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Non-ideal Flow and Reactor Characterization

Summary of Residence Time Distribution Properties and Most Important Models

Residence Time Distribution

In a reactor, C(t) is obtained by injecting pulse of tracer. From that:

Total amount of tracer injected is:

In a step tracer run:

Mean residence time:

Variance of the residence time distribution (RTD):

The square root of the variance, s, is called standard deviation.

Average residence time:

RTD in Ideal Reactors

For the plug flow reactor (PFR):

For the continuous stirred tank reactor (CSTR):

Tanks-in-series (TIS) Model

Dispersion Model

Bo < 0.01

Bo > 0.01, Closed-Closed Recipient

E(t) is only integrable by numerical methods.

Bo > 0.01, Open-Open Recipient

Designing E1 to the E(t) given by the first expression (valid for Bo < 0.01) and E2 to the one predicted by the open-open (o-o) assumption, the differences between these RTDs are small at low Bo, but at a high Bo number, E1 is not valid (for more details, consult spreadsheet "dispersion model Bo fitting.xls"):

Problem 1.1 A solution of potassium permanganate (KMnO4) was rapidly injected into a water stream flowing through a circular tube at a linear velocity of 35.70 cm/s. A photoelectric cell located 2.74 m downstream from the injection point was utilized to monitor the local concentration of KMnO4.

1. (a) By using the given effluent KMnO4 concentrations, calculate the average residence time of the fluid as well as the variance, E(t), F(t), and I(t).
2. (b) Determine the number of ideal tanks (nt), the variance, the dispersion number, and the Peclet number.

Solution to Problem 1.1

For details refer the Wiley website at http://www.wiley-vch.de/ISBN9783527354115

1. (a) To solve this problem, we should use a spreadsheet. First, data given in the statement are introduced, and then we should do the following calculations: t (s) C(t) C(t)dt E(t) t·E(t) t·E(t)·?t (t - tm)2·E(t) (t - tm)2·E(t)·?t F(t) I(t) 0 0 0 0.000 0.000 0.000 0.000 0.000 0.000 1.000 2 11 11 0.014 0.029 0.029 1.149 1.149 0.029 0.971 4 53 64 0.069 0.275 0.304 3.345 4.494 0.166 0.834 6 64 117 0.083 0.498 0.773 2.055 5.399 0.332 0.668 8 58 122 0.075 0.602 1.100 0.666 2.721 0.482 0.518 10 48 106 0.062 0.623 1.224 0.059 0.725 0.607 0.393 12 39 87 0.051 0.607 1.230 0.053 0.112 0.708 0.292 14 29 68 0.038 0.527 1.134 0.344 0.397 0.783 0.217 16 22 51 0.029 0.457 0.983 0.720 1.065 0.840 0.160 18 16 38 0.021 0.374 0.830 1.024 1.744 0.882 0.118 20 11 27 0.014 0.285 0.659 1.162 2.186 0.911 0.089 22 9 20 0.012 0.257 0.542 1.419 2.581 0.934 0.066 24 7 16 0.009 0.218 0.475 1.540 2.959 0.952 0.048 26 5 12 0.006 0.169 0.387 1.464 3.004 0.965 0.035 28 4 9 0.005 0.145 0.314 1.504 2.968 0.975 0.025 30 2 6 0.003 0.078 0.223 0.939 2.443 0.981 0.019 32 2 4 0.003 0.083 0.161 1.147 2.086 0.986 0.014 34 2 4 0.003 0.088 0.171 1.375 2.522 0.991 0.009 36 1 3 0.001 0.047 0.135 0.812 2.187 0.994 0.006 38 1 2 0.001 0.049 0.096 0.947 1.759 0.996 0.004 40 1 2 0.001 0.052 0.101 1.093 2.040 0.999 0.001 42 1 2 0.001 0.054 0.106 1.248 2.341 S(C(t)dt) = 771tm = 10.98 ss2 = 46.88 s2

   We can calculate time increment as the difference between time in two experimental data; in the third column, concentration and time increment are multiplied. For a better calculation, instead of calculating C(t)·?t for each time directly, we can use the average value of C(t) between two consecutive data. This is called Simpson's rule for evaluating areas graphically. The sum of the values in this column would give the area of the C(t) curve, so in the fourth column, the following relationship is applied to calculate E(t):

   In the next columns, the average value of E(t) between two consecutive data is multiplied by time, by time increment, and the sum would represent mean time, as we have:

   In a similar way, variance is calculated:

   Finally, we can calculate F(t) and I(t) according to their definitions:

   We can check the form of the graphs showing the distributions:

2. (b) From the calculated parameters, it is easy to find the number of tanks in the TIS model:

   For the dispersion model to be applied, first we should assume Bo < 0.01, and then:

   We obtain Bo = 11.24 » 0.01, so this assumption is not valid.

   Assuming now...