Difference between revisions of "Isothermal reactor design - 2013"
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| course_notes_alt = Course notes | | course_notes_alt = Course notes |
Revision as of 14:11, 15 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.
11 February 2013 (06A)
Example 5.6 |POLVER05_0 |1
# Differential equations
d(y) / d(W) = -alpha/(2*y) * (1+eps*X)
y(0) = 1.0
d(X) / d(W) = -rAdash / FA0
X(0) = 0
# Constants
FA0 = 0.1362 # [mol/s]
kdash = 0.0074 # [mol/(kg catalyst . s)]
alpha = 0.0367 # [1/kg]
eps = -0.15 # [-]
# Algebraic equations
rAdash = -kdash * (1-X)/(1+eps*X) * y
flow_ratio = (1 + eps*X)/y
# Initial and final values for independent variable:
W(0) = 0
W(f) = 20