10.491: Integrated Chemical Engineering

Prof. Noah Lotan

Problem Set 3

May 8, 2000

 

 

 

Solution:

 

Problem 1

 

From lecture, for the case in which

-         the body is modeled as a true CSTR

-         the total concentration of the feed stream is always constant

-         the feed stream is 100% B at t<0

-         the feed stream is 100% D at t≥0

we have

 

                                                                (1)

                                                       (2)

     

Plotting equations (1) and (2) for the three pairs of F and V given in the problem statement gives

 

 

From the figure it can be seen that as the value of F/V increases, the replacement of B in the body by D occurs at a faster rate. This is because in these cases the residence time of the tank is reduced, with B and D requiring less time to pass through the body.

 

Problem 2

 

Solving equation (2) for t gives

 

                                                          (3)

 

Using data set 1, with V=4 l, F=2 l/h, and CD/CB,0=CD,tox/CB,0=0.7/1=0.7, in equation (3) gives t=2.41 h. This will be the time required for the concentration of D to initially reach a toxic concentration in the body.

 

In the next step of the process, because D is being flushed from the system, it will show a profile exactly like that of B in Problem 1. We can then solve equation (1) for t and replace the concentration terms with those representing D, giving

 

                                                                 (4)

 

Solving equation (4) with the values V=4 l, F=2 l/h, and CD/CD,0=CD,eff/CD,tox=0.2/0.7= 0.29, gives t=2.48 h. This will be the time required for the concentration of D to drop from the minimum toxic concentration to the minimum effective concentration.

 

The final calculation again involves an increasing trend in D, so equation (3) will be used. Solving equation (3) with V=4 l, F=2 l/h, and CD/CD,0=(CD,tox-CD,eff)/(CD,in-CD,eff)= 0.5/0.8=0.625, gives t=1.96 h. This will be the time required for the concentration of D to increase from the minimum effective concentration back up to the minimum toxic concentration.

 

All other sections of the treatment protocol will simply involve repeating the second and third steps described above. From the calculated times, and those calculated using the values of data set 2, we can then fully describe the process.

 

 

Data Set 1

Data Set 2

t1

2.41 h

1.20 h

t2

4.89 h

1.44 h

t3

6.85 h

1.61 h

t4

9.33 h

1.85 h

t5

11.29 h

2.02 h

 

From the table it can be seen that as the therapeutic range of the drug narrows, the administration schedule requires more frequent changes from feeds of 100% B to those of 100% D and back again.

 

Problem 3

 

From lecture, for the case in which

-         the body is modeled as two true CSTRs in series

-         the total concentration of the feed stream is always constant

-         the feed stream is 100% B at t<0

-         the feed stream is 100% D at t≥0

we have

 

                                                    (5)

                                          (6)

 

Plotting equations (5) and (6) for the two-CSTR case of F=2 l/h and V=2 l, and the one-CSTR case of F=2 l/h and V=4 l, gives


 

 


From the figure it can be seen that for the case of a two-CSTR model, a longer period of time than in the case of the one-CSTR model is required for the replacement of B in the body by D. At later times, however, this replacement occurs at a faster rate with the two-CSTR model than with the one-CSTR model. The initial lag is caused by the extra time needed for changes in the concentrations of the first of the two CSTRs to affect the concentrations of the second tank. The later higher rate is caused by the smaller residence time of the second tank compared to that of the single tank of the one-CSTR model.