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
• Polymath code to solve the example in class

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