Isothermal reactor design - 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.
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
25 February 2013 (07A)
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 |
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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
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asdasd
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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)