Next, we multiply the first equation by 0.4 and subtract that result from the second equation to provide
We now divide the last equation by 0.8333 to obtain the solution for the unknown molar flow rate,
At this point, we begin the procedure of “back substitution” which requires that Eq. 4‐96b be substituted into Eq. 4‐96a in order to obtain the final solution for the two molar flow rates.
The procedure leading from Eqs. 4‐93 to the solution given by Eqs. 4‐97 is trivial for a pair of equations; however, if we were working with a five component system the algebra would be overwhelmingly difficult and a computer solution would be required.
4.8 Multiple Units
When more than a single unit is under consideration, some care is required in the choice of control volumes, and the two‐column distillation unit illustrated in Figure 4‐10 provides an example. In that figure we have indicated that all the
Figure 4‐10. Two‐column distillation unit
mass fractions in the streams entering and leaving the two‐column unit are specified, i.e., the problem is over‐specified and we will need to be careful in our degree of freedom analysis. In addition to the mass fractions, we are also given that the mass flow rate to the first column is 1000 lb / h . On the basis of this m
information, we want to predict the following:
1. The mass flow rate of both overhead (or distillate) streams (streams #2 and #3).
2. The mass flow rate of the bottoms from the second column (stream #4).
3. The mass flow rate of the feed to the second column (stream #5).
We begin the process of constructing control volumes by making the primary cuts shown in Figure 4‐11. Those cuts have been made where information is given (stream #1) and information is required (streams #2, #3, #4, and #5). In order to join the primary cuts to form control volumes, we are forced to construct two control volumes. We first form a control volume that connects three primary cuts (streams #1, #2 and #5) and encloses the first column of the two‐column