Type of Reactor |
Characteristics |
Continuously Stirred Tank Reactor (CSTR) |
Arranged as one long reactor or many short reactors in a tube bank ; no radial variation in reaction rate (concentration); concentration changes with length down the reactor |
Kinds of Phases Present |
Usage |
Advantages |
Disadvantages |
1. Primarily Gas Phase |
1. Large Scale 2. Fast Reactions 3. Homogeneous Reactions 4. Heterogeneous Reactions 5. Continuous Production 6. High Temperature |
1. High Conversion per Unit Volume 2. Low operating (labor) cost) 3. Continuous Operation 4. Good heat transfer |
1. Undesired thermal gradients may exist 2. Poor temperature control 3. Shutdown and cleaning may be expensive |
General Mole Balance Equation
At steady state-
Differentiating, that gives-
For single reactions in terms of conversion-
The differential form of the PFR mole balance is-
The integral form is-
EXAMPLE
Gas Phase Elementary Reaction Additional Information only A fed P_{0} = 8.2 atm T_{0} = 500 K C_{A0} = 0.2 mol/dm^{3} k = 10 dm^{3}/mol-s v_{o} = 25 dm^{3}/s
PFR Mole Balance: Rate Law: Stoichiometry: Gas: T = T_{0}, P = P_{0} Combine: For X = 0.9: V = 45.3 dm^{3}
Given -r_{A} as a function of conversion, one can size any type of reactor. The volume of a PFR can be represented as the shaded area in the Levenspiel Plot shown below.
The integral to calculate the PFR volume can be evaluated using a method such as Simpson's One-Third Rule (pg 925):
Numerical Evaluation of Integrals
NOTE: The intervals ( ) shown in the sketch are not drawn to scale. They should be equal.
EXAMPLE
Determine X_{e} for a PFR with no pressure drop, P = P_{0}
Given that the system is gas phase and isothermal, determine the reactor volume when X = 0.8 X_{e}.
Reaction Additional Information C_{A0} = 0.2 mol/dm^{3}
K_{C} = 100 dm^{3}/molk = 2 dm^{3}/mol-min
F_{A0} = 5 mol/minFirst calculate X_{e}:
One could then use Polymath to determine the volume of the PFR. The corresponding Polymath program is shown below.
Using Polymath
Algorithm Steps Polymath Equations Mole Balance d(X)/d(V) = -rA/FA0 Rate Law rA = -k*((CA**2)-(CB/KC)) Stoichiometry CA = (CA0*(1-X))/(1+eps*X) CB = (CA0*X)/(2*(1+eps*X)) Parameter Evaluation eps = -0.5 CA0 = 0.2 k = 2 FA0 = 5 KC = 100 Initial and Final Values X_{0} = 0 V_{0} = 0 V_{f} = 500 Polymath Screen Shots
Equations
Plot of X vs. V
Results in Tabular Form
A volume of 94 dm^{3} (rounding up from slightly more than 93 dm^{3}) appears to be our answer.