Nothing Special   »   [go: up one dir, main page]
Scribd Logo
Upload

Welcome to Scribd!

  • 26 views

    Linear Regression Fitting of A Nonlinear Expression (Example 1)

    Uploaded by

    XheikhKaleem
    This document describes using linear regression to fit nonlinear rate data from a first-order chemical reaction to determine the rate constant k. It provides sample concentration and time data, transforms the concentration data to linearize it, performs linear regression via two methods to determine the slope and intercept, and calculates k as the negative slope. It also outputs the standard error and correlation coefficient of the regression line.

    Copyright:

    © All Rights Reserved
    Download as PPTX, PDF, TXT or read online from Scribd

    Linear Regression Fitting of A Nonlinear Expression (Example 1)

    Uploaded by

    XheikhKaleem
    26 views6 pages
    This document describes using linear regression to fit nonlinear rate data from a first-order chemical reaction to determine the rate constant k. It provides sample concentration and time data, transforms the concentration data to linearize it, performs linear regression via two methods to determine the slope and intercept, and calculates k as the negative slope. It also outputs the standard error and correlation coefficient of the regression line.

    Original Description:

    numerical methods

    Original Title

    Linear Regression Example

    Copyright

    © © All Rights Reserved

    Available Formats

    PPTX, PDF, TXT or read online from Scribd

    Did you find this document useful?

    Is this content inappropriate?

    This document describes using linear regression to fit nonlinear rate data from a first-order chemical reaction to determine the rate constant k. It provides sample concentration and time data, transforms the concentration data to linearize it, performs linear regression via two methods to determine the slope and intercept, and calculates k as the negative slope. It also outputs the standard error and correlation coefficient of the regression line.

    Copyright:

    © All Rights Reserved
    Download as PPTX, PDF, TXT or read online from Scribd
    Download as pptx, pdf, or txt
    26 views6 pages

    Linear Regression Fitting of A Nonlinear Expression (Example 1)

    Uploaded by

    XheikhKaleem
    This document describes using linear regression to fit nonlinear rate data from a first-order chemical reaction to determine the rate constant k. It provides sample concentration and time data, transforms the concentration data to linearize it, performs linear regression via two methods to determine the slope and intercept, and calculates k as the negative slope. It also outputs the standard error and correlation coefficient of the regression line.

    Copyright:

    © All Rights Reserved
    Download as PPTX, PDF, TXT or read online from Scribd
    Download as pptx, pdf, or txt
    of 6

    Linear Regression Fitting of a

    Nonlinear Expression (Example 1)

    Consider the first order rate expression for a chemical


    reaction:
    r  kC A

    For a constant volume batch reactor the mass balance is:

    dCA
      kC A
    n

    dt
    Rearranging the expression and then integrating from
    an initial concentration of CA0 to a final concentration CA
    gives:

    CA
    dCA t


    CA0 CA
       kdt
    0

    CA
    ln  kt or C A  C A 0e  kt

    C A0
    Based on the data below find the rate constant k for the first order reaction:

    CA (gmole/liter) Time (sec)


    0.100 0
    0.0892 500
    0.0776 1000
    0.0705 1500
    0.0603 2000
    0.0542 2500
    0.0471 3000
    Ca=[0.1; 0.0892; 0.0776; 0.0705; 0.0603; 0.0542; 0.0471];
    t=0:500:3000
    t=t'
    n=7; %data points
    y=log(Ca./Ca(1));%transform y data
    x=t

    a1=(n*sum(x.*y)-sum(x)*sum(y))/(n*sum(x.^2)-sum(x)^2);
    a0=sum(y)/n-a1*sum(x)/n;

    fprintf('\nThe slope of the regression line equals %.6f\n',a1);


    fprintf('\nThe intercept of the regression line equals %.6f\n',a0);
    fprintf('\nThus k equals %.7f\n',-a1);
    S=sum((a0+a1.*x-y).^2);
    stderror=sqrt(S/(n-2));
    fprintf('\nThe standard error of the regression line is %.6f\n',stderror);

    Sb=sum((y-sum(y)/n).^2);
    r=sqrt((Sb-S)/Sb);
    fprintf('\nThe correlation coefficient of the regression line equals %.6f\n\n',r);
    We could also solve example using polyfit

    Ca=[0.1; 0.0892; 0.0776; 0.0705; 0.0603; 0.0542; 0.0471];


    t=0:500:3000
    t=t'
    n=7; %data points
    y=log(Ca./Ca(1));%transform y data
    x=t
    fprintf('\nComparing the results with the use of the polyfit function \n');

    coeff=polyfit(x, y, 1);
    fprintf('\nThe slope of the regression line equals %.6f\n',coeff(1));
    fprintf('\nThe intercept of the regression line equals %.6f\n',coeff(2));
    fprintf('\nThus k equals %.7f\n',-coeff(1));

    You might also like