Difference between revisions of "Collection and analysis of rate data - 2013"
From Introduction to Reactor Design: 3K4
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#F-statistic: 284.2 on 1 and 4 DF, p-value: 7.259e-05 | #F-statistic: 284.2 on 1 and 4 DF, p-value: 7.259e-05 | ||
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+ | #. :math:`\displaystyle \int{ \frac{1}{(1-X)^2} \,dX} =` | ||
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So we got a higher \(R^2\) on the second order model. The reaction rate can be considered second order. What is the reaction rate constant, \(k_A\) for this system? | So we got a higher \(R^2\) on the second order model. The reaction rate can be considered second order. What is the reaction rate constant, \(k_A\) for this system? |
Latest revision as of 17:36, 23 January 2017
Class date(s): | 28 February to 06 March | ||||
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Textbook references
- F2011: Chapter 7
- F2006: Chapter 5
Suggested problems
F2011 | F2006 |
---|---|
Problem 7-7 (a) | Problem 5-6 (a) |
Problem 7-8 (a) | Problem 5-7 (a) |
Problem 7-15 | Not in this edition |
Class materials
28 February 2013 (07C)
04 March 2013 (08A)
06 March 2013 (08B-1)
Source code: R
R code for the example covered in class:
t <- c(0, 50, 100, 150, 200, 250) # [minutes]
# or you can write t <- seq(0, 250, 50)
CA <- c(154, 114, 87, 76, 70, 58) # [mol/m^3]
# Plot the data first. See the expected trend. It it decreasing nonlinearly, so it is either first or second order (but not zeroth order)
plot(t, CA)
grid()
# Assume it is a first order, rA = -k CA
# So plot log(CA0/CA) against t
plot(t, log(CA[1]/CA))
grid()
# Does not quite look like a straight line, but let us fit a least-squares model anyway
mod.first <- lm( log(CA[1] / CA) ~ t + 0) # force intercept to be zero
summary(mod.first)
#Call:
#lm(formula = log(CA[1]/CA) ~ t + 0)
#
#Residuals:
# 1 2 3 4 5 6
# 1.874e-16 8.988e-02 1.493e-01 7.361e-02 -5.502e-02 -7.784e-02
#
#Coefficients:
# Estimate Std. Error t value Pr(>|t|)
#t 0.0042174 0.0002555 16.51 1.49e-05 ***
#---
#Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
#Residual standard error: 0.09474 on 5 degrees of freedom
#Multiple R-squared: 0.982, Adjusted R-squared: 0.9784
#F-statistic: 272.5 on 1 and 5 DF, p-value: 1.489e-05
# Now assume a second order system, rA = -k CA^2
# So plot 1/CA against t
plot(t, 1/CA)
grid()
abline(mod.first)
# Definitely more linear. Let us fit the least squares model to check:
mod.second <- lm( I(1/CA) ~ t)
summary(mod.second)
#Call:
#lm(formula = I(1/CA) ~ t)
#
#Residuals:
# 1 2 3 4 5 6
#-2.751e-04 -5.219e-05 6.146e-04 2.227e-04 -7.051e-04 1.951e-04
#
#Coefficients:
# Estimate Std. Error t value Pr(>|t|)
#(Intercept) 6.769e-03 3.692e-04 18.33 5.21e-05 ***
#t 4.111e-05 2.439e-06 16.86 7.26e-05 ***
#---
#Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
#Residual standard error: 0.0005101 on 4 degrees of freedom
#Multiple R-squared: 0.9861, Adjusted R-squared: 0.9826
#F-statistic: 284.2 on 1 and 4 DF, p-value: 7.259e-05
So we got a higher \(R^2\) on the second order model. The reaction rate can be considered second order. What is the reaction rate constant, \(k_A\) for this system?
Source code: Using MATLAB or Octave
t = [0, 50, 100, 150, 200, 250]; % [minutes]
CA = [154, 114, 87, 76, 70, 58]; % [mol/m^3]
% Plot the data first. See the expected trend. It it decreasing nonlinearly, so
% it is either first or second order (but not zeroth order)
figure
plot(t, CA, 'o')
grid on
% Some template code for the linear regression
x = t;
y = log(CA(1)./CA);
n = numel(y);
% Assume it is a first order, rA = -k CA; so plot log(CA0/CA) against t
figure
plot(x, y, 'o')
grid on
% Does not quite look like a straighTSt line, but let us fit a least-squares model anyway
% Fit the model y = b x + 0 (i.e. force the intercept to zero)
X = [x(:)];
y = y(:); % make sure it is a column vector
b = inv(X'*X)*X'*y; % recall this from the 3E4 course
%b = regress(y, X); % only if you have the Statistics Toolbox in MATLAB
% Plot the data along with the fitted line:
figure
plot(x, y, 'o')
hold('on')
grid('on')
plot(x, X*b, 'r')
xlabel('x values')
ylabel('y values')
legend({'Original data', 'Fitted line'}, 'Location', 'Best')
% Additional calculations
predicted = X*b;
resids = y - predicted; % resids = e = y - Xb
RSS = resids' * resids; % residual sum of squares
TSS = sum((y - mean(y)).^2); % total sum of squares
R2_first_order = 1 - RSS/TSS
% Now try a second order model: -rA = k CA^2; so plot 1/CA against t
figure
plot(t, 1./CA, 'o')
grid('on')
% Definitely more linear. Let us fit the least squares model,
% 1/CA = 1/CA0 + kt, or y = b0 + b1 x, where y = 1/CA and x = t, and there is an intercept
% Some template code for the linear regression
x = t;
y = 1./CA;
n = numel(y);
X = [ones(n,1) x(:)];
y = y(:); % make sure it is a column vector
b = inv(X'*X)*X'*y; % recall this from the 3E4 course
% b = regress(y, X); % only if you have the Statistics Toolbox in MATLAB
% Plot the data along with the fitted line:
figure
plot(x, y, 'o')
hold('on')
grid('on')
plot(x, X*b, 'r')
xlabel('x values')
ylabel('y values')
legend({'Original data', 'Fitted line'}, 'Location', 'Best')
% Additional calculations
predicted = X*b;
resids = y - predicted; % resids = e = y - Xb
RSS = resids' * resids; % residual sum of squares
TSS = sum((y - mean(y)).^2); % total sum of squares
R2_second_order = 1 - RSS/TSS