Isothermal reactor design - 2013

Class date(s):	04 February to 14 February
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• F2011: Chapter 5 and 6
• F2006: Chapter 4

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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

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

Polmath

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