Difference between revisions of "Steady-state nonisothermal reactor design - 2013"
From Introduction to Reactor Design: 3K4
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| date = 18 March | | date = 18 March | ||
| dates_alt_text = | | dates_alt_text = | ||
− | | vimeoID1 = | + | | vimeoID1 = FnsyFP1QoP0 |
− | | vimeoID2 = | + | | vimeoID2 = 8YVhVL5dVIU |
− | | vimeoID3 = | + | | vimeoID3 = GJxYdIZ-nCI |
− | | vimeoID4 = | + | | vimeoID4 = -kMr6kk54Pk |
− | | vimeoID5 = | + | | vimeoID5 = yLXZSyJK05E |
− | | vimeoID6 = | + | | vimeoID6 = bAPUtqIWAsI |
− | | vimeoID7 = | + | | vimeoID7 = 6JIZAhz7Y6o |
− | | vimeoID8 = | + | | vimeoID8 = xy6N_-Ob4GA |
− | | vimeoID9 = | + | | vimeoID9 = gaEvAyo29Yw |
| course_notes_PDF = | | course_notes_PDF = | ||
| course_notes_alt = Course notes | | course_notes_alt = Course notes |
Latest revision as of 14:35, 6 January 2017
Class date(s): | 18 March | ||||
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Textbook references
- F2011: Chapter 11 and 12
- F2006: Chapter 8
Suggested problems
F2011 | F2006 |
---|---|
Problem 12-6 | Problem 8-5 |
Problem 12-15 (a) | Problem 8-16 (a) |
Problem 12-16 (a) | Problem 8-18 (a) |
Problem 12-24 (set up equations) | Problem 8-26 (set up equations) |
Class materials
- 18 March 2013 (10A): Audio and video recording of the class.
- 20 March 2013 (10B): Audio and video recording of the class; and the handout used in class.
- 21 March 2013 (10C): Audio and video recording of the class; and the revised handout
- 25 March 2013 (11A): Audio and video recording of the class.
- 27 March 2013 (11B): Audio and video recording of the class.
- 28 March 2013 (11C): Audio and video recording of the class.
- 01 April 2013 (12A): Audio and video recording of the class.
- 03 April 2013 (12B): Audio and video recording of the class.
- 04 April 2013 (12C-part1): Audio and video recording of the class.
Source code
Example on 21 March (class 10C)
pfr_example.m
function [d_depnt__d_indep, CA] = pfr_example(indep, depnt, param)
% Dynamic balance for the reactor
%
% indep: the independent ODE variable, such as time or length
% depnt: a vector of dependent variables
% Returns d(depnt)/d(indep) = a vector of ODEs
% Assign some variables for convenience of notation
X = depnt(1);
% Constants. Make sure to use SI for consistency
FT0 = param.FT0; % mol/s. Note how we use the "struct" variable to access the total flow
FA0 = 0.9 * FT0; % mol/s
T1 = 360; % K
T2 = 333; % K
E = 65700; % J/mol
R = 8.314; % J/(mol.K)
HR = -6900; % J/(mol of n-butane)
CA0 = 9300; % mol/m^3
k_1 = 31.1/3600; % 1/s (was 1/hr originally)
K_Cbase = 3.03; % [-]
% Equations
T = 43.3*X + param.T_0; % derived in class, from the heat balance
k1 = k_1 * exp(E/R*(1/T1 - 1/T)); % temperature dependent rate constant
KC = K_Cbase * exp(HR/R*(1/T2 - 1/T)); % temperature dependent equilibrium constant
k1R = k1 / KC; % reverse reaction rate constant
CA = CA0 * (1 - X); % from the stoichiometry (differs from Fogler, but we get same result)
CB = CA0 * (0 + X);
r1A = -k1 * CA; % rate expressions derived in class
r1B = -r1A;
r2B = -k1R * CB;
r2A = -r2B;
rA = r1A + r2A; % total reaction rate for species A
n = numel(depnt);
d_depnt__d_indep = zeros(n,1);
d_depnt__d_indep(1) = -rA / FA0;
driver_ode.m
% Integrate the ODE
% -----------------
% The independent variable always requires an initial and final value:
indep_start = 0.0; % m^3
indep_final = 5.0; % m^3
% Set initial condition(s): for integrating variables (dependent variables)
X_depnt_zero = 0.0; % i.e. X(V=0) = 0.0
% Other parameters. The "param" is just a variable in MATLAB.
% It is called a structured variable, or just a "struct"
% It can have an arbitrary number of sub-variables attached to it.
% In this example we have two of them.
param.T_0 = 330; % feed temperature [K]
param.FT0 = 163000/3600; % mol/s (was kmol/hour originally)
% Integrate the ODE(s):
[indep, depnt] = ode45(@pfr_example, [indep_start, indep_final], [X_depnt_zero], odeset(), param);
% Deal with the integrated output to show interesting plots
X = depnt(:,1);
T = 43.3.*X + param.T_0; % what was the temperature profile?
rA_over_FA0 = zeros(numel(X), 1); % what was the rate profile?
C_A = zeros(numel(X), 1); % what was the concentration profile?
for i = 1:numel(X)
[rA_over_FA0(i), C_A(i)] = pfr_example([], X(i), param);
end % the above can be done more efficiently in a single line, but the code would be too confusing
% Plot the results
f=figure;
set(f, 'Color', [1,1,1])
subplot(2, 2, 1)
plot(indep, X); grid
xlabel('Volume, V [kg]', 'FontWeight', 'bold')
ylabel('Conversion, X [-]', 'FontWeight', 'bold')
subplot(2, 2, 2)
plot(indep, T); grid
xlabel('Volume, V [kg]', 'FontWeight', 'bold')
ylabel('Temperature profile [K]', 'FontWeight', 'bold')
subplot(2, 2, 3)
plot(indep, C_A); grid
xlabel('Volume, V [kg]', 'FontWeight', 'bold')
ylabel('Concentration C_A profile [K]', 'FontWeight', 'bold')
subplot(2, 2, 4)
plot(indep, rA_over_FA0); grid
xlabel('Volume, V [kg]', 'FontWeight', 'bold')
ylabel('(Reaction rate/FA0) profile [1/m^3]', 'FontWeight', 'bold')
% Now plot one of the most important figures we saw earlier in the course:
% F_A0 / (-rA) on the y-axis, against conversion X on the x-axis. This plot
% is used to size various reactors.
% The material leaves the reactor at equilibrium; let's not plot
% that far out, because it distorts the scale. So plot to 95% of
% equilibrium
f = figure; set(f, 'Color', [1,1,1])
index = find(X>0.95 * max(X), 1);
plot(X(1:index), 1./rA_over_FA0(1:index)); grid
xlabel('Conversion, X [-]', 'FontWeight', 'bold')
ylabel('FA0/(-r_A) profile [m^3]', 'FontWeight', 'bold')
% Updated code to find the optimum inlet temperature:
temperatures = 300:3:400;
conv_at_exit = zeros(size(temperatures));
i = 1;
for T =
param.T_0 = T; % feed temperature [K]
[indep, depnt] = ode45(@pfr_example, [indep_start, indep_final], ...
[X_depnt_zero], odeset(), param);
conv_at_exit(i) = depnt(end, 1);
i = i + 1;
end
plot(temperatures, conv_at_exit); grid;
xlabel('Inlet temperature')
ylabel('Conversion at reactor exit')