CSTR Stability

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We have seen in the text that exothermic reactions in a CSTR can lead to unstable operating conditions. There are two approaches to testing for stability, the first is graphical and is conceptually easier to grasp and the second is the mathematically rigorous. You should become familiar with both and be able to apply the rigorous test, since it will work in all situations. Both approaches require the simultaneous solution of the material and energy balances to establish that temperature and composition exiting the reactor. The first method, which employs the Van Heerden diagram, examines the rate of energy removal and generation about this operating temperature. The second method evaluates the unsteady-state material and energy balances at the steady-statesolution to determine if small excursions in temperature or composition always restore the system to the steady-state values.

To illustrate the concepts and to get an intuitive feel for CSTR stability, we will use an example based on propylene oxide (PO) hydrolysis to propylene glycol (PG) [adapted from Fogler]. The reaction (1)

(1)

is conducted in a large excess of water (W) and with a methanol (ME) co-solvent. Under the conditions we will examine

The CSTR parameters and the inlet conditions are

Base Case

Thermodynamic data are tabulated. You will need to assume the heat capacity does not change with temperature.

Thermodynamic Properties

 

The material and energy balances for this single reaction system reduce to

(2)

 

(3)

 

The solutions to these two algebraic equations for the Base Case conditions is

 

The graphical test for stability (a Van Heerden Diagram) entails substituting c_PO from Equation 2 into Equation 3

 

(4)

and resorting Equation 4 to give

(5)

Recognizing that we need only consider stability for exothermic reactions, we choose to rewrite the last term in Equation 5

(6)

The two sides of Equation 6 are renamed

 

Note the simultaneous solution to Equations 2 and 3 is also the solution to Equation 6, i.e.

 

Run the simulation for the Base Case to generate curves for Q_removaland Q_generation. Notice they intersect at T=560.31R. This CSTR condition is stable because the slope of the removal curve is greater than the slope of the generation curve. Using the plots can you describe why it is stable?

Now let's consider the rigorous test of stability. Following the development in the text we must evaluate the partial derivatives of the unsteady-state material and energy balances for a STR at the steady-state solution. These values are the elements of the Jacobian of the linearized equations. We will let propylene oxide be component A.

The material balance is

and the partial derivatives are

The energy balance is

and the partial derivatives are

The solution (T=530.31 and C_A=0.0892) is stable if the eigenvalues of the Jacobian have negative real values, which can be found numerically with Octave using the command lines

A=[a₁₁, a₁₂; a₂₁, a₂₂]

Eigenvalues_of_the_Jacobian = eig(A)

You could also solve the determinant of the following

which is a quadratic equation in l, viz .

For the Base Case, which is illustrated in the simulation output,

and the eigenvalues are are -6.6681 and -3.0163, so the solution is stable.

Now explore the effects of changing some operating paramters that cold be affected by plant conditions. Consider what would happen if the cooling fluid medium temperature changed. For this example we have set it to a constant, such as would be found for boiling a fluid. If the pressure on the boiling side changed, the temperature T_a would change. Change this by 20 degrees and see how the temperature and composition in the reactor change. Could you have predicted these trends? Next consider that the cooling line gets fouled so the overall heat reansfer coefficient decreases by 30%. Test the effect of changing U by up to 30%. (Note if you change these variables by too much the algebraic equation solver will not converge for the initial guess in the program.)