Exercise: Material balance with formation or consumption: Stoichiometry given

Let’s write out all of the relevant equations that would allow us to solve for the unknowns.

The general mole balance equation is

\[\begin{split}\sum_{\substack{\text{input}\\\text{streams}}} \dot n_{in} + r_{\text{formation, A}}= \sum_{\substack{\text{output}\\\text{streams}}} \dot n_{out} + r_{\text{consumption, A}}\end{split}\]

Applying this general equation for each species in our system, we have

species \(\text{A}\):

\[\dot n_{\text{A, stream 1}} + \dot n_{\text{A, stream 2}} + r_{\text{formation, A}} = \dot n_{\text{A, stream 3}} + r_{\text{consumption, A}}\]

species \(\text{B}\):

\[\dot n_{\text{B, stream 1}} + \dot n_{\text{B, stream 2}} + r_{\text{formation, B}} = \dot n_{\text{B, stream 3}} + r_{\text{consumption, B}}\]

species \(\text{C}\):

\[\dot n_{\text{C, stream 1}} + \dot n_{\text{C, stream 2}} + r_{\text{formation, C}} = \dot n_{\text{C, stream 3}} + r_{\text{consumption, C}}\]

species \(\text{NR}\):

\[\dot n_{\text{NR, stream 1}} + \dot n_{NR,\text{stream 2}} + r_{\text{formation, NR}} = \dot n_{\text{NR, stream 3}} + r_{\text{consumption, NR}}\]

Eliminating irrelevant terms, we get

\[ \begin{align}\begin{aligned}\dot n_{\text{A, stream 1}} = \dot n_{\text{A, stream 3}} + r_{\text{consumption, A}}\\\dot n_{\text{B, stream 2}} = r_{\text{consumption, B}}\\r_{\text{formation, C}} = \dot n_{\text{C, stream 3}}\\\dot n_{\text{NR, stream 1}} + \dot n_{\text{NR, stream 2}} = \dot n_{\text{NR, stream 3}}\end{aligned}\end{align} \]

At this point, we have 4 equations and 10 unknowns.

What else do we know?

From the stoichiometry, we have 2 equations:

\[ \begin{align}\begin{aligned}r_{\text{consumption, B}} = \frac{3}{2} r_{\text{consumption, A}}\\r_{\text{formation, C}} = \frac{1}{2} r_{\text{consumption, A}}\end{aligned}\end{align} \]

And from the givens in the problem, we have 4 more equations:

\[ \begin{align}\begin{aligned}\dot n_{A,\text{stream 1}} = \frac{(x_{A} \, \dot m)_{\text{stream 1}}}{MW_{A}}\\\dot n_{B,\text{stream 2}} = \frac{(x_{B} \, \rho \, \dot V)_{\text{stream 2}}}{MW_{B}}\\\dot n_{NR,\text{stream 1}} = \frac{[(1-x_{A}) \, \dot m]_{\text{stream 1}}}{MW_{NR}}\\\dot n_{NR,\text{stream 2}} = \frac{[(1-x_{B}) \, \rho \, \dot V]_{\text{stream 2}}}{MW_{NR}}\end{aligned}\end{align} \]

So, we now have the same number of equations as unknowns. Provided there are no peculiarities with the formulation or values, we should be able to solve for all of the unknowns in the system.

To do this we could go through these equations systematically and use algebra to find each unknown.

Or, we could set up a 10x10 matrix system and solve them all at once numerically. This is one of the things you will learn in your CBE MATLAB class and do frequently in your material and energy balance course (CBE 201).

Our system, in a form amenable to computer-assisted solution is

\[ \begin{align}\begin{aligned}\dot n_{\text{A, stream 1}} - \dot n_{\text{A, stream 3}} - r_{\text{consumption, A}} = 0\\\dot n_{\text{B, stream 2}} - r_{\text{consumption, B}} = 0\\r_{\text{formation, C}} - \dot n_{\text{C, stream 3}} = 0\\\dot n_{\text{NR, stream 1}} + \dot n_{\text{NR, stream 2}} - \dot n_{\text{NR, stream 3}} = 0\\r_{\text{consumption, B}} - \frac{3}{2} r_{\text{consumption, A}} = 0\\r_{\text{formation, C}} - \frac{1}{2} r_{\text{consumption, A}} = 0\\\dot n_{A,\text{stream 1}} = \frac{(x_{A} \, \dot m)_{\text{stream 1}}}{MW_{A}}\\\dot n_{B,\text{stream 2}} = \frac{(x_{B} \, \rho \, \dot V)_{\text{stream 2}}}{MW_{B}}\\\dot n_{NR,\text{stream 1}} = \frac{[(1-x_{A}) \, \dot m]_{\text{stream 1}}}{MW_{NR}}\\\dot n_{NR,\text{stream 2}} = \frac{[(1-x_{B}) \, \rho \, \dot V]_{\text{stream 2}}}{MW_{NR}}\end{aligned}\end{align} \]