# Isothermal reactor design - 2013

Class date(s): 04 February to 14 February

• F2011: Chapter 5 and 6
• F2006: Chapter 4

### 07 February 2013 (05C)

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

### 25 February 2013 (07A)

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

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')


### Python

asdasd


### Polymath

d(FA)/d(V) = rA
d(FB)/d(V) = rB - RB
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
RB = kDiff * CB
rA = -k * (CA - CB * CC / KC)
rB = -rA
rC = -rA