The autocatalytic reaction in a loop reactor

You should know the reactor equation and it's derivation (from the TFR) from a material balance for the splitting nodes, take 3 slides for repetition !
The task is to find the optimal recirculation factor R which yields a minimum space time for the given reaction data, i.e. mainly for the 'demanded' conversion (e.g. here: 99%). The 'normal' procedure is to solve the task graphically, as shown within our example in the archive of exercises, or alternatively in the book of Jens Hagen, 'Chemische Reaktionstechnik'. You have to find a line CD in the graph that makes two partial areas equal (if you want to understand why, you have to look at the reactor equation and the graph, it is simply the 'transposing' of the formulas to the areas in the graph. One hint: the area under the function 1/r from U_A,1 to U_A is equal to the area of the rightangle 'foot-point U_A,1', B, C, 'head-point over U_A,1' - just when CD creates two equal areas (as shown in the graph). And a further hint: the ratio of the lines OB to O'foot-point U_A,1' is (R+1)/R !! this means that '1' in the graph is not the value 1 but the 'unity' !!

My alternative proposal for the solution is to create a plot space-time versus R-factor with Hp-Vee: see snapshot. With the aid of the marker you can readout the minimum value.

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