Question : A slug of dye is placed in the feed stream to a stirred reac...

A slug of dye is placed in the feed stream to a stirred reaction vessel operating at a steady state. The dye concentration in the effluent stream was monitored as a function of time to generate the data in the table below. Times are measured relative to the instant at which the dye was injected.

Time, t Tracer concentration
(s) (g/m^{3})
0 0
120 6.5
240 12.5
360 12.5
480 10
600 5
720 2.5
840 1
960 0
1080 0

Determine the average residence time of the fluid and the F(t) curve for this system.

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For a constant-density system, concentrations are directly proportional to weight fractions. Thus, equation (11.1.5) becomes

 

F(t)=\frac{\int_{0}^{t}C_{E}\Phi _{m}dt}{\int_{0}^{\infty }C_{E}\Phi _{m}dt}

 

where C_{E} is the tracer concentration in the effluent.

 

One must replace the integrals by finite sums to be able to make use of the data given. Thus

 

F(t)=\frac{\sum_{t=0}^{t=t}(C_{E}\Phi _{m}\Delta t)}{\sum_{t=0}^{t=\infty }(C_{E}\Phi _{m}\Delta t)}

 

The data are reported at evenly spaced time increments, and the mass flow rate is invariant for steady state operation. Thus

 

F(t)=\frac{\sum_{t=0}^{t=t}(C_{E})}{\sum_{t=0}^{t=\infty }(C_{E})}

 

The average residence time is given by equation (11.1.2). Combination of this equation with equations (11.1.5) and (11.1.6) gives

 

\bar{t}=\frac{\int_{t=0}^{t=\infty }tw_{E}\Phi _{m}dt}{\int_{t=0}^{t=\infty }w_{E}\Phi _{m}dt}

 

In terms of the data and the aforementioned assumptions this equation becomes

 

\bar{t}=\frac{\sum_{t=0}^{t=\infty }(tC_{E})}{\sum_{t=0}^{t=\infty }(C_{E})}

 

Table I11.1 contains a workup of the data in terms of the analysis above. In the more general case one should be sure to use appropriate averaging techniques or graphical integration to determine both F(t) and \bar{t}. When there is an abundance of data, plot it, draw a smooth curve, and integrate graphically instead of using the strictly numerical procedure employed above. On the basis of the tabular entries,

 

\bar{t}=\frac{18,720}{50.0}=374.4 s

 

 

 

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