Difference between revisions of "Isothermal reactor design - 2013"

Revision as of 00:55, 26 February 2013

Class date(s):

04 February to 14 February



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F2011: Chapter 5 and 6
F2006: Chapter 4

04 February 2013 (05A)

General problem solving strategy for reactor engineering
Audio and video recording of the class

06 February 2013 (05B)

The Ergun equation derivation
Audio and video recording of the class

07 February 2013 (05C)

Notes used during the class
The spreadsheet with the Ergun equation example. Use it to try
- different lengths of reactor
- different catalyst particle sizes
- different pipe diameters
- gas properties (e.g. density)

to see the effect on pressure drop in the packed bed.

Audio and video recording of the class

11 February 2013 (06A)

Audio and video recording of the class
Codes to solve the example in class are available on the page software for integrating ODEs.

14 February 2013 (06C): midterm review

Audio and video recording of the class

25 February 2013 (07A)

Audio and video recording of the class

The example covered in class is based on example 4-8 in F2006 and example 6-2 in F2011. <rst> <rst-options: 'toc' = False/> <rst-options: 'reset-figures' = False/> The 3 ODE's are:

.. math::

\dfrac{dF_A}{dV} &= r_A\\ \dfrac{dF_B}{dV} &= r_B - R_B \\ \dfrac{dF_C}{dV} &= r_C

where :math:`-r_A = r_B = r_C` and :math:`-r_A = k\left(C_A - \dfrac{C_B C_C}{K_C} \right)`, and :math:`R_B = k_\text{diff}C_B`.

:math:`k = 0.01\,\text{s}^{-1}`
:math:`k_\text{diff} = 0.005\,\text{s}^{-1}`
:math:`K_C = 50\,\text{mol.m}^{-3}`

We derived earlier in the course that

.. math:: C_A = C_\text{TO}\left(\dfrac{F_A}{F_T}\right)\left(\dfrac{P}{P_0}\right)\left(\dfrac{T_0}{T}\right)

Assuming isothermal and isobaric conditions in the membrane:

.. math:: C_A = C_\text{T0}\left(\dfrac{F_A}{F_T}\right)

where :math:`F_T = F_A + F_B + F_C` and :math:`C_\text{T0} = \dfrac{P_0}{RT_0}`. Similar equations can be written for :math:`C_B` and :math:`C_C`.

Using all of the above derivations, we can set up our numerical integration as shown below. </rst>

MATLAB Python

In a file called membrane.m:

function d_depnt__d_indep = membranem(indep, depnt)
 
% Dynamic balance for the packed bed reactor (PBR); demo problem class 05C
% 
%    indep: the independent ODE variable, such as time or length
%    depnt: a vector of dependent variables
% 
%    X = depnt(1) = the conversion
%    y = depnt(2) = the pressure ratio = P/P_0 = y
% 
%    Returns d(depnt)/d(indep) = a vector of ODEs
 
% Assign some variables for convenience of notation
FA = depnt(1);
FB = depnt(2);
FC = depnt(3);

% Constants
kDiff = 0.005;  % s^{-1}
k = 0.01;       % s^{-1}
KC = 50;        % mol.m^{-3}
P0 = 830600;    % Pa
T0 = 500;       % K
R  = 8.314;     % J/(mol.K)

% Algebraic equations
FT = FA + FB + FC;
CT0 = P0 / (R * T0);
CA = CT0 * FA / FT;
CB = CT0 * FB / FT;
CC = CT0 * FC / FT;
RB = kDiff * CB;
rA = -k * (CA - CB * CC / KC);
rB = -rA;
rC = -rA;
    
% Output from this ODE function must be a COLUMN vector, with n rows
n = numel(depnt);
d_depnt__d_indep = zeros(n,1);
d_depnt__d_indep(1) = rA;
d_depnt__d_indep(2) = rB - RB;
d_depnt__d_indep(3) = rC;

In a separate file (any name), for example: ode_driver.m, which will "drive" the ODE solver:

% Integrate the ODE
% -----------------
 
% The independent variable always requires an initial and final value:
indep_start = 0.0;  % m^3
indep_final = 0.4;  % m^3
 
% Set initial condition(s): for integrating variables (dependent variables)
FA_depnt_zero = 0.25;   % i.e. FA(V=0) = 15 mol/min = 0.25 mol/s
FB_depnt_zero = 0.0;    % i.e. FB(V=0) = 0 mol/s
FC_depnt_zero = 0.0;    % i.e. FC(V=0) = 0 mol/s

% Integrate the ODE(s):
[V, depnt] = ode45(@membranem, [indep_start, indep_final], [FA_depnt_zero, FB_depnt_zero, FC_depnt_zero]);
 
% Plot the results:
clf;
plot(V, depnt(:,1), 'b')
grid('on')
hold('on')
plot(V, depnt(:,2), 'g')
plot(V, depnt(:,3), 'r')
xlabel('Reactor volume, V [m^3]')
ylabel('F_A, F_B and F_C')
legend('F_A', 'F_B', 'F_C')

File:Plots-MATLAB.png

asdasd

File:Plots-Python.png

and in Polymath:

d(FA)/d(V) = rA
d(FB)/d(V) = rB - kDiff * CB
d(FC)/d(V) = rC

FA(0) = 0.25  # mol/s
FB(0) = 0.0   # mol/s
FC(0) = 0.0   # mol/s

# Independent variable	
V(0) = 0
V(f) = 0.4  #m^3

# Constants
kDiff = 0.005  # s^{-1}
k = 0.01       # s^{-1}
KC = 50        # mol.m^{-3}
P0 = 830600    # Pa
T0 = 500       # K
R  = 8.314     # J/(mol.K)

# Algebraic equations
FT = FA + FB + FC
CT0 = P0 / (R * T0)
CA = CT0 * FA / FT
CB = CT0 * FB / FT
CC = CT0 * FC / FT
rA = -k * (CA - CB * CC / KC)
rB = -rA
rC = -rA