The steady-state solution is obtained when dCA/dt = 0 and dT/dt = 0, that is

(1s)

(2s)

To solve these two equations, all parameters and variables except for two (CA and T) must be specified. Given numerical values for all of the parameters and variables we can use Newtonís method (chapter 3) to solve for the steady-state values of CA and T. For convenience, we use an ësí subscript to denote a steady-state value (so we solve for CAs and Ts).

In this module we will study the set of parameters shown for case 2 conditions in Table 1. Cases 1 and 3 are left as exercises for the reader.

Table 1. Reactor Parameters
 parameter case 1 case 2 case 3 F/V, hr-1 1 1 1 k0, hr-1 14,825*3600 9,703*3600 18,194*3600 (-DH), kcal/kgmol 5215 5960 8195 E, kcal/kgmol 11,843 11,843 11,843 rcp, kcal/(m3oC) 500 500 500 Tf, oC 25 25 25 CAf, kgmol/m3 10 10 10 UA/V, kcal/(m3oC hr) 250 150 750 Tj, oC 25 25 25

Solution for Case 2 parameters using fsolve and cstr_ss.m

The function m-file for the steady-state equations is cstr_ss.m and is shown in Appendix 1. The command to run this file is

x = fsolve('cstr_ss',x0);

where x0 is a vector of the initial guesses and x is the solution. Before issuing this command the reactor parameters must be entered in the global parameter vector CSTR_PAR. We find that different initial guesses for the concentration and temperature lead to different solutions.

When choosing initial guesses for a numerical algorithm, it is important to use physical insight about the possible range of solutions. For example, since the feed concentration of A is 10 kgmol/m3 and the only reaction consumes A, the possible range for the concentration of A is 0 < CA < 10. Also, it is easy to show that a lower bound for temperature is 298 K, which would occur if there was no reaction at all, since the feed and jacket temperatures are 298 K. Notice also that there should be a correlation between concentration and temperature. If the concentration of A is high, this means that not much reaction has occurred so little energy has been released by reaction and therefore the temperature will not be much different than the feed and jacket temperatures.

Guess 1 - High concentration (low conversion), Low temperature. Here we consider an initial guess of CA = 9 and T = 300 K.

x = fsolve('cstr_ss',[9;300]);

x =

8.5636

311.1710

so the steady-state solution for guess 1 is = , that is, high concentration (low conversion) and low temperature.

Guess 2 - Intermediate concentration and temperature.

x = fsolve('cstr_ss',[5;350])

x =

5.5179

339.0971

so the steady-state solution for guess 2 is = .

Guess 3 - Low concentration and high temperature.

x = fsolve('cstr_ss',[1;450])

x =

2.3589

368.0629

so the steady-state solution for guess 3 is = , that is, low concentration (high conversion) and high temperature.

The results are summarized in Table 2.

Table 2. Guesses and solutions using fsolve.
 Guess and Solution Guess 1 Guess 2 Guess 3 x0(1), CA guessed 9 5 1 x0(2), T guessed 300 350 450 x(1), CA solution 8.564 5.518 2.359 x(2), T solution 311.2 339.1 368.1

Other initial guesses do not lead to any other solutions, so we see that there are three possible solutions for this set of parameters. In section 7 we show how to use physical insight to determine the number of steady-state solutions for this problem.