company realizes that the response is not between the highest and the lowest level of factor so they add the range that ⍴ can be the response value. For the calculation, we have k=3 for number of level factor and then for the coding of we obtain :⍴

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2. Determine The Regression Coefficients

The process of determining the regression coefficient is carried out by matrix operations with the formula:

or can be obtained using minitab, STAT > DOE > Response Surface > Analyze Response Surface Design (Term : Linear) so that the ANOVA is obtained as follows:

Data ANALYSIS

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4. Determine The Regression Coefficients

The step size for the first factor is determined by the researcher so we use the step size x1=1. Based on this value, we get Δx2=- 1.476 and Δx3=-0.797.

3. Find The Mean and Standard Deviation

To find the average value and standard deviation of each factor can be known using the following equation,

mean = (high level+low level)/2

standard deviation = (high level-low level)/2

● X1

mean =0.5

standard deviation =0.1

● X2

mean =150

standard deviation =50

● X3 mean =4

standard deviation =1

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5. Transformation of Coding Variable to Natural Variable The equation used is

Data ANALYSIS

6. The Path of Steepest Ascent

After doing the previous steps, we get the following path of steepest ascent.

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B. Second Order Response Surfaces

1. Determine The Regression Coefficients

● Matrix b

● Matrix B

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2. Determine Characteristic of Stationary Point First steps is calculate eigenvalue:

3. Determine Optimum Response Optimum Response calculation:

● Xs result

● Matrix x

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Results & Discussion

Data ANALYSIS

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From the calculation using minitab, the estimation model of first order is : Y = 66,67 + 11,99 x1 - 17,70 x2 - 9,56 x3

While the second order is :

Y = 67,35 + 11,99 x1 - 17,70 x2 - 9,56 x3 - 4,16 x1^2 + 5,33 x2^2 - 4,00 x3^3 - 5,07 x1x2 - 5,10 x1x3 + 10,22 x2x3

Because the model is conducted in the first order and second order, then to choose a model that optimizes the response, we must analyze the ANOVA and test the fit model hypothesis.

Model Estimation Parameter

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ANOVA and Fit Model Hypothesis Test

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● ANOVA for the first order Fit Model Hypothesis Test

H0 : Linear model is not significant H1 : Linear model is significant

Alpha : 0,05

Statistics Test : P value=0,000

Rejection : Reject of H null if P value<alpha Decision : reject H null

Conclusion : Linear model is significant Lack of Fit Test

H0 : There is no lack of fit/lack of fit meaningless H1 : there is a lack of fit

Alpha : 0,05

Statistics Test : P value=0,051

Rejection : Reject of H null if P value<alpha Decision : Don't reject H null

Conclusion : There is no lack of fit/ lack of fit meaningless

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ANOVA and Fit Model Hypothesis Test

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● ANOVA for the second order Fit Model Hypothesis Test

H0 : Quadratic model is not significant H1 : Quadratic model is significant

Alpha : 0,05

Statistics Test : P value=0,013

Rejection : Reject of H null if P value<alpha Decision : reject H null

Conclusion : Quadratic model is significant Lack of Fit Test

H0 : There is no lack of fit/lack of fit meaningless H1 : there is a lack of fit

Alpha : 0,05

Statistics Test : P value=0,051

Rejection : Reject of H null if P value<alpha Decision : Don't reject H null

Conclusion : There is no lack of fit/ lack of fit meaningless

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Determination of Stationary Points and Characteristics

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Based on the result calculation of Xs the stationary point of the factor is Xs1=0,32364 Xs2=1,41939 and Xs3=0,41195. For the characteristic for the stationary point, we have to determine the eigenvalue of matrix B.

The result of eigenvalue of B is � 1= -7,059 2= -4,251 � 3= 8,479

Because the sign of eigenvalues is different, so the characteristics of a stationary point is a saddle point or we cannot determine either the stationary point is maximum or minimum.

Determine of Optimum Response

Using the equation of Ys, we may find the predicted response at the stationary point as Ys=54,75949.

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SURFACE PLOT

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CONTOUR PLOT

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Concl u

sion

Data ANALYSIS

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Stationary points of the factor to optimize the response are

Xs1=0.32364, Xs2=1,41939 and Xs3=0.4119. To know the

characteristics of this stationary point, we have eigenvalues that have

different sign λ1=-7.059, λ2=-4.251, and λ3=8,479. Because of that,

we cannot determine whether the characteristics are minimum or

maximum. So the characteristics of a stationary point is a saddle

point. For the Respon Optimum, we got a value Ys=54.7594954,76.

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Do you have any questions?

THAN KS!

Data ANALYSIS

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