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> 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}`

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

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

# 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)