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.

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.

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.