Batch Reactor Problem

BACKGROUND CONCEPTS

The design equation for a batch reactor, in the limit of constant volume, is

The rate of production of each component Rj is

If only one reaction is needed to describe the chemical transformations, the design equation can be integrated to give an analytical solution. If more than one linearly independent reaction is needed to describe the chemical transformations, a set of design equations will need to be solved numerically. This set of equations will be no greater than the number of components, and it could be less.

To illustrate the analytical solution consider the isothermal liquid phase reaction discussed in lecture

The initial conditions for the batch reactor are cAo = 2 gmole/liter, cBo = 5 gmole/liter and cCo = 0. Let's assume we want to know the time it takes for 95% of the limiting reactant A to react. The design equation for component A

(1)

cannot be integrated until the concentration of B is expressed in terms of the concentration of A. (We can only have one dependent variable.) By using linear combinations of the design equation it is always possible to express the concentrations in terms of one component when only one reaction is present. Start by writing a design equation for each component in Equation 1.

(2)

(3)

The sum of Equation 2 and –0.5 times Equation 3 gives

therefore the argument in the parentheses is a constant. Using the initial condition to define the value of the constant

After inserting the initial concentration values

Equation 1 can now be written in term of one independent variable

Separating variables, integrating between the limits of (cAo,0) and (cA,tbatch) gives

which can be rearranged to give

If k = 0.2 liter/gmole-hr, it takes 7.84 hours for 95% of A to react.

    This simple example shows that the concentration of A depends on the initial conditions, the stoichiometry of the reaction, and the kinetics. The stoichiometry dependence leads to the constant 2 that appears with cA or cAo. The initial condition dependence leads to the constant of 1 that is added to cA. The kinetics dependence is in the value for the rate constant k. More generally, we could write for any component in an isothermal batch reactor

(4)

If any of these change, the exact functional dependence in Equation 4 will change. The changes may necessitate reevaluating the integral. Alternately, a numerical solution can be explored and the parameters can be easily changed to determine their effects.

    This simple reaction problem can be solved by the simultaneous solution of two initial value ordinary differential equations (ODEs). In this case we solve

subject to the initial conditions of cAo and cBo.

PROBLEM STATEMENT

    We will want to explore how changing the initial conditions or the value of the rate constant changes the time for the fractional conversion of the limiting reactant to reach 95%. Run the program to generate the base case. This case has cAo = 2 gmole/liter, cBo = 5 gmole/liter and cCo = 0, and k = 0.2 liter/gmole-hr. As you run the program, you will be prompted to input different values for the initial concentrations of A and B, the rate constant, and the integration time. You will see that as the conditions are changed, the time to reach 95% conversion will change and it may be necessary to increase the integration time to view this on the plot.

1. Study the reference case. Can you explain why the curve for the concentration of C appears to be a mirror image of the curve for A?

2. Keeping the rate constant equal to 0.2 liter/gmole-hr and the concentration of A at 2 gmole/liter, examine what happens as the concentration of B is increased and then decreased. If B is present in excess, why does it take longer for the 95% conversion limit to be reached as the initial concentration of B is decreased from 5 gmole/liter?

3. Now explore what happens as the initial concentration of B is less than a stoichiometric amount (less than 2 x cAo). How come the curves for A and B cross?

4. Now explore what happens as the value of the rate constant is changed. Can you explain why increasing the value has the effect it does? Since the rate constant changes exponentially with temperature, small changes in the temperature can cause profound changes in the rate of the reaction. We will revisit this issue of temperature in the discussion of nonisothermal reactors.

Page developed by Tom Leach with direction from Dr. John G. Ekerdt
last modified Saturday, February 19, 2000